NCERT Exemplar Solutions for Class 10 Maths Chapter 13 Volume And Surface Area Of Solids

Volume And Surface Area Of Solids

The objects which occupy space have three dimensions) arc called solids.

Volume And Surface Area Of Solids Cuboid And Cube

A cuboid is a rectangular solid having six faces all of which are rectangles. The cuboid has six faces. The opposite faces are all congruent. Two adjacent faces meet in a line.

This line is called the edge of the cuboid. There are 12 edges of a cuboid. The point where three adjacent edges meet is called the vertex of the cuboid. There are 8 vertices of a cuboid.

The three edges that meet on the vertex of a cuboid are called its length, breadth and height. A cube is a cuboid in which all the edges are of equal length and every one of the six faces is a square.

Total Surface (Surface Area) of a Cuboid and a Cube

The sum of the area of the faces of the cuboid is called its surface. The areas of opposite faces are equal, so the total surface of the A cuboid is

2lb + 2bh + 2hl = 2 (bh+ hl +lb)

For a cube, l = b = h = a (say)

Surface of a cube = 2(a.a + a.a + a.a) = 6a2

Volume And Surface Area Of Solids Total Surface Of A Cubiod And A Cube

The Length of the Diagonal of a Cuboid and a Cube

Diagonal of a floor \(D^{\prime} B^{\prime}=\sqrt{l^2+b^2}\). So, if we want to find the length of the longest rod DB’, then we make a new right-angled A treating, DB’ as hypotenuse, base as D’B’ and height as DD’ which is h. So, the length of the diagonal of a cuboid DB’.

= \(\sqrt{\left(D^{\prime} B^{\prime}\right)^2+\left(D D^{\prime}\right)^2}=\sqrt{l^2+b^2+h^2}\)

For a cube, l = b = h = a (say)

∴ The length of the diagonal of a cube = \(\sqrt{a^2+a^2+a^2}=a \sqrt{3}\)

Lateral Surface Area of a Cuboid and a Cube

  • Lateral surface area of a cuboid = 2(l + b)h. [CSA = TSA- (area of top and area of bottom)]
  • Lateral surface area of a cube = 4a2, where a = edge of the cube. It is also called the curved surface area or area of 4 walls.

The volume of a Cuboid and a Cube

The volume of any solid figure is the amount of space enclosed within its bounding faces.

The volume of a cuboid of length l units, breadth b units and height h units are lbh cubic units.

= (length x breadth x height) cubic units

= (area of the base x height) cubic units.

For a cube,

Volume of cube = a3 cubic units,

where, the length of the edge of the cube is a unit.

Note: The measurements of volume are as follows:

1 cm3 = 103 cubic mm

1 m3 = 1000 litre = 1 kl

Volume And Surface Area Of Solids Cylinder

If a rectangle is revolved about one of its sides as its axis, the solid so formed is called a right circular cylinder.

The side AD about which the rectangle ABCD revolves is called the height of the cylinder.

Volume And Surface Area Of Solids Right Circular Cylinder

The line CD is called the generating line because when it revolves round AB, it generates the cylinder.

If you cut the hollow cylinder along AD and spread the piece, it becomes a rectangle whose one side is AD and the other side is AB which is equal to the circumference of a circle whose radius is the radius of the cylinder.

Volume And Surface Area Of Solids Cylinder

Hence, Side AB = 2πr

If h is the height of the cylinder, the area of ABCD

= AB x AD = 2πrh

Thus, the area of the curved surface of the cylinder

= 2πrh = perimeter of base x height

Whole Surface of a Cylinder

The whole surface area of a cylinder

= Curved surface + Area of the base + Area of the top

= 2πrh + πr2 + nr2 = 2πrh + 2πr2 = 2πr(h + r)

Volume of a Cylinder

The volume of a cylinder = Area of the base x Height

= nr2 x h = nr2h

Note: If the top and bottom are removed from the total surface area, then only the curved surface area remains.

Volume And Surface Area Of Solids Hollow Cylinder

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.

For a hollow cylinder of height h and with external and internal radii R and r respectively, we have

Volume of the material = Exterior volume- Interior volume
= πR2h- πr2h = πh (R2– r2)

The curved surface of the hollow cylinder
= External surface + Internal surface = 2πRh + 2πrh = 2πh(R + r)

The total surface area of the hollow cylinder
= Curved surface + 2(Area of base rings)
= (2πRh + 2πrh) + 2(πR2– πr2) = 2πh(R + r) + 2π(R2 – r2)
= 2π(R + r) (h+R-r)

Volume And Surface Area Of Solids Right Circular Cone

A right circular cone is a solid generated by the revolution of a right-angled triangle about sides containing the right angle as the axis.

Let CAB be a right-angled triangle, right-angled at A. The hypotenuse CB revolves around the side CA as the fixed axis. The hypotenuse CB will generate the curved surface of a cone.

The radius of the circular base is called the radius of the cone. It is usually denoted by ‘r’.

The point C is called the vertex of the cone.

The length CA of the axis is called the height of the cone. It is usually ‘denoted by ‘h’.

The hypotenuse CB is called the generating line of the cone and its length is called the slant height. It is usually denoted by l.

Volume And Surface Area Of Solids Right Circular Cone

The volume of Right Circular Cone

Volume of a cone = \(\frac{1}{3}\) (area of the base) x height’

i.e., \(V=\frac{1}{3} \pi r^2 h\)

where r = radius of the base and h = height.

Slant height of right circular cone = \(l=\sqrt{h^2+r^2}\)

The curved surface of the Right Circular Cone

Curved surface of a cone = \(\pi r l=\pi r \sqrt{h^2+r^2}\)

where r = radius of the base, l = slant height of the cone.

Total Surface of Right Circular Cone

Total surface of the cone = Area of the base + Area of the curved surface

i.e., total surface of the cone = πr2 + πrl = πr(r + l).

Volume And Surface Area Of Solids Sphere

When a circle is revolved about its diameter, the solid thus formed is a sphere. Let AB be the diameter of a circle ADB and 0 be its centre.

If the circle ADB revolves around AB, point D takes different positions and a closed solid is formed.

A sphere may also be defined as a solid bounded by a closed surface, all points on which are at a constant distance from a fixed point.

Curved Surface Area and Volume of a Sphere

The curved surface area of a sphere = 4πr2, where r is the radius of the sphere.

The volume of sphere = \(\frac{4}{3} \pi r^3\)

Volume And Surface Area Of Solids Curved Surface Area And Volume Of A Sphere

Volume And Surface Area Of Solids Hemisphere

A plane through the centre of a sphere divides the sphere into two equal parts. Each part is called a hemisphere.

If r is the radius of the hemisphere, then

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

Curved surface area = 2πr2

Total surface area = 2πr2 + πr2 = 3πr2.

Volume And Surface Area Of Solids Hemisphere

Volume And Surface Area Of Solids Spherical Shell

A spherical shell having external radius R and internal radius r, then we have

Volume of material = \(\frac{4}{3} \pi R^3-\frac{4}{3} \pi r^3=\frac{4}{3} \pi\left(R^3-r^3\right)\)

Volume And Surface Area Of Solids Solved Examples

Question 1. A tent of cloth is cylindrical upto 1 m in height and conical above it of the same radius of base. If the diameter of the tent is 6 m and the slant height of the conical part is 5 m, find the cloth required to make this tent.

Solution:

Diameter of base 2r = 6 m

⇒ \(r=\frac{6}{2}=3 \mathrm{~m}\)

Height of cylindrical path h = 1 m

The slant height of conical part l = 5 m

Cloth required in tent = 2πrh + πrl

= πr (2h + l)

= \(\frac{22}{7} \times 3(2 \times 1+5)=66 \mathrm{~m}^2\)

Volume And Surface Area Of Solids A Tent Of Cloth Is Cylindrical

Question 2. The base and top of a right circular cylindrical drum are hemispherical. The diameter of the cylindrical part is 14 cm and the total height is 30 cm. Find the total surface area of the drum.

Solution:

Here 2r = 14

⇒ r = 7 cm

Height of cylindrical part h = 30 – 2 x 7 = 16 cm

The total surface area of the drum = 2 x the Curved surface of the hemisphere + the Curved surface of the cylinder

= 2 x 2πr2 + 2πrh

= 2πr(2r + h)

= \(2 \times \frac{22}{7} \times 7(2 \times 7+16)\)

= 44 x 30 = 1320 cm2

Volume And Surface Area Of Solids Circular Drum Are Hemispherical

Question 3. Three cubes, each with an 8 cm edge, are joined end to end. Find the total surface area of the resulting cuboid.

Solution:

As is clear from the adjoining figure;

the length of the resulting cuboid = 3 x 8 cm = 24 cm

Its width = 8 cm and its height = 8 cm

i.e., l = 24 cm, b = 8 cm and h = 8 cm

∴ The total surface area of the resulting cuboid

= 2(1 x b + b x h + h x l)

= 2 (24 x 8 + 8 x 8 + 8 x 24) cm2 = 896 cm2

Volume And Surface Area Of Solids Three Cubes

Question 4. A rectangular sheet of tin 58 cm x 44 cm is to be made into an open box by cutting off equal squares from the corners and folding up the flaps. What should be the volume of box if the surface area of box is 2452 cm2?

Solution:

Let a square of x cm from each corner be removed from a rectangular sheet.

So, length of box = (58 – 2x) cm

breadth of box = (44 – 2x) cm

and height of box = x cm

∴ Surface area of open box = 2 (lb + bh + hl)- lb

⇒ 2[(58 – 2x) (44 – 2x) + x (44 – 2x) + x (58 – 2x)]- (58 – 2x)(44 – lx) = 2452

⇒ (58- 2x) (44- 2x) + 2x(44- 2x) + 2x (58 – 2x) = 2452

⇒ (29 – x) (22 – x) + x (22 – x) + x (29 – x) = 613

⇒ 638 – 51x + x2 + 22x – x2 + 29x – x2 = 613

⇒ x2 = 25

⇒ x = 5 cm (x = – 5 is not admissible)

∴ Volume of box = x(58 – 2x) (44- 2x) = 5(58- 10)(44 – 10)

= 5 x 48 x 34 cm3 = 8160 cm3

Volume And Surface Area Of Solids A Rectangular Sheet

Question 5. The radii of two right circular cylinders are in the ratio 2 : 3 and their heights are in the ratio 5:4. Calculate the ratio of their curved surface areas and also the ratio of their volumes.

Solution:

Let the radii of the two cylinders be 2r and 3r respectively and their heights be 5h and 4h.

∴ \(\frac{\text { Curved surface area of 1st cylinder }\left(S_1\right)}{\text { Curved surface area of 2nd cylinder }\left(S_2\right)}=\frac{2 \pi \times 2 r \times 5 h}{2 \pi \times 3 r \times 4 h}\)

i.e., \(\frac{S_1}{S_2}=\frac{5}{6}\) or S1: S2 = 5:6

⇒ \(\frac{\text { Volume of 1st cylinder }\left(V_1\right)}{\text { Volume of 2nd cylinder }\left(V_2\right)}=\frac{\pi \times(2 r)^2 \times 5 h}{\pi \times(3 r)^2 \times 4 h}\)

i.e., \(\frac{V_1}{V_2}=\frac{5}{9}\) or V1:V2= 5:9

Question 6. The volume and surface area of a solid hemisphere are numerically equal. What is the diameter of the hemisphere?

Solution:

We have,

Volume of hemisphere = Surface area of hemisphere

⇒ \(\frac{2}{3} \pi r^3=3 \pi r^2\) = 2r = 9

Hence, the diameter of the hemisphere = 9 units

Question 7. The curved surface area of a cone of height 8 m is 1 88.4 m2. Find the volume of a cone.

Solution:

πrl = 188.4

⇒ \(r l=\frac{188.4}{3.14}=60\)

⇒ r2l2 = 3600

⇒ r2(h2 + r2) = 3600

⇒ r2(64 + r2) = 3600

⇒ r4 + 64r2 – 3600 = 0

⇒ (r2+ 100) (r2– 36) = 0

∴ r2 = -100 or r2 = 36

⇒ r = 6 (∵ r2 = -100 is not possible)

∴ Volume of a cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times 3.14 \times 36 \times 8=301.44 \mathrm{~m}^3\)

Volume And Surface Area Of Solids Curved Surface Of A Cone

Question 8. A conical tent is required to accommodate 157 persons, each person must have 2 m2 of space on the ground and 15 m3 of air to breathe. Find the height of the tent; also calculate the slant height.

Solution:

1 person needs 2 m2 of space.

∴ 157 persons needs 2 x 157 m space on the ground

∴ r2 = 2 x 157

∴ \(r^2=\frac{2 \times 157}{3.14}=100\) = r = 100

Also, 1 person needs 15 m of air.

∴ 157 persons need 15 x 157 m3 of air.

∴ \(\frac{1}{3} \pi r^2 h=15 \times 157\)

⇒ \(h=\frac{15 \times 157 \times 3}{3.14 \times 100}=22.5 \mathrm{~m}\)

∴ l2 =h2 + r2 = (22.5)2 + (10)2 = 606.25

∴ \(l=\sqrt{606.25}=24.62 \mathrm{~m}\)

 

Volume And Surface Area Of Solids A Conical Tent

Question 9. The radius and height of a solid right circular cone are in the ratio of 5:12. If its volume is 314 cm find its total surface area. [Take π = 3.14]

Solution:

Let the radius of the cone = 5x

∴ Height of cone = 12r

∴ l2 = (5r)2 + (12x)2 = 169 x2

∴ \(l=\sqrt{169 x^2}=13 x\)

It is given that volume = 314 cm2

∴ \(\frac{1}{3} \pi(5 x)^2(12 x)=314\)

⇒ \(\frac{1}{3} \times 3.14 \times 25 \times 12 \times x^3=314\)

⇒ \(x^3=\frac{314 \times 3}{3.14 \times 25 \times 12}=1\)

∴ x = 1

∴ Radius r = 5 x 1 = 5 cm

Height h = 1 2 x 1 = 1 2 cm

and slant height l= 13 x I = 13 cm

Now, total surface area of cone = πr(l + r) = 3.14 x 5 (13 + 5) = 3.14 x 5 x 18

= 282.60 cm2

Hence, the total surface area of a cone is 282.60 cm.

Volume And Surface Area Of Solids Solid Right Circular Cone

Question 10. If h, C, and V respectively are the height, the curved surface area and the volume of a cone. Prove that 3πVh3 – C2h2 + 9V2 = 0.

Solution:

⇒ \(C=\pi r l, V=\frac{1}{3} \pi r^2 h, l^2=h^2+r^2\)

L.H.S. = 3πVh3 – C2h2 + 9V2

= \(3 \pi\left(\frac{1}{3} \pi r^2 h\right) h^3-(\pi r l)^2 h^2+9\left(\frac{1}{3} \pi r^2 h\right)^2\)

= \(\pi^2 r^2 h^4-\pi^2 r^2 h^2\left(h^2+r^2\right)+9 \times \frac{1}{9} \pi^2 r^4 h^2\)

= π2r2h4 – π2r2h4 – π2r4h2 + π2r4h2 = 0 = R.H.S. Hence Proved.

Question 11. A cone of equal height and equal base is cut off from a cylinder of height 24 cm and base radius 7 cm. Find the total surface and volume of the remaining solid.

Solution:

Here r = 7 cm, h = 24 cm

The volume of remaining solid = Volume of the cylinder – Volume of a cone

= \(\pi r^2 h-\frac{1}{3} \pi r^2 h=\frac{2}{3} \pi r^2 h\)

= \(\frac{2}{3} \times \frac{22}{7} \times 7 \times 7 \times 24=2464 \mathrm{~cm}^3\)

Now, l2 = h2 + r2 = (24)2 + (7)2

= 576 + 49 = 625

⇒ l = 25 cm

∴ Total surface area of remaining solid

= Curved surface of cylinder + Curved surface of cone + Area of top

= 2πrh +πrI + πr2

= πr(2h + l + r)

= \(\frac{22}{7} \times 7(2 \times 24+25+7)=22 \times 80=1760 \mathrm{~cm}^2\)

Volume And Surface Area Of Solids A Cone Of Equal Height And Equal Base Is Cut Off From A Cylinder

Question 12: From a wooden cubical block of edge 7 cm, the largest possible right conical piece is cut out whose base is on one of the faces of the cube. Calculate:

  1. The volume of the wood left in the block, and
  2. the total surface area of the block left. (\(\text { Take } \pi=\frac{22}{7}\))

Solution:

Initial volume of the block = a3 = 73 cm3

The base of the largest cone will touch the sides of the base of the cube, and the height will be equal to the length of the edge of the cube.

∴ For the cone, r = 3.5 cm and h = 7 cm.

1. The volume of the cone = \(\frac{1}{3} \pi r^2 h=\frac{\pi}{3} \cdot\left(\frac{7}{2}\right)^2 \cdot 7 \mathrm{~cm}^3\)

∴ volume of the wood left = \(\left\{7^3-\frac{1}{3} \cdot \frac{22}{7} \cdot\left(\frac{7}{2}\right)^2 \cdot 7\right\} \mathrm{cm}^3\)

= \(\left(7^3-\frac{11 \times 7^2}{3 \times 2}\right) \mathrm{cm}^3=7^2\left(7-\frac{11}{6}\right) \mathrm{cm}^3\)

= \(\frac{49 \times 31}{6} \mathrm{~cm}^3=\frac{1519}{6} \mathrm{~cm}^3=253 \frac{1}{6} \mathrm{~cm}^3\)

2. The total surface area of the wood left

= Total surface area of the cube- Area of the base of the cone + Curved surface area of the cone.

= 6a2 – πr2 + πrl

= \(\left\{6 \times 7^2-\frac{22}{7} \cdot\left(\frac{7}{2}\right)^2+\frac{22}{7} \cdot \frac{7}{2} \cdot \sqrt{7^2+\left(\frac{7}{2}\right)^2}\right\} \mathrm{cm}^2\)

= \(\left(6 \times 49-\frac{11 \times 7}{2}+11 \cdot 7 \cdot \frac{\sqrt{5}}{2}\right) \mathrm{cm}^2=\left\{294+\frac{77}{2}(\sqrt{5}-1)\right\} \mathrm{cm}^2\)

= \(\left(294+\frac{77}{2} \times 1.24\right) \mathrm{cm}^2=341.74 \mathrm{~cm}^2\)

Volume And Surface Area Of Solids A Wooden Cubical Block

Question 13. A sphere is inscribed in a cylinder such that the sphere touches the cylinder. Show that the curved surface is equal to the curved surface of the cylinder.

Solution:

A cylinder circumscribed a sphere is shown.

Here, the radius of the sphere = radius of the cylinder = r

Height of cylinder h = 2r

Now, the curved surface of the sphere = 4πr2

and the curved surface of sphere of cylinder = 2πrh = 2πr(2r) = 4πr2

Therefore, the curved surface of the sphere and the curved surface of the cylinder are equal.

Hence proved.

Volume And Surface Area Of Solids A Sphere

Question 14. The largest possible cube Is made from a wooden sphere of radius 6√3 cm. Find the surface area of the cube.

Solution:

Here, a diagonal of the cube will be the diameter of the sphere

∴ length of a diagonal of the cube = 2 x 6√3 cm = 12√3 cm

If an edge of the cube is a then a diagonal of the cube = √3 a

= 12√3 cm

∴ a = 12 cm.

∴ The surface area of the cube = 6a2 = 6.(12)2 cm2 = 864 cm2

Volume And Surface Area Of Solids Surface Area Of A Cube

Question 15. A hemisphere is inscribed in a cylinder and a cone is inscribed in the hemisphere. The cone vertex lies in the centre of the upper circular part of the cylinder. Show that: \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{2}\)x Volume of hemisphere = Volume of cone

Solution:

Let radius of cone = radius of cylinder = radius of hemisphere = r

∴ Height of cone = Height of cylinder = r

Now, the volume of cone = \(\frac{1}{3} \pi r^2 \cdot(r)=\frac{1}{3} \pi r^3\) → (1)

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

⇒ \(\frac{1}{2}\)x Volume of hemisphere = \(\frac{1}{3} \pi r^3\) → (2)

and Volume of cylinder = πr2(r) = r3

⇒ \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{3} \pi r^3\) → (3)

From equations (1), (2) and (3)

⇒ \(\frac{1}{3}\) x Volume of cylinder = \(\frac{1}{2}\)x Volume of hemisphere

= Volume of cone

Hence proved.

Volume And Surface Area Of Solids A Hemisphere

Question 16. Water in a canal, 5.4 m wide and 1.8 m deep is flowing at a speed of 25 km/hr. How much area can it irrigate in 40 minutes, if 10 cm of standing water is required for irrigation?

Solution:

Width of canal = 5.4 m

Height of canal = 1.8 m

In one hour, water is moving upto 25 km = 25000 m

∴ We treat it as the length of the canal.

∴ Volume of water in canal in 1 hour = (25000 x 5.4 x 1.8)m3

⇒ Volume of water in the canal in 40 minutes = 243000 m3

∴ Volume of water in canal in 60 minutes = \(\frac{243000}{60} \times 40 \mathrm{~m}^3\)

= 162000 m3

This water can irrigate a field upto the height of 10 cm = 0. 1 m

∴ Volume = Area = Area x height

⇒ 162000 = Area x 0.1

⇒ Area of field = \(\frac{162000}{0.1} \mathrm{~m}^2=1620000 \mathrm{~m}^2\) (∵Note: 1 Hectare = 1000 m)

= \(\frac{1620000}{10000} \text { hectare }\)

= 162 hectare.

So, 162 hectare area can be irrigated.

Conversion Of Solid Form One Shape To Another And Mixed Problems

When a solid is converted into another without any loss of material then its volume remains the same.

If a larger solid is converted into smaller solids then the number of smaller solids

= \(\frac{\text { Volume of larger solid }}{\text { Volume of } 1 \text { smaller solid }}\)

Conversion Of Solid Form One Shape To Another And Mixed Problems Solved Examples

Question 1. A solid metallic cuboid of dimensions 9 m x 8 m x 2 m is melted and recast into solid cubes of edge 2m. Find the number of cubes so formed.

Solution:

Volume And Surface Area Of Solids A Solid Metallic Cuboid

Let x cubes each of edge 2 m are formed to melt a cuboid of dimensions 9 x 8 x 2.

So, \(\text { number of cubes }=\frac{\text { Volume of cuboid }}{\text { Volume of } 1 \text { cube }}\)

∴ \(x=\frac{9 \times 8 \times 2}{2}=72\)

So, 72 cubes will be formed

Question 2. Three cubes of metal whose edges are in the ratio 3:4 are melted down into a single cube whose diagonal is 12√3 cm, Find the edges of the three cubes.

Solution:

The ratio in the edges = 3:4:5

Let edges be 3x, 4x and 5x respectively.

∴ Volumes of three cubes will be 27x3, 64x3 and 125x3 in cm3 respectively.

Now, sum of the volumes of these three cubes = 27x3 + 64x3 + 125 x3 = 216 x3 cm3

Let the edge of the new cube be a cm.

Diagonal of new cube = a√3 cm

a√3 = 12√3

Volume of new cube = (12)3 = 1728 cm3

Now by the given condition

216x3 = 1728

⇒ x3 = 8

⇒ x = 2

∴ \(\left.\begin{array}{rl}
\text { Edge of 1 cube } & =3 \times 2=6 \mathrm{~cm} \\
\text { Edge of 2 cube } & =4 \times 2=8 \mathrm{~cm} \\
\text { Edge of 3 cube } & =5 \times 2=10 \mathrm{~cm}
\end{array}\right\}\).

Question 3. A cube of metal with a 2.5 cm edge is melted and cast into a rectangular solid whose base is 1.25 cm by 0.25 cm. Assuming no loss in melting find the height of the solid. Also, find the gain in the surface area.

Solution:

The volume of the cube = (edge)3 = (2.5)3 cu. cm

Area of the base of rectangular solid = 1 .25 x 0.25 sq. cm

∴ Height of solid = \(\frac{\text { Volume }}{\text { Area of base }}=\frac{(2.5)^3}{1.25 \times 0.25}=50 \mathrm{~cm}\)

Surface area of the cube = 6 x (edge)2 = 6 x (2.5)2 = 37.5 sq. cm

Surface area of the solid = 2(lb + bh + hl) = 2(50 x 1.25 + 1.25 x 0.25 + 50 x 0.25)

= 2(62.5 + 0.3125 + 12.5) = 150.625 sq. cm

∴ Gain in surface area = 150.625 – 37.50 = 113.125 sq. cm

Question 4. A granary is in the shape of a cuboid of size 8m x 6m x 3m. If a bag of grain occupies a space of 0.65 m3, how many bags can be stored in the granary?

Solution:

The size of the granary is 8 m x 6 m x 3 m,

Volume of granary = 8 x 6 x 3 = 144m3

The volume of one bag of grain = 0.65 m3

The number of bags which can be stored in the granary

= \(\frac{\text { Volume of granary }}{\text { Volume of each bag }}=\frac{144}{0.65}=221.54 \text { or } 221 \text { bags. }\)

Question 5. A cylinderical bucket 28 cm in diameter and 72 cm high is full of water. The water is emptied into a rectangular tank 66 cm long and 28 cm wide. Find the height of the water level in the tank.

Solution:

Let the height of the water level in the tank = x m,

then according to the problem

πr2h = l x b x x

or \(\frac{22}{7} \times 14 \times 14 \times 72=66 \times 28 \times x\)

or \(x=\frac{\frac{22}{7} \times 14 \times 14 \times 72}{66 \times 28}=24 \mathrm{~cm}\)

Question 6. A rectangular container whose base is a square of side 15 cm stands on a horizontal table and holds water upto 3 cm from the top. When a cube is placed in the water and is completely submerged, the water rises to the top and 54 cm of water overflows. Calculate the volume of the cube and its surface area.

Solution:

Volume of the cube submerged = Volume of water that fills 3 cm height of the container + Volume of water that overflows

= 15 x 15 x 3 + 54 = 729 cm3

If the side of the cube submerged = x cm

Its volume = x3 cm3

∴ x3 = 729 = 9 x 9 x 9

⇒ x = 9 cm.

∴ The side of the cube = 9 cm

And its surface area = 6 x (side) =6 x 9 x 9 = 486 cm2

Question 7. A solid spherical ball of iron with a radius of 6 cm is melted and recast into three solid spherical balls. The radii of the two of balls are 3 cm and 4 cm respectively, determine the diameter of the third ball.

Solution:

Let the radius of the third ball = r cm

∴ The volume of three balls formed = Volume of the ball melted

⇒ \(\frac{4}{3} \pi(3)^3+\frac{4}{3} \pi(4)^3+\frac{4}{3} \pi(r)^3=\frac{4}{3} \pi(6)^3\)

⇒ 27 + 64 + r3 = 216

⇒ r3 = 125, i.e., r = 5 cm

∴ The diameter of the third ball = 2 x 5 cm = 10 cm

Question 8. 50 circular plates each of radius 7 cm and thickness 0.5 cm are placed one above the other to form a solid right circular cylinder. Find

  1. The total surface area and
  2. The volume of the cylinder so formed

Solution:

The height of the cylinder formed by placing 50 plates = 50 x 0.5 = 25 cm

Radius of cylinder formed = Radius of plate = 7 cm

1. Total surface area of cylinder = 2πrh + 2πr2

= \(\left[2 \times \frac{22}{7} \times 7 \times 25+2 \times \frac{22}{7} \times(7)^2\right] \mathrm{cm}^2\)

= (1100 + 308) cm2 = 1408 cm2

2. Volume of cylinder = \(\pi r^2 h=\frac{22}{7} \times(7)^2 \times 25=3850 \mathrm{~cm}^3\)

Question 9. A rectangular paper of 22 cm x 12 cm is folded in two different ways and forms two cylinders.

  1. Find the ratio of the volumes of two cylinders.
  2. Find the difference in the volumes of the two cylinders.

Volume And Surface Area Of Solids Two Cylinders

Solution:

When the paper is folded along a 12 cm side,

then height of cylinder h1 = 22 cm and 2πr1 = 12

⇒ \(r_1=\frac{12}{2 \pi}=\frac{6}{\pi} \mathrm{cm}\)

∴ Volume = \(V_1=\pi r_1^2 h_1=\pi \times\left(\frac{6}{\pi}\right)^2 \times 22=\frac{792}{\pi}=\frac{792 \times 7}{22}=252 \mathrm{~cm}^3\)

When the paper is folded along a 22 cm side,

then height of cylinder h2 = 12 cm and 2πr2 = 22

⇒ \(2 \times \frac{22}{7} \times r_2=22\)

⇒ \(r_2=\frac{7}{2} \mathrm{~cm}\)

∴ Volume V2 = r22h2

= \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2 \times 12\)

= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12=462 \mathrm{~cm}^3\)

  1. Ratio of volumes, V1:V2 = 252 : 462 = 6: 11
  2. Difference in volume = (462 – 252) cm3 = 210 cm3

Volume And Surface Area Of Solids When The Paper Is Folded

Question 10. A semicircle of radius 17.5 cm is rotated about its diameter. Find the curved surface of the generated solid.

Solution:

The solid generated by a circle rotated about its diameter is a sphere.

Now, the radius of the sphere r = 17.5 cm.

and its curved surface = \(4 \pi r^2=4 \times \frac{22}{7} \times 17.5 \times 17.5=3850 \mathrm{~cm}^2\)

Question 11. A sphere of radius 6 cm is melted and recast into a cone of height 6 cm. Find the radius of the cone.

Solution:

Radius of sphere = 6 cm

∴ Volume of sphere = \(\frac{4}{3} \pi(6)^3=288 \pi \mathrm{cm}^3\)

Let the radius of cone = r

Height of cone = 6 cm

∴ Volume of cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi r^2 \times 6=2 \pi r^2\)

Given that, Volume of cone = Volume of a sphere

⇒ 2nr2 – 288 7r

⇒ r2= 144

⇒ r = 12

Therefore, the radius of the cone = 12 cm

Question 12. The height and radius of the base of a metallic cone are 27 cm and 16 cm respectively. It is melted and recast into a sphere. Find the radius and curved surface of the sphere.

Solution:

Height of cone h = 27 cm

The radius of cone r = 16 cm

∴ Volume of cone = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi(16)^2 \times 27 \mathrm{~cm}^3\)

Let the radius of the sphere = R

∴ Volume of sphere = \(\frac{4}{3} \pi R^3\)

Now, Volume of sphere = Volume of cone

⇒ \(\frac{4}{3} \pi R^3=\frac{1}{3} \pi(16)^2 \times 27\)

⇒ \(R^3=\frac{16^2 \times 27}{4}=4^3 \times 3^3\)

R = 4 x 3 = 12cm

and curved surface of sphere = 4R2 = 4(12)2 = 576 cm2

Question 13. The radius of a metallic sphere is 60 mm. It is melted and recast into a wire of diameter 0.8 mm. Find the length of the wire.

Solution:

Radius of sphere = 60 mm = 6 cm

∴ Volume of sphere = \(\frac{4}{3} \pi(6)^3=288 \pi \mathrm{cm}^3\)

Let, the length of wire = l cm

Radius of wire \(r=\frac{0.8}{2} \mathrm{~mm}=0.4 \mathrm{~mm}=\frac{0.4}{10} \mathrm{~cm}=\frac{4}{100} \mathrm{~cm}\)

∴ Volume of wire = \(\pi r^2 l=\pi\left(\frac{4}{100}\right)^2 \cdot l\)

Now, Volume of wire = Volume of a sphere

⇒ \(\pi \times \frac{4}{100} \times \frac{4}{100} \times l=288 \pi\)

⇒ \(l=\frac{288 \times 100 \times 100}{4 \times 4}=180000 \mathrm{~cm}=1800 \mathrm{~m}\)

Question 14. A wire of diameter 3 mm is wound around a cylinder whose height is 1 2 cm and radius 5 cm so as to cover the curved surface of the cylinder completely. Find

  1. Length of the wire.
  2. Mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution:

1. Let the wire wounded around the cylinder complete n revolutions.

Diameter (width) of wire = 3 m- 0.3 cm

So, the whole height of the cylinder = 0.3 n cm

But the whole height of the cylinder = 12 cm

∴ \(0.3 n=12 \Rightarrow n=\frac{12}{0.3}=\frac{120}{3}=40\)

So, 40 revolutions are completed to wound the wire completely on the cylinder.

In 1 revolution, the length of the wire = 2 πr

∴ In 40 revolutions, the length of the wire = 40 x 2πr

= \(80 \times \frac{22}{7} \times 5 \mathrm{~cm}=1257.14 \mathrm{~cm}\)

= 12.57 m.

2. Now, radius of wire = \(\frac{0.3}{2}=0.15 \mathrm{~cm}\)

Volume of wire = area of cross section x length of wire

= π(0.15)2 x 1257.14 = 88.898 cm3

∴ Mass of wire = volume x density

= 88.898 x 8.88 g= 789.41 g

Hence, the length of the wire = 12.57 m

and mass of wire = 789.41 g

Volume And Surface Area Of Solids A Wire Of Diameter

Question 15. A metallic cylinder of diameter 16 cm and height 9 cm is melted and recast into a sphere of diameter 6 cm. How many such spheres can be formed?

Solution:

For the cylinder,

Radius = \(\frac{16}{2} = 8cm[latex]

Height = 9cm

∴ Volume of cylinder = JI(8)2(9) = 576K cm3

Diameter of sphere = 6 cm

∴ Radius of sphere = [latex]\frac{6}{2}=3 cm\)

Now, volume of one sphere = \(\frac{4}{3} \pi(3)^3=36 \pi \mathrm{cm}^3\)

∴ Number of spheres formed = \(\frac{\text { Volume of cylinder }}{\text { Volume of one sphere }}=\frac{576 \pi}{36 \pi}=16\)

Question 16. A solid metallic right circular cone of height 6.75 cm and radius of the base 12 cm is melted and two solid spheres formed from it. If the volume of one of the spheres is 8 times that of the other, find the radius of the smaller sphere.

Solution:

Let the radius of the smaller sphere be r and the radius of the larger sphere be R.

image

So, \(\frac{1}{3} \pi(12)^2 \times 6.75=\frac{4}{3} \pi r^3+\frac{4}{3} \pi R^3\)

⇒ (12)2 x 6.75 = 4(r3 + R3) → (1)

Now, the volume of the larger sphere = 8 x the Volume of the smaller sphere

⇒ \(\frac{4}{3} \pi R^3=8 \times \frac{4}{3} \pi r^3\)

R3 = 8r3 = R = 2r → (2)

From equations (1) and (2), we get

12 x 12 x 6.75 = 4(r3 + 8r3)

⇒ \(9 r^3=\frac{12 \times 12 \times 6.75}{4}\)

⇒ \(r^3=\frac{12 \times 12 \times 6.15}{4 \times 9}=27=(3)^3\)

r = 3 cm.

Hence, the radius of a smaller sphere = 3 cm.

Volume And Surface Area Of Solids A Solid Metallic Right Circular Cone

Question 17. Water is flowing at the rate of 3 km an hour through a circular pipe of 20 cm internal diameter into a circular cistern of diameter 10 m and depth 2m. In how much time will the cistern be filled?

Solution:

Internal diameter of pipe = 20 cm

∴ Internal radius of pipe = 10 cm = \(\frac{1}{10}m\)

Rate of flow of water = 3 km/h

⇒ Rate of flow of water = \(\frac{3 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{sec}\)

Diameter of cistern = 10 m

Radius of cistern = 5 m

Height of cistern = 2 m

Water discharge by pipe in 1 sec

= π x r2 x flow of water

= \(\pi \times\left(\frac{1}{10}\right)^2 \times \frac{3 \times 1000}{60 \times 60} \mathrm{~m}^3 / \mathrm{sec}\)

Volume of cistern = π x 52 x 2 m3

Time taken to fill the cistern = \(\frac{\text { Volume of cistern }}{\text { Volume of water discharge in }1 \mathrm{sec}}\)

= \(\frac{\pi \times 25 \times 2}{\pi \times \frac{1}{100} \times \frac{3 \times 1000}{60 \times 60}} \mathrm{sec}=\frac{25 \times 2 \times 100 \times 60 \times 60}{3 \times 1000} \mathrm{sec}\)

= \(\frac{5 \times 60 \times 60}{3} \sec =\frac{5 \times 60 \times 60}{3 \times 60 \times 60} \text { hours }=\frac{5}{3} \text { hours }\)

Question 18. The volume of a sphere is 288 π cm3. 27 small spheres can be formed with this sphere. Find the radius of a small sphere.

Solution:

Volume of 27 small spheres = Volume of one big sphere = 288π

Volume of 1 small sphere = \(\frac{288}{27} \pi\)

\(\frac{4}{3} \pi r^3=\frac{32}{3} \pi\) where r = radius of small sphere

r3 = 8

r = 2 cm

Question 19. Find:

  1. The lateral or curved surface area of a closed cylindrical petrol storage tank is 4.2 m in diameter and 4.5 m high.
  2. How much steel was actually used, if 1/12 of the steel actually used was wasted in making the tank?

Solution:

1. Curved surface area = \(2 \pi r h=2 \times \frac{22}{7} \times \frac{4.2}{2} \times 4.5=59.4 \mathrm{~m}^2\)

2. Total steel used = Total surface area (Assuming thickness = 0 m)

= 2πr(r + h)

= \(2 \times \frac{22}{7} \times \frac{4.2}{2}\left(\frac{4.2}{2}+4.5\right)\)

= \(\frac{44}{7} \times 2.1 \times \frac{13.2}{2}=22 \times 0.3 \times 13.2=87.12 \mathrm{~m}^2\)

Let the actual steel used be x m2

= \(\frac{11}{12} x=87.12\)

= \(x=\frac{87.12 \times 12}{11}=95.04 \mathrm{~m}^2\)

Question 20. In one fortnight (15 days) of a given month, there was a rainfall of 11.10 cm in a river valley. If the area of the valley is 97280 km2, show that the total rainfall of a day was approximately equivalent to the addition (sum) to the normal water of three rivers earth 1072 km long, 7.5 m wide and 3m deep. (Round off the volumes upto one place of decimal).

Solution:

Area of the valley = 97280 km2 = 9.728 x 104 km2

= 9.728 x 1010 m2

∴ Volume (amount) of rainfall in the valley in one fortnight (15 days)

= \(9.728 \times 10^{10} \times \frac{11.10}{100} \mathrm{~m}^3=107.98 \times 10^8 \mathrm{~m}^3\)

∴ Amount of rainfall in 1 day = \(\frac{107.98 \times 10^8}{15} \mathrm{~m}^3\)

= 7.198 x 108 m3 = 7.2 x 108m3 → (1)

Now, length of each river = 1072 km = 1072000 m

breadth of each river = 75 m

and depth of each river = 3 m

∴ Amount (volume) of water in each river = 1072000 x 75 x 3 m3

∴ Volume of water in 3 such rivers = 3 x 1.072 x 106 x 75 x 3 m3

= 7.236 x 108 m3 = 7.2 x 108 m3 → (2)

From (1 ) and (2), we can say that total rainfall in a day is approximately the same as the volume of 3 rivers.

Question 21. The dimensions of a solid iron cuboid are 4.4 m x 2.6 m x 1.0 m. It is melted and recast into a hollow cylindrical pipe of 30 cm inner radius and thickness of 5 cm. Find the length of the pipe.

Answer:

Length of solid iron cuboid = 4.4 m

Breadth of solid iron cuboid = 2.6 m

Height of solid iron cuboid = 1.0 m

∴ Volume of solid cuboid = 4.4 x 2.6 x 1 = 11.44 m3

= 11.44 x 100 x 100 x 100 cm3

This cuboid is melted and recast into a hollow cylinder.

∴ The volume of a cuboid = Volume of a hollow cylinder

⇒ 11.44 x 106 = πR2h – πr2h

⇒ 11.44 x 106 = πh(R2– r2) [∵ r = 30 cm, width = 5 cm, R = 35 cm]

⇒ \(11.44 \times 10^6=\frac{22}{7} \times h(R-r)(R+r)\)

⇒ 11 .44 x 106 x 7 = 22 x h (35 – 30) (35 + 30)

⇒ \(h=\frac{11.44 \times 10^6 \times 7}{22 \times 5 \times 65} \mathrm{~cm}=\frac{11440000 \times 7}{22 \times 5 \times 65} \mathrm{~cm}\)

= 11200 cm = 112 m

Hence, the length of the pipe = 112 m

Volume And Surface Area Of Solids The Dimension Of A Solid Iron Cubiod

Question 22. A tank measures 2 m long, 1.6 m wide and 1 m depth. Water is there upto 0.4 m in height. Brides measuring 25 cm x 14 cm x 10 cm are put into the tank so that water may come upto the top. Each brick absorbs water equal \(\frac{1}{7} \text { th }\) to that of its own volume. How many bricks will be needed?

Solution:

Let x bricks be needed

∴ Volume of x bricks = x X 0.25 X 0.1 4 X 0.1 m3

The volume of water absorbed by bricks

= \(=\frac{1}{7} x \times 0.25 \times 0.14 \times 0.1\)

∴ Remaining water in the tank = \(2 \times 1.6 \times 0.4-\frac{1}{7} \times x \times 0.25 \times 0.14 \times 0.1\)

Volume And Surface Area Of Solids A Tanl

Now, the Volume of water in the tank + Volume of x bricks

= Volume of tank

⇒ \(\left(2 \times 1.6 \times 0.4-\frac{x}{7} \times 0.25 \times 0.14 \times 0.1\right)+x \times 0.25 \times 0.14 \times 0.1\) = 2 x 1.6 x 1

⇒ \(\frac{6 x}{7} \times 0.25 \times 0.14 \times 0.1=2 \times 1.6 \times 0.6\)

⇒ \(x=\frac{2 \times 1.6 \times 0.6 \times 7}{6 \times 0.25 \times 0.14 \times 0.1}=\frac{2 \times 16 \times 6 \times 7 \times 1000}{6 \times 25 \times 14 \times 1}=640\)

Volume And Surface Area Of Solids A Brick

Question 23. A metallic solid sphere of radius 10.5 cm is melted and recast into smaller solid cones, each of radius 3.5 cm and height 3 cm. How many cones will be made?

Solution:

Volume And Surface Area Of Solids A Metallic Solid Sphere

Let n cones be recast from the sphere.

∴ The sum of volumes of n cones = volume of a sphere

⇒ \(n\left[\frac{1}{3} \pi(3.5)^2 \times 3\right]=\frac{4}{3} \pi(10.5)^3\)

⇒ \(n=\frac{4 \times 10.5 \times 10.5 \times 10.5}{3.5 \times 3.5 \times 3}=126\)

Hence, 126 cones will be made.

Frustum Of A Right Circular Cone

Frustum: If a right circular cone is cut by a plane parallel to the base of the cone then the portion between the plane and base is called the frustum of the cone.

How To Find The Volume And Surface Area Of A Bucket

Let a bucket of height h and radii of upper and lower ends be r1 and r2 respectively.

Now we shall find three parts:

  1. The slant is the height of the bucket.
  2. Curved surface area and total surface area of the bucket.
  3. The volume of the bucket or capacity in litres.

Proof :

1. Let the slant height of the bucket be l

Now, draw DC ⊥ AB

l2=h2 + (r1 – r2)2

⇒ \(l=\sqrt{h^2+\left(r_1-r_2\right)^2}\)

Volume And Surface Area Of Solids Alant Height Of A Bucket

2. Let EO = H and DO = L

ΔABO ~ ΔDEO

∴ \(\frac{A B}{D E}=\frac{B O}{E O}=\frac{A O}{D O}\)

⇒ \(\frac{r_1}{r_2}=\frac{h+H}{H}=\frac{l+L}{L}\)

⇒ \(\frac{r_1}{r_2}=\frac{h+H}{H} \quad \text { and } \quad \frac{r_1}{r_2}=\frac{l+L}{L}\)

⇒ Hr1 = hr2 + Hr2 and Lr1 = lr2 + Lr2

⇒ H(r1– r2) = hr2 ⇒ L(r1 – r2) = lr2

⇒ \(H=\frac{h r_2}{r_1-r_2} \quad \text { and } \quad L=\frac{l r_2}{r_1-r_2}\)

Volume And Surface Area Of Solids Total And Curved Surfaces Of Bucket

Curved surface area of bucket = C.S.A of larger cone- C.S.A of smaller cone

= \(\pi r_1(L+l)-\pi r_2 L=\pi r_1\left[\frac{l r_2}{r_1-r_2}+l\right]-\pi r_2\left(\frac{l r_2}{r_1-r_2}\right)\)

= \(\pi r_1 l\left(\frac{r_2+r_1-r_2}{r_1-r_2}\right)-\frac{\pi r_2^2 l}{r_1-r_2}\)

= \(\frac{\pi l}{r_1-r_2}\left(r_1^2-r_2^2\right)=\frac{\pi l\left(r_1+r_2\right)\left(r_1-r_2\right)}{\left(r_1-r_2\right)}\)

= \(\pi l\left(r_1+r_2\right) \text { where } l=\sqrt{h^2+\left(r_1-r_2\right)^2}\)

Total surface area = C.S.A. of bucket + Area of the smaller circle

= πl(r1 + r2) + πr22

3. Volume of bucket = Volume of larger cone – Volume of smaller cone

= \(\frac{1}{3} \pi l_l\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi r_1^2(H+h)-\frac{1}{3} \pi r_2^2 \cdot H\)

= \(\frac{1}{3} \pi r_1^2\left[\frac{h r_2}{r_1-r_2}+h\right]-\frac{1}{3} \pi r_2^2\left(\frac{h r_2}{r_1-r_2}\right)\)

= \(\frac{1}{3} \pi r_1^2 h\left(\frac{r_2+r_1-r_2}{r_1-r_2}\right)-\frac{1}{3} \frac{\pi h r_2^3}{r_1-r_2}\)

= \(\frac{1}{3} \frac{\pi h}{r_1-r_2}\left(r_1^3-r_2^3\right)\)

= \(\frac{1}{3} \frac{\pi h}{r_1-r_2}\left(r_1-r_2\right)\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi h\left(r_1^2+r_2^2+r_1 r_2\right)\)

Note: Actually, the bucket is a frustum cone made of cutting the bucket by a plane parallel to the base.

Frustum Of A Right Circular Cone Solved Examples

Question 1. A cone is divided into two parts by drawing a plane through the mid-point of its axis, parallel to its base, Compare the volume of the two parts.

Solution:

We can solve this using similarity.

Let r and li be the radius and height of a cone OAB

Let OE = \(\frac{h}{2}\)

As OED and OFB are similar

∴ \(\frac{O E}{O F}=\frac{E D}{F B} \frac{h / 2}{h}=\frac{E D}{r}\)

⇒ \(E D=\frac{r}{2}\)

Now volume of cone OCD = \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \pi \times\left(\frac{r}{2}\right)^2 \times \frac{h}{2}=\frac{\pi r^2 h}{24}\)

and Volume of cone OAB = \(\frac{1}{3} \times \pi \times r^2 \times h=\frac{\pi r^2 h}{3}\)

∴ \(\frac{\text { Volume of part } O C D}{\text { Volume of part } C D A B}=\frac{\frac{\pi r^2 h}{24}}{\frac{\pi r^2 h}{3}-\frac{\pi r^2 h}{24}}=\frac{\frac{1}{24}}{\frac{1}{3}-\frac{1}{24}}=\frac{\frac{1}{24}}{\frac{8-1}{24}}=\frac{1}{7}\)

Volume And Surface Area Of Solids A Cone Is Divided Into Two Parts Of Volume

Question 2. The height of a right circular cone is trisected by two planes drawn parallel to the base. Show that the ratio of volumes of the three portions starting from the top is in the ratio 1:7:19.

Solution:

Since height is trisected, therefore by basic proportionality theorem, base radii of three cones VCD, VA’B’ and VAB are also in the ratio 1: 2 : 3.

( ∵ VL : VM : VN = r2 : r1 : r = 1:2:3)

Let volume of cone VCD = V = \(\frac{1}{3} \pi r_2{ }^2 h\)

∴ Volume of cone VA’B’ = \(\frac{1}{3} \pi r_1^2(2 h)=\frac{1}{3} \pi\left(2 r_2\right)^2(2 h)=8 \times \frac{1}{3} \pi r_2{ }^2 h=8 V\)

and volume of cone VAB = \(\frac{1}{3} \pi r^2(3 h)=\frac{1}{3} \pi\left(3 r_2\right)^2(3 h)=27 \times \frac{1}{3} \pi r_2^2 h=27 \mathrm{~V}\)

∴ Ratio of volumes of 3 portions

= Volume(VCD) : Volume(CD£’A’) : Volume(A’B’BA)

= V: 8V – V: 27V – 8V = 1: 7: 19

Hence Proved.

Volume And Surface Area Of Solids Height Of A Right Circular Cone

Question 3. The radii of the faces of a frustum of a cone are 3 cm and 4 cm and its height is 5cm. Find its volume.

Solution:

Here r = 3 cm, R = 4 cm and h = 5 cm

Volume of frustum of cone = \(\frac{1}{3} \pi l\left(R^2+r^2+R r\right) \text { cu. units }\)

= \(\frac{1}{3} \times \frac{22}{7} \times 5\left(4^2+3^2+4 \times 3\right)\)

= \(\frac{1}{3} \times \frac{110}{7}(16+9+12)\)

= \(\frac{110}{21} \times 37=\frac{4070}{21} \mathrm{~cm}^3\)

Volume And Surface Area Of Solids Faces Of A Frustum Of A Cone

Question 4. The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume is \(\frac{1}{27}\) of the volume of the given cone, at what height above the base, the section has been made?

Solution:

Let OA – h ⇒ AB = 30- h

and let AC = r, BD = R

ΔOAC ~ ΔOBD,

∴ \(\frac{h}{30}=\frac{r}{R} \Rightarrow r=\frac{h R}{30}\)

Now, Volume of smaller cone = \(\frac{1}{27}\) x Volume of larger cone

⇒ \(\frac{1}{3} \pi r^2 h=\frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)

⇒ \(\frac{1}{3} \pi\left(\frac{h R}{30}\right)^2 \cdot h=\frac{1}{27} \times \frac{1}{3} \pi R^2 \times 30\)

⇒ \(\frac{h^3}{30^2}=\frac{30}{27} \Rightarrow h^3=\frac{30^3}{3^3} \Rightarrow h=10\)

∴ Reqired height = 30 – 10 = 20 cm.

1 kl = 1 m3

1000 l = 100 x 100 x 100 cm3

1 litre = 1000 cm3

= \(1 \mathrm{~cm}^3=\frac{1}{1000} \text { litre }\)

Volume And Surface Area Of Solids Small Cone Is Cut Off At The Top By A Plane Parallel To Base

Question 5. The radii of the ends of a bucket 30 cm high are 21 cm and 7 cm. Find its capacity in litres and the amount of sheets required to make this bucket. (\(\text { Take } \pi=\frac{22}{7}\))

Solution:

Although we can solve this problem by using similar triangles, here we are using the direct formula.

Volume of bucket = \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 30\left(21^2+21 \times 7+7^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 30 \times 637=20020 \mathrm{~cm}^3=\frac{20020}{1000} \text { lit. }\)

⇒ Capacity = 20.02 lit

Area of sheet = C.S.A. of bucket + Area of base

= πl(r1 + r2) + r22 = π[l(r1 + r2) + r22 (∵ \(l=\sqrt{h^2+\left(n_1-r_2\right)^2}\) = \(\sqrt{900+14^2}\) = \(\sqrt{1096}\) = 33.106)

= \(\frac{22}{7}[33.106(28)+49]\)

= 3067.32 = 3067cm2 (approx.)

Question 6. A bucket is 32 cm in diameter at the top and 20 cm in diameter at the bottom. Find the capacity of the bucket in litres if it is 21 cm deep. Also, find the cost of the tin sheet used in making the bucket at the rate of ₹ 1.50 per sq dm.

Solution:

Here R = 16 cm, r = 10 cm and h = 21 cm

Volume of frustum of a cone = \(\frac{\pi h}{3}\left(R^2+r^2+R r\right)\)

= \(\frac{22}{7} \times \frac{1}{3} \times 21\left(16^2+10^2+16 \times 10\right)\)

= 22(256 + 100 + 160)

= 22 x 516 = 11352cm3

= \(\frac{11352}{1000} \text { litres }=11.352 \text { litres }\)

Now for slant height l of frustum

l2 = h2 + (R-r)2

l2 = 212 + (16 – 10)2

l2 = 441 + 36

l2 = 477

∴ \(l=\sqrt{477}=21.84 \mathrm{~cm}\)

Now S.A of bucket = πl(R + r) + πr2

= \(\frac{22}{7} \times 21.84 \times(16+10)+\frac{22}{7} \times 10^2\)

= \(\frac{22}{7}(21.84 \times 26+100)\)

= \(\frac{22}{7} \times 667.84=2098.92 \mathrm{~cm}^2=20.99 \mathrm{dm}^2\)

Cost of sheet @ ₹ 1 .50 per sq. dm. = 20.99 x 1.50 = ₹ 31.49

Volume And Surface Area Of Solids A Bucket Of A Diameter

Volume And Surface Area Of Solids Exercise 13.1

Question 1. 2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

Solution:

Given,

volume of cube = 64 cm3

(side)3 = 64

(side)3 = 43

side = 4 cm

Side of cube = 4 cm

Volume And Surface Area Of Solids Cubes

A cuboid is formed by joining two cubes together as shown.

∴ For cuboid

length l = 4 + 4 = 8 cm

height h = 4 cm

Now, the total surface area of a cuboid

= 2(l.b + b.h + l.h)

= 2 (8 x 4 + 4 x 4 + 8 x 4)

= 2 (32 + 16 + 32) = 160 cm2

Question 2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Solution:

In the adjoining figure, a cylinder is surmounted on the hemisphere.

Diameter of hemisphere 2r = 14 cm

⇒ Radius of hemisphere r = 7 cm

The radius of cylinder r = 1 cm

Now, the total height of the vessel = 13 cm

⇒ h + r = 13 cm

⇒ h = 13 – 7 = 6 cm

Height of cylinder, h = 6 cm

The inner surface area of the cylinder = 2πrh

Inner curved surface area of hemisphere = 2πr2

Inner surface area of vessel = 2πrh + 2πr2

= 2πr(h + r)

= \(2 \times \frac{22}{7} \times 7 \times(6+7)\)

= 44 x 13 = 572 cm2

Volume And Surface Area Of Solids A Vessel Is In The Form Of A Hollow Hemisphere

Question 3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Solution:

In the adjoining figure, a cone of the same cross-section is surmounted on the hemisphere.

The radius of base of cone r = 3.5 cm

⇒ Radius of hemisphere r = 3.5 cm

Total height of toy = 15.5 cm

⇒ h + r = 15.5

⇒ h = 15.5 – 3.5 = 12 cm

∴ Height of cone h = 12 cm

Now, from l2 = h2 + r2

l2 = (12)2 + (3.5)2

= 144 + 12.25

= 156.25

⇒ \(l=\sqrt{156.25}=12.5 \mathrm{~cm}\)

∴ The curved surface area of the cone = πrl

and curved surface area of hemisphere = 2πr2

So, the total surface area of the toy = πrl + 2πr2

= πr (l + 2r)

= \(\frac{22}{7} \times 3.5 \times(12.5+2 \times 3.5)\)

= 11 x 19.5 = 214.5cm2

Volume And Surface Area Of Solids Total Surface Of Area Of The Toy

Question 4. A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Solution:

The base of the hemisphere is on the upper face of a cube of edge 7 cm,

∴ Maximum diameter of hemisphere = edge of the cube

= 7 cm

⇒ 2r = 7 ⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

Now, the total surface area of the solid

= total surface area of cube + curved surface of the hemisphere – an area of the base of the hemisphere

= 6 x (side)2 + 2πr2– πr2

= 6 x (side)2 + πr2

= \(=6 \times 7 \times 7+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)

= 294 + 38.5 = 332.5 cm2

Volume And Surface Area Of Solids A Cubical Block

Question 5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Solution:

Let the side of the cube = a

Diameter of hemisphere = side of the cube

⇒ 2r = a = \(r=\frac{a}{2}\)

Now, the surface area of remaining solid = total surface of the hemisphere – an area of the base of the hemisphere

= 6a2 + 2πr2– πr2

= 6a2 + πr2

= \(6 a^2+\pi\left(\frac{a}{2}\right)^2=\frac{24 a^2+\pi a^2}{4}\)

= \(\frac{a^2(24+\pi)}{4} \text { square units }\)

Volume And Surface Area Of Solids A Cubical Wooden Block

Question 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Solution:

Diameter of capsule = 5 mm

2r = 5 mm

r = 2.5 mm

∴ The radius of the cylindrical part = radius of the hemisphere = 2.5 mm

Length of capsule = 14 mm

⇒ h + 2r = 14 mm

⇒ h = 14 – 2r = 14 – 5 = 9 mm

Now the surface area of the capsule = 2 x curved surface of hemisphere + curved surface of the cylinder

= 2 x 2πr2 + 2πrh = 2πr (2r + h)

= \(2 \times \frac{22}{7} \times 2.5 \times(5+9)\)

= \(\frac{110}{7} \times 14=220 \mathrm{~mm}^2\)

Volume And Surface Area Of Solids A Medicine Capsule

Question 7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m2.

Solution:

In the adjoining, a tent is shown in which a cone is surmounted by a cylinder.

Diameter of cylindrical part 2r = 2. 1 m

⇒ \(r=\frac{2.1}{2} \mathrm{~m}\)

Height of cylindrical part h = 4 m

Radius of conical part \(r=\frac{2.1}{2} \mathrm{~m}\)

The slant height of conical part l = 2.8 m

Now, the area of canvas required to form a tent = curved surface of the cylindrical part + curved surface of the conical part

= 2πrh + πrl

= πr (2h + l)

= \(\frac{22}{7} \times \frac{2.1}{2} \times(2 \times 4+2.8)\)

= 3.3 x 10.8 = 35.64 m2

Volume And Surface Area Of Solids A Tent Is In The Shape Of A Cylinder

Question 8. From a solid cylinder whose height is 2.4 cm and diameter is 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.

Solution:

Diameter of cylinder 2r = 1.4 cm

⇒ r = 0.7 cm

∴ The radius of the cylinder = radius of the cone = r = 0.7 cm

Height of cylinder = height of cone = h = 2.4 cm

If the slant height of the cone is l, then

l2 = h2 + r2= (2.4)2 + (0.7)2

= 5.76 + 0.49 = 6.25

⇒ \(l=\sqrt{6.25}=2.5 \mathrm{~cm}\)

Surface area of remaining solid

= area of base of cylinder + curved surface of cylinder + curved surface of cone

= πr2 + 2πrh + πrl

= πr (r + 2h + l)

= \(\frac{22}{7} \times 0.7 \times(0.7+2 \times 2.4+2.5)\)

= 2.2 x 8 = 17.6 cm2

Volume And Surface Area Of Solids A Solid Cylinder

Question 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Solution:

The total surface area of the article

= curved surface of cylinder + 2 x curved surface of a hemisphere

= 2πrh + 2 x 2πr2

= 2πr (h+2r)

= \(2 \times \frac{22}{7} \times 3.5 \times(10+2 \times 3.5)\)

= 22 x 17 = 374 cm2

Volume And Surface Area Of Solids A Wooden Article

Volume And Surface Area Of Solids Exercise 13.2

Question 1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Solution:

Radius of hemisphere = radius of cone = r = 1 cm

Height of cone h = radius of cone = 1 cm

Volume of hemisphere = \(\frac{2}{3} \pi r^3\)

Volume of cone = \(\frac{1}{3} \pi r^2 h\)

∴ The volume of solid = volume of hemisphere + volume of a cone

= \(\frac{2}{3} \pi r^3+\frac{1}{3} \pi r^2 h\)

= \(\frac{2}{3} \pi(1)^3+\frac{1}{3} \pi(1)^2(1)\)

= \(\frac{2}{3} \pi+\frac{1}{3} \pi=\pi \mathrm{cm}^3\)

Question 2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Solution:

The model is shown in the given below.

Diameter 2r = 3 cm

⇒ \(r=\frac{3}{2} \mathrm{~cm}\)

Height of each cone h = 2 cm

Let the height of the cylinder = H

∴ H + h + h = 12 cm

⇒ H + 2 + 2 = 12

⇒ H = 12 – 4 = 8 cm

The volume of model = volume of cylinder + 2 x volume of a cone

= \(\pi r^2 H+2 \times \frac{1}{3} \pi r^2 h\)

= \(\pi r^2\left(H+\frac{2 h}{3}\right)\)

= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times\left(8+\frac{2 \times 2}{3}\right)\)

= \(\frac{22}{7} \times \frac{3}{2} \times \frac{3}{2} \times \frac{28}{3}=66 \mathrm{~cm}^3\)

Air contained in model = 66 cm3

Volume And Surface Area Of Solids Cylinder With Two Cones

Question 3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length of 5 cm and a diameter of 2.8 cm.

Solution:

For one gulab jamun,

Diameter 2r = 2.8 cm

⇒ r = 1.4 cm

Height of cylindrical part h = 5 – r – r

= 5-1.4 – 1.4 = 2.2 cm

Volume of one gulab jamun

= volume of cylindrical part + 2 x volume of hemispherical part

= \(\pi r^2 h+2 \times \frac{2}{3} \pi r^3\)

= \(\pi r^2\left(h+\frac{4 r}{3}\right)\)

= \(\frac{22}{7} \times 1.4 \times 1.4 \times\left(2.2+\frac{4 \times 1.4}{3}\right)\)

= 25.051 cm3

⇒ Volume of 45 gulab jamuns = 45 x 25.051

= 1127.295 cm3

Volume of sugar syrup in 45 gulab jamuns

= 30% of 1127.295

= \(1127.295 \times \frac{30}{100} \mathrm{~cm}^3\)

= 338.1885 cm3 ≈ 338 cm3

Volume And Surface Area Of Solids A Gulab Jamun

Question 4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Solution:

Volume of cuboid = 15 x 10 x 3.5 cm3

= 525 cm3

For conical depression

r = 0.5 cm and h = 1.4 cm

∴ The volume of one cavity

= \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \times \frac{22}{7} \times 0.5 \times 0.5 \times 1.4\)

= \(\frac{1.1}{3} \mathrm{~cm}^3\)

⇒ Volume of four depression= \(4 \times \frac{1.1}{3} \mathrm{~cm}^3\)

= 1.467 cm3

Now, the volume of wood used in the pen stand

= volume of cuboid – volume of four depression

= (525 – 1 .467) cm3 = 523.533 cm3

Volume And Surface Area Of Solids The Shape Of A Cuboid With Four Conical

Question 5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

The radius of cone r = 5 cm

and height h = 8 cm

⇒ Volume of cone = \(\frac{1}{3} \pi r^2 h\)

The volume of water filled in the cone

= \(\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi \times(5)^2 \times 8\)

= \(\frac{200}{3} \pi \mathrm{cm}^3\)

The volume of water flows out on dropping lead shots in it.

= \(\frac{1}{4} \times \frac{200 \pi}{3}=\frac{50 \pi}{3} \mathrm{~cm}^3\)

The radius of shot R = 0.5 cm

Volume of one lead shot = \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi(0.5)^3 \mathrm{~cm}^3\)

= \(\frac{\pi}{6} \mathrm{~cm}^3\)

Now, number of shots

= \(\frac{\text { volume of water flows out }}{\text { volume of one shot }}\)

= \(\frac{50 \pi / 3}{\pi / 6}=\frac{50 \pi}{3} \times \frac{6}{\pi}=100\)

Question 6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)

Solution:

For the first cylinder,

Diameter 2r = 24 cm

⇒ r = 12 cm

Height h = 220 cm

∴ Volume = πr2h = π x 12 x 12 x 220 cm3

= 31680 π cm3

For the second cylinder,

Radius R = 8 cm

Height H = 60 cm

Volume = πr2H = π x 8 x 8 x 60

= 3840π cm3

Volume of the pole = (31680π + 3840π) cm3

= 35520π cm3

∴ Weight of pole = 35520π x 8 g

= 35520 x 3.14 x 8 g

= 892262.4 g = 892.2624 kg

= 892.26 kg

Question 7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Solution:

The radius of cylinder r = 60 cm and

height h = 1 80 cm

∴ Volume of cylinder = πr2h

= π x 60 x 60 x 180 cm3

= 648000π cm3

The radius of cone R = 60 cm

and height H = 120 cm

∴ Volume of cone = \(\frac{1}{3} \pi R^2 H\)

= \(\frac{1}{3} \pi \times 60 \times 60 \times 120 \mathrm{~cm}^3\)

= 144000 cm3

The volume of solid formed from the cone and hemisphere

= (144000π + 144000π) cm3

= 288000π cm3

⇒ Volume of water displaced by this solid

= 288000π cm3

∴ The volume of the remaining water in a cylinder

= (648000π – 288000π) cm3

= 360000π cm3

= \(360000 \times \frac{22}{7} \mathrm{~cm}^3\)

= \(1131428.57 \mathrm{~cm}^3=\frac{1131428.57}{1000000} \mathrm{~m}^3\)

= 1.131 m3 (approx.)

Volume And Surface Area Of Solids Radius Of Cylinder

Question 8. A spherical glass vessel has a cylindrical neck 8 cm long, and 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution:

Radius of cylindrical part \(r=\frac{2}{2}=1 \mathrm{~cm}\)

and height h = 8 cm

∴ The volume of the cylindrical part = nr2h

= π(1)2(8) = 8πcm3

Radius of spherical part, \(R=\frac{8.5}{2}=\frac{17}{4} \mathrm{~cm}\)

∴ The volume of the spherical part,

= \(\frac{4}{3} \pi R^3=\frac{4 \pi}{3} \times \frac{17}{4} \times \frac{17}{4} \times \frac{17}{4} \mathrm{~cm}^3\)

= \(\frac{4913}{48} \pi \mathrm{cm}^3\)

∴ Volume of cylinder = \(\left(8 \pi+\frac{4913 \pi}{48}\right) \mathrm{cm}^3\)

= \(\frac{384 \pi+4913 \pi}{48} \mathrm{~cm}^3[/latex

= [latex]\frac{5297 \pi}{48} \mathrm{~cm}^3=\frac{5297}{48} \times 3.14 \mathrm{~cm}^3\)

= 346.51 cm3

So the answer 345 cm3 of a child is not correct.

∴ Correct volume of cylinder = 346.51 cm3

Volume And Surface Area Of Solids A Spherical Glass Vessel

Volume And Surface Area Of Solids Exercise 13.3

Question 1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solution:

The radius of sphere R = 4.2 cm

The radius of cylinder r = 6 cm

Let the height of the cylinder = h

Now, the volume of the cylinder = volume of a sphere

⇒ \(\pi r^2 h=\frac{4}{3} \pi R^3\)

⇒ \(h=\frac{4 R^3}{3 r^2}=\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6} \mathrm{~cm}\)

= 2.744 cm

∴ Height of cylinder = 27.44 cm

Question 2. Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Solution:

Let r1 = 6 cm, r2 = 8 cm and r3 = 10 cm

Let the radius of a bigger solid sphere = R

The volume of a bigger solid volume

= sum of volumes of three given spheres

⇒ \(\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^3+\frac{4}{3} \pi r_3^3\)

⇒ \(R^3=r_1^3+r_2^3+r_3^3\)

= 63 + 83+ 102

= 216 + 512 + 1000 = 1728 = 122

⇒ R = 12 cm

∴ The radius of the new solid sphere = 12 cm

Question 3. A 20 m deep well with a diameter of 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Solution:

Radius of well, \(r=\frac{7}{2} m\)

and depth h = 20 m

Let the height of the platform be H metre.

∴ The volume of the platform = the volume of the well.

⇒ 22 x 14 x H = πr2h

⇒ \(22 \times 14 \times H=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 20\)

⇒ \(H=\frac{22 \times 7 \times 7 \times 20}{7 \times 2 \times 2 \times 22 \times 14}=2.5 \mathrm{~m}\)

∴ Height of platform = 2.5 m

Question 4. A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

Solution:

Diameter of well 2r = 3 m

⇒ \(r=\frac{3}{2}=1.5 \mathrm{~m}\)

and depth h = 14 m

∴ The volume of earth taken out from the well = nr2h

= \(\frac{22}{7} \times 1.5 \times 1.5 \times 14=99 \mathrm{~m}^3\)

Now, the outer radius of the well, R = 1.5 + 4 = 5.5m

∴ Area of the ring of platform = π (R2 – r2)

= \(\frac{22}{7}\left[(5.5)^2-(1.5)^2\right]\)

= \(\frac{22}{7} \times 7 \times 4=88 \mathrm{~m}^2\)

Let the height of the embankment = H

∴ 88 x H = 99

⇒ \(H=\frac{99}{88}=\frac{9}{8}=1.125 \mathrm{~m}\)

Height of embankment = 1.1 25 m

Question 5. A container shaped like a right circular cylinder having a diameter of 12 cm and a height of 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

Answer:

The radius of the cylindrical container

∴ \(r=\frac{12}{2} \mathrm{~cm}=6 \mathrm{~cm}\)

and height h= 15 cm

∴ Volume = πr2h = 71 x 6 x 6 x 15

= 540 π cm3

⇒ The total volume of ice cream = 540cm3

The radius of cone = radius of hemisphere = R

⇒ \(R=\frac{6}{2}=3 \mathrm{~cm}\)

Height of cone, H = 12 cm

The volume of ice cream in one cone + hemisphere

= \(\frac{1}{3} \pi R^2 H+\frac{2}{3} \pi R^3=\frac{1}{3} \pi R^2(H+2 R)\)

= \(\frac{1}{3} \pi \times 3 \times 3 \times(12+2 \times 3)=54 \pi\)

Now, number of cones = \(\frac{\text { Total volume of ice cream }}{\text { Volume of ice cream in one cone }}\)

= \(\frac{540 \pi}{54 \pi}=10\)

Question 6. How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?

Solution:

Let the number of silver coins = n

Radius of coin \(r=\frac{1.75}{2} \mathrm{~cm}=\frac{7}{8} \mathrm{~cm}\)

Height h = 2 mm = \(\frac{2}{10} \mathrm{~cm}=\frac{1}{5} \mathrm{~cm}\)

The volume of one coin = πr2h

= \(\frac{22}{7} \times \frac{7}{8} \times \frac{7}{8} \times \frac{1}{5}=\frac{77}{160} \mathrm{~cm}^3\)

∴ Volume of n coins = \(\frac{77 n}{160} \mathrm{~cm}^3\)

Volume of cuboid = 5.5 x 10 x 3.5 = 192.5 cm3

Now, the volume of n coins = volume of a cuboid

⇒ \(\frac{77 n}{160}=192.5 \Rightarrow n=\frac{192.5 \times 160}{77}\)

⇒ n = 400

∴ Number of silver coins = 400

Question 7. A cylinder bucket, 32 cm high and with radius of base of 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Solution:

For cylindrical buckets,

Radius r = 18 cm

Height h = 32 cm

∴ The volume of sand = volume of the bucket

= πr2h = π x 18 x 18 x 32 cm3

= 10368π cm3

For conical heap,

Let Raidus = R

Height H = 24 cm

∴ Volume of conical heap = \(\frac{1}{3} \pi R^2 H=\frac{1}{3} \pi R^2 \times 24=8 \pi R^2\)

Now, the volume of conical heap = volume of sand

⇒ 8πR2 = 10368π ⇒ R2 = 1296

⇒ R = 36 cm

∴ l2 = H2 + R2 = 242 + 362

= 576 + 1296 = 1872

⇒ \(l=\sqrt{1872}=12 \sqrt{13} \mathrm{~cm}\)

Question 8. Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

Solution:

Speed of water in canal = 10 km/h

= \(\frac{10 \times 1000}{60} \mathrm{~m} / \mathrm{min}\)

= \(\frac{500}{3} \mathrm{~m} / \mathrm{min}\)

Width of canal = 6 m and depth = 1.5 m

Now, the canal will transfer the water equal to the volume of a cuboid of dimensions.

⇒ \(6 \mathrm{~m} \times 1.5 \mathrm{~m} \times \frac{500}{3} m\) in 1 minute.

∴ Volume of water transfer in 30 minutes

= \(30 \times 6 \times 1.5 \times \frac{500}{3}=45000 \mathrm{~m}^3\)

If the depth of the irrigating region = 8 cm

= \(\frac{8}{100} \mathrm{~m}\) then

area x depth = 45000

⇒ \({Area} \times \frac{8}{100}=45000\)

⇒ \(\text { Area }=\frac{45000 \times 100}{8}=562500 \mathrm{~m}^2\)

Therefore, the area of the region irrigated by the canal in 30 minutes = 562500 m2

Question 9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

Solution:

Diameter of tank 2r = 10 m

⇒ r = 5 m

and depth h = 2m

∴ Volume of tank = r2h = (5)2 x 2 = 50 m3

Again, diameter of pipe = 2R = 20 cm

⇒ \(R=10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}=\frac{1}{10} \mathrm{~m}\)

Speed of water in pipe = 3 km/h

= \(\frac{3 \times 1000}{60} \mathrm{~m} / \mathrm{min}\)

= 50m/min

Now the pipe will transfer the water into the tank in 1 minute equal to the volume of a cylinder of radius \(\frac{1}{10}\) m and length 50 m.

∴ Time taken to fill the tank = \(\frac{\text { volume of cylindrical tank }}{\text { volume of water transfer in tank in 1 minute }}\)

= \(\frac{50 \pi}{\pi \times\left(\frac{1}{10}\right)^2 \times 50}=100 \text { minutes }\)

∴ Time taken to fill the tank completely = 100 minutes

Volume And Surface Area Of Solids Exercise 13.4

Question 1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

Solution:

The diameters of the frustum of the cone are 4 cm and 2 cm.

∴ Radius r1 = 2 cm and r2 = 1 cm

∴ Height of glass h = 14 cm

∴ The volume of a glass of the shape of a frustum of a cone

= \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \times \frac{22}{7} \times 14\left[(2)^2+2 \times 1+(1)^2\right]\)

= \(\frac{44}{3}[4+2+1]\)

= \(\frac{44 \times 7}{3}=\frac{308}{3} \mathrm{~cm}^3\)

= \(=102 \frac{2}{3} \mathrm{~cm}^3\)

So, capacity of glass = = \(102 \frac{2}{3} \mathrm{~cm}^3\)

Volume And Surface Area Of Solids A Drinking Glass

Question 2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

Solution :

The slant height of the frustum of cone l = 4 cm.

Circumference of one end 2πr1 = 18 cm

∴ πr1 = 9 cm

Circumference of other ends 2πr2 = 6 cm

∴ πr2 = 3 cm

Curved surface area of frustum = π(r1 + r2) l

= (πr1 + πr2) l

= (9 + 3) x 4

= 48 cm2

Therefore, the curved surface area of the frustum of the cone = 48 cm2.

Question 3. A fez, tire cap used by the Turks, is shaped like the frustum of a cone. If its radius on the open side is 10 cm, its radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.

Volume And Surface Area Of Solids Frustum Of A Cone

Solution:

The cap is in the form of a frustum of a cone whose slant height is l = 15 cm.

Radius r1 = 10 cm and radius r2= 4 cm

∴ Curved surface of cap = π(r1 + r2) l

= \(\frac{22}{7}(10+4) \times 15=660 \mathrm{~cm}^2\)

Area of the closed end of the cap

= \(\pi r_2^2=\frac{22}{7} \times(4)^2 \mathrm{~cm}^2\)

= \(\frac{352}{7} \mathrm{~cm}^2=50 \frac{2}{7} \mathrm{~cm}^2\)

∴ Total canvas used in cap = Curved surface of cap + area of closed-end

= \(\left(660+50 \frac{2}{7}\right) \mathrm{cm}^2\)

= \(710 \frac{2}{7} \mathrm{~cm}^2\)

Therefore, the area of material used for making cap

∴ \(710 \frac{2}{7} \mathrm{~cm}^2\)

Question 4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 1 6 cm with radii of its lower arid upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ₹ 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs ₹ 8 per 100 cm2. (Take π = 3.14)

Solution:

The vessel is in the shape of a frustum of a cone whose height is h = 16 cm.

And radius of upper end r1 = 20 cm and radius of lower end r2 = 8 cm

Then, the volume of the vessel = volume of the frustum

= \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

= \(\frac{1}{3} \pi \times 16 \times\left[(20)^2+20 \times 8+(8)^2\right]\)

= \(\frac{16}{3} \pi \times 624 \mathrm{~cm}^3=3328 \pi \mathrm{cm}^3\)

= 3328 x 3.14 cm3 = 10449.92 cm3 [∴ π = 3.14]

The milk required to fill the vessel is 10449.92 cm3 or 10.450 litre.

Then, cost of milk at ₹ 20 per litre = 20 x ₹ 10.45 = ₹ 209

The sheet will be used to make the curved surface and base of the vessel.

Then, the area of the base of the vessel = nr22

= 3.14 x (8)2 = 3.14 x 64 = 200.96 cm2

The slant height of the vessel

⇒ \(l=\sqrt{h^2+\left(r_1-r_2\right)^2}=\sqrt{(16)^2+(20-8)^2}\)

= \(\sqrt{256+144}=\sqrt{400}=20 \mathrm{~cm}\)

Then, curved surface of vessel = π (r1 + r2) l

= 3.14 (20 + 8) x 20 cm2

= 3.14 x 28 x 20 cm2

= 1758.4 cm2

∴ Area of sheet used in vessel

= (1758.4 + 200.96) cm2

= 1959.36 cm2

Cost of the sheet at the rate of ₹ 8 per 1 00 cm2

= \(₹ \frac{8}{100} \times ₹ 1959.36=₹ 156.7488\)

= ₹ 156.75

∴ Cost of milk = ₹ 209

and cost of sheet = ₹ 156.75

Question 5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter \({1}{16}\) cm, find the length of the wire.

Solution:

In the given, the diameter of the base of the cone is A’OA and the vertex is V. The angle of the vertex is A’VA = 60°, and then the semi-vertical angle of the cone is α = 30°.

Height of cone = 20 cm

Volume And Surface Area Of Solids Height Of Cone

Then, in right ΔOAV,

⇒ \(\tan \alpha=\frac{O A}{O V} \Rightarrow \tan 30^{\circ}=\frac{r_1}{20} \Rightarrow \frac{1}{\sqrt{3}}=\frac{r_1}{20}\)

⇒ \(r_1=\frac{20}{\sqrt{3}} \mathrm{~cm}\)

∵ ΔVO’B and ΔVOA are similar.

∴ \(\frac{V O^{\prime}}{V O}=\frac{O^{\prime} B}{O A} \Rightarrow \frac{10}{20}=\frac{r_2}{r_1} \Rightarrow \frac{r_2}{r_1}=\frac{1}{2}\)

⇒ \(r_1=2 r_2 \Rightarrow r_2=\frac{1}{2} r_1=\frac{1}{2} \times \frac{20}{\sqrt{3}}=\frac{10}{\sqrt{3}} \mathrm{~cm}\)

and height of frustum \(\dot{h}=\frac{1}{2}\) x height of cone

= 10 cm

Then, volume of frustum = \(\frac{1}{3} \pi h\left(r_1^2+r_1 r_2+r_2^2\right)\)

⇒ \(\frac{1}{3} \pi(10)\left[\left(\frac{20}{\sqrt{3}}\right)^2+\frac{20}{\sqrt{3}} \times \frac{10}{\sqrt{3}}+\left(\frac{10}{\sqrt{3}}\right)^2\right]\)

⇒ \(\frac{1}{3} \pi 10\left[\frac{400}{3}+\frac{200}{3}+\frac{100}{3}\right]\)

⇒ \(\frac{7000}{9} \pi \mathrm{cm}^3\)

Diameter of cylindrical wire = \(\frac{1}{16}\) cm

Radius of wire \(r=\frac{1}{32} \mathrm{~cm}\) cm

Let the length of the wire drawn be l cm.

Then, the volume of wire = πr2l

⇒ \(\pi \times \frac{1}{32} \times \frac{1}{32} \times l=\frac{\pi}{1024} l \mathrm{~cm}^3\)

∵ The wire is drawn from the frustum of the cone.

∴ The volume of wire = volume of the frustum

⇒ \(\frac{\pi}{1024} l=\frac{7000}{9} \pi\)

⇒ \(l=\frac{7000 \pi}{9} \times \frac{1024}{\pi} \mathrm{cm}^3\)

= \(\frac{70}{9} \times 1024 \mathrm{~m}\)

= \(\frac{71680}{9} \mathrm{~m}=7964.44 \mathrm{~m}\)

Therefore, the length of the wire = 7964.44 m

Volume and Surface Area of Solids Exercise 13.5

Question 1. A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.

Solution:

The diameter of the cylinder = 10 cm and

height of cylinder = 12 cm

∴ Circumference of cylinder

= π x diameter = π x 10 = 10π cm

∴ Length of wire used in one round about a cylinder

= 10 cm

∵ The length of the cylinder is 12 cm or 120 mm. When one round of wire is wound on the cylinder then it covers 3mm length of the cylinder.

When two rounds of wire are wound on the cylinder then it covers the (2 x 3) mm length of the cylinder.

Volume And Surface Area Of Solids Length Of Cylinder

When three rounds of wire are wound on the cylinder then it covers (3×3) mm length of the cylinder.

When four rounds of wire are wound on the cylinder then it covers the (4×3) mm length of the cylinder.

The number of wounds of wire to cover the cylinder = \(\frac{120}{3} = 40\)

Length of required wire to wound 40 rounds on cylinder

= 40 x 10 π = 400 π cm

= 400 x 3.14 cm = 1256 cm(approx.)

= 12.56 m

So, the required length of wire = 12.56 m

The volume of wire = length x area of the wire

= \(1256 \times \pi \frac{d^2}{4} \quad\left[d=3 \mathrm{~mm}=\frac{3}{10} \mathrm{~cm}\right]\)

= \(1256 \times 3.14 \times \frac{9}{100 \times 4}\)

= \(\frac{314 \times 3.14 \times 9}{100}=88.74 \mathrm{~cm}^3\)

and mass of wire = 88.74 x 8.88 g

= 788.01 g = 0.788 kg

Question 2. A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose the value of π as found appropriate.)

Solution:

In right ΔABC, ∠B = 90°, AB = 4 cm,

BC = 3 cm

Then, area of \(\triangle A B C=\frac{3 \times 4}{2}=6 \mathrm{~cm}^2\)

Hypotenuse \(A C=\sqrt{A B^2+B C^2}\)

= \(\sqrt{(4)^2+(3)^2}=\sqrt{25}=5\)

BOB’ is perpendicular to AC, if BO = r, then area of \(\triangle A B C=\frac{A C \times B O}{2}=\frac{5}{2} B O=\frac{5}{2} r\)

Then, \(\triangle A B C=\frac{A C \times B O}{2}=\frac{5}{2} B O=\frac{5}{2} r\)

Then, \(\frac{5}{2} r=6\) (∵\(\frac{5}{2} r\)and 6 both are area of AABC)

∴ \(r=\frac{6 \times 2}{5}=2.4 \mathrm{~cm}\)

Now, the radius r = 2.4 cm of the double cone formed by rotating the right ΔABC.

Then, volume of double cone (two cones) = volume of cone (A, BB’) + volume of cone (C, BB’)

= \(\frac{1}{3} \pi r^2(A O)+\frac{1}{3} \pi r^2(O C)\)

= \(\frac{1}{3} \pi r^2(A O+O C)\)

= \(\frac{1}{3} \pi r^2(A C)\) [ AO + OC = AC]

= \(\frac{1}{3} \pi \times(2.4)^2 \times 5=9.6 \pi \mathrm{cm}^3\)

= 9.6 x 3.14 cm3 (π = 3.14)

= 30.144 cm3

and surface area of the double cone (both cones)

= curved surface of the cone (A, BB’) + curved surface of the cone (C, BB’)

= πr (AB) + πr (BC) = nr (AB +BC)

= 3.14 x 2.4 x (4 +3) =3.14 x 2.4 x 7

= 52.75 cm2

Therefore, the volume of the double cone = 30.144 cm3 and surface area = 52.75 .cm2 (approximately).

Volume And Surface Area Of Solids A Right Triangle

Question 3. A cistern, internally measuring 150cm x 120cm x 110cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm x 7.5 cm x 6.5 cm?

Solution:

Volume of cistern

= 150 x 120 x 110 cm3

= 1980000 cm3

The volume of water filled in cistern = 129600 cm3

Volume of each brick = 22.5 x 7.5 x 6.5 cm3

= 1096.875 cm3

Let on placing x bricks, the water rises upto the brim in the cistern.

Then, volume of x bricks = 1096.875 x cm3

Then, volume of absorbs water by bricks = \(1096.875 x \times \frac{1}{17}=\frac{1096.875 x}{17} \mathrm{~cm}^3\)

Then, the volume of remaining water in cistern = \(\left(129600-\frac{1096.875 x}{17}\right) \mathrm{cm}^3\)

Now, volume of x bricks + volume of water in cistern = volume of cistern

∴ \(1096.875 x+129600-\frac{1096.875 x}{17}\) = 1980000

or \(1096.875 x-\frac{1096.875 x}{17}\) = 1980000 – 129600

or \(1096.875 x\left(1-\frac{1}{17}\right)=1850400\)

or \(1096.875 x=\frac{1850400 \times 17}{16}\)

or \(x=\frac{1850400 \times 17}{16 \times 1096.875}\)

= 1792.4 = 1792 (approximately)

Therefore, the number of bricks placed in the cistern is 1792 (approximately).

Question 4. In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km2, show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

Solution:

The volume of each river

= 10721cm x 75 m x 3 m

= 1072 x 75 x 3 x 1000 m3

= 241200000 m3

∴ The volume of total water in three rivers

= 3 x 241200000 m3

∴ Total water of rivers = 723600000 m3

∴ Area of valley = 7280 km2

= 7280 x (1000)2 m2

= 7280000000 m

∴ Volume of rainwater

= \(7280000000 \times \frac{10}{100} \mathrm{~m}^3\) (∵\(10 \mathrm{~cm}=\frac{10}{100} \mathrm{~m}\))

= 728000000 m3

These two volumes are not equal.

So, it is clear that the given data given in the question are incorrect.

Question 5. An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, the diameter of the cylindrical portion is 8 cm and the diameter portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel.

Volume And Surface Area Of Solids Oil Funnel

Solution:

Height of cylindrical part h = 10 cm

Total height of funnel = 22 cm

∴ Height of frustum of cone (H) = 22 – 10 = 12 cm

Upper radius of frustum of cone = \(R_1=\frac{18}{2}=9 \mathrm{~cm}\)

Lower radius of frustum of cone = \(R_2=\frac{8}{2}=4 \mathrm{~cm}\)

The radius of cylindrical part r = 4 cm

The curved surface of the cylindrical part = 2 πrh

= 2π x 4 x 10 = 80 π cm2

The slant height of the frustum of a cone

⇒ \(l=\sqrt{H^2+\left(R_1-R_2\right)^2}\)

= \(\sqrt{(12)^2+(9-4)^2}=\sqrt{144+25}\)

= \(\sqrt{169}=13 \mathrm{~cm}\)

So, the total surface area of the funnel

∴ The curved surface of the frustum of a cone

= π (R1+ R2) l

= π (9 + 4) x 13 = 169 π cm2

∴ The curved surface area of the cylindrical part + curved surface area of the frustum of a cone

= 80 π + 169π = 249π cm2

= \(249 \times \frac{22}{7} \mathrm{~cm}^2\)

= \(\frac{5478}{7}=782 \frac{4}{7} \mathrm{~cm}^2\)

Therefore, area of tin sheet used in funnel = \(782 \frac{4}{7} \mathrm{~cm}^2\)

Question 6. Derive the formula for the curved surface area and total surface area of the frustum of a cone.

Solution:

Let for the (V, AB), V is the vertex, r2 the base radius and l2 the slant height. A cone (V, CD) is cut off from this cone from a point O’ below h1 from the vertex V of this cone, parallel to the base whose’ radius is r1 and slant height is l1.

Draw the perpendicular DE from point D to the base.

Volume And Surface Area Of Solids Total Surface Area Of The Frustum Of A Cone

In ΔVO’D and ΔDEB,

∠VO’D = ∠DEB (VO and DE both are perpendicular to the base)

∠VDO’ = ∠DBE (the bases of two cones are parallel to each other)

∴ ΔVO’D and ΔDEB are similar

\(\frac{V D}{B D}=\frac{O^{\prime} D}{E B}\)

or \(\frac{l_1}{l}=\frac{O^{\prime} D}{O B-O E}=\frac{O^{\prime} D}{O B-O^{\prime} D}\)

while BD = l = slant height of the frustum

⇒ \(\frac{l_1}{l}=\frac{r_1}{r_2-r_1} \Rightarrow l_1=\left(\frac{r_1}{r_2-r_1}\right) l\) → (1)

The curved surface area of a frustum

= curved surface area of cone (V, AB) – curved surface area of cone ( V, CD)

= πr2l2 -πr1l1 = πr2(l1 + BD) – πr1l1

= πr2l1 +πr2 (BD) – πr1l1

= π(r2-r1) l1 + πr2 l

= \(\pi\left(r_2-r_1\right)\left(\frac{r_1}{r_2-r_1}\right) l+\pi r_2 l\) [from eqn. (1)]

= π r1l + πr2l

So, curved surface area of frustum = π (r1 + r2) l

Hence Proved

And total surface area of a frustum

= curved surface + area of first end + area of second end

= π(r1 + r2) / + πr12 + πr22

= π(r1 + r2)l + π (r12 + r22)

Hence Proved.

Question 7. Derive the formula for the volume of the frustum of a cone.

Solution:

From the last question, for the cone (V, AB), height = h2 and radius = r2

∴ Volume of cone (V, AB) = \(\frac{1}{3} \pi r_2^2 h_2\)

and volume of cone (V, CD) = \(\frac{1}{3} \pi r_1^2 h_1\)

∴ Volume of frustum = volume of cone (V, AB) – volume of cone (V, CD)

∴ Volume of frustum (V)

= \(\frac{1}{3} \pi r_2^2 h_2-\frac{1}{3} \pi r_1^2 h_1\) → (1)

∴ h2 = VO’= VO’ + O’O = h1 + h

∴ Put h2 = h1 +h in eqn. (1),

Volume of frustum V = \(\frac{1}{3} \pi r_2^2\left(h_1+h\right)-\frac{1}{3} \pi r_1^2 h_1\)

Volume of frustum V = \(\frac{1}{3} \pi\left(r_2^2-r_1^2\right) h_1+\frac{1}{3} \pi r_2^2 h\) → (2)

In similar ΔVO’D and ΔDEB,

\(\frac{h_1}{h}=\frac{r_1}{r_2-r_1} \quad \Rightarrow \quad h_1=\left(\frac{r_1}{r_2-r_1}\right) h\)

Put \(h_1=\left(\frac{r_1}{r_2-r_1}\right) h\) in eqn. (2),

⇒ \(V=\frac{1}{3} \pi\left(r_2^2-r_1^2\right) \frac{\eta_1}{\left(r_2-r_1\right)} h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_2+r_1\right) r_1 h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2\right) h+\frac{1}{3} \pi r_2^2 h\)

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2+r_2^2\right) h\)

Therefore, the volume of the frustum of the cone

= \(\frac{1}{3} \pi\left(r_1^2+r_1 r_2+r_2^2\right) h\) Hence Proved.

Volume And Surface Area Of Solids Multiple Choice Questions

Question 1. A surah is a combination of:

  1. A sphere and a cylinder
  2. A hemisphere and a cylinder
  3. Two hemispheres
  4. A cylinder and a cone

Answer: 1. A sphere and a cylinder

Question 2. A glass is generally of the shape of:

  1. A cone
  2. A frustum of a cone
  3. A cylinder
  4. A sphere

Answer: 2. A frustum of a cone

Question 3. A plummet is a combination of:

  1. A cone and a cylinder
  2. A hemisphere and a cone
  3. A frustum of a cone and a cylinder
  4. A sphere and a cylinder

Answer: 2. A hemisphere and a cone

Question 4. An iron piece in the shape of a cuboid of dimensions 49 cm x 33 cm x 24 cm is melted and recast into a solid sphere. The radius of the sphere is:

  1. 21 cm
  2. 23 cm
  3. 25 cm
  4. 19 cm

Answer: 1. 21 cm

Question 5. While converting a shape of a solid into another shape, the volume of the new shape:

  1. Increases
  2. Decreases
  3. Remains same
  4. Becomes twice.

Answer: 3. Remains same

Question 6. The ratio of the surface of two spheres is 16:9. The ratio of their volumes is:

  1. 3:4
  2. 64: 27
  3. 27: 64
  4. 4:3

Answer: 2. 64: 27

Question 7. The diameter of a sphere exactly inscribed in a right circular cylinder of radius r cm and height h cm (h > 2r) is:

  1. r cm
  2. 2r cm
  3. h cm
  4. 2h cm

Answer: 2. 2r cm

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

Arithmetic Progression Introduction

In daily life, there are so many examples which follow a certain pattern.

  1. In a fixed deposit scheme, the amount becomes. 1 item of itself after every year. The maturity amount of ₹10,000 after 1, 2, 3 and 4 years will be ₹11,000, ₹12,100, ₹13,310,14.641
  2. When a person is offered a job with a monthly salary of ₹18,000 and with an annual increment of ₹400, His salary for the succeeding years will be ₹18,000, ₹18,400, ₹ 18,800 per mon.
    • In the first example, the succeeding terms are obtained by multiplying with 1.1. In the second example, the succeeding terms are obtained by adding 400.
    • Now, we will discuss all those patterns in which the succeeding terms are obtained by adding a fixed number to the preceding terms.

Sequence

A number of things that come one after another form a sequence. It may be possible that we do not have a formula to find the nth term of the sequence, but still, we know about the next term.

Sequence For example: 2, 3, 5, 7, 11,13, 17, 19, …is a sequence of prime numbers.

We all know about the next number of the sequence but we do not have any formula to calculate the next number.

  • The numbers present in the sequence are called the terms of the sequence.
  • The nth term of the sequence can be represented by Tn or an. It is called the general term.
  • A sequence which has finite terms is called a finite sequence.

Progression

Those sequences whose terms follow certain patterns are called progressions. In progression, each term (except the first) is obtained from the previous one by some rule.

The difference between a progression and a sequence is that a progression has a specific formula to calculate its nth term, whereas a sequence can be based on a logical rule like a group of prime numbers, which does not have any formula associated with it.

The numbers 2, 4, 6, 8, 10, 12 … form a progression as its nth term Tn =2n while 2,3,5,7,11,13,17,19,23,29,31,… is a sequence of numbers, as there is no formula to find the next number. It is a group of prime numbers.

Series

If T1, T2, T3,…, Tn form a progression, then T1, T2, T3,…, Tn is called its corresponding series. In other words, if all the terms of a progression are connected by ‘+’ started

Series For example:1-2 + 3- 4+5-6 ……… is a series. The first term is 1, the second term is -2, the third term is 3, the fourth is -4 and so on. Its nth is n(-1) n+1

NCERT Exemplar Solutions for Class 10 Maths Chapter 5 Arithmetic Progression

Series Solved Examples

Question 1. The nth of a sequence is an = 2n+1. Find its first four terms.
Solution:

Given

The nth of a sequence is an = 2n+1.

an=2n+1

put n= 1,2,3,4, we get

a1 = 2×1+3 = 5

a2 = 2×2+3 = 7

a3 = 2×3+3 = 9

a4 = 2×4+3 = 11

∴ The first four terms of the sequence are 5,7,9,11.

Question 2. The nth term of a sequence is an=n2+5. Find its first terms.
Solution:

Given

The nth term of a sequence is an=n2+5

an=n2+5

put n=1,2,3, we get

a1 = 12 + 5 = 6

a2 = 22 + 5 = 9

a3 = 32 + 5 = 14

∴ The first four terms of the sequence are 6,9,14.

Question 3. Fibonacci sequence is defined as follows: a1 = a2 = 1 and an = an-2 + an-1 , where n > 2. Find third, fourth and fifth terms
Solution:

Given

Fibonacci sequence is defined as follows: a1 = a2 = 1 and an = an-2 + an-1 , where n > 2.

a1 = a1 = 1

⇒ \(a_n=a_{n-2}+a_{n-1}, n>2\)

put n = 3, we get

⇒ \(a_3=a_1+a_2=1+1=2\)

put n = 4, we get

⇒ \(a_4=a_2+a_3=1+2=3\)

put it = 5, we get

⇒ \(a_5=a_3+a_4=2+3=5\)

Question 4. A sequence is defined as follows: a1 = 3, an = 2an-1 + 1, where n > 1. Find \(\frac{a_{n+1}}{a_n}\) for n=1,2,3.
Solution:

Given

A sequence is defined as follows: a1 = 3, an = 2an-1 + 1, where n > 1.

a1=3

⇒ \(a_n=2 a_{n-1}+1 \quad \text { where } n>1\)

put n= 2, we get

⇒ \(a_2=2 a_1+1=2 \times 3+1=7\)

put n = 3, we get

⇒ \(a_3=2 a_2+1=2 \times 7+1=15\)

put n = 4, we get

\(a_4=2 a_3+1=2 \times 15+1=31\)

Now, for n = 1

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_2}{a_1}=\frac{7}{3}\)

For n = 2,

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_3}{a_2}=\frac{15}{7}\)

For n = 3,

⇒ \(\frac{a_{n+1}}{a_n}=\frac{a_4}{a_3}=\frac{31}{15}\)

Arithmetic Progression

An arithmetic progression (A.P.) is a list of numbers in which each term is obtained by adding a fixed number to the preceding term, except the first term.

or

If the difference of any two consecutive terms of a progression is the same (constant), it is called arithmetic progression.

For example: a, a + d, a + 2d, a + 3d, ….

Common Difference

The difference between two consecutive terms (i.c., any term — preceding term) of an arithmetic progression is called a common difference.

It is denoted by ‘d’

∴ d = a2-a1 = a3-a2= … = an-an-1.

General Term Of Arithmetic Progression

Let the first term and common difference of an arithmetic progression be ‘a’ and ‘b’ respectively.

∴ Arithmetic progression is a, a +d, a + 2d, a + 3d,…..

Here, first term = a = a+(1-1)d

Second term = a+d = a+(2-1)d

Third term = a+2d = a+(3-1)d
.
.
.
.

nth term = a+(n-1)d

∴ an=a+(n-1)d

nth term of a progression is called its general term.

nth Term From The End Of An Arithmetic Progression

Let the first term, common difference and last term of an arithmetic progression be a, d and / respectively.

Arithmetic progression is a, a + d, a + 2d,…….l – 2d, l-d,l

Here, the First term from the end -l = l – (1 -1)d

Second term from the end = l – d =l-(2-l)d

Third term from the end = l – 2d = l – (3 – 1 )d

.

.

.

nth term from the end =l-(n- 1 )d.

Arithmetic Progression Solved Examples

Question 1. For the following A.P., write the first term and common difference: 3, 1,-1,-3, …
Solution:

3, 1,-1,-3,…

Here, first term a = 3

Common difference = 1 – 3 = -2

Question 2. Write the first four terms of the A.P., when the first term V and the common difference ‘d’ are given as follows:

a = 10, d = 5

Solution:

a = 10, d = 5

a1 = 10

a2 =10 + 5=15

a3 =15 + 5 = 20

a4 = 20 + 5 = 25.

∴ First 4 terms are 10, 15, 20, 25.

Question 3. Find the 18th term of the A.P. 4, 7, 10, …
Solution:

Given

A.P. 4, 7, 10

Here, a = 4,

d = 7 – 4 = 10-7 = 3,

n = 18

∴ an = a + (n – 1 )d

a18 = 4 + (18 – 1) × 3 =4 + 51=55

∴ 18th term of the given A.P. = 55.

Question 4. What is the common difference of an A.P. in which a21-a7 = 84?
Solution:

Given

a21-a7 = 84

Let the first term of A.P. be ‘a’ and common difference be ‘d’.

Since, a21-a7 =84

∴ (a+20d) -(a-6d) = 84

⇒ 14d = 84

⇒ d= 6

Hence, the common difference is 6.

Question 5. Which term of the A.P. 3, 8, 13, 18,….is 88?
Solution:

Given

A.P. 3, 8, 13, 18,….is 88

Here, a = 3,

A = 8-3 = 13 -8 = 5

Let an = 88

⇒ 3 + (n – 1)5 = 88

⇒ 3 + 5K – 5 = 88

⇒ 5n – 2 = 88

⇒ 5n = 90

⇒ n= 18

The 18th term of the given A.P. is 88.

Question 6. Which term of the A.P. 90, 87, 84,…is zero?
Solution:

Given

90, 87, 84,…

Here, a = 90,

d = 87 – 90 = -3

Let an = 0

⇒ 90 + (n – 1) (-3) = 0

⇒ 90 – 3n + 3 = 0

⇒ -3n = -93

⇒ n= 31

∴ 31st term of the given A.P. is zero.

Question 7. If 2, a, b, c, d, e,f and 65 form an A.P., find the value of e.
Solution:

Given

2, a, b, c, d, e,f and 65 form an A.P.

Here, T1 = 2, T8 = 65, T6 = e

Let the common difference be D

T8 = 65 ⇒  T1 + (8 – 1)D = 65

⇒ 2+7D = 65  ⇒ D = 9

∴ e = T6 = T1 + (6 – 1 )D = 2 + 5(9) = 47

Hence, the value of e is 47.

Question 8. Which term of the A.P. \(10,9 \frac{1}{3}, 8 \frac{2}{3}\) ….. is the first negative term?
Solution:

Here, a = 10,

⇒ \(d=9 \frac{1}{3}-10=8 \frac{2}{3}-9 \frac{1}{3}=-\frac{2}{3}\)

Let an<0

⇒ \(10+(n-1)\left(-\frac{2}{3}\right)<0\)

⇒ \(\frac{30-2 n+2}{3}<0\)

⇒ 32-2n<0

⇒  32 < 2n

⇒ 2n>32

⇒ n>16

⇒ n = 17, 18, 19, (all are negative terms)

∴ First negative term = 17th term

Now, a17=a+(17-1) d

⇒ \(10+16\left(-\frac{2}{3}\right)=10-\frac{32}{3}=\frac{-2}{3}\)

Thus, 17th term which is \(-\frac{2}{3}\) is the first negative term.

Question 9. The 18th term of an A.P. exceeds its 10th term by 8. Find the common difference.
Solution:

The 18th term of an A.P. exceeds its 10th term by 8.

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

⇒ a18 = a10 + 8

⇒ a18-a10= 8

⇒ {a +(18-1)d}-{a + (10- 1)d} = 8

⇒ [a + 17d)-{a + 9d) = 8

⇒ a + 17d-a-9d = 8

⇒ 8d = 8

⇒ d = 1

∴ Common difference = 1.

Example 10. The 26th, 11th and last term of an A.P. are 0, 3 and \(-\frac{1}{5}\) respectively. Find the common difference and the number of terms.
Solution:

The 26th, 11th and last term of an A.P. are 0, 3 and \(-\frac{1}{5}\) respectively.

Let the first term, common difference and number of terms in the A.P. be a, d and n respectively.

Given that

⇒ a26 = 0

⇒ a+25d = 0 ……(1)

⇒ a = -25d

⇒ a11 = 3 a+10d = 3

⇒ -25d+10d = 3 [from (1)]

⇒ -15d = 3

⇒ d = \(-\frac{1}{5}\)

put d = \(-\frac{1}{5}\) in equation (1) we get

⇒ \(a=-25\left(-\frac{1}{5}\right)=5\) and \(a_n=-\frac{1}{5}\)

⇒ \(5+(n-1)\left(-\frac{1}{5}\right)=-\frac{1}{5}\)

⇒ \((n-1)\left(-\frac{1}{5}\right)=-\frac{1}{5}-5=-\frac{26}{5}\)

⇒ n- 1=26

⇒ n= 27

No. of terms in the A.P. = 27.

Question 11. If the 9th term of an A.P. is zero, prove that its 29th term is twice its 19th term.
Solution:

The 9th term of an A.P. is zero

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

a9 = 0

a + 8d = 0

a = -8d

Now, a29 = a+28d = -8d+28d

= 20d

T19 = a+18d = -8d + 18d

= 10d

⇒ 2T19 = 20d

∴ T29 = 2T19

Question 12. The nth term of a sequence is 3n + 5. Show that it is an A.P.
Solution:

Given

The nth term of a sequence is 3n + 5.

Here, an = 3n+5

an-1 = 3(n-1) +5

3n-3+5 = 3n+2

Now, an-an-1 = (3n+5)-(3n+2) = 3

which does not depend on n i.e., it is constant.

∴ Given sequence is in A.P.

Question 13. For what value of k will k + 9, 2k- 1 and 2k +7 are the consecutive terms of an A.P?
Solution:

Since k + 9, 2k- 1 and 2k + 7are in A.P.

So, there must be a common difference.

i.e., (2k- 1) – (k + 9) = (2k + 7) – (2k- 1)

⇒ k- 10 = 8

⇒ k= 18

The value of k= 18.

Question 14. Find how many integers between 200 and 500 are divisible by 8.
Solution:

Integers between 200 and 500 which are divisible by 8 are as follows :

∴ 208, 216, 224, 232, …, 496

This forms an A.P. whose first term = 208, common difference = 8 and last term = 496. Let there be n terms.

∴ an = 496

⇒ a+ (n -1)<d = 496

⇒ 208 + (n- 1)8 = 496

⇒ (n- 1)8 = 496-208 = 288

⇒ \(n-1=\frac{288}{8}\)

∴ n-1 = 36

∴ n = 37

Hence, 37 integers are between 200 to 500 which are divisible by 8.

Question 15. If m times the with term of an A.P. is equal to n times the mth term and m ≠ n, show that its (m + n)th term is zero.
Solution:

Let the first term and the common difference of the A.P. be ‘a and ‘d’ respectively.

Given that

m.am = n.an

⇒ \( m\{a+(m-1) d\}=n\{a+(n-1) d\}\)

⇒ \(a m+\left(m^2-m\right) d=a n+\left(n^2-n\right) d\)

⇒ \(a(m-n)+\left\{\left(m^2-n^2\right)-m+n\right\} d=0\)

⇒ \(a(m-n)+\{(m-n)(m+n)-1(m-n\} d=0\)

⇒ \(a(m-n)+(m-n)(m+n-1) d=0\)

⇒ \((m-n)\{a+(m+n-1) d\}=0\)

⇒ \(a+(m+n-1) d=0\) (…m=n)

⇒ \(a_{m+n}=0\)

∴ (m + n)th term of the given A.P. is zero.

Question 16. If the term of an A.P. is \(\frac{1}{n}\) and its nth term be \(\frac{1}{m}\) then shows that its (nm)th term is 1.
Solution:

Let the first term and the common difference of the A.P. be ‘a’ and ‘d’ respectively

∴ \(a_m=\frac{1}{n} \quad \Rightarrow \quad a+(m-1) d=\frac{1}{n}\) ….(1)

and \(a_n=\frac{1}{m} \quad \Rightarrow \quad a+(n-1) d=\frac{1}{m}\)

Subtract equation (2) from equation (1), we get

Arithmetic Progression If The mth Term Of AP

⇒ \((m-n) d=\frac{m-n}{n m}\)

⇒ \(d=\frac{1}{m n}\)

put \(d=\frac{1}{m n}\)

⇒ \( a+(m-1) \cdot \frac{1}{m n}=\frac{1}{n}\)

⇒ \(a+\frac{1}{n}-\frac{1}{m n}=\frac{1}{n}\)

⇒ \(a=\frac{1}{m n}\)

⇒ \(a_{m n}=a+(m n-1) d\)

⇒ \(\frac{1}{m n}+(m n-1) \cdot \frac{1}{m n}=\frac{1}{m n}+1-\frac{1}{m n}=1\)

∴ (mn)th term of the A.P. = 1

Question 17. Find the 7th term from the end of the A.P. : 3, 8, 13, 18,….98
Solution:

Here, l = 98,

d = 8-3 = 13-8 = 5,

n = 7

∴ 7th term from the end =l- (7 – 1)d

= 98 – 6 x 5

= 98 – 30

= 68

Alternative Method :

Write the A.P. in reverse order 98, …18, 13,8,3

Here, a = 98

d = 3-8 = 8- 13 = -5

a7 = a + (7 – 1)d

= 98 + 6 x (-5) = 98- 30 = 68

which is the 7th term from the end of the given A.P.

Sum Of n Terms Of An A.P.

Let the first term and common difference of an A.P. be ‘a’ and ‘d’          respectively. Let the A.P. contain ‘n’ terms and the last term be ‘l’.

∴ l = a + (n-1)d ….(1)

Now. a sum of n terms

Sn = a + (a + d) + + (1-d)+l …..(2)

In reverse order

Sn = l + (l + d) + + (a-d)+a …..(3)

Adding equations (2) and (3), we get

2Sn = (a + l) + (a + l) + …… + (a+l) + (a +l) (no.of terms = n)

2Sn = n(a+l)

⇒ \(S_n=\frac{n}{2}(a+l)\)

⇒ \(S_n=\frac{n}{2}[a+a+(n-1) d]\)

From eq.(1)

⇒ \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

Sum Of n Terms Solved Examples

Question 1. Find the sum of n terms of the series

⇒ \(\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots\)

Solution:

We are given the series

⇒ \(S=\left(4-\frac{1}{n}\right)+\left(4-\frac{2}{n}\right)+\left(4-\frac{3}{n}\right)+\ldots \text { to } n \text { terms }\)

⇒ \((4+4+4+\ldots \text { to } n \text { terms })-\left(\frac{1}{n}+\frac{2}{n}+\frac{3}{n}+\ldots \text { to } n \text { terms }\right)\)

⇒ \(4 n-\frac{1}{n}(1+2+3+\ldots \text { to } n \text { terms })\)

⇒ \(4 n-\frac{1}{n}\left[\frac{n}{2}\{2 \times 1+(n-1) \times 1\}\right] \quad( a=1, d=1)\)

⇒ \(4 n-\frac{1}{n}\left[\frac{n}{2}(n+1)\right]=4 n-\frac{n+1}{2}=\frac{8 n-n-1}{2}=\frac{1}{2}(7n-1)\)

Hence, the required sum of n terms is \(\frac{1}{2}(7 n-1) \text {.}\)

Question 2. Find the sum of the following series : 5 + (-41) + 9 +(-39) + 13 + (-37)+ 1 7 + … + (-5) + 81 + (-3)
Solution:

Let 5 = 5 + (-41) + 9 + (-39)+ 13 + (-37) + 17 + … + (-5) + 81 + (-3)

= (5 + 9+ 13 + 17 + … + 81) – (41 + 39 + 37 + … + 3)

= S1-S2

where, ,S1 = 5 + 9+ I3 + 17 + … + 81

It is an A.P. with a1 = 5,d = 9- 5 = 4

Let there be n terms.

∴ an = 81

⇒ a1 + (n- 1)d = 81

⇒ 5 + (n – 1)4 = 81

⇒ (n-1)4 = 76

⇒ n-1= 19

⇒ n = 20

∴ S1 = Sum of 20 terms with first term 5 and last term 81

⇒ \(\frac{20}{2}[5+81]=10 \times 86\)

=860

and S2 = 41 + 39 + 37 + … + 3

It is also an A.P. with a1 = 41, d = 39 – 41 = 2

Let there be n terms.

∴ an = 3

⇒ a1 + (n- 1)d = 3

⇒ 41 + (n – 1)(-2) = 3

⇒ (n- 1)(-2)=-38

⇒ n- 1 = 19

⇒ n=20

∴ S2 = Sum of 20 terms with first term 41 and last term 3

⇒ \(\frac{20}{2}[41+3]\)

⇒ 10×44 = 440

From equation (1),

S = S1 – S2

⇒ S = 860- 440 = 420

Question 3. The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 150.

Find the number of terms and the common difference of the A.P.

Solution:

Given

The first term of an A.P. is 5, the last term is 45 and the sum of all its terms is 150.

Let a number of terms = n, the first term is a = 5 and the last term is l = 45.

We have, Sn = 150

⇒ \(\frac{n}{2}[a+l]=150\)

⇒ \(\frac{n}{2}(5+45)=150\)

n = 6

Now, l = 45 i.e., nth term = 45

⇒ a + (n-1)d = 45

⇒ 5 + (6- 1)d = 45

⇒ 5d = 40

⇒ \(d=\frac{40}{5}\)

= 8

Hence, number of terms = 6 and common difference = 8

Question 4. Solve for x: 5+ 13 + 21 +……. +x = 2139
Solution:

Given

5+ 13 + 21 +……. +x = 2139

Here, a = 5,d= 13-5 = 8

Let Tn=x

∴ Sn = 2139

∴ \(\frac{n}{2}[2 \times 5+(n-1) 8]=2139\)

⇒ n(5 + 4n – 4)= 2139

⇒ 4n2 + n- 2139 = 0

⇒ 4n2 – 92n + 93n -2139 = 0

⇒ (n – 23) (4n + 93) = 0

∴ n = 23 or \(n=-\frac{93}{4}\)

since n > 0, we have n = 23

x = Tn = a + (n- 1 )d = 5 + (23 – 1 )(8)

x = 181

Question 5. How many terms of an A.P. -6, \(\frac{-11}{2}\),-5 are needed to give the sum -25? Explain the double answer.
Solution:

Here, a= -6, \(d=\frac{-11}{2}-(-6)=\frac{1}{2}\)

Let -25 be the sum of it terms of this A.P. (n ∈ N)

Using \(S_n=\frac{n}{2}[2 n+(n-1) d]\)

⇒ \(-25=\frac{n}{2}\left[2(-6)+(n-1)\left(\frac{1}{2}\right)\right]\)

⇒ \(-50=n\left(-12+\frac{n-1}{2}\right)\)

⇒ \(-50=n\left(\frac{n-25}{2}\right)\)

⇒ -100 = n2– 25n

⇒ n2-25n+ 100 = 0

⇒ (n – 5) (n – 20) = 0

∴ n = 5, 20

Both the values of n are natural and therefore, admissible.

Explanation of Double Answer:

S20 = \( -6-\frac{11}{2}-5-\frac{9}{2}-4-\frac{7}{2}-3-\frac{5}{2}-2-\frac{3}{2}-1-\frac{1}{2} +0+\frac{1}{2}+1+\frac{3}{2}+2+\frac{5}{2}+3+\frac{7}{2} \)

⇒ \(-6-\frac{11}{2}-5-\frac{9}{2}-4\)

⇒ S5

Question 6. Find the sum of all odd numbers lying between 100 and 200.
Solution:

The series formed by odd numbers lying between 100 and 200 is 101 + 103 + 105 + … + 199

Here, a = 101,

d= 103- 101 = 105- 103 = 2

Let, an= 199

⇒ 101 + (n- 1)2= 199

⇒ (n- 1)2 = 98

⇒ n-1 = 49

⇒ n = 50

Now, \(S_{50}=\frac{50}{2}(101+199)\)

=7500

The sum of all odd numbers lying between 100 and 200 =7500

Question 7. If an = 3 – 4n, show that a1,a2,a3, … form an A.P. Also find S20.
Solution:

Given

an = 3 – 4n

an = 3 – 4n

an-1 = 3-4(n-1)

= 3-4n+4=7-4n

∴ an-an-1 = (3-4n)-(7-4n) = -4n

Which does not depend on ‘n’ i.e., the difference of two consecutive terms is constant.

∴ Given sequence is in A.P.

Now, a1 = a = 3 – 4(1 ) = -1 ,

d = -4

∴ \( S_{20}=\frac{20}{2}[2 a+(20-1) d]\)

⇒ \(=10[2(-1)+19(-4)]\)

=-780

Question 8. If the sum of the first 6 terms of an A.P. is 36 and that of the first 1 6 terms is 256, find the sum of the first 10 terms.
Solution:

Given

The sum of the first 6 terms of an A.P. is 36 and that of the first 1 6 terms is 256

Let the first term and common difference of A.P. be ‘a’ and ‘d’ respectively.

S6 = 36

⇒ \(\frac{6}{2}[2 a+(6-1) d]=36\)

2a + 5d= 12 ……(1)

5,6 = 256

⇒ \(\frac{16}{2}[2 a+(16-1) d]=256\)

2a + 15d = 32 …..(2)

Subtract eq. (1) from eq. (2), we get

⇒ \(\begin{array}{r}
2 a+15 d=32 \\
2 a+5 d=12 \\
-\quad-\quad- \\
\hline 10 d=20
\end{array}\)

d = 2

put d = 2 in eq. (1), we get

2a + 5(2) = 12

⇒ 2a = 2

⇒ a = 1

Now, the sum of first 10 terms 510 = \(\frac{10}{2}[2 a+(10-1) d]=5[2(1)+9(2)]\)

= 100

The sum of first 10 terms = 100

Question 9. The sum of the first V terms of an A.P. whose first term is 8 and the common difference is 20, is equal to the sum of the first 2, i terms of another A.P. whose first term is -30 and the common difference is 8. Find ‘n’.
Solution:

Given

The sum of the first V terms of an A.P. whose first term is 8 and the common difference is 20, is equal to the sum of the first 2, i terms of another A.P. whose first term is -30 and the common difference is 8.

For first A.P., a = 8, d = 20

∴ \( S_n=\frac{n}{2}[2(8)+(n-1)(20)]\)

⇒ \(n(8+10 n-10)=10 n^2-2 n \)

For second A.P., a = -30, d = 8

⇒ \(S_{2 n}=\frac{2 n}{2}[2(-30)+(2 n-1)(8)]\)

⇒ \(n(-60+16 n-8)=16 n^2-68n\)

Question 10. If the pth term of an A.P. is x and the qth term is y. Show that the sum of first (p + q) terms is \(\frac{p+q}{2}\left\{x+y+\frac{x-y}{p-q}\right\}\)
Solution:

Given

The pth term of an A.P. is x and the qth term is y.

Let first term = n and common difference = d

Now, Tp = x a + (p- 1)d = x …(1)

and Tq + a(q- 1)d =y …(2)

Subtracting eq. (2) from eq. (1), we get

(p – q)d = x- y

⇒ \(d=\frac{x-y}{p-q}\)

Now, if you put the value of d in eq. (1 ) or (2), and find a then it will be a very tedious job. So, don’t find ‘a we will simplify especially:

⇒ \(S_{p+q}=\frac{p+q}{2}[2 a+(p+q-1) d]\)

Bifurcate the terms inside the bracket as

⇒ \(S_{p+q}=\frac{p+q}{2}[\{a+(p-1) d\}+\{a+(q-1) d\}+d]\)

⇒ \(S_{p+q}=\frac{p+q}{2}\left[x+y+\frac{x-y}{p-q}\right]\) [From (1) ,(2) and (3)]

Question 11. The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to 30th term is 1 : 3. Calculate the first term and 1 3th term of the A.P.
Solution:

Given

The sum of the first six terms of an A.P. is 42. The ratio of its 10th term to 30th term is 1 : 3.

Let the first term and common difference of A.P. be V and ‘d’ respectively.

Given that,

⇒ \(\frac{a_{10}}{a_{30}}=\frac{1}{3}\)

⇒ \(\frac{a+9 d}{a+29 d}=\frac{1}{3}\)

⇒ 3a + 27d = a + 29d

⇒ 2a = 2d

⇒ a=d

and, S6 = 42

⇒ \(\frac{6}{2}(2 a+5 d)=42\)

3(2 d+5 d)=42

d = 2

a = 2

a13 = a+12d

= 2 + 12 × 2

= 26

The first term and 13th term of the A.P is 2 and 26.

Question 12. The ratio between the sum of fird=st n terms of two A.P.’s are in the ratio (7n-5): (5n+17). Find the ratio of their 10th terms.
Solution:

Given

The ratio between the sum of fird=st n terms of two A.P.’s are in the ratio (7n-5): (5n+17).

Let a1 and d1, be the first term and common difference of first A.P and let a2 and d2 be the first term and common difference of other A.P.

∴ \( \frac{\frac{n}{2}\left[2 a_1+(n-1) d_1\right]}{\frac{n}{2}\left[2 a_2+(n-1) d_2\right]}=\frac{7 n-5}{5 n+17}\)

⇒ \(\frac{2 a_1+(n-1) d_1}{2 a_2+(n-1) d_2} =\frac{7 n-5}{5 n+17}\)

⇒ \(\frac{a_1+\left(\frac{n-1}{2}\right) d_1}{a_2+\left(\frac{n-1}{2}\right) d_2}=\frac{7 n-5}{5 n+17}\)

Replace \(\frac{n-1}{2}\) by 9

⇒ \(\frac{n-1}{2}\)

⇒ n = 19

∴ From eq. (1)

⇒ \(\frac{a_1+9 d_1}{a_2+9 d_2}=\frac{7(19)-5}{5(19)+17}\)

⇒ \(\frac{T_{10}}{T_{10}^*}=\frac{128}{112}=\frac{8}{7}\)

∴ \(T_{10}: T_{10}^*=8: 7\)

Question 13. If the sum of the first ‘m’ terms of an A.P. be V and the sum of its first ‘n’ terms is ‘m’. then show that the sum of its first (m + n) terms is -(m +n).
Solution:

Given

The sum of the first ‘m’ terms of an A.P. be V and the sum of its first ‘n’ terms is ‘m’.

Let the first term and common difference of the A.P. be ‘a’ and ‘d’ respectively.

Given that,

Sm = n

⇒ \(\frac{m}{2}[2 a+(m-1) d] =n \)

⇒ \(2 a m+\left(m^2-m\right) d =2 n\) ….(1)

and Sn =m

⇒ \(\frac{n}{2}[2 a+(n-1) d] =m\)

⇒ \(2 a n+\left(n^2-n\right)d =2 m\) …..(2)

Subtract eq. (2) from eq. (1), we get

⇒ 2a(m – n) + {(m2 – m2) – (m – n))d = 2(n – m)

⇒ 2a(m – n) + {(m – n) (m + n) – (m – n)}d = -2(m – n)

⇒ 2d(m – n) + (m – n) (m + n – 1)d = -2(m – n)

⇒ 2a + (m + n – l)d = -2 …..(3)

Now, \(S_{m+n}=\frac{m+n}{2}\{2 a+(m+n-1) d\}\)

⇒ \(=\frac{m+n}{2}(-2)=-(m+n)\)

Hence proved.

Question 14. If the sum of first n, 2n and 3n terms of an A.P. be S1, S2 and S3 respectively, then prove that: S3 = 3(S2 – S1)
Solution:

Given

The sum of first n, 2n and 3n terms of an A.P. be S1, S2 and S3 respectively,

Let the first term and common difference of the A.P. be V and ld’ respectively.

S1 = sum of first ‘n’ terms

⇒ \(S_1=\frac{n}{2}[2 a+(n-1) d]\) ….(1)

S2 = sum of first ‘2n terms

⇒ \(\frac{2 n}{2}[2 a+(2 n-1) d]\)

⇒ \(S_2=\frac{n}{2}[4 a+(4 n-2) d]\) ….(2)

and S3 = sum of first ‘3n’ terms

⇒ \(S_3=\frac{3 n}{2}[2 a+(3 n-1) d]\) …..(3)

⇒ \(S_2-S_1=\frac{n}{2}[4 a+(4 n-2) d]-\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{n}{2}[4 a+(4 n-2) d-2 a-(n-1) d]=\frac{n}{2}[2 a+(3 n-1) d]\)

⇒ \(3\left(S_2-S_1\right)=\frac{3 n}{2}[2 a+(3 n-1) d]=S_3\)

Hence proved

Question 15. If the ratio of the sum of the first m and n terms of an A.P. is m2:n2, show that the ratio of its mth and nth terms is (2m – 1) : (2n – 1).
Solution:

Given

The ratio of the sum of the first m and n terms of an A.P. is m2:n2

⇒ \(\frac{S_m}{S_n}=\frac{\frac{m}{2}[2 a+(m-1) d]}{\frac{n}{2}[2 a+(n-1) d]}=\frac{m^2}{n^2} \)

⇒ \(\frac{2 a+(m-1) d}{2 a+(n-1) d}=\frac{m}{n}\)

⇒ \(\frac{a+\left(\frac{m-1}{2}\right) d}{a+\left(\frac{n-1}{2}\right) d}=\frac{m}{n} \ldots\)

Caution

Some students write as :

put \(\frac{m-1}{2}=m-1 \quad \Rightarrow \quad m-1=2 m-2\)

m-2m = -2+1

m = 1

which is wrong.

Here has mixed up the m’s of \(\frac{m-1}{2} \text { and } m-1\)

We want, \(\frac{T_m}{T_n} \text { i.e., } \frac{a+(m-1) d}{a+(n-1) d}\)

So, we replace \(\frac{m-1}{2}\) as m-1

i.e., m-1 as 2(m – 1)

⇒ m – 1 as 2m – 2

⇒ m as 2m -2+1

i.e., replace m by 2m – 1

Similarly, replace n by 2n – 1 in eq. (1), we get

∴ \( \frac{a+(m-1) d}{a+(\dot{n}-1) d}=\frac{2 m-1}{2 n-1}\)

i.e., \(\frac{T_m}{T_n}=\frac{2 m-1}{2 n-1}\)

Question 16. An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429. Find the A.P.
Solution:

Given

An A.P. consists of 37 terms. The sum of the three middlemost terms is 225 and the sum of the last three terms is 429.

Total number of terms (n) = 37, which is odd

∴ Middle term = \(\frac{37+1}{2}\) th term = 19th term

So, 3 middle most terms are 18th, 19th and 20th

∴ T18 +T19 + = 225

⇒ a + 17d + a+ 18d + a+ 19d- 225

⇒ 3a + 54d = 225

⇒ a=18d = 75 ……(1)

Also, a sum of the last 3 terms = 429

T35 + T36 + T37 = 429

⇒ a + 34d + a + 35d + a + 36d = 429

⇒ 3a + 105d = 429

⇒ a + 35d = 143 ……(2)

Solving, (1) and (2), we get d- 4 and a = 3

∴ Required A.P. is a, a + d, a + 2d, a + 3d,…

i.e., 3, 3 + 4, 3 + 2(4), 3 + 3(4)

i.e., 3, 7, 11, 15, …

The A.p = 3, 7, 11, 15, …

Question 17. 25 trees are planted in a straight line 5 metres apart from each other. To water them the gardner must bring water for each tree separately from a well 10 metres from the first tree in line with trees. Find the distance he will move in order to water all the trees beginning with the first if he starts from the well.
Solution:

Arithmetic Progression 25 Trees Are Planted In A Straight Line 5 Metre Apart From Each Other

Distance Covered by Gardner from well to well :

⇒ \(\underbrace{10+10}_{\text {well-1-well }}+\underbrace{(10+5)+(10+5)}_{\text {well-2-well }}+\underbrace{(10+2 \times 5)+(10+2 \times 5)}_{\text {well-3-well }}\)

⇒ \(+\underbrace{(10+23 \times 5)+(10+23 \times 5)}_{\text {well-24-well }}+\underbrace{(10+24 \times 5)}_{\text {well to } 25 \text { th tree }}\)

2[10 + (10 + 5) + (10 + 2×5) + ……+(10 + 23×5)] + (10 + 24×5)

⇒ \(2 \underbrace{[10+15+20+\ldots+125]}_{24 \text { terms }}+(10+120)\)

⇒ \(2 \times \frac{24}{2}[10+125]+130=24 \times 135+130=3240+130=3370 \mathrm{~m}\)

Question 18. A child puts one five-rupee coin of her savings in the piggy bank on the first day. She increases her savings by one five-rupee coin daily. If the biggy bank can hold 1 90 coins of five-rupees in all, find the number of days she can continue to put the five-rupee coins into it and find the total money she saved. Write your views on the habit of saving.
Solution:

Child’s savings day-wise are,

Arithmetic Progression Childs Savings Day Wise Are

We can have at most 190 coins

i.e., 1+2 + 3+ 4 + 5 + …to n terms =190

⇒ \(\frac{n}{2}[2 \times 1+(n-1) 1]=190\)

⇒ n(n + 1) = 380

⇒ n2 +n- 380 = 0

⇒ (n + 20) (n- 19) = 0

⇒ n = -20

or n = 19

But many coins cannot be negative

∴ n = 19

So, number of days =19

Total money she saved = 5+10+15+20 + … upto 19 terms

⇒ \(\frac{19}{2}[2 \times 5+(19-1) 5]\)

⇒ \(\frac{19}{2}[100]=\frac{1900}{2}\)

=950

So, number of days = 19

and total money she saved = ₹950

Arithmetic Mean Of Two Numbers

If three numbers are in A.P., then the middle term is called the arithmetic mean of the remaining two numbers.

Let A be the arithmetic mean of a and b.

∴ a, A, b are in A.P.

⇒ A – a = b – A

⇒ 2A = a +b

⇒ A= \(\frac{a+b}{2}\)

Selection Of Continuous Terms Of A.P.

  1. Three consecutive terms in A.P. a— d, a, a + d (common difference is d)
  2. Four consecutive terms in A.P. a — 3d, a — d, a+ d, a + 3d (common difference is 2d)
  3. Five consecutive terms in A.P. a — 2d, a — d, a, a + d, a + 2d (common difference is d)

Selection Of Continuous Terms Of A.P. Solved Examples

Question 1. Find the value of x for which (x + 2), 2x, (2x + 3) are three consecutive terms of A.P. 
Solution:

Given

(x + 2), 2x, (2x + 3) are three consecutive terms of A.P.

∴ 2x = \(\frac{(x+2)+(2 x+3)}{2}\)

⇒ 4x = 3x + 5

x = 5

The value of x = 5.

Question 2. Split 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.
Solution:

Let three parts be a- d, a, a + d.

∴ a-d + a+ a+ d = 207

⇒ 3a = 207

⇒ a = 69

and (a-d)-a = 4623

⇒ (69-d)69 = 4623

⇒ 69 – d = 67

⇒ d = 2

∴ a-d = 69 – 2 = 67

a= 69

a+d = 692 = 71

⇒ three parts are 67,69,71

Question 3. The angles of a triangle are in A.P. The greatest angle is twice the least. Find all the angles of the triangle.
Solution:

Given

The angles of a triangle are in A.P. The greatest angle is twice the least.

Let angles of the triangle be a – d, a, a + d

∴ a – d+a+a + d = 180°

⇒ 3a = 1 80°

⇒ a = 60°

and a + d = 2(a – d)

⇒ 60° + W = 2(60°- d)

⇒ 60° + d = 120° -2d

⇒ 3d = 60°

⇒ d = 20°

∴ a- d = 60°- 20° = 40°

a +d = 60° + 20° = 80°

∴ Three angles of triangle are 40°, 60°, 80°

Question 4. The angles of a quadrilateral are in A.P. and their common difference is 10°. Find the angles.
Solution:

Given

The angles of a quadrilateral are in A.P. and their common difference is 10°.

Let the angles of the quadrilateral be

a, a + 10°, a + 20°, a + 30°   (∵ common difference is 10°)

∴ a + (a + 10°) + (a + 20°) + (a + 30°) = 360°

⇒ 4a + 60° = 360°

⇒ 4a = 300°

⇒ a = 75°

∴ a+ 10°= 75°+ 10° = 85°

a + 20° = 75° + 20° = 95°

a + 30° = 75° + 30°= 105°

Hence, the angles are 75°, 85°, 95°, 105°.

Alternative Method :

Let the four angles of a quadrilateral are

a – 3d, a – d, a + d and a + 3d

∴ Here the common difference is 2d (remember)

∴ (a – 3d) + (a-d) + (a + d) + (a + 3d) = 360°

4a = 360°

⇒ a = 90°

the common difference is given to be 10°

i.e., 2d = 10°

⇒ d = 5°

Four angles are a – 3d = 90°- 3(5°) = 75°,

a – d 90°- 5° = 85°,

a + d = 90° + 5° = 95°,

a + 3d = 90° + 3(5°) = 105°

Question 5. There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m -7)th mean is 1: 4. Find the value of m.
Solution:

Given

There are m arithmetic means between 5 and -16 such that the ratio of the 7th mean to the (m -7)th mean is 1: 4

Let A1, A2,…….Am be m arithmetic means between 5 and -16.

∴ 5,A1, A2, A3,…..Am, -16 are in A.P.

∴ \(T_{m+2}=-16\)

⇒ \(5+(m+1) d=-16\)

⇒ \(d=\frac{-21}{m+1}\)

⇒ \(\frac{A_7}{A_{m-7}}=\frac{1}{4}\)

⇒ \(\frac{T_8}{T_{m-6}}=\frac{1}{4}\)

⇒\(\frac{5+7 d}{5+(m-7) d}=\frac{1}{4}\)

⇒ \(\frac{5+7\left(\frac{-21}{m+1}\right)}{5+(m-7)\left(\frac{-21}{m+1}\right)} \frac{1}{4}\)

⇒ \(20-\frac{588}{m+1}=5-\frac{21(m-7)}{m+7}\)

⇒ 20m + 20- 588 = 5m +5- 21m + 147

⇒ 36m = 720

⇒ m = 20

The value of m = 20.

Question 6. The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.
Solution:

Given

The digits of a positive number of three digits are in A.P. and their sum is 15. The number obtained by reversing the digits is 594 less than the original number.

Let the 3 digits in A.P. at units, tens and hundredth places be a – d, a and a + d.

According to the first condition,

(a -d) +a + (a +d) = 15

⇒ 3a = 15

a = 5 …(1)

The number is (a – d) + 10a + 100 (a + d)….(1)

i.e., 111+ 99d …(2)

The number; on reversing the digits is

(a + d) + 1 0a + 100(A — d) i.e., 111+ 99d

∴ According to the 2nd condition,

111a- 99d = (11la + 99d) – 594

⇒ 198d = 594

⇒ d = 3 …(3)

∴ Required number = 111a + 99d [from (2)]

= 111(5) + 99(3) [from (1) and (3)]

= 555 + 297

= 852

Required number = 852

Arithmetic Progression Exercise 5.1

Question 1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

  1. File taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
  2. The amount of air present in a cylinder when a vacuum pump removes — the
    air remaining in the cylinder at a time.
  3. The cost of digging a well after every metre of digging, when it costs ₹150 for the first metre rises by ₹50 for each subsequent metre.
  4. The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.

Solution :

1. Fare of first kilometre = ₹15 

Fare of 2 kilolnetre = ₹(15 + 8) = ₹23

Fare of 3 kilometre = ₹(23 + 8) = ₹31

Fare of 4 kilometre = ₹(3 1 + 8) = ₹39

Now, a1 = 15, a2 = 23, a3 = 31, a4 = 39

a2 -a1 = 23 – 15 = 8

a3-a2 = 31 -23 = 8

a4 – a3 = 39 – 31 = 8

∴ The difference between two consecutive terms is constant.

∴ The taxi fare after each kilometre is in A.P

2. Let the initial volume of air in the cylinder = V

In the first pump,

air remove = \(\frac{V}{4}\)

Remaining air = \(V-\frac{V}{4}=\frac{3 V}{4}\)

In the second pump,

air remove = \(\frac{1}{4} \times \frac{3 V}{4}=\frac{3 V}{16}\)

Remaining air = \(\frac{3 V}{4}-\frac{3 V}{16}=\frac{9 V}{16}\)

Now, \(a_1=V, a_2=\frac{3 V}{4}, a_3=\frac{9 V}{16}\)

∴ \(a_2-a_1=\frac{3 V}{4}-V=-\frac{V}{4}\)

and \(a_3-a_2=\frac{9 V}{16}-\frac{3 V}{4}=-\frac{3 V}{16}\)

∵ a2– a1 ≠ a3– a2

∴ The volumes of air are not in A.P.

3. The cost of digging of first metre = ₹150

The cost of digging of 2 metres = ₹.(150 + 50)

= ₹200

The cost of digging of 3 metres

= ₹ (150 + 50 + 50)

= ₹250

The cost of digging of 4 metres

= ₹(150 + 50 + 50 + 50)

= ₹300

Now, a1=₹150, a2 = 200,a3 = ₹250, a4 = ₹300

∴ a2-a1 = ₹200-₹150 = ₹50

a3-a2= ₹250 -₹200 = ₹50

a4-a3= ₹300- ₹250 = ₹50

∵  The difference between two consecutive terms is constant.

The costs of digging each metre are in A.P.

4. Principal P = ₹10,000;

rate of interest R = ₹8%

Amount after 1 year,

⇒ \(A_1=P\left(1+\frac{R}{100}\right)^1=10,000\left(1+\frac{8}{100}\right)^1\)

⇒ \(=10,000 \times \frac{108}{100}=₹ 10,800\)

Amount after 2 years,

⇒ \(A_2=P\left(1+\frac{R}{100}\right)^2=10,000\left(1+\frac{8}{100}\right)^2\)

⇒ \(=10,000 \times \frac{108}{100} \times \frac{108}{100}=₹ 11664\)

Amount after 3 years

⇒ \(A_3=P\left(1+\frac{R}{100}\right)^3=10,000\left(1+\frac{8}{100}\right)^3\)

⇒ \(=10,000 \times \frac{108}{100} \times \frac{108}{100} \times \frac{108}{100}\)

= ₹12597.12

Now, A2-A1 = 11664- 10800 = 864

A3-A2= 12597.12- 11664 = 933.12

The difference between two consecutive terms is not the same.

∴ Amounts are not in A.P.

Question 2. Write the first four terms of the A.P., when the first term a and the common difference d are given as follows :

  1. a= 10, d= 10
  2. a =- 2, d = 0
  3. a = 4,d = -3
  4. a=-1, d=\(\frac{1}{2}\)

Solution:

1. a = 10, d = 10

First term = a = 10

Second term =a + d= 10 + 10 = 20

Third term = a + 2d = 10 + 2 × 10 = 30

Fourth term = a + 3d = 1 0 + 3 × 1 0 = 40

∴ The first four terms of A.P. are 10, 20, 30, 40

2. a = -2, d = 0

First term = a = -2

Second term = a + d = -2 + 0 = -2

Third term = a + 2d = -2 + 2 × 0 = -2

Fourth term = a + 3d = -2 + 3 × 0=-2

∴ The first four terms of A.P. are -2, -2, -2, -2

3. a= 4, d = -3

First term = a = 4

Second term = a + d = 4 + (-3) = 1

Third term = a + 2d = 4 + 2(-3) = -2

Fourth term = a + 3d = 4 + 3(-3) =-5

∴ The first four terms of A.P are 4, 1, -2, -5

4. a=-1, d = \(\frac{1}{2}\)

First term =a = -1

Second term = a + d = \(-1+\frac{1}{2}=-\frac{1}{2}\)

Third term= a + 2d = \(-1+2 \times \frac{1}{2}=0\)

Fourth term= a + 3d = \(-1+3 \times \frac{1}{2}=\frac{1}{2}\)

∴ First four terms of A.P. are = \(-1,-\frac{1}{2}, 0, \frac{1}{2}\)

5. a =- 1.25, d = -0.25

First term = a = -1.25

Second term = a+d =-1.25 + (-0.25) = -1.50

Third term =a + 2d = -1.25 + 2(-0.25)

= -1.75

Fourth term = a + 3d =- 1.25 + 3(-0.25.)

= -2.00

∴ The first four terms of A.P. are

= -1.25, -1.50, -1.75, -2.00

Question 3. For the following A.Ps., write the first term and the common difference :

  1. 3, 1,-1, -3,…
  2. -5, -1,3, 7,…
  3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)
  4. 0.6, 17,2.8,3.9,…

Solution:

1. 3, 1, -1, -3,…

First term a = 3

Common difference d = 1-3 = (-1)- 1 = -2

2. -5,-1, 3, 7,…

First term a = -5

Common difference d = -1-(-5) = 3-(-1) = 4

3. \(\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots\)

First term a = \(\frac{1}{3}\)

Common difference d = \(\frac{5}{3}-\frac{1}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)

4. 0.6, 17,2.8,3.9,…

First term a = 0.6

Common difference d= 1.7 – 0.6

= 2.8-17=1.1

Question 4. Which of the following are A.P.s? If they form an A.P., find the common difference d and write three more terms.

  1. \(2,4,8,16,\)
  2. \(2, \frac{5}{2}, 3, \frac{7}{2},\)
  3. \(-1.2,-3.2,-5.2,-7.2,\)
  4. \(-10,-6,-2,2,\)
  5. \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2},\)
  6. \(0.2,0.22,0.222,0.2222,\)
  7. \(0,-4,-8,-12,\)
  8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},\)
  9. \(1,3,9,27,\)
  10. \(a, 2 a, 3 a, 4 a,\)
  11. \(a, a^2, a^3, a^4,\)
  12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32},\)
  13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12},\)
  14. \(1^2, 3^2, 5^2, 7^2,\)
  15. \(1^2, 5^2, 7^2, 73,\)

Solution:

1. 2,4, 8, 16,…

Here, a1 = 2, = 4, = 8, = 16

a2– a1 = 4- 2 = 2

a3-a2 = 8- 4 = 4

∵ a2-a1 ≠ a3-a2

∴ Given sequence is not an A.P.

2. \(2, \frac{5}{2}, 3, \frac{7}{2},\)

Here, \(a_1=2, a_2=\frac{5}{2}, a_3=3, a_4=\frac{7}{2}\)

⇒ \(a_2-a_1=\frac{5}{2}-2=\frac{1}{2}\)

∴ \(a_3-a_2=3-\frac{5}{2}=\frac{1}{2}\)

⇒ \(a_4-a_3=\frac{7}{2}-3=\frac{1}{2}\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=\frac{1}{2}\)

∴ Given sequence is an A.P. and d = \(\frac{1}{2}\)

Now, fifth term \(a_5=a_4+d=\frac{7}{2}+\frac{1}{2}=4\)

Sixth term \(a_6=a_5+d=4+\frac{1}{2}=\frac{9}{2}\)

Seventh term \(a_7=a_6+d=\frac{9}{2}+\frac{1}{2}=5\)

∴ Next three terms of A.P. = 4, \(\frac{9}{2}\),5

3. -1.2, -3.2, -5.2, -7.2, …

Here a1 = -1.2, a2 =-3.2, a3 = -5.2, a4 =-7.2

∴ a2-a1 =(-3.2) -(-1.2) =-3.2+ 1.2= -2

a3 -a2 = (-5-2) – (-3.2) =-5.2 + 3.2 =-2

a4 – a3 = (-7.2) – (-5.2) = -7.2 + 5.2 =- 2

∵ a2 – a1 = a3 – a2 = a4 – a3 = -2

∴ Given sequence is A.P. and d = -2

Now, fifth term a5= a4 + d = -7.2 + (-2) = -9.2

Sixth term a6 = a5 + d = -9.2 + (-2) = -1 1 .2

Seventh term a7 =a6+ d = -11.2 + (-2) =-13.2

∴ Next three terms of A.P. =- 9.2, -1 1.2, -13.2

4. -10, -6, -2,2,…

Here a1 = -10, a2 =-6, a3 =-2, a4 = 2

∴ a2-a1=(-6)-(-10)=-6+ 10 = 4

a3 – a2 = (-2) – (-6) =-2 + 6 = 4

a4 – a3 = 2-(-2) = 2 + 2 = 4

∵ a2-a1 = a3 – a2 = a4 – a3 = 4

∴ Given sequence is A.P. and d = 4

Now, fifth term a5 = a4 + d = 2 + 4 = 6

Sixth term a6 = a5 + d = 6 + 4 = 10

Seventh term a7 = + d = 1 0 + 4 = 14

∴ Next three-term of A.P. = 6, 10, 14

5. \(3,3+\sqrt{2}, 3+2 \sqrt{2}, 3+3 \sqrt{2},\)

Here, \(a_1=3, a_2=3+\sqrt{2}, a_3=3+2 \sqrt{2},\)

⇒ \(a_4=3+3 \sqrt{2}\)

⇒ \(a_2-a_1=(3+\sqrt{2})-3=\sqrt{2}\)

∴ \(a_3-a_2=(3+2 \sqrt{2})-(3+\sqrt{2})=\sqrt{2}\)

⇒ \(a_4-a_3=(3+3 \sqrt{2})-(3+2 \sqrt{2})=\sqrt{2}\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=\sqrt{2}\)

∴ Given sequence is A.P. and d = \(\sqrt{2}\)

Now, fifth term \(a_5=a_4+d\)

⇒ \(3+3 \sqrt{2}+\sqrt{2}=3+4 \sqrt{2}\)

Sixth term \(a^6=a^5+d\)

⇒ \(3+4 \sqrt{2}+\sqrt{2}=3+5 \sqrt{2}\)

Seventh term \(a_7=a_6+d\)

⇒ \(3+5 \sqrt{2}+\sqrt{2}=3+6 \sqrt{2}\)

Next three terms of A.P.

⇒ \(3+4 \sqrt{2}, 3+5 \sqrt{2}, 3+6 \sqrt{2}\)

6. 0.2,0.22,0.222,0.2222, …

Here a1 = 0.2, a2 = 0.22, a3 = 0.222, a4 = 0.2222

∴ a2-a1=0.22-0.2 = 0.02

a3-a2 = 0.222 -0.22 = 0.002

∵ a2 – a1 = a3 – a2

∴ Given sequence is not an A.P.

7. 0, -4, -8, -12, …

Here, a1 = 0, a2 = -4, a3 = -8, a4 = -12

∴ a2 – a1 = -4 – 0 = – 4

a3-a2 =-8- (-4) =- 8 + 4 = -4

∵ a4 – a3 = -12 – (-8) =-12 + 8 =- 4

a2 – a1 = a3 – a2 = a4 – a3 = -4

∴ Given sequence is A.P. and d =-4

Now, fifth term a5 =a4 + d =-12 + (-4) =-16

Sixth term a6 = a5 + d = -16 + (-4) = -20

Seventh term a7 = a6 + d = -20 + (-4) = -24

∴ Next three terms of A.P. =-16, -20, -24

8. \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}, \ldots\)

Here, \( a_1=a_2=a_3=a_4=-\frac{1}{2}\)

∴ \(a_2-a_1=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

⇒ \(a_3-a_2=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

⇒ \(a_4-a_3=\left(-\frac{1}{2}\right)-\left(-\frac{1}{2}\right)=0\)

∴ a2– 1 = a3 – a2 = a4 – a3 = 0

∴ Given sequence is A.P. and d = 0

Now, fifth term \(a_5=a_4+d=\frac{-1}{2}+0=-\frac{1}{2}\)

Sixth term \(a_6=a_5+d=\frac{-1}{2}+0=-\frac{1}{2}\)

Seventh term \(a_7=a_6+d=\frac{-1}{2}+0=-\frac{1}{2}\)

∴ Next three terms of A.P. = \(-\frac{1}{2},-\frac{1}{2},-\frac{1}{2}\)

9. 1,3, 9,27,…

Here, a1 = 1, a2 = 3, a3 = 9, a4 = 27

∴ a2-a1 = 3-1=2

a3– a2 = 9- 3 = 6

∵ a2 -a1 ≠ a3– a2

∴ Given sequence is not an A.P.

10. a, 2a, 3a, 4a

Here, a1 = a, a2 = 2a, a2 = 3a, a4 = 4a

∴ a2-a1 =2a – a = a

a3 – a2 = 3a – 2a =a

a4 – a3= 4a – 3a = a

a2– a1 = a3 – a2 =a4– a3 = a

∴ Given sequence is A.P. and d = a

Now, fifth term a5 = a4 + d = 4a + a = 5a

Sixth term a6 = a5 + d = 5a + a = 6a

Seventh term a7 = a6 + d = 6a + a = 7a

∴ Next three terms of A.P. = 5A, 6a, 7a

11. a, a2, a3, a4 = ,…

Here, a1 = a, a2 = a2, a3 = a3, a4 = a4

∴ a2-a1 = a2 – a

a3 – a2 = a3-a2

a2– a1 ≠ a3– a2

∴ Given sequence is not an A.P.

12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \cdots\)

Here, \(=\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, 4 \sqrt{2}, \cdots\)

∴ \(a_1=\sqrt{2}, a_2=2 \sqrt{2}, a_3=3 \sqrt{2}, a_4=4 \sqrt{2}\)

⇒ \(a_2-a_1=2 \sqrt{2}-\sqrt{2}=\sqrt{2}\)

⇒ \(a_3-a_2=3 \sqrt{2}-2 \sqrt{2}=\sqrt{2}\)

⇒ \(a_4-a_3=4 \sqrt{2}-3 \sqrt{2}=\sqrt{2}\)

⇒ \(a_2-a_1=a_3-a_2=a_4-a_3\)

∴ Given sequence is A.P. and d = \(\sqrt{2}\)

Now, fifth term \(a_5=a_4+d=4 \sqrt{2}+\sqrt{2}=5 \sqrt{2}\)

Sixth term \(a_6=a_5+d=5 \sqrt{2}+\sqrt{2}=6 \sqrt{2}\)

Seventh term \(a_7=a_6+d=6 \sqrt{2}+\sqrt{2}=7 \sqrt{2}\)

∴ Next three terms of A.P. = \(5 \sqrt{2}, 6 \sqrt{2}, 7 \sqrt{2}\)

13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots\)

Here, \(\quad a_1=\sqrt{3}, a_2=\sqrt{6}, a_3=\sqrt{9}, a_4=\sqrt{12}\)

⇒ \(a_2-a_1=\sqrt{6}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)\)

⇒ \(a_3-a_2=\sqrt{9}-\sqrt{6}=\sqrt{3}(\sqrt{3}-\sqrt{2})\)

⇒ \(a_2-a_1 \neq a_3-a_2\)

∴ Given sequence is not an A.P.

14. \(1^2, 3^2, 5^2, 7^2, \ldots\)

Here, \(\quad a_1=1^2, a_2=3^2, a_3=5^2, a_4=7^2\)

∴ \(a_2-a_1=3^2-1^2=9-1=8\)

⇒ \(a_3-a_2=5^2-3^2\)

⇒ \(=25-9=16\)

∵ \(a_2-a_1\neq a_3-a_2\)

∴ Given sequence is not an A.P.

15. \(1^2, 5^2, 7^2, 73, \ldots\)

Here, \(a_1=1^2, a_2=5^2, a_3=7^2, a_4=73\)

∴ \(a_2-a_1=5^2-1^2=25-1=24\)

⇒ \(a_3-a_2=7^2-5^2=49-25=24\)

⇒ \(a_4-a_3=73-7^2=73-49=24\)

∵ \(a_2-a_1=a_3-a_2=a_4-a_3=24\)

∴ Given sequence is not an A.P. and d = 24

Now, fifth term a5 = a4 + d = 73 + 24 = 97

Sixth term a6= a5+d = 97 + 24

= 121 = 112

Seventh term a7=a6 + d= 121 + 24 = 145

∴ Next three terms of A.P. = 97, 112, 145

Arithmetic Progression Exercise 5.2

Question 1. Fill in the blanks in the following table given the first term. d the common difference and an nth term of the A.P. :

Arithmetic Progression The Common Difference An And nth Term Of The AP

Solution:

1. Here, a = 7, d = 3. n= 8

∴ an – a + (n -1)d

= 7 + (8 – 1) 3

= 7 + 7 × 3=28

Therefore, an =28

2. a =-18, n = 10, an = 0

∴ a + (n – 1)d = an

⇒ -18 + (10- 1)d = 0

⇒ 9d = 18

⇒ d = 2

3. d =-3, n= 1 8, an =-5

∴ a + (n – 1 )d = an

⇒ a + (18 – 1) (-3) =-5

⇒ a-51 =- 5

⇒ a = -5 + 51 =46

4. a = -18.9, d = 2.5, an = 3.6

∴ a + (n -1)d = an

-18.9 + (n -1) × 2.5 = 3.6

⇒ (n- 1) × 2.5 =3.6+18.9 = 22.5

⇒ \(n-1=\frac{22.5}{2.5}=9\)

⇒ n = 9+1 = 10

5. a = 3.5, d = 0, n= 105

∴ an = a + (n -1 )d

= 3.5 + (105 – 1) × 0

= 3.5

Question 2. Choose the correct choice in the following and justify :

  1. 30th term of the A.P.: 10, 7, 4, …, is
    1. 97
    2. 77
    3. -77
    4. -87
  2. 11th term of the A.P.: -3, 2 …, is
    1. 28
    2. 22
    3. -38
    4. \(-48 \frac{1}{2}\)

Solution:

1. Given A.P.: 10,7,4…

Mere n= 10,d= 7- 10 = 4- 7 = -3, n =30

∴ an =a( n – 1 )d

⇒  a30 = 10 + (30 – 1) (-3) = 10 – 87 – -77

2. Given A.P. : \(-3,-\frac{1}{2}, 2, \ldots\)

Here a = \(a=-3, d=-\frac{1}{2}-(-3)=2-\left(-\frac{1}{2}\right)=\frac{5}{2}\)

n = 11

∴ an = a + (n – 1 )d

⇒ \(a_{11}=-3+(11-1) \frac{5}{2}=-3+25=22\)

Question 3. In the following A.Ps., find the missing terms in the boxes :

Arithmetic Progression In The Following APs Find The Missing Terms In The Boxes

Solution:

1. Here, first term a = 2

Third term = 26

⇒ a + (3-1)d = 26 2 + 2d = 26

⇒ 2d = 24

⇒ d= 12

∴ Second term = a + d = 2 + 12 = 14

∴ Term of the box = 14

2. Second term =13

⇒ a + (2-1)d = 13 (where a = first term and d = common difference)

⇒ a+d =13 ….(1)

and Fourth term = 3

⇒ a + (4 – 1 )d= 3 ⇒ a + 3d = 3 ….(2)

Subtracting equation (1) from equation (2),

Arithmetic Progression Fourth Term Is 3 Subtracting Equation 1 And 2

⇒ d = -5

Put the value of d in equation (1)

a + (-5) = 13 ⇒ a = 13+5

= 18

Third term a2 + d = 13 + (-5) = 8

∴ Term of the boxes = 18. 8 respectively.

3. Here, first term a = 5

Fourth term \(a_4=9 \frac{1}{2}\)

⇒ \(a+3 d=\frac{19}{2}\)

⇒ \(5+3 d=\frac{19}{2}\)

⇒ \(3d=\frac{19}{2}-5=\frac{9}{2} \Rightarrow d=\frac{3}{2}\)

Now, second term \(a_2=a+d=5+\frac{3}{2}=\frac{13}{2}\)

Third term \(a_3=a_2+d=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8\)

∴ Term of the boxes = \(\frac{13}{2}, 8\)

4. Here, first term a = -4

Sixth term = 6

⇒ a + 5d = 6 ⇒ -4 + 5d = 6

⇒ 5d = 6+4= 10

⇒ d =2

∴ Second term a2 = a + d = -4 + 2 = -2

Third term a3 = a2 + d = -2 + 2 = 0

Fourth term a4 = a3+d = 0 + 2 = 2

Fifth term a5= a4 + d = 2 + 2 = A

∴ Term of the boxes = -2, 0, 2, 4 respectively.

5. Here, second term a2 = 38

⇒ a + d= 38

Sixth term a5 = -22

⇒ a + 5d = -22

Subtracting equation (1) from (2),

Arithmetic Progression Here Second Term A2 Is 38

d = -15

Put the value of d in equation (1),

a + (-15) =38

⇒ a = 38+ 15=53

Third term a3 = a2 + d = 38 + (-15) = 23

Fourth term a4 = a3+d = 23 + (-15) = 8

fifth term a5 = a4 + d = 8 + (-15) -7

∴ Term of the boxes = 53, 23, 8, -7 respectively.

Question 4. Which term of the A.P. : 3, 8, 13, 18,… is 78?
Solution:

Given A.P. : 3, 8, 13, 18,…

Here a = 3, d = 8- 3 = 13-8 = 5

Let an = 78 ⇒ a + (n – 1 )d = 78

⇒ 3 + (11 – 1)5 = 78

⇒ (n- 1)5 =78-3 = 75

⇒ \(n-1=\frac{75}{5}=15\)

⇒ n= 15+ 1 = 16

∴ The 16th term is 78.

Question 5. Find the number of terms in each of the following A.Ps. :

  1. 7, 13, 19,…, 205
  2. 18, \(15 \frac{1}{2}\), 13…..-47

Solution:

Given A.P.: 7, 13, 19,…, 205

a=7

d= 13-7= 19- 13 = 6

Let an = 205

⇒ a + (n- l)d = 205

⇒ 7 + (n – 1)6 = 205

⇒ (n- 1)6 = 205-7 = 198

⇒ \(n-1=\frac{198}{6}=33\)

⇒ n = 33 + 1 = 34

∴ Number of terms in given A.P. = 34

2. Given A.P.:18, \(15 \frac{1}{2}\), 13…..-47

Here, a= 18

⇒ \(d=15 \frac{1}{2}-18=13-15 \frac{1}{2}=-2 \frac{1}{2}=\frac{-5}{2}\)

Let \(a_n=-47\)

⇒ \(a+(n-1) d=-47\)

⇒ \(18+(n-1)\left(\frac{-5}{2}\right)=-47\)

⇒ \((n-1)\left(\frac{-5}{2}\right)=-47-18=-65\)

⇒ \(n-1=(-65)\left(-\frac{2}{5}\right)=26\)

⇒ n = 26 + 1 = 27

∴ Number of terms in given A.P. = 27

Question 6. Check whether -150 is a term of the A.P. 1 1,8, 5. 2….
Solution:

Given A.P: 1 1, 8, 5, 2,…

⇒ a= 11, d=8- 11 =5- 8 = -3

Let an = -150

⇒ a + (n- 1 )d = -150

⇒ 11 + (n – 1) (-3) = -150

⇒ 11 – 3n + 3 = -150

⇒ 14 + 150 = 3

⇒ \(n=\frac{164}{3}=54 \frac{2}{3}\)

∵ The value of n is not a whole number.

∴ -150 is not a term of a given A.P.

Question 7. Find the 31st term of an A.P. whose 11th term is 38 and the 16th term is 73.
Solution:

Given

In AP 11th term is 38 and the 16th term is 73

Let a be the first term and d the common difference of the A.P.

Now, a11 = 38

⇒ n +(11 – 1)d = 38

⇒ a+ 10 = 38 ….(1)

and a16 = 73

⇒ a+ (16- 1)d = 73

⇒ a+15d=73 …..(2)

Subtracting equation (1) from (2),

Arithmetic Progression Find The 31st Term Of An AP

Put the value of d in equation (1),

a+ 10 × 7 = 38

⇒ a=38- 70 =-32

Now, the 31st term of A.P.

a31 = n + (31 – 1)d

=- 32 + 30 × 7

= -32 + 210= 178

The 31st term of an A.P = 178.

Question 8. An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:

Given

An A.P. consists of 50 terms of which 3rd term is 12 and the last term is 106.

Let a be the first term and d the common difference of A.P.

∴ a3 = 12 ⇒ a + (3 -1)d = 12

⇒ a+2d = 12 ….(1)

and last term= 50th term = 106

⇒ a + (50-1)d = 106

⇒ a + 49 d = 106 …..(2)

Subtracting equation (1) from (2)

Arithmetic Progression An AP Consists Of 50 Terms Of Which 3rd Term Is 12 And The Last Term Is 106

⇒ d = 2

Put the value of d in equation (1),

a + 2 × 2 = 12

a = 12-4 = 8

Now, 29th term = a + (29 – 1)d = 8 + 28 × 2

= 8 + 56 = 64

The 29th term = 64

Question 9. If the 3rd and the 9th terms of an A.P. are 4 and- 8 respectively, which term of this A.P. is zero?
Solution:

Given

The 3rd and the 9th terms of an A.P. are 4 and- 8 respectively,

Let a be the first term and d the common difference of A.P.

a3 = 4 ⇒ a + (3-1)d = 4

a + 2d = 4…..(1)

a9 = -8 ⇒ a + (9- 1 )d = -8

a + 8d = -8 ……(2)

Subtracting equation (1) from (2),

Arithmetic Progression If The 3rd And The 9th Terms Of AP

⇒ d = -2

Put the value of d in equation (1),

⇒ a + 2(-2) = 4

⇒ a = 4 + 4 =8

Now, let an = 0 a + (n- 1)d =0

⇒ 8 + (n- l)(-2) =0

⇒ 8- 2n + 2 = 0

⇒ -2n =-10

⇒ n =5

∴ 5th term of the progression is zero.

Question 10. The 17th term of an A.P. exceeds its 10th term by 7. Find the common difference.
Solution:

Given

The 17th term of an A.P. exceeds its 10th term by 7.

Let a be the first term and d the common difference of A.P.

∴ a17 =a10 + 7

⇒ a + (17- 1)d = a + (10- 1)d + 7

16d-9d = 7 ⇒ 7d =7

⇒ d=1

Common difference of progression = 1

Question 11. Which term of the A.P. : 3, 15, 27, 39, … will be 132 more than its 54th term?
Solution:

Given A.P. : 3, 15, 27, 39…

a = 3, d=15-3 = 27-15=12

∴ a54 = a+ (54 -1)d = 3 + 53 × 12

= 3 + 636 = 639

Let an =a54 + 132

⇒ a + (n – 1)d =639+ 132

⇒ 3 + (n -1)12 =771

⇒ (n- 1)12 =771 -3 = 768

⇒ \((n-1)=\frac{768}{12}=64\)

⇒ n = 64 + 1 = 65

∴ Required term = 65th term.

Question 12. Two A.P.s have the same common difference. The difference between their 100th terms is 1 00, what is the difference between their 1,000th terms?
Solution:

Given

Two A.P.s have the same common difference. The difference between their 100th terms is 1 00

Let the first term be a and the common difference be d of first A.P.

Let the first term be A and the common difference be D of the second A.P.

100th term of first progression = + (100 -1)d

= a + 99d

100th term of the second progression

=A + (100- 1)d

=A + 99d

Difference of 100th terms of two progression

= 100

⇒ (a + 99d) – (A + 99d) = 100

⇒ a + 99d-A-99d = 100

⇒ a-A = 100 ……(1)

Again, the 1000th term of the first progression

= a + (1000-1)d

= a + 999d

1000th term of the second progression

=A + (1000-1)d

=A + 999d

Difference of 1000th terms of two progressions

= (a + 999d) – (A + 999d)

= a + 999d-A-999d

= a – A = 100 [from equation (1)]

Question 13. How many three-digit numbers by 12. are divisible by 7?
Solution:

Three digit numbers: 100, 101 102, …, 999

Three-digit numbers divisible by 7: 105, 112, 119,…, 994

Here, a = 105, d= 1 12 – 105 = 1 19 – 1 12 = 7

Let an = 994

⇒ a + (n-1)d =994 ⇒ 105 + (n -1)7 = 994

⇒ (n- 1)7 =994- 105 = 889

⇒ \(n-1=\frac{889}{7}=127\)

⇒ n= 127+ 1 = 128

∴ Number of 3-digit numbers divisible by 7.

= 128

Question 14. How many multiples of 4 lie between 10 and 250?
Solution: The multiples of 4 between 10 and 250 are :

12, 16, 20, …,248

Here a=12,d= 16-12 = 20-16 = 4

an =248

⇒ a + (n- 1)d =248

⇒12 + (n- 1)4=248

⇒ (n- 1)4 =248- 12 = 236

⇒ n-1 = 59

⇒ n = 59 + 1 = 60

∴ Multiples of 4 between 10 and 250 = 60

Question 15. For what value of n, are the nth terms of two A.Ps?: 63,65,67,… and 3, 10, 17,… equal?
Solution:

First A.P.: 63, 65, 67…

Here a = 63

d = 65 – 63 = 67- 65 = 2

an = a + (n – 1 )d

= 63 + (n – 1)2 = 63 + 2n – 2

= 2n + 61

Second A.P. 3, 10, 17, …

Here A =3,D= 10-3= 17- 10 = 7

An =A + (n -1)D = 3 + (n – 1)7

= 3 + 7n-7 = 7n-4

According to the problem, an =An

⇒ 2n + 61 = 7n – 4

⇒ n -7n = -4 – 61

⇒ -5n = -65

⇒ n = 13

∴ The 13th terms of given progressions are equal.

Question 16. Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
Solution:

Let the first term be a and common difference d of the A.P.

According to the problem,

a7 – a5 =12

⇒ (a + 6d) – (a + 4d) = 12

⇒ a+6d-a-4d = 12

⇒ 2d = 12

⇒ d = 6

⇒ a3 = 16

⇒ a + (3-1)d = 16

⇒ a + 2 × 6 = a + (3-1)d = 16

⇒ a + 2 × 6 = 16

⇒ a = 16 – 12 = 4

Now A.P.: 4, 4 + 6, 4 + 2 × 6,…

= 4, 10, 16, …

Question 17. Find the 20th term from the last term of the A.P. : 3, 8, 13, …. 253.
Solution:

Given A.P. : 3, 8, 13,…, 253

Here, last term l = 253, = 8- 3 = 13-8 = 5

20th term from the end =l-(n-1)d

= 253 – (20- 1) × 5

= 253- 19 × 5 = 253-95 = 158

∴ 20th term from the end of the progression

= 158

The 20th term from the last term of the A.P = 158

Question 18. The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the
Solution :

Given

The sum of the 4th and 8th terms of an A.P. is 24 and the sum of the 6th and 10th terms is 44.

Let the first term be a and common difference d of the A.P.

∴ a4 + a8 =24 ⇒ a + 3d + a + 7d = 24

⇒ 2a + 10d =24

⇒ a+2d = 12 …(1)

and a6 + a10 = 44 ⇒ a + 5d + a + 9d = 44

⇒ 2a+14d =44

⇒ a + 7d =22

Subtracting equation (1) from (2),

Arithmetic Progression Let First Term Be A And Common Difference D Of The AP Subtracting Equation 1 from 2

⇒ d = 5

Put the value of d in equation (1),

a + 5 × 5= 12

⇒ a= 12-25 =-13

∴ Second term = a + d = -13 + 5 =-8

Third term-a + 2d =-13+2×5 = -3

So, the first three terms of A.P. are -13, -8,-3.

Question 19. Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?
Solution:

Given

Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year.

Salary in first year = ₹5000 Salary in second year = ₹5000 + ₹200 = ₹5200 Salary in third year = ₹5200 + ₹200 = ₹5400 Progression formed from the salary of each year ₹5000, ₹5200, ₹5400, …

Here, a2-a1 = 5200 -5000 = 200

a3-a2= 5400 -5200 = 200

a2 – a1 = -a2

⇒ The above progression is an A.P.

a = ₹5000, d = ₹200

Let in the nth year, the salary becomes ₹7000

Let in the nth year, the salary becomes ₹7000.

∴ an =7000

⇒ a + (n-1)d =7000

⇒ 5000 + (n- 1)200 =7000

⇒ (n- 1)200 =7000-5000

⇒ (n- 1)200 =2000

⇒ \(n-1=\frac{2000}{200}=10\)

⇒ n= 10+ 1 = 11

∴ In the 11th year, the salary of Subba Rao will be ₹7000.

Question 20. Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75, find n.
Solution:

Given

Ramkali saved ₹5 in the first week of the year and then increased her weekly savings by ₹1.75. If in the nth week, her weekly savings become ₹20.75

Saving of the first week = ₹5

∵ ₹1.75 is increasing in the savings of every week. The saving of every week from an A.P, in which.

a = ₹5 and d = ₹1.75

Let the saving in the nth week = ₹20.75

⇒ a + (n – 1) d = 20.75

⇒ 5 + (w- 1) (1.75) = 20.75

⇒ (n-1) (1.75) =20.75 -5 = 15.75

⇒ \(n-1=\frac{15.75}{1.75}=9\)

⇒ n=9 + 1 = 10

n= 10

Arithmetic Progression Exercise 5.3

Question 1. Find the sum of the following A.Ps.:

  1. 2, 7, 12, …, to 10 terms.
  2. -37, -33, -29, …, to 12 terms
  3. 0.6, 1.7, 2.8, …, to 100 terms.
  4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10},\)…, to 11 terms.

Solution:

1. 2, 7, 12…..to 10th term

Here, a = 2,d = 7 -2= 12-7 = 5,n= 10

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{10}=\frac{10}{2}[2 \times 2+(10-1) \times 5]\)

= 5(4 + 45) = 5×49 = 245

2. -37, -33, -29,…, to 12 terms

Here a =-37

d = -33- (-37) =-29- (-33) = 4, n = 12

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times(-37)+(12-1) \times 4]\)

= 6(-74 + 44) = 6 x (-30) = -180

3. 0.6, 1.7, 2.8,…, to 100 terms

Here a = 0.6, d = 1.7 – 0.6 = 2.8 – 1.7 = 1.1 n= 100

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{100}=\frac{100}{2}[2 \times 0.6+(100-1) \times 1.1]\)

= 50[1.2 + 108.9] = 50×110.1

= 5505

4. \(\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots, \text { to } 11 \text { terms }\)

Here, \(a=\frac{1}{15}, d=\frac{1}{12}-\frac{1}{15}=\frac{1}{10}-\frac{1}{12}=\frac{1}{60}, n=11\)

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_n=\frac{11}{2}\left[2 \times \frac{1}{15}+(11-1) \times \frac{1}{60}\right]\)

⇒ \(=\frac{11}{2}\left[\frac{8+10}{60}\right]=\frac{11}{2} \times \frac{18}{60}=\frac{33}{20}\)

Question 2. Find the sums given below :

  1. 7 +\(10 \frac{1}{2}\) + 14+ … + 84
  2. 34 + 32 + 30 + … + 10
  3. -5 + (-8) + (-11) + …(-230)

Solution:

1. 7 +\(10 \frac{1}{2}\) + 14+ … + 84

Here a = 7, d = \(10 \frac{1}{2}\) = \(14-10 \frac{1}{2}\)

⇒ \(3 \frac{1}{2}=\frac{7}{2}\)

Let an = 84

⇒ a + (n-1)d = 84 => \(7+(n-1) \frac{7}{2}=84\)

⇒ \( (n-1) \frac{7}{2}=84-7=77\)

⇒ \(n-1=77 \times \frac{2}{7}=22\)

⇒ \(n=22+1=23\)

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_{23}=\frac{23}{2}\left[2 \times 7+(23-1) \times \frac{7}{2}\right]\)

⇒ \(\frac{23}{2}[14+77]=\frac{23 \times 91}{2}\)

⇒ \(\frac{2093}{2}\)

⇒ \(1046 \frac{1}{2}\)

2. 34 + 32 + 30 + … + 10

Here a = 34, = 32- 34 = 30 – 32 = -2

Let an = 10 ⇒ a + (n – 1)d = 10

⇒ 34 + (n-1)(-2) =10

⇒ (n- 1) (-2) = 10-34

⇒ (n-1) (-2) =-24

⇒ n-1 =12

⇒ n=12+1 = 13

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{13}=\frac{13}{2}[2 \times 34+(13-1)(-2)]\)

⇒ \(\frac{13}{2}[68-24]\)

⇒ \(\frac{13}{2} \times 44=286 \quad\)

3. -5 + (_8) + (-11)… + (-230)

Here a =-5, d =-8- (-5) =-11 – (-8) =-3

Let an = -230 ⇒ a + (n-1)d = -230

⇒ -5 + (n – 1) (-3) = -230

⇒ (n- 1) (-3) =-230 + 5

⇒ (n – 1)(-3) = -225

⇒ n- 1 = 75

⇒ n = 75 + 1 = 76

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{76}=\frac{76}{2}[2 \times(-5)+(76-1)(-3)]\)

= 38(-10 – 225)

= 38 × (-235) = -8930

Question 3. In an A.P. :

  1. Given a = 5, d = 3, an = 50, find ii and Sn.
  2. Given a = 7, a13 = 35, find d and S13.
  3. Given a12 = 37, d = 3, find a and S12.
  4. Given a3 = 15, S10 = 125, findd and a10.
  5. Given d = 5, S9 = 75, find a and a9.
  6. Given a = 2, d = 8, Sn = 90, find n and an.
  7. Given a = 8, an = 62, Sn = 210, find n and d.
  8. Given an = 4, n = 2, Sn = -14, find n and a.
  9. Given a = 3, n = 8, S = 192, find d.
  10. Given l = 28, S = 144, and there are a total 9 terms. Find a

Solution:

1. a = 5,d = 3, an = 50

⇒ a + (n -1)d = 50

⇒ 5 + (n-1)3 =50

⇒  (n-1)3 = 45

⇒ a-1= 15

⇒ n = 16

Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{16}{2}[2 \times 5+(16-1) \times 3]\)

⇒ \(8(10+45)=8 \times 55=440\)

n = 16 and Sn = 440

2. a = 7

a13= 35

⇒ a + (13 – 1)d = 35

⇒ 7 + 12d = 35

⇒ 12d =35-7 = 28

⇒ \(d=\frac{28}{12}=\frac{7}{3}\)

and, from \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{13}=\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right]\)

⇒ \(\frac{13}{2}[14+28]=\frac{13}{2} \times 42=273\)

∴ \(d=\frac{7}{3} \text { and } S_{13}=273 \quad\)

3. d = 3

a12 = 37

a+(12-1)d = 37

a + 11 × 3 = 37

a = 37 – 33 = 4

and, from \( S_n=\frac{n}{2}[2 n+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times 4+(12-1) \times 3]\)

⇒ \(6[8+33]=6 \times 41=246\)

∴ \(a=4, S_{12}=246\)

4. Let the first term be a and the common difference is d.

a3 = 15 ⇒ a + (3 -1)d= 15

⇒ a+2d = 15 …(1)

and S10= 125

⇒ \(\frac{10}{2}[2 a+(10-1) d]=125\)

⇒ 5[2a + 9d] = 125

⇒ 2a + 9d = 25 …(2)

Multiply equation (1) by 2 and subtracting from equation (2),

Arithmetic Progression The First Term Of AP Be A And Common Difference D Multiply Equation 1 by 2

d = -1

Put the value of d in equation (1)

⇒ a + 2 × – 1 = 15

⇒ a-2= 15

⇒ a = 15+2= 17

a10 = a+ (10-1)d

= 17 + 9(-1) = 17-9 = 8

∴ d = -1, a10 = 8

5. d = 5

and S9 = 75

⇒ \(\frac{9}{2}[2 a+(9-1) \times 5]=75\)

⇒ \( 2 a+40=\frac{75 \times 2}{9}\)

⇒ \(a +20=\frac{75}{9} \Rightarrow a+20=\frac{25}{3}\)

⇒ \(a=\frac{25}{3}-20=\frac{25-60}{3}\)

⇒ \(a=\frac{-35}{3}\)

and \(a_9=a+8 d=\frac{-35}{3}+8 \times 5=\frac{-35}{3}+40\)

⇒ \(\frac{-35+120}{3}=\frac{85}{3}\)

∴ \(a=\frac{-35}{3} \text { and } a_9=\frac{85}{3}\)

6. a = 2, d= 8

Sn = 90

⇒ \( \frac{n}{2}[2 a+(n-1) \cdot d]=90\)

⇒ \(\frac{n}{2}[2 \times 2+(n-1) \cdot 8]=90\)

⇒ \(\frac{4n}{2}[1+(n-1) \cdot 2]=90\)

⇒ \( n(2 n-1)=\frac{90 \times 2}{4}\)

⇒ \(2 n^2-n=45\)

⇒ \(2 n^2-n-45=0\)

⇒ \(2 n^2-10 n+9 n-45=0\)

⇒ \(2 n(n-5)+9(n-5)=0\)

⇒ \((n-5)(2 n+9)=0\)

⇒ \(n-5=0 \quad \text { or } 2 n+9=0\)

⇒ \(n =5 \quad \text { or } \quad n=-\frac{9}{2}\)

⇒ \(n =-\frac{9}{2} \text { is not possible.}\)

∴ n = 5

Now, an = a + (n-1)d

a5 = 2 + (5 – 1) × 8 = 2 + 32 = 34

n = 5 and an = 34

7. a = 8, an = 62

Sn= 210

⇒ \(\frac{n}{2}\left(a+a_n\right)=210\)

⇒ \(\frac{n}{2}(8+62)=210\)

⇒ \(n=\frac{210 \times 2}{70}=6\)

⇒ \(a_n=62\)

⇒ \(a+(n-1) d=62\)

⇒ 8 + (6-1)d= 62 => 5d = 62 – 8 = 54

⇒ \(d=\frac{54}{5}\)

∴ \(n=6, d=\frac{54}{5}\)

8. an = 4, d = 2 and Sn=-14

an = 4

⇒ a + (n -1)d = 4

⇒ a + (n – 1).2 = 4

⇒ a + 2n – 2 =4

⇒ a + 2n = 6

and Sn = -14 …(1)

⇒ \( \frac{n}{2}\left(a+a_n\right)=-14\)

⇒ \(\frac{n}{2}(a+4)=-14\)

⇒ \(\frac{n}{2}(6-2 n+4)=-14 \quad \text { from eqn. (1) }\)

⇒ \(\frac{n}{2}(10-2 n)=-14\)

⇒ \(n(5-n)=-14\)

⇒ \(5 n-n^2=-14\)

⇒ \(0=n^2-5 n-14\)

⇒ \(n^2-7 n+2 n-14=0\)

⇒ \( n(n-7)+2(n-7)=0\)

⇒ \((n-7)(n+2)=0\)

⇒ \(n-7=0 \text { or } n+2=0\)

⇒ \(n=7 \text { or } \quad n=-2\)

⇒ \(n=-2 \text { is not possible.}\)

∴ n=7

From equation (1)

a + 2 × 7 = 6

a = 5

a = 6 – 14= -8

∴ n = 7 and a = -8

9. \(a=3, n=8 \text { and } S=192\)

⇒ \(S=192\)

⇒ \(\frac{n}{2}[2 a+(n-1) d]=192\)

⇒ \(\frac{8}{2}[2 \times 3+(8-1) d]=192\)

⇒ \(4(6+7 d)=192\)

⇒ \(24+28 d=192\)

⇒ \(28 d=192-24=168\)

⇒ \(d=\frac{168}{28}=6\)

∴ d=6

10. l=28, S =144, n=9

S=144

⇒ \(\frac{n}{2}(a+l)=144 \Rightarrow \frac{9}{2}(a+28)=144\)

⇒ \(a+28=\frac{144 \times 2}{9}=32\)

⇒ a=32-28=4

∴ a=4

Question 4. How many terms of the A.P.: 9, 17, 25,… must be taken to give a sum of 636?
Solution:

Given A.P: 9. 17. 25….

Here a = 9,d= 17-9 = 25-17 = 8

Let Sn=636

⇒ \(\frac{n}{2}[2 a+(n-1)] d=636\)

⇒ \(\frac{n}{2}[2 \times 9+(n-1) \cdot 8]=636\).

⇒ \(n[9+(n-1) \cdot 4]=636\)

⇒ \(n(9+4 n-4)=636\)

⇒ \(n(4 n+5)=636\)

⇒ \(4 n^2+5 n-636=0\)

⇒ \(4 n^2+53 n-48 n-636=0\)

⇒ \( n(4 n+53)-12(4 n+53)=0\)

⇒ \((4 n+53)(n-12)=0\)

⇒ \(4 n+53=0 \text { or } n-12=0\)

⇒ \(n=-\frac{53}{4} \text { or } n=12\)

but \(n=-\frac{53}{4}\) is not possible.

∴ n = 12

Therefore, the number of terms = 12

Question 5. The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.
Solution:

Here a = 5

Let the number of terms = n

⇒ \(a_n=45 \text { and } S_n=400\)

⇒ \(S_n=400 \Rightarrow \frac{n}{2}\left(a+a_n\right)=400\)

⇒ \(\frac{n}{2}(5+45)=400\)

⇒ \(n=\frac{400 \times 2}{50}=16\)

⇒ \(a_n=45\)

⇒ \(a+(n-1) d=45\)

⇒ \(5+(16-1) d=45\)

⇒ \(15 d=45-5=40\)

⇒ \(d=\frac{40}{15}=\frac{8}{3}\)

∴ \(n=16 \text { and } d=\frac{8}{3}\)

The number of terms and the common difference 16 and\(\frac{8}{3}\)

Question 6. The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?
Solution:

Given

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9

Let the number of terms = n

a= 17, d = 9

an = 350 ⇒ a + {n- l)d = 350

⇒ \( 17+(n-1) \cdot 9=350\)

⇒ \((n-1)9=350-17=333\)

⇒ \(n-1=\frac{333}{9}=37\)

⇒ \(n=37+1=38\)

Now, from \( S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{38}=\frac{38}{2}(17+350)\)

⇒ \(19 \times 367=6973\)

∴ n=38 and Sn=6973

Question 7. Find the sum of the first 22 terms of an A.P. in which d = 7 and the 22nd term is 149.
Solution:

d =7

a22=149

⇒ \(a+(22-1) d=149\)

⇒ \(a+21 \times 7=149\)

⇒ \(a+147=149\)

⇒ \(a=149-147=2\)

⇒ \(S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{22}=\frac{22}{2}\left(a+a_{22}\right)=11(2+149)\)

= 11 × 151 = 1661

The sum of the first 22 terms of an A.P. = 1661

Question 8. Find the sum of the first 51 terms of an A.P. whose second and third terms are 14 and 18 respectively.
Solution :

Given

Second and third terms of an A.P are 14 and 18 respectively

Let the first form of A.P. be a and the common difference be d.

Now, \(a_2=14 \quad \Rightarrow \quad a+d=14 \quad \ldots(1)\)

and \(a_3=18 \quad \Rightarrow a+2 d=18 \quad \ldots(2)\).

Subtracting equation (1) from (2)

Arithmetic Progression The First Term Of AP Be A And Common Difference D From Equation 1 And 2

Put the value of d in equation (l),

a+4 = 14

a = 14-4 = 10

Now, from the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{51}=\frac{51}{2}[2 \times 10+(51-1) \times 4]\)

⇒ \(\frac{51}{2}(20+200)\)

⇒ \(\frac{51}{2} \times 220=5610\)

∴ The sum of 51 terms = 5610

Question 9. If the sum of first 7 terms of an A.P is 49 and that of 1 7 terms is 289, find the sum of first n terms.
Solution:

Given

The sum of first 7 terms of an A.P is 49 and that of 1 7 terms is 289

Let the first term of A.P. be a and common difference be d.

S7 = 49

⇒ \( \frac{7}{2}[2 a+(7-1) d]=49 \Rightarrow \frac{1}{2}[2 a+6 d]=7\)

⇒ \(a+3 d=7\)

⇒ \(S_{17}=289\)

⇒ \(\frac{17}{2}[2 a+(17-1) d]=289\)

⇒ \(\frac{1}{2}[2 a+16 d]=17 \Rightarrow a+8 d=17 \ldots(2)\)

Subtracting equation (1) from (2),

Arithmetic Progression The First Term Of AP Be A And Common Difference D

d = 2

Put the value of d in equation (1),

a + 3×2=7

⇒ a + 6=7

⇒ a =7 -6=1

Now, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{n}{2}[2 \times 1+(n-1) \cdot 2]\)

⇒\(n(1+n-1)=n^2\)

∴ The sum of n terms of A.P. = n2

Question 10. Show that a1, a2, an A.P. where a„ is defined as below :

  1. an = 3 + 4n
  2. an = 9 – 5n

Also, find the sum of the first 15 terms in each case.

Solution:

the nth term of the sequence,

an=3+4n

Put n = 1\(a_1=3+4 \times 1=3+4=7\)

Put n = 2\(a_2=3+4 \times 2=3+8=11\)

Put n = 3\(a_3=3+4 \times 3=3+12=15\)

Now, \(a_2-a_1=11-7=4\)

⇒ \(a_3-a_2=15-11=4\)

⇒ \( a_2-a_1=a_3-a_2=4\)

∴ The difference between two consecutive terms of the sequence is constant.

So, the sequence is A.P.

Now, d = 4 and a = 7

A sum of first 15 terms

⇒ \(frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}[2 \times 7+(15-1) \times 4]\)

⇒ \(\frac{15}{2}[14+56]=\frac{15}{2} \times 70=525\)

2. \(a_n=9-5 n\)

Put n = 1, \(a_1=9-5 \times 1=9-5=4\)

Put n = 2, \(a_2=9-5 \times 2=9-10=-1\)

Put n = 3, \(a_3=9-5 \times 3=9-15=-6\)

Now, \(a_2-a_1=-1-4=-5\)

⇒ \(a_3-a_2=6-(-1)=-6+1=-5\)

⇒ \(a_2-a_1=a_3-a_2=-5\)

∴ The difference between two consecutive terms of the sequence is constant.

So, the sequence is A.P.

Now d =-5,a = 4

∴ The sum of the first 15 terms

⇒ \(\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}[2 \times 4+(15-1)(-5)]\)

⇒ \(\frac{15}{2}(8-70)=\frac{15}{2} \times(-62)=-465\)

Question 11. If the sum of the first n terms of an A.P. is 4n – n2, what is the first term (that is S1)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th and the nth terms.
Solution:

Here Sn = 4n – n2

Put n = 1

⇒ \(S_1=4 \times 1-(1)^2=4-1=3\)

Put n = 2

⇒ \(S_2=4 \times 2-2^2=8-4=4\)

Second term \(a_2=S_2-S_1=4-3=1\)

Put n = 3

⇒ \(S_3=4 \times 3-3^2=12-9=3\)

⇒ \(a_3=S_3-S_2=3-4=-1\)

Put n = 9

⇒ \(S_9=4 \times 9-9^2=36-81=-45\)

Put n = 10

⇒ \( S_{10}=4 \times 10-10^2=40-100=-60\)

⇒ \(a_{10}=S_{10}-S_9=(-60)-(-45)\)

= -60+45 = -15

Replace n by (n – 1)

⇒ \(S_{n-1}=4(n-1)-(n-1)^2\)

⇒ \(4 n-4-\left(n^2-2 n+1\right)\)

⇒ \(4 n-4-n^2+2 n-1=6 n-n^2-5\)

⇒ \(a_n=S_n-S_{n-1}\)

⇒ \(\left(4 n-n^2\right)-\left(6 n-n^2-5\right)\)

⇒ \(4 n-n^2-6 n+n^2+5\)

⇒ 5-2 n

Question 12. Find the sum of the first 40 positive integers divisible by 6.
Solution:

The progression formed the positive integers divisible by 6

6, 12, 1 8, 24, … to 40 terms

Here a = 6, = 6, n = 40

From the formula, \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{40}=\frac{40}{2}[2 \times 6+(40-1) \times 6]\)

= 20(12 + 234) = 20 x 246 = 4920

The sum of 40 terms = 4920

Question 13. Find the sum of the first 15 multiples of 8.
Solution:

The first 15 multiples of 8 are 8, 16, 24, …, to 15 terms

Here, a = 8, d = 16 — 8 = 24 — 16 = 8, n= 15

From the formula \( S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{15}=\frac{15}{2}[2 \times 8+(15-1) \times 8]\)

⇒ \(\frac{15}{2} \times 8[2+14]\)

= 60×16 = 960

The sum of the first 15 multiples of 8 = 960

Question 14. Find the sum of the odd numbers between 0 and 50.
Solution:

Odd numbers between 0 and 50 are 1,3,5…..49.

Here, a= 1,d =3- 1 =5-3=2

Let, an = 49

⇒ 1+(n-1).2 = 49

⇒ 1 + 2n – 2 = 49

⇒ 2n – 2 = 49

⇒ 2n = 49 + 1 = 50

⇒ n = 25

Now, from the formula \(S_n=\frac{n}{2}\left(a+a_n\right)\)

⇒ \(S_{25}=\frac{25}{2}(1+49)\)

⇒ \(\frac{25}{2} \times 50=625\)

The sum of the odd numbers between 0 and 50 is 625.

Question 15. A contract on a construction job specifies a penalty for delay of completion beyond a certain date as follows: ₹200 for the first day, ₹250 for the second day, ₹300 for the third day, etc., the penalty for each succeeding day being? 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days?
Solution:

Penalty for delay of first day = ₹200

Penalty for delay of second day = ₹250

Penalty for delay of third day = ₹300

This progression is an A.P.

Here a = 200, d = 250 – 200 = 300 – 250 = 50, n = 30

From the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \( S_{30}=\frac{30}{2}[2 \times 200+(30-1) \times 50]\)

=15(400+1450)

=15×1850=27750

The contractor will pay the penalty of ₹27750.

Question 16. A sum of ₹700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is ₹20 less than its preceding prize, find the value of each of the prizes.
Solution:

Let first prize = ₹a

Common difference d =- 20, n = 7

Given, S7 = 700

⇒ \(\frac{7}{2}[2 a+(7-1)(-20)]=700\)

⇒ \({[2 a-120]=200}\)

⇒ 2a =200+ 120

⇒ 2a =320

⇒ a = 160

∴ a2 =a + d= 160-20= 140

⇒ \(a_3=a_2+d=140-20=120\)

⇒ \(a_4=a_3+d=120-20=100\)

⇒ \(a_5=a_4+d=100-20=80\)

⇒ \(a_6=a_5+d=80-20=60\)

⇒ \(a_7=a_6+d=60-20=40\)

Prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹ 60 and ₹40.

Question 17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, would be the same as the class, in which they are studying, There are three sections of each class. How many trees will be planted by the students?
Solution:

Arithmetic Progression There Are Three Sections Of Each Class.

The sequence so formed : 3, 6, 9, … is an A.P.

Here, a = 3,d = 6- 3 = 9- 6 = 3

and n= 12

Now, from the formula \(S_n=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(S_{12}=\frac{12}{2}[2 \times 3+(12-1) \times 3]\)

= 6(6 + 33) = 6 x 39 = 234

Total trees planted = 234

Question 18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with the centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,…. What is the total length of such a spiral made up of thirteen consecutive semicircles? ( Take \(\pi=\frac{22}{7}\))

Arithmetic Progression A spiral Is Made Up Of Successive Semicircles

Solution:

The radius of first semicircle r1 = 0.5 cm.

The radius of second semicircle r2 =1.0 cm

The radius of third semicircle r3 = 1.5 cm

.
.
.
.
.
.
.

This sequence is an A.P.

Here a = 0.5 and d = 1.0 – 0.5 = 0.5, n = 13

Now, the length of the spiral is made of 1 3 consecutive semicircles.

⇒ \(\pi r_1+\pi r_2+\pi r_3+\ldots \text { to } 13 \text { terms }\)

⇒ \(\pi\left[r_1+r_2+r_3+\ldots 13\right.\)

⇒ \(\frac{22}{7} \times \frac{13}{2}[2 a+(13-1) \cdot d]\)

⇒ \(\frac{143}{7}[2 \times 0.5+12 \times 0.5]\)

⇒ \(\frac{143}{7} \times 7=143 \mathrm{~cm}\)

The total length of such a spiral made up of thirteen consecutive semicircle 143cm.

Question 19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 1 9 in the next row, 1 8 in the row next to it and so on (see figure). In how many rows are the 200 logs placed and how many logs are in the top row?

Arithmetic Progression 200 Logs Are Stacked

Solution:

No. of logs in lowest row = 20.

Starting from the bottom.

Logs in first row = 20

Logs in the second row =19

Logs in the third row = 18

This sequence is an A.P. in which

a = 20, d = 19-20 =-l

Let no. of rows = n

∴ Sn=200

⇒ \(\frac{n}{2}[2 a+(n-1) \cdot d]=200\)

⇒ \(\frac{n}{2}[2 \times 20+(n-1) \cdot(-1)]=200\)

⇒ \(n(40-n+1)=400\)

⇒ n(41-n)=400

⇒ \(41 n-n^2=400\)

⇒ \(0=n^2-41 n+400\)

⇒ \(n^2-25 n-16 n+400=0\)

⇒ \(n(n-25)-16(n-25)=0\)

⇒ \((n-25)(n-16)=0\)

⇒ \(n-25=0 \text { or } n-16=0\)

⇒ \(n=25 \text { or } n=16\)

∴ n = 25th

a25 =n + (25-1)d = 20 + 24(-l)

= -4 which is not possible.

∴ n= 16

n16 =n + (16-1)d = 20 + 1 5(-1)

= 20-15 = 5

So total rows = 1 6

and no. of logs in upper row = 5

Question 20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

Arithmetic Progression There Are Ten Potatoes In The Line

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket drops it In, and continues in the same way until all the potatoes arc in the bucket. What is the total distance the competitor has to run?
Solution:

Distance of the first potato from the first bucket = 5 m

Distance of the second potato from a bucket

= 5 + 3 = 8 m

Distance of the third potato from the bucket

= 8 + 3 = 11 cm

Once start from a bucket, pick up a potato and run back with it, drop it in the bucket.

Distance covered to drop the potatoes in the bucket.

= 2×5m,2×8m,2×11m, …

= 10 m, 16 m, 22 m,

Here a = 10, d = 16 – 10 = 22 – 16 = 6, n = 10

Distance covered to drop n potatoes in a bucket

⇒ \(\frac{n}{2}[2 a+(n-1) d]\)

Distance covered to drop 10 potatoes in a bucket

⇒ \(\frac{10}{2}[2 a+(10-1) d]\)

⇒ \(5[2 \times 10+9 \times 6]=5(20+54)\)

⇒ \(5 \times 74=370 \mathrm{~m}\)

The total distance the competitor has to run 370 meters,

Arithmetic Progression Exercise 5.4 (Optional)

Question 1. Which term of the A.P.: 121, 117, 113…..is its first negative term?

[Hint: Find n for an < 0]

Solution:

Given, A.P.: 121, 117, 113, …

Here a= 121,d= 117- 121 = 113- 117 = -4

Let \(a_n<0 \quad \Rightarrow a+(n-1) d<0\)

⇒ \(121+(n-1)(-4)<0\)

⇒ \(121-4 n+4<0\)

⇒ \(-4 n<-125\)

⇒ \(n>\frac{125}{4} \quad \Rightarrow \quad n>31 \frac{1}{4}\)

n = 32, 33, 34…

∴ The first negative term = 32nd term.

Question 2. The sum of the third and the seventh terms of an A.P. is 6 and their product is 8. Find the sum of the first sixteen terms of the A.P.
Solution:

Let the first term of A.P. be a and the common difference be d.

∴ Third term a3 = a + 2d

and Seventh term a7 = a + 6d

According to the problem,

⇒ \( a_3+a_7=6\)

⇒ \(a_3 a_7=8\)

⇒ \(a_3\left(6-a_3\right)=8 \text { [from equation }(1)]\)

⇒ \(6 a_3-a_3^2=8\)

⇒ \(0=a_3^2-6 a_3+8\)

⇒ \(a_3^2-4 a_3-2 a_3+8=0\)

⇒ \(a_3\left(a_3-4\right)-2\left(a_3-4\right)=0\)

⇒ \(\left(a_3-4\right)\left(a_3-2\right) =0\)

⇒ \(a_3-4=0 \text { or } \quad a_3-2=0\)

⇒ \(a_3=4 \text { or } \quad a_3=2\)

If a3 = 4 then from equation (1)

Now, \( a_7=6-4=2\)

a+2 d=4 ….(2)

a+6 d=2 ….(3)

On subtracting

-4d = 2

⇒ \(d=-\frac{1}{2}\)

Put the value of d in equation (2),

⇒ \(a+2\left(-\frac{1}{2}\right)=4\)

a-1=4 ⇒ a = 4+1=5

⇒ \(S_{16}=\frac{16}{2}[2 a+(16-1) d]\)

⇒ \(8\left[2 \times 5+15\left(-\frac{1}{2}\right)\right]\)

⇒ \(8\left(10-\frac{15}{2}\right)\)

⇒ \(8 \times \frac{5}{2}=20\)

If a3 = 2 then from equation (1), a7 = 4

Now, a + 2d = 2 …..(4)

a + 6d = 4 …..(5)

On subtracting

-4d =-2

d = \(\frac{1}{2}\)

Put the value of d in equation (4)

⇒ \(a+2 \times \frac{1}{2}=2 \Rightarrow a=2-1=1\)

⇒ \(S_{16}=\frac{16}{2}[2 a+(16-1) d]\)

⇒ \(8\left[2 \times 1+15 \times \frac{1}{2}\right]\)

⇒ \(8\left(2+\frac{15}{2}\right)\)

⇒ \(8 \times \frac{19}{2}=76\)

The sum of the first sixteen terms of the A.P = 76.

Question 3. A ladder has rungs 25 cm apart. (see figure). The rungs decrease uniformly in length from 45cm at the bottom to 25cm at the top. If the top and the bottom rungs are 2\(\frac{1}{2}\) m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = \(\frac{250}{25}+1\)]

Arithmetic Progression Horizontal Distance Between First And Last Rung

Solution:

Horizontal distance between first and last rung = 2\(\frac{1}{2}\) m = 250cm

and distance between two consecutive rungs

Number of rungs in ladder \(=\frac{250}{25}+1=10+1=11\)

Now, the length of the first rung a = 25 cm

Length of last rung l = 45 cm

Length of wood used in 11 rungs

⇒ \(\frac{11}{2}(a+l)=\frac{11}{2}(25+45)\)

⇒ \(11 \times 35=385 \mathrm{~cm}\)

The length of the wood required for the rungs 385cm.

Questionfrom4. The1 tohouses49. Showofa row that is numbered there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.

Hint: \(S_{x-1}=S_{49}-S_x\)

Solution:

Numbers mark on houses: 1, 2, 3, …47, 48, 49

x is a number such that Sum of the numbers before x = sum of the numbers after x

1 +2 + 3 + …+(x-1)

= (x+ 1) + (x + 2) + … + 49

⇒ \(S_{x-1}=S_{49}-S_x\)

⇒ \( \frac{x-1}{2}[1+x-1]=\frac{49}{2}[1+49]-\frac{x}{2}(1+x)\)

⇒ \(\frac{x^2-x}{2}=1225-\frac{x^2+x}{2}\)

⇒ \(\frac{x^2-x}{2}+\frac{x^2+x}{2}=1225\)

⇒ \(\frac{x^2-x+x^2+x}{2}=1225\)

⇒ \(x^2=(35)^2\)

x=35

The value of x =35.

Question 5. A small terrace at a football ground comprises of 1 5 steps each of which is 50 m long and built of solid concrete.

Each step has a rise of \(\frac{1}{4}\)m and a tread of \(\frac{1}{2}\)m (see figure). Calculate the total volume of concrete required to build the terrace.

[Hint: Volume of concrete required to build first step = \(\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{~m}^3\)]

Arithmetic Progression Volume Of Concrete Required To Build

Solution:

Given, the length of each step=50m and breadth is \(\frac{1}{2}\)m

The number of steps are 15 and the height of each step from the ground from an A.P. is as follows:

⇒ \(\frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}, \frac{5}{4}, \frac{6}{4}, \frac{15}{4}\)

So, the volume of concrete used in the first step

⇒ \(50 \times \frac{1}{2} \times \frac{1}{4}=\frac{50}{8} \mathrm{~m}^3\)

The volume of concrete used in the second step

⇒ \(50 \times \frac{1}{2} \times \frac{2}{4}=\frac{100}{8} \mathrm{~m}^3\)

The volume of concrete used in the third step

⇒ \(50 \times \frac{1}{2} \times \frac{3}{4}=\frac{150}{8} \mathrm{~m}^3\)

The volume of concrete used in the fourth step

⇒ \(50 \times \frac{1}{2} \times \frac{4}{4}=\frac{200}{8} \mathrm{~m}^3\)

So, the volume of total concrete

⇒ \(\frac{50}{8}+\frac{100}{8}+\frac{150}{8}+\frac{200}{8}+\ldots+\text { to } 15 \text { term }\)

⇒ \(a=\frac{50}{8}\)

⇒ \(d=\frac{100}{8}-\frac{50}{8}=\frac{50}{8} \text { and } n=15\)

Therefore, the total volume of concrete

⇒ \(V=\frac{n}{2}[2 a+(n-1) d]\)

⇒ \(\frac{15}{2}\left[2 \times \frac{50}{8}+(15-1) \frac{50}{8}\right]\)

⇒ \(\frac{15}{2} \times \frac{50}{8}[2+15-1]\)

⇒ \(\frac{15}{2} \times \frac{50}{8} \times 16=15 \times 50=750 \mathrm{~m}^3\)

Therefore, the volume of concrete used in the terrace = 750 m3

Arithmetic Progression Multiple Choice Questions

Question 1. The sum of the first 6 multiples of 3 is :

  1. 55
  2. 60
  3. 63
  4. 65

Answer: 3. 63

Question 2. The sum of 10 terms of the progression 5, 11, 17, … is :

  1. 300
  2. 320
  3. 280
  4. 240

Answer: 2. 320

Question 3. 8 times the 8th term of an A.P. is equal to 12 times the 12th term. Its 20th term is

  1. 20
  2. 0
  3. -20
  4. None of these

Answer: 2. 0

Question 4. The first two terms of an A.P. are 2 and 7. Its 18th term is :

  1. 87
  2. 92
  3. 82
  4. None of these

Answer: 1. 87

Question 5. How many terms are there in A.P. 42, 63, 84 … 210?

  1. 7
  2. 8
  3. 10
  4. 9

Answer: 4. 9

Question 6. In an A.P., d =-4,n = 7,an = 7, then the value of a is:

  1. 6
  2. 7
  3. 28
  4. 30

Answer: 3. 28

Question 7. The common difference between the two arithmetic progressions are same. If their first terms are 2 and 10 respectively, then the difference between their 5th terms is :

  1. 8
  2. 2
  3. 10
  4. 6

Answer: 1. 8

NCERT Exemplar Solutions for Class 10 Maths Chapter 4 Quadratic Equation

Quadratic Equation

A quadratic equation In the variable x Is the equation of the form ax2+bx+c=0, where a, b, c are real numbers, a ≠ 0. For example.,

  1. 3x2 + 5x -1=0
  2. 3x-x2 + 1 =0

Roots Of Quadratic Equation

A real number a is called a root of the quadratic equation ax2 + bx + c = 0, a= 0 if
2 + bα + c = 0

i.e., x = α satisfies the equation ax2 + bx + c = 0

or x = α is a solution of the equation ax2 + bx + c = 0

The roots of a quadratic equation ax2 + bx + c = 0 are called zeroes of the polynomial ax2 + bx + c.

Solution Of A Quadratic Equation By Factorisation Method

Consider the quadratic equation ax2 + bx + c = 0, a≠0

Let it be expressed as a product of two linear expressions (factors) namely (px + q) and (rx + s) where p, q, r, s are real numbers and p ≠ 0, r ≠ 0, then

ax2 + bx + c = 0 ⇒ (px + q)(rx + s) = 0

⇒ px + q = 0  or  rx + s = 0

⇒ \(x=-\frac{q}{p}\)  or  \(x=-\frac{s}{r}\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 4 Quadratic Equation

Solution Of A Quadratic Equation Solved Examples

Question 1. Which of the following are the solutions of 2x2 – 5x – 3 = 0?

  1. x = 2
  2. x = 3
  3. \(x=-\frac{1}{2}\)

Solution:

The given equation is 2x2 – 5x – 3 = 0

1. On substituting x = 2 in the given equation

L.H.S. = 2×22 – 5×2-3 = 8-10-3 = -5 ≠ R.H.S.

∴ x = 2 is not a solution of 2x2 – 5x – 3 = 0

2. On substituting x = 3 in the given equation

L.H.S. = 2×32– 5×3-3 = 18 -15- 3 = 0 = R.H.S. 2

∴ x = 3 is a solution of 2×2 – 5x – 3 = 0

3. On substituting x = \(-\frac{1}{2}\) in the given equation

L.H.S = \(2 \times\left(\frac{-1}{2}\right)^2-5 \times\left(\frac{-1}{2}\right)-3\)

⇒ \(2 \times \frac{1}{4}+5 \times \frac{1}{2}-3\)

⇒ \(\frac{1}{2}+\frac{5}{2}-3=\frac{1+5-6}{2}\)

=0 = R.H.S

∴ x = \(-\frac{1}{2}\) is a solution of 2×2-5x-3 = 0

Question 2. If x = 2 and x = 3 are roots of the equation 3x2-mx+ 2n = 0, then find the values of m and n.
Solution:

Since, x = 2 is a solution of 3x2 – mx + 2n = 0

∴ 3 × (2)2 -m×2 + 2n = 0

⇒ 12 -2m + 2n=0

⇒ -m + n = -6 ……(1)

Again, x = 3 is a solution of 3x2 – mx + 2n = 0

∴ 3 × (3)2 – m × 3 + 2n = 0

⇒ 27 – 3m + 2n = 0

⇒ -3m + 2n = -27 ……(2)

On multiplying equation (1) by 2 and subtracting from (2) we get

-m =-15 or m= 15

On substituting the value of m in equation (1) we get

-15 + n =-6

⇒ n= -6+15

⇒ n=9

Hence,The values of m = 15 and n = 9

Question 3. Solve the following quadratic equation : (3x-5)(2x + 3) = 0
Solution:

Given equation is (3x – 5)(2x + 3) = 0

⇒ 3x -5=0 or 2x + 3 = 0

⇒ 3x = 5 or 2x = -3

⇒ \(x=\frac{5}{3}\)  or  \(x=-\frac{3}{2}\)

Here, x = \(\frac{5}{3}\) and x = \(-\frac{3}{2}\) are the solutions.

Question 4. Find the roots of the following quadratic equation by factorization: 2z2 + az – a2 = 0
Solution:

Given equation is 2z2 + az – a2 =0

⇒ 2z2 + (2a – a) z -a2 = 0

⇒ 2z2 + 2az – az-a2 = 0

⇒ 2z(z + a) -a(z + a) = 0

⇒ (z + a)(2z – a) = 0

⇒ z + a = 0 or 2z-a = 0

when z + a = 0 ⇒ z = – a

and 2z – a = 0 ⇒ z = \(\frac{a}{2}\)

Hence, the roots of the equation are -a and \(\frac{a}{2}\)

Question 5. Solve the following quadratic equations by factorization:

  1. 4-11x = 3x2
  2. \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Solution:

1. Given equation is 4 -11x = 3x2

⇒ 3x2+11x-4 = 0

⇒ 3x2+(12-1)-4 = 0

⇒ 3x2+12x-x-4 = 0

⇒ 3x(x+4)(x+4)

⇒ (3x-1)(x+4)=0  or x+4 = 0

when 3x+1 = 0

⇒ \(x=\frac{1}{3}\)

and x + 4 = 0

x = -4

Hence,\(\frac{1}{3}\) and- 4 are roots of equation

2. Given equation is \(x^2-\frac{11}{4} x+\frac{15}{8}=0\)

Multiplying both sides by 8, we get

⇒ 8x2– 22x + 15 = 0

⇒ 8x2 – (12 + 10)x+ 15 = 0

⇒ 8x2– 12x- 10x+ 15 = 0

⇒ 4x(2x- 3) – 5(2x- 3) = 0

⇒ (2x-3)(4x-5) = 0

∴ either 2x- 3 = 0  or 4x- 5 = 0

2x = 3 or 4x = 5

⇒ \(x=\frac{3}{2}\) or \(x=\frac{5}{4}\)

Hence \(x=\frac{3}{2}\) and \(x=\frac{5}{4}\) are the roots of given equation.

Question 6. Solve the following quadratic equation :

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

Solution:

Given equation is

⇒ \(x^2-(1+\sqrt{2}) x+\sqrt{2}=0\)

⇒ \(x^2-x-\sqrt{2} x+\sqrt{2}=0\)

⇒ \(x(x-1)-\sqrt{2}(x-1)=0\)

⇒ \((x-1)(x-\sqrt{2})=0\)

x-1 = 0 or \(x-\sqrt{2}=0\)

when x-1 = 0 ⇒ x = 1

and \(x-\sqrt{2}=0\) ⇒ \(\sqrt{2}\)

Hence, 1 and \(\sqrt{2}\) are roots of the equation.

Question 7. Solve the following quadratic equation: a2b2x2 + b2x- a2x-1 = 0
Solution:

Given equation is

a2b2x2 + b2x- a2x-1=0

b2x(a2x + 1)-1 (a2x + 1 ) = 0

(a2x+1)(b2x- 1) = 0

a2x + 1 = 0 or b2x-1=0

when a2x+1=0 ⇒ \(x=-\frac{1}{a^2}\)

and b2x- 1=0 ⇒ \(x=\frac{1}{b^2}\)

Hence, \(-\frac{1}{a^2} \text { and } \frac{1}{b^2}\) are roots equation.

Question 8. Solve the following quadratic equation: 4x2– 2(a2 + b2)x + a2b2=0
Solution.

Given equation is

4x2 -2(a2 + b2)x + a2b2 = 0

4x2– 2a2x- 2b2x + a2b2 = 0

2x(2x- a2)-b2(2x -a2) = 0

(2x-2)(2x-b2) = 0

2x-a2 = 0

or 2x-b2 = 0

when 2x-a2 = 0 ⇒ \(x=\frac{a^2}{2}\)

and 2x-b2 = 0 ⇒ \(x=\frac{b^2}{2}\)

Question 9. Solve the following equation :

⇒ \(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3},(x \neq 4,3)\)

Solution:

⇒ \( \frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\)

⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)

⇒ \(\frac{2 x^2-6 x+2 x^2-5 x-8 x+20}{x^2-3 x-4 x+12}=\frac{25}{3}\)

⇒ \(\frac{4 x^2-19 x+20}{x^2-7 x+12}=\frac{25}{3}\)

⇒ \(3\left(4 x^2-19 x+20\right)=25\left(x^2-7 x+12\right)\)

⇒ \(12 x^2-57 x+60=25 x^2-175 x+300\)

⇒ \(25 x^2-175 x+300-12 x^2+57 x-60=0\)

⇒ \(13 x^2-118 x+240=0\)

⇒ \(13 x^2-(78+40) x+240=0\)

⇒ \(13 x^2-78 x-40 x+240=0\)

⇒ 13x(x-6)-40(x-6) = 0

⇒ (x-6)-409x-6) = 0

⇒ x-6 = 0 or 13x-40 = 0

⇒ x-6 0  or \(x=\frac{40}{13}\)

Hence, 6 and \(\frac{40}{13}\) are roots of the equation.

Question 10. Solve the following equation:

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5,(x \neq-3,1)\)

Solution:

Given equation is

⇒ \(2\left(\frac{2 x-1}{x+3}\right)-3\left(\frac{x+3}{2 x-1}\right)=5\)…..(1)

Let \(\frac{2 x-1}{x+3}=y\)

Hence, \(\frac{x+3}{2 x-1}=\frac{1}{y}\)

Now from equation (1)

⇒ \(2 y-\frac{3}{y}=5\)

⇒ \(2 y^2-3=5 y\)

⇒ \(2 y^2-5 y-3=0\)

⇒ \(2 y^2-(6-1) y-3=0\)

⇒ \(2 y^2-6 y+1 y-3=0\)

⇒ 2y(y-3)+1(y-3)=0

⇒ (2y+1)(y-3)=0

⇒ 2y+1=0 or y-3=0

when 2y + 1 = 0 ⇒ y = \(-\frac{1}{2}\)

and y-3 = 0 ⇒ y = 3

Substituting values of y in equation (2)

when y = \(-\frac{1}{2}\)

⇒ \(\frac{2 x-1}{x+3}=-\frac{1}{2}\)

⇒ 2(2x-1) = -1(x + 3)

⇒ 4x- 2 = -x- 3

⇒ 5x = -1

⇒ x = \(-\frac{1}{5}\)

when y = 3

⇒ \(\frac{2 x-1}{x+3}=3\)

⇒ 2x- 1 = 3(x + 3) ⇒ 2x- 1 = 3x+9

⇒ -x = 10        ⇒           x =-10

Hence, x = -10 or x = \(-\frac{1}{5}\) are roots of the equation.

Question 11. Solve the equation:

⇒\(\frac{a}{x-b}+\frac{b}{x-a}=2 \quad(x \neq b, a)\)

Solution:

Given equation is \(\frac{a}{x-b}+\frac{b}{x-a}=2\)

⇒ \(\frac{a}{x-b}+\frac{b}{x-a}=1+1\)

⇒ \(\frac{a}{x-b}-1+\frac{b}{x-a}-1=0\)

⇒ \(\frac{a-x+b}{x-b}+\frac{b-x+a}{x-a}=0\)

⇒ \((a+b-x)\left(\frac{1}{x-b}+\frac{1}{x-a}\right)=0\)

a+b -x= 0 or \(\frac{1}{x-b}+\frac{1}{x-a}=0\)

when a +b-x= 0 ⇒  x=a +b

and when  \(\frac{1}{x-b}+\frac{1}{x-a}=0\) ⇒ \(\frac{x-a+x-b}{(x-b)(x-a)}=0\)

⇒ 2x-a-b = 0

⇒ 2x= a + b

⇒ \(x=\frac{a+b}{2}\)

Alternatively,

⇒ \(\frac{1}{x-b}=-\frac{1}{x-a}\)

⇒ \(x-b=a-x\)

⇒ \(2 x=a+b\)

⇒ \(x=\frac{a+b}{2}\)

Hence, x = a + b and \(x=\frac{a+b}{2}\) arc roots of the equation.

Quadratic Formula

The roots of the quadratic equation ax2 + bx+ c= 0 where a ≠ 0 can be obtained by using the formula.

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

Proof: Given ax2 + bx + c = 0

Dividing each term by a, we get

⇒ \(x^2+\frac{b}{a} x+\frac{c}{a}=0\)   (∵ a ≠ 0)

⇒ \(x^2+\frac{b}{a} x=-\frac{c}{a}\)  (transposing the constant)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e. }\left(\frac{b}{2 a}\right)^2\) on both sides, we get

⇒ \(x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\left(\frac{b}{2 a}\right)^2-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}-\frac{c}{a}\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}\)

Taking square root on both sides, we get

⇒ \(x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)  (try to understand ‘±’)

⇒ \(x=\frac{-b}{2 a} \pm \frac{\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\) or ⇒ \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

This is known as “Sridharacharya Formula” or “quadratic formula”.

Discriminant

For the quadratic equation ax2 + bx + c = 0, the expression D = b2– 4ac is called the discriminant. Roots of ax2 + bx + c = 0 are real, only when b2 – 4ac > 0, otherwise they are imaginary (not real).

Sum Of Roots And Product Of Roots

We know that the two roots of a quadratic equation ax2 + bx + c = 0 are

⇒ \(\alpha=\frac{-b+\sqrt{D}}{2 a} \quad \text { and } \quad \beta=\frac{-b-\sqrt{D}}{2 a}\) where D=b2- 4acis called the discriminant.

∴ Sum of roots :

⇒ \(\alpha+\beta=\frac{-b+\sqrt{D}}{2 a}+\frac{-b-\sqrt{D}}{2 a}=\frac{-b+\sqrt{D}-b-\sqrt{D}}{2 a}=\frac{-2 b}{2 a}=\frac{-b}{a}\)

∴ Sum of roots = \(\frac{-b}{a}=-\frac{\text { Coeff. of } x}{\text { Coeff. of } x^2}\)

Product of roots :

⇒ \(\alpha \beta=\frac{-b+\sqrt{D}}{2 a} \times \frac{-b-\sqrt{D}}{2 a}\)

⇒ \(\frac{(-b)^2-(\sqrt{D})^2}{4 a^2}=\frac{b^2-D}{4 a^2}=\frac{b^2-\left(b^2-4 a c\right)}{4 a^2}=\frac{4 a c}{4 a^2}=\frac{c}{a}\)

∴ Product of roots : \(\frac{c}{a}=\frac{\text { Constant term }}{\text { Coeff. of } x^2}\)

Sum Of Roots And Product Of Roots Solved Examples

Question 1. Find the roots of the following quadratic equation, if they exist by the method of completing the square.

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Solution:

Given equation is

⇒ \(3 x^2+4 \sqrt{3} x+4=0\)

Dividing both sides by 3

⇒ \(x^2+\frac{4 \sqrt{3}}{3} x+\frac{4}{3}=0\)

⇒ \(x^2+\frac{4}{\sqrt{3}} x=\frac{-4}{3}\)

Adding \(\left(\frac{\text { coefficient of } x}{2}\right)^2 \text { i.e., }\left(\frac{2}{\sqrt{3}}\right)^2=\frac{4}{3}\) on both sides

⇒ \(x^2+\frac{4}{\sqrt{3}} x+\frac{4}{3}=-\frac{4}{3}+\frac{4}{3}\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)^2=0\)

⇒ \(\left(x+\frac{2}{\sqrt{3}}\right)=0 \quad \text { and } \quad\left(x+\frac{2}{\sqrt{3}}\right)=0\)

⇒ \(x=\frac{-2}{\sqrt{3}} \quad \text { and } \quad x=\frac{-2}{\sqrt{3}}\)

Hence, roots of the equation are \(\frac{-2}{\sqrt{3}} \text { and } \frac{-2}{\sqrt{3}}\)

Question 2. Find roots of the following quadratic equations by using the quadratic formula, if they exist.

  1. 3x2 +x-4 = 0
  2. 3x2 +x+ 4 = 0

Solution:

1. Given equation is 3x2 +x-4 = 0

On comparing with ax2 + bx+ c = 0, we get

a = 3, b = 1 and c = -4

∴ Discriminant, D = b2 – 4ac

D = (1)2 – 4 × 3 × (-4)

D = 1 +48

D = 49 > 0

Hence, the given equation has two real roots.

∴ \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{49}}{6}=\frac{-1 \pm 7}{6}=\frac{6}{6}\)

⇒ \(\text { or } \frac{-8}{6}\)

⇒ \(1 \text { or } \frac{-4}{3}\)

⇒ \(x=1, \frac{-4}{3}\) are roots of the equation.

2. Given equation is 3x2 +x+ 4 = 0

On comparing with ax2 + bx + c = 0, we get

a = 3, b = 1 and c = 4

∴ Discriminant, D – b2 – 4ac

D = (1)2 – 4 × 3 × 4

D = 1 – 48

D = -47 < 0

Hence, the equation has no real roots.

Question 3. Find roots of the equation by quadratic formula : x+x -(a + 2) (a+1) = 0
Solution:

The given equation is x2 + x- (a + 2) (a +1) = 0

Comparing it with Ax2 + Bx + C = 0, we get

A = 1, B = 1 and C = -(a + 2) (a+1)

∴ \(x=\frac{-B \pm \sqrt{B^2-4 A C}}{2 A}\)

⇒ \(x=\frac{-1 \pm \sqrt{1^2-4 \times 1 \times[-(a+2)(a+1)]}}{\cdots}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4\left(a^2+3 a+2\right)}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{1+4 a^2+12 a+8}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{4 a^2+12 a+9}}{2}\)

⇒ \(x=\frac{-1 \pm \sqrt{(2 a+3)^2}}{2}\)

⇒ \(x=\frac{-1 \pm(2 a+3)}{2}\)

⇒ \(x=\frac{-1+2 a+3}{2} \text { and } \frac{-1-2 a-3}{2}\)

⇒ \(x=\frac{2 a+2}{2} \text { and } \frac{-2 a-4}{2}\)

⇒ x = (a + 1) and -(a + 2) are roots of the equation.

Question 4. Solve the following equation by the method of completing the square :

⇒ \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Solution:

We have, \(4 \sqrt{3} x^2+5 x-2 \sqrt{3}=0\)

Dividing each term by \(4 \sqrt{3}\) we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x-\frac{2 \sqrt{3}}{4 \sqrt{3}}=0\)

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x=\frac{1}{2}\)

Adding \(\left(\frac{1}{2} \text { coefficient of } x\right)^2 \text { i.e., }\left(\frac{5}{8 \sqrt{3}}\right)^2\) to both sides, we get

⇒ \(x^2+\frac{5}{4 \sqrt{3}} x+\left(\frac{5}{8 \sqrt{3}}\right)^2=\left(\frac{5}{8 \sqrt{3}}\right)^2+\frac{1}{2}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25}{192}+\frac{1}{2} \Rightarrow\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{25+96}{192}\)

⇒ \(\left(x+\frac{5}{8 \sqrt{3}}\right)^2=\frac{121}{192}\)

Taking the square root of both sides, we get

⇒ \(x+\frac{5}{8 \sqrt{3}}= \pm \frac{11}{8 \sqrt{3}}\)

∴ \(x=-\frac{5}{8 \sqrt{3}} \pm \frac{11}{8 \sqrt{3}}=\frac{-5 \pm 11}{8 \sqrt{3}}\)

⇒ \(x=\frac{-5+11}{8 \sqrt{3}} \quad \text { or } \quad x=\frac{-5-11}{8 \sqrt{3}}\)

⇒ \(x=\frac{3}{4 \sqrt{3}} \quad \text { or } \quad x=\frac{-2}{\sqrt{3}}\)

Hence x = \(\frac{3}{4 \sqrt{3}} \text { or } x=\frac{-2}{\sqrt{3}}\) are the solutions of given equation.

Question 5. Solve: x2 + x- (a + 2) (a + 1) = 0 by

  1. Factorisation
  2. Method of completing the square

Solution:

1. By factorisation :

We have x2 +x- (a + 2) (a + 1) = 0

⇒ x2+x× 1 -{a + 2) (a+ 1) = 0

⇒ x2 +x[(a + 2) — (a + 1)] — (a + 2) (n + 1) = 0

⇒ x2 +x(a + 2) -x(a + 1) – (a + 2) (a + 1) = 0

⇒ x[x+ (a + 2)] – (a + 1)[x+ (a + 2)] = 0

⇒ [x+(a + 2)] [x-(a+ 1)] = 0

∴ either x+ (a + 2) = 0  or x- (a + 1) = 0

⇒ x = -(a + 2) or x=(a+ 1)

2. By the method of completing the square:

We have x2 + x- (a + 2) (a + 1) = 0

⇒ x2 +x=(a + 2) (a + 1)

Adding \(\left(\frac{1}{2}\right)^2\) on both sides, we get

⇒ \(x^2+x+\left(\frac{1}{2}\right)^2=a^2+3 a+2+\left(\frac{1}{2}\right)^2\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\frac{4 a^2+12 a+9}{4}\)

⇒ \(\left(x+\frac{1}{2}\right)^2=\left(\frac{2 a+3}{2}\right)^2\)

Taking the square root of both sides, we get

⇒ \(x+\frac{1}{2}= \pm \frac{2 a+3}{2}\)

∴ \(x=\frac{-1}{2} \pm \frac{2 a+3}{2}=\frac{-1 \pm(2 a+3)}{2}\)

∴ \(x=\frac{-1+2 a+3}{2}\text { or }x=\frac{-1-(2 a+3)}{2}\)

⇒ \(x=\frac{2(a+1)}{2}\text { or }x=\frac{-2(a+2)}{2}\)

⇒ \(x=(a+1)\text { or }x=-(a+2)\)

Question 6. Let f(x) = 3x2 – 5x- 1. Then solve f(x) = 0 by

  1. Factoring the quadratic
  2. Using the quadratic formula
  3. Completing the square and then rewrite f(x) in the formv4(x±B)2 ± C.

Solution:

f(x) = 3x2– 5x- 1

f(x) = 0

3x2-5x- 1=0

1. The given quadratic equation cannot be fully factorised using real integers. So, it is better to solve this equation by any other method.

2. 3x-5x- 1 =0

Compare it with ax2 +bx + c = 0, we get

a = 3, b = -5,c =-1

∴ Let two roots of this equation are

⇒ \(\alpha=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)+\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\frac{5+\sqrt{37}}{6}\)

and

⇒ \(\beta=\frac{-b-\sqrt{b^2-4 a c}}{2 a}\)

⇒ \(\frac{-(-5)-\sqrt{(-5)^2-4(3)(-1)}}{2 \times 3}\)

⇒ \(\beta=\frac{5-\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} { and } \frac{5-\sqrt{37}}{6}\)

3. \(3 x^2-5 x-1=0\)

⇒ \(x^2-\frac{5}{3} x-\frac{1}{3}=0\)

⇒ \(x^2-\frac{5}{3} x+\ldots \ldots=\frac{1}{3}+\ldots \ldots\)

⇒ \(x^2-\frac{5}{3} x+\left(\frac{5}{6}\right)^2\)

⇒ \(\frac{1}{3}+\left(\frac{5}{6}\right)^2\)

⇒ \(\left[\text { adding }\left(\frac{\text { Coeff. of } x}{2}\right)^2 \text { on both sides }\right]\)

⇒ \(\left(x-\frac{5}{6}\right)^2=\frac{1}{3}+\frac{25}{36} \Rightarrow\left(x-\frac{5}{6}\right)^2=\frac{12+25}{36}\)

∴ \(\left(x-\frac{5}{6}\right)^2=\left(\frac{\sqrt{37}}{6}\right)^2\)

⇒ \(x-\frac{5}{6}= \pm \frac{\sqrt{37}}{6}\)

∴ Two values of x are \(\frac{5+\sqrt{37}}{6} \text { and } \frac{5-\sqrt{37}}{6} \text {. }\)

Now, f(x) = \(3 x^2-5 x-1=3\left(x^2-\frac{5}{3} x\right)-1\)

⇒ \(3\left(x^2-\frac{5}{3} x+\frac{25}{36}-\frac{25}{36}\right)-1\)

⇒ \(3\left(x-\frac{5}{6}\right)^2-\frac{25}{12}-1=3\left(x-\frac{5}{6}\right)^2-\frac{37}{12}\)

which is of the form A(x- B)2– C, where A = 3, B = \(\frac{5}{6}\), C = \(\frac{37}{12}\)

Nature Of Roots Of A Quadratic Equation

The nature of roots of a quadratic equation ax2 +bx + c = 0 depends on the value of its discriminant (D). i.e., upon b2 – 4ac.

If a, b, and c are real numbers and a ≠ 0 then discriminant D = b2-4ac.

The value of discriminant affects the nature of roots in the following ways :

The roots of a quadratic equation are :

  1. Real: When D > 0 i.e. (D > 0 or D = 0) (when quadratic equation can be expressed as in real linear factors, D > 0)
  2. No Real (Imaginary): When D < 0
  3. Real and Distinct: When D > 0
  4. Real and Equal (Coincident): When D = 0

In this case each equal root will \(\left(\frac{-b}{2 a}\right)\)

Remember:

  1.  ax-b>0 ⇒ \(x>\frac{b}{a} \text {, if } a>0 \text { and } x<\frac{b}{a} \text {, if } a<0\)
  2. x2– a2 > 0 ⇒ x<-a or x>a
  3. x2– a2 = 0 ⇒ x= —a or x =  a
  4. x2– a2 < 0 ⇒ x<a or x> -a ⇒ -a< x < 0
  5. (x-a) (x-b)>0,a<b ⇒ x<a or x>b
  6. (x- a) (x- b) < 0, a < b ⇒ a<x<b

Solved Examples

Question 1. Find the value of k so that the equation 2x2 – 5x + k = 0 has two equal roots.
Solution:

The given equation is 2x2 – 5x + K = 0

Comparing with ax2 + bx + c = 0, we get

a = 2,b = -5 and c = k

The equation will have two equal roots if

D = 0

D = b2– 4ac = 0

or (-5)2 -4×2×K = 0

25 – 8K = 0

⇒ \(k=\frac{25}{8}\)

The value of k =\(\frac{25}{8}\)

Question 2. The equation 3x2 – 12x + (n – 5) = 0 has repeated roots. Find the value of n.
Solution:

Given equation is 3x2– 12x + (n – 5) = 0

Comparing with ax2 +bx + c = 0, we get

a = 3, b = -12 and c = n- 5

The equation will have repeated (two equal) roots if

Discriminant (D) = 0

∴ D =b2-4ac=0

or (-12)2– 4 × 3 × (n- 5) = 0

⇒ 144 -12n + 60 = 0

⇒ 204- 12n = 0

⇒ -12n = -204

⇒ n = 17

Hence, the value of n is 17

Question 3. Find the value of k for which the equation 2 + k(2x +k- l) + 2 = 0 has real and equal
roots.
Solution:

Given equation is

x2 + k(2x + k – 1) + 2 = 0

x2 + 2kx +k(k- 1) + 2 = 0

x2 + 2kx + (k2 -k + 2) = 0

Comparing with ax2 + bx + c = 0,we get

a = 1, b = 2k and c = k2 – k + 2

For real and equal roots,

Discriminant (D) = 0

∴ D =b2-4ac=0

or (2k)2 – 4(1)(k2 -k + 2)= 0

4k2 – 4(k2 -k + 2) = 0

k2– (k2 -k + 2) = 0

K-2 = 0

k = 2.

Hence, the value of k is 2

Question 4. Find the value of p, for which one root of the quadratic equation px2 – 14x + 8 = 0 is 6 times the other.
Solution:

Given equation is px2 – 14x + 8 = 0

Let a and p be two roots of quadratic equation px2– 14x + 8 = 0, such that β = 6a

∴ Sum of roots = \(\frac{-b}{a}\)

⇒ \(\alpha+6 \alpha=-\frac{(-14)}{p}\)

⇒ \(7 \alpha=\frac{14}{p} \Rightarrow \alpha=\frac{2}{p}\)….(1)

Product of roots = \(\frac{c}{a}\)

⇒ \(\alpha \cdot 6 \alpha=\frac{8}{p}\)

⇒ \(6 \alpha^2=\frac{8}{p} \Rightarrow \alpha^2=\frac{8}{6 p}\) ….(2)

From equations (1 ) and (2), we get

⇒ \(\left(\frac{2}{p}\right)^2=\frac{8}{6 p} \Rightarrow \frac{4}{p^2}=\frac{4}{3 p}\)

P2 = 3P

p2 – 3p = 0

p(p-3) = 0 (don’t cancel p on both sides)

Either p = 0 or p = 3.

But p = 0 is not possible, as on putting, p = 0 in the given equation, we don’t have a quadratic equation and therefore we cannot get two roots.

Hence, P = 3

Alternatively,

Let one root of the quadratic equation px2 – 14x + 8 = 0 is a.

∴ pα2 – 14α + 8 =0 ….(1)

∴ Other root of the equation will be 6a.

p(6α)2– 14(6α) + 8 =0

36pα2– 84α + 8=0

9pα2 – 21α + 2 =0 ….(2)

Solving equations (1) and (2) by cross-multiplication method.

⇒ \(\frac{\alpha^2}{-14(2)-8(-21)}=\frac{\alpha}{8(9 p)-2 p}=\frac{1}{p(-21)-9 p(-14)}\)

⇒ \(\frac{\alpha^2}{-28+168}=\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \( \frac{\alpha^2}{140}=\frac{1}{105 p}\text { and }\frac{\alpha}{70 p}=\frac{1}{105 p}\)

⇒ \(\alpha^2=\frac{140}{105 p}=\frac{4}{3 p}\text { and }\alpha=\frac{70 p}{105 p}=\frac{2}{3}\)

⇒ \(\left(\frac{2}{3}\right)^2=\frac{4}{3 p} \quad \Rightarrow \quad \frac{4}{9}=\frac{4}{3 p} \quad \Rightarrow \quad 3 p=9\)

p = 3

The value of p = 3

Question 5. The equation x2+2(m-1)x+ (m + 5) 0 has real and equal roots. Find the value of m.
Solution:

Given equation is x2+ 2 (m – 1 )x + (m+ 5) = 0

Comparing with ax2 + bx + c = 0, we get

a = 1, b = 2(m -1), c = m + 5

The equation will have two real and equal roots if

Discriminant (D) = 0

∴ D = b2 – 4ac= 0

or [2(m-1)]2-4 × 1 × (m + 5) = 0

4(m2 + 1 – 2m) – 4(m + 5) = 0

4m2 + 4- 8m – 4m – 20 = 0

4m2– 12m, – 16 = 0

m2– 3m, -4 = 0

m2 – 4m, + m, – 4 = 0

m(m, – 4) + 1 (m – 4) = 0

(m + 1) (m – 4) = 0

m = -1 or m = 4

Hence, the value(s) of m are -1 and 4.

Question 6. If -4 is a root of the equation x2 + px- 4 = 0 and the equation x2 + px + q = 0 has coincident roots, find the values of p and q.
Solution:

Since -4 is a root of x2 + px- 4 = 0

Hence, (-4) will satisfy the equation.

Therefore, (-4)2 +p(-4) -4=0

16 – 4p – 4 = 0

-4p +12 = 0

-4p = -12 .

p = 3 …….(1)

Given that, x2 + px + q = 0 has coincident roots.

D = b2 – 4ac = 0

D = p2-4×1×q=0

p2-4q = 0

32 -4q =0 [from (1)]

9 – 4q = 0

-4q = -9

q = \(\frac{9}{4}\)

Hence, the values of p = 3 and q = \(\frac{9}{4}\)

Question 7. Prove that both roots of the equation (x -a) (x- b) + (x- b) (x- c) + {x- c) (x-a) = 0 are real but they are equal only when a =b =c.
Solution:

The given equation may be written as

3x2 – 2(a + b + c)x+ (ab + bc + ac) = 0

∴ Discriminant D = B2 – 4AC

D = \(4(a+b+c)^2-4 \times 3(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2+2 a b+2 b c+2 a c\right)-12(a b+b c+a c)\)

D = \(4\left(a^2+b^2+c^2-a b-b c-a c\right)\)

D = \(2\left(2 a^2+2 b^2+2 c^2-2 a b-2 b c-2 a c\right)\)

D = \(2\left[a^2+b^2-2 t b+b^2+c^2-2 b c+c^2+a^2-2 a c\right]\)

D = \(2\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \geq 0\)

∵ \((a-b)^2 \geq 0,(b-c)^2 \geq 0 \text { and }(c-a)^2 \geq 0\)

Hence, both roots of the equation are real.

For equal root, we must have D = 0

⇒ (a – b)2 + (b- c)2 + (c- a)2 = 0

⇒ a-b =0, b-c = 0, c-a = 0

⇒ a = b,b = c,c = a

⇒ a – b =c

Hence, roots are equal only when a=b = c

Question 8. Find the positive values of k for which the equations x2 + kx + 64 = 0 and x2 – 8x + k = 0 nail both have real roots:
Solution:

Given equations are

x2 + kx + 64 = 0 …(1)

and x2-8x + k= 0 …(2)

Let D1 and D2 be discriminants of equations (1) and (2) respectively, then

D1 = k2 – 4 x 64 or  D1 = k2 – 656

and D2 = (-8)2 – 4k or D2 = 64-4K

Both equations will have real roots, if

D1 ≥ 0 and D2 ≥ 0

⇒ K2 -256 ≥ 0 and 64- 4k ≥ 0

⇒ k2 ≥ 256 and64 ≥ 4k

⇒ k ≥ 16 and K ≤ 16

k = 16

Hence, both equations will have real roots, when k = 16.

The positive values of k = 16.

Question 9. Find the value(s) of k for which the given quadratic equations have real and distinct roots:

  1. 2x2 + kx + 4 = 0
  2. 4x2-3kx+ 1=0
  3. kx2 + 6x + 1 = 0
  4. x2-kx+ 9 = 0

Solution:

1. The given equation is 2x2 + kx + 4 = 0

Comparing with ax2 +bx + c = 0, we get

a-2, b=k and c = 4

∴ D = b2– 4ac ≥ 0 for real and distinct roots.

Therefore, D = k2– 4 × 2 × (4) ≥ 0

⇒ k2 – 32 ≥ 0

⇒ k2 ≥ 32

⇒ \(k \leq-4 \sqrt{2} \text { and } k \geq 4 \sqrt{2}\)

2. The given equation is 4x2 – 3kx + 1 – 0

Comparing with ax2 + bx + c = 0, wc get

a = 4, b = -3k and c = 1

∴ D = b2 – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-3k)2 – 4 × 4 × 1 ≥ 0

9k2 -16>0 ⇒ 9k2 > 16

⇒ \(k^2 \geq \frac{16}{9}\)

⇒ \(k \leq-\frac{4}{3} \quad \text { and } \quad k \geq \frac{4}{3}\)

3.  The given equation is kx2 + 6x+1 = 0

Comparing with ax2 +bx + c = 0, we get

a = k, b = 6 and c = 1

D = b2 – 4ac > 0 for real and distinct roots

Therefore, D = (6)2– 4 × k × 1 ≥ 0

36-4K ≥ 0

36 ≥ 4K

k ≤ 9

4. The given equation is

x2 -kx + 9 = 0

Comparing with ax2 + bx + c = 0, we get

a = 1, b = -k and c = 9

∴ D = b2 – 4ac ≥ 0 for real and distinct roots

Therefore, D = (-k)2– 4 × 1 × 9 ≥ 0

⇒ K2 – 36 ≥ 0

⇒ k ≤ -6 or K ≥ 6

Question 10. If roots of the equation (1 + m2)x2 + 2mcx + (c2 – a2) = 0 are equal, prove that : c2 = a2(1+m2)
Solution:

We have,

(1 + m2)x2 + 2mcx + (c2 – a2) = 0

It has equal roots, if

D = 0

B2-4AC =0

⇒ (2mc)2 – 4(1 + m2) (c2 – a2) =0

⇒ 4m2c2 – 4(c2 – a2 + m2c2– m2a2) = 0

⇒ m2c2 -c2 + a2 – m2c2 + m2a2 = 0

⇒ c2 = a2+ m2a2 = a2(1 +m2)

Word Problems Based On Quadratic Equations

To solve the word problem, first translate the words of the problem into an algebraic equation, then solve the resulting equation.

For solving a word problem based on a quadratic equation adopt the following steps:

Step 1: Read the statement of the problem carefully.

Step 2: Represent the unknown quantity of the problem by a variable.

Step 3: Translate the given statement to form an equation in terms of variables.

Step 4: Solve the equation.

Quadratic Equations Solved Examples

Question 1. The sum of a number and its reciprocal is \(\frac{10}{3}\), find the number(s).
Solution:

Let the number be x

∴ According to a given statement

⇒ \(x+\frac{1}{x}=\frac{10}{3}\)

⇒ \(\frac{x^2+1}{x}=\frac{10}{3}\)

⇒ \(3 x^2+3=10 x\)

⇒ \(3 x^2-10 x+3=0\)

⇒ \(3 x^2-(9+1) x+3=0\)

⇒ \(3 x^2-9 x-x+3=0\)

⇒ 3x(x-3)- l(x-3) = 0

⇒ (3x-1) (x-3) = 0

when 3x- 1 = 0  \(x=\frac{1}{3}\)

and when x-3 = 0,  x = 3

Hence, the number(s) are 3 and \(\frac{1}{3}\)

Question 2. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution:

Let the larger part be x. Then, the smaller part = 16 – x

According to a given statement

⇒ 2x2-(16-x)2 = 164

⇒ 2x2 – (256 +x2 – 32x) = 164

⇒ 2x2 – 256- x2 + 32x- 164 = 0

⇒ x2 + 32x- 420 = 0

⇒ x2 + 42x- 10x- 420 = 0

⇒ x(x + 42)- 10(x + 42) = 0

⇒ (x + 42) (x- 10) = 0

⇒ x = -42 or x =10

⇒ x = 10

Hence, the required parts are 10 and 6.

Question 3. The sum of squares of three consecutive natural numbers is 149, Find the numbers.
Solution:

Given

The sum of squares of three consecutive natural numbers is 149,

Let three consecutive natural numbers be a-, (x +1) and (x +2) respectively.

According to the given condition

⇒ \(x^2+(x+1)^2+(x+2)^2=149\)

⇒ \(x^2+\left(x^2+1+2 x\right)+\left(x^2+4+4 x\right)=149\)

⇒ \(3 x^2+6 x+5=149\)

⇒ \(3 x^2+6 x-144=0\)

⇒ \(x^2+2 x-48 =0\)

⇒ \(x^2+8 x-6 x-48=0\)

⇒ x(x+8)-6(x+8)=0

⇒ (x+8)(x-6)=0

⇒ x = -8 and x = 6

⇒ x = 6 (x = -8 is not a natural number)

Hence, the required natural numbers are 6, (6 + 1), (6 + 2) = 6, 7, 8 respectively.

Question 4. A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places. Find the number.
Solution:

Given

A two-digit number is such that the product of its digits is 8. When 63 is subtracted from the number digits interchange their places.

Let the digit at the unit place be x and the digit at ten’s place be y

∴ Number = x × 1 + 10 × y

= x+ 10y.

After reversing the order of digits, the reversing number =y + 10x

According to the first condition,

xy = 8 ….(1)

According to the second condition,

⇒ (x+10y) – 63 -y + 10ar

⇒ 9y- 9x = 63

⇒ y -x = 7

⇒ y =x + 7 ….(2)

∴ from (1) and (2), we get

x (x + 7) = 8

⇒ x2 +7x- 8 = 0

⇒ (x + 8) (x- 1) = 0

∴ x= 1 or x = -8

If \(\left.\begin{array}{l}
x=1 \\
y=1+7=8
\end{array}\right\} \quad \Rightarrow \text { number }=1+10(8)=81\)

If \(\left.\begin{array}{rl}
x=-8 \\
y=-8+7=-1
\end{array}\right\}\) not possible as digits cannot be negative.

So, required number = 81

Question 5. The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\), find the fraction.
Solution:

Given

The denominator of a fraction is one more than twice the numerator. If the sum of the 16 fractions and its reciprocal is \(2 \frac{16}{21}\)

Let, the fraction be \(\frac{x}{y}\) where numerator is.v, then denominator = 2x + y

According to a given statement

⇒ \(\frac{x}{2 x+1}+\frac{2 x+1}{x}=2 \frac{16}{21}\)

Now, let \(\frac{x}{2 x+1}=a\)

⇒ \(a+\frac{1}{a}=\frac{58}{21} \Rightarrow \frac{a^2+1}{a}=\frac{58}{21}\)

⇒ \(21 a^2+21=58 a\)

⇒ \(21 a^2-58 a+21=0\)

⇒ \(21 a^2-49 a-9 a+21=0\)

⇒ \(7 a(3 a-7)-3(3 a-7)=0\)

⇒ \((7 a-3)(3 a-7)=0\)

⇒ \(a=\frac{3}{7} \quad \text { or } \quad a=\frac{7}{3}\)

when, \(a=\frac{3}{7} \Rightarrow \frac{x}{2 x+1}=\frac{3}{7}\)

⇒ 7x = 6x + 3

⇒ x = 3

when, \(a=\frac{7}{3} \Rightarrow \frac{x}{2 x+1}=\frac{7}{3}\)

⇒ 3x = 14x + 7

⇒ -11x = 7

⇒ \(\frac{-7}{11}\)

x = 3   \(\left(x=\frac{-7}{11} \text { is not a natural number }\right)\)

Hence, required fraction is \(\frac{x}{2 x+1}=\frac{3}{7} \text {.}\)

Question 6. The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse, find the sides of the triangle.
Solution:

Given

The hypotenuse of a right-angled triangle is 6 meters more than twice the shorter side. If the third side is 2 meters less than the hypotenuse

Let, the length of the shortest side be x meters

then hypotenuse = (2x + 6) metres

the third side = (2x + 6 – 2) metres

= (2x + 4) metres

Now, using Pythagoras theorem (2x+6)2=x2 + (2x + 4)2

⇒ (4x2 + 24x + 36)=x2 + (4x2+ 16x + 16)

⇒ x2 – 8x- 20 = 0

⇒ x2 – 10x+2x- 20 = 0

⇒ x(x-10)+2(x+2) = 0

⇒ (x+10)(x+2) = 0

⇒ x = 10 or x = -2

⇒ x = 10

So, the length of the shortest side = 10 meters

length of the hypotenuse = (2 × 10 + 6) = 26 metres

and length of the third side = (2 × 10 + 4) = 24 metres

Hence, the sides of the triangle are 10 m, 24 m, and 26 m.

Question 7. The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son. Find their present ages.
Solution:

Given

The age of a father is twice the square of the age of his son. Eight years hence, the age of his father will be 4 years more than 3 times the age of the son.

Let, the present age of the son be x years.

Hence, the age of father = 2×2

8 years hence, the age of son = (x + 8) years

and the age of father = (2x2 + 8) years

According to a given statement

⇒ \(2 x^2+8=3(x+8)+4\)

⇒ \(2 x^2+8=3 x+24+4\)

⇒ \(2 x^2-3 x-20=0\)

⇒ \(2 x^2-8 x+5 x-20=0\)

⇒ \(2 x(x-4)+5(x-4)=0\)

⇒ (2 x+5)(x-4)=0

⇒ \(x=\frac{-5}{2}\) and x = 4

⇒ x = 4

Therefore, the present age of the son is 4 years and present age of the father is 2 × 42 = 32 years.

Question 8. Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank, then how much time will each tap take to fill the tank?
Solution:

Given

Two taps running together can fill a tank in 3 \(3 \frac{1}{13}\) hours. If one tap takes 3 hours more 13 than the other to fill the tank

Quadratic Equations Two Taps Running Together

Let the time taken by TapI to fill a tank = x hrs

∴ The time taken by TapII to fill a tank = (x + 3) hrs

and time taken by both to fill a tank =  \(3 \frac{1}{13}=\frac{40}{13} \mathrm{hrs}\)

∴ Tap 1’s 1 hr work = \(\frac{1}{x}\)

Tap 2’s 1 hr work = \(\frac{1}{x+3}\)

and (Tap1 + Tap 2)’s 1hr work = \(\frac{13}{40}\)

⇒ \(\frac{1}{x}+\frac{1}{x+3}=\frac{13}{40}\)

⇒ \(\frac{x+3+x}{x(x+3)}=\frac{13}{40}\)

⇒ \(40(2 x+3)=13 x(x+3)\)

⇒ \(80 x+120=13 x^2+39 x\)

⇒ \(13 x^2-41 x-120=0\)

⇒ \(13 x^2-65 x+24 x-120=0\)

⇒ \(13 x(x-5)+24(x-5)=0\)

⇒ (x-5)(13 x+24)=0

∴ Either x-5 = 0 or 13x+ 24 = 0

x = 5 or \(x=\frac{-24}{13}\)

But time cannot be negative, so we reject x = \(\frac{-24}{13}\)

x = 5

Hence, time taken by Tap 1 = 5 hrs

and time taken by Tap 2 = (5 + 3) hrs = 8 hrs.

Question 9. A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes, find how much time B alone takes to complete the work.
Solution:

Given

A takes 6 hours less than B to complete a work. If together they complete the work in 13 hours 20 minutes,

Let, B alone complete the work in* hours, then A alone will complete the work in (x- 6) hours.

⇒ \(\frac{1}{x-6}+\frac{1}{x}=\frac{3}{40}\)    \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\left(13 \mathrm{hr}+20 \mathrm{~min}=\frac{40}{3} \mathrm{hrs} .\right)\)

⇒ \(\frac{x+x-6}{(x-6) x}=\frac{3}{40}\)

⇒ \(3 x^2-18 x=80 x-240\)

⇒ \(3 x^2-98 x+240=0\)

⇒ \(3 x^2-90 x-8 x+240=0\)

⇒ \(3 x(x-30)-8(x-30)=0\)

⇒ \((x-30)(3 x-8)=0\)

x = 30

⇒ \(x=\frac{8}{3}\)

x = 30   \(\left(\text { if } x=\frac{8}{3} \text {, then } x-6 \text { in negative }\right)\)

B alone will take 30 hours to complete the work.

Question 10. An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.
Solution:

Given

An airplane takes 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed.

Let the usual speed of an airplane be x km/hr. Given, distance = 1200 km

Time taken for journey of 1200 km = \(\frac{1200}{x} \text { hours }\)

⇒ \(\left(\text { Time }=\frac{\text { Distance }}{\text { Speed }}\right)\)

When the speed is increased by 100 km/hr, time taken for the same journey = \(\frac{1200}{x+100} \text { hours. }\)

According to the given condition

⇒ \(\frac{1200}{x}-\frac{1200}{x+100}=1\)

⇒ \(\frac{1200(x+100)-1200 x}{x(x+100)}=1\)

⇒ \(1200 x+120000-1200 x=x^2+100 x\)

⇒ \(x^2+100 x-120000=0\)

⇒ \(x^2+400 x-300 x-120000=0\)

⇒ \(x(x+400)-300(x+400)=0\)

⇒ \((x+400)(x-300)=0\)

⇒ x = -400 or x = 300

⇒ x = 300 km/hr

∴ The usual speed = 300 km/hr.

Question 11. A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes. Determine the speed of water.
Solution:

Given

A motor boat whose speed is 15 km/hr in still water goes 30 Ion downstream and comes back in a total of 4 hours 30 minutes.

Let, the speed of water be x km/hr.

Given the speed of a motor boat in still water is 15 km/hr.

Therefore, its speed downstream is (15 + x) km/hr and the speed upstream is (15- x) lon/hr.

Time taken for going 30 Ion downstream = \(\frac{30}{15+x} \text { hours. }\)

Time taken for going 30 Ion upstream = \(\frac{30}{15-x} \text { hours. }\)

According to the given condition

⇒ \(\frac{30}{15+x}+\frac{30}{15-x}=4+\frac{30}{60}\)

⇒ \(\frac{30(15-x)+30(15+x)}{(15+x)(15-x)}=\frac{9}{2}\)

⇒ \(\frac{450-30 x+450+30 x}{225-x^2}=\frac{9}{2}\)

⇒ \(900 \times 2=9\left(225-x^2\right)\)

⇒ \(1800=2025-9 x^2\)

⇒ \(9 x^2=225 \text { or } x^2=25\)

⇒ \(x= \pm 5\)

⇒ x=5

Hence, the speed of water is 5 Km/hr.

Question 12. A dealer sells an article for ₹24 and gains as much percent as the price of the article. Find the cost price of the article.
Solution:

Given

A dealer sells an article for ₹24 and gains as much percent as the price of the article.

Let, the C.P. of article be ₹x

Then, gain = x %

∴ \(\text { S.P. }=\frac{100+\text { gain } \%}{100} \times \text { C.P. } \quad\left(\text { or S.P. }=\text { C.P. }+ \text { C.P. } \times \frac{x}{100}\right)\)

⇒ \(24=\frac{100+x}{100} \times x\)

⇒ \(2400=100 x+x^2\)

⇒ \(x^2+100 x-2400=0\)

⇒ \(x^2+120 x-20 x-2400=0\)

⇒ \(x(x+120)-20(x+120)=0\)

⇒ \((x-20)(x+120)=0\)

⇒ x = 20 or x = -120

⇒ x = 20

Hence, the cost of the article is ₹20.

Question 13. A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per meter?
Solution:

Given

A piece of cloth costs ₹200. If the piece was 5 m longer and each meter of cloth cost 2 less, the cost of the piece would have remained unchanged.

Let, the length of the piece be A meters

Since the cost of A meters of cloth = ₹200

⇒ Cost of each metre of cloth = \(₹ \frac{200}{x}\)

New length of cloth = (x + 5)m

New cost of each metre of cloth = \(₹ \frac{200}{x+5}\)

Now, given \(\frac{200}{x}-\frac{200}{x+5}=2\)

⇒ \(\frac{200(x+5)-200 x}{x(x+5)}=2 \quad \Rightarrow \quad \frac{200 x+1000-200 x}{x^2+5 x}=2\)

⇒ \(1000=2 x^2+10 x\)

⇒ \(2 x^2+10 x-1000=0\)

⇒ \(x^2+5 x-500=0\)

⇒ \(x^2+25 x-20 x-500=0\)

⇒ \(x(x+25)-20(x+25)=0\)

⇒ \((x-20)(x+25)=0\)

x = 20 or x = -25

Since length cannot be negative

Hence, x = 20 m

and the original rate per metre = \(₹ \frac{200}{20}=₹ 10\)

Question 14. Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10. Find how many students went for a picnic.
Solution:

Given

Some students planned a picnic. The budget for the food was ₹480. As eight of them failed to join the picnic, the cost of the food for each member increased by ₹10.

Let, no. of students who planned the picnic = x

Given, the budget for food = ₹480

∴ Share of each student = \(₹ \frac{480}{x}\)

Since eight of them failed to join the picnic

∴ No. of students went for picnic = (x- 8)

∴ Share of each student = \(₹ \frac{480}{x-8}\)

According to the given condition

⇒ \(\frac{480}{x-8}-\frac{480}{x}=10\)

⇒ \(\frac{480 x-480(x-8)}{(x-8) x}=10\)

⇒ \(\frac{480 x-480 x+3840}{x^2-8 x}=10\)

⇒ \(10 x^2-80 x=3840\)

⇒ \(x^2-8 x-384=0\)

⇒ \(x^2-24 x+16 x-384=0\)

⇒ \(x(x-24)+16(x-24)=0\)

⇒ (x- 24) (x + 16) =0 ⇒  x = 24    or  x = -16

Since no. of students cannot be negative

Hence, x = 24

No. of students who went for picnic = x- 8 = 24-8=16 students.

Quadratic Equation Exercise 4.1

Question 1. Check whether the following are quadratic equations:

  1. \((x+1)^2=2(x-3)\)
  2. \(x^2-2 x=(-2)(3-x)\)
  3. (x-2)(x+1) =(x-1)(x+3)
  4. \((x-3)(2 x+1)=x(x+5)\)
  5. \((2 x-1)(x-3)=(x+5)(x-1)\)
  6. \(x^2+3 x+1=(x-2)^2\)
  7. \((x+2)^3=2 x\left(x^2-1\right)\)
  8. \(x^3-4 x^2-x+1=(x-2)^3\)

Solution:

1. \((x+1)^2=2(x-3)\)

⇒ x2 + 2x + 1 = 2x- 6

⇒ x2 + 7 =0

The highest power of the variable x in it is 2.

∴ The given equation is a quadratic equation.

2. \(x^2-2 x=(-2)(3-x)\)

x2 -2x =- 6 + 2x

x2-2x-2x+6 =0

x2 -4x + 6 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

3. (x-2)(x+1) =(x-1)(x+3)

⇒ x(x + 1 ) -2 (x + 1 ) = x(x + 3) – 1 (x + 3)

⇒ x2 +x-2x- 2 =x2 + 3x-x-3

⇒ x2 – x- 2 =x2 + 2x- 3

⇒ x2 +x-2- x2 +3 = 0

⇒ -3x+1 = 0

The highest power of the variable x is not 2, in it.

∴ Given equation is not a quadratic equation.

4. (x-3) (2x+ 1) = x(x + 5)

x(2x+1)-3 (2x + 1) =x2 + 5X

2x2 + x- 6x- 3 – x2 – 5x =0

x2– 10x – 3 =0

The highest power of the variable x in it is 2.

∴ Given equation is a quadratic equation.

5. (2x-1)(x-3)=(x +5)(x-1)

2x(x-3)-1(x-3) = x(x-1) + 5(x- 1)

2x2 – 6x-x + 3 = x2 -x+ 5x- 5

2x2 – 7x + 3 = x2 + 4x- 5

2x2– 7x + 3 -x2 – 4x + 5 = 0

x2– 11x+ 8 = 0

The highest power of variable x is 2 in it.

∴ Given equation is a quadratic equation.

6. x2 + 3x + 1 = (x- 2)2

x2 + 3x + 1 =x2 – 4x + 4

x2 + 3x+ 1 -x2 + 4x- 4 = 0

7x – 3 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is not a quadratic equation.

7. (x + 2)3 = 2x(x2 – 1 )

⇒ x3 + 3.x.2 .(x + 2) + 23 = 2x3 – 2x

⇒ x3+ 6x2 + 12x + 8 – 2x3 + 2x =0

⇒ -x3+6x2+14x+8 = 0

The highest power of the variable x is not 2 in it.

∴ Given equation is a quadratic equation.

8. \(x^3-4 x^2-x+1=(x-2)^3\)

⇒ \(x^3-4 x^2-x+1=x^3-3 x \cdot 2(x-2)-2^3\)

⇒ \(x^3-4 x^2-x+1=x^3-6 x^2+12 x-8\)

⇒ \(x^3-4 x^2-x+1-x^3+6 x^2-12 x+8=0\)

⇒ \(2 x^2-13 x+9=0\)

Question 2. Represent the following situations in the form of quadratic equations :

  1. The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of tire plot.
  2. The product of two consecutive positive integers is 306. We need to find the integers.
  3. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
  4. A train travels a distance of 480 1cm at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.

Solution:

1. Let the breadth of the plot =x meter

∴ Length of plot = (2x + 1) metre

Now, from length x breadth = area

(2x + 1) × x =528

⇒ 2x2 + x =528

⇒ 2x2 + x- 528 =0

which is the required quadratic equation.

2. Let two consecutive positive integers are x and x+ 1

∵ The product of two consecutive positive integers = 306

∴ x(x+1) = 360

⇒ x2+x = 306

⇒ x2+x-306 = 0

which is the required quadratic equation.

3. Let the present age of Rohan = x years

∴ Present age of Rohan’s mother

= (x + 26), years

After 3 years,

Rohan’s age = (x + 3) years

The age of Rohan’s mother = (x + 26 + 3) years

= (x + 29) years

According to the problem,

After 3 years, the product of their ages = 360

(x + 3) (x + 29) = 360

x(x + 29) + 3(x + 29) = 360

x2 + 29x + 3x + 87 = 360

x2 + 32x + 87 – 360 = 0

x2 + 32x- 273 = 0

which is the required quadratic equation.

4. Let the speed of train = A km/hr

Distance = 480 1cm

∴ Time taken to cover a distance of 480 km

⇒ \(\frac{480}{x} \mathrm{hrs}\)

If, the speed of the train = (x- 8) Km/hr  then the time is taken to cover 480 Km distance

⇒ \(\frac{480}{x-8} \mathrm{hrs}\)

According to the problem,

⇒ \(\frac{480}{x-8}-\frac{480}{x}=3\)

⇒ \(\frac{480 x-480(x-8)}{x(x-8)}=3\)

⇒ 480x- 480x + 3840 = 3x(x- 8)

⇒ 3840 = 3(x2 – 8x)

⇒ 1280 =x2– 8x

⇒ 0= x2 -8x -1280

⇒ x2 -8x- 1280 = 0

which is the required quadratic equation.

Quadratic Equation Exercise 4.2

Question 1. Find the roots of the following quadratic equations by factorisation:

  1. x2-3x-10 = 0
  2. 2x2+x-6 = 0
  3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0\)
  4. \(2 x^2-x+\frac{1}{8}=0\)
  5. 100x2– 20x + 1=0

Solution:

1. x2-3x-10 = 0

⇒ x2-5 x+2 x-10=0

⇒ x(x-5)+2(x-5)=0

⇒ (x-5)(x+2)=0

⇒ 5=0 or x+2=0

⇒ x=5 or x=-2

∴ Roots of given quadratic equation = 5, -2

2. 2 x^2+x-6=0

⇒ 2 x^2+4 x-3 x-6=0

⇒ 2 x(x+2)-3(x+2)=0

⇒ (x+2)(2 x-3)=0

⇒ x+2 =0 or 2x-3 = 0

⇒ x = -2 or 2x = 3

⇒ x =-2 or x = \(\frac{3}{2}\)

∴ Roots of given quadratic equation = -2, \(\frac{3}{2}\)

3. \(\sqrt{2} x^2+7 x+5 \sqrt{2}=0 \)

⇒ \(\sqrt{2} x^2+5 x+2 x+5 \sqrt{2}=0\)

⇒ \((\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5)=0\)

⇒ \((\sqrt{2} x+5)(x+\sqrt{2})=0\)

⇒ \(\sqrt{2} x+5=0 \quad \text { or } x+\sqrt{2}=0\)

⇒ \(\sqrt{2} x=-5 \text { or }x=-\sqrt{2}\)

⇒ \(x=\frac{-5}{\sqrt{2}} \text { or }x=-\sqrt{2}\)

∴ Roots of given quadratic equation

⇒ \(\frac{-5}{\sqrt{2}},-\sqrt{2} \text {. }\)

4. \(2 x^2-x+\frac{1}{8}=0\)

⇒ \(16 x^2-8 x+1=0\)

⇒ \(16 x^2-4 x-4 x+1=0\)

⇒ \(4 x(4 x-1)-1(4 x-1)=0\)

⇒ (4x-1)(4 x-1)=0

⇒ 4x-1 = 0  4x- 1 = 0

⇒ 4x= 1 or 4x= 1

⇒ \(x=\frac{1}{4} \quad \text { or } \quad x=\frac{1}{4}\)

∴ Roots of given quadratic equation = \(\frac{1}{4}, \frac{1}{4} \text {.}\)

5. \(100 x^2-20 x+1=0\)

⇒ \(100 x^2-10 x-10 x+1=0\)

⇒ 10x(10 x-1)-1(10 x-1)=0

⇒ (10x-1)(10 x-1)=0

⇒ 10x-1=0 or 10 x-1=0

⇒ 10x=1 or 10 x=1

⇒ \(x=\frac{1}{10} \text { or } \quad x=\frac{1}{10}\)

∴ Roots of given quadratic equation \(\frac{1}{10}, \frac{1}{10}\)

Question 2. Represent the following situations mathematically:

  1. John and Jiwanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
  2. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was 750. We would like to find out the number of toys produced on that day.

Solution : 

1. Let the number of marbles initially with John =x

∴ Initially, the number of marbles with Jiwanti = 45-x

After losing 5 marbles each,

Remaining marbles with John =x- 5

Remaining marbles with Jiwanti = 45 -x- 5

= 40-x

According to the problem,

Product of remaining marbles with them =124

(x- 5) (40-x) = 124

⇒ x(40-x) -5(40-x) = 124

⇒ 40x-x2– 200 + 5x= 124

⇒ -x2 + 45x- 200 = 124

⇒ 0= 124 +x2– 45x + 200

⇒ x2 – 45x + 324 = 0

⇒ x2– 36x -9x + 324 = 0

⇒ x(x- 36) -9(x- 36) = 0

⇒ (x-36) (x-9) = 0

⇒ x-36 = 0 or x-9 = 0

⇒ x = 36 or x = 9

If x = 36 then 45-x = 45-36 = 9

If x = 9 then 45-x = 45-9 = 36

∴ Marbles with John = 36

and marbles with Jiwanti = 9

or

Marbles with John = 9

and marbles with Jiwanti = 36

2. Let the number of toys = x

Cost of each toy =  ₹(55 -x)

Cost of x toys = ₹(55 – x) x-

₹(55x-x2)

According to the problem,

⇒ 55x -x2 = 750

⇒ 0 =x2-55X+750

⇒ x2 – 30x- 25x + 750 =0

⇒ x(x- 30) -25 (x-30) =0

⇒ (x-30) (x-25) =0

⇒ x-30 = 0 or x- 25 = 0

⇒ x = 30 or x = 25

∴ Number of toys produced = 25 or 30.

Question 3. Find two numbers whose sum is 27 and whose product is 182.
Solution:

Let one number = x

∴ Second number =27 -x

According to the problem,

Product of two numbers = 182

⇒ x (27-x) = 182

⇒ 27x -x2 = 182

⇒ 0 =x2 – 27x+ 182

⇒ x2– 13x- 14x+ 182 =0

⇒ x(x- 13) -14 (x- 13) =0

⇒ (x- 13) (x- 14) =0

⇒ x-13 = 0 or x-14 =0

⇒ x= 13 or x = 14

If x = 13, then 27-x = 27-13 = 14

If x = 14, then 27 -14 = 13

Therefore,numbers=(13 and 14)or(14and 13).

Question 4. Find two consecutive positive integers, a sum of whose squares is 365.
Solution:

Given

The sum of whose squares is 365

Let two consecutive positive integers be x and x + 1.

According to the problem,

x2 + (x+1)2 =365

x2 + x2 + 2x + 1 =365

2x2 + 2x + 1 – 365 = 0

2x2 + 2x- 364 = 0

x2+x- 182 =0

x2+14x-13x-182 = 0

=x (x + 14) (x- 13) =0

x + 14=0 or x-13=0

x = -14 or x = 13

x is a positive integer,

∴ neglecting x = -14,

x = 13

x+1 = 13+1 = 14

Therefore, required positive integers = 13 and 14.

Question 5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Solution:

Given

The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm

Let the base of the right triangle = x cm

∴ Its height = (x- 7) cm

Given, the hypotenuse of the right triangle = 13cm

From Pythagoras theorem, in a right triangle (base)2 + (height)2 = (hypotenuse)2

⇒ x2 + (x- 7)2 = 132

⇒ x2+x2– 14x + 49 = 169

⇒ 2x2 – 14x + 49 — 169 =0

⇒ 2x2 -14x- 120 =0

⇒ x2 – 7x- 60 =0

⇒ x2– 12x + 5x -60 =0

⇒ x(x- 12) + 5(x- 12) =0

⇒ (x- 12) (x+ 5) =0

⇒ x-12 =0 or x + 5= 0

⇒ x = 12 or x = -5

but the value of x cannot be negative.

∴ Neglecting

x = -5

x = 12

⇒ x-7 =12-7 = 5

Therefore, the other two sides of the triangle =12 cm and 5 cm.

Question 6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90, find the number of articles produced and the cost of each article.
Solution:

Given

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was ₹90,

Let the number of pottery articles produced in a day = x

∴ Cost of each article = ₹(2x+ 3)

⇒ Cost of x articles = ₹(2x + 3)x

According to the problem,

(2x + 3) x = 90

⇒ 2x2 + 3x = 90

⇒ 2x2 + 3x-90 =0

⇒ 2x2 + 15x- 12x- 90 = 0

⇒ x(2x+ 15) -6(2x+ 15) =0

⇒ (2x+ 1 5) (x -6) =0

⇒ 2x + 15 = 0 or x -6 = 0

⇒ \(x=-\frac{15}{2} \text { or } \quad x=6\)

but x = \(-\frac{15}{2}\) is not possible

∴ x = 6

⇒ 2x + 3 = 2×6 + 3=15

Therefore, the number of pottery articles in a day = 6, and the cost of each article =₹15.

Quadratic Equation Exercise 4.3

Question 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square :

  1. 2x2 – 7x + 3 = 0
  2. 2x2 + x- 4 = 0
  3. \(4 x^2+4 \sqrt{3} x+3=0\)
  4. 2x2 +x + 4 = 0

Solution:

1. 2x2 – 7x + 3 = 0

⇒ \(x^2-\frac{7}{2} x+\frac{3}{2}=0\)

⇒ \( {\left[x^2-2 x \cdot \frac{7}{4}+\left(\frac{7}{4}\right)^2\right]+\frac{3}{2}-\left(\frac{7}{4}\right)^2 }=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2+\frac{24-49}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\frac{25}{16}=0\)

⇒ \(\left(x-\frac{7}{4}\right)^2-\left(\frac{5}{4}\right)^2=0\)

⇒ \(\left(x-\frac{7}{4}-\frac{5}{4}\right)\left(x-\frac{7}{4}+\frac{5}{4}\right)=0\)

⇒ \((x-3)\left(x-\frac{1}{2}\right)=0\)

⇒ \(x-3=0\text { or }x-\frac{1}{2}=0\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of given equation = \(3, \frac{1}{2}\)

2. \(2 x^2+x-4=0\)

⇒ \(x^2+\frac{1}{2} x-2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2-2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{32+1}{16}\right)=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\frac{33}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2-\left(\frac{\sqrt{33}}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}-\frac{\sqrt{33}}{4}\right)\left(x+\frac{1}{4}+\frac{\sqrt{33}}{4}\right)=0\)

⇒ \(x+\frac{1}{4}-\frac{\sqrt{33}}{4}=0 \text { or } x+\frac{1}{4}+\frac{\sqrt{33}}{4}=0\)

⇒ \(x=\frac{\sqrt{33}}{4}-\frac{1}{4} \quad \text { or } \quad x=-\frac{\sqrt{33}}{4}-\frac{1}{4}\)

⇒ \(x=\frac{\sqrt{33}-1}{4} \quad \text { or } \quad x=-\left(\frac{\sqrt{33}+1}{4}\right)\)

∴ Roots of a given equation

⇒ \(\frac{\sqrt{33}-1}{4},-\left(\frac{\sqrt{33}+1}{4}\right)\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

⇒ \(x^2+\sqrt{3} x+\frac{3}{4}=0\)

⇒ \(x^2+2 x \cdot \frac{\sqrt{3}}{2}+\left(\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(\left(x+\frac{\sqrt{3}}{2}\right)^2=0\)

⇒ \(x+\frac{\sqrt{3}}{2}=0 \text { or }x+\frac{\sqrt{3}}{2}=0\)

⇒ \(x=-\frac{\sqrt{3}}{2}\text { or }x=-\frac{\sqrt{3}}{2}\)

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. \(2 x^2+x+4=0\)

⇒ \(x^2+\frac{1}{2} x+2=0\)

⇒ \(x^2+2 x \cdot \frac{1}{4}+\left(\frac{1}{4}\right)^2+2-\left(\frac{1}{4}\right)^2=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2+\frac{32-1}{16}=0\)

⇒ \(\left(x+\frac{1}{4}\right)^2=\frac{-31}{16}\)

⇒ \(x+\frac{1}{4}=\sqrt{-\frac{31}{16}}\) which is an imaginary number.

∴ Roots of given equation does not exist.

Question 2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Solution:

1. 2x2 – 7x + 3 = 0

On comparing with ax2 +bx + c,

a = 2,b=-7,c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-7) \pm \sqrt{(-7)^2-4(2)(3)}}{2(2)}\)

⇒ \(\frac{7 \pm \sqrt{49-24}}{4}=\frac{7 \pm \sqrt{25}}{4}=\frac{7 \pm 5}{4}\)

⇒ \(x=\frac{7+5}{4}\text { or }x=\frac{7-5}{4}\)

⇒ \(x=3\text { or }x=\frac{1}{2}\)

∴ Roots of equation = \(3, \frac{1}{2}\)

2. 2x2 +x- 4 = 0

On comparing with ax2 + bx + c = 0

a = 2,b = 1,c = -4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(-4)}}{2 \times 2}=\frac{-1 \pm \sqrt{1+32}}{4}\)

⇒ \(\frac{-1 \pm \sqrt{33}}{4}\)

⇒ \(x=\frac{-1+\sqrt{33}}{4} \text { or } \frac{-1-\sqrt{33}}{4}\)

∴ Roots of equation = \(\frac{-1+\sqrt{33}}{4}, \frac{-1-\sqrt{33}}{4}\)

3. \(4 x^2+4 \sqrt{3} x+3=0\)

On comparing with ax2 +bx + c = 0

a = 4, b = \(4 \sqrt{3}\), c = 3

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-4 \sqrt{3} \pm \sqrt{(4 \sqrt{3})^2-4 \times 4 \times 3}}{2 \times 4}\)

⇒ \(\frac{-4 \sqrt{3} \pm \sqrt{48-48}}{8}=\frac{-4 \sqrt{3} \pm \sqrt{0}}{8}\)

⇒ \(-\frac{4 \sqrt{3}}{8}=-\frac{\sqrt{3}}{2}\)

Number of roots of a quadratic equation = 2

∴ Roots of given equation = \(-\frac{\sqrt{3}}{2},-\frac{\sqrt{3}}{2}\)

4. 2x2 + x + 4 = 0

On comparing with ax2, + bx + c = 0

a = 2,b=1,c = 4

Now, \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-1 \pm \sqrt{(1)^2-4(2)(4)}}{2 \times 2}\)

⇒ \(\frac{-1 \pm \sqrt{1-32}}{4}=\frac{-1 \pm \sqrt{-31}}{4}\)

⇒ \(\sqrt{-31}\)  is an imaginary number,

∴ the values for are imaginary.

So, the real roots of the given equation do not exist.

Question 3. Find the roots of the following equations:

  1. \(x-\frac{1}{x}=3, x \neq 0\)
  2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

Solution:

1. \(x-\frac{1}{x}=3, x \neq 0\)

⇒ \(\frac{x^2-1}{x}=3\)

⇒ \(x^2-1=3 x \Rightarrow x^2-3 x-1=0\)

On comparing with ax2 + bx + c = 0

a= 1, b = -3, c =-1

x = \(\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\)

⇒ \(x=\frac{-(-3) \pm \sqrt{(-3)^2-4(1)(-1)}}{2(1)}=\frac{3 \pm \sqrt{9+4}}{2}\)

⇒ \(\frac{3 \pm \sqrt{13}}{2}\)

⇒ \(x=\frac{3+\sqrt{13}}{2} \text { or } \quad x=\frac{3-\sqrt{13}}{2}\)

Therefore, the roots of given equations

⇒ \(\frac{3+\sqrt{13}}{2}, \frac{3-\sqrt{13}}{2} \text {. }\)

2. \(\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}, x \neq-4,7\)

⇒ \(\frac{(x-7)-(x+4)}{(x+4)(x-7)}=\frac{11}{30}\)

⇒ \(\frac{x-7-x-4}{x(x-7)+4(x-7)}=\frac{11}{30}\)

⇒ \(\frac{-11}{x^2-7 x+4 x-28}=\frac{11}{30}\)

⇒ \(\frac{-1}{x^2-3 x-28}=\frac{1}{30}\)

⇒ \(x^2-3 x-28=-30\)

⇒ \(x^2-3 x-28+30=0\)

⇒ \(x^2-3 x+2=0\)

⇒ \(x^2-2 x-x+2=0\)

⇒ x(x-2)-1(x-2)=0

⇒ (x-2)(x-1)=0

⇒ x-2 = 0 or x-1 = 0

⇒ x = 2 or x = 1

∴ Roots of given equation = 2,1

Question 4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\). Find his present age.
Solution:

Given

The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is \(\frac{1}{3}\).

Let the present age of Rehman = x years

3 years before, Rehman’s age = (x- 3) years

After 5 years, Rehman’s age = (x + 5) years

According to the problem, \(\frac{1}{x-3}+\frac{1}{x+5}=\frac{1}{3}\)

⇒ \(\frac{(x+5)+(x-3)}{(x-3)(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x(x+5)-3(x+5)}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+5 x-3 x-15}=\frac{1}{3}\)

⇒ \(\frac{2 x+2}{x^2+2 x-15}=\frac{1}{3}\)

⇒ \(x^2+2 x-15=6 x+6\)

⇒ \(x^2+2 x-15-6 x-6=0\)

⇒ \(x^2-4 x-21=0\)

⇒ \(x^2-7 x+3 x-21=0\)

⇒ x(x-7)+3(x-7)=0

⇒ (x-7)(x+3)=0

⇒ x-7=0 or  x+3=0

⇒ x=7 or  x=-3

but the age cannot be negative.

∴ x = 7

⇒ Rehman’s age = 7 years.

Question 5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:

Given

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210.

Let marks obtained by Shefali in Mathematics =x

∴ Her marks in English = (30- x)

If, marks in Mathematics = x + 2

marks in English =30-x-3 = 27-x

then (x + 2) (27 -x) =210

⇒ x(27 -x) + 2(27 -x) =210

⇒ 27x -x2 + 54 – 2x = 210

⇒ 25x-x2 + 54 =210

⇒ 0 = x2– 25x- 54 + 210

⇒ x2 – 25x + 156 =0

⇒ x2– 13x- 12x+ 156 =0

⇒ x(x -13) – 12(x – 13) =0

⇒ (x- 13) (x- 12) =0

⇒ x -13 = 0 or x-12=0

⇒ x -13 or x-12

If x = 13 then 30-x = 30-13 = 17

If x = 12 then 30-x = 30-12 = 18

So, for Shefali

Maries in Mathematics = 13

and marks in English = 17

or

Maries in Mathematics = 12

and marks in English = 18

Question 6. The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the field.
Solution:

Given

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side,

Let the smaller side of the rectangular field =xmetre = breadth

Larger side = (x + 30) metre = length

diagonal = (x + 60) metre

Now, (length) + (breadth) = (diagonal)

⇒ x2 + (x+ 30)2 = (x+60)2

⇒ x2 +x2 + 60x + 900 =x2 + 120x + 3600

⇒ x2 +x2 + 60x + 900 -x2 – 120x- 3600 = 0

⇒ x2– 60x -2700 = 0

⇒ x2– 90x + 3x -2700 =0

⇒ x(x- 90) + 30(x- 90) = 0

⇒ (x-90) (x + 30) =0

⇒ x-90 =0 or x+30 = 0

⇒ x = 90 or x = -30

but the side cannot be negative.

⇒ x = 90

⇒ x+30 = 90+30 = 120

Therefore, sides of the rectangular field = 120 m and 90 m.

Question 7. The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:

Given

The difference of the squares of two numbers is 1 80. The square of the smaller number is 8 times the larger number.

Let smaller number =x

∴ Larger number × 8 = x2

⇒ Larger number = \(\frac{x^2}{8}\)

According to the problem, (larger number)2 – (smaller number)2 =180

⇒ \(\left(\frac{x^2}{8}\right)^2-x^2=180\)

⇒ \(\left(\frac{y}{8}\right)^2-y=180\)

⇒ \(\text { where } x^2=y \text { (say) }\)

⇒ \(\frac{y^2}{64}-y=180\)

⇒ \(y^2-64 y=11\)

⇒ \(y^2-64 y-11520=0\)

⇒ \(y^2-144 y+80 y-11520=0\)

⇒ \(y(y-144)+80(y-144)=0\)

⇒ \((y-144)(y+80)=0\)

⇒ \(y-144=0 \text { or } y+80=0\)

⇒ \(y=144 \text { or } y=-80\)

⇒ \(x^2=144 \text { or } x^2=-80\)

x2 =- 80 is not possible.

∴ \(x^2=144\)

⇒ \(x= \pm 12\)

⇒ \(x=12 \quad \text { or } \quad x=-12\)

⇒ \(\frac{x^2}{8}=\frac{144}{8}=18\)

Therefore, a number are 12, 18 or -12, 18.

Question 8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:

Given

A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey.

Let the speed of the train =x km/hr

∴ Time taken to cover 360 km distance = \(\frac{360}{x} \mathrm{~km}\)

If the speed of train = (x+ 5) km/hr

then time taken to cover 360 km distance = \(\frac{360}{x+5} \mathrm{hr}\)

According to the problem, \(\frac{360}{x}-\frac{360}{x+5}=1\)

⇒ \(\frac{360(x+5)-360 x}{x(x+5)}=1\)

⇒ \(\frac{360 x+1800-360 x}{x^2+5 x}=1\)

⇒ \(\frac{1800}{x^2+5 x}=1\)

⇒ \(x^2+5 x=1800\)

⇒ \(x^2+5 x-1800=0\)

⇒ \(x^2+45 x-40 x-1800=0\)

⇒ x(x+45)-40(x+45)=0

⇒ (x+45)(x-40)=0

⇒ x+45=0  or x-40=0

⇒ x=-45  or  x=40

but speed cannot be negative.

∴ Speed of train = 40 km/hr.

Question 9. Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:

Given

Two water taps together can fill a tank in \(9 \frac{3}{8}\)hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately.

Let the time taken to fill the tank by the tap of smaller diameter =x hours

Time taken to fill the tank by the tap of larger diameter = (x- 10) hours

Now, work done by tap of smaller diameter in 1 hour = \(\frac{1}{x}\)

and work done by tap of larger diameter in 1hour = \(\frac{1}{x}+\frac{1}{x-10}\)

Work done by two taps in 1 hour = \(\frac{1}{x}+\frac{1}{x-10}\)

According to the problem, both taps fill the Time taken by the passenger train to cover 132 1cm tank in \(9 \frac{3}{8}=\frac{75}{8} \)

∴ \(\left(\frac{1}{x}+\frac{1}{x-10}\right) \times \frac{75}{8}=1\)

⇒ \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\)

⇒ \(\frac{x-10+x}{x(x-10)}=\frac{8}{75}\)

⇒ \(\frac{2 x-10}{x^2-10 x}=\frac{8}{75}\)

⇒ \(8 x^2-80 x=150 x-750\)

⇒ \(8 x^2-80 x-150 x+750=0\)

⇒ \(8 x^2-230 x+750=0\)

⇒ \(4 x^2-115 x+375=0\)

⇒ \(4 x^2-100 x-15 x+375=0\)

⇒ 4 x(x-25)-15(x-25)=0

⇒ (x-25)(4 x-15)=0

⇒ x-25=0  or  4 x-15=0

⇒ x-25 or \(x=\frac{15}{4}\)

but \(x=\frac{15}{4}\) is not possible because both taps fill the tank in \(9 \frac{3}{4}\) hours.

⇒ x = 25 and x- 10 = 25 – 10= 15

Therefore, the tap with a smaller diameter fills the tank in 25 hours and the tap with a larger diameter fills the tank in 15 hours.

Question 10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Solution:

Given

An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train

Let the average speed of passenger train =x km/hr.

∴ Average speed of express train = (x+11) km/hr

Time taken by passenger train to cover 1321cm distance = \(\frac{132}{x} \mathrm{hr}\)

and time taken by express train to cover 1321cm distance = \(\frac{132}{x+11} \mathrm{hr}\)

According to the problem,

⇒ \(\frac{132}{x}-\frac{132}{x+11}=1\)

⇒ \(\frac{132(x+11)-132 x}{x(x+11)}=1\)

⇒ \(\frac{132 x+1452-132 x}{x^2+11 x}=1\)

⇒ \(x^2+11 x=1452\)

⇒ \(x^2+11 x-1452=0\)

⇒ \(x^2+44 x-33 x-1452=0\)

⇒ x(x+44)-33(x+44)=0

⇒ (x+44)(x-33)=0

⇒ x+44=0  or  x-33=0

⇒ x=-44  or  x=33

but the speed cannot be negative.

∴ x = 33

⇒ x + 11 =33+ 11 =44

Average speed of passenger train = 33 km/hr

an average speed of express train = 44 km/hr

Question 11. The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.
Solution:

Given

The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m

Let side of a square = x metre

Perimeter of this square = 4x metre

∴ Perimeter of second square = (4x + 24) metre

⇒ Side of second square = \(\frac{4 x+24}{4}\) = (x + 6) metre

Now, area of first square =x2 m2 and area of second square = (x+ 6)2 m2

According to the problem,

⇒ \((x+6)^2+x^2=468\)

⇒ \(x^2+12 x+36+x^2-468=0\)

⇒ \(2 x^2+12 x-432=0\)

⇒ \(x^2+6 x-216=0\)

⇒ \(x^2+18 x-12 x-216=0\)

⇒ x(x+18)-12(x+18)=0

⇒ (x+18)(x-12)=0

⇒ x+18=0   or   x-12=0

⇒ x=-18   or x=12

but the side of a square cannot be negative.

∴ x= 12

⇒ x+ 6 = 12 + 6= 18

Therefore, sides of squares = 12 m and 18 m

Quadratic Equation Exercise 4.4

Question 1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :

  1. \(2 x^2-3 x+5=0\)
  2. \(3 x^2-4 \sqrt{3} x+4=0\)
  3. \(2 x^2-6 x+3=0\)

Solution:

1. \(2 x^2-3 x+5=0\)

On comparing with ax2 + bx + c = 0

a = 2, b=-3,c = 5

Discriminant D = b2 – 4ac = (-3)2 – 4(2) (5)

= 9-40 =-31

∵ D is negative

∴ Roots of the equation are imaginary.

Here, the real roots of the equation do not exist.

2. \(3 x^2-4 \sqrt{3} x+4=0\)

On comparing with ax2 + bx + c = 0

a = 3,b =\(-4 \sqrt{3}\), c= 4

Discriminant D = b2– 4ac

⇒ \((-4 \sqrt{3})^2-4(3)(4)\)

= 48-48 = 0

The roots of given equation are real and equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{4 \sqrt{3} \pm 0}{2 \times 3}=\frac{2}{\sqrt{3}}\)

Roots of given equation = \(\frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

3. 2x2 – 6x + 3=0

On comparing with ax2 + bx + c = 0

a = 2, b = – 6, c = 3

∴ Discriminant D = b2– 4ac = (-6)2– 4 (2) (3)

= 36-24 = 12 > 0

∵ D is positive

Roots of the given equation are real and unequal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}\)

⇒ \(x=\frac{6 \pm \sqrt{12}}{2 \times 2}=\frac{6 \pm 2 \sqrt{3}}{4}\)

⇒ \(x=\frac{3 \pm \sqrt{3}}{2}\)

⇒ \(\frac{3+\sqrt{3}}{2} \text { or } \frac{3-\sqrt{3}}{2} \text {. }\)

Question 2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

  1. 2x2+kx+ 3 = 0
  2. k x(x- 2) + 6 = 0

Solution:

2x2+kx+ 3 = 0

On comparing with ax2 +bx + c = 0

a = 2, b=k, c = 3

DiscriminantD = b2– 4ac = k2 – 4 x 2 x 3

= k2 -24

Given that, the roots are equal.

∴ D = 0  ⇒ k2 – 24 = 0

⇒ k2 = 24

⇒ K = \(\pm \sqrt{24}= \pm 2 \sqrt{6}\)

2.  kx(x -2) + 6 = 0

kx2 – 2kx+6 = 0

On comparing with a2 + bx + c = 0

a = k, b = -2k, c = 6

∴ Discriminant D = b2– 4ac = (-2k)2 – 4k(6)

= 4k2– 24k = 4k(k- 6)

Given that the roots are equal.

∴ D = 0

⇒ 4k(k – 6) = 0

⇒ k – 6 = 0

⇒ k = 6

Question 3. Is It possible to design a rectangular mango grove whose length Is twice its breadth, and the area is 800 m2? If so, find its length and breadth.
Solution:

Let the breadth of mango grove

= x metre

∴ Length = 2x metre

Area of grove = length x breadth

= (2x)(x) = 2x2

According to the problem,

⇒ 2x2 = 800

⇒ x2 = 400

⇒ x = ± 20

but the breadth cannot be negative.

∴ x = 20

2x = 2 × 20 = 40

Therefore, a grove is possible, and its length = 40m and breadth = 20 m.

Question 4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.
Solution:

Let the present age of one friend =  x years

∴ Present age of second friend = (20- x) years

4 years ago, the age of first friend = (x- 4) years

The age of second friend = 20 – x- 4

= (16- x) years

According to the problem,

(x-4) (16 -x) =48

⇒ x(1 6 – 4) -4(16 – x) =48

⇒ 16x – x2– 64 + 4x = 48

⇒ 20x-x2 -64 = 48

⇒ 0 = 48 – 20x + x2 + 64

x2-20x+ 112=0 ….(1)

On comparing with ax2 + bx + c = 0

a = 1, b = -20, c = 112

Now, discriminant D=b2 – 4ac

= (- 20)2 – 4 (1) (112)

= 400 -448 =-48 <0

Roots of equation (1) are imaginary.

Therefore, the given condition is not possible.

Question 5. Is it possible to design a rectangular park of perimeter 80 m and an area of 400 m2? If so, find its length and breadth.
Solution:

Let the length of the park = x meter

Now, 2 (length + breadth) = perimeter

2(x + breadth) = 80 metre

x+ breadth = 40 metre

breadth = (40 – x) metre

According to the problem,

Area of park= 400 m2

x(40 – x) = 400

40x-x2 = 400

0 = x2 – 40x + 400

x2 – 40x + 400 = 0

On comparing with ax2 +bx+ c = 0

a = 1, b = -40, c = 400

Discriminant D = b2 – 4ac

= (-40)2 – 4(1)(400)

= 1600- 1600 = 0

⇒ Roots are equal

Now, \(x=\frac{-b \pm \sqrt{D}}{2 a}=\frac{40 \pm 0}{2 \times 1}=20\)

So, such a park is possible.

Its length = breadth = 20m.

Quadratic Equation Multiple Choice Questions

Question 1. Which of die following equations has the root 3?

  1. \(x^2+x+1\)
  2. \(x^2-4 x+3=0\)
  3. \(3 x^2+x-1=0\)
  4. \(x^2+9=0\)

Answer: 2. \(x^2-4 x+3=0\)

Question 2. The sum of roots of the equation 5x2 – 3x + 2 = 0 is

  1. \(\frac{3}{5}\)
  2. \(-\frac{3}{5}\)
  3. \(\frac{2}{5}\)
  4. \(-\frac{2}{5}\)

Answer: 1. \(\frac{3}{5}\)

Question 3. The quadratic equation \(2 x^2-\sqrt{5} x+1=0\) has

  1. Two distinct real roots
  2. Two equal roots
  3. No real root
  4. More than two real roots

Answer: 3. No real root

Question 4. One root of the equation x2 + k x + 4 = 0 is -2. The value of k is :

  1. -2
  2. 2
  3. -4
  4. 4

Answer: 4. 4

Question 5. The value of k for which die roots of equation 2kx2 – 6x + 1 = 0 is equal, is :

  1. \(-\frac{9}{2}\)
  2. \(\frac{9}{2}\)
  3. 9
  4. -9

Answer: 2. \(\frac{9}{2}\)

Question 6. Which of the following is a quadratic equation?

  1. (x + 2)2 =x2– 5x + 3
  2. x3 +x2 = (x- 1)3
  3. 3x2 + 1 = (3x- 2) (x + 5)
  4. 5x- 7 = 1 + x

Answer: 2. x3 +x2 = (x- 1)3

Question 7. If the product of roots of the equation 5×2- 3x + /{ = 0 is 2, then the value of k is:

  1. 1
  2. 2
  3. 5
  4. 10

Answer: 4. 10

Question 8. The discriminant of quadratic equadon 3xx3 +x2 = (x- 1)3– 6x + 4 = 0 is :

  1. 12
  2. 13
  3. -12
  4. \(3 \sqrt{6}\)

Answer: 3. -12

Question 9. The roots of the quadratic equation x2– 4 = 0 are :

  1. ± 0.2
  2. ±1
  3. ± 2
  4. ±4

Answer: 3. ± 2

Question 10. The discriminant of equation \(3 x^2-2 x+\frac{1}{3}=0\) will be:

  1. 3
  2. 2
  3. 1
  4. 0

Answer: 4. 0

Question 11. If the roots of the quadratic equation 3x2 – 12x + m = 0 are equal, then the value of m will be :

  1. 4
  2. 7
  3. 9
  4. 12

Answer: 4. 12

NCERT Exemplar Solutions for Class 10 Maths Chapter 2 Polynomials

Polynomials

An Expression of the form p(x)= a0+a0x+a0x0…..+a0xn

For example.,

5x+1 is a polynomial of degree in x. Here, a0,a1,a2,…..an are real numbers.

2x2-x-1 is apolynomial in x of degree 2.

y3-2y2+y+5 is a polynomial in y of degree 3.

5z4-3z+1 is a polynomial in z of degree 4.

Constant Polynomial

A polynomial of degree zero is called a constant polynomial.

For example., p(x) = -3, q(x) = 2,f(x) = \(\sqrt{2}\) etc., are constant polynomials. (These are independent of variable)

Zero Polynomial

It is also a constant polynomial with a particular constant value of 0.

So,f(x) = 0 is a zero polynomial. Its degree is not defined, as we cannot say definitely about its degree. Forms of zero polynomial may be

f(x) = 0 = 0 . x2 – 0 .x + 0 = 0 . x5 – 0 . x2= 0 . J-5 + 0. x9 + 0. x4 – 0 . x2 + 0., etc.

So, we cannot say anything about the degree of a zero polynomial.

Linear Polynomial

A polynomial of degree 1 is called a linear polynomial.

It is of the form

p(x) = ax + b,

where a ≠ 0

For example., 3x + 5, 5 – 2x, etc.

Quadratic Polynomial

A polynomial of degree 2 is called a quadratic polynomial. It is of the form

p(x) = ax2 +bx + c, where a ≠ 0

x2 + 5x + 1, 3x2 – x + 1, 5 – x2 etc.

NCERT Exemplar Solutions For Class 10 Maths Chapter 2 Polynomials

Cubic Polynomial

A polynomial of degree 3 is called a cubic polynomial.

It is of the form

p(x) = ax3 +bx2 +cx + d, where a ≠ 0

For example., x3 – x + 1, 5x2 – 4x2 – 2x + 1 etc.

Bi-quadratic Polynomial (or Quartic)

A polynomial of degree 4 is called a biquadratic polynomial.

It is of the form

p(x) = ax4 + bx3 + cx2 + dx + e, a ≠ 0

For example., 2x4 -x2 + 1, 2- 3x +x2 + 4x2 -x4 etc.

Note

Some other names for polynomials are:

Polynomials Some Other Names Of Polynomials Are

Value Of A Polynomial At A Given Point

Let p(x) be a polynomial in x and a is any real number. Then the value obtained by putting x = a in the polynomial p(x) is called the value of the polynomial p(x) at x = a.

This value is denoted by p(a).

P(x) = x2 +x- 1,

then p(5) = 52+5-1= 25+5 = 29

Zeroes Of A Polynomial

A really a is called a zero of a polynomial P(x), if P(α) = 0

or, in other words,

zero of a polynomial is the real value of a variable that vanishes the whole polynomial i.e., the value of a variable that makes the whole polynomial zero.

For example., if P(x) = x2-3x+2

p(2) = 22-3×2+2 = 4-6+2 = 0

∴ 2 is a zero of p(x)

To find the zero/es of a polynomial p(x), put the polynomial equation p(x) = 0.

For example,

Find the zeroes of polynomial p(x) = x2 – 5x – 6

We know that zero of a polynomial is the value of a variable by which p(x) = 0.

⇒ x2 – 5x – 6 = 0 ⇒ (x – 6) (x + 1) = 0

∴ either x – 6 = 0 or x+1=0

⇒ x = 6 or x= -1

So, 6 and -1 are two values of x which make the value of polynomial zero. (You can check on putting these values directly They will make p(x) = 0

Hence, 6 and -1 are the zeroes of the given polynomial.

2. Find the zeroes of polynomial p(x) = x2 – 4x + 4

For zeroes, p(x) = 0

⇒ x2-4x +4 = 0

⇒ (x – 2)2 = 0x – 2 = 0

⇒ x = 2

or x2-4x + 4 = 0 ⇒ (x – 2)(x – 2) = 0

⇒ x = 2, 2

It means, in this case, we get the repeated zeroes. But, we shall say the zero of this polynomial is 2 not the 2 and 2. To say 2 and 2 has no sense. So, if a polynomial has 2 or 3 or more repeated zeroes, we will say only one zero it has.

3. Find the zeros of polynomial p(x)_ = ax2+bx+c, a≠0

To find the zeroes, we put p(x) = 0

∴ \(a x^2+b x+c=0\)

⇒ \(x^2+\frac{b x}{a}+\frac{c}{a}=0\)  (dividing both sides by a)

⇒ \(x^2+\frac{b}{a} x+\underbrace{\frac{b^2}{4 a^2}-\frac{b^2}{4 a^2}}_{\begin{array}{c}
\text { Adding and subtracting } \\
\text { the same quantity }
\end{array}}+\frac{c}{a}=0\)

⇒ \(\left[\text { adding and subtracting }\left(\frac{\text { coff. of } x}{2}\right)^2\right]\)

⇒ \( \left(x+\frac{b}{2 a}\right)^2-\left(\frac{b^2}{4 a^2}-\frac{c}{a}\right)=0 \)

⇒ \(\left(x+\frac{b}{2 a}\right)^2-\left(\frac{b^2-4 a c}{4 a^2}\right)=0\)

⇒ \(\left(x+\frac{b}{2 a}\right)^2-\left(\frac{\sqrt{b^2-4 a c}}{2 a}\right)^2=0\)

(every number is the square of its square root)

⇒ \(\left(x+\frac{b}{2 a}+\frac{\sqrt{b^2-4 a c}}{2 a}\right)\left(x+\frac{b}{2 a}-\frac{\sqrt{b^2-4 a c}}{2 a}\right)=0\left[a^2-b^2=(a+b)(a-b)\right]\)

⇒ \(x=\frac{-b-\sqrt{b^2-4 a c}}{2 a} \text { or } x=\frac{-b+\sqrt{b^2-4 a c}}{2 a}\)

There are two zeroes of a given polynomial.

From the above 3 examples, we observe some very special points :

Zeroes are the values of x when p(x) orj; = 0. It means for example (6, 0) and (-1, 0) must be the points on the polynomial as well as on the x-axis (because the y-coordinate is zero) i.e., the intersection point of the polynomial and x-axis.

Conversely, we can say that if we draw the graph of a polynomial then the x-coordinate or abscissa of the intersection point of the curve and x-axis will give us the zeroes of the polynomial.

From the example, we notice that repeated zeroes will occur only when the polynomial is a perfect square. We still say it as only one zero. We also get only one zero from a linear polynomial, then what is the difference between linear polynomial and quadratic polynomial when we get only one zero in both cases?

So, our conclusion in this case is that a linear polynomial will be a straight line that will cut the x-axis at one point, and the x-coordinate of that point is the zero of that polynomial while a quadratic polynomial in this case of repeated zeroes will not cut the x-axis but it will touch the x-axis and turns thereafter. This is the basic difference between both types of polynomials in the case of one zero.

For example, we see that if x = 6 is a zero of the polynomial then, (x – 6) is one factor of the polynomial. If x = -1 is a zero of the polynomial then (x+ 1) is also one factor of polynomial.

So, if x = α and x = β are the zeroes of a polynomial then, necessarily (x – α) and (x – β) are the factors of that polynomial.

So, if α and β are the zeroes of a polynomial then the polynomial will be In the form (x-α) (x-β). Is it correct 100%? Perhaps, not, Why?

If we find the zeroes of x2-5x-6, we get 6 and -1 as zeros, if we find the zeros of 2x2-10x-12 i.e., 2(x2-5x-6) we also get 6 and -1 as zeros. If we find zeros of 3x2-15x-18 i.e., 3(x2-5x-6). we also get 6 and -1 as the zeroes. If we find the zero of

⇒ \(-\frac{1}{3} x^2+\frac{5}{3} x+2 \text { i.e… }-\frac{1}{3}\left(x^2-5 x-6\right)\)

also get the zeroes as 6 and -1.

Then, if it is given that a and |3 arc the zeroes of a quadratic polynomial then the quadratic polynomial will be k (x – α) (x – β), where k may be any non-zero real number.

Now, let us study the curves (polynomials) graphically.

Geometrical Meaning Of Zeroes Of A Polynomial

Case 1: Consider the first-degree polynomial p(x) = ax + b, a≠0. We know that the graph of y = ax + b is a straight line for example., consider the equation y = 2x – 3. This line passes through the point (1, -1) and (2, 1). Draw the graph of this line as shown. It crosses the x-axis at a point this linear equation has exactly one zero namely \(\frac{3}{2}\).

Thus, for the polynomial p(x) = ax + b, (a≠0) zero of p(x) is \(x=-\frac{b}{a}\)

Case 2: Consider the second-degree polynomial p(x) = ax + bx + c, a ≠ 0.

Consider an example say p(x) = x2 – 5x + 6. Let us see the graph of y = x2 -5x + 6.

Polynomials Consider The Second Degree Polynomial

Draw the graph of this line as shown.

Polynomials The First Degree Polynomial

This graph intersects the j-axis at two points namely (2, 0) and (3, 0). In fact the graph of y = ax2 + bx + c, a ≠ 0 has one of the two shapes either open upwards like when a > 0 or open downward like A when a < 0. These curves are called parabolas.

Polynomials The Second Degree Polynomial

1. When the graph cuts the x-axis at two points A and A’.

Polynomials X Axis At Two Points

Here, the x coordinates of A and A’ are two zeroes of the quadratic polynomial.

2. When the graph cuts the x-axis at one point i.e., at two coincident points.

Polynomials X Axis At One Point

The x coordinate of A is the only zero of the quadratic polynomial.

3. When the graph is either completely above the x-axis or completely below the x-axis i.e., it does not cut the x-axis at any point.

Polynomials X Axis At Any Point

The quadratic polynomial has no zero in this case.

Case 3: Consider the third-degree polynomial p(x) = ax3 + bx2 + cx + d, a ≠ 0.

It can have at most 3

zeroes, depending upon the point of intersection of y = ax3 +bx2+cx+d and x-axis.

Relation Between Zeroes And Coefficients Of A Quadratic Polynomial

Consider the quadratic polynomial

p(x) – ax2 +bx + c, a ≠ 0

Let α, and β be the zeroes of p(x), then (x – α) and (x – β) will be the factors of p(x).

ax2 + bx + c = k(x – α) (x – β), k ≠ 0

= k [x2 – (α + β)x + αβ]

= kx2 – k(α + β)x + Kαβ

On comparing the coefficients of like powers on both sides, we get

k = a

⇒ \(-k(\alpha+\beta)=b \Rightarrow-a(\alpha+\beta)=b \Rightarrow \alpha+\beta=\frac{-b}{a}\)

⇒ \(k \alpha \beta=c \Rightarrow \quad a \alpha \beta=c \Rightarrow \alpha \beta=\frac{c}{a}\)

∴ \(\text { sum of zeroes }=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2},\)

⇒ \(\text { product of zeroes }=\frac{\text { constant term }}{\text { coefficient of } x^2}\)

If a and p are the zeroes of a quadratic polynomial p(x), then

p(x) = k[x2 – (α + β)x + αβ], k ≠ 0

Quadratic Polynomial Solved Examples

Question 1. Find zeroes of the polynomial x – 3x + 2 and verify the relation between its zeroes and coefficients.
Solution:

Given x – 3x + 2

Let p(x) =x2-3x + 2 = x2-2x-x + 2

= x(x-2)-1 (x-2) = (x-2) (x- 1)

∴ P(x) = 0

⇒ (x- 2)(x – 1) = 0

⇒ x – 2 = 0 or x – 1 = 0

⇒ x = 2 or x – 1

∴ Zeroes of p(x) are 2 and 1

Now, Sum of zeroes = 2 + 1 = 3 = \(-\frac{-3}{1}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

and product of zeroes = (2) (1) = 2 = \(\frac{2}{1}=\frac{\text { constant term }}{\text { coefficient of } x^2}\)

Question 2. Find zeroes of the quadratic \(\sqrt{3} x^2-8 x+4 \sqrt{3}\) and verify the relation between the zeroes and coefficients.
Solution:

Given \(\sqrt{3} x^2-8 x+4 \sqrt{3}\)

Let p(x) = \(\sqrt{3} x^2-8 x+4 \sqrt{3}=\sqrt{3} x^2-6 x-2 x+4 \sqrt{3}\)

\(\sqrt{3} x(x-2 \sqrt{3})-2(x-2 \sqrt{3})=(x-2 \sqrt{3})(\sqrt{3} x-2)\)

∴ p(x) = 0

⇒ \((x-2 \sqrt{3})(\sqrt{3} x-2)\) = 0

⇒ \(x-2 \sqrt{3}=0\)

or \(\sqrt{3} x-2=0\)

⇒ \(x=2 \sqrt{3}\)

or \(x=\frac{2}{\sqrt{3}}\)

∴ Zeroes of p(x) are \(2 \sqrt{3} \text { and } \frac{2}{\sqrt{3}}\)

Now, sum of zeros = \(=2 \sqrt{3}+\frac{2}{\sqrt{3}}\)

⇒ \(\frac{6+2}{\sqrt{3}}\)

⇒ \(\frac{8}{\sqrt{3}}\)

⇒ \(-\frac{(-8)}{\sqrt{3}}\)

⇒ \(-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

and product of zeros = \((2 \sqrt{3})\left(\frac{2}{\sqrt{3}}\right)=4\)

⇒ \(\frac{4 \sqrt{3}}{\sqrt{3}}\)

⇒ \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

Question 3. Find zeroes of the polynomial x2 – 4 and verify the relation between /crocs and coefficients.
Solution:

Given x2 – 4

Let p(x) = x2 – 4 = x2 – 22 = (x – 2) (x + 2)

∴ p(x) = 0

(x-2)(x + 2) = 0

x – 2 = 0 or x + 2 = 0

x = 2 or x = -2

∴ Zeroes of p(x) are 2 and -2

Now, sum of zeroes = 2 + (-2) = 0 = \(-\frac{0}{1}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^2}\)

and product of zeroes = (2) (-2) = -4 = \(\frac{-4}{1}=\frac{\text { constant term }}{\text { coefficient of } x^2}\)

Question 4. Find a quadratic polynomial, the sum of whose zeroes is 5 and their product is 6. Hence, find the zeroes of the polynomial.
Solution:

Given

The sum of whose zeroes is 5 and their product is 6

Let a and (3 be the zeroes of the polynomial p(x).

Given that α + β = 5 and αβ = 6

Now, p(x) – x2 – (α + β)x + αβ = x2 – 5x ÷ 6 = x2 – 3x – 2x ÷ 6

= x(x – 3) – 2 (x – 3) = (x – 3) (x – 2)

There may be so many different polynomials which satisfy the given condition. Actually the general quadratic polynomial will be k(x2 – 5x + 6), where k = 0

∴ p(x) = 0

⇒ (x – 2)(x – 3) = 0

⇒ (x – 2) = 0 or (x – 3) = 0

⇒ x = 2 or x = 3

Zeroes are 2 and 3

zeroes of the polynomial 2 and 3.

Question 5. If the product of zeroes of the polynomial (ax2 – 6x – 6) is 4. find the value of a.
Solution:

Given polynomial = ax2 – 6x – 6

product of zeroes = \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

⇒ \(4=\frac{-6}{a}\)

4a = -6

⇒ \(a=-\frac{6}{4}\)

⇒ \(=-\frac{3}{2}\)

The value of a =\( -\frac{3}{2}\)

Question 6. If x = \(\frac{2}{3}\) and .v = -3 are zeroes of the quadratic polynomial ax2+7x+b, then find the values of a and b.
Solution:

Given

x = \(\frac{2}{3}\) and .v = -3 are zeroes of the quadratic polynomial ax2+7x+b

Let p(x) = ax2 + 7x + b

∵ \(x=\frac{2}{3}\) and x = -3 are zeroes of p(x)

⇒ \(p\left(\frac{2}{3}\right)=0\)

⇒ \(a\left(\frac{2}{3}\right)^2+7\left(\frac{2}{3}\right)+b=0\)

⇒ \(\frac{4 a}{9}+\frac{14}{3}+b=0\)

⇒ \(b=-\frac{4 a}{9}-\frac{14}{3}\)

and p(-3) = 0

⇒ a(-3)2+7(-3)+b = 0

⇒ \( 9 a-21-\frac{4 a}{9}-\frac{14}{3}=0\)

⇒ \(\frac{81 a-4 a}{9}=\frac{14}{3}+21=\frac{14+63}{3}\)

⇒ \(\frac{77 a}{9}=\frac{77}{3}\)

a=3

For equation (1)

⇒ \(b=\frac{-4 \times 3}{9}-\frac{-14}{3}=-6\)

a=3, b=-6

The values of a and b are 3 and -6.

Question 7. If one zero of the polynomial (a2 + 9)x2 + 13x+6a is reciprocal of the other, find the value of a.
Solution:

Given

one zero of the polynomial (a2 + 9)x2 + 13x+6a is reciprocal of the other

Let p(x) = (a2 + 9)x2 +13x+ 6a

Let α and \(\frac{1}{\alpha}\) be zeroes of p(x) then,

product of zeroes = \(\frac{\text { constant term }}{\text { coefficient of } x^2}\)

⇒ \(\alpha \cdot \frac{1}{\alpha}=\frac{6 a}{a^2+9}\)

⇒ \(I=\frac{6 a}{a^2+9}\)

⇒ a2+9 = 6a

⇒ a2-6a+9 = 0

⇒ (a-3)2 = 0

⇒ a-3 = 0

a = 3

The value of a is 3.

Relation Between Zeroes And Coefficients Of A Cubic Polynomial

Consider a cubic polynomial.

p(x) = ax3 + bx2 + cx + d, a ≠ 0

Let α, β, γ be zeroes of p(x), then (x – α), (x – β), (x – γ) will be the factors of p(x).

∴ ax3 + bx2 + cx + d = k(x – α)(x- β)(x – γ)

Comparing, we get

k = a

⇒ \(-k(\alpha+\beta+\gamma)=b \quad \Rightarrow \quad \alpha+\beta+\gamma=-\frac{b}{a}\)

(k=a)

⇒ \(k(\alpha \beta+\beta \gamma+\gamma \alpha)=c \quad \Rightarrow \quad \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\)

⇒ \(-k \alpha \beta \gamma=d \quad \Rightarrow \quad \alpha \beta \gamma=-\frac{d}{a}\)

p(x) = \(x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma .\)

Division Algorithm For Polynomials

If(x) and g(x) are two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x), such that

f(x) = q{x) x g(x) + r(x).

where r(x) = 0 or deg {r(A)} < deg {g(x)}.

Division Algorithm Solved Examples

Question 1. Verify that 1,2,3 is the zeros of the cubic polynomial p(x) = x3-6x2+11x-6 and verify the relation between its zeroes and coefficients.
Solution:

Given p(x) = x3-6x2+11x-6

Here,

p(x)=\(x^3-6 x^2+11 x-6\)

p(1)=\(1^3-6(1)^2+11(1)-6=1-6+11-6=0\)

p(2)=\(=2^3-6(2)^2+11(2)-6=8-24+22-6=0\)

p(3)=\(=3^3-6(3)^2+11(3)-6=27-54+33-6=0\)

and

∴ 1,2 and 3 are zeroes of p(x).

Now, \(\alpha+\beta+\gamma\)

⇒ \(1+2+3=6\)

⇒ \(-\frac{-6}{1}\)

⇒ \(-\frac{\text { coefficient of } x^2}{\text { coefficient of } x^3}\)

αβ + βγ+ γα = (1 )(2) + (2)(3) + (3)(1) = 2 + 6 + 3= 11

⇒ \(\frac{11}{1}\)

⇒ \(\frac{\text { coefficient of } x}{\text { coefficient of } x^3}\)

⇒ \(\text { and } \alpha \beta y=(1)(2)(3)=6=-\frac{-6}{1}=-\frac{\text { constant term }}{\text { coefficient of } x^3}\)

Question 2. Find a cubic polynomial whose zeroes are \(\frac{1}{2},-\frac{3}{2}\) and 2
Solution:

Given

a cubic polynomial whose zeroes are \(\frac{1}{2},-\frac{3}{2}\) and 2

Let \(\alpha=\frac{1}{2}, \beta=-\frac{3}{2} \text { and } \gamma=2\)

∴ \(\alpha+\beta+\gamma=\frac{1}{2}-\frac{3}{2}+2=1\)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=\left(\frac{1}{2}\right)\left(-\frac{3}{2}\right)+\left(-\frac{3}{2}\right)(2)+\left(\frac{1}{2}\right)(2)=\frac{-3}{4}-3+1=-\frac{11}{4}\)

⇒ \(\alpha \beta \gamma=\left(\frac{1}{2}\right)\left(-\frac{3}{2}\right)(2)=-\frac{3}{2}\)

Cubic polynomial = \(x^3-(\alpha+\beta+\gamma) x^2+(\alpha \beta+\beta \gamma+\gamma \alpha) x-\alpha \beta \gamma\)

⇒ \(x^3-x^2-\frac{11}{4} x-\frac{3}{2} \approx 4 x^3-4 x^2-11 x-6\)

The cubic polynomial is \(x^3-x^2-\frac{11}{4} x-\frac{3}{2} \approx 4 x^3-4 x^2-11 x-6\)

Question 3. Divide 2x2+x-5 by x + 2 and verify the division algorithm.
Solution:

Given  2x2+x-5 and x + 2

Polynomials The Division Algorithm

Now, quotient = 2x- 3, remainder = 1

dividend = 2x2 + x- 5 and divisor = x+ 2

and quotient x division + remainder = (2x-3)(x+2) +1

\(2 x^2+4 x-3 x-6+1=2 x^2+x-5\)

= division

Question 4. If the polynomial \(\left(x^4+2 x^3+8 x^2+12 x+18\right)\) is divided by another polynomial (x2+5), the remainder comes out to be (px+q). Find the values of p and q.
Solution:

Given \(\left(x^4+2 x^3+8 x^2+12 x+18\right)\) is divided by another polynomial (x2+5)

Let f(x) = \(x^4+2 x^3+8 x^2+12 x+18\) and g(x) = x2+5

On dividing f(x) by g(x)

Polynomials Find The Values Of P And Q

∵ the remainder is given px+q.

∴ p=2 and q=3

The values of p and q 2 and 3.

Question 5. What real number should be subtracted from the polynomial (3x3 + 10x2 – 14x + 9) so that (3x- 2) divides it exactly?
Solution:

Given (3x3 + 10x2 – 14x + 9)

On dividing (3x3 + 10x2 – 14x + 9) by (3x- 2)

Polynomials Real Number Should Be Subtracted From The Polynomial

∵ the remainder = 5

∴ required number = 5

The real number should be subtracted from the polynomial is 5.

Question 6. If 2 is a zero of the polynomial x3-2x2-x+2, then find its other zeroes.
Solution:

Given

2 is a zero of the polynomial x3-2x2-x+2

Let p(x) = x3-3x3-x+2

∵ x=2 is a zero of p(x)

(x-2) is a factor of p(x)

Polynomials If 2 Is A Zero Of The Polynomial

∴ p(x) = \(x^3-2 x^2-x \div 2=(x-2)\left(x^2-1\right)=(x-2)\left(x^2-1^2\right)\)

= (x-2)(x-1)(x 1)

Now, p(x) = 0

⇒ (x-2)(x-1)(x 1) = 0

⇒ x-2 = 0 or x-1 = 0  or x  1 =0

⇒ x=2 or x=1 or x=-1

Hence, other zeroes are 1 and -1.

Question 7. Obtain all other zeroes of (x4 + 4x3– 2x2– 20x- 15) if two of its zeroes are \(\sqrt{5} \text { and }-\sqrt{5}\)
Solution:

Given

(x4 + 4x3– 2x2– 20x- 15) if two of its zeroes are \(\sqrt{5} \text { and }-\sqrt{5}\)

Let p(x) = x4 + 4x3– 2x4– 20x- 15

\(\sqrt{5} \text { and }-\sqrt{5}\) are zeroes of p(x)

∴ \(x-\sqrt{5} \text { and } x+\sqrt{5}\) are factors of p(x).

So, \((x-\sqrt{5})(x+\sqrt{5})=x^2-5\) is a factor of p(x)

Polynomials If Two Of Its Zeroes Are Root 5 And Minus Root 5

∴ p(x) = \(x^4+4 x^3-2 x^2-20 x-15=\left(x^2-5\right)\left(x^2+4 x+3\right)\)

⇒ \(\left(x^2-5\right)\left[x^2+x+3 x+3\right]=\left(x^2-5\right)[x(x+1)+3(x+1)]\)

⇒ \(\left(x^2-5\right)(x+1)(x+3)\)

∴ The other zeroes are given by

x+1 = 0 or x+3 = 0

⇒ x = -1 or x = -3

Hence, other zeroes are -1 and -3

Question 8. Find zeroes of the polynomial p(x) = x3 – 9x2 + 26x – 24, if it is given that the product of its two zeroes is 8.
Solution:

Given polynomial p(x) = x3 – 9x2 + 26x – 24

Let α, β, γ be zeroes of the given polynomial p(x), such that αβ = 8 …….(1)

⇒ \(\alpha+\beta+\gamma=-\frac{(-9)}{1}=9\) ……….(2)

⇒ \(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{26}{1}=26\) …..(3)

⇒ \(\alpha \beta \gamma=-\frac{(-24)}{1}=24\) ………..(4)

From (1) and (4)

8γ = 24

γ = 3

Put γ = 3 in (2), we get α+β = 6 ……..(5)

⇒ \((\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta\)

⇒ \(=(6)^2-4(8)\)

⇒ \(=36-32\)

⇒ \(=4\)

⇒ \(\alpha-\beta= \pm 2\) …..(6)

Solving (5) and (6), we get

a= 4, b = 2  or a=2, b = 4 and r = 3

So, zeroes are 2,3 and 4

Question 9. Find the common zeroes of the polynomials x3 +x2– 2x- 2 and x3– x2 – 2x + 2.
Solution:

Given x3 +x2– 2x- 2 and x3– x2 – 2x + 2

For the common zeroes, first, we find the H.C.F. of given polynomials by Euclid’s division method (long division method)

Polynomials First We Find The HCF Of Given Polynomials By Euclids Division Method

Hence, the H.C.F. of the given polynomials is (x2 – 2). Thus, the common zeroes of the given polynomials are the zeroes of x2– 2 i.e. \((x+\sqrt{2})(x-\sqrt{2})\)

So, zeroes of \(\sqrt{2} \text { and }-\sqrt{2}\).

Polynomials Exercise 2.1

Question 1. The graphs o(y = p(x) are given below, for some polynomials p{x). Find the number of zeroes of p(x), in each case.

Polynomials Find The Number Of Zeroes Of P Of X

Polynomials Find The Number Of Zeroes Of P Of X.

Solution:

1. The graph of the polynomial p(x) does not intersect the x-axis at any point.

∴ Number of its zeroes is zero.

2. The graph of the polynomial p(x) intersects the x-axis at one point.

∴ Number of its zeroes is one.

3. The graph of the polynomial p(x) intersects the x-axis at three points.

∴ Number of its zeroes is 3.

4. The graph of the polynomial p(x) intersects the x-axis at two points.

∴ Number of its zeroes is 2

5. The graph of the polynomial p(x) intersects the x-axis at four points.

∴ Number of its zeroes is 4.

6. The graph of the polynomial p(x) intersects the x-axis at three points.

∴ Number of its zeroes is 3.

Polynomials Exercise 2.2

Question 1. Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:

  1. x2 – 2x- 8
  2. 4s2 – 4s + 1
  3. 6x2 – 3 – 7x
  4. 4u2 + 8u
  5. t2– 15
  6. 3x2 -x- 4

Solution:

Let p(x) =x2 – 2x- 8

= x2-4x + 2x-8

= x(x- 4) + 2(x- 4)

= (x- 4)(x + 2)

The zeroes of p(x) will be given by x- 4 = 0 and x + 2 = 0.

x- 4 = 0 ⇒ x = 4

x + 2 = 0 ⇒ x = -2

∴ Zeroes of p(x) = 4,-2

Now, from p(x) = x2 – 2x- 8

a = 1 , b = -2, c = -8

∴ \(-\frac{b}{a}=-\frac{(-2)}{1}=2\)

= 4+(-2) = sum of zeroes

and \(\frac{c}{a}=\frac{-8}{1}=-8\)

= 4(-2) = product of zeroes

∴ Relations between the zeroes of polynomial and the coefficients are true.

2. Let p(s) = 4s2 – 4s +1 = 4s2-2s-2s +1

= 2s(2s-1) -1 (2s-1)

= (2s-1)(2s-1)

The zeroes ofp(s) will be given by 2s -1 = 0 and 2s – 1 = 0.

∴ 2s – 1 = 0 \(s=\frac{1}{2}\)

Therefore, zeroes of p(s) = \(\frac{1}{2}, \frac{1}{2}\)

Now, from p(s) = 4s2 – 4s + 1

a = 4, b =-4, c = 1

⇒ \(-\frac{b}{a}=-\frac{(-4)}{4}=1\)

⇒ \(\frac{1}{2}+\frac{1}{2}\)

=  sum of zeroes

and \(\frac{c}{a}=\frac{1}{4}=\frac{1}{2} \times \frac{1}{2}\) = Product zeroes.

Relations between the zeroes of polynomial and the coefficients are true.

3. Let p(x) = 6x2 – 3 -7x = 6x2 – 7x- 3

= 6x2– 9x + 2x- 3

= 3x(2x-3) + 1(2x- 3)

= (2r- 3)(3x+ 1)

The zeroes of p(x) will be given by 2x- 3 = 0 and 3x+ 1=0.

∴ 2x-3 = 0 ⇒ \(x=\frac{3}{2}\)

and 3x + 1 = 0 ⇒ \(x=-\frac{1}{3}\)

Therefore, zeroes of p(x) = \(\frac{3}{2},-\frac{1}{3}\)

Now, from p(x) =  6×2- 7x- 3

a = 6, b=-7, c = -3

⇒ \(-\frac{b}{a}=-\frac{(-7)}{6}=\frac{7}{6}=\frac{3}{2}+\left(-\frac{1}{3}\right)\)

= sum of zeroes.

⇒ \(\frac{c}{a}=\frac{-3}{6}=-\frac{1}{2}=\frac{3}{2} \times\left(-\frac{1}{3}\right)\)

= product of zeroes.

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

4. Let p(u)= 4u2 + 8u = 4u(u + 2) Zeroes of p(u) will be given by 4u = 0 and u + 2 = 0.

∴ 4u = 0 ⇒ u = 0

and u + 2 = 0  ⇒ u = -2

Therefore, zeroes of p(u) = 0, -2.

Now, from p(u) = 4u2 + 8u, a = 4, b = 8, c = 0

∴ \(-\frac{b}{a}=-\frac{8}{4}=-2=0+(-2)=\text { sum of zeroes. }\)

and \(\frac{c}{a}=\frac{0}{4}=0=0(-2)=\text { product of zeroes. }\)

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

5.  Let p(t) = t2 – 15

⇒ \(t^2-(\sqrt{15})^2=(t-\sqrt{15})(t+\sqrt{15})\)

The zeroes of p{t) will be given by

⇒ \(t-\sqrt{15}=0\) and \(t-\sqrt{15}=0\)

Now, \(t-\sqrt{15}=0\) \(\Rightarrow \quad t=\sqrt{15}\)

and \(t-\sqrt{15}=0\) \(\Rightarrow t=-\sqrt{15}\)

Zeroes of p(t) = \(\sqrt{15} \text { and }-\sqrt{15}\)

From p(t) = t2 – 15

a = 1, b = 0, c =-l5

⇒ \(-\frac{b}{a}=-\frac{0}{1}=0=\sqrt{15}+(-\sqrt{15})\)

= sum of zeroes.

and \(\frac{c}{a}=\frac{-15}{1}=-15=(\sqrt{15})(-\sqrt{15})\)

= product of zeroes.

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

6.  Let p(x)  = \(3 x^2-x-4=3 x^2-4 x+3 x-4\)

⇒ \(x(3 x-4)+1(3 x-4)\)

⇒ \((3 x-4)(x+1)\)

The zeroes of p(x) will be given by 3x – 4 = 0 and x + 1 = 0.

Now, 3x- 4 = 0 ⇒ \(x=\frac{4}{3}\)

and x+ 1 = 0 ⇒ A =-l

Zeroes of p(x) = \(\frac{4}{3}\) and -1

From p(x) = 3x2-x-4

a = 3, b = -1, c = -4

⇒ \(-\frac{b}{a}=-\frac{(-1)}{3}=\frac{1}{3}=\frac{4}{3}+(-1)\)

= sum of zeroes.

and \(\frac{c}{a}=\frac{-4}{3}=\frac{4}{3} \times(-1)=\text { sum of product. }\)

Therefore, the relations between the zeroes of the polynomials and the coefficients are true.

Question 2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:

  1. \(\frac{1}{4},-1\)
  2. \(\sqrt{2}, \frac{1}{3}\)
  3. \(0, \sqrt{5}\)
  4. 1,1
  5. \(-\frac{1}{4}, \frac{1}{4}\)
  6. 4,1

Solution:

1. Let the zeroes of the polynomial be α and β.

∴ \(\alpha+\beta=\frac{1}{4} \text { and } \alpha \beta=-1\)

∴ quadratic polynomial = (x- α) (x- β)

= x2 – (α+ β)x + αβ

= \(x^2-\frac{1}{4} x-1=\frac{1}{4}\left(4 x^2-x-4\right)\)

∴ Required polynomial = 4x2 – x- 4.

2. Let the zeroes of the polynomial be a and p.

∴ \(\alpha+\beta=\sqrt{2} \quad \text { and } \quad \alpha \beta=\frac{1}{3}\)

∴ quadratic polynomial = (x – α)(x – β)

⇒ \(x^2-(\alpha+\beta) x+\alpha \beta=x^2-\sqrt{2} x+\frac{1}{3}\)

⇒ \(\frac{1}{3}\left(3 x^2-3 \sqrt{2} x+1\right)\)

∴ Required polynomial = \(3 x^2-3 \sqrt{2} x+1\)

3. Let the zeroes of the polynormal be a and p

∴ \(\alpha+\beta=0 \text { and } \alpha \beta=\sqrt{5}\)

∴ quadratic polynomial = (x – α)(x – β)

⇒ \(x^2-(\alpha+\beta) x+\alpha \beta\)

⇒ \(x^2-0(x)+\sqrt{5}=x^2+\sqrt{5}\)

4.  Let the zeroes of the polynomial be a and p.

α + β = 1 and αβ = 1

∴ quadratic polynomial = (x- a)(x- P)

= x2 – (α + β)x + αβ

x2– x + 1

5.  Let the zeroes of the polynomial be a and p.

∴ \(\alpha+\beta=-\frac{1}{4} \text { and } \alpha \beta=\frac{1}{4}\)

Now, quadratic polynomial = (x – α)(x – β)

x2 – (α + β)x + αβ

⇒ \(\alpha+\beta=-\frac{1}{4} \text { and } \alpha \beta=\frac{1}{4}\)

Therefore, required polynomial = 4x2 + x + 1

6. Let the zeroes of the polynomial be α and β.

∴ α + β = 4 and αβ = 1

Now, quadratic polynomial = (x- α)(x- β)

= x2 — (α + β)x + αβ

= x2-4x+ 1.

Polynomials Exercise 2.3

Question 1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following :

  1. p(x) =x3– 3x2 + 5x-3, g(x) =x2-2
  2. p(x) = x4 – 3x2 + 4x + 5, g(x) =x2+1 -x
  3. p(x) = x4 – 5A + 6, g(x) = 2 -x2

Solution:

Polynomials Divide The Polynomial P Of X And G Of X

Now, P(x) = g(x). q(x) + r(x)

∴ quotient q(x) = x- 3

and remainder r(x) = 7x- 9

Polynomials Divide The Polynomial P Of X By The Polynomial Of G Of X 2

Now, P(x) = g(x) f(x) + r(x)

∴ quotient q(x) = x2 +x-3

and remainder r(x) = 8

Polynomials Divide The Polynomial P Of X By The Polynomial Of G Of X 3

Now, p(x) = g(x) q(x) + r(x)

∴ quotient q(x) = -x2 – 2

and remainder r(x)= -5x + 10

Question 2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

  1. t2-3, 2t4 + 3t3– 2t2– 9t- 12
  2. x2 + 3x + 1, 3x4 + 5x3– 7x2 + 2x + 2
  3. x3– 3x + 1, x5 – 4x3 + x2 + 3x +1

Solution:

Polynomials Dividing The First And Second Polynomial 1

Now, p(t) g(t) + (t) + r(t)

∴ quotient q(t) = 2t2 + 3t + 4 and remainder r(x)= 0.

∵ The remainder is zero.

∴ t2 – 3 is a factor of 2t4 + 3t3 -2t2 – 9t – 12.

Polynomials Dividing The First And Second Polynomial 2

Now, p= g(x) q(x) + r(x)

∴ quotient q(x) = 3x2 – 4x + 2 and remainder r(x) = 0

∵ The remainder is zero.

∴ x2 + 3x + 1 is a factor of 3x4 + 5x3– 7x2 + 2x + 2.

Polynomials Dividing The First And Second Polynomial 3

Now, P(x) = g(x) q(x) + r(A)

∴ quotient q(x) = x2– 1 and remainder r(x) = 2.

∵ The remainder is not zero.

∴ x2– 3x + 1 is not a factor x5 – 4x3 + x2 + 3x + 1 .

Question 3. Obtain all other zeroes of 3×4 + 6×3 – 2×2 – 10x – 5 if two of its zeroes are \(\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}\)
Solution:

Given 3x4 + 6x3 – 2x2 – 10x – 5

Two of its zeroes are \(\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}\)

Let the remaining two zeroes be α and β.

Given zeroes are \(\sqrt{\frac{5}{3}} \text { and }-\sqrt{\frac{5}{3}}\)

Quadratic polynomial from zeroes.

⇒ \(\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)\)

⇒ \(x^2-\frac{5}{3}=\frac{1}{3}\left(3 x^2-5\right)\)

∴ quadratic polynomial = 3x2 – 5

Now, (x- α)(x- β)(3x2 – 5)

⇒ \(3 x^4+6 x^3-2 x^2-10 x-5\)

⇒ \((x-\alpha)(x-\beta)=\frac{3 x^4+6 x^3-2 x^2-10 x-5}{3 x^2-5}\)

Polynomials Quadratic Polynomial

∴ (x- α)(x- β) = x2 + 2x + 1

= (x+ 1)(x + 1)

∴ α = -1, β =-1

⇒ Remaining zeroes =-1,-1.

Question 4. On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x- 2 and -2x + 4, respectively. Find g(x).
Solution:

Given

x3 – 3x2 + x + 2

Let p(x) = x3 – 3x2 + x + 2

quotient q(x) =x – 2 and remainder r(x)= -2x + 4

Now, p(x) = g(x) . q(x) + r(x)

⇒ g(x)-q(x) = p(x)- r(x)

= (x3 – 3x2 + x + 2) – (-2x + 4)

= x3 – 3x2 + x + 2 + 2x- 4

= x3 – 3x2 + 3x- 2

⇒ g(x) = \(\frac{x^3-3 x^2+3 x-2}{x-2}\) (∵ q (x) = x-2)

Polynomials On Dividing Polynomial G Of X

Therefore, g(x) = x2-x+1

Question 5. Give examples of polynomials p(x), g(A), q(x) and r(x), which satisfy the division algorithm and

  1. Deg P(x) = deg q(x)
  2. Deg q(x) = deg r(x)
  3. Deg r(x) = 0

Solution:

1. deg p(x) = deg q(x)

We know that

p(x) = g(x) q(x) + r(x)

∴ degree of g(x) = zero

Let p(x) = 2x3 + 6x2 + 2x- 1

and g(x) = 2

Polynomials Give Examples Of Polynomials 1

∴ q(x) = x3 + 3x2 + x and r(x) = -1

So, p(x) = 2x3 + 6x2 + 2x- 1, = 2,

q(x) = x3 + 3x2 + x, r(x) = -1

degree q(x) = degree r(x)

We know that

p(x) = q(x) +r(x)

If q(x) and r(x) are polynomials of the first degree then a degree of p(x) must be 1 more than the degree: of g(x)

Let p(x) =  x3 + 3x2 + 2x + 5 and g(x) = x2 – 1

Polynomials Give Examples Of Polynomials 2

∴ q(x) = x + 3 and r(x) = 6x + 5

so, p(x) = x3 + 3x2 + 2x + 5, g(x) =x2-1

q(x) = x + 3, r(x) = 6x + 5

3. degree r(x) = 0

For this, the degree of g(x) must be 1.

Let p(x) = 2x3 – 3×2 +x+ 4 and g(x) = x- 1 .

Polynomials Give Examples Of Polynomials 3

∴ q(x) = 2x2 – x and r(x) = 4

Therefore, p(x) = 2x3 – 3x2 + x + 4, g(x) = x- 1,

q(x) = 2x2 – x, r(x) = 4

Polynomials Exercise 2.4 (Optional)

Question 1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case :

  1. \(2 x^3+x^2-5 x+2 ; \frac{1}{2}, 1,-2\)
  2. \(x^3-4 x^2+5 x-2 ; 2,1,1\)

Solution:

1. Let p(x) = \(2 x^3+x^2-5 x+2\)

∴ \(p\left(\frac{1}{2}\right)=2\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^2-5\left(\frac{1}{2}\right)+2\)

⇒ \(\frac{1}{4}+\frac{1}{4}-\frac{5}{2}+2=0\)

⇒ \(\frac{1}{2}\)is a zero of p(x).

Again ,p(1) = 2(1)3 + (1)2 – 5(1) + 2

=2 + 1 – 5 + 2 = 0

⇒ 1 is a zero of p(A).

Again p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2

= -16 + 4+10 + 2 = 0

⇒ -2 is a zero of p(x)

Now, in p(x) = 2x3 +x2 -5x + 2

a = 2,b = 1 ,c = -5, d = 2

⇒ \(-\frac{b}{a}=-\frac{1}{2}=\frac{1}{2}+1+(-2)=\text { sum of zeroes }\)

⇒ \(\frac{c}{a}=\frac{-5}{2}=\frac{1}{2} \times 1+1 \times(-2)+\frac{1}{2} \times(-2)\)

= sum of the product of zeroes taken two at a time.

and \(-\frac{d}{a}=\frac{-2}{2}=-1\)

⇒ \(\frac{1}{2} \times 1 \times(-2)=\text { product of zeroes}\)

Therefore the relations between the zeroes of the polynomial and the coefficients are true.

2. Let p(x) = x3 – 4x2 + 5x -2

p(2) = (2)3 – 4(2)2 + 5(2)-2

= 8-16+10-2 = 0

⇒ 2 is a zero of p(x).

Again P(1) = (1)3 – 4(1)2 + 5(1)-2

= 1-4 + 5- 2 = 0

= 1 is a zero of p(x).

Now, for p(x) = x3 – 4x2 + 5x- 2

a= 1, b =-4, c = 5, d = -2

∴ \(-\frac{b}{a}=-\frac{(-4)}{1}=4=2+1+1=\text { sum of zeroes. }\)

⇒ \(\frac{c}{a}=\frac{5}{1}=5=2 \times 1+1 \times 1+1 \times 2\)

= sum of the product of zeroes taken two at a time.

and \(-\frac{d}{a}=-\frac{(-2)}{1}=2=2 \times 1 \times 1\)

= product of zeroes.

Therefore the relations between the zeroes of the polynomial and the coefficients are true.

Question 2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, and -14 respectively.
Solution:

Given

The sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, and -14 respectively

Let the zeroes be α, β, and γ.

∴ α + β + γ = 2

αβ + βγ + γα = -7 and αβγ = -14

∴ Cubic polynomial = (x-α)(x-β)(x-γ)

= x3 – (α+ β + γ)x2+ (αβ + βγ + γα)x- αβPγ

= x3-2x2-7x+ 14

Question 3. If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Solution:

Given

x3 – 3x2 + x + 1

a – b, a, a + b, are zeroes of polynomial.

Let p(x) = x2 – 3x2 +x + 1

⇒ A= 1, B =-3, C = 1,D = 1

sum of zeroes = \(-\frac{B}{A}\)

a-b + a + a + b = \(-\frac{(-3)}{1}\)

3a = 3

a= 1

and product of zeroes = \(-\frac{D}{A}\)

⇒ \((a-b) \cdot a \cdot(a+b)=-\frac{1}{1}\)

⇒ \(a\left(a^2-b^2\right)=-1\)

⇒ \(1-b^2=-1\)

⇒ \(b^2=2\)

⇒ \(b= \pm \sqrt{2}\)

∴ \(a=1, b= \pm \sqrt{2}\)

Question 4. If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x- 35 are \(2 \pm \sqrt{3}\) , find other zeroes.
Solution:

Given

x4 – 6x3 – 26x2 + 138x- 35

Two zeroes of a polynomial are \(2+\sqrt{3} \text { and } 2-\sqrt{3} \text {. }\)

Let the remaining two zeroes be a and p

∴ \((x-\alpha)(x-\beta)(x-2-\sqrt{3})(x-2+\sqrt{3})\)

⇒ \(x^4-6 x^3-26 x^2+138 x-35\)

⇒ \(\quad(x-\alpha)(x-\beta)\left[(x-2)^2-(\sqrt{3})^2\right]\)

⇒ \(x^4-6 x^3-26 x^2+138 x-35\)

⇒ \(\quad(x-\alpha)(x-\beta)\left(x^2-4 x+4-3\right)\)

⇒ \(x^4-6 x^3-26 x^2+138 x-35\)

⇒ \(\quad(x-\alpha)(x-\beta)\)

⇒ \(\quad\frac{x^4-6 x^3-26 x^2+138 x-35}{x^2-4 x+1}\)

Polynomials If Two Zeroes Of The Polynomial

∴ (x -α) (x- β) = x2 – 2x- 35 = x2– 7x + 5x- 35

= x(x- 7) + 5(x- 7)

= (x- 7)(x + 5)

The other zeroes are α = 7, β = -5

Question 5. If the polynomial x4 – 6x3 + 16x2-25x+10 is divided by another polynomial x2-2x + k. the remainder comes out to be x + a, find k and a.
Solution:

Given

x4 – 6x3 + 16x2-25x+10 and x2-2x + k

Let p(x) = x4 – 6x3 + 16x2 – 25x + 10 divisor g(x) = x2 -2x + k and remainder r(x) = x + a

Polynomials If The Polynomial Is Divided By Another Polynomial

According to the problem,

(2k – 9)x +(10-8k + k2) = x + a Comparing the coefficient of x

2k-9= 1

⇒ 2k = 1 + 9= 10

⇒ k = 5

Comparing the constant terms

10 – Sk + k2 = a

⇒ 10-40 + 25 = a  (put K = 5)

⇒ a = -5

So, k = 5 and a =-5

Polynomials Multiple Choice Questions

Question 1. If one zero of the polynomial 3x2 + x- k is 3 then the value of k is :

  1. -30
  2. -24
  3. 30
  4. 24

Answer: 3. 30

Question 2. A polynomial with zeroes 2 and -3 is:

  1. x2 -x- 6
  2. x2 + x- 6
  3. x2 – 6
  4. x2 + 6

Answer: 2. x2 + x- 6

Question 3. The number of polynomials with zeroes 4 and 3, is :

  1. 1
  2. 2
  3. 3
  4. Infinite

Answer: 4. Infinite

Question 4. The zeroes of x2 + 6a + 5 are:

  1. Both positive
  2. Both negative
  3. Both equal
  4. One zero is zero

Answer: 2. Both negative

Question 5. If two zeroes of the polynomial ax3 + bx2 + cx + d are zero, then third zero is :

  1. \(-\frac{b}{a}\)
  2. \(\frac{b}{a}\)
  3. \(\frac{c}{a}\)
  4. \(-\frac{c}{a}\)

Answer: 1. \(-\frac{b}{a}\)

Question 6. If x6– 1 is divided by a polynomial of third degree, the maximum degree of the remainder can be:

  1. 0
  2. 1
  3. 2
  4. 3

Answer: 3. 2

Question 7.  The product of zeros of ax2 + bx + c, a≠0 is:

  1. \(-\frac{b}{a}\)
  2. \(\frac{b}{a}\)
  3. \(-\frac{c}{a}\)
  4. \(\frac{c}{a}\)

Answer: 4. \(\frac{c}{a}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 7 Co-Ordinate Geometry

Class 10 Maths Chapter 7 Co-Ordinate Geometry

(1) Vertices of the given triangle are

Co-Ordinate Geometry

  • Introduction: Co-ordinate geometry is the branch of mathematics in which geometry is studied algebraically, i.e., In which geometrical figures (as points, lines etc.) are studied using equations.
  • Rene Descartes (1596-1665), a French mathematician was the first who introduced the Co-ordinate Geometry or Analytical Geometry or Cartesian Geometry.
  • In class IX, we have studied how to locate the position of a point on a plane. Now, we will study to find the distance between two points, the section formula and the area of a triangle.

NCERT Exemplar Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

Distance Between Two Points

Let X’OX and Y’OY be coordinate axes. P and Q are two points in this cartesian plane with coordinates (X1, Y1) and (X2, Y2) respectively.

Co Ordinate Geometry Distance Between Two Objects

PL and QM are perpendiculars from P and Q respectively to the X-axis. PN is perpendicular from P to QM.

Now, O L =x1, P L=y1

O M =x2, Q M=y1

P N =L M

=O M-O L=x2-x1

P N=L M

and Q N =Q M-M N=Q M-P L

=y2 – y1

⇒ \(\triangle P Q N\) is a right-angled triangle.

⇒ \(P Q^2=P N^2+Q N^2\)

⇒ \(P Q^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2\)

⇒ \(P Q=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

Distance Between Origin And Point \(x_1, y_1\)

Distance between origin and point \(\left(x_1, y_1\right)=\sqrt{\left(x_1-0\right)^2+\left(y_1-0\right)^2}=\sqrt{x_1^2+y_1^2}\)

Condition Of Collinear Points On The Basis Of Distance

If the sum of any two distances is equal to the third distance, then the three points will be collinear.

Solved Examples

Example 1. Find the distance between the following points :

  1. (3, 4) and (5, 2)
  2. (0, 2) and (4, – 1)
  3. (a, 2a) and (- a, – 2a)
  4. (4, – 3) and (- 6, 5)

Solution.

(1) Distance between the points (3,4) and (5,2)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(5-3)^2+(2-4)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) units

(2) Distance between the points (0,2) and (4,-1)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(4-0)^2+(-1-2)^2}\)

=\(\sqrt{(4)^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}\)=5 units

(3) Distance between the points (n, 2a) and (-r,-2 a)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(-a-a)^2+(-2 a-2 a)^2}\)

=\(\sqrt{(-2 a)^2+(-4 a)^2}=\sqrt{4 a^2+16 a^2}=\sqrt{20 a^2}=2 \sqrt{5} a \)units

(4) Distance between the points (4,-3) and (-6,5)

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(-6-4)^2+(5+3)^2}\)

=\(\sqrt{(-10)^2+(8)^2}=\sqrt{100+64}=\sqrt{164}=2 \sqrt{41}\)units

Example 2. Find the distance between the points (5, 8) and (- 3, 2).

Solution:

Distance between the points (5, 8) and (- 3, 2).

=\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt{(-3-5)^2+(2-8)^2}\)

=\(\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=\sqrt{100}\)=10 units

The distance between the points =10 units

Example 3. Find the distance of the point (a cos \(\theta\), a sin \(\theta\)) from the origin.

Solution:

Distance between the points {a cos \(\theta\), a sin θ) and origin (0, 0)

= \(\sqrt{(a \cos \theta-0)^2+(a \sin \theta-0)^2}\)

= \(\sqrt{a^2 \cos ^2 \theta+a^2 \sin ^2 \theta}=\sqrt{a^2\left(\cos ^2 \theta+\sin ^2 \theta\right)}\)

= \(\sqrt{a^2} (\cos ^2 \theta+\sin ^2 \theta=1)\)

= a units

The distance of the point = a units

Example 4. Find the distance of the point (3, 4) from the origin.

Solution:

Distance of the point (3, 4) to the origin

= \(\sqrt{(3-0)^2+(4-0)^2}=\sqrt{9+16}=\sqrt{25}\)=5 units

The distance of the point (3, 4) from the origin=5 units

Example 5. If the distance between the points (x, 2) and (6, 5) is 5 units, find the value of x.

Solution:

Distance between the points (x, 2) and (6, 5)

=\(\sqrt{(6-x)^2+(5-2)^2}=\sqrt{x^2-12 x+36+9}=\sqrt{x^2-12 x+45}\)

Given that, \(\sqrt{x^2-12 x+45}\)=5

⇒ \(x^2-12 x+45=25\)

⇒ \(x^2-12 x+20=0\)

⇒ \(x^2-2 x-10 x+20=0\)

x(x-2)-10(x-2)=0

(x-2)(x-10)=0

x-2=0 or x-10=0

x = 2 or x = 10

The value of x 2 or 10

Example 6. If the distances of P(x,y) from A(5, 1) and B(-1,5) are equal, then prove that 3x = 2y.

Solution:

Since P(x,y) is equidistant from A(5, 1) and B(-1, 5).

PA = PB

⇒ \(\sqrt{(x-5)^2+(y-1)^2}=\sqrt{(x+1)^2+(y-5)^2}\) (by using distance formula)

Squaring both sides, we get

⇒ \((x-5)^2+(y-1)^2=(x+1)^2+(y-5)^2\)

⇒ \(x^2-10 x+25+y^2-2 y+1=x^2+2 x+1+y^2-10 y+25\)

-10 x-2 y+26=2 x-10 y+26

-10 x-2 x=-10 y+2 y

12 x=8 y

3 x = 2 y

Hence Proved.

Example 7. Prove that the points (5, -2), (-4, 3) and (10, 7) are the vertices of an isosceles right-angled triangle.

Solution:

Given

(5, -2), (-4, 3) and (10, 7)

Let the points are A (5, – 2), B (- 4, 3) and C (10, 7).

Therefore,\(A B^2=(-4-5)^2+(3+2)^2=(-9)^2+(5)^2=81+25=106\)

⇒ \(B C^2=(10+4)^2+(7-3)^2=(14)^2+(4)^2\)=196+16=212

\(A C^2=(10-5)^2+(7+2)^2=(5)^2+(9)^2\)=25+81=106

A B=A C=\(\sqrt{106}\)

and \(A B^2+A C^2=B C^2\)

⇒ \(\triangle A B C\) is an isosceles right-angled triangle.

Hence Proved.

Example 8. Prove that the points (a, a),(-a,-a) and \((-a \sqrt{3}, a \sqrt{3})\) are the vertices of an equilateral triangle.

Solution:

Given

(a, a),(-a,-a) and \((-a \sqrt{3}, a \sqrt{3})\)

Let the points are A(a, a), B(-a,-a) and \(C(-a \sqrt{3}, a \sqrt{3})\).

A B=\(\sqrt{(-a-a)^2+(-a-a)^2}=\sqrt{(-2 a)^2+(-2 a)^2}\)

= \(\sqrt{4 a^2+4 a^2}=\sqrt{8 a^2}=2 \sqrt{2} a\)

B C=\(\sqrt{(-a \sqrt{3}+a)^2+(a \sqrt{3}+a)^2}\)

=\(\sqrt{3 a^2+a^2-2 \sqrt{3} a^2+3 a^2+a^2+2 \sqrt{3} a^2}=\sqrt{8 a^2}=2 \sqrt{2} a\)

C A=\(\sqrt{(-a \sqrt{3}-a)^2+(a \sqrt{3}-a)^2}\)

=\(\sqrt{3 a^2+a^2+2 \sqrt{3} a^2+3 a^2+a^2-2 \sqrt{3} a^2}\)

=\(\sqrt{8 a^2}=2 \sqrt{2} a\)

A B =B C = C A

∴ \(\triangle A B C\) is an equilateral triangle.

Hence Proved.

Example 9. Prove that the points (2, – 1), (4, 1), (2, 3) and (0, 1) are the vertices of a square.

Solution:

Given

(2, – 1), (4, 1), (2, 3) and (0, 1)

Let the points are A(2, – 1), B(4, 1), C(2, 3) and D(0, 1).

⇒ \(A B^2 =(4-2)^2+(1+1)^2=4+4=8\)

A B =\(\sqrt{8}=2 \sqrt{2}\)

⇒ \(B C^2 =(2-4)^2+(3-1)^2=(-2)^2+(2)^2\)=4+4=8

B C =\(\sqrt{8}=2 \sqrt{2}\)

⇒ \(C D^2 =(0-2)^2+(1-3)^2=(-2)^2+(-2)^2\)=4+4=8

C D =\(\sqrt{8}=2 \sqrt{2}\)

⇒ \(D A^2 =(2-0)^2+(-1-1)^2=(2)^2+(-2)^2\)=4+4=8

D A =\(\sqrt{8}=2 \sqrt{2}\)

Co-ordinate Geometry

⇒ \(A C^2 =(2-2)^2+(3+1)^2\)=0+16=16

A C =\(\sqrt{16}=4\)

and \(B D^2 =(0-4)^2+(1-1)^2=16+0=16\)

B D =\(\sqrt{16}=4\)

Now, A B=B C=C D=D A and A C=B D

⇒  A B C D is a square.

Example 10. Show that the points A(- 3, 3), B(7, – 2) and C(l, 1) are collinear.

Solution:

Given

A(- 3, 3), B(7, – 2) and C(l, 1)

A B =\(\sqrt{(7+3)^2+(-2-3)^2}=\sqrt{(10)^2+(-5)^2}\)

=\(\sqrt{100+25}=\sqrt{125}=5 \sqrt{5}\)

B C =\(\sqrt{(1-7)^2+(1+2)^2}=\sqrt{(-6)^2+(3)^2}\)

= \(\sqrt{36+9}=\sqrt{45}=3 \sqrt{5}\)

and \(A C =\sqrt{(1+3)^2+(1-3)^2}=\sqrt{(4)^2+(-2)^2}\)

= \(\sqrt{16+4}=\sqrt{20}=2 \sqrt{5}\)

Now, A C+B C=2 \(\sqrt{5}+3 \sqrt{5}=5 \sqrt{5}=A B\)

Points A, B and C are collinear.

Example 11. Show that the points (9, – 2), (- 5, 12) and (- 7, 10) lie on that circle whose centre is the point (1,4).

Solution:

Given

(9, – 2), (- 5, 12) and (- 7, 10)

Let the given points are A (9, -2), B (-5, 12) and C (- 7, 10),

If point‘O’is (1, 4), then

O A =\(\sqrt{(1-9)^2+(4+2)^2}=\sqrt{(-8)^2+(6)^2}\)

=\(\sqrt{64+36}=\sqrt{100}=10\)

O B =\(\sqrt{(1+5)^2+(4-12)^2}=\sqrt{(6)^2+(-8)^2}\)

= \(\sqrt{36+64}=\sqrt{100}=10\)

O C =\(\sqrt{(1+7)^2+(4-10)^2}=\sqrt{(8)^2+(-6)^2}\)

= \(\sqrt{64+36}=\sqrt{100}=10\)

OA = OB = OC

Point ‘O’ is equidistant from the points A, B and C.

Point (1,4) is the centre of that circle at which the points (9, -2), (-5, 12) and (- 7, 10) lie.

Hence Proved.

Example 12. In the given figure, \(\triangle\)ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.

Solution:

Given

In the given figure, \(\triangle\)ABC is an equilateral triangle of side 3 units.

Since B is at a distance of 3 units from A on the X-axis in the positive direction, so B will be 5 units away from the origin.

Co Ordinate Geometry The Coordinates Of The Other Two Vertices

So, B = (5, 0)

Let M be the mid-point of AB

⇒ \(A M=\frac{1}{2} A B=\frac{3}{2}\) units

AC = 3 Units

In right \(\triangle C M A\), by Pythagoras theorem

Co Ordinate Geometry The Coordinates Of The C By Using Pythagoras Theroem

⇒ \(C M^2 =A C^2-\Lambda M^2\)

= \(9-\frac{9}{4}=\frac{27}{4}\)

⇒  \(C M =\sqrt{\frac{27}{4}}=\frac{3 \sqrt{3}}{2}\)

So, the coordinates of C are (O M, M C)

C=(O A+A M, M C) .

C=\(\left(2+\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)=\left(\frac{7}{2}, \frac{3 \sqrt{3}}{2}\right)\)

Example 13. The coordinates of two vertices of an equilateral triangle are (0, 0) and (3,\(\sqrt{3}\)). Find the coordinates of the third vertex of the triangle.

Solution:

Given

The coordinates of two vertices of an equilateral triangle are (0, 0) and (3,\(\sqrt{3}\)).

Let the \(\triangle\) ABC be an equilateral triangle in which the coordinates of points B and C are (0, 0) and (3,3, \(\sqrt{3})\) respectively.

Let the coordinates of the third vertex A be (x,y).

Co Ordinate Geometry The Co Ordinates Of The Third Vertex Of The Triangle

In equilateral \(\triangle\)ABC

A B=A C=B C

⇒ \(A B^2=A C^2=B C^2\)

⇒ \(A B^2=A C^2\)

⇒ \((x-0)^2+(y-0)^2=(x-3)^2+(y-\sqrt{3})^2 \quad B(0,0)\)

⇒ \(x^2+y^2=x^2-6 x+9+y^2-2 \sqrt{3} y+3\)

⇒ \(6 x+2 \sqrt{3} y=12 \quad \Rightarrow \quad 3 x+\sqrt{3} y=6 \)

⇒ \(\sqrt{3} x+y=2 \sqrt{3} \quad y=2 \sqrt{3}-\sqrt{3} x\)

and \(A B^2=B C^2\)

⇒ \((x-0)^2+(y-0)^2=(3-0)^2+(\sqrt{3}-0)^2\)

⇒ \(x^2+y^2=9+3\)

⇒ \(x^2+(2 \sqrt{3}-\sqrt{3} x)^2=12\)

⇒ \(x^2+12+3 x^2-12 x=12\)

∴ \(4 x^2-12 x\)=0

4 x(x-3)=0

x=0 or x-3=0

x=0 or x=3

Put these values in eq. (1)

x=0, then y=2\( \sqrt{3}-0=2 \sqrt{3}\)

x=3 , y=2\( \sqrt{3}-3 \sqrt{3}=-\sqrt{3}\)

Co-ordinates of third vertex =(0,2 \(\sqrt{3}\)) or (\(3,-\sqrt{3})\)

Example 14. What point on the X-axis is equidistant from (7, 6) and (-3, 4)?

Solution:

Given

(7, 6) and (-3, 4)

We know that the y-co-ordinate of a point on the X-axis is always 0. So, let a point on the X-axis be P(x, 0) and let two given points be A(7, 6) and B(-3, 4).

According to the condition,

P A=P B

⇒ \( \sqrt{(x-7)^2+(0-6)^2}=\sqrt{(x+3)^2+(0-4)^2}\)

Squaring both sides, we have

⇒ \(x^2-14 x+49+36 =x^2+6 x+9+16\)

Required point is (3,0) 20 x =60 \(\Rightarrow x=3\)

Example 15. Find the equation of the set of all points which are twice as far from (3, 2) as from (1, 1).

Solution:

Given

(3, 2) and (1, 1)

Let P(x,y) be a point and let A(3, 2) and B = (1, 1) be two other points on the plane, such that

P A=2 P B

⇒ \(\sqrt{(x-3)^2+(y-2)^2}=2 \sqrt{(x-1)^2+(y-1)^2}\)

Squaring both sides, we have

⇒ \(x^2-6 x+9+y^2-4 y+4=4\left(x^2-2 x+1+y^2-2 y+1\right)\)

∴ \(3 x^2+3 y^2-2 x-4 y-5=0\) Which is the required equation.

Example 16. If A(2, 2), B(-2, -2), C(-2\(\sqrt{3}\),2\(\sqrt{3}\)) and D(-4 – 2\(\sqrt{3}\),4 + 2\(\sqrt{3}\)) are the co-ordinates of 4 points. What can be said about these four points?

Solution:

Given

(2, 2), B(-2, -2), C(-2\(\sqrt{3}\),2\(\sqrt{3}\)) and D(-4 – 2\(\sqrt{3}\),4 + 2\(\sqrt{3}\)) are the co-ordinates of 4 points.

A B=\(\sqrt{(-2-2)^2+(-2-2)^2}=4 \sqrt{2}\) units .

B C=\(\sqrt{(-2+2 \sqrt{3})^2+(-2-2 \sqrt{3})^2}\)

=\(\sqrt{4+12-8 \sqrt{3}+4+12+8 \sqrt{3}}=4 \sqrt{2}\) units

C D=\(\sqrt{(-2 \sqrt{3}+4+2 \sqrt{3})^2+(2 \sqrt{3}-4-2 \sqrt{3})^2}\)

= \(\sqrt{16+16}=4 \sqrt{2}\) units

A C=\(\sqrt{(2+2 \sqrt{3})^2+(2-2 \sqrt{3})^2}\)

=\(\sqrt{4+12+8 \sqrt{3}+4+12-8 \sqrt{3}}=4 \sqrt{2}\) units

A D=\(\sqrt{(2+4+2 \sqrt{3})^2+(2-4-2 \sqrt{3})^2}\)

=\(\sqrt{36+12+24 \sqrt{3}+12+4+8 \sqrt{3}}=\sqrt{64+32 \sqrt{3}}\) units

B D=\(\sqrt{(-2+4+2 \sqrt{3})^2+(-2-4-2 \sqrt{3})^2}\)

=\(\sqrt{4+12+8 \sqrt{3}+36+12+24 \sqrt{3}}=\sqrt{64+32 \sqrt{3}}\) units

Here, AB = BC = CD = AC and also, AD = BD

So, in first view, it seems to be the vertices of a square.

(But),Here, AB, BC, CD and DA are not equal, (the order of A, B, C and D must be cyclic in the case of the square).

Also, AD and BD are equal but they cannot be diagonals. So, they do not form a square. Actually, A, B and D lie on a circle with C as the centre (as CA = CB = CD i.e., C is equidistant from A, B and D).

To Divide A Line Segment In a Given Ratio

Internal Division

The coordinates of a point which divides the line segment joining the points A(x1,y1) and B(x2,y2) in the ratio m: n internally are

⇒ \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)

Co Ordinate Geometry To Divide A Line Segment In Given Ratio By Internal Division

Proof: Let X’OX and Y’OYare the coordinate axes and A (x1,y1) and B (x2,y2) be any two points in this cartesian plane. Let P(x,y) be any point on line segment AB which divides AB in the ratio m: n.

⇒ \(\frac{A P}{B P}=\frac{m}{n}\)

AL, PN and BM are perpendiculars from, P and B respectively on X-axis.

AH and PIC are perpendiculars from and P respectively to PN and BM.

Now, OL = x1, OM = x2, ON = x

AL = y1, BM = y2> PN =y

AH = LN = ON -OL=x-x

PH = PN – NH = PN – AL =y -y1

Pk = NM = OM – ON =x2-x

and BIC = BM – ICM = BM – PN =y2 -y

Here, \(\triangle\)AHP and \(\triangle\)PKB are similar.

⇒  \(\frac{A H}{P K}=\frac{P H}{B K} =\frac{A P}{P B}\)

⇒  \(\frac{x-x_1}{x_2-x}=\frac{y-y_1}{y_2-y}=\frac{m}{n}\)

⇒  \(\frac{x-x_1}{x_2-x} =\frac{m}{n}\) and \(\frac{y-y_1}{y_2-y}=\frac{m}{n}\)

⇒  \(\frac{x-x_1}{x_2-x} =\frac{m}{n} \quad \Rightarrow \quad n x-n x_1=m x_2-m x\)

⇒  \(m x+n x =m x_2+n x_1 \quad \Rightarrow \quad(m+n) x=m x_2+n x_1\)

x =\(\frac{m x_2+n x_1}{m+n}\)

Similarly,\(\frac{y-y_1}{y_2-y}=\frac{m}{n} \quad \Rightarrow \quad n y-n y_1=m y_2-m y\)

⇒  \(m y+n y=m y_2+n y_1 \quad \Rightarrow \quad(m+n) y=m y_2+n y_1\)

y=\(\frac{m y_2+n y_1}{m+n}\)

Therefore, the co-ordinates of point P=\(\left(\frac{m x_2+m x_1}{m+n}, \frac{m y_2+m y_1}{m+n}\right)\)

Co-ordinates of the Mid-point of a Line Segment

The co-ordinates of the mid-point of the line segment joining the points A(x1,y1) and (x2,y2) are \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Co Ordinate Geometry Co Ordinate Midline Of A Segment

Proof: Let P (x, y) be the mid-point of the line joining the points A (x1,y1) and B (x2, y2)

Ap : PB = 1:1

x =\(\frac{1\left(x_2\right)+1\left(x_1\right)}{1+1}=\frac{x_1+x_2}{2}\)

y =\(\frac{1\left(y_2\right)+1\left(y_1\right)}{1+1}=\frac{y_1+y_2}{2}\)

Co-ordinates of point P =\(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\)

Solved Examples

Example 1. Find the coordinates of a point which divides the line segment joining the points (5, 1) and (-10, 11) in the ratio 2 : 3 internally.

Solution:

Given

(5, 1) and (-10, 11)

Let the coordinates of the required point he (x,y).

Here, \(\quad\left(x_1, y_1\right)=(5,1)\) and \(\left(x_2, y_2\right)=(-10,11)\)

and \(\left(x_1, y_1\right)\) =(5,1) and \(\quad\left(x_2, y_2\right)=(-10,11)\)

m: n = 2: 3

x =\(\frac{m x_2+n x_1}{m+n}=\frac{2(-10)+3(5)}{2+3}=\frac{-20+15}{5}=\frac{-5}{5}=-1\)

y =\(\frac{m y_2+n y_1}{m+n}=\frac{2(11)+3(1)}{2+3}=\frac{22+3}{5}=\frac{25}{5}=5\)

Co-ordinates of required point =(-1,5)

Example 2. Suppose point P lies on the line segment joining points A{-3, 4) and B(- 2, – 6) such that 2AP=3BP then, find the coordinates of point P.

Solution:

A{-3, 4) and B(- 2, – 6) such that 2AP=3BP

Given That, 2 AP = 3 BP

⇒  \(\frac{A P}{B P}=\frac{3}{2}\)

m : n = 3 : 2

(x1, y1) = (- 3, 4) and (x2,y2) = (-2,- 6)

Now, let the coordinates of point P are (x,y)

x=\(\frac{3(-2)+2(-3)}{3+2}=\frac{-12}{5}\)

and y=\(\frac{3(-6)+2(4)}{3+2}=-2\)

Co-ordinates of point P=\(\left(\frac{-12}{5},-2\right)\)

Example 3. Find the co-ordinates of the mid-point of the line segment joining the points A(3, -5) and B(1, 1 )

Solution:

Co-ordinates of the mid-point of AB = \((-1 + 3 5 + (-1)^ \left(\frac{3+1}{2}, \frac{-5+1}{2}\right)=(2,-2)\)

Example 4. The coordinates of the endpoints of a diameter are (-1, 5) and (3, -1). Find the coordinates of the centre and the radius of the circle.

Solution:

Given

The coordinates of the endpoints of a diameter are (-1, 5) and (3, -1).

Let the coordinates of the end points of diameter AB be A (-1, 5) and .8 (3, -1).

Co-ordinates of the centre P = Co-ordinates of mid-point of AB

= \(\left(\frac{-1+3}{2}, \frac{5+(-1)}{2}\right)=(1,2)\)

and radius of circle = length of P A

= \(\sqrt{(1+1)^2+(2-5)^2}=\sqrt{4+9}=\sqrt{13}\) units

Example 5. The coordinates of the mid-point of the line joining the points A and B are (2, -3). If the coordinates of point A are (-3, 4), then find the coordinates of point B.

Solution:

Given

The coordinates of the mid-point of the line joining the points A and B are (2, -3). If the coordinates of point A are (-3, 4),

Let the co-ordinates of point B=\(\left(x_2, y_2\right)\)

Given that,\(\frac{-3+x_2}{2}=2 \quad \Rightarrow \quad-3+x_2=4\)

⇒ \(x_2=7\) and \(\frac{4+y_2}{2}=-3\)

4+y_2=\(-6 \quad \Rightarrow \quad y_2=-10\)

Co-ordinates of point B=(7,-10)

Example 6. Find the ratio in which the X-axis divides the line segment joining the points (8, 5) and (-3,-7).

Solution:

Let the X-axis divide the join of points (8, 5) and (- 3, – 7) in the ratio k: 1.

We know that at X-axis

y = 0

⇒ \(\frac{k \cdot y_2+1 \cdot y_1}{k+1} =0\)

Co Ordinate Geometry The Ratio Of X Axis Divides The Line Segment Joining The Points

⇒ \(\frac{k(-7)+1(5)}{k+1}\) =0

-7 k+5 =0

k = \(\frac{5}{7}\)

Required ratio =5: 7

Example 7. In what ratio does the point \(\left(\frac{24}{11}, y\right)\) divide the line segment joining the points P(2,-2) and Q(3, 7)? Also, find the value of y.

Solution:

Let \(M\left(\frac{24}{11}, y\right)\) divide the line segment joining the points.

P(2,-2) and Q(3,7) in the ratio k: 1.

⇒ \(\frac{24}{11}=\frac{k(3)+1(2)}{k+1}\) (by using section formula)

⇒ \(11(3 k+2)=24(k+1) \quad \Rightarrow \quad 33 k+22=24 k+24\)

33 k-24 k=24-22 \(\quad \Rightarrow \quad 9 k=2\)

⇒ \(k=\frac{2}{9}\)

Required ratio =k: 1

i.e.,\(\frac{2}{9}: 1\)

i.e.,2: 9 internally.

Required ratio = 2:9

Example 8. The co-ordinates of the vertices of \(\triangle\)ABC are A(3, 2), B( 1, 4) and C(-1, 0). Find the length of the median drawn from point A.

Solution:

Given

The co-ordinates of the vertices of \(\triangle\)ABC are A(3, 2), B( 1, 4) and C(-1, 0).

Let AP be the median drawn from vertex A.

The midpoint of BC is P.

Co Ordinate Geometry The Length Of Median Drawn From The Point A

Now, the coordinates of P

=\(\left(\frac{1+(-1)}{2}, \frac{4+0}{2}\right)=(0,2)\)

A P =\(\sqrt{(3-0)^2+(2-2)^2}\)

=\(\sqrt{9+0}=3 units\)

Example 9. Find the coordinates of the points of trisection of the line joining the points (3, -2) and (-3, -4).

Solution:

Let P (a, b) and Q (c, d) trisect the line joining the points A(3, -2) and B(-3, -4).

Co Ordinate Geometry The Coordinates Of The Points Of Trisection Of The Line Segment

Now, point P (a, b) divides the line AB in the ratio 1:2

a =\(\frac{1(-3)+2(3)}{1+2}=\frac{3}{3}=1\)

b =\(\frac{1(-4)+2(-2)}{1+2}=-\frac{8}{3}\)

Therefore, coordinates of point P=\(\left(1,-\frac{8}{3}\right)\)

Q(c, d) divides the line A B in the ratio 2: 1.

c=\(\frac{2(-3)+1(3)}{2+1}=-1\)

d=\(\frac{2(-4)+1(-2)}{2+1}=\frac{-10}{3}\)

Therefore, co-ordinates of Q=\(\left(-1, \frac{-10}{3}\right)\)

Co-ordinates of points of trisection of A B=\(\left(1, \frac{-8}{3}\right) and \left(-1, \frac{-10}{3}\right)\)

Example 10. If two adjacent vertices of a parallelogram are (3, 2) and (-1, 0) and telagona intersect at (2, -5), then find the coordinates of the other two vertices.

Solution:

Given

Two adjacent vertices of a parallelogram are (3, 2) and (-1, 0) and telagona intersect at (2, -5)

Let two adjacent vertices of a parallelogram be A = (3, 2) and B(-1,0).

Co Ordinate Geometry Two adjacent Vertices Of A Parallelogram And The Diagonals Intersect

Let the coordinates of the other two vertices be C(X1, y1) and D(x2, y2)

We know that diagonals of a parallelogram bisect each other.

Mid-point of AC and the mid-point of BD are the same i.e., point O(2, -5).

⇒ \(\frac{3+x_1}{2}=2 and \frac{2+y_1}{2}\)=-5

⇒ \(x_1=1 and y_1=-12 \Rightarrow C \equiv\left(x_1, y_1\right) \equiv(1,-12)\)

Also, \(\frac{x_2-1}{2}=2 and \frac{y_2+0}{2}=-5\)

x2=5 and y2=-10 ⇒ D = (5,-10)

Hence, the remaining vertices are (1,-12) and (5,-10).

Example 11. The coordinates of three consecutive vertices of a parallelogram are (-1,0), (3, 1) and (2, 2). Find the coordinates of the fourth vertex of the parallelogram.

Solution:

Given

The coordinates of three consecutive vertices of a parallelogram are (-1,0), (3, 1) and (2, 2).

Let A(-1, 0), B (3, 1), C (2, 2) and D (x,y) be the vertices of a parallelogram ABCD.

We know that the diagonals of a parallelogram bisect each other.

Co-ordinates of the mid-point of AC = Co-ordinates of the mid-point of BD.

⇒  \(\left(\frac{-1+2}{2}, \frac{0+2}{2}\right)=\left(\frac{3+x}{2}, \frac{1+y}{2}\right)\)

⇒ \(\left(\frac{1}{2}, 1\right)=\left(\frac{3+x}{2}, \frac{1+y}{2}\right)\)

⇒ \(\frac{1}{2}=\frac{3+x}{2} and \quad 1=\frac{1+y}{2}\)

1=3+x and 2=1+y

x=-2 and y=1

Co-ordinates of fourth vertex =(-2,1)

Example 12. In which ratio the line y-x+2 = 0 Divides the line segment joining the points (3, -1) and (8,9)

Solution:

Let the ratio the line y-x+2 = 0 Divides the line segment joining the points (3, -1) and (8,9) in the ratio k: 1

Co-ordinates of P \(\equiv\left(\frac{8 k+3}{k+1}, \frac{9 k-1}{k+1}\right)\)

Co Ordinate Geometry Ratio Of The Line Divides The Line Segment Joining The Points

but this point P lies on the line y-x+2=0

⇒\(\frac{9 k-1}{k+1}-\frac{8 k+3}{k+1}+2\) =0

9 k-1-8 k-3+2 k+2 =0

3 k =2

k =\(\frac{2}{3}\)

Required ratio =\(\frac{2}{3}: 1=2: 3\)

Example 13. Find a point on the line through A{5, – 4) and B(-3, 2), that is, twice as far from A as from B.

Solution:

Given

A{5, – 4) and B(-3, 2)

Let P(x,y) be a point on AB such that

Co Ordinate Geometry A Point Of The Line Through Twice As Far From A As B

PA = 2PB

P A=2 P B

⇒  \(\frac{P A}{P B}=\frac{2}{1} \quad \Rightarrow \quad P A: P B=2: 1\)

So, by using the section formula for internal division,

x=\(\frac{m x_2+n x_1}{m+n}, y=\frac{m y_2+n y_1}{m+n}\)

x=\(\frac{2(-3)+1(5)}{2+1} \quad \Rightarrow \quad x=-\frac{1}{3}\)

and y=\(\frac{2(2)+1(-4)}{2+1} \Rightarrow \quad y=0\)

Required point =\(\left(-\frac{1}{3}, 0\right)\)

But, This is not the end of this question.

Think: Is it not possible that P(x, y) divides AB externally in the ratio 2: 1?

So, by using the section formula for external division,

x=\(\frac{2(-3)-1(5)}{2-1} \quad \Rightarrow x=-\frac{11}{1} \quad \Rightarrow \quad x=-11\)

and \( y=\frac{2(2)-1(-4)}{2-1} \Rightarrow y=\frac{8}{1} \quad \Rightarrow \quad y=8\)

So, the coordinates of p are (-11,8) also.

Hence, required points are \(\left(-\frac{1}{3}, 0\right)\) and (-11,8).

Example 14. Find the centroid of the triangle whose vertices are A(-1, 0), B{5, -2) and C(8, 2).

Solution:

Given

A(-1, 0), B{5, -2) and C(8, 2)

Centroid, the point where the medians of a triangle intersect, divides each median in the ratio 2: 1. Let AD be the median and G{x,y) be the centroid of \(\triangle\)ABC.

Co Ordinate Geometry The Centroid Of The Triangle Of The Vertices

D is the mid-point of BC

D =\(\left(\frac{5+8}{2}, \frac{-2+2}{0}\right)\) (mid-point formula)

=\(\left(\frac{13}{2}, 0\right)\)

Now, G(x, y) divides the line segment joining A(-1,0) and \(D(\frac{13}{2}\), 0) internally in the ratio 2: 1.

So, by using section formula,

x=\(\frac{2\left(\frac{13}{2}\right)+1(-1)}{2+1} \quad \Rightarrow \quad x=4\)

y=\(\frac{2(0)+1(0)}{2+1} \quad \Rightarrow \quad y=0\)

and Centroid of \(\triangle A B C\)=(4,0)

Example 15. A line intersects the Y-axis and X-axis at points P and Q respectively. If (2, -5) is the mid-point of PQ, then find the coordinates of P and Q.

Solution:

Given

A line intersects the Y-axis and X-axis at points P and Q respectively. If (2, -5) is the mid-point of PQ

We know that at Y-axis, the x-coordinate is zero and at X-axis, y-coordinate is zero.

Co Ordinate Geometry A Line Intersect The Y Axis And X Axis At The Points

So, let P = (0,y)

and Q = (x, 0)

Since M(2, -5) is the mid-point of PQ.

By using the mid-point formula,

\(\frac{0+x}{2} =2, \quad \frac{y+0}{2}\)=-5

x =4, y=-10

P = (0,-10) and Q=(4,0)

Example 16. Point P(h, k) divides a line segment between the axes in the ratio 1: 2. Find the lengths (intercepts) on the axes made by this line segment. Also, find the area of the triangle formed by the line segment and the axes.

Solution:

Given

Point P(h, k) divides a line segment between the axes in the ratio 1: 2.

Let AB be the line segment joining A (0,y) and B(x, 0) between the axes.

Co Ordinate Geometry The area Of The Triangle Formed By The Line Segment And The Axes

P(h, k) divides the line segment in the ratio 1: 2.

Now, question arises that whether PA : PB = 1 : 2 or PB : PA = 1:2

The answer is that we always take the former part of the ratio towards the X-axis and the latter part of the ratio towards the Y-axis.

So, here we will take PB: PA = 1: 2

By using the section formula,

h=\(\frac{1(0)+2(x)}{1+2} \Rightarrow h=\frac{2 x}{3} \quad \Rightarrow \quad x=\frac{3 h}{2}\)

and k=\(\frac{1(y)+2(0)}{1+2} \Rightarrow k=\frac{y}{3} \Rightarrow y\)=3 k

So, length of intercept on the X-axis =O B=x=\(\frac{3 h t}{2}\)

and the length of intercept on the Y-axis =O A=y=3 k.

Area of \(\triangle A O B=\frac{1}{2} \times O B \times O A=\frac{1}{2} \times \frac{3 h}{2} \times 3 k=\frac{9}{4}\) hk square units

Area Of Triangle:

If the coordinates of the vertices of a triangle are [x1,y1), (x2,y2) and (x3,y3), then the area of a triangle (A) is given by

⇒ \(\Delta=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

Co Ordinate Geometry The Co Ordinates Of The Vertices Of The Triangle

or \(\Delta=\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]\)

Proof: Let X’OX and Y’OY be co-ordinate axes.

Let the co-ordinates of the vertices of \(\triangle\)ABC are A (x1,y1), B (x2,y2) and C (x3,y3). Draw the 1 perpendicular AM, BN and CL from A, B and C. Y-axis.

□ ALMLC, □ CLNB and □ AMNB are trapezium.

Now, the Area of AABC

= Area of □ AMLC + Area of □ CLNB – Area of □ AMNB

= \(\frac{1}{2}(A M+L C) \times M L+\frac{1}{2}(C L+B N) \times L N-\frac{1}{2}(A M+B N) \times M N\)

=\(\frac{1}{2}\left(y_1+y_3\right)\left(x_3-x_1\right)+\frac{1}{2}\left(y_3+y_2\right)\left(x_2-x_3\right)-\frac{1}{2}\left(y_1+y_2\right)\left(x_2-x_1\right)\)

= \(\frac{1}{2}\left[\left(y_1+y_3\right)\left(x_3-x_1\right)+\left(y_3+y_2\right)\left(x_2-x_3\right)-\left(y_1+y_2\right)\left(x_2-x_1\right)\right]\)

=\(\frac{1}{2}\left[y_1 x_3-y_1 x_1+y_3 x_3-y_3 x_1+y_3 x_2-y_3 x_3+y_2 x_2-y_2 x_3-y_1 x_2+y_1 x_1-y_2 x_2+y_2 x_1\right] \)

=\(\frac{1}{2}\left[x_1 y_2-x_1 y_3+x_2 y_3-x_2 y_1+x_3 y_1-x_3 y_2\right]\)

=\(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

=\(\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_1\right)-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]\)

If the area of a triangle is negative then we neglect the negative sign, because the area of a triangle is always positive.

If it is given that the area of a triangle is 10 then it will be taken ± 10 for calculations to evaluate the value (s) of unknown terms.

Collinear Points

Three points A (x1, y1 )B (x2,y2) and C (x3, y3) are collinear if Area of \(\triangle\)ABC = 0

Note: If the order of description of the boundary is anticlockwise, then the area is considered to be positive, but if the order of description is clockwise, then the area is considered to be negative.

Co Ordinate Geometry Collinear Points In Anticlockwise And Clockwise Direction

Remark: To move from A to 6 and then from B to C, we are moving in a clockwise direction. So the above area comes out to be negative.

Co Ordinate Geometry Collinear Points In Clockwise And Anticlockwise Direction

Now, if we take the points in an anticlockwise direction, as A(2,1), C(6,3) and B(4,5) then

area =\(\frac{1}{2}[2(3-5)+6(5-1)+4(1-3)]\)

= \(\frac{1}{2}(-4+24-8)\)=+6 square units.

Solved Examples

Example 1. Find the area of the triangle, whose vertices are (2,1), (4,5) and (6, 3).

Solution:

Given

Vertices are (2,1), (4,5) and (6, 3)

Area of triangle = \(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

= \(\frac{1}{2}[2(5-3)+4(3-1)+6(1-5)]=\frac{1}{2}(4+8-24)\)=-6

But the area of a triangle cannot be negative

Area of triangle = 6 square units

Example 2. Find the area of a triangle, whose vertices are (2, 3), (7, 5) and (-7, -5).

Solution:

Given

Vertices are (2, 3), (7, 5) and (-7, -5)

  • Area of triangle = =\(\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\)]
  • = \(\frac{1}{2}[2(5+5)+7(-5-3)-7(3-5)]=\frac{1}{2}(20-56+14)\)=-11
  • But the area of a triangle cannot be negative
  • Area of triangle = 11 square units

Example 3. Find the area of the triangle, whose vertices are (a, c +a), (a,c) and (-a, c, -b).

Solution:

Given

Vertices are (a, c +a), (a,c) and (-a, c, -b)

  • Area of triangle =\(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)
  • = \(\frac{1}{2}[a(c-c+a)+a(c-a-c-a)-a(c+a-c)]\)
  • = \(\frac{1}{n}\left[a^2-2 a^2-a^2\right]=-a^2\)
  • But the area of a triangle cannot be negative
  • Area of triangle = \(a^2\) square units

Example 4. Prove that the points (6, 4) (4, 5) and (2, 6) are collinear.

Solution:

Given

(6, 4) (4, 5) and (2, 6)

Area of triangle = \(\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\)

⇒  \(\frac{1}{2}[6(5-6)+4(6-4)+2(4-5)]=\frac{1}{2}[-6+8-2]=0\)

Therefore, the given points are collinear.

Example 5. If the points A{x,y), B( 1, 4) and C(-2, 5) are collinear, then show that x + 3y = 13.

Solution:

Given points are collinear

Area of triangle = 0

⇒ \(\frac{1}{2}[x(4-5)+1(5-y)-2(y-4)]\) =0

-x+5-y-2 y+8 =0

x+3 y =13

Example 6. For what value of T, the points (k, 1), (1,-1) and (11,4) are collinear?

Solution:

The given points will be collinear if the area of the triangle = 0

⇒ \(\frac{1}{2}[k(-1-4)+1(4-1)+11(1+1)]\) =0

-5 k+3+22 =0

5 k =25

k =5

Example 7. If a \(\neq\) b \(\neq 0\), prove that the points \(\left(a, a^2\right),\left(b, b^2\right)\),(0,0) will not be collinear.

Solution:

Let the 3 points A=\(\left(a, a^2\right), B=\left(b, b^2\right)\) and C=(0,0) form a triangle ABC.

Area of triangle A B C = \(\frac{1}{2}\left|a\left(b^2-0\right)+b\left(0-a^2\right)+0\left(a^2-b^2\right)\right|\)

= \(\frac{1}{2}\left[a b^2-a^2 b\right]=\frac{1}{2} a b(b-a) \neq 0\)

Since a\((\triangle A B C) \neq 0\).

So, \(\triangle A B C\) will be formed.

Therefore 3 points A, B and C will not be collinear.

Hence Proved.

Example 8. If (x,y) is any point on the line segment joining the points (a, 0) and (0, b), then prove that \(\frac{x}{a}+\frac{y}{b}\)=1

Solution:

Given three points are collinear

Area of \(\triangle\)=0

⇒ \(\frac{1}{2}|x(0-b)+u(b-y)+0(y-0)|\)=0

-bx + ab – ay=0

Divide both sides by ab

–\(\frac{x}{a}+1-\frac{y}{b}=0 \quad \Rightarrow \quad \frac{x}{a}+\frac{y}{b}\)=1

Example 9. If P be a point equidistant from points A(3, 4) and B( 5, -2) and area \(\triangle\)PAB is 10 square units, then find the coordinates of point P.

Solution:

Let the coordinates of point P be (x,y).

Now, PA = PB

⇒ \(\sqrt{(x-3)^2+(y-4)^2}=\sqrt{(x-5)^2+(y+2)^2}\)

⇒ \(x^2+9-6 x+y^2+16-8 y=x^2+25-10 x+y^2+4+4 y^{\prime}\)

4 x-12 y=4

x-3 y=1  → Equation 1

and area of Δ=10

⇒ \(\frac{1}{2}[x(4+2)+3(-2-y)+5(y-4)]= \pm 10\)

⇒ \(6 r-6-3 y+5 y-20= \pm 20\)

6x + 2y – 26= \(\pm\) 20

6 x+2 y-26=20 or 6 x+2 y-26=-20

3 r+y=23  → Equation 2

or 3 r+y=3  →  Equation 3

From eqs. (1) and (2) From eqs. (1) and (3)

x=7, y=2 , x=1, y=0

Required co-ordinates =(7,2) or (1,0)

Example 10. The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is \(\left(\frac{7}{2}, y\right)\), find the value of y.

Solution :

Areat of a \(\triangle A B C\)=5 sq units (given)

Co Ordinate Geometry The Value Of Y By Using The Vertices

⇒ \(\frac{1}{2}\left|2(-2-y)+3(y-1)+\frac{7}{2}(1+2)\right|=5\)

⇒ \(\left|-4-2 y+3 y-3+\frac{21}{2}\right|=10\)

⇒ \(y+\frac{7}{2}= \pm 10\)

⇒ \(y+\frac{7}{2}=10\)

y=\(10-\frac{1}{7}\)or \(y=-10-\frac{1}{7}\)

y=\(\frac{13}{2}\) or y=\(-\frac{27}{2}\)

Hence, the value of y can be \(\frac{13}{2}\) and \(-\frac{27}{2}\).

Co-Ordinate Geometry Miscellaneous Examples

Example 1. If the point P(k – 1, 2) is equidistant from the points A(3, K) and B(k, 5), find the values of k.

Solution:

Given that. P A=P B

⇒ \((k-1-3)^2+(2-k)^2=(k-1-k)^2+(2-5)^2\)

⇒ \((k-4)^2+(2-k)^2=(-1)^2+(-3)^2\)

⇒ \(k^2-8 k+16+4+k^2-4 k=1+9\)

⇒ \(2 k^2-12 k+10=0\)

⇒ \(k^2-6 k+5=0\)

⇒ \(k^2-5 k-k+5=0\)

⇒ \(k(k-5)-1(k-5)=0\)

(k-5)(k-1)=0

k-5=0 or k-1=0

k=5 or k=1

The values of k is 5 or 1.

Example 2. Find the point on X-axis which is equidistant from the points (5, -2) and (-3, 2).

Solution:

Let the required point on the X-axis be P(x, 0) and the given points be (5, -2) and B(—3, 2).

Now, given that

P A =P B

⇒ \(P A^2=P B^2\)

⇒ \((x-5)^2+(0+2)^2 =(x+3)^2+(0-2)^2\)

⇒ \(x^2-10 x+25+4 =x^2+6 x+9+4\)

-16 x =-\(16 \quad \Rightarrow x\)=1

The required point is (1, 0).

Example 3. Points and B(5, 7) lie on a circle uadi centre 0(2, -3y). Find the values of y. Hence, find the die radius of the die circle.

Solution:

Given

Points and B(5, 7) lie on a circle uadi centre 0(2, -3y).

Here, OA = OB (radii of a circle)

O A =O B

O A² =O B²

Co Ordinate Geometry The Radius Of The Circle By Using The Value Y

⇒ \((2+1)^2+(-3 y-y)^2=(2-5)^2+(-3 y-7)^2\)

⇒ \(9+16 y^2=9+9 y^2+42 y+49\)

⇒ \(7 y^2-42 y-49=0\)

⇒ \(y^2-6 y-7=0\)

⇒ \(y^2-7 y+y-7=0\)

⇒ y(y-7)+1(y-7)=0

⇒ y-7=0 or y+1=0

y=7 or y=-1

Now, the co-ordinates of centre O are either (2, -21) or (2, 3) when the centre is 0(2, -21), then radius = OB

= \(\sqrt{(2-5)^2+(-21-7)^2}\)

= \(\sqrt{9+784}=\sqrt{793}\) units

When centre is 0(2, 3), then

radius = OB

= \(\sqrt{(2-5)^2+(3-7)^2}\)

= \(\sqrt{9+16}=\sqrt{25}\)=5 units

The die radius of the die circle =5 units

Example 4. The points A(-1, 7), B{p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B. Find the value of p.

Solution:

Given

The points A(-1, 7), B{p, 3) and C(7, 3) are the vertices of a right triangle, right-angled at B.

In \(\triangle\)ABC

⇒ \(\angle\)B = 90°

⇒ \(A B^2+B C^2=A C^2\)

⇒ \((p-4)^2+(3-7)^2+(p-7)^2+(3-3)^2=(7-4)^2+(3-7)^2\)

⇒ \(p^2-8 p+16+p^2-14 p+49+0=9\)

⇒ \(2 p^2-22 p+56=0\)

⇒ \(p^2-11 p+28=0\)

⇒ \(p^2-7 p-4 p+28=0\)

⇒ \(p(p-7)-4(p-7)=0\)

⇒ \((p-7)(p-4)=0\)

p-7=0 or p-4=0

p=7 or p=4

when p=7, then the points B and C coincide and so no triangle is formed.

⇒ \(p \neq 7\)

Hence, p=4

Example 5. Find the coordinates of the point of trisection of the line segment joining the points A(-5, 6) and B(4, -3).

Solution:

Let P and Q be the points of trisection of AB, and then P divides AB in the ratio 1:2.

Co-ordinates of point P \(\equiv\left(\frac{-5 \times 2+4 \times 1}{1+2}, \frac{6 \times 2+1 \times-3}{1+2}\right) \equiv(-2,3)\) and point Q divides A B in the ratio 2: 1.

Co-ordinates of point Q=\(\left(\frac{-5 \times 1+4 \times 2}{2+1}, \frac{6 \times 1+2 \times-3}{1+2}\right)\)=(1,0)

The coordinates of the point of trisection are (-2,3) and (1,0).

Example 6. Find the ratio in which the point P(x, 2) divides the line segment joining the points. A(12, 5) and B(4, -3). Also, find the value of. v.

Solution:

Let point P(x, 2) divide AB in the ratio k: 1.

x=\(\frac{12 \times 1+k \times 4}{k+1}\) and 2=\(\frac{5 \times 1+k \times-3}{k+1}\)

Now, 2=\(\frac{5-3 k}{k+1}\)

2 k+2=\(-3 k+5 \quad \Rightarrow \quad 5 k=3\)

k=\(\frac{3}{5}\)

Required ratio =3: 5

and x=\(\frac{4 k+12}{k+1}=\frac{4 \times \frac{3}{5}+12}{\frac{3}{5}+1}=\frac{12+60}{3+5}=9\)

x=9

Example 7. Find The lengths of the medians AD and BE of \(\triangle\)ABC whose vertices are A( 7, 3),B(5, 3) and C(3, -1 ).

Solution:

Co-ordinates of mid-point D of BC

= \((\frac{3+5}{2}\), \(\frac{-1+3}{2})=(4,1)\)

Co Ordinate Geometry The Length Of The Medians And Their Vertices Of Triangle ABC

Ciomodinates of mid-point E of AC :

= \(\left(\frac{3+7}{2}, \frac{-1-3}{2}\right)=(5,-2)\)

A D=\(\sqrt{(7-4)^2+(-3-1)^2}=\sqrt{9+16}=\sqrt{25}\)=5 units

and \(B E=\sqrt{(5-5)^2+(3+2)^2}=\sqrt{0+25}=\sqrt{25}=5 units\)

Example 8. If the points A(-1, -4), B(b,c) and C(5, -1) are collinear and 2b + c = 4, find the values of b and c.

Solution:

Points A(- 1, -4), B(b, c) and C(5, -1) are collinear

area of \(\triangle\)ABC = 0

-1(c + 1)+ b(-1 -2 + 4) + 5(-4-c) = 0

– c – 1 + 3b – 20 -5c = 0

3b – 6c = 21

b -2c – 7 …(1)

Given that, 2b + c = 4 (2)

from eqs. (1) and (2), we get

b = 3 and c = -2

Co-Ordinate Geometry Exercise 7.1

Question 1. Find the distance between the following pairs of points :

  1. (2, 3), (4,1)
  2. (-5, 7), (-1,3)
  3. (a, b), {-a, -b)

Solution :

(1) Distance between the points (2, 3) and (4, 1)

= \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)

= \(\sqrt{(4-2)^2+(1-3)^2}\)

= \(\sqrt{(2)^2+(-2)^2}\)

= \(\sqrt{(4+4)}=\sqrt{8}=2 \sqrt{2}\) units

(2) Distance between the points (-5, 7) and (-1, 3)

= \(\sqrt{(-1+5)^2+(3-7)^2}\)

= \(\sqrt{(4)^2+(-4)^2}\)

= \(\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) units

(3) Distance between the points (a, b) and {-a, -b).

= \(\sqrt{(-a-a)^2+(-b-b)^2}\)

= \(\sqrt{(-2 a)^2+(-2 b)^2}\)

= \(\sqrt{4 a^2+4 b^2}=\sqrt{4\left(a^2+b^2\right)}\)

=2 \(\sqrt{a^2+b^2}\) units

Question 2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B where town A is located 36 km east and 15 km north of town

Solution :

Distance between the points (0, 0) and (36, 15)

= \(\sqrt{(36-0)^2+(15-0)^2}\)

= \(\sqrt{1296+225}\)

=\(\sqrt{1521}\)=39 units

We can find the distance between two cities. The coordinates of the given cities in the cartesian coordinate system are A = (0,0) and B (36, 15).

The distance between these points is AB = 39 1cm.

Question 3. Determine if the points (1, 5), (2, 3) and (- 2, -11) are collinear.

Solution :

Let the given points are A(l, 5), B{2, 3) and C(-2,-ll).

A B =\(\sqrt{(2-1)^2+(3-5)^2}\)

=\(\sqrt{1+4}=\sqrt{5}\) units

B C =\(\sqrt{(-2-2)^2+(-11-3)^2}\)

=\(\sqrt{16+196}=\sqrt{212}\) units

C A =\(\sqrt{(-2-1)^2+(-11-5)^2}\)

=\(\sqrt{9+256}=\sqrt{265}\) units

Now, AB+B C \(\neq\) C A

Given points are not collinear.

Question 4. Check whether (5, -2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.

Solution :

Let the given points are A(5, -2), B{6, 4) and C(7, -2).

Now, A B =\(\sqrt{(6-5)^2+(4+2)^2}\)

=\(\sqrt{1+36}=\sqrt{37}\) units

B C =\(\sqrt{(7-6)^2+(-2-4)^2}\)

=\(\sqrt{1+36}=\sqrt{37}\) units

In a \(\triangle\)ABC,AB = BC

⇒ \(\triangle\)ABC is an isosceles triangle.

Therefore, the given points are the vertices of an isosceles triangle.

Question 5. In a classroom, 4 friends are seated at points A, B, C and D as shown in the figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using the distance formula, find which of them is correct

Co Ordinate Geometry Square Graph

Solution :

In the figure, the coordinates of A, B, C and D are (3, 4), (6, 7), (9, 4) and (6, 1) respectively.

A B =\(\sqrt{(6-3)^2+(7-4)^2}\)

= \(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} units\)

B C =\(\sqrt{(9-6)^2+(4-7)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)units

C D =\(\sqrt{(6-9)^2+(1-4)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) units

D A =\(\sqrt{(3-6)^2+(4-1)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) units

A C =\(\sqrt{(9-3)^2+(4-4)^2}\)

=\(\sqrt{36+0}=\sqrt{36}=6\)units

B D =\(\sqrt{(6-6)^2+(1-7)^2}\)

=\(\sqrt{0+36}=\sqrt{36}=6 units\)

Now, AB =BC = CD = DA and AC = BD

ABCD is a square.

So, Champa is correct.

Question 6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:

  1. (-1,-2). (1,0), (-1,2), (-3,0)
  2. (-3,5). (3, 1), (0,3), (-1,-4)
  3. (4, 5). (7. 6), (4, 3), (1,2)

Solution:

(1) Let points be A (-1, -2), (1,0), C (-1,2) and D (-3,0).

A B =\sqrt{(1+1)^2+(0+2)^2}

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} units\)

B C =\(\sqrt{(-1-1)^2+(2-0)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2} units\)

C D =\(\sqrt{(-3+1)^2+(0-2)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\)units

D A =\(\sqrt{(-1+3)^2+(-2-0)^2}\)

=\(\sqrt{4+4}=\sqrt{8}=2 \sqrt{2}\) units

A C =\(\sqrt{(-1+1)^2+(2+2)^2}\)

=\(\sqrt{0+16}=\sqrt{16}\)=4 units

B D=\(\sqrt{(-3-1)^2+(0-0)^2}\)

=\(\sqrt{16+0}=\sqrt{16}\)=4 units

AB = BC = CD = DA and AC = BD

ABCD is a square.

Therefore, given points are the vertices of a square.

(2) Let given points are A(-3, 5), B(3, 1), C(0, 3) and D(-l, -4).

A B =\(\sqrt{(3+3)^2+(1-5)^2}\)

=\(\sqrt{36+16}=\sqrt{52}\)

=\(2 \sqrt{13}\)units

B C =\(\sqrt{(0-3)^2+(3-1)^2}\)

=\(\sqrt{9+4}=\sqrt{13} units\)

A C =\(\sqrt{(0+3)^2+(3-5)^2}\)

=\(\sqrt{9+4}=\sqrt{13}\) units

Now, BC + A C =\(\sqrt{13}+\sqrt{13}\)

=2 \(\sqrt{13}\)=A B

A, B and C are collinear.

Therefore, no quadrilateral will be formed from the given points.

(3) Let given points are A(4, 5), B(7, 6), C(4, 3) and D( 1,2).

A B=\(\sqrt{(7-4)^2+(6-5)^2}\)

=\(\sqrt{9+1}=\sqrt{10}\) units

B c =\(\sqrt{(4-7)^2+(3-6)^2}\)

=\(\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\) units

C D =\(\sqrt{(1-4)^2+(2-3)^2}\)

=\(\sqrt{9+1}=\sqrt{10}\) units

=\(\sqrt{(4-1)^2+(5-2)^2}\)

=\(\sqrt{9+9}=\sqrt{18}\)

=3 \(\sqrt{2}\) units

A C =\(\sqrt{(4-4)^2+(3-5)^2}\)

=\(\sqrt{0+4}=\sqrt{4}\)=2 units

B D =\(\sqrt{(1-7)^2+(2-6)^2}\)

=\(\sqrt{36+16}=\sqrt{52}\)

=2 \(\sqrt{13}units\)

AB = CD, AD = BC and AC \(\neq\) BD

ABCD is a parallelogram.

Therefore, given points are the vertices of a parallelogram.

Question 7. Find the point on the A-axis which is equidistant from (2, -5) and (-2, 9).

Solution :

Let any point of the A-axis be P(x, 0). The distance of this point from the points A(2, -5) and B(-2, 9) are equal.

PA = PB

P A^2 =P B^2

⇒ \((x-2)^2+(0+5)^2 =(x+2)^2+(0-9)^2\)

⇒ \(x^2-4 x+4+25 =x^2+4 x+4+81\)

⇒ \(x^2-4 x-x^2-4 x =4+81-4-25\)

-8x = 56

x = -7

Coordinates of required point = (-7, 0).

Question 8. Find the values of y for which the distance between the points P(2, – 3) and Q(10,jV is 10 units.

Solution :

According to the problem, PQ = 10

⇒ \(P^2=100 \)

⇒ \((10-2)^2+(y+3)^2=100\)

⇒ \(64+(y+3)^2=100\)

⇒ \((y+3)^2=100-64\)

⇒ \((v+3)^2=36\)

⇒ \((y+3)= \pm 6\)

y+3=6 or y+3=-6

y=3 or y=-9

Required values of y = 3, – 9

Question 9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of .r. Also, find the distances QR and PR.

Solution :

Given that, Q is equidistant from P and R.

QP = QR

Q P = Q R

⇒ \(Q P^2=Q R^2\)

⇒ \((5-0)^2+(-3-1)^2=(x-0)^2+(6-1)^2\)

⇒ \(25+16=x^2+25\)

⇒ \(x^2=6 \Rightarrow x= \pm 4\)

⇒ \(Q R=\sqrt{( \pm 4-0)^2+(6-1)^2}\)

= \(\sqrt{16+25}=\sqrt{41}\) units

and P R =\(\sqrt{(5 \pm 4)^2+(-3-6)^2}\)

= \(\sqrt{(5+4)^2+81}\) or \(\sqrt{(5-4)^2+81}\)

= \(\sqrt{81+81}\) or \(\sqrt{1+81}\)

= \(\sqrt{162}\) or \(\sqrt{82}\)

P R =9 \(\sqrt{2} units\) or \(\sqrt{82} units\).

Question 10. Find a relation between x and; such that the point (x,y) is equidistant from the point (3, 6) and (- 3, 4).

Solution :

Given that, the distance of point {x,y) to (3, 6) = distance of point {x,y) to (-3, 4).

⇒ \(\sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}\)

⇒ \((x-3)^2+(y-6)^2\)

= \((x+3)^2+(y-4)^2\)

⇒ \(x^2-6 x+9+y^2-12 y+36\)

=\(x^2+6 x+9+y^2-8 y+16\)

⇒ \(0=x^2+6 x+9+y^2-8 y+16-x^2 +6 x-9-y^2+12 y-36\)

12x- + 4y – 20 = 0

3x + y – 5 = 0

3x + y = 5

Co-Ordinate Geometry Exercise 7.2

Question 1. Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.

Solution:

Let A=(-1,7),B=(4,-3)

Let p(x, y) divide AB in the ratio 2: 3.

A(-1,7), P(x, y),B(4,-3)

⇒ \((x_1, y_1)=(-1,7),(x_2, y_2)=(4,-3)\)

and m: n=2: 3

Now, x=\(\frac{m x_2+m x_1}{m+n}\)

x=\(\frac{2 \times 4+3 \times(-1)}{2+3}=\frac{8-3}{5}=\frac{5}{5}\)=1

and y=\(\frac{m y_2+n y_1}{m+n}\)

y=\(\frac{2 \times(-3)+3 \times 7}{2+3}=\frac{-6+21}{5}=\frac{15}{5}\)=3

Coordinates of required point =(1,3).

Question 2. Find the coordinates of the points of trisection of the line segment joining (4,-1) and (-2,-3).

Solution :

Given (4,-1) and (-2,-3)

Let A = (4, – 1) and B = (-2, – 3).

Let P and Q trisect the line segment AB.

Co Ordinate Geometry The Coordinates Of The Points Of Trisection Of The Line Segment

AP:PB= 1:2 and

AQ:QB = 2 : 1

For point P, x=\(\frac{m x_2+n x_1}{m+n}\)

x=\(\frac{1(-2)+2(4)}{1+2}=\frac{-2+8}{3}=\frac{6}{3}=2\)

and y=\(\frac{m r_2+n y_1}{m+n}\)

⇒ \(y=\frac{1(-3)+2(-1)}{1+2}=\frac{-3-2}{3}=\frac{-5}{3}\)

Coordinates of point P \(\equiv\left(2, \frac{-5}{3}\right)\)

For point Q, \(x^{\prime}=\frac{m_1 x_2+n_1 x_1}{m_1+n_1}\)

⇒ \(x^{\prime}=\frac{2(-2)+1(4)}{2+1}=\frac{-4+4}{3}=0\)

and \(y^{\prime}=\frac{m_1 y_2+n_1 y_1}{m_1+n_1}\)

⇒ \(y^{\prime}=\frac{2(-3)+1(-1)}{2+1}=\frac{-6-1}{3}=\frac{-7}{3}\)

Coordinates of point \(Q=\left(0, \frac{-7}{3}\right)\)

Question 3. To conduct Sports Day activities, in your rectangular-shaped school ground ABCD, lines have been drawn with chalk powder at a distance of lm each. 100 flower pots have been placed at a distance of lm from each other along AD, as shown in the figure. Niharika runs \(\frac{1}{4}\)th the distance AD on the 2nd line and posts a green flag.

Preet runs \(\frac{1}{5}\) through the distance AD on the eighth line and posts a red flag. What is the distance between both flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?

Co Ordinate Geometry The Distance Beyween The Two Persons From The Rectangular Shaped Ground

Solution :

100 flower pots are placed on side AD at a distance of 1 m from each other.

AD = 100 m

Position of Niharika’s flag = distance of \(\frac{1}{4}\)th part of AD in second row

⇒ \(\left(2,100 \times \frac{1}{4}\right)=(2,25)\)

Position of Preet’s flag = distance of \(\left(\frac{1}{5}\right)\) th part of AD in eighth row

= \(\left(8,100 \times \frac{1}{5}\right)=(8,20)\)

Distance between flags

= \(\sqrt{(8-2)^2+(20-25)^2}\)

= \(\sqrt{36+25}=\sqrt{61} \mathrm{~m}\)

Coordinates of the mid-point of (2,25) and (8,20)

= \(\left(\frac{2+8}{2}, \frac{25+20}{2}\right)=\left(5, \frac{45}{2}\right)\)

So, Rashmi should post her flag in the 5th row along AD at a distance of \(\frac{45}{2} \mathrm{~m}\).

Question 4. Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (- 1, 6).

Solution :

Given

The line segment joining the points (-3, 10) and (6, – 8) is divided by (- 1, 6)

Let the points (-1, 6) divide the line segment joining the points (-3, 10) and (6, -8) in the ratio m: 1.

⇒ \(-1=\frac{m(6)+1(-3)}{m+1}\)

and \(6=\frac{m(-8)+1(10)}{m+1}\)

⇒ \(-m-1=6 m-3 and 6 m+6=-8 m+10\)

-7 m=-2 and 14 m=4

m=\(\frac{2}{7}\) and m=\(\frac{4}{14}=\frac{2}{7}\)

Required ratio = 2: 7.

Question 5. Find the ratio in which the line segment joining A (1, -5) and (-4, 5) is divided by the Ar-axis. Also, find the coordinates of the point of division.

Solution:

Given

(1, -5) and (-4, 5)

A=\((1,-5)=\left(x_1, y_1\right)\)

and \(B=(-4,5)=\left(x_2, y_2\right)\)

Let the x-axis divide the line segment A B at point (x, 0) in the ratio m: 1.

Now, 0=\(\frac{m \cdot y_2+1 \cdot y_1}{m+1}\)

⇒ \(0=m(5)+1(-5) \Rightarrow m=1\)

m: 1=1: 1

Required ratio =1: 1

Now, \(x=\frac{m x_2+1 \cdot x_1}{m+1}=\frac{1(-4)+1(1)}{1+1}\)

= \(\frac{-4+1}{2}=\frac{-3}{2}\)

Coordinates of point of division =\(\left(-\frac{3}{2}, 0\right)\)

Question 6. If (1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y-0

Solution:

Given

(1, 2), (4,y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order

Let the vertices of the parallelogram ABCD are A = (1, 2), B = (4, y), C = (x, 6) and D = (3, 5).

We know that the diagonals of a parallelogram bisect each other.

Coordinates of mid-point of AC = Coordinates of mid-point of BD

⇒ \(\left(\frac{1+x}{2}, \frac{2+6}{2}\right)=\left(\frac{4+3}{2}, \frac{y+5}{2}\right)\)

⇒ \(\frac{1+x}{2}=\frac{4+3}{2} and \frac{2+6}{2}=\frac{y+5}{2}\)

1+x=7 and 8=y+5

x=6 and y=3

Question 7. Find the coordinates of point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Solution :

Given

AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4)

Coordinates of center O = (2, -3)

Coordinates of point B = (1,4)

Let coordinates of point A = (h, k)

Now, coordinates of the mid-point of AB = coordinates of O

⇒ \(\left(\frac{h+1}{2}, \frac{k+4}{2}\right)=(2,-3)\)

⇒ \(\frac{h+1}{2}=2\) and \(\frac{k+4}{2}=-3\)

h+1=4 and k+4=-6

h=3 and k=-10

Coordinates of point A=(3,-10) .

 Question 8. If A and B are (- 2, -2) and (2, -4) respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the line segment AB.

Solution :

Given

A and B are (- 2, -2) and (2, -4) respectively

A P=\(\frac{3}{7} A B\)

7 A P =3 A B\( \quad \Rightarrow \quad 7 A P=3(A P+B P)\)

7 A P =3 A P+3 BP

4 A P =3 B P \(\quad \Rightarrow \quad \frac{A P}{B P}=\frac{3}{4}\)

⇒ \(A B: B P=3: 4 \quad \Rightarrow \quad m: n=3: 4\)

and \(A A =(-2,-2)=\left(x_1, y_1\right)\)

B =\((2,-4)=\left(x_2, y_2\right)\)

Let coordinates of point P=(x, y).

x=\(\frac{m x_2+n x_1}{m+n}\)

x=\(\frac{3 \times 2+4 \times(-2)}{3+4}=\frac{6-8}{7}=\frac{-2}{7}\)

and y=\(\frac{m y_2+n y_1}{m+n}\)

⇒ \(y=\frac{3 \times(-4)+4 \times(-2)}{3+4}\)

=\(\frac{-12-8}{7}=\frac{-20}{7}\)

Coordinates of point P=\(\left(\frac{-2}{7}, \frac{-20}{7}\right)\)

Question 9. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Solution:

Given

A(-2, 2) and B(2, 8)

Co Ordinate Geometry The Coordinates Of The Points Divide The Line segment Joining Into Four Equal Parts

Coordinates of mid-point Q of AB

=\(\left(\frac{-2+2}{2}, \frac{2+8}{2}\right)=(0,5)\)

Coordinates of mid-point P of AQ

=\(\left(\frac{-2+0}{2}, \frac{2+5}{2}\right)=\left(-1, \frac{7}{2}\right)\)

Coordinates of mid-point R of QB

=\(\left(\frac{0+2}{2}, \frac{5+8}{2}\right)=\left(1, \frac{13}{2}\right)\)

Coordinates of the points dividing A and B into equal parts are

∴ \(\left(-1, \frac{7}{2}\right),(0,5),\left(1, \frac{13}{2}\right)\)

Question 10. Find the area of a rhombus in its vertices as (3, 0), (4, 5), (— 1, 4) and diagonals (-2.-1) taken in order. [Hint : Area of a rhombus \(\frac{1}{2}\)(product of its diagonals)]

Solution :

Given

(3, 0), (4, 5), (— 1, 4) and diagonals (-2.-1)

Let the vertices of a rhombus be in the following order :

A =(3,0), B=(4,5),

C =(-1,4), D=(-2,-1)

A C=\(\sqrt{(-1-3)^2+(4-0)^2}\)

= \(\sqrt{16+16}=\sqrt{32}=4 \sqrt{2}\) units

B D =\(\sqrt{(-2-4)^2+(-1-5)^2}\)

= \(\sqrt{36+36}=\sqrt{72}=6 \sqrt{2}\) units

Now, area of rhombus =\(\frac{1}{2} \times\) product of diagonals

= \(\frac{1}{2} \times A C \times B D\)

= \(\frac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}\)

=24 sq. units

The area of a rhombus =24 sq. units

Co-Ordinate Geometry Exercise 7.3

Question 1. Find the area of the triangle whose vertices are :

  1. (2, 3), (-1,0), (2,-4)
  2. (-5-1), (3,-5), (5, 2)

Solution :

(1) Vertices of the given triangle are

⇒ \(\left(x_1, y_1\right)=(2,3)\),

⇒ \(\left(x_2, y_2\right)=(-1,0)\),

⇒ \(\left(x_3, y_3\right)=(2,-4)\)

Area of triangle =\(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1)\)\(-(x_2 y_1+x_3 y_2+x_1 y_3)]\)

= \(\frac{1}{2}[\{2 \times 0+(-1) \times(-4)+2 \times 3\}\)-{(-1) \times 3+2 \times 0+2 \times(-4)\}][/latex]

= \(\frac{1}{2}[(0+4+6)-(-3+0-8)]\)

= \(\frac{1}{2}(10+11)=\frac{21}{2}\)

Area of triangle =\(\frac{21}{2}\) sq. units

(2) Vertices of triangle are

⇒ \(\left(x_1, y_1\right)=(-5,-1),\left(x_2, y_2\right)=(3,-5),\left(x_3, y_3\right)=(5,2)\)

=\(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1\) \(-(x_2 y_1+x_3 y_2+x_1 y_3)]\)

=\(\frac{1}{2}[\{(-5) \times(-5)+3 \times 2+5 \times(-1)\}-\{3 \times(-1)+5 \times(-5)+(-5) \times 2\}]\)

=\(\frac{1}{2}[(25+6-5)-(-3-25-10)]\)

=\(\frac{1}{2}(26+38)=32\)

Area of triangle =32 sq. units

Question 2. In each of the following find the value of ‘k’, for which the points are collinear.

  1. (7,-2), (5, 1), (3, k)
  2. (8,1),(k,-4),(2,-5)

Solution :

(1) Let the given points are as follows:

A=\(\left(x_1, y_1\right)=(7,-2), \quad B=\left(x_2, y_2\right)=(5,1)C=\left(x_3, y_3\right)=(3, k)\)

A, B and C are collinear.

Area of \(\triangle A B C\)=0

⇒ \(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1) – (x_2 y_1+x_3 y_2+x_1 y_3)]\)=0

⇒ \((x_1 y_2+x_2 y_3+x_3 y_1)\)

⇒ \(-(x_2 v_1+x_2 v_2+x_1 y_3)\) =0

⇒ \({7 \times 1+5 \times k+3 \times(-2)\}-\{5 \times(-2)+3 \times 1+7 \times k}\) =0

(7+5 k-6)-(-10+3+7 k) =0

(1+5 k)-(-7+7 k) =0

1+5 k+7-7 k =0

8-2 k=0 \(\Rightarrow k =4\)

(2) Let the given points are as follows:

A \(\equiv\left(x_1, y_1\right)=(8,1), B=\left(x_2, y_2\right)=(k,-4)\),

C=\(\left(x_3, y_3\right)=(2,-5)\)

A, B and C are collinear

Area of \(\triangle A B C\)=0

⇒ \(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1)\).

⇒ \(\left.-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right]\)=0

⇒ \(\left(x_1 y_2+x_2 y_3+x_3 y_1\right)\)

– \(\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\)=0

⇒ \({8 \times(-4)+k \times(-5)+2 \times 1}\)

⇒ \({k \times 1+2 \times(-4)+8 \times(-5)}\)=0

⇒ \((-32-5 k+2)-(k-8-40)\)=0

(-30-5 k)-(k-48)=0

-30-5 k-k+48=0

-6 k+18=0 \(\Rightarrow k=3\)

Question 3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Solution :

Given

(0, -1), (2, 1) and (0, 3)

Let vertices of \(\triangle\)ABC are

A=\(\left(x_1, y_1\right)=(0,-1), B=\left(x_2, y_2\right)=(2,1)\)

and C=\(\left(x_3, y_3\right)=(0,3)\).

Area of \(\triangle A B C\)

= \(\frac{1}{2}[(x_1 y_2+x_2 y_3+x_3 y_1\)

⇒ \(\left.-\left(x_2 y_1+x_3 y_2+x_1 y_3\right)\right\}\)

=\(\frac{1}{2}[\{0 \times 1+2 \times 3+0 \times(-1)\}\)

⇒ \(-\{2 \times(-1)+0 \times 1+0 \times 3\}]\)

=\(\frac{1}{2}[(0+6+0)-(-2+0+0)]\)

=\(\frac{1}{2}(6+2)\)=4 sq. units

Coordinates of mid-point P of AB

= \(\left(\frac{0+2}{2}, \frac{-1+1}{2}\right)=(1,0)\)

Coordinates of mid-point Q of BC

=\(\left(\frac{2+0}{2}, \frac{1+3}{2}\right)=(1,2)\)

Coordinates of mid-point R of CA

=\(\left(\frac{0+0}{2}, \frac{3-1}{2}\right)=(0,1)\)

Area of \(\triangle P Q R\)

= \(\frac{1}{2}[(1 \times 2+1 \times 1+0 \times 0) -(1 \times 0+0 \times 2+1 \times 1)]\)

= \(\frac{1}{2}[(2+1+0)-(0+0+1)\)

= \(\frac{1}{2}(3-1)\)=1 sq. unit

Now,\(\frac{\text { area of } \triangle P Q R}{\text { area of } \triangle A B C}=\frac{1}{4}\)=1: 4

Question 4. Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3,-2) and (2,3).

Solution:

Given

(-4, -2), (-3, -5), (3,-2) and (2,3)

Let the vertices of □ ABCD in order are as follows :

A=\(\left(x_1, y_1\right)=(-4,-2)\)

B=\(\left(x_2, y_2\right)=(-3,-5)\)

C=\(\left(x_3, y_3\right)=(3,-2)\)

D=\(\left(x_4, y_4\right)=(2,3)\)

Area of  Square A B C D

= \(\frac{1}{2}\left[\left(x_1 y_2+x_2 y_3+x_3 y_4+x_4 y_1\right)\right.\)

–\((x_2 y_1+x_3 y_2+x_4 y_3+x_1 y_4)]\)

= \(\frac{1}{2}[\{(-4) \times(-5)+(-3) \times(-2)+3 \times 3 +2 \times(-2)\}-\{(-3) \times(-2)+3 \times(-5)\) +\(2 \times(-2)+(-4) \times 3\}]\)

= \(\frac{1}{2}[(20+6+9-4)-(6-15-4-12)]\)

= \(\frac{1}{2}(31+25)\)=28 sq. units

Question 5. The median of a triangle divides it into two triangles of equal areas. Verify this result for \(\triangle\)ABC whose vertices are A(4, -6), B(3, -2) and C(5, 2).

Solution:

Conlinates of mid-point D of BC’

=\(\left(\frac{3+5}{2}, \frac{-2+2}{2}\right)=(1,0)\)

Area of \(\triangle ABD\)

=\(|\begin{array}{c} \frac{1}{2}\,\{4 \times(-2)+3 \times(1)+4 \times(-6)\} -\{3 \times(-6)+4 \times(-2)+4 \times(0) \mid \end{array}\|\)

=\(\left|\frac{1}{2}\right|(-8+0-2.4)-(-18-8+(0)||\)

=\(\left|\frac{1}{2}(-32+20)\right|=|-3|\)=3 sq. units

Area of \(\triangle ACD\) е:

\( =\begin{array}{r}
\frac{1}{2}, 1(4 \times 0+4 \times 2+5 \times(-6) \mid \\
-14 \times(-6)+5 \times 0+4 \times 2) \mid
\end{array}|\)

=  \(|\frac{1}{2}(0+8-3(0)-(-24+0+8)|\)

=\( \frac{1}{2}(-22+|(0)|=|-3|\)=3sq. units

⇒ I\(\triangle ABD\)I= \(\triangle ACD\)

Therefore, the median AD divides it into two triangles of equal areas. Hence Proved.

Co-Ordinate Geometry Exercise 7.4

Question 1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, – 2) and B(3, 7).

Solution :

Let the line 2x + y – 4 =0 divides the line segment joining the points A (2, -2) and P (3, 7) in the ratio m: 1 and the coordinates of the point of division are (x,y).

x=\(\frac{m(3)+1(2)}{m+1}=\frac{3 m+2}{m+1}\)

and y=\(\frac{m(7)+1(-2)}{m+1}=\frac{7 m-2}{m+1}\)

Point (x, y), lies on the line 2 x+y-4=0

⇒ \(2\left(\frac{3 m+2}{m+1}\right)+\frac{7 m-2}{m+1}-4=0\)

6 m+4+7 m-2-4(m+1)=0

13 m+2-4 m-4=0

9 m=2 \(\quad \Rightarrow \quad m=\frac{2}{9}\)

Required ratio =2: 9

Question 2. Find a relation between x and y if the points (x, y), (1,2) and (7, 0) are collinear.

Solution :

Given

Points (x, y), (1,2) and (7, 0) are collinear.

Area of \(\Delta\)=0

⇒ \(\frac{1}{2}[\{x \cdot 2+1 \cdot 0+7 \cdot y\}\)

–\(\{1 \cdot y+7 \cdot 2+0 \cdot x\}]\)=0

(2 x+7 y)-(y+14)=0

2 x+7 y-y-14=0

x+3 y-7=0

which is the required relation between x and y.

Question 3. Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Solution :

Given

(6, -6), (3, -7) and (3, 3)

Let the points A(6, -6), B(3, -7) and C(3, 3) lie on the circumference of a circle whose centre is P(h,k)

P A=P B=P C (radii of the circle )

⇒ \(P A^2=P B^2=P C^2\)

Now, vP A^2=P B^2[/latex]

Now, \(P A^2=P B^2\)

⇒ \((h-6)^2+(k+6)^2 =(h-3)^2+(k+7)^2\)

⇒ \(h^2-12 h+36+k^2 +12 k+36\)

= \(h^2-6 h+9+k^2+14 k+49\)

-12 h+6 h+12 k-14 k+72-58=0

-6 h-2 k+14=0

3 h+k=7

and \(P B^2=P^{\prime} C^2\)

⇒ \((h-3)^2+(k+7)^2=(h-3)^2+(k-3)^2\)

⇒ \((k+7)^2=(k-3)^2\)

⇒ \(k^2+14 k+49=k^2-6 k+9\)

14 k+6 k=9-49

20 k=-40

k=-2

Put the value of k in equation (1),

3 h-2 =7

h =3

Coordinates Of the centre of the circle = (3,2)

Question 4. The two opposite vertices of a MJU.UV ;uv (-1, 2) aiui (3, 2). Find the coordinates of the other two vertices.

Solution :

Given

The two opposite vertices of a MJU.UV ;uv (-1, 2) aiui (3, 2).

Let two opposite vertices of square ABCD be A(-1,2) and C(3, 2) which are known.

Let the coordinates of vertex B be (h, k).

Now, AB = BC (sides of the square)

⇒\(A B^2=B C^2\)

⇒ \((h+1)^2+(k-2)^2=(h-3)^2+(k-2)^2\)

⇒ \((h+1)^2=(h-3)^2\)

⇒ \(h^2+2 h+1=h^2-6 h+9\)

2 h+6 h=9-

8 h=8

h=1

⇒ \(Again, \quad A B^2+B C^2=A C^2 \quad\left( \angle B=90^{\circ}\right)\)

⇒ \((h+1)^2+(k-2)^2+(h-3)^2+(k-2)^2\)

=\((3+1)^2+(2-2)^2\)

⇒ \((1+1)^2+k^2-4 k+4+(1-3)^2+k^2-4 k+4\)

=16+0

⇒ \(4+2 k^2-8 k+8+4=16\)

⇒ \(2 k^2-8 k=0 \quad \Rightarrow 2 k(k-4)=0\)

k=0 or k=4

Therefore, the remaining two vertices of the square = (1,0) or (1,4)

Question 5. The Class X students of a secondary school in Krishinagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.

Co Ordinate Geometry The Students Are To Sow Seeds Of Flowering Plants On The Remaining Area Of The Plot

  1. What will be the coordinates of the vertices of A BQR if C is the origin?
  2. Also, calculate the areas of the triangles in these cases. What do you observe?

Solution :

(1) If A is the origin, then

P = (4. 6), Q=(3, 2), R=(6, 5)

Area of \(\triangle P Q R\)

=\(\frac{1}{2}[(4.2+3.5+6.6)-(3.6+6.2+4.5)] \)

=\(\frac{1}{2}[(8+15+36)-(18+12+20)]\)

=\(\frac{1}{2}(59-50)=\frac{9}{2}\) sq. units

(2) If C is the origin, then

P =(-12,-2), Q=(-13,-6),

R =(-10,-3)

Area of \(\triangle P Q R\)

= \(\frac{1}{2}[\{(-12)(-6)+(-13)(-3)+(-10)(-2)\}\)

–\(\{(-13)(-2)+(-10)(-6)+(-12)(-3)\}]\)

= \(\frac{1}{2}[(72+39+20)-(26+60+36)]\)

= \(\frac{1}{2}(131-122)=\frac{9}{2}\) sq. units

Area Of the Triangle is both in some cases

Question 6. The vertices of a \(\triangle A B C\) are A (4,6), B(I, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{4}\).

Calculate the area of the \(\triangle A D E\) AB AC 4 and compare it with the area of \(\triangle A B C\).

Solution:

Given

The vertices of a \(\triangle A B C\) are A (4,6), B(I, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively,

A=(4,6), B=(1,5), C=(7,2)

Co Ordinate Geometry The Area Of Triangle ADE And Compare With The Area Of Triangle ABC

Area of \(\triangle A B C\)

= \(\frac{1}{2}[\{4.5+1.2+7.6\}\)

–\(\{1.6+7.5+4.2\}]\)

= \(\frac{1}{2}[(20+2+42) \quad \overbrace{B(1.5)}^{A(4,6)} -(6+35+8)] \)

= \(\frac{1}{2}(64-49)=\frac{15}{2}\) sq. units

Given, \(\quad \frac{A D}{A B}=\frac{1}{4} \quad \Rightarrow \quad \frac{A D}{A D+B D}=\frac{1}{4}\)

4 A D=A D+B D

3 A D=B D \(\quad \Rightarrow \quad \frac{A D}{B D}=\frac{1}{3}\)=1: 3

Point d divides B in the ratio 1 :  3. Therefore, the coordinates of point D

=\(\left(\frac{1 \times 1+3 \times 4}{1+3}, \frac{1 \times 5+3 \times 6}{1+3}\right)\)

=\(\left(\frac{1+12}{4}, \frac{5+18}{4}\right)=\left(\frac{13}{4}, \frac{23}{4}\right)\)

Again,\(\frac{A E}{A C} =\frac{1}{4} \quad \Rightarrow \quad \frac{A E}{A E+E C}=\frac{1}{4}\)

4 A E =A E+E C \(\Rightarrow 3 A E=E C\)

⇒ \(\frac{A E}{E C} =\frac{1}{3}=1: 3\)

Point E divides A-C in the ratio 1: 3. Therefore, the coordinates of point E

= \(\left(\frac{1 \times 7+3 \times 4}{1+3}, \frac{1 \times 2+3 \times 6}{1+3}\right)\)

= \(\left(\frac{7+12}{4}, \frac{2+18}{4}\right)=\left(\frac{19}{4}, 5\right)\)

Now, A = (4,6), D=\((\frac{13}{4}, \frac{23}{4}), E \equiv(\frac{19}{4}, 5\)

Area of \(\triangle A D E\)

= \(\frac{1}{2} {\left[\left\{4 \cdot \frac{23}{4}+\frac{13}{4} \cdot 5+\frac{19}{4} \cdot 6\right\}\right.}\)

–\({\frac{13}{4} \cdot 6+\frac{19}{4} \cdot \frac{23}{4}+4.5}]\)

=\(\frac{1}{2}[(23+\frac{65}{4}+\frac{114}{4})-(\frac{78}{4}+\frac{437}{16}+20)]\)

⇒ \(\frac{1}{2}\left[\frac{92+65+114}{4}-\frac{312+437+320}{16}\right] \)

= \(\frac{1}{2}\left[\frac{271}{4}-\frac{1069}{16}\right]=\frac{1}{2}\left(\frac{1084-1069}{16}\right)\)

= \(\frac{15}{32}\) square units

Now, \(\frac{\text { Area of } \triangle A D E}{\text { Area of } \triangle A B C}=\frac{15 / 32}{15 / 2}=\frac{1}{16}\)=1: 16

Co-Ordinates Multiple Choice Questions

Question 1. The distance of the point (3,4) from the X-axis is :

  1. 3 units
  2. 4 units
  3. 5 units
  4. 1 unit

Answer:

Question 2. The distance of the point (8, -6) from the origin is :

  1. 10 units
  2. 6 units
  3. 8 units
  4. 14 units

Answer:

Question 3.  The perimeter of a triangle with vertices (0, 0), (-3, 0) and (0, ‘1) is:

  1. 14 units
  2. 7 units
  3. 1 unit
  4. 12 Units

Answer:

Question 4. The points (-1. 0), (-1, 0) and (0, It) are the vertices of:

  1. a right-angled triangle
  2. an isosceles triangle
  3. an equilateral triangle
  4. a scalene triangle

Question 5. A(7, 0), B(4, 0) and C(S, 4) are the vertices of \(\triangle ABC\). The area of this triangle is :

  1. 14
  2. 28
  3. 8
  4. 6

Answer:

Question 6. (A-2, 3), B(6, 7) and C(8, 3) are three vertices of a parallelogram ABCD. The coordinates of vertex D are :

  1. (0,1)
  2. (0,-1)
  3. (c) (-1, 0)
  4. D(1,0)

Answer:

Question 7. The perpendicular bisector of the line segment joining the point (1,5) and (4, 6) intersects the Y-axis at point:

(0,13)

  1. (0,-13)
  2. (0, 12)
  3. (13, 0)

Answer:

Question 8. The mid-point of line segment joining the points A(-4, 2) and B( 5, 6) is \(P\left(\frac{a}{8}, 4\right)\). Then the value of A is

  1. -8
  2. -4
  3. 2
  4. 4

Question 9. The distance of the point (-3, 5) from the Y-axis is :

  1. -3
  2. 2
  3. 5
  4. None of these

Answer:

Question 10. The distance between two points (2, 3) and (4, 1) will be :

  1. 2
  2. 2 \(\sqrt{3}\)
  3. 2\( \sqrt{2}\)
  4. 3

Answer:

Question 11. The distance between the points P(2, -3) and (3(10, y) is 10 units. The value of y will be

  1. -3, 9
  2. -9, 3
  3. 9, 3
  4. -9, 2

Answer:

Question 12. A point on X-axis, equidistant from the points A(2, -5) and 2, 9) will be :

  1. (-7, 0)
  2. (-6, 0)
  3. (-2, 0)
  4. (2, 0)

Question 13. The distance between the points (-1, -3) and (5, 2) is :

  1. \(\sqrt{61}\) units
  2. \(\sqrt{37} units\)
  3. \(\sqrt{17}\) units
  4. 3 units

Answer:

NCERT Exemplar Solutions for Class 10 Maths Chapter 12 Area Related To Circles

Area Related To Circles

Area Related To Circles Introduction

We are familiar with the shape of a circle. The circumference of the wheels of cars, and coins appears as a circle.

A circle can be defined as “A figure with an outline, every point on this outline is at the same constant distance from a certain point inside the figure, which is called the centre”.

NCERT Exemplar Solutions for Class 10 Maths Chapter 12 Area Related To Circles

Area Related To Circles Area And Circumference Of A Circle

Let r be the radius of the circle. The distance travelled once around a circle is its perimeter, usually called its circumference. Now

Area of circle = πr2

Circumference of circle = 2πr

Area Related To Circles Area And Circumference Of A Circle

π(pi) is a fixed irrational number whose approximate value is \(3.1416 \text { as } \frac{22}{7} \text { or } \frac{355}{113}\) (sometimes)

Diameter: A chord of a circle passing through its centre is called the diameter of the circle.

Diameter = 2 x radius

Area Related To Circles Formulae Of Circle

1. Area of circle = πr2
2. Circumference of a circle = 2πr

Area Related To Circles Area Of Circle

3. Area of semicircle = \(\frac{1}{2} \pi r^2\)

Area Related To Circles Area Of Semicircle

4. Perimeter of semicircle = (πr + 2r)

5. For a ring having outer radius = R and inner radius = r
Area of ring = π(R2 – r2)

Area Related To Circles Ring Having Outer Radius

6. For rotation of the hands of a clock

  1. Angle described by minute hand in 60 minutes = 360°
  2. Angle described by hour hand in 12 hours = 360°

Area Related To Circles Rotation Of The Hands Of A Clock

7. For rotating wheels

  1. Distance moved by a wheel in 1 rotation = its circumference
  2. Number of rotation in unit time = \(\frac{\text { distance moved in unit time }}{\text { circumference }}\)

Area Related To Circles Rotating Wheels

Area Related To Circles Solved Examples

Question 1. The circumference of a field is 220 m. Find

  1. Its radius
  2. Its area.

Solution:

Circumference of circle = 220 m

2πr = 220

⇒ \(r=\frac{220 \times 7}{2 \times 22}=35 \mathrm{~m}\)

Area of circle = \(\pi r^2=\frac{22}{7} \times 35 \times 35=3850 \mathrm{~m}^2\)

Question 2. Find the area of a circular park whose radius is 4.5 m.

Solution:

Area of circular park = r2

⇒ Area = \(\text { Area }=\frac{22}{7} \times 4.5 \times 4.5\)

Area = 63.63 m2

Question 3. The area of a circular plot is 346.5 m2. Calculate the cost of fencing the plot at the rate of ₹ 6 per metre.

Solutions:

Area of plot = 346.5 m2

⇒ πr2 = 346.5 m2 ⇒ \(r^2=\frac{346.5 \times 7}{22}\)

⇒ r2 = 110.25 ⇒ r = 10.5 m

Circumference of plot = \(2 \pi r=2 \times \frac{22}{7} \times 10.5=66 \mathrm{~m}\)

Cost of fencing = Circumference x Cost of fencing per metre

= ₹ 66 x 6 = ₹ 396

Question 4. The diameter of a cycle wheel is 28 cm. How many revolutions will it make in moving 13.21cm?

Solution:

Distance travelled by the wheel in one revolution = \(2 \pi r=\frac{22}{7} \times 28=88 \mathrm{~cm}\)

Total distance travelled by the wheel = 13.2 x 1000 x 100 cm

Number, of revolutions made by the wheel = \(\frac{\text { total distance }}{\text { circumference }}\)

= \(\frac{13.2 \times 1000 \times 100}{88}=15000 \text { revolutions }\)

Question 5. The circumference of a circle exceeds the diameter by 16.8 cm. Find the radius of the circle.

Solution:

Let the radius of the circle be r.

Diameter = 2r

Circumference of circle = 2πr

Using the given information, we have

2πr = 2r + 16.8

⇒ \(2 \times \frac{22}{7} \times r=2 r+16.8\)

⇒ 44r = 14r + 16.8 x 7

⇒ 30r = 117.6

∴ \(r=\frac{117.6}{30}=3.92 \mathrm{~cm}\)

Question 6. Find the area of a ring whose outer and inner radii are respectively 20cm and 1 5cm.

Solution:

Outer radius R = 20 cm

Inner radius r = 15 cm

∴ Area of ring = π(R2 – r2)

⇒ Area of ring = \(\frac{22}{7}\left[(20)^2-(15)^2\right]=\frac{22}{7}(400-225)=\frac{22}{7} \times 175 \mathrm{~cm}^2\)

= 22 x 25 = 550 cm2.

Question 7. A race track is in the form of a ring whose inner circumference is 352 m and the outer circumference is 396 m. Find the width of the track.

Solution:

Let R and r be the outer and inner radii of the circle.

The width of the track = (R – r) cm

Now, 2πr = 352 ⇒ \(2 \times \frac{22}{7} \times r=352\)

⇒ \(r=\frac{352 \times 7}{2 \times 22}=7 \times 8=56 \mathrm{~m}\)

Again, 2πR = 396 ⇒ \(2 \times \frac{22}{7} \times R=396\)

⇒ \(R=\frac{396 \times 7}{2 \times 22}=7 \times 9=63 \mathrm{~m}\)

∴ R = 63 m, r = 56 m

Width of the track = (R- r) m = (63 – 56) m = 7 m.

Area Related To Circles Outer And Inner Radii Of The Circle

Question 8. Two circles touch internally. The sum of their areas is 116π cm2 and the distance between their centres is 6 cm. Find the radii of the circles.

Solution:

Let two circles with centres O’ and O having radii R and respectively touch each other at P.

It is given that OO’ = 6

R – r = 6 ⇒ R = 6 + r → (1)

Also, πR2 + πr2 = 116π (given)

π(R2 + r2) = 116π

⇒ R2 + r2= 116 → (2)

∴ From equations (1) and (2), we get

(6 + r)2 + r2 = 116

⇒ 36 + r2 + 12r + r2= 116 ⇒ 2r2 + 12r – 80 = 0

⇒ r2 + 6r – 40 = 0 = (r + 10) (r – 4) = 0

∴ r = -10 or r = 4

But the radius cannot be negative. So, we reject r = -10

∴ r = 4 cm

∴ R = 6 + 4 = 10cm [from(1)]

Hence, the radii of the two circles are 4 cm and 10 cm.

Area Related To Circles Two Circles Touch Internally

Question 9. The radius of a wheel of a bus is 45 cm. Determine its speed in kilometres per hour, when its wheel makes 315 revolutions per minute.

Solution:

The radius of the wheel of the bus = 45 cm

∴ Circumference of the wheel = 2ar

= \(2 \times \frac{22}{7} \times 45=\frac{1980}{7} \mathrm{~cm}\)

∴ Distance covered by the wheel in one revolution = \(\frac{1980}{7} \mathrm{~cm}\)

Distance covered by the wheel in 315 revolutions = \(315 \times \frac{1980}{7}\)

= 45 x 1 980 = 89 100 cm

= \(\frac{89100}{1000 \times 100} \mathrm{~km}=\frac{891}{1000} \mathrm{~km}\)

∴ Distance covered in 60 minutes or 1 hr =\(\frac{891}{1000} \times 60=\frac{5346}{100}=53.46 \mathrm{~km}\)

Hence, speed of bus = 53.46 km/hr

Question 10. Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the adjoining figure. Find the area of the shaded region.

Solution:

Required area = (area of larger semicircle with radius 4.5 cm) – (area of two smaller semicircles with radius of each \(\frac{3}{2}\) cm and a circle with radius \(\frac{4.5}{2}\) cm + (area of smaller semicircle with radius \(\frac{3}{2}\) cm)

= \(\frac{1}{2} \pi(4.5)^2-\left[2 \times \frac{1}{2} \pi\left(\frac{3}{2}\right)^2+\pi\left(\frac{4.5}{2}\right)^2\right]+\left[\frac{1}{2} \pi\left(\frac{3}{2}\right)^2\right]\)

= \(\frac{1}{2} \pi(4.5)^2-\pi\left(\frac{3}{2}\right)^2-\pi\left(\frac{4.5}{2}\right)^2+\frac{1}{2} \pi\left(\frac{3}{2}\right)^2\)

= \(\frac{1}{4} \pi\left[2 \times(4.5)^2-9-(4.5)^2+\frac{9}{2}\right]=\frac{1}{4} \pi \times 4.5[2 \times 4.5-2-4.5+1]\)

= \(\frac{1}{4} \pi \times 4.5(3.5)=\frac{1}{4} \times \frac{22}{7} \times 4.5 \times 3.5=12.375 \mathrm{~cm}^2\).

Area Related To Circles Three Semicircles

Question 11. In the adjoining figure, find the area of the shaded region. (Use π = 3.14)

Solution:

Diameter BD = \(\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10 \mathrm{~cm}\)

Radius = \(]frac {10}{2}\) = 5 cm

Area of circle = πr2 = 3. 1 4 x 52 = 3. 14 x 25 = 78.50 cm2

Area of rectangle ABCD = 8 x 6 = 48 cm2

Hence, area of shaded region = 78.50- 48 = 30.5 cm2

Area Related To Circles Area Of Shaded Region

Question 12. A park is of the shape of a circle of diameter 7 m. It is surrounded by a path of width 0.7 m. Find the expenditure of cementing the path, if its cost is ₹ 110 per sq. m.

Solution:

The radius of the circular park = 3.5 m

There is a path of width 0.7 m.

So, the radius of the external circle R = 3.5 + 0.7

⇒ R = 4.2 cm

Area of path = πR2 – πr2 = n(R- r) (R + r)

= \(\frac{22}{7}(4.2-3.5)(4.2+3.5)\)

= \(\frac{22}{7} \times 0.7 \times 7.7=16.94 \mathrm{~m}^2\)

Now, the cost of cementing 1 m2 of path = ₹ 110

Cost of cementing 16.94 m2 of path = ₹ ( 110 x 16.94) = ₹ 1863.40

Area Related To Circles Shape Of A Circle Of Diameter

Question 13. Find the area of the region between the two concentric circles, if the length of a chord of the outer circle just touching the inner circle at a particular point on it is 10 cm. \(Take, \pi=\frac{22}{7}\)

Solution:

Let the chord AB touch the inner circle at C and let 0 be the centre of both the circles, then

OC = r, OA = R and AB = 10 cm

Now, since OC ⊥ AB (∵ radius through the point of contact is perpendicular to the tangent)

∴ C is the mid-point of AB (∵ ⊥ drawn from the centre to the chord, bisects the chord)

⇒ \(A C=\frac{1}{2} A B=\frac{1}{2} \times 10=5 \mathrm{~cm}\)

Now, in right ΔOCA,

OA2 = OC2 +AC2 (by Pythagoras theorem)

R2 – r2 = 25 → (1)

The required area of the region between two concentric circles

= πR2 – πr2 = π(R2– r2) – 25π [from(1)]

= \(25 \times \frac{22}{7}=78.57 \mathrm{~cm}^2\)

Area Related To Circles Area Of The Region Between The Two Concentric Circles

Question 14. In the adjoining figure, CM = 5 cm, RB = 9 cm, CD ⊥ AB, O is the centre of the larger circle and K is the centre of the smaller circle. Find the area of the shaded region.

Solution:

Let the radius of the larger circle be R and the radius of the smaller circle is r.

Since, CM = 5 cm

∴ MO = R – 5

Also OB = R and RB = 9

∴ OR = R – 9

In ΔAOM,

∠1 + ∠2 = 90° → (1)

In ΔAMR,

∠2 + ∠3 = 90° (the angle in a semicircle is the right angle) → (2)

From eqs. (1) and (2), we get

∠1 + ∠2 = ∠2 + ∠3

⇒ ∠1 = ∠3

In Δs AMO and MOR

∠1 = ∠3 (just proved)

∠4 = ∠5 (each 90°)

∴ ΔAMO ~ ΔMRO (AA corollary)

∴ \(\frac{M O}{R O}=\frac{A O}{M O}\)

⇒ \(\frac{R-5}{R-9}=\frac{R}{R-5}\)

⇒ (R – 5)2 = R (R-9) = R2 + 25 – 10R = R2 – 9R

⇒ R = 25 cm

Now, 2R – 9 = 2r ⇒ 2r = 41

⇒ r = 20.5 cm

Required area = πR2 – πr2 = π(R + r) (R – r)

= \(\frac{22}{7} \times 45.5 \times 4.5 = 643.5 \mathrm{~cm}^2\)

Area Related To Circles Area Of Larger And Smaller Circles

Question 15. A square of the largest area is cut out of a circle. What % of the area of the circle is lost as trimmings?

Solution:

Let the radius of the circle be r units.

∴ Area of circle = πr2 sq. units

Let ABCD be the largest square.

∴ Length of diagonal = 2r = side √2

side = \(\frac{2 r}{\sqrt{2}}=\sqrt{2} r\)

∴ Area (square ABCD) = (side)2 = (√2r)2 = 2r2

∴ Area of circle lost by cutting out of a square of largest area = 2r2

∴ Required percentage of the area of circle lost = \(\frac{2 r^2}{\pi r^2} \times 100=\frac{200}{\pi} \%\).

Area Related To Circles A Square Of The Largest Area Is Cut Out Of A Circle

Question 16. In a circular table covering of radius 32 cm, a design (shade) is formed leaving an equilateral triangle ABC in the middle as shown in the adjacent figure. Find the area of the shaded region.

Solution:

Since ABC is an equilateral triangle.

∴ ∠A = 60°

⇒ ∠BOC = 2 x ∠BAC = 2 x 60° = 120° (degree measure of an arc is twice the angle subtended by it in an alternate segment)

Draw OM ⊥ BC

So, we can prove

ΔOMB ≅ ΔOMC (R.H.S. congruency)

∴ ∠BOM = ∠COM = 60° (c.p.c.t.)

In the right ΔOMB, we have

⇒ \(\sin 60^{\circ}=\frac{B M}{O B}\)

∴ \(\frac{\sqrt{3}}{2}=\frac{B M}{32} \quad = \quad B M=16 \sqrt{3} \mathrm{~cm}\)

∴ BC = 2 x 16√3 = 32√3 cm

∴ Area of shaded region = Area of circle- ar(AABC)

= \(\pi r^2-\frac{(\text { side })^2 \sqrt{3}}{4}=\pi \times(32)^2-(32 \sqrt{3})^2 \times \frac{\sqrt{3}}{4}\)

= \((32)^2\left(\frac{22}{7}-\frac{3 \sqrt{3}}{4}\right)=1024\left(\frac{22}{7}-\frac{3 \sqrt{3}}{4}\right) \mathrm{cm}^2\)

Area Related To Circles Equilateral Triangle

Area Of Sector And Segment Of A Circle

Sector: The shaded region (shown in the figure) OAXB is called a sector of the circle. Its boundary consists of arc AXB and two radii OA and OB. This sector has an angle θ, subtended at the centre of the circle by the arc AXB.

The region bounded by two radii of a circle and intercepted by them is called a sector of the circle.

Area Related To Circles Area Of Sector And Segment Of A Circle

When θ < 180°, arc AB is a minor arc and when θ > 180°, arc AB is a major arc.

Now for sector AOB with ∠AOB = θ°, then the length of minor arc AB \(2 \pi r \times \frac{\theta}{360^{\circ}} \text { i.e., } \frac{\pi r \theta}{180^{\circ}}\)

(Actually, 2πr is the distance covered in travelling the whole circumference in which an angle of 360 is formed at the centre. But for the length of the arc AXB, we do not rotate 360°, here we needed only 0 part of 360°)

∴ \(l=\frac{2 \pi r \theta}{360}\)

and the area of sector is \(\pi r^2 \times \frac{\theta}{360^{\circ}}\)

∴ \(A=\frac{\pi r^2 \theta}{360^{\circ}}\)

Relation between Length of Arc and Area of Sector

Area of sector \(A=\pi r^2 \times \frac{\theta}{360^{\circ}}\)

= \(\pi r \times \frac{\theta}{360^{\circ}} \times r=\frac{1}{2} \times 2 \pi r \times \frac{\theta}{360} \times r=\frac{1}{2} l r\)

Segment of a Circle: A segment of a circle is defined as the part of a circle bounded by a chord and the circumference. The segment containing the major arc is the major segment while the segment containing the minor arc is a minor segment.

Area of minor segment = Area of the sector- an area of ΔOAB.

= \(\left(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\right) \text { sq. unit }\)

Area Related To Circles Segment Of A Circle

Area of major segment = Area of circle- Area of minor segment

Perimeter of sector = \(\frac{\pi r \theta}{180^{\circ}}+2 r\)

Perimeter of minor segment = \(\frac{\pi r \theta}{180^{\circ}}\) + Length of chord AB

Area of semicircle = \(\frac{\pi r^2}{2}\)

Perimeter of semicircle = \(\pi r+2 r\)

Area of quarter circle = \(\frac{\pi r^2}{4}\)

Perimeter of quarter circle = \(\frac{\pi r}{2}+2 r\)

Area Related To Circles Area Of Major Segment

Area Related To Circles Solved Questions And Answers

Question 1. The perimeter of a semi-circular protractor is 32.4 cm. Calculate:

  1. The radius of the protractor in cm,
  2. the arc of the protractor in cm2.

Solution:

Let the radius of the protractor be r cm.

Perimeter of semicircle protractor = (πr + 2r)cm

∴ r(π + 2) = 32.4

⇒ \(r\left(\frac{22}{7}+2\right)=32.4\)

⇒ \(r \times \frac{36}{7}=32.4 \quad ⇒ \quad r=\frac{32.4 \times 7}{36} ⇒ r=6.3 \mathrm{~cm}\)

Hence, the radius of the protractor = 6.3 cm

Area of semi-circular protractor = \(\frac{1}{2} \pi r^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times 6.3 \times 6.3=62.37 \mathrm{~cm}^2\)

Hence, the area of the protractor = 62.37 cm2

Question 2. The minute hand of a clock is √2I cm long. Find the area described by the minute hand on the face of the clock between 6 a.m. and 6.05 a.m.

Solution:

In 60 minutes, the minute hand of a clock move through an angle of 360°.

∴ In 5 minutes hand will move through an angle = \(\frac{360^{\circ}}{60} \times 5=30^{\circ}\)

Now, r = √21 cm and θ = 30°

The area of sector described by the minute hand between 6 a.m. and 6.05 a.m.

= \(\frac{\pi r^2 \theta}{360^{\circ}}=\frac{22}{7} \times(\sqrt{21})^2 \times 30^{\circ} \times \frac{1}{360^{\circ}}=5.5 \mathrm{~cm}^2\)

Question 3. In the adjoining figure, calculate:

  1. The length of minor arc ACB
  2. Area of the shaded sector.

Solution:

Here, θ = 150°, r = 14 cm

Area Related To Circles Adjoining Triangle

1. Length of minor arc = \(\frac{\pi r \theta}{180^{\circ}}\)

= \(\frac{22}{7} \times 14 \times 150^{\circ} \times \frac{1}{360^{\circ}}\)

= 36.67 cm

2. Area of shaded sector = \(\frac{\pi r^2 \theta}{360^{\circ}}=\frac{22}{7} \times(14)^2 \times 150^{\circ} \times \frac{1}{360^{\circ}}=256.67 \mathrm{~cm}^2\)

Question 4. A chord AB of a circle of radius 10 cm makes a right angle at the centre of the circle. Find the area of the major and minor segments. (Use π = 3.14).

Solution:

Area of sector OAB = \(\frac{\pi r^2 \theta}{360^{\circ}}=3.14 \times 10^2 \times \frac{90^{\circ}}{360^{\circ}}=78.5 \mathrm{~cm}^2\)

Now, area of \(\triangle A O B=\frac{1}{2} \times(10)^2=50 \mathrm{~cm}^2\)

Area of the minor segment = (78.5 – 50) cm2 = 28.5 cm2

Area of the major segment = Area of the circle – Area of the minor segment

= (3.14 x 102 – 28.5) cm2 = (314 – 28.5) cm2 = 285.5 cm2

Area Related To Circles Area Of Major And Minor Segment

Question 5. In the adjoining figure, the side of the square is 28 cm and the radius of each circle is half of the length of the side of the square where O and O’ are the centres of the circle. Find the area of the shaded region.

Answer:

Side of square = 28 cm

and the radius of each circle = 14 cm

Required area = area of square excluding the two sectors + area of two circles

= \((28)^2-2\left[\pi(14)^2 \times \frac{90^{\circ}}{360^{\circ}}\right]+2\left[\pi(14)^2\right]\)

= \((28)^2-2 \times \frac{22}{7} \times 14 \times 14 \times \frac{1}{4}+2 \times \frac{22}{7} \times 14 \times 14\)

= 784 – 308 + 1232 = 1708 cm2

Area Related To Circles Area Of Square Excluding The Two Sectors And Two Circles

Question 6. In the adjoining figure, ABCD is a trapezium with AB || DC, AB = 18 cm, DC = 32 cm and the distance between AB and DC is 14 cm. If arcs of equal radii 7 cm taking A, B, C and D as centres, have been drawn, then find the area of the shaded region.

Solution:

Area of the shaded region

= area of trapezium – area of four sectors

= \(\begin{aligned}
\frac{1}{2} \times 14(32+18)-\left[\pi(7)^2 \times \frac{\angle A}{360^{\circ}}+\right. & \pi(7)^2 \times \frac{\angle B}{360^{\circ}} \left.+\pi(7)^2 \times \frac{\angle C}{360^{\circ}}+\pi(7)^2 \times \frac{\angle D}{360^{\circ}}\right]
\end{aligned}\)

= \(=7 \times 50-\frac{\pi \times(7)^2}{360^{\circ}}(\angle A+\angle B+\angle C+\angle D)\)

= \(=350-\frac{22}{7} \times \frac{7 \times 7}{360^{\circ}} \times 360^{\circ}\) (∵ sum of all the four angles of a quad. is 360°)

= 350 – 154 = 196 cm2

Hence, the area of the shaded region is 196 cm2.

Area Related To Circles Trapezium

Question 7. In the adjoining figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flower bed is the point of intersection O of the diagonals of the square lawn, find the sum of areas of the lawn and the flower beds.

Solution:

We know that the diagonals of a square bisect each other perpendicularly.

∴ ∠DOC = 90°

Also diagonal BD = side √2 = 56√2m

∴ \(O D=\frac{1}{2} \times B D=28 \sqrt{2} \mathrm{~m}\)

∴ Area of \(\triangle D O C=\frac{1}{2} \times 28 \sqrt{2} \times 28 \sqrt{2}=(28)^2 m^2\)

Area of sector ODCO = \(\pi(O D)^2 \times \frac{\theta}{360^{\circ}}\)

= \(\pi(28 \sqrt{2})^2 \times \frac{90^{\circ}}{360^{\circ}}=\frac{11 \times 28 \times 28}{7} \mathrm{~m}^2\)

= 1232 m2

∴ Area of 1 flower bed = Area of sector- Area of A

= 1232 -(28)2 = 448 m2

∴ Area of 2 flowers beds = 2 x 448 = 896 m2

Area of square lawn = (56)2 = 3136 m2

∴ Required area = 896 + 3136 = 4032 m2

Area Related To Circles Two Circular Flower Beds

Question 8. In the adjoining, ABCD is a square of side 10 cm and two A semicircles with side of the square as diameter. A quarter circle is also seen with a side of the square as a radius. Find the area of the square region excluding the shaded part (Deepak).(Takeπ = 3.14)

Area Related To Circles A Square Of Semicircles With Side Of The Square As Diameter

Solution:

First of all, we will find the area of 1 and 2 i.e., Batti of Deepak, draw OM ⊥ AB and ON ⊥ BC.

ar(1) = ar(sector NBKO) – ar(ABNO)

= \(\pi(5)^2 \times \frac{90^{\circ}}{360^{\circ}}-\frac{1}{2} \times 5 \times 5\)

ar (2) = ar (1)

⇒ \(\operatorname{ar}(\mathrm{1}+\mathrm{2})=2\left(\frac{25 \pi}{4}-\frac{25}{2}\right)=\frac{25}{2}(\pi-2)\)

∴ \(\frac{25}{2}(3.14-2)=14.25 \mathrm{~cm}^2\)

Now, area of deepak = ar(quadrant APCB) – [ar(semicircle 4 + 1 + 2) + ar(semicircle 5 + 1 + 2) – 2 x ar (1 + 2)]

= \(\pi(10)^2 \times \frac{90}{360}-\left[2 \times \frac{\pi(5)^2}{2}-2 \times 14.25\right]\)

25π – (25π – 28.5) = 28.5 cm2

Area of unshaded portion = Area of the square- Area of deepak

= (10)2– 28.5 = 100 – 28.5 = 71.5 cm2

Area Related To Circles A Square Of Semicircles With Side Of The Square As Diameter.

Question 9. If the hypotenuse of an isosceles right triangle is 7√2 cm, find the area of the circle inscribed in it.

Solution:

Let AB = BC = x cm

∴ In right ΔABC, by Pythagoras theorem

x2 + y2 = (7√2)2

⇒ 2x2 = 98 = x2 = 49 ⇒ x = 7 cm

∴ ar(ABC) = ar(AOB)+ ar(BOC) + ar(COA)

⇒ \(\frac{1}{2} \times x \times x=\frac{1}{2} \times x \times r+\frac{1}{2} \times x \times r+\frac{1}{2} \times 7 \sqrt{2} \times r\)

⇒\(r=\frac{7}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{7(2-\sqrt{2})}{2}\)

⇒ \(7 \times 7=7 \times r+7 \times r+7 \sqrt{2} \times r \Rightarrow r=\frac{7}{2+\sqrt{2}}\)

Area of circle = πr2

= \(\frac{22}{7} \times \frac{49(2-\sqrt{2})^2}{4}=\frac{77}{2} \times(4+2-4 \sqrt{2})\)

= 77(3 – 2√2) cm2

Area Related To Circles Hypotenuse Of An Isosceles Right Triangle

Question 10. In the adjoining figure, two concentric circles with centre O have radii of 21 cm and 42 cm. If ∠AOB = 60°, find the area of the shaded region. (\(\text { Use } \pi=\frac{22}{7}\))

Solution:

Required area = area of a larger circle with a radius of 42 cm – an area of a smaller circle with a radius of 21 cm – (area of unshaded portion CDBA)

= π(42)2 – π(21 )2 – [(area of the larger sector with radius 42 cm) – (area of the smaller sector with radius 21 cm)]

= \(\pi(42)^2-\pi(21)^2-\left[\pi(42)^2 \times \frac{60}{360}-\pi(21)^2 \times \frac{60}{360}\right]\)

= \(\pi(42)^2-\pi(21)^2-\frac{1}{6} \pi(42)^2+\frac{1}{6} \pi(21)^2=\frac{5}{6} \pi(42)^2-\frac{5}{6} \pi(21)^2\)

= \(\frac{5}{6} \pi \times(21)^2[4-1]=\frac{5}{6} \times \frac{22}{7} \times 21 \times 21 \times 3=3465 \mathrm{~cm}^2\)

Area Related To Circles Two Concentric Circles

Question 11. In the adjoining figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.

Solution:

Since O is the centre of the circle.

∴ BC is the diameter of the circle.

∴ ∠CAB = 90° (angle in a semicircle is a right angle)

Now in right ΔACB, by Pythagoras Theorem,

BC2 = AC2 + AB2 = (24)2 + (7)2 = 576 + 49 = 625

∴ BC = = 25 m

∴ Diameter of circle = 25 m

∴ Radius of circle = \(\frac{25}{2}\) = 12.5 m

∴ OC = OD = 12.5 m (each radii)

∴ Area of shaded region = area of circle – ar (ΔABC) – ar (sector COD)

= \(\pi(12.5)^2-\frac{1}{2} \times A C \times A B-\pi(O C)^2 \times \frac{90^{\circ}}{360^{\circ}}\)

= \(\frac{22}{7} \times 12.5 \times 12.5-\frac{1}{2} \times 24 \times 7-\frac{22}{7} \times 12.5 \times 12.5 \times \frac{1}{4}\)

= 491.07 – 84 – 122.768 = 284.302 = 284.30 m2 (approx.)

Area Related To Circles Centre Of The Circle

Area Related To Circles Exercise 12.1

Question 1. The radii of the two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has a circumference equal to the sum of the circumferences of the two circles.

Solution:

Here, r1 = 19 cm and r2 = 9cm

Let the radius of the new circle = R cm

Given that,

Circumference of new circle = sum of the circumference of given two circles

2πR = 2πr1 + 2πr2

R = r1 + r2 = 19 + 9 = 28 cm.

Question 2. The radii of the two circles are 8 cm and 6 cm respectively. Find the radius of the circle having an area equal to the sum of the areas of the two circles.

Solution:

Here, r1 = 8 cm and r2 = 6 cm

Let the radius of the new circle = R cm

Given that,

Area of new circle = sum of areas of given circles

⇒ πR2 = πr2 + πr2

⇒ R2 = r21 + r22 = 82 + 62

= 64 + 36 = 100

⇒ R = 10 cm

Question 3. The depicts an archery target marked with its five scoring regions from the centre outwards as Gold, Red, Blue, Black and White. The diameter of the region representing the Gold score is 21cm and each of the other bands is 10.5 cm wide. Find the area of each of the five scoring regions.

Solution:

The diameter of the Gold circle = 21 cm

∴ Radius of Gold circle = 10.5 cm

Now, the area of the Gold Circle

= \(\frac{22}{7} \times 10.5 \times 10.5 \mathrm{~cm}^2\)

= 346.5 cm2

Given, the width of each band = 10.5 cm

∴ Radius of Red circle = (10.5 + 10.5) = 21cm

Now, area of Red ring = [(external radius)2 – (internal radius)2]

= \(\frac{22}{7}\left[(21)^2-(10.5)^2\right]\)

= \(\frac{22}{7} \times(441-110.25)\)

= \(\frac{22}{7} \times 330.75\)

= 1039.5 cm2

Again, radius of Blue circle = (21 + 10.5)cm = 31.5 cm

∴ Area of Blue ring

= \(\frac{22}{7}\left(31.5^2-21^2\right)\)

= \(\frac{22}{7} \times(992.25-441)\)

= \(\frac{22}{7} \times 551.25\)

= 1732.5 cm2

Again, the radius of the Black Circle

= (31.5 + 10.5)cm

= 42cm

∴ Area of Black circle

= \(\frac{22}{7} \times\left(42^2-31.5^2\right)\)

= \(\frac{22}{7} \times(1764-992.25)\)

= \(\frac{22}{7} \times 771.75=2425.5 \mathrm{~cm}^2\)

Again, radius of white circle = (42 + 10.5) cm

∴ Area of White circle

= \(=\frac{22}{7} \times\left(52.5^2-42^2\right)\)

= \(\frac{22}{7} \times(2756.25-1764)\)

= \(\frac{22}{7} \times 992.25=3118.5 \mathrm{~cm}^2\)

Area Related To Circles Archery Target

Question 4. The wheels of a car are of diameter 80 cm each. How many complete revolutions does each wheel make in 10 minutes when the car is travelling at a speed of 66 km per hour?

Solution:

The diameter of the wheel of a car, 2r= 80 cm

⇒ r = 40 cm

∴ Distance covered in one revolution

= circumference

= \(2 \pi r=2 \times \frac{22}{7} \times 40=\frac{1760}{7} \mathrm{~cm}\)

Now, speed of car = 66 km/hr

= \(\frac{66 \times 1000 \times 100}{60} \mathrm{~cm} / \mathrm{min}\)

∴ Distance covered in 10 minutes

= \(\frac{66 \times 1000 \times 100}{60} \times 10 \mathrm{~cm}\)

= 11 x 1000 x 100 cm

Now, the number of revolutions made by the wheel

= \(\frac{\text { Total distance covered }}{\text { Distance covered by wheel in 1 revolution }}\)

= \(\frac{11 \times 1000 \times 100}{1760 / 7}\)

= \(=\frac{11 \times 1000 \times 100 \times 7}{1760}=4375 \text { Ans }\)

Question 5. Tick the correct answer in the following and justify your choice: if the perimeter and the area of a circle are numerically equal, then the radius of the circle is:

  1. 2 units
  2. units
  3. 4 units
  4. 7 units

Solution:

1. 2 units

Let r be the radius of the circle.

Given that, in numerical form, area of circle = perimeter of circle

⇒ πr2 = 2πr ⇒ r = 2 units

Area Related To Circles Exercise 12.2

Question 1. Find the area of a sector of a circle with a radius of 6 cm if the angle of the sector is 60°.

Solution:

Here, the radius of the circle, r = 6 cm

The angle of the sector, θ= 60°

∴ Area of sector = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6 \mathrm{~cm}^2\)

= \(\frac{132}{7} \mathrm{~cm}^2 \text { or } 18.86 \mathrm{~cm}^2\)

Question 2. Find the area of a quadrant of a circle whose circumference is 22 cm.

Solution:

Circumference of the circle, 2πr = 22

⇒ \(2 \times \frac{22}{7} \times r=22 \quad \Rightarrow \quad r=\frac{7}{2} \mathrm{~cm}\)

Now, the area of the quadrant of a circle

⇒ \(\frac{1}{4} \pi r^2=\frac{1}{4} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\)

∴ \(\frac{77}{8} \mathrm{~cm}^2\)

Question 3. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.

Solution:

Length of minute hand of clock = 14 cm

∴ The radius of the circle, r = 14 cm

∵ The angle subtended by minute hand in 60 min = 360°

∴ Angle subtended by minute hand in 1 minute = \(\frac{360^{\circ}}{60^{\circ}}=6^{\circ}\)

∴ The angle subtended by minute hand in 5 minutes = 30°

∴ From the formula,

area of sector of circle = \(\frac{\theta \pi r^2}{360^{\circ}}\)

= \(30^{\circ} \times \frac{22 \times(14)^2}{7 \times 360^{\circ}}\)

= \(\frac{22 \times 14 \times 2}{12}\)

= \(\frac{616}{12}=\frac{154}{3} \mathrm{~cm}^2\)

Question 4. A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

  1. Minor segment
  2. Major sector

Solution:

Given, the radius of the circle, AO = 10 cm.

The perpendicular is drawn from the centre of the circle to the chord of the circle which bisects this chord.

Area Related To Circles Radius Of The Circle

AD = DC

And ∠AOD = ∠COD

= 45°

∴ ∠AOC = ∠AOD + ∠COD

= 45° + 45° = 90°

In the right ΔAOD,

⇒ \(\sin 45^{\circ}=\frac{A D}{A O} \quad \Rightarrow \quad \frac{1}{\sqrt{2}}=\frac{A D}{10}\)

AD = 5√2 cm

and \(\cos 45^{\circ}=\frac{O D}{A O} \Rightarrow \frac{1}{\sqrt{2}}=\frac{O D}{10}\)

⇒ OD = 5√2 cm

Now, AC = 2AD

2 x 5√2 = 10√2 cm

Now, the area of ΔAOC

= \(\frac{1}{2} A C \times O D\)

= \(\frac{1}{2} \times 10 \sqrt{2} \times 5 \sqrt{2}=50 \mathrm{~cm}^2\)

Now the area of the sector

= \(\frac{\theta \pi r^2}{360^{\circ}}=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2\)

= \(\frac{314}{4}=78.5 \mathrm{~cm}^2\)

1. Area of minor segment AEC

= area of sector OAEC – area of AOC

= 78.5 – 50 = 28.5 cm2

2. Area of major sector OAFGCO

= area of circle- area of sector OAEC

= πr2– 78.5 = 3.14 x (10)2 – 78.5

314 – 78.5 = 235.5cm2

Question 5. In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:

  1. The length of the arc
  2. The area of the sector formed by the arc
  3. The area of the segment is formed by the corresponding chord.

Solution:

Here, the radius of circle r = 21 cm

The angle subtended by are at the centre, θ = 60°

1. Length of arc

= \(l=\frac{\theta}{360^{\circ}} \times 2 \pi r\)

= \(\frac{60^{\circ}}{360^{\circ}} \times 2 \times \frac{22}{7} \times 21=22 \mathrm{~cm}\)

2. Area of sector formed by the arc

= \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 21 \times 21\)

= 231 cm2

3. Area of segment formed by the corresponding chord

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(231-\frac{1}{2} \times 21 \times 21 \times \sin 60^{\circ}\)

= \(\left(231-\frac{441 \sqrt{3}}{4}\right) \mathrm{cm}^2\)

Question 6. A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.

(Use π = 3.14 and √3 = 1.73)

Solution:

Here, the radius of the circle, r = 15 cm

The angle subtended by a chord at the centre, θ = 60°

∴ Area of the corresponding minor segment

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(r^2\left(\frac{\pi \theta}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

= \(15 \times 15\left(\frac{3.14 \times 60^{\circ}}{360^{\circ}}-\frac{1}{2} \sin 60^{\circ}\right)\)

= \(225\left(\frac{3.14}{6}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

= \(225\left(\frac{3.14}{6}-\frac{1.73}{4}\right)=20.4375 \mathrm{~cm}^2\)

Now, the area of the circle = πr2

= 3.14 x 15 x 15 = 706.5 cm2

∴ Area of major segment = area of circle – an area of the minor segment

= (706.5 – 20.4375) cm2

= 686.0625 cm2.

Question 7. A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and √3 = 1.73)

Solution:

Here, the radius of the circle, r = 12 cm

The angle subtended by the chord at the centre, θ = 120°

Area of the corresponding minor segment

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(r^2\left(\frac{\pi \theta}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

= \(12 \times 12 \times\left(\frac{3.14 \times 120^{\circ}}{360^{\circ}}-\frac{1}{2} \times \sin 120^{\circ}\right)\)

= \(144\left(\frac{3.14}{3}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

= \(144\left(\frac{3.14}{3}-\frac{1.73}{4}\right)\)

= 88. 44 cm2

Question 8. A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m long rope. Find:

  1. The area of that part of the field in which the horse can graze.
  2. The increase in the grazing area, if the rope were 10 m long instead of 5m. (Use π = 3.14)

Area Related To Circles A Horse Is Tied To A Peg

Solution:

Side of square field = 15 m

The radius of the circle, r = 5m

The angle formed by chord, θ = 90°

1. Area of that part of the field where the horse can graze the grass

= \(\frac{\theta}{360^{\circ}} \times \pi r^2=\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times 5^2\)

= 19.625 m2

2. If the length of the rope is 10 m then the increase in the area of grazing.

= \(\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times\left(10^2-5^2\right)\)

= \(\frac{1}{4} \times 3.14 \times 75\)

= 58.875 m2.

Question 9. A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown. Find:

  1. The total length of the silver wire required.
  2. The area of each sector of the brooch.

Area Related To Circles A Brooch Is Made With Silver Wire In The Form Of A Circle

Solution:

1. Diameter of circular brooch,

2r = 35 mm

= \(r=\frac{35}{2} \mathrm{~mm}\)

Length of required silver wire = 2πr

= \(2 \times \frac{22}{7} \times \frac{35}{2} \mathrm{~mm}=110 \mathrm{~mm}\)

Now, the length of 5 diameters = 5 x 2r

= \(5 \times 2 \times \frac{35}{2}=175 \mathrm{~mm}\)

Length of total wire = (110 + 175) mm

= 285 mm

2. For each sector,

angle, \(\theta=\frac{360^{\circ}}{10^{\circ}}=36^{\circ}\)

∴ Area = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7} \times \frac{35}{2} \times \frac{35}{2}\)

= \(\frac{385}{4} \mathrm{~cm}^2\)

Question 10. An umbrella has 8 ribs which are equally spaced. Assuming an umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Area Related To Circles Umbrella To Be A Flat Circle

Solution:

The radius of the umbrella,

r = length of ribs of umbrella = 45 cm

Here, number of sectors = 8

∴ Angle of sector = \(\theta=\frac{360^{\circ}}{8}=45^{\circ}\)

Now, the area between two ribs = area of the sector

= \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{45^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 45 \times 45 \mathrm{~cm}^2\)

= \(\frac{22275}{28} \mathrm{~cm}^2\)

Question 11. A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.

Solution:

Given, the length of the wiper blade

r = 25 cm = r(say)

The angle formed by this blade, θ =115°

∴ Area cleaned by a blade = area of sector formed by blade

= \(\frac{\theta \pi r^2}{360^{\circ}}=115^{\circ} \times \frac{22}{7 \times 360^{\circ}} \times(25)^2\)

= \(\frac{23 \times 22}{7 \times 72} \times 625=\frac{23 \times 11 \times 625}{7 \times 36}\)

= \(\frac{158125}{252} \mathrm{~cm}^2\)

∴ Total area cleaned by two blades

= 2 x area cleaned by a blade

= \(\frac{2 \times 158125}{252}=\frac{158125}{126} \mathrm{~cm}^2\)

Question 12. To warn ships of underwater rocks, a lighthouse spreads a red-coloured light over a sector of angle 80° to a distance of 16.51cm. Find the area of the sea over which the ships are warned. (Use π = 3.14)

Solution:

Given, the angle of the sector, θ = 80°

And distance or radius, r = 16.5 km

Area of sector

= \(\frac{\theta \pi r^2}{360^{\circ}}=\frac{80^{\circ} \times 3.14 \times(16.5)^2}{360^{\circ}}\)

= \(\frac{2 \times 3.14 \times 272.25}{9}=\frac{1709.73}{9}=189.97 \mathrm{~km}^2\) which is the area of the sea over which the ships are warned.

Question 13. A round table cover has six equal designs as shown. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ₹ 0.35 per cm2. (Use √3 = 1.7)

Area Related To Circles A Round Table Cover Has Six Equal Designs

Solution:

The radius of the table cover, r = 28 cm

Number of designs formed on table cover = 6

∴ The angle formed by each chord at the centre

= \(\frac{360^{\circ}}{6}=60^{\circ}\)

Now, the area of segment

= \(\frac{\pi r^2 \theta}{360^{\circ}}-\frac{1}{2} r^2 \sin \theta\)

= \(r^2\left(\frac{\pi}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

⇒ Area of 6 segments

= \(6 r^2\left(\frac{\pi}{360^{\circ}}-\frac{1}{2} \sin \theta\right)\)

= \(6 \times 28 \times 28\left(\frac{60^{\circ}}{360^{\circ}} \times \frac{22}{7}-\frac{1}{2}-\sin 60^{\circ}\right)\)

= \(6 \times 28 \times 28\left(\frac{11}{21}-\frac{1}{2} \times \frac{\sqrt{3}}{2}\right)\)

= \(=6 \times 28 \times 28\left(\frac{11}{21}-\frac{1.73}{4}\right)=464.8 \mathrm{~cm}^2\)

Now, the total cost of making the design

= ₹ 0.35 x 464.8 = ₹ 162.68

Question 14. The area of a sector of angle p (in degrees) of a circle with radius R is

  1. \(\frac{p}{180} \times 2 \pi R\)
  2. \(\frac{p}{180} \times \pi R^2\)
  3. \(\frac{p}{360} \times 2 \pi R\)
  4. \(\frac{p}{720} \times 2 \pi R^2\)

Solution:

4. \(\frac{p}{720} \times 2 \pi R^2\)

Area of sector = \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{p}{360^{\circ}} \times \pi R^2=\frac{p}{720} \times 2 \pi R^2\)

Area Related To Circles Exercise 12.3

Question 1. Find the area of the shaded region in the figure, if PQ= 24 cm, PR = 7 cm and O is the centre of the circle.

Area Related To Circles The Area Of The Shaded Region

Solution:

Here, PQ = 24 cm and PR = 7 cm

∵ The angle in a semicircle is a right angle.

∴ ∠QPR = 90°

In ΔPQR,

QR2 = PQ2 + PR2 = 242 + 72 = 625

QR = 25 cm

If r is the radius of the circle, then

2r = 25 cm

= \(r=\frac{25}{2} \mathrm{~cm}\)

Area of shaded region = area of semicircle – area of ΔPQR

= \(\frac{1}{2} \pi r^2-\frac{1}{2} \times P Q \times P R\)

= \(\frac{1}{2} \times \frac{22}{7} \times \frac{25}{2} \times \frac{25}{2}-\frac{1}{2} \times 24 \times 7\)

= \(\frac{6875}{28}-84=\frac{4523}{28} \mathrm{~cm}^2\)

Question 2. Find the area of the shaded region, if the radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠AOC = 40°.

Area Related To Circles Radii Of The Two Concentric Circles With Centre

Solution:

Here, OA = 14 cm and OB = 7 cm

The angle of the sector, θ = 40°

∴ Area of the shaded portion

= \(\frac{\theta}{360^{\circ}} \times \pi \times\left(O B-O A^2\right)\)

= \(\frac{40^{\circ}}{360^{\circ}} \times \frac{22}{7} \times\left(14^2-7^2\right)\)

= \(\frac{1}{9} \times \frac{22}{7} \times 147=51.33 \mathrm{~cm}^2\)

Question 3. Find the area of the shaded region, if ABCD is a square of side 14 cm and APD and BPC are semicircles.

Area Related To Circles Areas Of Two Semicircles

Solution:

Side of square = 14

Area of square = 14 x 14 cm2 = 196 cm2

Diameter of each semicircle 2r = side of square = 14 cm

⇒ r = 7 cm

Area of one semicircle = \(\frac{1}{2} \pi r^2\)

Area of two semicircles = \(2 \times \frac{1}{2} \pi r^2=\pi r^2\)

= \(\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2\)

Area of shaded portion = area of the square – the sum of areas of two semicircles

= (196 – 154) cm2 = 42 cm2

Question 4. Find the area of the shaded region given, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as the centre.

Area Related To Circles Each Side Of Equilateral Triangle

Solution:

Each side of the equilateral triangle = 12 cm

and each angle = 60°

Now, the area of an equilateral triangle

= \(\frac{\sqrt{3}}{4} \times(\text { side })^2=\frac{\sqrt{3}}{4} \times 12 \times 12 \mathrm{~cm}^2\)

= 36√3 cm2

The radius of the circle, r = 6 cm

Angle of major sector, 0 = 360° – 60° = 300°

∴ Area of major sector

= \(\frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(\frac{300^{\circ}}{360^{\circ}} \times \frac{22}{7} \times 6 \times 6=\frac{660}{7} \mathrm{~cm}^2\)

Now, the area of the shaded region = area of the equilateral triangle + area of a major sector.

= \(\left(36 \sqrt{3}+\frac{660}{7}\right) \mathrm{cm}^2\)

Question 5. From each A corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown. Find the area of the remaining portion of the square.

Area Related To Circles A Square

Solution:

Side of square = 4 cm

∴ Area of square = (side)2 = 42 = 16 cm2

The radius of a quadrant of a circle, r = 1 cm

∴ Area of four quadrants

= \(4 \times \frac{1}{4} \pi r^2\)

= \(\frac{22}{7} \times 1 \times 1=\frac{22}{7} \mathrm{~cm}^2\)

For the circle inside the square diameter, 2R = 2 cm ⇒ R = 1 cm

Area of this circle = \(\pi R^2=\frac{22}{7} \times 1 \times 1=\frac{22}{7} \mathrm{~cm}^2\)

Now, area of shaded portion = \(16-\left(\frac{22}{7}+\frac{22}{7}\right)=\frac{68}{7} \mathrm{~cm}^2\).

Question 6. In a circular table covered of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown. Find the area of the design.

Area Related To Circles An Equilateral Triangle

Solution:

ΔABC is an equilateral triangle.

Area Related To Circles An Equilateral Triangle..

∴ ∠A = ∠B = ∠C = 60°

AD is perpendicular to BC.

∴ OA, OB, and OC are the radii of the circle.

Given that,

OA = OB = OC = 32 cm

∠BOC = 2

∠BAC = 2 x 60° = 120°

∴ \(\angle B O D=\frac{1}{2} \times 120^{\circ}=60^{\circ}\)

In ΔBOD

⇒ \(\sin 60^{\circ}=\frac{B D}{O B} ⇒ \frac{\sqrt{3}}{2}=\frac{B D}{32}\)

BD = 16√3 cm

BC = 2 BD = 32√3 cm

Area of ΔABC = \(\frac{\sqrt{3}}{4} \times B C^2\)

= \(\frac{\sqrt{3}}{4} \times(32 \sqrt{3})^2=768 \sqrt{3} \mathrm{~cm}^2\)

Area of circle = π(OB)2

= \(\frac{22}{7} \times 32 \times 32=\frac{22528}{7} \mathrm{~cm}^2\)

Now, the area of the shaded portion

= area of a circle – an area of ΔABC

= \(\left(\frac{22528}{7}-768 \sqrt{3}\right) \mathrm{cm}^2\).

Question 7. In the ABCD is a square of side 14 cm. With centres A. B. C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Area Related To Circles Areas Of Four Quadrants

Solution:

Side of square ABCD = 14 cm

∴ Area of square = (14)2 =196 cm2

Radius of first quadrant \(r=\frac{14}{2}=7 \mathrm{~cm}\)

∴ The sum of area’s four quadrants

= \(4 \times \frac{1}{4} \pi r^2=\pi r^2\)

= \(\frac{22}{7} \times 7 \times 7=154 \mathrm{~cm}^2\)

Now, the area of the shaded portion = area of the square – the sum of areas of four quadrants

= 196 = 154 = 42 cm2

Question 8. depicts a racing track whose left and right ends are semicircular. The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find:

  1. The distance around the track along its inner edge.
  2. The area of the track.

Area Related To Circles Depicts A Racking Track Left And Right Ends Are Semicircular

Solution:

Hence OB O’C = \(\frac{60}{2}=30 \mathrm{~m}\)

Area Related To Circles Depicts A Racking Track Left And Right Ends Are Semicircular.

AB = CD = 10m

∴ OA = O’D = (30 + 10) m = 40 m

1. Distance covered in one round along the inner edges of the path.

= BC + EH + 2 x circumference of semicircle

= \(106+106+2 \times \frac{1}{2} \times 2 \pi \times 30\)

= \(212+2 \times \frac{22}{7} \times 30=\frac{2804}{7} \mathrm{~m}\)

2. Inner radius r = OB = 30 m

Outer radius R = OA = 40 m

Area of path = 2 x area of ABCD + 2 x area of semicircular rings

= \(=2 \times 106 \times 10+\frac{1}{2} \times \pi\left(R^2-r^2\right)\)

= \(2120+\frac{22}{7} \times\left(40^2-30^2\right)\)

= 2120 + 2200 = 4320 m2

Question 9. In the figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Area Related To Circles Diameters Of A Circle

Solution:

The diameter of the smaller circle

OD = OA = 7 cm

∴ Radius r = \(\frac{7}{2}=3.5 \mathrm{~cm}\)

∴ Area of smaller circle = nr2

= \(\frac{22}{7} \times 3.5 \times 3.5 \mathrm{~cm}^2=38.5 \mathrm{~cm}^2\)

Area of semicircle OAQCPBO

= \(\frac{1}{2} \pi(O A)^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times 7 \times 7=77 \mathrm{~cm}^2\)

Area of ΔABC = = \(\frac{1}{2} \times A B \times O C\)

= \(\frac{1}{2} \times 14 \times 7=49 \mathrm{~cm}^2\)

Now the area of the shaded portion = area of the smaller circle + area of OAQCPBO – an area of ABC

= 38.5 + 77 –  49 = 66. cm2

Question 10. The area of an equilateral triangle 11 ABC is 17320.5 cm2. With each vertex of the triangle as the centre, a circle is drawn with a radius equal to half the length of the side of the triangle. Find the area of the shaded region. (Use π= 3.24 and √3 = 1.73205).

Area Related To Circles Area Of An Equilateral Triangle

Solution:

Let the radius of each circle = r

Side of equilateral triangle = r + r = 2r

Now, area of equilateral triangle = \(\)

Given that, √3r2 = 17320.5 ⇒ 1.73205r2 = 17320.5

⇒ r2 = 10000 ⇒ r = 100 cm

Each angle of an equilateral triangle, θ = 60° Now, the area of sectors of three circles

= \(3 \times \frac{\theta}{360^{\circ}} \times \pi r^2\)

= \(3 \times \frac{60^8}{360^{\circ}} \times 3.14 \times(100)^2\)

= 15700 cm2

Area of shaded region = area of equilateral ABD = area of three sectors

= (17320.5 – 15700) cm2

= 1620.5 cm2

Question 11. On a square handkerchief, nine circular designs each of a radius of 7 cm are made. Find the area of the remaining portion of the handkerchief.

Area Related To Circles A Square Handkerchief Nine Circular Designs

Solution:

Given, the radius of each circle, r = 7 cm

Diameter of circle, d = 14 cm = 42 cm (∵ diameter = 2 x radius)

Three horizontal circles touch each other.

Length of square = 3 x 14 cm = 42 cm

Now area of a circle = πr2 = (7)2

= \(\frac{22}{7} \times(7)^2=154 \mathrm{~cm}^2\)

∴ Area of 9 circles = 9 x 154 = 1386 cm2

Now, area of square ABCD = (side)2

= (42)2 = 1764 cm2

∴ Area of the remaining part of the handkerchief

= 1764 – 1386 = 378 cm2

Question 12. The OACB is a quadrant of a circle with centre O and a radius of 3.5 cm. If OD = 2 cm. find the area of the

  1. Quadrant OACB
  2. Shaded region.

Area Related To Circles A Quadrant Of A Circle

Solution:

Radius of quadrant AOB. r= 3.5 cm

1. Area of quadrant OACB

= \(\frac{1}{4} \pi^2\)

= \(\frac{1}{4} \times \frac{22}{7} \times 3.5 \times 3.5=\frac{77}{8} \mathrm{~cm}^2\)

2. Area of ΔOBD

= \(\frac{1}{2} \times O B \times O D\)

= \(\frac{1}{2} \times 3.5 \times 2=3.5 \mathrm{~cm}^2=\frac{7}{2} \mathrm{~cm}^2\)

Area of the shaded portion

= area of quadrant OACB – area of OBD

= \(\left(\frac{77}{8}-\frac{7}{2}\right) \mathrm{cm}^2=\frac{49}{8} \mathrm{~cm}^2\)

Question 13. In the square O.ABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Area Related To Circles Each Side Of A Square

Solution:

Each side of square OABC = 20 cm

∴ Area of OABC = (20)2 = 400 cm2

Area of OABC = (20)2 = 400 cm2

In ΔOAB,

OB2 = OA2 ~ AB2 = 202 + 202 = 800

⇒ OB = 20√2cm

which is the radius of quadrant OPBQ.

Area of quadrant OPBQ

= \(\frac{1}{4} \pi(O B)^2=\frac{1}{4} \times 3.14 \times(20 \sqrt{2})^2\)

= 628 cm2

Now, the area of the shaded portion

= area of quadrant OPBD – area of square OABC

= (628 – 400) cm2 = 228 cm2

Question 14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O . If ∠AOB = 30°, find the area of the shaded region.

Area Related To Circles Arcs Of Two Concentric Circles

Solution:

Here Q = 30°

Area of the shaded portion

= Area of sector OAB – area of sector OCD

= \(\frac{\theta}{360^{\circ}} \times \pi(O B)^2-\frac{\theta}{360^{\circ}} \times \pi(O C)^2\)

= \(\frac{\theta}{360^{\circ}} \times \pi\left[(O B)^2-(O C)^2\right]\)

= \(\frac{30^{\circ}}{360^{\circ}} \times \frac{22}{7} \times\left(21^2-7^2\right)\)

= \(\frac{1}{12} \times \frac{22}{7} \times 392 \mathrm{~cm}^2=\frac{308}{3} \mathrm{~cm}^2\)

Question 15. The ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Area Related To Circles Quadrant Of A Circle

Solution:

For the quadrant ABC,

Radius r = 1 4 cm

∴ Area of quadrant ABC

= \(\frac{1}{4} \pi r^2\)

= \(\frac{1}{4} \times \frac{22}{7} \times 14 \times 14=154 \mathrm{~cm}^2\)

Area of \(\triangle A B C=\frac{1}{2} \times A B \times A C\)

= \(\frac{1}{2} \times 14 \times 14=98 \mathrm{~cm}^2\)

In right ABC,

BC2 = AC2 + AB2 = 142 + 142 = 392

BC = 14√2 cm

∴ For the semicircle formed on BC

Radius \(R=\frac{B C}{2}=\frac{14 \sqrt{2}}{2}=7 \sqrt{2} \mathrm{~cm}\)

and area of semicircle

= \(\frac{1}{2} \pi R^2\)

= \(\frac{1}{2} \times \frac{22}{7} \times(7 \sqrt{2})^2=154 \mathrm{~cm}^2\)

Now, the area of the shaded portion

= area of ΔABC + area of semicircle – area of quadrant ABC

= (98 + 154 – 154) cm2 = 98 cm2

Question 16. Calculate the area of the designed region in the figure common between the two quadrants of circles of radius 8 cm each.

Area Related To Circles Common Between The Two Quadrants Of Circles

Solution:

The radius of each of quadrants ABMD and BNDC, r = 8 cm

Area Related To Circles Radius Of Each Quadrants

∴ Area of two quadrants

= \(2 \times \frac{1}{4} \pi r^2\)

= \(2 \times \frac{1}{4} \times \frac{22}{7} \times 8 \times 8\)

= \(\frac{704}{7} \mathrm{~cm}^2\)

and area of square ABCD = AB2 = 82 = 64 cm2

Now, the area of the shaded portion

= area of two quadrants – area of square ABCD

= \(\left(\frac{704}{7}-64\right) \mathrm{cm}^2\)

= \(\frac{256}{7} \mathrm{~cm}^2\)

Area Related To Circles Multiple Choice Questions And Answers

Question 1. If the sum of the circumference of two circles of radii R1 and R2is equal to the circumference of a circle of radius R, then:

  1. R1 + R2 = R
  2. R1 + R2 > R
  3. R1 + R2 < R
  4. None of these

Answer: 1. R1 + R2 = R

Question 2. If the sum of the areas of two circles of radii and R1 and R2 is equal to the area of a circle of radius R, then:

  1. R1 + R2 = R
  2. R1 + R2 < R
  3. R12 + R22 = R2
  4. R12 + R22 < R2

Answer: 3. R12 + R22 = R2

Question 3. If the area of a circle is 154 cm2, then its circumference is:

  1. 11 cm
  2. 22 cm
  3. 44 cm
  4. 66 cm

Answer: 3. 44 cm

Question 4. The area of the largest triangle inscribed in a semicircle of radius r is:

  1. r2
  2. \(\frac{1}{2} r^2\)
  3. 2r2
  4. r2√2

Answer: 1. r2

Question 5. The area of the largest square inscribed in a circle of radius 8 cm is:

  1. 256 cm2
  2. 64 cm2
  3. 128 cm2
  4. 32 cm2

Answer: 3. 128 cm2

Question 6. The area largest circle inscribed in a square of side 4 cm is:

  1. 16π cm2
  2. 8π cm2
  3. 6π cm2
  4. 4π cm2

Answer: 2. 8ir cm2

Question 7. If the circumference of a circle and the perimeter of a square are equal, then the ratio of their areas is:

  1. 22:7
  2. 14:11
  3. 7:22
  4. 11:14

Answer: 2. 14:11

Question 8. In the adjoining figure, the perimeter of sector OAB is:

Area Related To Circles The Perimeter Of A Sector

  1. \(\frac{64}{3} \mathrm{~cm}\)
  2. 26 cm
  3. \(\frac{64}{5} \mathrm{~cm}\)
  4. 19 cm

Answer: 1. \(\frac{64}{3} \mathrm{~cm}\)

Question 9. The length of the minute hand of a clock is 14 cm. The area swept by hand in one minute will be:

  1. 10.26 cm2
  2. 10.50 cm2
  3. 10.75 cm2
  4. 11.0 cm2

Answer: 1. 10.26 cm2

NCERT Exemplar Solutions For Class 10 Maths Chapter 10 Circles

Circles

  • Circle: A circle is a collection of all points in a plane which are at the same
  • Centre: The fixed point from which all points in a plane are at the same constant distance is called the centre.
  • Radius: The distance between the centre and circumference of a circle is called the radius.
  • Chord: A straight line segment joining two points on a circle is called a chord of the circle.
  • Secant: A straight line which intersects a circle in two distinct points is called a secant to the circle.
  • Tangent: A straight line meeting a circle only at one point is called a tangent to the circle at that point.
  • Point Of Contact: The point where the straight line touches (or meets at only one point) the circle is called its point of contact.
  • Concentric Circles: Circles having the same centre are said to be concentric circles.

NCERT Exemplar Solutions For Class 10 Maths Chapter 10 Circles

Theorem 1:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Given: A circle with centre O and a tangent AB at a point P of the circle.

To Prove: OP \(\perp\) AB

Circles The Tangent At Any Point Of A Circle Is Perpendicular To The Radius

Construction: Take a point Q other than P on AB. Join OQ.

Proof: Q is a point on the tangent AB, other than the point of contact P.

Q lies outside the circle.

Let OQ interest the circle at R.

Then, OR < OQ

But, OP = OR

Therefore, OP < OQ

Thus, OP is shorter than any other line segment joining O to any point of AB.

It means OP is the shortest distance among all the lines drawn from O to. the point on the tangent other than the point of contact.

Also, we know that perpendicular distance is the shortest distance So, \(\perp\)

i.e., the radius through the point of contact is perpendicular to the tangent.

Theorem 2:

A line drawn through the end of a ratlins and perpendicular to it is a tangent to the circle.

Given: A circle with centre O in which OP is a radius and AB is a line through P such that OP \(\perp\) AB.

Circles A Line Drawn Through The End Of A Radius And Perpendicular

To Prove: AB is a tangent to the circle at the point P.

Construction: Take a point Q different from P, on AB. Join OQ.

Proof: We know that the perpendicular distance from a point to a line is the shortest distance between them.

  • OP \(\perp\) AB
  • OP is the shortest distance from O to AB.
  • OP < OQ
  • Q lies outside the circle.
  • Thus, every point on AB other than P, lies outside the circle.
  • AB meets the circle at point P only. Hence, AB is the tangent to the circle at the point P.

An Important Result of the Above Theorem

If two circles touch internally or externally, the point of contact lies in a straight line through their centres.

Given: Two circles with centre O and O’ which touch each other at P.

To Prove : P lies on the straight line 00′ i.e., the line joining the centres.

Circles Touch Externally And Internally

Construction: Join OP, and O’P and draw a common tangent PT to the two circles at point P.

Proof: When Circles Touch Externally :

  • ∠1 = 90° …(1) (radius through the point of contact is perpendicular to the tangent)
  • ∠2 = 90° (same reason) …(2)
  • ∠1 + ∠2 = 90° + 90° [from (1) and (2)]
  • ∠l+∠2=180°
  • OPO’ is a straight line. (L.P.A.) …(3)
  • When Circles Touch Internally :
  • ∠OPT = ∠O’PT = 90° (radius through the point of contact is perpendicular to the tangent)
  • O’OP is a straight line. (O’P, OP are both 1 to PT at the same point P and only one

∴ \(\perp\) can be drawn to a line through one point on it) …(4) From both (3) and (4), we conclude that P lies on the straight line OO’ i.e., P lies on the straight line joining the centre of the circles.

Theorem 3:

The lengths of tangents drawn from an external point to a circle are equal.

Given: Two tangents AP and AQ are drawn from point A to a circle with centre O.

Circles The Lengths Of Tangents Drawn From An External Point To A Circle

To Prove: AP = AQ

Construction: Join OP, OQ and OA.

Proof: AP is a tangent at P and OP is the radius through P.

OP\(\perp\) AP

Similarly, OQ \(\perp\) AQ

In the right \(\triangle\) OPA and \(\triangle\) OQA

OP = OQ

OA = OA

OPA = ∠OQA

⇒ \(\triangle\) OPA = \(\triangle\) OQA

Hence, AP = AQ

Corollary : (A) If two tangents are drawn from an external point then

  • They subtend equal angles at the centre, and
  •  they are equally inclined to the line segment joining the centre to that point, (or, tangents are equally inclined at the centre).

Given: A circle with centre O and a point A outside it. Also, AP and AQ are the two tangents to the circle.

To Prove : \(\angle\)AOP = \(\angle\) AOQ and \(\angle\) OAP = \(\angle\) OAQ

Circles A Circle With Centre O And A Point A

Proof : In \(\triangle\)AOP and \(\triangle\)AOQ, we have

AP = AQ (tangents from an external point)

OP = OQ (radii of the same circle)

OA = OA (common)

⇒ \(\triangle\)AOP = \(\triangle\)AOQ (by SSS congruence)

Hence, \(\angle\)AOP = \(\angle\)AOQ and \(\angle\)OAP = \(\angle\)OAQ

(2) Prove that the tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.

Given: Let AB be a chord of the given circle. PA and PB are the tangents at endpoints A and B.

To Prove : \(\angle\)5 = \(\angle\)6

Proof: Since OA and OB are the radii of a circle.

Circles The Tangent Drawn At The End Points Of A Chord Of A Circle

⇒ \(\angle\)1 = \(\angle\)2 …(1 )(each 90°, as radius through point of contact is 1 to the tangent)

Also in \(\triangle\)OAB,

Since OA = OB

⇒ \(\angle\)3 = \(\angle\)4 (angles opposite to equal sides are equal)

Subtracting equation (2) from equation (1), we get

⇒ \(\angle\)1 – \(\angle\)3 = \(\angle\)2 – \(\angle\)4

⇒ \(\angle\)5 = \(\angle\)6

Hence, tangents drawn at the endpoints of a chord of a circle make equal angles with the chord.

Solved Examples

Example 1. Find the length of tangent drawn to a circle of radius 6 cm, from a point at a distance of 10 cm from the centre.

Solution:

Since the tangent is perpendicular to the radius through the point of contact.

Circles The Length Of Tangent Drawn To Circle Of Radius

  • \(\angle O T P=90^{\circ}\)
  • In the right triangle OTP, we have
  • \(O P^2 =O T^2+P T^2\)
  • \(10^2 =6^2+P T^2\)
  • \(P T^2 =100-36=64\)
  • P T =8 cm
  • Hence, the length of the tangent is 8 cm.

Question 2. AP is tangent to circle O at point P. What is the length of OP?

Solution:

Let the radius of the given circle be r.

Circles Radius Through Point Of Contact Is Perpendicular To The Tangent

OP = OB = r

OA = 2 + r, OP = r, AP =4

⇒ \(\angle OPA\) = 90° (radius through the point of contact is perpendicular to the tangent)

In right \(\triangle O P A\),

⇒ \(O A^2 =O P^2+A P^2\)

⇒ \((2+r)^2 =r^2+(4)^2\)

⇒ \(4+r^2+4 r =r^2+16\)

⇒ \(4 r =12 \quad \Rightarrow \quad r=3\)

O P = 3 cm.

Example 3. If the angle between two tangents drawn from an external point P to a circle of radius V and centre O, is 60°, then find the length of OP.

Solution:

PA and PB are two tangents from an external point P such that

Circles The Angle Between Two Tangents Drawn From An External Point

⇒ \(\angle\)APB = 60°

⇒ \(\angle\)OPA = \(\angle\)OPB = 30° (tangents are equally inclined at the centre)

Also, \(\angle\)OAP = 90° (radius through the point of contact is perpendicular to the tangent)

Now in right \(\triangle\)OAP,

⇒ \(\sin 30^{\circ} =\frac{O A}{O P}\)

⇒ \(\frac{1}{2} =\frac{a}{O P} \quad \Rightarrow \quad O P\)=2 a units

Example 4. In the adjoining figure, PQ is a chord of a circle and is the tangent atP such that \(\angle\)QPT = 60°. Find \(\angle\)PRQ.

Solution:

Circles In The Adjoining The Chord Of A Circle And Tangent

Join OP and OQ. Take any point S on the circumference in the alternate segment. Join SP and SQ.

Since OP \(\perp\) PT (radius through the point of contact is 1 to the tangent

⇒ \(\angle 2+\angle 1=90^{\circ}\)

⇒ \(\angle 2+60^{\circ}=90^{\circ}\)

⇒ \(\angle 2=90^{\circ}-60^{\circ}=30^{\circ}\) (given)

But O P=O Q (each radii)

⇒ \(\angle 2=\angle 3\) (angles opposite to equal sides are equal)

But O P=O Q

⇒ \(\angle 2=\angle 3\) (angles opposite to equal sides are equal)

Now in \(\triangle P O Q\),

⇒ \(\angle 2+\angle 3+\angle 4 =180^{\circ}\)

⇒ \(30^{\circ}+30^{\circ}+\angle 4 =180^{\circ}\)

⇒ \(\angle 4 =120^{\circ}\)

⇒ \(\angle 5 =\frac{1}{2} \times \angle 4\) (angle sum property) [from (1) and (2)]

(The degree measure of an area is twice the angle subtended by it in an alternate segment)

⇒ \(\angle 5=\frac{1}{2} \times 120^{\circ}\)

⇒ \(\angle 5 =60^{\circ}\)

∴ Also,\(\angle 5+\angle 6 =180^{\circ}\)

⇒ \(60^{\circ}+\angle 6 =180^{\circ}\)

∴ \(\angle 6 =\angle P R Q=120^{\circ}\)

Example 5. In the given figure two circles touch each other at point C. Prove that the common tangent to the circle at C bisects the common tangent at P and Q.

Solution:

In the given figure, PR and CR are both tangents drawn to c the same circle from an external point R

Circles Two Circles Touch Each Other At A Point

PR = CR …(1)

Also, QR and CR are both tangents drawn to the same circle from an external point R.

QR = CR -(2)

From ( 1 ) and (2) we get

PR = QR

R is the mid-point of PQ

i.e., the common tangent at C bisects the common tangents at P and Q

Example 6. Two circles of unequal radii neither touch nor intersect each other. Are the common tangents AB and CD always equal? If no, then give an explanation of it and if your answer is yes, then prove it.

Circles Two Circles OF Unequal Radii Neither Touch Nor Intersect Each Other

Solution:

Let the two tangents AB and CD on producing meet at P.

Since PA and PC are tangents from an external point P to the circle with centre O

PA=PC

Also, PB and PD are tangents from an external point P to the circle with centre O’.

Circles The Tangents From A External Point To The Circle With Centre

PB = PD …(2)

Subtracting (2) from (I), we get

PA – PB = PC- PD

AB = CD

So, the direct common tangents are of equal length

Example 7. In the adjoining figure, common tangents AB and CD to two circles intersect at P. Prove that AB = CD.

Solution:

Circles In Adjoining Common Tangents Of Two Circles Intersect

  • Since PA and PC are two tangents to a circle with centre O from an external point P.
  • PA =PC
  • Also, since PB and PD are two tangents to a circle with centre O’ from an external point P.
  • PB = PD
  • Adding (2) and ( 1 ), we get
  • PA + PB = PC + PD
  • AB = CD (AB and CD are two straight lines)
  • Hence Proved.

Example 8. In the given diagram, PQ and RS arc common tangents to the two circles with centres C and D. Find the length of PQ and hence the area of trapezium RSDC.

Solution:

Circles The Common Tangents To The Two Circles With Centers

Draw CM//PS so that DCMSP becomes a rectangle.

Now, we have CR = 2 cm, DS = 7cm and CD = 13 cm

DM = DS- MS

= DS- CR

= 7-2 = 5 cm

In right ADMC, by Pythagoras theorem,

Circle The Length Of A Common Tangents To Two Circles Are Always Same

⇒ \(C M^2 =C D^2-D M^2\)

=\((13)^2-(5)^2=(12)^2\)

RS = 12 cm (opposite sides of the rectangle are equal)

PQ = 12 cm (length of common tangents to two circles are always same)

Now, ar(Trapezium RSDC)

= \(\frac{1}{2} \times\) h sum of parallel sides)

= \(\frac{1}{2} \times C M(C R+D S)\)

= \(\frac{1}{2} \times 12(2+7)=54 \mathrm{~cm}^2\)

Example 9. AB is the diameter of a circle with centre O. AH and BK are perpendiculars from A and B to the tangent at a point P on the circle. Prove that AD + BK = AB.

Solution:

Let AH = x, BK=y and BM = z

Circles AB Is A Diameter Of A Circle With Centre

⇒ \(\triangle M B K \sim \triangle M A H\) (AA corollary)

⇒ \(\frac{B K}{A H}=\frac{B M}{A M} \quad \Rightarrow \quad \frac{y}{x}=\frac{z}{2 r+z}\)

⇒ \(2 r y+y z=x z \quad \Rightarrow \quad z(x-y)=2 r y\)

z=\(\frac{2 r y}{x-y}\)

Similarly, \(\triangle M B K \sim \triangle M O P\)

⇒ \(\frac{B K}{O P}=\frac{B M}{O M} \quad \Rightarrow \quad \frac{y}{r}=\frac{z}{z+r}\)

y z+y r=z r

z(r-y)=y r

z=\(\frac{y r}{r-y}\)

From (1) and (2), we get

⇒ \(\frac{2 y}{x-y} =\frac{y r}{r-y} \quad \Rightarrow \quad \frac{2}{x-y}=\frac{1}{r-y}\)

2 r-2 y =x-y

x+y 2 r

A H+B K =A B

Example 10. In the given figure, if AB =AC, prove that BE = EC.

Solution:

We know that lengths of tangents from an external point are equal.

Circles Length Of Tangents From An External Point Are Equal

AD =AF  → Equation 1

DB=BE    → Equation 2

EC =FC  →  Equation 3

Now, it is given that

AB=AC

AD + DB =AF + FC

AD + DB = AB + E C From 1 And 3

D B = E C

B E = E C [From (2), DB = BE]

∴ Hence proved

Example 11. In the given figure ABC is a right-angled triangle with AB = 6 K cm, and BC = 8 cm. A circle with a centre O has been inscribed inside the triangle. Find the radius of the circle.

Solution:

Circles A Circle With Centre Has Been Inscribed Inside The Triangle

Let x be the radius of the circle. In the right triangle ABC

⇒ \(A C^2 =A B^2+B C^2\) (by Pythagoras Th.)

⇒ \(A C^2 =6^2+8^2\)

⇒ \(A C^2\) =36+64

⇒ \(A C^2\) =100

A C =10

Now in quadrilateral OPBR

⇒ \(\angle B =\angle P=\angle R=90^{\circ}\) each

⇒ \(\angle R O P =90^{\circ}\)sum of all angles of a quadrilateral is 360°) (each radii)

and also OP = OR (each radii)

Hence, OPBR is a square with each side x cm.

Therefore, CR = (8 -x) and PA = (6 -x)

BP = RB = x cm

Since the tangents from an external point to a circle are equal in length

AQ = AP = (6-x) and CQ = CR = (8 -x)

Now, AC = AQ + CQ

10 = 6 -x + 8 -x

10 = 14 – 2

2x = 4 = 2 cm

Example 12. A circle is touching the side BC of a \(\triangle\) ABC at the point and touching/IB and AC produced at Q and li respectively. Prove that AQ = \(\frac{1}{2}\) (perimeter of \(\triangle\) ABC).

Solution:

Given : \(\triangle ABC\) and a circle which touches BC, AB and AC in P, Q and R respectively.

Proof: Since the length of the two tangents drawn from an external
point to a circle are is equal, therefore,

Circles A Circle Is Touching The Side At A Point

AQ = AR

BQ = BP

CP = CR

Now, perimeter of \(\triangle ABC\) = AB + BC + AC

= AB +BP + PC +AC

= AB + BQ + CR + AC [from (2) and (3)]

= AQ+AR = 2AQ [from (1)]

Perimeter of \(\triangle ABC\) = \(\frac{1}{2} \times \text { (perimeter of } \triangle A B C)\)

Example 13. In the given figure, PA and PB are tangents to the circle from an external point P. CD is another tangent touching the circle at Q. IfPA = 12 cm, QC = QD = 3 cm, then find PC + PD.

Solution:

Circle The Length Of The Two Tangents Drawn From An External Point To A Circle

Since the lengths of the two tangents drawn from an external point to a circle are equal,

PA = PB  → Equation (l)

CA = CQ  → Equation (2)

DB = DQ  →  Equation (3)

Now, PA = 12

PC + CA= 12 (given)

PC + CQ = 12  From 2

PC + 3 = 12 ⇒ PC= 9 cm  →  Equation 4

PB=PA = 12

PD +DB= 12

PD +DQ= 12

PD + 3 = 12

PD = 9 cm  → Equation 5

PC + PD = (9 + 9) cm = 18 cm [From 4 and 5]

Example 14. O is the centre of a circle of radius 5 cm. T is a point such that OT = 13 cm and OT intersects the circle at E. If AB is the tangent to the circle at E, find the length AB.

Circle O Is The Centre Of The Circle Of Radius

Solution:

Since, \(\angle O P T=90^{\circ}\) (radius through point of contact is \(\perp\) to the tangent)

In right \(\triangle O P T\),

⇒ \(OP^2+P T^2=O T^2\)

⇒ \(P T^2=O T^2-O P^2\)

⇒ \(P T^2=(13)^2-(5)^2=(12)^2\)

⇒ \(P T=12 \mathrm{~cm}\)

Let A P=x cm

At E=A P=x (lengths of tangents from an external point are equal)

A T=T P-A P=12-x

E T=O T-O E=13-5=8 cm

⇒ \(\angle A E T=90^{\circ}\)

Now; since \(\angle A E O=90^{\circ}\)(radius through point of contact is \(\perp\) to the tangent)

In right \(\triangle A E T\), by Pythagoras theorem,

⇒ \(A E^2+E T^2=A T^2\)

⇒ \(x^2+(8)^2=(12-x)^2\)

⇒ \(x^2+64=144+x^2-24 x \)

⇒ \(24 x=144-64=80\)

x=\(\frac{80}{24}=\frac{10}{3}\)

Similarly, B E=\(\frac{10}{3} \mathrm{~cm}\)

⇒ \(A B=A E+B E=\left(\frac{10}{3}+\frac{10}{3}\right) \mathrm{cm}=\frac{20}{3} \mathrm{~cm}\)

A B=\(\frac{20}{3} \mathrm{~cm}\)

Example 15. In the given figure, T is tangent to the circle with centre O such that O T=4 cm and \(\angle O T A=30^{\circ}\). Find the length of segment AT.

Circle A Tangent To The Circle With The Centre

Solution:

In right \(\triangle O A T\),

⇒ \(\cos 30^{\circ} =\frac{A T}{O T}\)

∴ \(\frac{\sqrt{3}}{2} =\frac{A T}{4} \quad A T=2 \sqrt{3} \mathrm{~cm}\)

Example 16. In the given figure, OP is equal to the diameter of the circle. Prove that \(\triangle ABP\) is an equilateral triangle.

Solution: 

Circles Op Is Equal To The Diameter Of The Circle Of An Equilateral Triangle

Let, \(\angle O P A=\angle O P B=\theta\)( tangents are equally inclined at the centre) and the radius of the circle be r.

Since, \(\angle 1=90^{\circ}\) (radius through point of contact is \(\perp\) to the tangent)

In right \(\triangle O A P\),

⇒ \(\sin \theta =\frac{O A}{O P}=\frac{r}{2 r}=\frac{1}{2}=\sin 30^{\circ}\)

⇒ \(\theta \Rightarrow \quad 30^{\circ} \quad \Rightarrow A P B=2 \theta=2 \times 30^{\circ}=60^{\circ}\)

Now, since PA = PB (length of tangents from an external point are equal)

⇒ \(\angle 2=\angle 3\) (angles opposite to equal sides are equal)

In \(\triangle M P B\).

⇒ \(\angle 2+\angle 3+\angle A P B=180^{\circ}\)

⇒ \(\angle 2+\angle 2+60^{\circ}=180^{\circ}\)

⇒ \(\angle 2=\angle 3=60^{\circ}\)

⇒ \(\angle 2+\angle 3+\angle A P B =180^{\circ}\) (angle sum property)

⇒ \(\angle 2+\angle 2+60^{\circ} =180^{\circ}\) [from (1) and (2)]

2 \(\angle 2 =120^{\circ} \quad \Rightarrow \quad \angle 2=60^{\circ}\)

⇒ \(\angle 2=\angle 3=60^{\circ}\)

So, all the angles of \(\triangle A P B are 60^{\circ}\).

∴ \(\triangle A P B\) is an equilateral triangle.

Example 17. If from an external point B of a circle with centre O, two tangents BC and BD are drawn such that \(\angle D B C=120^{\circ}\), prove that B C+B D=B O.

Solution:

⇒ \(\angle 1+\angle 2 =120^{\circ}\)

But \(\angle 1 =\angle 2\)

Circle An External Point B Of A Circle With Centre O

⇒ \(\angle 1+\angle 1 =120^{\circ}\)

2 \(\angle 1 =120^{\circ}\)

⇒ \(\angle 1 =60^{\circ}\)

Also, \(\angle O C B =90^{\circ}\) (radius through point of contact is \(\perp\) to the tangent)

Now, in right \(\triangle O C B\),

⇒ \(\cos 60^{\circ}=\frac{B C}{O B} \quad \Rightarrow \quad \frac{1}{2}=\frac{B C}{O B}\)

O B=B C+B C

O B=2 B C \(\quad \Rightarrow \quad O B=B C+B C\)

O B=B C+B D(length of tangents from an external point are equal)

∴ Hence Proved.

Example 18. In the adjoining figure, AB is a chord of length 9.6 cm of a circle with centre O and radius 6 cm. The tangents at A and B intersect at P. Find the length of PA.

Solution:

Join OP, OA and OB.

Circle In The Adjoining Figure, AB Is a Chord Of Length

Let PA = x cm and PM = y cm

P A =P B (length of tangents from an external point are equal)

P M =P M (common)

⇒ \(\angle 1 =\angle 2\) (tangents are equally inclined at the centre)

⇒ \(\triangle A M P \cong \triangle B M P\) (SAS congruency)

Circle The Length Of A PA With A Centre O

A M=M B=\(\frac{9.6}{2}=4.8 \mathrm{~cm}\)

and \(\angle P M A=\angle P M B\)

But \(\angle P M A+\angle P M B=180^{\circ}\)

⇒ \(\angle P M A=\angle P M B=90^{\circ}\)

Now, in right \(\triangle A M P\),

⇒ \(x^2=y^2+(4.8)^2\) (by Pythagoras theorem)

Also, in right \(x^2=y^2+(4.8)^2\)\triangle A M O\(x^2=y^2+(4.8)^2\),

⇒ \((4.8)^2+O M^2 =(6)^2\)

⇒ \(O M^2 =36-23.04=12.96\)

⇒ \(O M =\sqrt{12.96}=3.6 \mathrm{~cm}\)

Now, \(\angle O A P=90^{\circ}\) (radius through point of contact is \perp to the tangent)

In right \(\triangle M O P\),

⇒ \(O P^2 =O A^2+A P^2 \Rightarrow(y+3.6)^2=36+x^2\)

⇒ \(y^2+12.96+7.2 y =36+y^2+(4.8)^2\)

⇒ \(7.2 y =36+23.04-12.96 \quad \Rightarrow \quad 7.2 y=46.08 \)

y =\(\frac{46.08}{7.2}=6.4 \mathrm{~cm}\)

Put this value of y in equation (1),

⇒ \(x^2 =(6.4)^2+(4.8)^2=40.96+23.04=64\)

x = 8 cm

Hence, x=8 cm and y=6.4 cm

Example 19. The radii of two concentric circles are 1 3 cm and 8 cm. AB is the diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

Solution:

Produce BD to E which cuts the circle at E. Join AE and OD.

Since AB is the diameter of the bigger circle.

⇒  \(\angle A E B=90^{\circ}\) (angle in a semicircle is right angle)

Also, \(\angle O D B=90^{\circ}\)(radius through point of contact is \(\perp\) to the tangent)

Circle The Radii Of Two concentric Circles

Now, in \(\triangle B O D\) and \(\triangle B A E\)

⇒ \(\angle B =\angle B\) (common)

⇒ \(\angle O D B =\angle A E B\) (each \(90^{\circ}\) )

⇒ \(\Delta B O D =\triangle B A E\)

⇒ \(\frac{O D}{A E} =\frac{O B}{A B}\)(corresponding sides of similar triangles are proportional)

⇒ \(\frac{8}{A E}=\frac{r}{2 r} \quad \Rightarrow A E=16 \mathrm{~cm}\)

Since, \(O D \perp E B\)

D E=D B (\(\perp\) drawn from the centre to the chord bisects the chord)

In right \(\triangle O D B\),

⇒ \(D B^2 =O B^2-O D^2=(13)^2-(8)^2=169-64=105\)

D B =\(\sqrt{105} \mathrm{~cm}=E D\)

Now, in right \(\triangle A E D\), by Pythagoras theorem

⇒ \(A D^2 =A E^2+E D^2\)

⇒ \(A D^2 =(16)^2+105=256+105=361\)

∴ \(A D =\sqrt{361}\) i.e.. 19 cm

Hence, A D =19 cm.

Circles Exercise 10.1

Question 1. How many tangents can a circle have?

Solution :

Infinitely many tangents can be drawn on a circle.

Question 2. Fill in the blanks :

  1. A tangent to a circle intersects it in point (s).
  2. A line intersecting a circle in two points is called a
  3. A circle can have parallel tangents at the most.
  4. The common point of a tangent to a circle and the circle is called

Answer :

  1. one
  2. secant
  3. two
  4.  point of contact

Question 3. A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm, Length PQ is :

Circle A Tangent At A Point P Of A Circle Of Radius

  • 12 cm
  • 13 cm
  • 8.5 cm
  • \(\sqrt{119} \mathrm{~cm}\)

Solution : 4. \(\sqrt{119} \mathrm{~cm}\)

Here OP = 5 cm,

OQ = 12 cm In \(\triangle\)POQ

In \(\triangle P O Q \)

⇒ \(P Q^2 =O Q^2-O P^2\)

=\(12^2-5^2=119\)

P Q =\(\sqrt{119}\) cm

Question 4. Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.

Solution:

  1. Let the centre of the circle be O. The line AB lies outside the circle. Draw the perpendicular OM from O to AB.
  2. The perpendicular OM intersects the circle at P. Draw a line EPF from P parallel to AB.
  3. EF is the required tangent of the circle.
  4. Draw a line CD parallel to AB which intersects the circle at two points. It is the required secant of the circle.

Circle A Circle And Two Lines Parallel To A Given Line

Circles Exercise 10.2

Question 1. From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is :

  1. 7 cm
  2. 12 cm
  3. 15 cm
  4. 24.5 cm

Answer: 1.

Here, PQ = 24 cm and OQ = 25 cm

In \(\triangle O P Q\)

⇒ \(O P^2=O Q^2-P Q^2\)(from Pythagoras theorem)

Circle The Radius Of Circle From A Point

= \(25^2-24^2=625-576=49\)

OP = 7 cm

Radius of circle = 7 cm

Question 2. In the figure, if TP and TQ arc the Uvo tangents to a circle with centre O so that ZPOQ = 110°, then Z PTQ is equal to :

  1. 60°
  2. 70°
  3. 80°
  4. 90°

Answer: 2.

Here TP and TQ are the tangents to the circle.

Circle The Two Tangents To A Circle With The Centre

⇒ \(\angle\)OPT = \(\angle\)OQT = 90°

In □ OPTQ,

⇒ \(\angle\)OPT + \(\angle\)PTQ + \(\angle\)OQT + \(\angle\)POQ = 360°

90° + ZPTQ + 90° + 110° = 360°

⇒ \(\angle\)PTQ = 360° -290°

∴ \(\angle\)PTQ = 70°

Question 3. If tangents PA and PB from a point P to a circle with centre O are inclined to each other at an angle of 80°, then ∠POA is equal to :

  1. 50°
  2. 60°
  3. 70°
  4. 80°

Answer: 1. 50°

PA and PB are two tangents of the circle.

⇒ \(\angle\)PAO = \(\angle\)PBO = 90°

Circle The Tangents From A Point Of Circle With The Centre

Given : \(\angle\)APB =80°

Now, in □PAOB,

⇒ \(\angle\)AOB + \(\angle\)PAO + \(\angle\)APB + \(\angle\)PBO = 360°

⇒ \(\angle\)AOB + 90° + 80° + 90° = 360°

⇒ \(\angle\)AOB = 360° – 260° = 100°

Now, PO bisects \(\angle\)AOB.

⇒ \(\angle\)POA = – \(\angle\)AOB = – x 100° = 50°

Question 4. Prove that the tangents drawn at the ends of the diameter of a circle are parallel.

Solution :

Let AB be the diameter of a circle with centre O. PA and PB are the tangents to the circle at A points A and B respectively.

Circle The Tangents Drawn At The Ends Of A Diameter Of A Circle

Now Z\(\angle\)PAB = 90°

and \(\angle\)QBA = 90°

\(\angle\)PAB +QBA = 90° + 90° = 180°

PA || QB Hence Proved.

Question 5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Solution :

Given: A circle with centre O a tangent AQB and a perpendicular PQ is drawn from the point of contact Q to AB.

To Prove: The perpendicular PQ passes through the centre of the circle.

Proof: AQ is the tangent of the circle at point Q.

AQ will be the perpendicular to the radius of the circle.

PQ \(\perp\) AQ

The centre of the circle will lie on the line PQ.

Perpendicular PQ passes through the centre of the circle.

Hence Proved.

Question 6. The length of a tangent from point A at a distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.

Solution :

Let O be the centre of the circle and PQ is a tangent to the circle from point P.

Given that, PQ = 4 cm and OP = 5 cm

Circle The Length Of A Tangent From A Point At A Distance

Now,\(\angle O Q P=90^{\circ}\)

In \(\triangle O Q P\) ,

⇒  \(O Q^2=O P^2-P Q^2\)

=\(5^2-4^2\)

= 25-16 = 9

OQ = 3 cm

Radius of circle = 3 cm

Question 7. Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle.

Solution :

Here, we draw two circles C1 and C2 with radii = 3 cm and r2 = 5 cm c2 respectively.

Circles The Length Of The Chord Of The Larger Circle Touches The Smaller Circle

Now, we draw a chord AB which touches the circle C1 at D.

O is the centre of concentric circles.

Now we draw the perpendicular from O to AB which bisects AB at D.

i.e., AD = BD

In right \(\triangle O B D\),

⇒ \(O B^2=O D^2+D B^2\) (from Pythagoras theorem)

⇒ \(5^2 =3^2+D B^2\)

⇒ \(D B^2\) =25-9=16

DB = 4 cm

Length of chord = AB = 2BD

= 2 x 4 = 8 cm

Question 8. A quadrilateral ABCD is drawn to circumscribe a circle. Prove that AB + CD=AD + BC.

Circle An Quadrilateral ABCD Is Drawn To The Circumscribe A Circle

Solution :

The sides of quadrilateral ABCD touch the circle at P, Q, R and S as shown in the figure. We know that the tangents drawn from an external point to the circle are equal.

AP = AS, BP = BQ,

CR = CQ, DR = DS

On adding, AP + BP + CR + DR

= AS + BQ + CQ + DS

⇒ AB + CD = (AS + DS) + (BQ + CQ)

⇒ AB + CD = AD + BC

Hence Proved.

Question 9. In the figure, XY and X’ Y’ are two parallel tangents to a circle with centre O and another tangent AB with the point of contact C intersecting XY at A and X’ Y’ at B. Prove that \(\angle\)AOB = 90°.

Circle Two Parallel Tangents To A Circle With Centre O And Another Tangent

Solution :

The tangents from an external point to a circle are equal.

AP = AC

In \(\triangle A P O\) and \(\triangle A C O\),

AP = AC

AO = AO (common)

OP = OC (radii of a circle)

From S.S.S. congruency,

⇒ \(\triangle A P O \cong \triangle A C O\)

⇒ \(\angle P A O=\angle O A C\)

⇒ \(\angle P A C=2=\angle C A O\)

Similarly, we can prove that

⇒ \(\angle C B O =\angle O B Q\)

⇒ \(\angle C B Q =2 \angle C B O\)

⇒ \(Y \| X^{\prime} Y^{\prime}\)

⇒ \(\angle P A C+\angle Q B C=180^{\circ}\)

(sum of interior angles of the same side of a transversal is \(180^{\circ}\) )

2. \(\angle C A O+2 \angle C B O =180^{\circ}\)

⇒ \(\angle C A O+\angle C B O =90^{\circ}\)

In \(\triangle A O B\),

⇒  \(C A O+\angle C B O+\angle A O B=180^{\circ}\)

⇒ \(\angle C A O+\angle C B O=180^{\circ}-\angle A O B\) From equations (1) and (2)

⇒ \(180^{\circ}-\angle A O B=90^{\circ}\)

∴ \(\angle A O B=90^{\circ}\) Hence Proved.

Question 10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtends ey the line segment joining the points of contact at the centre.

Solution :

PA and PB are the tangents of the circle.

\(\angle O A P=\angle O B P=90^{\circ}\)

In O A P B

⇒  \(\Rightarrow \quad 90^{\circ}+\angle A P B+90^{\circ}+\angle A O B=360^{\circ}\)

⇒ \(\angle A P B+\angle A O B=180^{\circ}\)

⇒ \(\angle A P B\) and \(\angle A O B\) are supplementary.

Hence Proved.

Question 11. Prove that the parallelogram circumscribing a circle is a rhombus.

Circle The Parallelogram Circumscribing A Circle

Solution :

Let a parallelogram ABCD is given. Let the parallelogram touch the circle at points, P, Q, R and S.

AP and AS are the tangents drawn from an -external point A to the circle.

AP = AS …(1)

Similarly, BP = BQ _ (2)

CR = CQ (3)

DR = DS (4)

Adding equations (1), (2), (3) and (4),

AP + BP + CR+DR = AS + BQ + CQ + DS

(AP + BP) + (CR + DR)

= (AS+ DS) + (BQ + CQ)

AB + CD = AD+BC

AB + AB = AD + AD {CD = AB, BC = AD, opposite sides of a parallelogram)

2 AB = 2 AD

AB = AD

So, ABCD is a rhombus. (adjacent sides of a parallelogram are equal)

Hence Proved.

Question 12. A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively. Find the sides AB and AC.

Circle A Triangle ABC Is Drawn To The Circumscribe A Circle Of Radius

Solution :

Given : CD = 6 cm,

BD = 8 cm and radius = 4 cm

Join OC, OA and OB.

Circle The Tangents Drawn From An External Point To A Circle

We know that the tangents drawn from an external point to a circle are equal.

CD = CF = 6 cm and BD = BE = 8 cm

Let AF = AE = x cm c

In \(\triangle\)OCB,

area of triangle \(A_1 =\frac{1}{2} \times \text { base } \times \text { height }\)

= \(\frac{1}{2} \times C B \times O D\)

= \(\frac{1}{2} \times 14 \times 4=28 \mathrm{~cm}^2\)

In \(\triangle O C A\),

area of triangle

⇒ \(A_2 =\frac{1}{2} \times A C \times O F \)

=\(\frac{1}{2}(6+x) \times 4\)

=12+2 x

In \(\triangle O B A\).

area of triangle

⇒ \(A_3 =\frac{1}{2} \times A B \times O E\)

=\(\frac{1}{2}(8+x) \times 4\)

=16+2 x

Now, semiperimeter of triangle ABC,

s =\(\frac{1}{2}(A B+B C+C A)\)

s =\(\frac{1}{2}(x+6+14+8+x)\)

=14+x

Now, area of \(\triangle A B C\)

=\(A_1+A_2+A_3\)

=28+(12+2 x)+(16+2 x)

=56+4 x

From Heron’s formula,

Area of \(\triangle A B C\)

= \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{(14+x)(14+x-14)(14+x-x-6)} \quad(14+x-x-8)\)

= \(\sqrt{(14+x)(x)(8)(6)}\)

= \(\sqrt{(14+x) 48 x}\)

From equations (1) and (2),

⇒  \(\sqrt{(14+x) 48 x}=56+4 x\)

Squaring both sides,

⇒ \((14+x) 48 x =4^2(14+x)^2\)

3 x =14+x

⇒ \(2 x=14-\Rightarrow \quad x\) =7

A C=6+x=6+7 =13 cm

A B=8+x=8+7 =15 cm

Question 13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Solution :

We know that the tangents drawn R from an external point to a circle subtend equal angles at the centre.

⇒ \(\angle 1=\angle 2\) ,

⇒ \(\angle 3=\angle 4\) ,

⇒ \(\angle 5=\angle 6\)

and \(\angle 7=\angle 8\)

Circles The Opposite Sides Of A Quadrilateral Circumscribing A Circle

Now, \(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5+\angle 6+\angle 7+\angle 8 =360^{\circ}\)

⇒ \(2 \angle 2+2 \angle 3+2 \angle 6+2 \angle 7 =360^{\circ} \)

⇒ \((\angle 2+\angle 3)+(\angle 6+\angle 7) =180^{\circ}\)

⇒ \(\angle A O B+\angle C O D =180^{\circ}\)

Similarly, \(\angle B O C+\angle A O D =180^{\circ}\)

Hence Proved.

Circles Multiple Choice Questions

Question 1. If the angle between two radii of a circle is 1 10° then the angle between the tangents drawn at the ends of these radii is :

  1. 110°
  2. 100°
  3. 90°
  4. 70°

Answer: 4. 70°

Question 2. If two tangents PA and PB drawn from point P are of equal length 4 cm, then the radius of the die circle is :

Circle The Two Tangents Are Drawn From A Point Are Equal Length

  1. 1 cm
  2. 2 cm
  3. 4 cm
  4. 3 cm

Answer: 3. 4 cm

Question 3. In the adjoining figure, PA and PB are tangents to a circle with centre 0 such that \(angle\)APB = 40°. \(angle\)OAB is :

Circle The Tangent To A Circle With The Centre O

  1. 20°
  2. 40°
  3. 30°
  4. 15°

Answer: 1. 20°

Question 4. If two tangents of a circle of radius 6 cm are drawn such that the angle between them is 60 then the length of each tangent is :

  1. 2 \(\sqrt{3} \mathrm{~cm}\)
  2. 6 \(\sqrt{3} \mathrm{~cm}\)
  3. 3 cm
  4. 6 cm

Answer: 2. 6 \(\sqrt{3} \mathrm{~cm}\)

Question 5. Two tangents PQ and PR are drawn to a circle of radius 5 cm where P is 13 cm away from centre O. The area of quadrilateral PQOR is :

  1. \(60 \mathrm{~cm}^2\)
  2. \(30 \mathrm{~cm}^2\)
  3. \(65 \mathrm{~cm}^2\)
  4. \(32.5 \mathrm{~cm}^2\)

Answer: 1. \(60 \mathrm{~cm}^2\)

Question 6. In the adjoining figure, O is the centre of the circle, PQ is the chord and the tangent PR drawn from points on the circle makes a 50° angle from chord PQ, \(\angle\) POQ is :

Circle The Chord And The Tangent Are Drawn From A Point

  1. 90°
  2. 80°
  3. 100°
  4. 75°

Answer: 3. 100°

Question 7. The radii of two circles are 3 cm and 4 cm and both circles touch each other externally. The distance between their centres is :

  1. 1 cm
  2. 3 cm
  3. 5 cm
  4. 7 cm

Answer: 4. 7 cm

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions

Constructions Division Of A Line Segment

To divide a line segment (internally) in a given ratio m: n

Working Rule: (Internal division)

Draw a line segment AB of a given length.

Draw a ray AX making an acute angle XAB with AB.

Mark (m + n) points A1, A2, A3, …, Am+n on AX such that AA1 = A1A2 = A2A3 = … = Am+n-1 Am+n.

Join Am+n B.

Through Am, draw AmY || Am+n B (if m: n) meeting AB at Y. So, Y divides AB internally in the ratio m: n.

Constructions Meeting AB At Y

Through An, draw An1Z || Am+n B (if n: m) meeting AB at Z. So, Z divides AB internally in the ratio n: m.

Constructions Meeting AB At Z

In Short:

Sum (m +n) endpoint

first (m) Parallel (Y)  (say)

NCERT Exemplar For Class 10 Maths Chapter 11 Constructions

Constructions Solved Problems

Question 1. Determine a point which divides a line segment 7 cm long, internally in the ratio 2:3.

Solution:

Steps of Construction:

  1. Draw a line segment AB = 7 cm by using a ruler.
  2. Draw any ray malting an acute ZBAC with AB.
  3. Along AC, mark off (2 + 3) = 5 points A1, A2, A3, A4 and A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
  4. Join BA5
  5. Through A2 draw a line A2P parallel to A5B by making an angle equal to \(\angle A A_5 B\) at A2 intersecting AB at A point P.

The point P so obtained is the required point.

Constructions Meeting Acute Angle

Justification: In ΔA5B,

A2P || A5B (Construction)

∴ \(\frac{A A_2}{A_2 A_5}=\frac{A P}{P B}\) (by B.P theorem)

⇒ \(\frac{2}{3}=\frac{A P}{P B}\) (Construction)

⇒ AP : PB = 2: 3

i.e., P divides AB internally in the ratio 2 : 3.

Alternate Method:

Draw the line segment AB = 7 cm.

Draw any ray AC making an acute angle ∠BAC with AB.

Draw a ray BD parallel to Ac by making ∠ABC equal to angle ∠BAC.

Mark off 2 points A1 and A2 on AC and 3 points B1, B2, B3 on AD such that AA1 = A1A2 = BB1 = B1B2 = B2B3

Join B3A2, suppose it intersects AB at point P. Then, P is the required point.

Constructions Meeting Intersects AB At Point P

To Divide a Line Segment (Externally) in a Given Ratio m: n

Working Rule: (External Division)

Constructions Meeting To Divided A Line Segment

Question 2. Determine a point which divides a line segment 6 cm long externally in the ratio 5 : 3.

Solution:

Steps of Construction:

  1. Draw a line segment AB = 6 cm.
  2. Draw any ray making an acute ∠BAX with AB.
  3. Along AX, mark off (larger among the ratios) 5 points A1, A2, A3, A4, A5 such that AA1 = A1A2 = A2A3 = A3A4 = A4A5.
  4. Join the point A2 (5-3) with end point B.
  5. Draw a line parallel to A2B from A5 (larger among the ratios) which meets AB produced at P.

The point P, so obtained is the required point such that AP: BP = 5 : 3.

Constructions Meets AB Produced At P

Justification: In AA5P,

Since A2B || A5P, (Construction)

∴ \(\frac{A P}{B P}=\frac{A A_5}{A_2 A_5}\) (by B.P theorem)

⇒ \(\frac{A P}{B P}=\frac{5}{3}\) (Construction)

Question 3. Determine a point which divides a line segment 6 cm long externally in the ratio 3:5.

Solution:

Steps of Construction:

  1. Draw a line segment AB = 6 cm.
  2. Draw any ray making an acute ∠ABX with AB.
  3. Along BX, mark off (larger among the ratios) 5 points B1, B2, B3, B4 and B5 such that BB1 =B1B2 = B2B3 — B3B4 = B4B5.
  4. Join the point B2 (5 – 3) with endpoint A
  5. Draw a line parallel to B2A from B5 (larger among the ratios) which meets BA produced at P.

The point P so obtained is the required point such that AP: BP = 3:5.

Constructions Meets BA Produced At P

Justification: In ΔPBB5,

Since, B2A || B5P (Construction)

∴ \(\frac{B_2 B_5}{B B_5}=\frac{A P}{B P}\) (by B.P theorem)

⇒ \(\frac{3}{5}=\frac{A P}{B P}\) (Construction)

i.e., P divides AB externally in the ratio of 3:5

Constructions To Construct A Triangle Similar To A Given Triangle

Question 1. Construct a triangle similar to a given triangle ABC such that each of its sides is \(\frac{2}{3} \mathrm{rd}\) of the corresponding sides of the triangle ABC. It is given that AB = 4 cm, BC = 5 cm and AC = 6 cm.

Solutions:

Steps of construction:

  1. Take BC = 5cm and Construct ΔABC with BA = 4cm and CA = 6cm.
  2. Divide BC into three equal parts.
  3. Let C be a point on BC such that \(B C^{\prime}=\frac{2}{3} B C\)
  4. Draw A’C parallel to AC through C” intersecting BA at A’. ΔA’BC” is the required triangle.

Constructions Triangle

Question 2. Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then construct another triangle whose sides are \(\frac{3}{4}\) times the corresponding sides of the first triangle.

Solution:

Steps to Construction:

  1. Draw a line segment BC = 6 cm.
  2. Draw a perpendicular bisector of BC.
  3. From mid-point D of BC on perpendicular bisector mark DA = 4 cm, join AB and AC.
  4. Below BC make an acute angle ∠CBZ.
  5. Along BZ mark off four points B1, B2, B3 and B4 such that BB1 = B1B2 = B2B3 = B3B4.
  6. Join B4C.
  7. From B3 draw B3E parallel to B4C meeting BC at E.
  8. From E draw EF || CA meeting BA at F. Then, ΔFBE is the required triangle.

Constructions Isosceles Triangle

Question 3. Construct a quadrilateral ABCD with AB = 3 cm, AD = 2.7 cm, DB = 3.6 cm, ∠B =110° and BC = 4.2 cm. Construct another quadrilateral A’BC’D’ similar to quadrilateral ABCD so that diagonal BD’ = 4.8 cm.

Solutions:

Steps to construction:

  1. Draw a line segment BC = 4.2 cm
  2. At B, construct angle YBC = 110°
  3. With centre B and a radius equal to 3 cm, draw an arc-cutting BT at A.
  4. With centre A and a radius equal to 2.7 cm, draw an arc.
  5. With centre B and radius equal to 3.6 cm, draw another arc cutting the previous arc at D.
  6. Join AD, CD and BD. Then, ABCD is the required quadrilateral.
  7. Produce BD to D’ such that BD’ = 4.8 cm.
  8. From D’, draw a line parallel to DA which cuts BY at A’.
  9. From D’, draw a line parallel to DC which cuts BC produced at D’.

Then, □ - Wiktionary, the free dictionaryA’BC’D’ is the required quadrilateral similar to □ - Wiktionary, the free dictionaryABCD.

Constructions A Quadrilateral

Question 4. Construct a cyclic quadrilateral ABCD in which AB = 4.2 cm, BC = 5.5 cm, CA = 4.6 cm and AD = 3 cm. Also, construct a quadrilateral similar to □ - Wiktionary, the free dictionaryABCD whose sides are 1 .5 times the corresponding sides of □ - Wiktionary, the free dictionaryABCD.

Solution:

Steps to construction:

  1. Draw a line segment AB = 4.2 cm.
  2. With centre A and a radius equal to 4.6 cm, draw an arc.
  3. With centre B and radius equal to 5.5 cm, draw another arc cutting the previous arc at C.
  4. Join AC and BC.
  5. Draw the perpendicular bisectors of any two sides say AB and BC respectively of ΔABC. Let them intersect each other at O.
  6. Taking O as the centre and radius as OA or OB or OC, draw a circle. This is the circumcircle of ΔABC.
  7. With centre A and radius equal to 3 cm, cut an arc on the opposite side of B, to cut the circle at D.
  8. Join AD and CD. Then, □ - Wiktionary, the free dictionaryABCD is the required cyclic quadrilateral.
  9. Produce Ac to C’ such that \(A C^{\prime}=1.5 \times A C \text { i.e., }\left(1+\frac{1}{2}\right) A C \Rightarrow A C+\frac{1}{2} \times 4.6\) i.e., 2.3cm more.
  10. From C’, draw a line parallel to CD which meets AD produced at D’.
  11. From C’, draw a line parallel to CB which meets AB produced at B’. Then, □ - Wiktionary, the free dictionaryAB’C’D’ is the required quadrilateral similar to cyclic quadrilateral ABCD.

Constructions A Cyclic Quadrilateral

Constructions Of Tangents To A Circle

Question 1. Take a point O on the plane of the paper. With O as the centre draw a circle of radius 4 cm. Take a point on this circle and draw a tangent at P.

Solution:

Steps of Construction:

  1. Take a point O on the plane of the paper and draw a circle of a given radius of 4 cm.
  2. Take a point P on the circle and join OP.
  3. Construct ∠OPT = 90°.
  4. Produce TP to T’ to obtain the required tangent TPT’.

Constructions Tangent At P

Question 2. Draw a circle of radius 3 cm. Take a point P on it. Without using the centre of the circle, draw a tangent to the circle at point P.

Solution:

Steps of construction:

  1. Draw any chord PQ through the given point P on the circle.
  2. Take a point R on the circle and join P and Q to a point R.
  3. Construct ∠QPY = ∠PRQ and on the opposite side of the chord PQ.
  4. Produce YP to X to get YPX as the required tangent.

Constructions Tangent To The Circle At Point P

Question 3. Draw a circle of radius 2.5 cm. Take a point at a distance of 5 cm from the centre of the circle. From point P, draw two tangents to the circle.

Solution:

Steps of Construction:

  1. Take a point O in the plane of the paper and draw a circle of radius 2.5 cm.
  2. Mark a point P at a distance of 5.0 cm from the centre O and, join OP.
  3. Draw the right bisector of OP, intersecting OP at Q.
  4. Taking Q as a centre and OQ = PQ as the radius, draw a circle to intersect the given circle at T and T.
  5. Join PT and PT’ to get the required tangents.

Constructions Two Tangent To The Circle

Question 4. Draw a pair of tangents to a circle of radius 5 cm inclined to each other at an angle of 60°.

Solution:

Steps of Construction:

  1. Take a point O on the plane of the paper and draw a circle with centre O and radius OA = 5 cm.
  2. At O construct radii OA and OB such that ∠AOB equals 120° i.e., supplement of the angle between the tangents.
  3. Draw perpendiculars to OA and OB at A and B respectively suppose these perpendiculars intersect at P. Then, PA and PB are required tangents.

Constructions Pair Of Tangent To The Circle

Question 5. Draw a circle of radius 4 cm. Take a point P outside the circle. Without using the centre of the circle, draw two tangents to the circle from point P.

Solutions:

Steps of Construction:

  1. Draw a line segment of 4 cm.
  2. Take a point P outside the circle and draw a second PAB, intersecting the circle at A and B.
  3. Produce AP to C such that AP = CP.
  4. Draw a semi-circle with CB as the diameter.
  5. Draw PD ⊥ CB, intersecting the semi-circle at D.
  6. Widi P as centre and PD as radius draw arcs to intersect the given circle at T and T’.
  7. Join PT and PT’. Then, PT and PT’ are the required tangents.

Constructions Two Tangent Two The Circle From Point P

Constructions Exercise 11.1

Question 1. Draw a line segment of length 7.6 cm and divide it in the ratio 5:8. Measure the two parts.

Solution:

Steps of construction :

  1. Draw a line segment AB = 7.6 cm.
  2. Draw a ray AX which forms an acute angle from AB.
  3. Cut (8 + 5) = 13 equal marks on ray AX and mark them X1, X2, X3, X4, …, X13.
  4. Join X13 to B.
  5. Draw X5C || X13 B from X5 which meets AB at C.

So, point C divides the line segment AB in the ratio 5:8.

On measuring two line segments, we get AC = 4.7 cm, BC = 2.9 cm

Constructions Line Segment

Verification: In ΔABX13 and ΔACX5, CX5 || BX13

∴ \(\frac{A C}{C B}=\frac{A X_5}{X_5 X_{13}}=\frac{5}{8}\)

⇒ AC: AB = 5:8

Question 2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.

Solutions:

Steps of construction:

  1. Construct a ΔABC such that BC = 6 cm, AC = 5 cm and AB = 4 cm.
  2. Draw a ray BX such that ∠CBX is at an acute angle.
  3. Mark three points X1, X2 and X3 on BX such that BX1 = X1X2 = X2X3.
  4. Join X3 and C.
  5. Draw a line parallel to line X3C from X2 which intersects BC at C’.
  6. Draw a line parallel to line CA from C which meets BA at A’.

So, ΔA’B C’ is the required triangle.

Constructions Two Triangles

Verification: By construction

X3C || X2C’  ⇒ \(\frac{B X_2}{X_2 X_3}=\frac{B C^{\prime}}{C^{\prime} C}\)

but \(\frac{B X_2}{X_2 X_3}=\frac{1}{2} \quad ⇒ \quad \frac{B C^{\prime}}{C^{\prime} C}=\frac{2}{1}\)

⇒ \(\frac{C^{\prime} C}{B C^{\prime}}=\frac{1}{2}\)

Adding 1 on both sides,

⇒ \(\frac{C^{\prime} C}{B C^{\prime}}+1=\frac{1}{2}+1\)

⇒ \(\frac{C^{\prime} C+B C^{\prime}}{B C^{\prime}}=\frac{1+2}{2} = \frac{B C}{B C^{\prime}}=\frac{3}{2}\)

Now, in ΔBC’A’ and ΔBCA,

CA || C’A’

from A.A. similarity, ΔBC’A’ ∼ ΔBCA

⇒ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C} \quad\left[\text { each }=\frac{2}{3}\right]\)

Question 3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are \(\frac{7}{5}\) of the corresponding sides of the first triangle.

Solution:

Steps of construction:

  1. Draw a line segment BC = 5 cm
  2. Draw two arcs with centres B and C of radii 7 cm and 6 cm respectively which intersect each other at A.
  3. Join BA and CA. ΔABC is the required triangle.
  4. Draw a ray BX from B downwards, making an acute angle ∠CBX.
  5. Mark seven points B1, B2, B3, B4, B5, B6 and B7 on B8 such that BB1 = B1B2 = B2B3 = B3B4 = B4B5 = B5B6 = B6B7
  6. Join B5C and draw B7M || B5C from B7, which intersects the produced BC at M.
  7. Draw MN || CA from point M which intersects the produced BA at N.

Now ΔNBM is the required triangle whose sides are \(\frac{7}{5}\) of the sides of ΔABC.

Constructions A Triangle And Then Another Triangle

Justification:

By construction,

B7M || B5C

∴ \(\frac{B C}{C M}=\frac{5}{2}\)

Now, \(\frac{B M}{B C}=\frac{B C+C M}{B C}\)

= \(1+\frac{C M}{B C}=1+\frac{2}{5}=\frac{7}{5}\)

∴ \(\frac{B M}{B C}=\frac{7}{5}\)

and, MN || CA

∴ ΔABC ∼ ΔNBM

and \(\frac{N B}{A B}=\frac{B M}{B C}=\frac{M N}{C A}=\frac{7}{5}\)

Question 4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are \(1 \frac{1}{2}\) times the corresponding sides of the isosceles triangle.

Solutions:

Steps of construction:

  1. Draw the line segment BC = 8cm.
  2. Draw the perpendicular bisector OQ of BC which intersects BC at P.
  3. Take PA = 4 cm along PO.
  4. Join BA and CA. Now ΔABC is the required isosceles triangle.
  5. Draw a ray BX from B making acute angle ∠CBX.
  6. Mark three points B1, B2 and B3 on BX such that BB1 = B1B2 = B2B3
  7. Join B2C and draw B3N || B2C from B3 which intersects the Produced BC at N.
  8. Draw NM || CA from point N which intersects the produced BA at M.

Then ΔMBN is the required rectangle.

Constructions An Isosceles triangle

Justification:

∵ B3 || B2C (by construction)

∴ \(\frac{B C}{C N}=\frac{2}{1}\)

Now \(\frac{B N}{B C}=\frac{B C+C N}{B C}=1+\frac{C N}{B C}=1+\frac{1}{2}=1 \frac{1}{2}\)

Question 5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ABC = 60°. Then construct a triangle whose sides are \(\frac{4}{3}\) of the corresponding sides of the triangle ABC.

Solution:

Steps of construction:

  1. Construct a triangle ABC such that BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
  2. Draw a ray \(\overrightarrow{B X}\) such that CBX is an acute angle.
  3. Mark four points X1, X2, X3, and X4 on BX such that BX1 = X1X2 = X2X3 = X3X4.
  4. Join X4C.
  5. Draw X3C’ || X4C which intersects BC at C.
  6. Draw a line from C, parallel to CA which intersects BC at A’.

So, ΔA’BC’ is the required triangle.

Constructions Triangle ABC

Verification: By construction

X4C || X3C’ [from B.P.T.]

∴ \(\frac{B X_3}{B X_4}=\frac{B C^{\prime}}{B C} \text { but } \frac{B X_3}{B X_4}=\frac{3}{4}\) (by construction)

⇒ \(\frac{B C^{\prime}}{B C}=\frac{3}{4}\) → (1)

Now, CA || C’A’ (by construction)

ΔBC’A’ BCA [from A.A. similarity]

⇒ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}=\frac{3}{4}\) [from (1)].

Question 6. Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are \(\frac{4}{3}\) times the corresponding sides of ΔABC.

Solution:

Steps of construction:

  1. Construct a ΔABC such that BC = 7 cm, ∠B = 45° and ∠A = 105°.
  2. Draw a ray BX such that ∠CBX is at an acute angle.
  3. Mark four points X1, X2, X3 and X4 on BX such that:
    Bx1 = X1X2 = X2X3 = X3X4.
  4. Draw a line from X4 parallel to X3C which intersects BC produced at C’.
  5. Draw a line from C parallel to CA, that intersects BA produced at A’.
  6. Thus, ΔA’BC’ is the required triangle.

Constructions A Triangle Corresponding Sides Of ABC

Verification: By construction,

C’A’ || CA [from A.A. similarity]

ΔABC ∼ ΔA’BC’

⇒ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}\) → (1)

Again, by construction

X4C’ || X3C

∴ BX4C’ BX3C

⇒ \(\frac{B C^{\prime}}{B C}=\frac{B X_4}{B X_3}\)

but \(\frac{B X_4}{B X_3}=\frac{4}{3} \Rightarrow \frac{B C^{\prime}}{B C}=\frac{4}{3}\) → (2)

from (1) and (2),

∴ \(\frac{A^{\prime} B}{A B}=\frac{A^{\prime} C^{\prime}}{A C}=\frac{B C^{\prime}}{B C}=\frac{4}{3}\).

Question 7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of the given triangle.

Solution:

Steps of construction:

  1. Draw a line segment BC = 4 cm.
  2. Draw a line segment AB = 3 cm from B which makes a 90° angle from BC.]
  3. Join AC. ΔABC is the given right-angled triangle.
  4. Draw an acute angle ∠CBY from B downwards.
  5. Mark 5 points B1, B2, B3, B4 and B5 on BY such that BB1 = B1B2 = B2B3 = B3B4 = B4B5.
  6. Join B3C.
  7. Draw B5C’ || B3 C from B5, which meets the produced BC at C.
  8. Draw C’A’ || CA from C’ which meets the produced BA at A’

So, ΔA’BC’ is the required triangle.

Constructions A Right Triangle

Justification:

By construction, B5C’ || B3C

∴ \(\frac{B C}{C C^{\prime}}=\frac{3}{2}\)

Now, \(\frac{B C^{\prime}}{B C}=\frac{B C+C C^{\prime}}{B C}=1+\frac{C C^{\prime}}{B C}\)

= \(1+\frac{2}{3}=\frac{5}{3}\)

and, C’A’ = CA

∴ ΔABC ∼ ΔA’BC’

and \(\frac{A^{\prime} B}{A B}=\frac{B C^{\prime}}{B C}=\frac{A^{\prime} C^{\prime}}{C A}=\frac{5}{3}\)

Constructions Exercise 11.2

Question 1. Question 1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution:

Steps of construction:

  1. Mark a point O.
  2. Draw a circle with centre O and a radius of 6 cm.
  3. Mark a point P at a distance of 10 cm from the centre.
  4. Join O and P
  5. Bisects OP at point M.
  6. With the centre at point M, draw a circle with a radius MO or MP which intersects the given circle at A and B.
  7. Join PA and PB. So, PA and PB are two required tangents. On measuring PA = PB = 9.6 cm.

Constructions The Pair Of Tangent To The Circle

Verification: Join OA and OB. Since OP is a diameter.

∠OAP = 90º; ∠OBP = 90º [angle in semicircle]

Again OA and OB are the radii of a circle.

⇒ PA and PB are tangents to the circle.

Question 2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.

Solution:

Steps of construction:

  1. Draw two circles of radii 4 cm and 6 cm with a centre O.
  2. Mark a point P on a larger circle.
  3. Join O and P.
  4. Find the point M of the perpendicular bisector of.OP.
  5. With centre M and radius OM or PM draw a circle which intersects the smaller circle at A and B.
  6. Join A and P.

So, PA is the required tangent. On measuring PA = 4.5 cm.

Constructions A Tangent To A Circle Of Radius

Verification: Join O and A.

∠PAO = 90° [angle in semcircle]

PA ⊥ OA

∵ OA is the radius of the small circle.

∴ PA is a tangent of the smaller circle

Question 3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Solution:

Given : P and Q are two points on a diameter of the circle of radius 3 cm.

and OP = OQ = 7cm

We have to construct the tangents to the circle from P and Q.

Steps of construction:

  1. Draw a circle of radius 3 cm with centre O.
  2. Produce its diameter on both sides and take two points P and Q on it such that OP = OQ = 7 cm.
  3. Bisect OP and OQ. Let E and F be the mid-points of OP and OQ respectively.
  4. Draw a circle with centre E and radius OE, which intersects the given circle (0, 3) at M and N. Again draw a circle with centre F and radius OF which intersects the given circle at P’ and Q’.
  5. Join PM, PN, QP’ and QQ’. These are the required tangents from P and Q to the circle (0, 3).

Constructions Tangents To The Circle Of Two Points

Justification:

Join OM and ON. ∠OMP lies in the semicircle, so ∠OMP = 90°. OM is the radius of the circle, so MP is the tangent to the circle. Similarly PN, QP’ and OQ’ are also the tangents to the circle.

Question 4. Draw a pair of tangents to a circle of radius 5 cm inclined to each other at an angle of 60°.

Solution:

Steps of construction:

  1. Construct a circle with centre O and radius = 5 cm.
  2. Draw ∠AOB = 120°.
  3. Draw a perpendicular on OA from point A.
  4. Draw a perpendicular on OB from B.
  5. Both perpendiculars intersect each other at point C.

So, CA and CB are the required tangents to the circle, inclined at a 60° angle.

Constructions A Pair Of Tangents To A Circle

Verification:

In quadrilateral OACB, from angle sum property.

⇒ 120° + 90° + 90° + ∠ACB = 360°

⇒ 300° + ∠ACB = 360°

⇒ ∠ACB = 360°- 300° = 60°

Question 5. Draw a line segment AB of length S cm. Taking A as the centre, draw a circle of radius 4 cm and taking B as the centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Solution:

Steps of construction:

  1. Draw a line segment AB = 8 cm.
  2. Draw a circle of radius 4 cm taking A as the centre.
  3. Draw another circle of radius 3 cm talcing B as the centre.
  4. Draw the perpendicular bisector of AB and find the mid-point M of AB.
  5. Draw the circle with centre M and radius MA or MB which intersects the circle with centre A at P and Q and the circle with centre F at R and S.
  6. Join BP and FQ. So, BP and FQ are the required tangents on a circle with centre A from B.
  7. Now join RA and SA.

So, RA and SA are the tangents on a circle with centre B from A.

Constructions Tangents To Each Circle From The Centre Of The Other Circle

Verification: Join A and P.

∠APB = 90° => BP ⊥ AP

but AP is the radius of a circle with centre A.

⇒ AP is a tangent of the circle with centre A.

Similarly, BQ is also a tangent of the circle with centre A. Similarly AR and AS are the tangents of the circle with centre B.

Question 6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90° BD is the perpendicular from b on AC. The circle through B, C, and D is drawn. Construct the tangents from A to this circle.

Solution:

Steps of construction:

  1. Draw line segments AB = 6 cm and BC= 8 cm perpendicular to each other. Join AC. Now ΔABC is a right-angled triangle.
  2. Taking the mid-point F of BC as the centre and radius of 4 cm, draw a circle which passes through points B, C and D.
  3. Join AF and bisects A it. Let O be the midpoint of AF.
  4. The circle drawn with f centre O and radius OA intersects the given circle at B and M.
  5. Join AB and AM, which are the required tangents.

Constructions Right Triangle

Justification:

Join FM and FB. Now ∠AMF lies in a semicircle,

so ∠AMF = 90° ⇒ FM ⊥ AM

∵ FM is the radius of the circle, so AM is the tangent to the circle and F is the centre.

Similarly, AB is also the tangent to the circle with centre F.

Question 7. Draw a circle with the help of a bangle Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Solution:

Steps of construction:

  1. Draw a circle using a bangle.
  2. Draw two chords AP and MT. The perpendicular bisectors of AP and MT intersect each other at O, which is the centre of the circle.
  3. Take a point R outside the circle. Join OR and bisect it.
  4. Let Q be the mid-point of OR. The circle drawn with centre Q and radius OQ intersects the given circle at S and N.
  5. Join RS and RN, so RS and RN are the required tangents from R.

Constructions The Pair Of Tangents From This Point To The Circle

Justification:

Join OS and ON. ∠OSR lies in a semicircle, so

∠OSR = 90° ⇒ OS || SR

∵ OS is the radius of a circle with a centre of O, so SR is the centre of that circle whose centre is O. Similarly, RN is also the tangent to the circle with a centre of O.

NCERT Exemplar For Class 10 Maths Chapter 9 Some Applications Of Trigonometry

Some Applications Of Trigonometry

Practical Use Of Trigonometry

The main purpose of studying trigonometry is to determine the height of buildings, towers, telephone poles, trees, the width of the river, the distance of the ship from the lighthouse etc. Although it is not easy to measure them, we can determine these things by using knowledge of trigonometric ratios, before doing so, let us first discuss some necessary definitions.

NCERT Exemplar For Class 10 Maths Chapter 9 Some Applications Of Trigonometry

Line Of Sight: When an observer looks at an object then the line joining the observer’s eye to the object is called the line of sight.

Angle Of Elevation: When an observer sees an object situated in an upward direction, the angle formed by the line of sight with a horizontal line is called an angle of elevation.

Applications Of Trigonometry Angle Of Elevation

Angle Of Depression: When an observer sees an object .situated in a downward direction, the angle formed by the line of sight with a horizontal line is called an angle of depression.

Applications Of Trigonometry Angle Of Depression

In the adjoining above given, θ is the angle of depression of the object as seen from O.

Solved Examples

Example 1. The length of the shadow of a vertical pole is \(\frac{1}{\sqrt{3}}\) times its height. Show that the angle of elevation of the sun is 60°.

Solution:

Given:

The length of the shadow of a vertical pole is \(\frac{1}{\sqrt{3}}\) times its height.

Let PQ be a vertical pole whose height is \(\frac{h}{\sqrt{3}}\). Its shadow is OQ whose height is.

Let the angle of elevation of the sun be \(\angle\)POQ = 0

Applications Of Trigonometry The Angle Of Elevation Of Sun Is 60 Degrees

In \(\triangle\)POQ,

⇒ \(\tan \theta =\frac{P Q}{O Q}=\frac{h}{h / \sqrt{3}}=\sqrt{3}=\tan 60^{\circ}\)

⇒  \(\theta =60^{\circ}\)

The angle of elevation of the sun = 60°.

Example 2. If a tower 30 m high, casts a shadow 10 \(\sqrt{3}\) m long on the ground, then what is the angle of elevation of the sun?

Solution:

It is given that AB = 30 m be the height of the tower and BC = \(\sqrt{3}\) m its shadow on the ground.

Applications Of Trigonometry The Angle Of The Elevation Of Sun

Let \(\theta\) be the angle of elevation.

In a right triangle,

⇒  \(\tan \theta =\frac{A B}{B C}\)

= \(\frac{30}{10 \sqrt{3}}=\frac{3}{\sqrt{3}}=\sqrt{3}\)

= \(\tan 60^{\circ}\)

⇒  \(\theta =60^{\circ}\)

Hence, the angle of elevation of the sun= 60°

Example 3. A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

Solution:

Given:

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall

Let PR be a ladder of length 15 m and QR, a wall of height h.

Applications Of Trigonometry The Ladder Makes An Angle Then The Height Of The Wall

Given that \(\angle\)PRQ = 60°

In \(\triangle P Q R\),

⇒  \(\cos 60^{\circ} =\frac{h}{P R} \quad \Rightarrow \quad \frac{1}{2}=\frac{h}{15}\)

⇒  \(\Rightarrow \quad h =\frac{15}{2} \mathrm{~m}\)

Height of the wall =\(\frac{15}{2} \mathrm{~m}=7.5 \mathrm{~m}\)

Question 4. The shadow of a tower standing on a level plane is found to be 50 m longer when the sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Solution:

Given

The shadow of a tower standing on a level plane is found to be 50 m longer when the sun’s elevation is 30° than when it is 60°.

Let AB be a tower of height ‘h’ metres and BD and BC be its shadows when the angles of elevation of the sun are 30° and 60° respectively.

Applications Of Trigonometry The Shadow Of A Tower Standing On A Level Plane Then The Height Of The Tower

⇒  \(\angle A D B=30^{\circ}, \angle A C B=60^{\circ} \text { and } C D=50 \mathrm{~m}\)

Let BC = x metres.

In \(\triangle A B C\)

⇒  \(\tan 60^{\circ} =\frac{A B}{B C} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

⇒  \( x =\frac{h}{\sqrt{3}}\)

In \(\triangle A B D\)

⇒  \(\tan 30^{\circ}=\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x+50}\)

⇒  \(\sqrt{3} h=x+50 \Rightarrow \sqrt{3} h=\frac{h}{\sqrt{3}}+50\)

⇒  \(h=h+50 \sqrt{3} \quad \Rightarrow \quad 2 h=50 \sqrt{3} \quad \Rightarrow \quad h=25 \sqrt{3}\)

Height of the tower =\(25 \sqrt{3} \mathrm{~m}\)

Question 5. The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 40 m towards the tower, the angle of elevation becomes 60°. Find the height of the tower.

Solution:

Given

The angle of elevation of the top of a tower from a point on the ground is 30°. After walking 40 m towards the tower, the angle of elevation becomes 60°.

Let AB be a tower of height 7i’ metres. From points D and C on the ground, the angle of elevation of top A of the tower is 30° and 60° respectively.

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Tower From A Point

Given that CD = 40 m

Let BC = x metres

In \(\triangle A B C\)

⇒ \(\tan 60^{\circ} =\frac{A B}{B C} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

x =\(\frac{h}{\sqrt{3}}\)

In \(\triangle A B D\)

⇒ \(\tan 30^{\circ} =\frac{A B}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{40+x}\)

⇒ \(\sqrt{3} h =40+x \Rightarrow \sqrt{3} h=40+\frac{h}{\sqrt{3}}\)

⇒ \(3 h =40 \sqrt{3}+h \Rightarrow 2 h=40 \sqrt{3}\)

⇒ \(h =20 \sqrt{3}\)

Height of the tower =\(20 \sqrt{3} \mathrm{~m}\)

Question 6. The angle of elevation of the top of a tower from two points distant ‘s’ and ‘t’ from its foot are complementary. Prove that the height of the tower is \(\sqrt{st}\) .

Answer:

Given

The angle of elevation of the top of a tower from two points distant ‘s’ and ‘t’ from its foot are complementary.

Let BC be a tower of height ‘h’. Let AC = s and DC = t.

From points A and D, the angle of elevation of top B of the tower is complementary.

Applications Of Trigonometry The Angles Of Elevation Of The Top Of Tower From Two Points At A Distance

Let \(\angle B A C=\theta\)

⇒ \(\angle B D C=90^{\circ}-\theta\)

In \(\triangle B A C\)

⇒ \(\tan \theta=\frac{B C}{A C}=\frac{h}{s}\)

In \(\triangle B D C\)

⇒ \(\tan \left(90^{\circ}-\theta\right) =\frac{B C}{C D} \Rightarrow \cot \theta=\frac{h}{t}\)

⇒ \(\Rightarrow\frac{1}{\tan \theta} =\frac{h}{t} \quad \Rightarrow \frac{s}{h}=\frac{h}{t}\)

⇒ \(h^2 =s t \Rightarrow h=\sqrt{s t}\)

height of the tower =\(\sqrt{s t}\)

Example 7. A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 4 m from the root. Find the height of the tree before broken.

Solution:

Given

A tree is broken by the wind. The top struck the ground at an angle of 30° and at a distance of 4 m from the root.

Given that A B=4 m and \(\angle\) B A C=\(30^{\circ}\). Also C D=C A In \(\triangle \)A B C,

Applications Of Trigonometry A Tree Broken By The Wind Then The Height Of The Tree Before Broken

⇒  \(\tan 30^{\circ}=\frac{B C}{A B} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{B C}{4}\)

B C=\(\frac{4}{\sqrt{3}} \mathrm{~m}\)

and \(\quad \cos 30^{\circ}=\frac{A B}{A C} \quad \Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{4}{A C}\)

A C=\(\frac{8}{\sqrt{3}} \mathrm{~m} \Rightarrow C D=\frac{8}{\sqrt{3}} \mathrm{~m}( A C=C D)\)

Now the total height of the tree =B C+C D

= \(\frac{4}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{12}{\sqrt{3}}=4 \sqrt{3} \mathrm{~m}\)

The height of the tree before broken = \(4 \sqrt{3} \mathrm{~m}\)

Example 8. From the top of a tower 71 m high, the angles of depression of two objects, which are in line with the foot of the tower are \(\alpha\) and \(\beta(\beta>\alpha)\). Find the distance between the two objects.

Solution:

Given

From the top of a tower 71 m high, the angles of depression of two objects, which are in line with the foot of the tower are \(\alpha\) and \(\beta(\beta>\alpha)\).

Let AB be a tower of height ‘h’m. From the top A’ of the tower, the angle of depression of two objects D and C are ‘ \(\beta\) And \(\alpha\) respectively.

Applications Of Trigonometry The Top Of A Tower High And The Angles Of Depression Then The Distance Between The Two Objects

In \(\triangle A B C\)

⇒ \(\tan \beta=\frac{A B}{B C} =\frac{h}{B C}\)

B C =\(\frac{h}{\tan \beta}=h \cot \beta\)

In \(\triangle A B D\)

⇒ \(\tan \alpha =\frac{A B}{B D}=\frac{h}{B D}\)

B D =\(\frac{h}{\tan \alpha}=h \cot \alpha\)

Subtract equation (1) from (2), we get

⇒ \(B D-B C=h \cot \alpha-h \cot \beta\)

⇒ \(C D=h(\cot \alpha-\cot \beta)\)

⇒ \(C D=h(\cot \alpha-\cot \beta)\)

The distance between the objects =\(h(\cot \alpha-\cot \beta) \mathrm{m}\).

Example 9. Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 60°. If the height of the tower is 150 m, find the distance between the two men.

Solution:

Given

Two men are on opposite sides of a tower. They measure the angles of elevation of the top of the tower as 30° and 60°. If the height of the tower is 150 m

Lot CD bo a tower of height 1 50 m. Two men A and B are on the opposite sides of the tower.

Applications Of Trigonometry The Distance Between The Two Men

Given that, \(\angle D A C=60^{\circ}\) and \(\angle D B C=30^{\circ}\)

In \(\triangle D A C\)

⇒ \(\tan 60^{\circ} =\frac{D C}{A C} \quad \Rightarrow \quad \sqrt{3}=\frac{150}{A C}\)

⇒ \(A C =\frac{150}{\sqrt{3}}=50 \sqrt{3} \mathrm{~m}\)

In \(\triangle B C D\)

⇒ \(\tan 30^{\circ} =\frac{D C}{B C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{150}{B C} \Rightarrow B C=150 \sqrt{3} \mathrm{~m}\)

A B = A C+B C

= \((50 \sqrt{3}+150 \sqrt{3}) \mathrm{m}=200 \sqrt{3} \mathrm{~m}\)

Therefore the distance between two \(\mathrm{men}=200 \sqrt{3} \mathrm{~m}\)

Example 10. A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h, At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are a and p respectively. Prove that the height of the tower is \(\left(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\right)\)

Solution:

Given:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flagstaff of height h, At a point on the plane, the angles of elevation of the bottom and the top of the flagstaff are a and p respectively.

Let AB be a tower and BC be the flagstaff.

Let O be the observer.

Applications Of Trigonometry A Vertical Tower Stands On A Horizontal Plane

Now, \(\angle A O B=\alpha\), \(\angle A O C=\beta\) and B C=h

Let A B = H and O A = x

In \(\triangle O A B\)

\(\tan \alpha=\frac{A B}{O A}=\frac{H}{x} \quad \Rightarrow \quad x=\frac{H}{\tan \alpha}\) →  Equation 1

In \(\triangle O A C\)

⇒ \(\tan \beta=\frac{A C}{O A}=\frac{H+h}{x} \Rightarrow x=\frac{H+h}{\tan \beta}\)  → Equation 2

From equations (1) and (2), we get

⇒ \(\frac{H}{\tan \alpha} =\frac{H+h}{\tan \beta}\)

⇒ \(H \tan \beta =H \tan \alpha+h \tan \alpha\)

⇒ \(H(\tan \beta-\tan \alpha) =h \tan \alpha\)

H =\(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\)

Height of the tower =\(\frac{h \tan \alpha}{\tan \beta-\tan \alpha}\)

Example 11. The angle of elevation of the top of a vertical lower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 47′. Find the height of the tower.

Solution:

Given

The angle of elevation of the top of a vertical lower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 47′.

Let All he a vertical lower. From point C, the angle of elevation of the top of the tower is 60° and from point) as shown, the angle of elevation of the tower is 45°.

⇒ \(\angle A C B=60^{\circ}, \angle A D E=45^{\circ}\) and \(D C=10 \mathrm{~m}\)

In \(\triangle D E\),

Applications Of Trigonometry The Height Of The Tower From The Top

⇒ \(\tan 45^{\circ}=\frac{A E}{D E} \quad \Rightarrow \quad A E=D E \ldots(1)\left( \tan 45^{\circ}=1\right)\)

In \(\triangle A B C\),

⇒ \(\tan 60^{\circ} =\frac{A B}{B C} \quad \Rightarrow \quad \sqrt{3}=\frac{A B}{D E} \quad( B C=D E)\)

\(A B =\sqrt{3} D E\)

A E+B E =\(\sqrt{3} A E\)

⇒ \(B E =A E(\sqrt{3}-1)\)

A E =\(\frac{B E}{\sqrt{3}-1}=\frac{10}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

⇒ \(\frac{10(1.732+1)}{3-1}=5 \times 2.732=13.66\)

Now, A B =A E+B E=A E+C D

= \((13.66+10) \mathrm{m}=23.66 \mathrm{~m}\)

Height of the tower =23.66 m

Example 12. The angle of elevation of the top Q of a vertical towerPQ from a point Y- on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of the tower is 45°. Find the height of the tower PQ and the distance PX. (Use \(\sqrt{3}\)=1.73)

Solution:

Given

The angle of elevation of the top Q of a vertical towerPQ from a point Y- on the ground is 60°. From a point Y, 40 m vertically above X, the angle of elevation of the top Q of the tower is 45°.

Let height of tower PQ be h m and let P X = x m

Since, X Y=40=P Z

Q Z=P Q-P Z=h-40.

In right \(\triangle Q Z Y\),

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Vertical Tower

⇒ \(\tan 45^{\circ}=\frac{h-40}{x} \Rightarrow 1=\frac{h-40}{x}\)

⇒ \(h-40=x \Rightarrow x=h-40\)

In right \(\triangle Q P X\),

⇒ \(\tan 60^{\circ}=\frac{h}{x} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

x=\(\frac{h}{\sqrt{3}}\)

From equations (1) and (2), we get

h-40=\(\frac{h}{\sqrt{3}} \quad \Rightarrow h \sqrt{3}-40 \sqrt{3}=h\)

⇒ \(h(\sqrt{3}-1)=40 \sqrt{3} \Rightarrow h=\frac{40 \sqrt{3}}{\sqrt{3}-1}\)

h= \(\frac{40 \sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=20(3+\sqrt{3})\)

h = \(20(3+1.73)=20 \times 4.73=94.6 \mathrm{~m}\)

x =h-40

=94.6-40=54.6 m

Hence, the height of tower PQ is 94.6 m and the distance P X is 54.6 m.

Example 13. As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, change from 30° to 60°. Find the distance travelled by the ship during the period of observation.(Use \(\sqrt{3}\)=1.73)

Solution:

Given

As observed from the top of a lighthouse, 100 m high above sea level, the angles of depression of a ship, sailing directly towards it, change from 30° to 60°.

Let the height of the lighthouse be A B=100 m and let D be the ship which is sailing towards it. Also, let D C=x m.

Applications Of Trigonometry The Distance Travelled By The Ship During The Period Of Observation

Here, \(\angle C A D =60^{\circ}-30^{\circ}=30^{\circ}\)

⇒ \(\angle C D A =\angle C A D\)

D C=A C=x (angles opposite to equal sides are equal)

Now, in right \(\triangle A B C\),

⇒ \(\sin 60^{\circ}=\frac{A B}{A C} \quad \Rightarrow \quad \frac{\sqrt{3}}{2}=\frac{100}{D C}\)

⇒ \(D C=\frac{200}{\sqrt{3}}=\frac{200 \sqrt{3}}{3}=\frac{200 \times 1.73}{3}\)

⇒ \(D C=\frac{346}{3}\)=115.3 m

Hence, the distance travelled by ship is 115.3 m.

Question 14. An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively. Find the width of the river. (Use \(\sqrt{3}\)=1.73)

Solution:

Given

An aeroplane is flying at a height of 300 m above the ground. Flying at this height, the angles of depression from the aeroplane of two points on both banks of a river in opposite directions are 45° and 60° respectively.

Let an aeroplane P fly at a height of 300 m above the x-ground

PM = 300m

Depression angles are \(\angle X P A=45^{\circ}\) and \(\angle Y P B=60^{\circ}\).

Applications Of Trigonometry An Aeroplane Is Flying At A Height Above The Ground

Now, in right \(\triangle P M A\),(alternate angles)

⇒ \(\tan 45^{\circ}=\frac{P M}{A M} \quad \Rightarrow \quad 1=\frac{300}{A M} . \quad \Rightarrow \quad A M=300 \mathrm{~m}\)

In right \(\triangle P M B\),

⇒ \(\tan 60^{\circ}=\frac{P M}{M B} \quad \Rightarrow \quad \sqrt{3}=\frac{300}{M B} \quad \Rightarrow \quad M B=\frac{300}{\sqrt{3}}=100 \sqrt{3}\)

Width of river =A M+M B

= \((300+100 \sqrt{3}) \mathrm{m}=(300+100 \times 1.732) \mathrm{m}\)

=(300+173.2) m=473.20 m

Hence, the width of the river = 473.20 m

Example 15. A man observes a car from the top of a tower, which is moving towards the tower at a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes, find the time taken by the car now to reach the tower.

Solution:

Given

5. A man observes a car from the top of a tower, which is moving towards the tower at a uniform speed. If the angle of depression of the car changes from 30° to 45° in 12 minutes

Let AB = h m be the height of the tower.

Applications Of Trigonometry The Time Taken By The Car Now To Reach The Tower

Since the depression angle changes from 30° to 45° in 12 minutes, therefore, time taken from D to C = 12 min.

Let DC = x m and CB -y m

Now, in right \(\triangle A B C\),

⇒ \(\tan 45^{\circ}=\frac{h}{y} \Rightarrow 1=\frac{h}{y} \Rightarrow h=y\)

In right \(\triangle A B D\),

⇒ \(\tan 30^{\circ} =\frac{h}{x+y} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{h}{x+y}\)

⇒ \(x+y =h \sqrt{3}\)

x+y =y \(\sqrt{3}\)

⇒ \(y(\sqrt{3}-1) =x\)

Now time taken by car in moving \(x \mathrm{~m}=12 \mathrm{~min}\)

Time taken by car in moving \(y(\sqrt{3}-1) \mathrm{m}=12 \mathrm{~min}\)

Time taken by car in moving \(y \mathrm{~m}=\frac{12}{\sqrt{3}-1} \mathrm{~min}\)

= \(\frac{12(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \min =\frac{12(1.732+1)}{2}\)

=6 \(\times 2.732=16.39 \mathrm{~min}\)

Hence, required time =16.39 minutes.

The time taken by the car now to reach the tower =16.39 minutes.

Example 16. A bird is sitting on the top of an 80 m-high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remains at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°. Find the speed of flying of the bird.(Take, \(\sqrt{3}\)=1.73)

Solution:

Given

A bird is sitting on the top of an 80 m-high tree. From a point on the ground, the angle of elevation of the bird is 45°. The bird flies away horizontally in such a way that it remains at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30°.

Let P be the position of a bird at the height of 80 m with the angle of elevation 45° from A.

Let after 2 seconds, it reaches Q from where its elevation angle is 30°.

Applications Of Trigonometry The Speed Of Flying Of The Bird On The Top Of A High Tree

Now, in right \(\triangle P B A\),

⇒ \(\tan 45^{\circ} =\frac{P B}{A B} \quad \Rightarrow \quad 1=\frac{80}{A B}\)

A B = 80 m

In right \(\triangle Q C A\),

⇒ \(\tan 30^{\circ} =\frac{Q C}{A C} \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{80}{A C}\)

⇒ \(A C =80 \sqrt{3} \mathrm{~m}\)

BC =A C-A B

=80 \(\sqrt{3}-80 \quad=80(\sqrt{3}-1)\)

= \(80(1.732-1)=80 \times 0.732=58.56 \mathrm{~m}\)

Now, speed of bird =\(\frac{\text { Distance }}{\text { time }}=\frac{58.56}{2} \mathrm{~m} / \mathrm{sec}=29.28 \mathrm{~m} / \mathrm{sec}\) .

The speed of flying of the bird =\( 29.28 \mathrm{~m} / \mathrm{sec}\) .

Example 17. The angle of elevation of a cloud from a point 7z’ metres above a lake is a and the angle of depression of its reflection in the lake is p. Prove that the distance of the cloud from the point of observation is \(\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}\)

Solution:

Given:

The angle of elevation of a cloud from a point 7z’ metres above a lake is a and the angle of depression of its reflection in the lake is p.

Let AB be a lake. The angle of elevation of cloud P at point A on height TP from the lake is a and the angle of depression of the reflection F of cloud is \(\beta\).

Applications Of Trigonometry The Distance Of The Cloud From The Point Of Observation

⇒ \(\angle P C E=\alpha and \angle F C E=\beta\)

Let B P=F B=d

P E=B P-E B

P E=B P-A C=d-h

and F E=F B+B E=F B+A C=d+h

Let C E = x

In \(\triangle C E F\),

⇒ \(\tan \beta=\frac{E F}{C E} \quad \Rightarrow \quad \tan \beta=\frac{d+h}{x}\)  → Equation 1

In \(\triangle P E C\),

⇒ \(\tan \alpha=\frac{P E}{C E} \quad \Rightarrow \quad \tan \alpha=\frac{d-h}{x}\)  → Equation 2

Subtract equation (2) from equation (1), we get

⇒ \(\tan \beta-\tan \alpha=\frac{d+h}{x}-\frac{d-h}{x}=\frac{2 h}{x} \quad \Rightarrow \quad x=\frac{2 h}{\tan \beta-\tan \alpha}\)

In \(\triangle P E C\),

⇒ \(\cos \alpha=\frac{C E}{P C}=\frac{x}{P C} \quad \Rightarrow \quad P C=\frac{x}{\cos \alpha}=x \sec \alpha\)  Equation 3

PC = \(\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}\)

Hence, the distance of the cloud from the point of observation =\(\frac{2 h \sec \alpha}{\tan \beta-\tan \alpha}\)

Example 18. A man standing on the deck of a ship which is 14 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°. Find the distance of the hill from the ship and the height of the hill.

Solution:

Given

A man standing on the deck of a ship which is 14 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of the hill as 30°.

Let the height of the hill AB = h m and let the position of man at 14 m above the sea level is D.

Let B C=x \(\mathrm{~m} \Rightarrow D E=x \mathrm{~m}\)

In right \(\triangle D C B\),

Applications Of Trigonometry The Distance Of The Hill From The Ship And The Height Of The Hill

⇒ \(\tan 30^{\circ}=\frac{D C}{B C} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{14}{x} \quad \Rightarrow \quad x=14 \sqrt{3} \mathrm{~m}\)

In right \(\triangle A E D\),

⇒ \(\tan 60^{\circ} =\frac{A E}{D E} \quad \Rightarrow \quad \sqrt{3}=\frac{h-14}{x}\)

⇒ \(x \sqrt{3} =h-14 \quad \Rightarrow \quad 14 \sqrt{3} \times \sqrt{3}\)=h-14 [ from (1) ]

h =42+14=56 m

Hence, the die distance of the hill from the ship is x i.e., \(14 \sqrt{3} m\) and the height of the hill is 56 m.

Some Applications Of Trigonometry Exercise 9.1

Question 1. A circus artist is climbing a 20 m-long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.

Solution :

Given:

A circus artist is climbing a 20 m-long rope, which is tightly stretched and tied from the top of a vertical pole to the ground.

In \(\triangle A B C\),

Applications Of Trigonometry The Angle Made By The Rope With The Ground Level

⇒ \(\sin 30^{\circ} =\frac{A B}{A C}\)

⇒ \(\frac{1}{2}=\frac{A B}{20}\)

B =10

Height of pole = 10 m

Question 2. A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution :

Given

A tree breaks due to a storm and the broken part bends so that the top of the tree touches the ground making an angle of 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m.

Let the part CD of the tree BD break in the air and touch the ground at point A.

Applications Of Trigonometry A Tree Breaks Due To Storm And Broken Part And The Height Of The Tree

According to the problem,

AB = 8 m and \(\angle\) BAC = 30°

In \(\triangle A B C\),

⇒ \(\tan 30^{\circ} =\frac{B C}{A B} \quad \stackrel{A 0^{\circ}}{\longleftrightarrow} \mathrm{m}_8\)

⇒ \(\frac{1}{\sqrt{3}} =\frac{B C}{8}\)

⇒ \(B C =\frac{8}{\sqrt{3}} \mathrm{~m}\)

and \(\cos 30^{\circ} =\frac{A B}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{8}{A C}\)

⇒ \(A C =\frac{16}{\sqrt{3}} \mathrm{~m}\)

C D =\(\frac{16}{\sqrt{3}} \mathrm{~m}\) (A C=C D)

Now, the height of the tree = BC + CD

=\(\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}=8 \sqrt{3} \mathrm{~m}\)

The height of the tree =\( 8 \sqrt{3} \mathrm{~m}\)

Question 3. A contractor plans to install two slides for the children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution :

Given

A contractor plans to install two slides for the children to play in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground.

Let the slide for elder children be AC and for younger children be DE.

Applications Of Trigonometry The Length Of The Slide In Each Case

In \(\triangle A B C\),

⇒ \(\angle A B C=90^{\circ}\)

A B=3 m

⇒ \(\sin 60^{\circ}=\frac{A B}{A C} \Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{A C}\)

A C=\(\frac{6}{\sqrt{3}}=2 \sqrt{3} \mathrm{~m}\)

In \(\triangle B D E\),

⇒ \(\angle D B E =90^{\circ}\)

B E=1.5 m

⇒ \(\sin 30^{\circ} =\frac{B D}{D E}\)

⇒ \(\frac{1}{2} =\frac{1.5}{D E} \Rightarrow D E\) =3 m

Length of slide for elder children

=2\( \sqrt{3} \mathrm{~m}\)

and length of slide for younger children =3 m

Question 4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution :

Let AB be the tower.

The angle of elevation of the top of the tower from point C, 30m away from A is 30°.

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Tower From A Point On The Ground

In \(\triangle B A C\),

⇒ \(\tan 30^{\circ}=\frac{A B}{A C}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{A B}{30}\)

⇒ \(A B=\frac{30}{\sqrt{3}}=10 \sqrt{3} \mathrm{~m}\)

Height of the tower =\(10 \sqrt{3} \mathrm{~m}\)

Question 5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

Given

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°.

Let the height of kite A from the ground is 60 m and AC is the string.

Applications Of Trigonometry The Length OF The String, Assuming That There Is No Slack In The String

Given : \(\angle A C B=60^{\circ}\)

In \(\triangle A B C\),

⇒ \(\sin 60^{\circ}=\frac{60}{A C}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{60}{A C}\)

A C=\(\frac{120}{\sqrt{3}}=40 \sqrt{3} \mathrm{~m}\)

Length of string =40 \(\sqrt{3}\) m

Question 6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building

Solution:

Given

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building.

Let, the height of building AN = 30 m

Here, BM = height of boy = 1.5 m

DN = BM = 1.5 m

AD = AN -AD = 30- 1.5 = 28.5 m

Applications Of Trigonometry The Distance Walked Towards The Building

In \(\triangle A C D\),

⇒ \(\tan 60^{\circ} =\frac{A D}{C D} \quad \Rightarrow \quad \sqrt{3}=\frac{28.5}{C D}\)

⇒ \(C D =\frac{28.5}{\sqrt{3}}=9.5 \sqrt{3} \mathrm{~m}\)

In \(\triangle A B D\),

⇒ \(\tan 30^{\circ}=\frac{A D}{B D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{B D}\)

⇒ \(B D=28.5 \sqrt{3} \mathrm{~m}\)

Now, \(B C=B D-C D\)

=28.5 \(\sqrt{3}-9.5 \sqrt{3}=19 \sqrt{3} \mathrm{~m}\)

Distance walked by boy towards the building \(19 \sqrt{3} \mathrm{~m}\)

Question 7. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

Given

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively.

Let, CD be the height of the transmission tower.

Here, the height of the building

Applications Of Trigonometry The Height Of The Tower From A Point On The Ground

BC = 20 m

In \(\triangle A B C\),

⇒ \(\tan 45^{\circ} =\frac{B C}{A B}\)

=\(\frac{20}{A B}\)

A B =20 m

In \(\triangle A B D\),

⇒ \(\tan 60^{\circ}=\frac{B D}{A B} \quad \Rightarrow \sqrt{3}=\frac{B D}{20}\)

B D=20 \(\sqrt{3} m\)

B C+C D=20 \(\sqrt{3}\)

20+C D=20 \(\sqrt{3}\)

C D=20\((\sqrt{3}-1) \mathrm{m}\)

Height of transmission tower =20\((\sqrt{3}-1) \mathrm{m}\)

Question 8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:

Given

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point, the angle of elevation of the top of the pedestal is 45°.

Let AB be the statue of height 1.6 m at the top of pedestal BC.

⇒ \(\angle B D C =45^{\circ}\)

and \(\angle A D C =60^{\circ}\)

Let C D =x

and BC =h

In \(\triangle B C D\),

Applications Of Trigonometry The Height Of The Pedestal Of The Statue

⇒ \(\tan 45^{\circ} =\frac{B C}{C D}\)

1 =\(\frac{h}{x}\)

h =x (1)

In ACD.

⇒ \(\tan 60^{\circ} =\frac{A C}{C D}\)

⇒ \(\sqrt{3} =\frac{h+1.6}{x}\)

⇒ \(\sqrt{3} =\frac{h+1.6}{h}\) [from eqn. (1)]

⇒ \(\sqrt{3} h\) =h+1.6

h\((\sqrt{3}-1)\) =1.6

h =\(\frac{1.6}{\sqrt{3}-1}=\frac{1.6 \times(\sqrt{3}+1)}{3-1}\)

=0.8(1.732+1)=2.1856

Height of pedestal \(\approx 2.18 \mathrm{~m}\)

Question 9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building

Solution:

Given

The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high

Let AB be the tower and CD be the building.

Applications Of Trigonometry The Angle Of Elevation Of The Top Of A Building From The Foot Of The Tower

Here, the height of Tower AB is 50 m

⇒ \(\angle A C B=60^{\circ}\),

⇒ \(\angle D B C=30^{\circ}\)

In \(\triangle A B C\),

⇒ \(\sqrt{3}=\frac{50}{B C}\)

⇒ \(\tan 60^{\circ}=\frac{A B}{B C}\)

B C=\(\frac{50}{\sqrt{3}}\)

In \(\triangle B C D\),

⇒ \(\tan 30^{\circ}=\frac{C D}{B C} \quad \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{C D}{B C}\)

C D=\(\frac{B C}{\sqrt{3}}=\frac{50}{\sqrt{3} \times \sqrt{3}}=\frac{50}{3}\)

C D=16.67 m

Height of building =16.67 m

Question 10. Two poles of equal height are standing opposite each other on either side of the road, which is 80 m wide. them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the points from the poles.

Solution:

Given:

Two poles of equal height are standing opposite each other on either side of the road, which is 80 m wide. them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively.

Let two poles AB and CD of equal heights ‘h’ be on either side of a road 80 m broad.

Applications Of Trigonometry Two Poles Of Equal Heights Are Standing Opposite Each Other On Either Side Of The Road

At point E,

Given : \(\angle C E D=60^{\circ}\) and \(\angle A E B=30^{\circ}\)

Let D E=x

B E = 80-x

In \(\triangle C D E\),

⇒ \(\tan 60^{\circ} =\frac{C D}{D E} \quad \Rightarrow \quad \sqrt{3}=\frac{h}{x}\)

h = \(x\sqrt{3}\) Equation (1)

In \(\triangle A B E\)

⇒ \(\tan 30^{\circ}=\frac{A B}{B E} \quad \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{80-x}\)

⇒ \(\frac{1}{\sqrt{3}}=\frac{x \sqrt{3}}{80-x}\) [from Equation (1)]

3 \(\mathrm{r}=80-x \Rightarrow 4 x=80\)

x = 20

80-x=80-20=60 and h=20 \(\sqrt{3}\)

Height of each pole =20 \(\sqrt{3} \mathrm{~m}\)

Distance of two poles from point E=20 m and 60 m.

Question 11. A TV tower stands vertically on the bank of an A canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point D 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal.

Applications Of Trigonometry The Height Of The Tower And The Width Of The Canal

Solution:

Given

A TV tower stands vertically on the bank of an A canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point D 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°.

In \(\triangle B C\).

⇒ \(\tan 60^{\circ} =\frac{A B}{B C}\)

⇒ \(\sqrt{3} =\frac{A B}{B C}\)

⇒ \(A B =\sqrt{3} \cdot B C\)

In \(\triangle A B D\),

⇒ \(\tan 30^{\circ} =\frac{A B}{B D}\)

⇒ \(\frac{1}{\sqrt{3}} =\frac{\sqrt{3} B C}{B C+C D}\)

3 BC =B C+C D

2 BC =20

BC =10

Put in equation (1),

A B =\(10 \sqrt{3} \mathrm{~m}\)

Height of tower =10 \(\sqrt{3} \mathrm{~m}\)

and width of canal =10 m

Question 12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Solution :

Let CD be a building of height 7m and AB be a cable tower.

Given :

⇒ \(\angle A D E=60^{\circ}\) and \(\angle D B C=45^{\circ}\)

Applications Of Trigonometry The Height Of The Tower From The Top Of A High Building

In \(\triangle B C D\),

⇒ \(\tan 45^{\circ} =\frac{D C}{B C} \Rightarrow 1=\frac{7}{B C}\)

B C = 7 m

In \(\triangle A D E\),

⇒ \(\tan 60^{\circ}=\frac{A E}{D E} \Rightarrow \sqrt{3}=\frac{A E}{B C} \quad(D E=B C)\)

⇒ \(A E=\sqrt{3} B C=1.732 \times 7=12.124 \mathrm{~m}\)

⇒ \(A B=A E+B E=12.124+7=19.124 \mathrm{~m}\)

Height of tower =19.124 m

Question 13. As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of the two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships

Solution :

Given

As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of the two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse

Let AB be a lighthouse whose height is 75 m. The position of the two ships is at C and D.

Applications Of Trigonometry The Distance Between The Two Ships

In \(\triangle A B C\),

⇒ \(\tan 45^{\circ} =\frac{A B}{A C}\)

1 =\(\frac{75}{A C} \Rightarrow A C=75 \mathrm{~m}\)

In \(\triangle A B D\),

⇒ \(\tan 30^{\circ}=\frac{A B}{A D} \Rightarrow \frac{1}{\sqrt{3}}=\frac{75}{A D}\)

⇒ \(A D=75 \sqrt{3} \Rightarrow C D+A C=75 \sqrt{3}\)

⇒ \(C D=75 \sqrt{3}-75=75(\sqrt{3}-1)\)

Distance between two ships =\(75(\sqrt{3}-1)\) m

Question 14. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After elevation reduces to 30°. Find the distance travelled by the balloon during the interval.

Applications Of Trigonometry The Angle Of Elevation Of The Balloon From The Eyes OF The Girl

Solution:

Given

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After elevation reduces to 30°.

Let C be the position of the girl. The two positions of the balloon are A and P.

PD = AB = 88.2 – 1.2 = 87 m

In right \(\triangle A B C\),

Applications Of Trigonometry The Distance Travelled By The Balloon During The Interval

⇒ \(\tan 60^{\circ} =\frac{A B}{B C}\)

⇒ \(\sqrt{3}=\frac{87}{B C} \quad \Rightarrow B C=\frac{87}{\sqrt{3}}\)  → Equation 1

In right \(\triangle PDC\),

⇒ \(\tan 30^{\circ}=\frac{P D}{C D}\)

⇒ \(\frac{1}{\sqrt{3}} =\frac{87}{C D} \quad \Rightarrow C D=87 \sqrt{3}\)

Now, BD = C D – B C = 87

⇒ \(\sqrt{3}-\frac{87}{\sqrt{3}}=87\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)\)

=87 \(\times \frac{3-1}{\sqrt{3}}=\frac{2 \times 87}{\sqrt{3}} \mathrm{~m}\)

= \(\frac{2 \times 87}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2 \times 87 \times \sqrt{3}}{3}\)

= 58 \(\sqrt{3} \mathrm{~m}\)

Therefore, distance between two positions of balloons =58 \(\sqrt{3} m\)

Question 15. Question 15. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution :

Given

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°.

Let the height of the tower be AB and the two positions of the car be C and D.

Applications Of Trigonometry The Time Taken By The Car To Reach The Foot Of The Tower

⇒ \(\angle D B C=\angle D B X-\angle C B X=60^{\circ}-30^{\circ}=30^{\circ}\)

In \(\triangle B D C\),

⇒ \(\angle D B C =\angle D C B\)

C D =B D

⇒ \(\left({cach} 30^{\circ}\right)\)

(the sides opposite to equal angles are equal) In right \(\triangle B A D\),

⇒ \(\cos 60^{\circ}=\frac{A D}{D B} \Rightarrow \frac{1}{2}=\frac{A D}{D B} \Rightarrow D B=2 A D\)

CD = 2 AD

Now, time taken to cover distance CD = 6 sec.

Time taken to cover distance 2AD = 6 sec.

Time is taken to cover distance AD = 3 sec.

Question 16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution :

Given

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary.

Let AB be a tower of height h. Two points C and D are at 4 m and 9 m distances respectively from B.

Let, \(\angle A D C=\theta\)

Applications Of Trigonometry The Angles Of Elevation Of The Top Of Tower From Two Points At A Distance

⇒ \(\angle A C B=90^{\circ}-\theta\)

In \(\triangle A B D\),

⇒ \(\tan \theta=\frac{A B}{B D}=\frac{h}{9}\)  → Equation 1

In \(\triangle A B C\),

⇒ \(\tan \left(90^{\circ}-\theta\right)=\frac{A B}{B C}\)

⇒ \(\cot \theta=\frac{h}{4}\)

⇒ \(\frac{1}{\tan \theta}=\frac{h}{4} \quad \Rightarrow \frac{9}{h}=\frac{h}{4}\) [from equation 1 ]

⇒ \(h^2=36 \quad \Rightarrow h=6\)

Height of tower =6 m

Hence Proved.

NCERT Exemplar For Class 10 Maths Chapter 8 Introduction To Trigonometry

Introduction To Trigonometry

The word ‘trigonometry’ is derived from the Greek words: trigonon and metron. The word trigonon means a figure formed by three sides i.e., triangle, and metron means a measure. So, we can say that in trigonometry we solve the problems related to the sides and angles of a triangle.

NCERT Exemplar Solutions for Class 10 Maths Chapter 8 Introduction To Trigonometry

Identity

  1. An equation is a statement of equality between two expressions that is true for all values of the variable involved (taking into consideration the domain i.e., limitations of the variable), and is called an identity.
  2. An equation that involves trigonometric ratios of an angle and is true for all values of the angle is called a trigonometric identity.

Angle

An angle is considered as a figure obtained by rotating a given ray about its endpoint. The measure of an angle is the amount of rotation from its initial side to the terminal side. If the ray rotates in an anticlockwise direction then the angle will be positive. If the ray rotates in a clockwise direction then the angle will be negative.

Trigonometry An Angle Is Obtained A Given Ray

Trigonometric Ratios

The ratio of any two sides of a right-angled triangle is called trigonometric ratio. In the adjoining figure, \(\angle\)YAX = 0 is an acute angle. Consider a point C on ray AY. Draw perpendicular CB from C to AX.

⇒ \(\triangle\) ABC is a right-angled triangle in which

Trigonometry The Ratio Of Sides Of A Right Angle Triangle

⇒ \(\angle\)ABC = 90°

In \(\triangle\) ABC, let \(\angle\)BAC = 0

For \(\angle\)BAC = \(\theta\),

perpendicular P = BC

base B = AB

And hypotenuse H = AC

The ratio \(\frac{\text { perpendicular }}{\text { hypotenuse }}\) is called the sine of \(\theta\) and is written as \(\sin \theta\).

⇒ \(\sin \theta=\frac{P}{H}=\frac{B C}{A C}\)

The ratio \(\frac{\text { base }}{\text { hypotenuse }}\) is called the cosine of \(\theta\) and is written as \(\cos \theta\).

⇒ \(\cos \theta=\frac{B}{H}=\frac{A B}{A C}\)

The ratio \(\frac{\text { perpendicular }}{\text { base }}\) is called the tangent of \(\theta\) and is written as \(\tan \theta\).

∴ \(\tan \theta=\frac{P}{B}=\frac{B C}{A B}\)

Introduction to Trigonometry

The ratio \(\frac{\text { base }}{\text { perpendicular }}\) is called the cotangent of \(\theta\) and is written as \(\cot \theta\).

⇒ \(\cot \theta=\frac{B}{P}=\frac{A B}{B C}\)

The ratio \(\frac{\text { hypotenuse }}{\text { base }}\) is called the secant of \(\theta\) and is written as \(\sec \theta\).

⇒ \(\sec \theta=\frac{H}{B}=\frac{A C}{A B}\)

The ratio \(\frac{\text { hypotenuse }}{\text { perpendicular }}\) is called the cosecant of \(\theta\) and is written as \(cosec\theta\).

cosec \(\theta=\frac{H}{P}=\frac{A C}{B C}\)

Therefore, \(\sin \theta=\frac{1}{{cosec} \theta}=\frac{\text { perpendicular }}{\text { hypotenuse }}\), cosec \(\theta=\frac{1}{\sin \theta}=\frac{\text { hypotenuse }}{\text { perpendicular }}\)

⇒ \(\cos \theta=\frac{1}{\sec \theta}=\frac{\text { base }}{\text { hypotenuse }}\), \(\sec \theta=\frac{1}{\cos \theta}=\frac{\text { hypotenuse }}{\text { base }}\)

⇒ \(\tan \theta=\frac{1}{\cot \theta}=\frac{\sin \theta}{\cos \theta}=\frac{\text { perpendicular }}{\text { base }}\), \(\cot \theta=\frac{1}{\tan \theta}=\frac{\cos \theta}{\sin \theta}=\frac{\text { base }}{\text { perpendicular }}\)

Remember :

  •  \(\sin \theta \neq \sin \times \theta\)
  • \(\cos \theta \neq \cos \times \theta\)
  • \(\sin ^2 \theta =(\sin \theta)^2 \neq \sin ^2 \theta^2 \neq \sin ^2\)
  • \(cosec \theta =\frac{1}{\sin \theta}=(\sin \theta)^{-1} \neq \sin ^{-1} \theta\)

Perpendicular, Base, And Hypotenuse In A Right-Angled Triangle

See carefully the following right-angled triangles :

Trigonometry The Values Of Trigonometric Ratio In The Different Angles

Let us see the values of trigonometric ratios in the different figures in terms of AB, BC, and AC.

Remember:

  1. The side of the triangle at which the right angle (90°) and the given angle lie, is called the base.
  2.  The side opposite to the 90° angle is called the hypotenuse.
  3. The third remaining side is called the perpendicular.

Now,

In (1). \(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{B C}\)

In (2). \(\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{A C}{A B}\)

In (3). \(\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{B C}{A C}\)

In (4). cosec \(\theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}\)

In (5). \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{A C}{A B}\)

Solved Examples

Example 1. In \(\triangle\) ABC, \(\angle\) B = 90°, if AB = 5 cm, BC = 12 cm, then find the values of the following :

  1. sin A
  2. cos A
  3. cot A
  4. cosec C
  5. sec C
  6. tan C

Solution.

In \(\triangle\) A B C, from Pythagoras theorem

⇒ \(A C^2=A B^2+B C^2 =5^2+12^2\)

Trigonometry In Triangle ABC Form Pythagoras Theorem

=25+144=169

A C=13 cm

For \(\angle\) A, the base is AB, perpendicular is BC. For \(\angle C\), the base is BC, the perpendicular is AB, while the hypotenuse is the same i.e., AC for both angles.

  1. \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{12}{13}\)
  2. \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{5}{13}\)
  3. \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{5}{12}\)
  4. cosec C=\(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{A B}=\frac{13}{5}\)
  5. \(\sec C=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{B C}=\frac{13}{12}\)
  6.  \(\tan C=\frac{\text { perpendicular }}{\text { base }}=\frac{A B}{B C}=\frac{5}{12}\)

Question 2. In \(\triangle\) ABC, \(\angle B=90^{\circ}\) and \(\sin A=\frac{4}{5}\), then find the values of all other trigonometric ratios for \(\angle\) A.

Solution:

We know that, \(\sin A=\frac{4}{5}=\frac{\text { perpendicular }}{\text { hypotenuse }}\)

Now construct a \(\triangle\) A B C in which

Trigonometry In Triangle ABC The Values Of All Other Trigonometric Ratio

⇒ \(\angle B=90^{\circ}\), B C=4 k and A C=5 k .

In \(\triangle\) A B C, from Pythagoras theorem

⇒ \(A B^2+B C^2=A C^2\)

⇒ \(A B^2=A C^2-B C^2=(5 k)^2-(4 k)^2\)

=25 \(k^2-16 k^2=9 k^2\)

AB = 3k

Now,\(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{3 k}{5 k}=\frac{3}{5}\)

⇒ \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{4 k}{3 k}=\frac{4}{3}\)

cosec A=\(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{5 k}{4 k}=\frac{5}{4}\)

⇒ \(\sec A=\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\frac{5 k}{3 k}=\frac{5}{3}\)

∴ \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{3 k}{4 k}=\frac{3}{4}\)

Example 3. In \(\triangle\) PQR, \(\angle R=90^{\circ}\) and \(\tan \theta=\frac{5}{12}\) where \(\angle\) QPR = \(\theta\). Find the values of all other trigonometric ratios for \(\angle \theta\).

Solution:

We know that,

⇒ \(\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{5}{12}\)

Trigonometry The Values Of Trigonometric Ratio For Theta

Now, construct a \(\triangle P Q R\) in which

P R=12 k, Q R=5 k and \(\angle Q R P=90^{=}\). }

Let \(\angle Q P R=\theta\)

In \(\triangle\) PQR,

From Pythagoras theorem

⇒ \(P Q^2=P R^2+Q R^2=(12 k)^2+(5 k)^2=144 k^2+25 k^2=169 k^2\)

PQ = 13k

Now,\(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{Q R}{P Q}=\frac{5 k}{13 k}=\frac{5}{13}\)

⇒ \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{P R}{P Q}=\frac{12 k}{13 k}=\frac{12}{13}\)

⇒ \(cosec 0 =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{P(2}{Q R}=\frac{13 k}{5 k}=\frac{13}{5}\)

⇒ \(\sec \theta =\frac{\text { hypotenuse }}{\text { base }}=\frac{P Q}{P R}=\frac{13 k}{12 k}=\frac{13}{12}\)

⇒ \(\cot \theta =\frac{\text { basse }}{\text { perpendicular }}=\frac{P R}{Q R}=\frac{12 k}{5 k}=\frac{12}{5}\)

Example 4. In \(\triangle A B C\), \(\angle C=90^{\circ}\) and cosec A=\(\frac{13}{12}\) find the values of all other trigonometric ratios for \(\angle A\).

Solution:

We know that,

cosec A=\(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{13}{12}\)

Now, construct a \(\triangle A B C\) in which A B=13 k, B C=12 k and \(\angle A C B=90^{\circ}\).

Trigonometry The Value Of Other Trigonometric Ratio For Angle A

In \(\triangle A B C\), from Pythagoras theorem

⇒ \(A C^2+B C^2 =A B^2\)

⇒ \(A C^2 =A B^2-B C^2=(13 k)^2-(12 k)^2\)

=\(169 k^2-144 k^2=25 k^2\)

AC = 5 k

Now, \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A B}=\frac{12 k}{13 k}=\frac{12}{13}\)

⇒ \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A C}{A B}=\frac{5 k}{13 k}=\frac{5}{13}\)

⇒ \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A C}=\frac{12 k}{5 k}=\frac{12}{5}\)

⇒ \(\sec A=\frac{\text { hypotenuse }}{\text { base }}=\frac{A B}{A C}=\frac{13 k}{5 k}=\frac{13}{5}\)

∴ \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{5 k}{12 k}=\frac{5}{12}\)

Example 5. If cos A=\(\frac{1}{3}\), then find the values of sin A and tan A

Solution:

Construct a right-angled triangle ABC in which cos A=\(\frac{1}{3}=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}\)

Let AB = k and AC = 3 k

From Pythagoras theorem,

Trigonometry The Values Of Sin A And Tan A

From Pythagoras theorem, \(B C^2+A B^2 =A C^2\)

⇒ \(B C^2 =A C^2-A B^2=(3 k)^2-(k)^2\)

= \(9 k^2-k^2=8 k^2\)

BC = \(\sqrt{8 k^2}=2 \sqrt{2} k\)

The values of sin A and tan A are

⇒ \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{2 \sqrt{2} k}{3 k}=\frac{2 \sqrt{2}}{3}\)

and \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{2 \sqrt{2} k}{k}=2 \sqrt{2}\)

Example 6. In \(\triangle A B C\), \(\tan B=\sqrt{3}\) find the values of cosec B and cos B.

Solution:

Construct a right-angled triangle ABC in which

⇒ \(\tan B=\sqrt{3}=\frac{\sqrt{3}}{1}=\frac{\text { perpendicular }}{\text { base }}=\frac{A C}{A B}\)

Let \(A C=\sqrt{3}\) k and AB = k

From Pythagoras theorem

Trigonometry The Values Of Cosec B And Cos B

⇒ \(B C^2 =A B^2+A C^2\)

=\((k)^2+(\sqrt{3} k)^2\)

⇒ \(B C^2 =k^2+3 k^2=4 k^2\)

BC =2 k

The values of cosec B and cos B are

⇒ \(cosec B =\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{B C}{A C}=\frac{2 k}{\sqrt{3} k}=\frac{2}{\sqrt{3}}\)

and \(\cos B =\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{B C}=\frac{k}{2 k}=\frac{1}{2}\)

Example 7. If \(\cos \theta=\frac{4}{5}\), then find the value of \(\left(\sin \theta \cos \theta+\tan ^2 \theta\right) \)

Solution:

Given that, \(\cos \theta=\frac{4}{5}\)

Construct a right-angled \(\triangle A B C\) in which \(\angle B A C=90^{\circ}\), AC = 4 k and BC = 5k.

In \(\triangle ABC\),

Trigonometry Construction Of An Right Angle Triangle ABC

From Pythagoras theorem

⇒ \(A B^2+A C^2 =B C^2\)

⇒ \(A B^2 =B C^2-A C^2=(5 k)^2-(4 k)^2\)

= \(25 k^2-16 k^2=9 k^2\)

AB = 3k

and \(\sin \theta =\frac{A B}{B C}=\frac{3 k}{5 k}=\frac{3}{5}\)

⇒ \(\tan \theta =\frac{A B}{A C}=\frac{3 k}{4 k}=\frac{3}{4}\)

Now, \(\sin \theta \cos \theta+\tan ^2 \theta=\frac{3}{5} \cdot \frac{4}{5}+\left(\frac{3}{4}\right)^2=\frac{12}{25}+\frac{9}{16}\)

= \(\frac{192+225}{400}=\frac{417}{400}\)

The value of \(\left(\sin \theta \cos \theta+\tan ^2 \theta\right) \)= 417/400.

Example 8. If sec A=2, then find the value of \(\frac{1}{\cot A}+\frac{\cos A}{1+\sin A}\).

Solution:

Given that, \(\sec A=2=\frac{2}{1}\)

Construct a right-angled \(\triangle A B C\) in which

⇒ \(\angle A B C=90^{\circ}\), AB = k and AC = 2k .

In \(\triangle A B C\),

From Pythagoras theorem

Trigonometry Construction Of An Right Angle Triangle ABC From Pythagoras Theorem

⇒ \(A B^2+B C^2 = A C^2\)

⇒ \(B C^2 =A C^2-A B^2=(2 k)^2-(k)^2=4 k^2-k^2=3 k^2\)

⇒ \(B C=\sqrt{3} k\)

Now, \(\cot A=\frac{A B}{B C}=\frac{k}{\sqrt{3} k}=\frac{1}{\sqrt{3}}\)

⇒ \(\sin A=\frac{B C}{A C}=\frac{\sqrt{3} k}{2 k} =\frac{\sqrt{3}}{2}\) and \(\cos A=\frac{A B}{A C}=\frac{k}{2 k}=\frac{1}{2}\)

⇒ \(\frac{1}{\cot A}+\frac{\cos A}{1+\sin A}=\frac{1}{\frac{1}{\sqrt{3}}}+\frac{\frac{1}{2}}{1+\frac{\sqrt{3}}{2}}=\sqrt{3}+\frac{\frac{1}{2}}{\frac{2+\sqrt{3}}{2}}\)

= \(\sqrt{3}+\frac{1 \times(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}=\sqrt{3}+\frac{2-\sqrt{3}}{4-3}\)

= \(\sqrt{3}+2-\sqrt{3}=2\)

The value of \(\frac{1}{\cot A}+\frac{\cos A}{1+\sin A}\) = 2

Example 9. If \(\cot A=\frac{b}{a}\), then prove that:

Solution:

⇒ \(\frac{a \sin A-b \cos A}{a \sin A+b \cos A}=\frac{a^2-b^2}{a^2+b^2}\)

We have,

L.H.S. =\(\frac{a \sin A-b \cos A}{a \sin A+b \cos A} =\frac{a \frac{\sin A}{\sin A}-b \frac{\cos A}{\sin A}}{a \frac{\sin A}{\sin A}+b \frac{\cos A}{\sin A}}\)

=\(\frac{a-b \cot A}{a+b \cot A}=\frac{a-b \times \frac{b}{a}}{a+b \times \frac{b}{a}}\) (dividing Nr. and Dr. by sin A ) (\(\cot A=\frac{b}{a}\))

=\(\frac{\frac{a^2-b^2}{a}}{\frac{a^2+b^2}{a}}=\frac{a^2-b^2}{a^2+b^2}\)= R.H.S

Hence Proved.

Example 10. In the adjoining figure, AM = BM and \(\angle B=90^{\circ}\). If \(\angle B C M=\theta\), then find the values of the following :

  1. sin θ
  2. tan θ
  3. sec θ

Solution:

In \(\triangle A B C\),

From Pythagoras theorem

Trigonometry In The Adjoining The Values Of Triangle ABC

⇒ \(B C^2 =A C^2-A B^2=b^2-(2 a)^2=b^2-4 a^2\)

⇒ \(B C =\sqrt{b^2-4 a^2}\)

Now, \(B M=\frac{A B}{2}=a\)

In \(\triangle B C M\),

From Pythagoras theorem

⇒ \(C M^2 =B C^2+B M^2=\left(b^2-4 a^2\right)+a^2=b^2-3 a^2\)

⇒ \(C M =\sqrt{b^2-3 a^2}\)

⇒ \(\sin \theta=\frac{B M}{C M}=\frac{a}{\sqrt{b^2-3 a^2}}\)

⇒ \(\tan \theta=\frac{B M}{B C}=\frac{a}{\sqrt{b^2-4 a^2}}\)

∴ \(\sec \theta=\frac{C M}{B C}=\frac{\sqrt{b^2-3 a^2}}{\sqrt{b^2-4 a^2}}\)

Example 11. In the adjoining figure, \(\angle B C D=\angle A D B(each 90^{\circ} )\). If B C=3 cm and the length of the side opposite \(\angle C\) in \(\triangle B C D\) is 5 cm, then find the square root of the length of the side opposite to \(\angle D\) in MDB.

Solution:

Draw DE \(\perp A B\)

In right \(\triangle B C D\), by Pythagoras theorem,

Trigonometry The Square Root Of Length Side Opposite

In right \(\triangle A E D\), by Pythagoras theorem,

⇒ \(A D^2=A E^2+D E^2\)

⇒ \(y^2=x^2+9\)

In right \(\triangle A D B\), by Pythagoras theorem,

Trigonometry In Right Triangle ADB By Pythagoras Theorem

⇒ \(A B^2=A D^2+D B^2\)

⇒ \((x+4)^2=y^2+25\)

⇒\(x^2+8 x+16=\left(x^2+9\right)+25\)

8 x=18

x=\(\frac{18}{8}=2.25 \mathrm{~cm}\)

⇒\(A B=x+4=2.25+4=6.25 \mathrm{~cm}\)

⇒\(\sqrt{A B}=\sqrt{6.25}=2.5 \mathrm{~cm}\)

Signs Of The Trigonometric Ratios

Let a rotating line rotate \(\angle XOA\) = \(\theta\) in an anticlockwise direction, starting from its initial position OX. Here, PM is perpendicular from P to OX where P is a point on side OA.

Trigonometry Signs Of The Trigonometric Ratios

In the first quadrant, 0 is the acute angle.

Here, OM > 0, PM > 0, OP > 0

Now,

⇒ \(\sin \theta=\frac{P M}{O P}>0\) ,\(\cos \theta=\frac{O M}{O P}>0\)

⇒ \(\tan \theta=\frac{P M}{O M}>0\) ,\(\cot \theta=\frac{O M}{P M}>0\)

⇒ \(\sec \theta=\frac{O P}{O M}>0 , cosec \theta=\frac{O P}{P M}>0\)

Therefore, all trigonometric ratios for all angles in the first quadrant are positive.

Trigonometric Ratios Of Specific Angles

1. Trigonometric Ratios for 30° and 60°

Trigonometry Trigonometric Ratios Of Specific Angles

⇒ \(\triangle ABC\) is an equilateral triangle whose side is ‘2a’.

⇒ \(\triangle A B C\) is an equilateral triangle whose side is ‘ 2 a ‘.

⇒ \(\angle A B C=\angle A C B=\angle B A C=60^{\circ}\)

AD is perpendicular from A to BC.

and \(\angle BAD =\angle C A D=30^{\circ}\)

B D = CD = a.

In \(\triangle A B D\), from Pythagoras theorem

⇒ \(A D^2+B D^2=A B^2\)

⇒ \(A D^2+a^2 =(2 n)^2\)

⇒ \(A D^2 =3 a^2\)

A D =a \(\sqrt{3}\)

For \(30^{\circ}\), in \(\triangle A B D\)

Base AD = a \(\sqrt{3}\), perpendicular B D= a and hypotenuse AB = 2a

⇒ \(\sin 30^{\circ}=\frac{B D}{A B}=\frac{a}{2 a}=\frac{1}{2}\)

⇒ \(cosec 30^{\circ}=\frac{A B}{B D}=\frac{2 a}{a}=2\)

⇒ \(\cos 30^{\circ}=\frac{A D}{A B}=\frac{a \sqrt{3}}{2 a}=\frac{\sqrt{3}}{2}\)

⇒ \(\sec 30^{\circ}=\frac{A B}{A D}=\frac{2 a}{a \sqrt{3}}=\frac{2}{\sqrt{3}}\)

⇒ \(\tan 30^{\circ}=\frac{B D}{A D}=\frac{a}{a \sqrt{3}}=\frac{1}{\sqrt{3}}\)

⇒ \(\cot 30^{\circ}=\frac{A D}{B D}=\frac{a \sqrt{3}}{a}=\sqrt{3}\)

For \(60^{\circ}\), in \(\triangle A B D\)

⇒ \(\sin 60^{\circ}=\frac{A D}{A B}=\frac{a \sqrt{3}}{2 a}=\frac{\sqrt{3}}{2}\)

⇒ \(\cos 60^{\circ}=\frac{B D}{A B}=\frac{a}{2 a}=\frac{1}{2}\)

⇒ \(\tan 60^{\circ}=\frac{A D}{B D}=\frac{a \sqrt{3}}{a}=\sqrt{3}\)

⇒ \(cosec 60^{\circ}=\frac{A B}{A D}=\frac{2 a}{a \sqrt{3}}=\frac{2}{\sqrt{3}}\)

⇒ \(\sec 60^{\circ}=\frac{A B}{B D}=\frac{2 a}{a}=2\)

⇒ \(\cot 60^{\circ}=\frac{B D}{A D}=\frac{a}{a \sqrt{3}}=\frac{1}{\sqrt{3}}\)

2. Trigonometric Ratios for 450

In \(\triangle\)ABC, \(\angle\)ABC = 90° and \(\angle\)BAC = 45°.

Trigonometry Ratios Of 45 Degrees

Therefore, \(\angle\)ACB = 45°

Let AB = BC = a

From Pythagoras theorem

In \(\triangle A B C\), \(\angle A B C=90^{\circ}\) and \(\angle B A C=45^{\circ}\). Therefore, \(\angle A C B=45^{\circ}\)

Let AB = BC = a

From Pythagoras theorem

⇒ \(A C^2 =A B^2+B C^2\)

=\( a^2+a^2=2 a^2\)

A C =a \(\sqrt{2}\)

For A=\(45^{\circ}\),

⇒ \(\sin 45^{\circ}=\frac{B C}{A C}=\frac{a}{a \sqrt{2}}=\frac{1}{\sqrt{2}}\)

⇒ \(cosec 45^{\circ}=\frac{A C}{B C}=\frac{a \sqrt{2}}{a}=\sqrt{2}\)

⇒ \(\cos 45^{\circ}=\frac{A B}{A C}=\frac{a}{a \sqrt{2}}=\frac{1}{\sqrt{2}}\)

⇒ \(\sec 45^{\circ}=\frac{A C}{A B}=\frac{a \sqrt{2}}{a}=\sqrt{2}\)

⇒ \(\tan 45^{\circ}=\frac{B C}{A B}=\frac{a}{a}\)=1

⇒ \(\cot 45^{\circ}=\frac{A B}{B C}=\frac{a}{a}=1\)

3. Trigonometric Ratios for 0°

Trigonometry Ratio For Zero

In \(\triangle \)ABC, \(\angle\)BAC = 0 and \(\angle\)ABC = 90°.

For angle \(\theta\), base = AB, perpendicular = BC and hypotenuse = AC.

In \(\triangle\)ABC, it is clear that as the value of ‘\(\theta\)’ decreases, the length of BC decreases, and for 0 = 0°, BC = 0 and AC = AB. Therefore,

Introduction to Trigonometry

⇒ \(\sin 0^{\circ}=\frac{B C}{A C}=\frac{0}{A C}=0\)

⇒ \(cosec 0^{\circ}=\frac{A C}{B C}=\frac{A C}{0}=\infty\)

⇒ \(\cos 0^{\circ}=\frac{A B}{A C}=\frac{A B}{A B}=1\)

⇒ \(\sec 0^{\circ}=\frac{A C}{A B}=\frac{A B}{A B}=1\)

⇒ \(\tan 0^{\circ}=\frac{B C}{A B}=\frac{0}{A B}=0\)

⇒ \(\cot 0^{\circ}=\frac{A B}{B C}=\frac{A B}{0}=\infty\)

4. Trigonometric Ratios for 90°

In \(\triangle\)ABC, it is clear that as the value of ‘0’ increases, the length of AB decreases, and for 0 = 90°, AB = 0 and AC = BC.

Trigonometry Ratio For 90 Degrees

Therefore, \(\sin 90^{\circ}=\frac{B C}{A C}=\frac{A C}{A C}=1\)

⇒ \(cosec 90^{\circ}=\frac{A C}{B C}=\frac{B C}{B C}=1\)

⇒ \(\cos 90^{\circ}=\frac{A B}{A C}=\frac{0}{A C}=0 \sec 90^{\circ}=\frac{A C}{A B}=\frac{A C}{0}=\infty\)

∴ \(\tan 90^{\circ}=\frac{B C}{A B}=\frac{B C}{0}=\infty \cot 90^{\circ}=\frac{A B}{B C}=\frac{0}{B C}=0\)

Trigonometry Values Of Trigonometric Ratios

Trigonometry Solved Examples

Example 1. Evaluate : \(\sin ^2 60^{\circ} \tan 45^{\circ}-\cos ^2 45^{\circ} \sec 60^{\circ}\)

Solution:

We know that,

⇒ \(\sin 60^{\circ}=\frac{\sqrt{3}}{2}, \tan 45^{\circ}=1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}, \sec 60^{\circ}=2\)

⇒ \(\sin ^2 60^{\circ} \tan 45^{\circ} -\cos ^2 45^{\circ} \sec 60^{\circ} \)

= \(\left(\frac{\sqrt{3}}{2}\right)^2(1)-\left(\frac{1}{\sqrt{2}}\right)^2(2)=\frac{3}{4}-\frac{1}{2} \times 2=\frac{3}{4}-1=\frac{3-4}{4}=-\frac{1}{4}\)

\(\sin ^2 60^{\circ} \tan 45^{\circ}-\cos ^2 45^{\circ} \sec 60^{\circ}\) = -1/4.

Example 2. Evaluate : \(\cos 60^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}\)

Solution:

We know that,

⇒ \(\cos 60^{\circ} =\frac{1}{2}, \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \sin 60^{\circ}=\frac{\sqrt{3}}{2}, \sin 30^{\circ}=\frac{1}{2}\)

⇒ \(\cos 60^{\circ} \cos 30^{\circ}+ \sin 60^{\circ} \sin 30^{\circ}\)

= \(\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\sqrt{3}+\sqrt{3}}{4}=\frac{2 \sqrt{3}}{4}=\frac{\sqrt{3}}{2}\)

\(\cos 60^{\circ} \cos 30^{\circ}+\sin 60^{\circ} \sin 30^{\circ}\) = 1/2

Example 3. Show that : \(\cos 60^{\circ}=2 \cos ^2 30^{\circ}-1\)

Solution:

L.H.S. =\(\cos 60^{\circ}=\frac{1}{2}\)

R.H.S. = \(2 \cos ^230^{\circ}-1=2\left(\frac{\sqrt{3}}{2}\right)^2-1=2\left(\frac{3}{4}\right)-1=\frac{3}{2}-1\)

=\(\frac{3-2}{2}=\frac{1}{2}\)

L.H.S. = R.H.S. Hence Proved

Example 4. If \(A=15^{\circ}\), then find the value of \(\sec 2 A\).

Solution: 

A =\(15^{\circ} \Rightarrow 2 A=2 \times 15^{\circ}=30^{\circ} \)

⇒ \(\sec 2 A =\sec 30^{\circ}=\frac{2}{\sqrt{3}}\)

The value of \(\sec 2 A\) =\( \frac{2}{\sqrt{3}}\)

Example 5. If \(\sin x=1\), then find the value of \(\tan \frac{x}{3}\).

Solution:

⇒ \(\sin x =1\)

⇒ \(\sin x =\sin 90^{\circ} \Rightarrow x =90^{\circ}\)

⇒ \(\frac{x}{3} =\frac{90^{\circ}}{3}=30^{\circ}\)

∴ \(\tan \frac{x}{3}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}\)

The value of \(\tan \frac{x}{3}\) =[latec]\frac{1}{\sqrt{3}}[/latex]

Example 6. If \(\sin (A+B)=\frac{\sqrt{3}}{2}\) and \(\cos (A-B)=\frac{\sqrt{3}}{2}\), then find the values of A and B Given that, \(\sin (A+B)=\frac{\sqrt{3}}{2}\)

Solution:

⇒ \(\sin (A+B)=\sin 60^{\circ} \quad \Rightarrow \quad A+B=60^{\circ}\) Equation 1

and \(\cos (A-B)=\frac{\sqrt{3}}{2}\)

⇒ \(\cos (A-B)=\cos 30^{\circ} \quad \Rightarrow \quad A-B=30^{\circ}\) Equation 2

Adding equations (1) and (2)

⇒ \(A+B =60^{\circ}\)

2 A =\(90^{\circ} \quad\Rightarrow \quad A-B =30^{\circ} \)

A = \(45^{\circ}\)

Put A=\(60^{\circ}\) in equation (1)

⇒ \(45^{\circ}+B=60^{\circ} \quad \Rightarrow \quad B=60^{\circ}-45^{\circ}=15^{\circ}\)

The values of A and B are

⇒ \(A=45^{\circ}\) and \(B=15^{\circ}\)

Example 7.  In an acute-angled AABC, if tan (A + B – C) = 1 and see (B + C – A) = 2, then find the value of cos (45 – 3A).

Solution:

Solution. We have.

tan (A + B – C) = 1 = tan 45°

=* A + B- C = 45° …(1)

Also sec (B + C – A) = 2 = sec 60°

B + C – A = 60° …(2)

Adding equations (1) and (2), we get

25 = 105° ⇒ 5 = 52.5° …(3)

Subtracting equation (1) from equation (2), we get

2C – 2A = 15° ⇒ C-A = 7.5° …(4)

We know that A + 5 + C = 180°

⇒ A + C = 180° – 52.5° [from (3)]

⇒ A + C= 127.5° …(5)

⇒ -A + C = 7.5° …(4)

Adding equations (4) and (5), we get

2C = 135° ⇒ C = 67.5°

A = 127.5°-67.5° = 60°

cos (45 – 3A) = cos (4 x 52.5° – 3 x 60°) = cos (210° – 180°) = cos 30° = \(\frac{\sqrt{3}}{2}\)

The value of cos (45 – 3A) = \(\frac{\sqrt{3}}{2}\)

Example 8. If sin A= cos A, then evaluate \(\tan A+\sin ^2 A+1\).

Solution:

We know that, in A = cos A, then

Now, A =\(45^{\circ}\)

⇒ \(\tan A+\sin ^2 A+1 =\tan 45^{\circ}+\sin ^2 45^{\circ}+1\)

= \(1+\left(\frac{1}{\sqrt{2}}\right)^2+1=2+\frac{1}{2}=\frac{5}{2}\)

\(\tan A+\sin ^2 A+1\) = \(\frac{5}{2}\)

Example 9. Find the value of \(\left(\theta_1+\theta_2\right)\) if

Solution:

⇒ \(\tan \left(\theta_1+\theta_2\right)=\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \cdot \tan \theta_2}\)

where, \(\tan \theta_1=\frac{1}{2}\) and \(\tan \theta_2=\frac{1}{3}\).

⇒ \(\tan \left(\theta_1+\theta_2\right) =\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1 \cdot \tan \theta_2}=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\)

= \(\frac{\frac{3+2}{6}}{1-\frac{1}{6}}=\frac{\frac{5}{6}}{\frac{5}{6}}=1=\tan 45^{\circ}\)

∴ \(\theta_1+\theta_2 =45^{\circ}\)

The value of \(\left(\theta_1+\theta_2\right)\) = 45°

Example 10. Evaluate : cos 1° cos 2° cos 3° … cos 179°

Solution:

cos 90° whose value is zero lies in between cos 1° cos 2° cos 3°.., cos 179°

cos 1° cos 2° cos 3°… cos 179°

= cos 1° cos 2° cos 3° …. cos 90° cos 179°

= cos 1° cos 2° cos 3° …. X 0 X cos 179°

= 0 (0 x finite number = 0)

cos 1° cos 2° cos 3° … cos 179° = 0

Question 11. In the adjoining figure, a right-angled triangle $A B C$ is shown in which AM = CM = 3 m. If \(\angle A C M=15^{\circ}\), then find A C:

Solution:

Here, AM = CM

⇒ \(\angle A C M = \angle C A M\) (opposite angles of equal sides)

Trigonometry In Adjoining A Right Angled Triangle ABC

⇒ \(\angle C A M=15^{\circ}\)

⇒ \(\angle A C B=90^{\circ}-15^{\circ}=75^{\circ}\)

and \(\angle B C M=\angle A C B-\angle A C M=75^{\circ}-15^{\circ}=60^{\circ}\)

In \(\triangle B C M\),

⇒ \(\cos (\angle B C M)=\frac{B C}{C M}\)

⇒ \(\cos 60^{\circ}=\frac{B C}{3}\)

⇒ \(\frac{1}{2}=\frac{B C}{3}\)

⇒ \(B C=\frac{3}{2} \mathrm{~m}\)

and \(\sin (\angle B C M)=\frac{B M}{C M}\)

⇒ \(\sin 60^{\circ}=\frac{B M}{3}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{B M}{3}\)

⇒ \(B M=\frac{3 \sqrt{3}}{2} \mathrm{~m}\)

⇒ \(A B=A M+B M=\left(3+\frac{3 \sqrt{3}}{2}\right) \mathrm{m}\)

Now, in \(\triangle A B C\), from Pythagoras theorem

⇒ \(A C^2 =A B^2+B C^2=\left(3+\frac{3 \sqrt{3}}{2}\right)^2+\left(\frac{3}{2}\right)^2=9+\frac{27}{4}+9 \sqrt{3}+\frac{9}{4}\)

= \(18+9 \sqrt{3}=9(2+\sqrt{3})=\frac{9}{2}(4+2 \sqrt{3})\)

AC =\(\sqrt{\frac{9(4+2 \sqrt{3})}{2}}=\frac{3(\sqrt{3}+1)}{\sqrt{2}} \mathrm{~m}\)

Relation Between Trigonometric Ratios

1. \(\sin \theta \times cosec\theta=1\)

  • \(\sin \theta=\frac{1} {cosec \theta}\)
  • \(cosec \theta=\frac{1}{\sin \theta}\)

2. \(\cos \theta \times \sec \theta=1\)

  •  \(\cos \theta=\frac{1}{\sec \theta}\)
  •  \(\sec \theta=\frac{1}{\cos \theta}\)

3. \(\tan \theta \times \cot \theta=1\)

  • \(\tan \theta=\frac{1}{\cot \theta}\)
  •  \(\cot \theta=\frac{1}{\tan \theta}\)

4. \(\tan \theta=\frac{\sin \theta}{\cos \theta}\)

5. \(\cot \theta=\frac{\cos \theta}{\sin \theta}\)

If the value of a trigonometric function is known, then we can find the values of other trigonometric functions. We can use Pythagoras’ theorem and the above results for it.

Trigonometric Identities

We know about the algebraic equations.

The algebraic equation satisfies a particular value of the variable but in trigonometry, the equation can satisfy all values of the variable, such equations are called trigonometric identities.

  1.  \(\sin ^2 \theta+\cos ^2 \theta=1\)
  2.  \(\sec ^2 \theta=1+\tan ^2 \theta\)
  3. \(cosec^2 \theta=1+\cot ^2 \theta\)

Identity 1. \(\sin ^2 \theta+\cos ^2 \theta\)=1

Proof : Let \(\triangle\)ABC is a right-angled triangle in which \(\angle\)ABC = 90°

Trigonometry Identities

Let \(\angle\)ACB = \(\theta\)

For this angle ‘\(\theta\)’

AB = perpendicular

BC = base

AC = hypotenuse

From Pythagoras theorem,

⇒ \(A B^2+B C^2=A C^2\)

Now,\(\sin \theta=\frac{A B}{A C}\) and \(\cos \theta=\frac{B C}{A C}\)

Divide each term in equation (1) by A C^2

⇒ \(\frac{A B^2}{A C^2}+\frac{B C^2}{A C^2}=\frac{A C^2}{A C^2}\)

⇒ \(\left(\frac{A B}{A C}\right)^2+\left(\frac{B C}{A C}\right)^2=1 \quad \Rightarrow \quad \sin ^2 \theta+\cos ^2 \theta\)=1

Identity 2. \(\sec ^2 \theta=1+\tan ^2 \theta\)

Proof: In \(\triangle M B C\),

⇒ \(A B^2+B C^2=A C^2\)

Now,\(\tan \theta=\frac{A B}{B C}\) and \(\sec \theta=\frac{A C}{B C}\)

Divide each term by \(B C^2\) in equation (1)

⇒ \(\frac{A B^2}{B C^2}+\frac{B C^2}{B C^2}=\frac{A C^2}{B C^2}\)

⇒ \(\left(\frac{A B}{B C}\right)^2+1=\left(\frac{A C}{B C}\right)^2 \Rightarrow \tan ^2 \theta+1=\sec ^2 \theta\)

Identity 3. \(cosec^2 \theta=1+\cot ^2 \theta\)

Proof : In \(\triangle A B C\),

⇒ \(A B^2+B C^2=A C^2\)

Now,\(\cot \theta=\frac{B C}{A B}\) and \(cosec \theta=\frac{A C}{A B}\)

Divide each term in equation (1) by \(A B^2\)

⇒ \(\frac{A B^2}{A B^2}+\frac{B C^2}{A B^2}=\frac{A C^2}{A B^2} \)

⇒ \(1+\left(\frac{B C}{A B}\right)^2=\left(\frac{A C}{A B}\right)^2 \Rightarrow 1+\cot ^2 \theta={cosec}^2 \theta\)

Alternate Proof: Identity (2) \(\sec ^2 \theta=1+\tan ^2 \theta\) and (3)\(cosec^2 \theta=1+\cot ^2 \theta\) can be proved with the help of identity (1) \(\sin ^2 \theta+\cos ^2 \theta\)=1.

Proof of \(\sec ^2 \theta=1+\tan ^2 \theta\) :

From identity (1)

⇒ \(\sin ^2 \theta+\cos ^2 \theta\)=1

Divide each term by \(\cos ^2 \theta\)

⇒ \(\frac{\sin ^2 \theta}{\cos ^2 \theta}+\frac{\cos ^2 \theta}{\cos ^2 \theta}=\frac{1}{\cos ^2 \theta}\)

⇒ \(left(\frac{\sin \theta}{\cos \theta})^2+1=\left(\frac{1}{\cos \theta}\right)^2\)

⇒ \(\tan ^2 \theta+1 =\sec ^2 \theta \Rightarrow \sec ^2 \theta=1+\tan ^2 \theta\)

Proof of \(cosec^2 \theta=1+\cot ^2 \theta\) :

From identity ( 1 ) \(\sin ^2 \theta+\cos ^2 \theta\)=1

Divide each term by \(\sin ^2 \theta\)

⇒ \(\frac{\sin ^2 \theta}{\sin ^2 \theta}+\frac{\cos ^2 \theta}{\sin ^2 \theta}=\frac{1}{\sin ^2 \theta}\)

⇒ \(1+\left(\frac{\cos \theta}{\sin \theta}\right)^2=\left(\frac{1}{\sin \theta}\right)^2\)

⇒ \(1+\cot ^2 \theta= cosec^2 \theta\)

∴ \(cosec^2 \theta=1+\cot ^2 \theta\)

Other Form of the Above Identities :

1. \(\sin ^2 \theta+\cos ^2 \theta=1\)

  • \(\cos ^2 \theta=1-\sin ^2 \theta\)
  •  \(\sin ^2 \theta=1-\cos ^2 \theta\)

2. \(\sec ^2 \theta=1+\tan ^2 \theta\)

  • \(\tan ^2 \theta=\sec ^2 \theta-1\)
  • \(\sec ^2 \theta-\tan ^2 \theta=1\)

3. \(cosec^2 \theta=1+\cot ^2 \theta\)

  •  \(\cot ^2 \theta= cosec^2 \theta-1\)
  •  \(cosec^2 \theta-\cot ^2 \theta\)=1

Trigonometric Solved Examples

Example 1. Simplify : \(\left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta)\)

Solution:

⇒ \(\left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta) \)

⇒ \(=\left(1+\tan ^2 \theta\right)\left[(1)^2-(\sin \theta)^2\right]=\left(1+\tan ^2 \theta\right)\left(1-\sin ^2 \theta\right)\)

= \(\sec ^2 \theta \cdot \cos ^2 \theta \quad \text { (using the identities } \sec ^2 \theta=1+\tan ^2 \theta \text { and } \sin ^2 \theta+\cos ^2 \theta=1 \text { ) }\)

= \(\frac{1}{\cos ^2 \theta} \cdot \cos ^2 \theta\)=1

\(\left(1+\tan ^2 \theta\right)(1-\sin \theta)(1+\sin \theta)\) = 1

Example 2. Prove that : \(\cos ^2 \theta \cdot cosec\theta+\sin \theta= cosec \theta\)

Solution:

L.H.S. =\(\cos ^2 \theta \cdot cosec \theta+\sin \theta=\cos ^2 \theta \cdot \frac{1}{\sin \theta}+\sin \theta\)

= \(\frac{\cos ^2 \theta}{\sin \theta}+\frac{\sin \theta}{1}=\frac{\cos ^2 \theta+\sin ^2 \theta}{\sin \theta}=\frac{1}{\sin \theta}\)

= cosec \(\theta\)= R.H.S.

\(\cos ^2 \theta \cdot cosec\theta+\sin \theta= cosec \theta\)

Question 3. Prove that : \(\sec ^4 \theta-\tan ^4 \theta=1+2 \tan ^2 \theta\)

Solution:

L.H.S. =\(\sec ^4 \theta-\tan ^4 \theta=\left(\sec ^2 \theta\right)^2-\left(\tan ^2 \theta\right)^2\)

=\(\left(\sec ^2 \theta+\tan ^2 \theta\right)\left(\sec ^2 \theta-\tan ^2 \theta\right)=\left(1+\tan ^2 \theta+\tan ^2 \theta\right)\left(1+\tan ^2 \theta-\tan ^2 \theta\right)\)

=\(\left(1+2 \tan ^2 \theta\right)(1)=1+2 \tan ^2 \theta= R.H.S\).

Hence Proved

\(\sec ^4 \theta-\tan ^4 \theta=1+2 \tan ^2 \theta\)

Question 4. Prove that : \(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta+\sin \theta \cos \theta}=\cot \theta\)

Answer:

L.H.S. =\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta+\sin \theta \cos \theta}=\frac{\left(1-\sin ^2 \theta\right)+\cos \theta}{\sin \theta(1+\cos \theta)}=\frac{\cos ^2 \theta+\cos \theta}{\sin \theta(1+\cos \theta)}\)

=\(\frac{\cos \theta(\cos \theta+1)}{\sin \theta(1+\cos \theta)}=\frac{\cos \theta}{\sin \theta}=\cot \theta\)= R.H.S.

\(\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta+\sin \theta \cos \theta}=\cot \theta\)

Example 5. If \(\tan \theta=\frac{4}{3}\) , then find the value of \(\frac{3 \sin \theta-2 \cos \theta}{3 \sin \theta+5 \cos \theta}\) .

Solution:

⇒ \(\frac{3 \sin \theta-2 \cos \theta}{3 \sin \theta+5 \cos \theta}=\frac{3 \frac{\sin \theta}{\cos \theta}-2 \frac{\cos \theta}{\cos \theta}}{3 \frac{\sin \theta}{\cos \theta}+5 \frac{\cos \theta}{\cos \theta}}\)

=\(\frac{3 \tan \theta-2}{3 \tan \theta+5}=\frac{3 \times \frac{4}{3}-2}{3 \times \frac{4}{3}+5}=\frac{4-2}{4+5}=\frac{2}{9}\)

The value of \(\frac{3 \sin \theta-2 \cos \theta}{3 \sin \theta+5 \cos \theta}\) =\( \frac{2}{9}\)

Example 6. Prove that : \((\sec A+\tan A)(1-\sin A)=\cos A\)

Solution:

L.H.S. =\((\sec A+\tan A)(1-\sin A)\)

= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)=\frac{(1+\sin A)}{\cos A}(1-\sin A)\)

= \(\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A}=\cos A\) = R.H.S.

\((\sec A+\tan A)(1-\sin A)=\cos A\)

Example 7. Prove that : \({cosec A-\cot A=\frac{1}{cosec} A+\cot A}\)

Solution:

L.H.S. = \(cosec A-\cot A=(cosec A-\cot A) \cdot \frac{(cosec A+\cot A)}{(cosec A+\cot A)}\)

= \(\frac{cosec^2 A-\cot ^2 A}{cosec A+\cot A}=\frac{1}{cosec A+\cot A}\)= R.H.S.

Hence Proved.

\({cosec A-\cot A=\frac{1}{cosec} A+\cot A}\)

Example 8. Prove that: \(\frac{\sec A+1}{\tan A}=\frac{\tan A}{\sec A-1}\)

Solution:

L.H.S. =\(\frac{\sec A+1}{\tan A}=\frac{\sec A+1}{\tan A} \times \frac{\sec A-1}{\sec A-1}\)

[divide numerator and denominator by (sec A-1)]

= \(\frac{\sec ^2 A-1}{\tan A(\sec A-1)}=\frac{\tan ^2 A}{\tan A(\sec A-1)}\)

= \(\frac{\tan A}{\sec A-1}\)= R.H.S.

Hence Proved.

\(\frac{\sec A+1}{\tan A}=\frac{\tan A}{\sec A-1}\)

Example 9. Prove that : \(\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)= cosec \theta+\sec \theta\)

Solution:

L.H.S. =\(\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)\)

=\(\sin \theta(1+\tan \theta)+\cos \theta\left(1+\frac{1}{\tan \theta}\right)\)

= \(\sin \theta(1+\tan \theta)+\cos \theta\left(\frac{\tan \theta+1}{\tan \theta}\right)\)

=\((1+\tan \theta)\left[\sin \theta+\frac{\cos \theta}{\left(\frac{\sin \theta}{\cos \theta}\right)}\right]=(1+\tan \theta)\left(\sin \theta+\frac{\cos ^2 \theta}{\sin \theta}\right)\)

=\(\left(1+\frac{\sin \theta}{\cos \theta}\right)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta}\right)=\left(1+\frac{\sin \theta}{\cos \theta}\right)\left(\frac{1}{\sin \theta}\right)\)

⇒ \(=\frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta} \times \frac{1}{\sin \theta}\)

= \(cosec \theta+\sec \theta\)= R.H.S.

\(\sin \theta(1+\tan \theta)+\cos \theta(1+\cot \theta)= cosec \theta+\sec \theta\)

Example 10. If \(\sec \theta+\tan \theta=p\),then prove that : \(\frac{p^2-1}{p^2+1}=\sin \theta\)

Solution:

L.H.S. =\(\frac{p^2-1}{p^2+1}=\frac{(\sec \theta+\tan \theta)^2-1}{(\sec \theta+\tan \theta)^2+1}=\frac{\sec ^2 \theta+\tan ^2 \theta+2 \sec \theta \tan \theta-1}{\sec ^2 \theta+\tan ^2 \theta+2 \sec \theta \tan \theta+1}\)

= \(\frac{\left(\sec ^2 \theta-1\right)+\tan ^2 \theta+2 \sec \theta \tan \theta}{\sec ^2 \theta+\left(\tan ^2 \theta+1\right)+2 \sec \theta \tan \theta}\)

= \(\frac{\tan ^2 \theta+\tan ^2 \theta+2 \sec \theta \tan \theta}{\sec ^2 \theta+\sec ^2 \theta+2 \sec \theta \tan \theta}\)

= \(\frac{2 \tan ^2 \theta+2 \sec \theta \tan \theta}{2 \sec ^2 \theta+2 \sec \theta \tan \theta}=\frac{2 \tan \theta(\tan \theta+\sec \theta)}{2 \sec \theta(\sec \theta+\tan \theta)}\)

= \(\frac{\tan \theta}{\sec \theta}=\frac{\sin \theta}{\cos \theta} \times \frac{\cos \theta}{1}\)

= \(\sin \theta\) = R.H.S.

Hence Proved

\(\frac{p^2-1}{p^2+1}=\sin \theta\)

Alternative Method : We have \(\sec \theta+\tan \theta=p\)

⇒ \(\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\frac{p}{1}\Rightarrow \quad \frac{1+\sin \theta}{\cos \theta}=\frac{p}{1}\)

Squaring both sides, we get

⇒ \(\frac{(1+\sin \theta)^2}{\cos ^2 \theta}=\frac{p^2}{1}\)

⇒ \(\frac{(1+\sin \theta)^2}{1-\sin ^2 \theta}=\frac{p^2}{1}\){ (using identity } \(\sin ^2 \theta+\cos ^2 \theta=1 \text { ) }\)

⇒ \(\frac{(1+\sin \theta)^2}{(1+\sin \theta)(1-\sin \theta)}=\frac{p^2}{1} \quad \Rightarrow \quad \frac{1+\sin \theta}{1-\sin \theta}=\frac{p^2}{1}\)

Applying componendo and dividends, we get

⇒ \(\frac{2}{2 \sin \theta}=\frac{p^2+1}{p^2-1}\)

∴ \(\frac{1}{\sin \theta}=\frac{p^2+1}{p^2-1} \Rightarrow \quad \sin \theta=\frac{p^2-1}{p^2+1}\)

Example 11. Prove that : \(\frac{\sec \theta+1-\tan \theta}{\tan \theta+1-\sec \theta}=\frac{\sin \theta}{1-\cos \theta}\)

Solution:

Note: In such type of questions, it is better to write \(\sec ^2 \theta-\tan ^2 \theta or cosec^2 \theta-\cot ^2 \theta\) in only numerator.

If in R.H.S. the single term in either numerator or denominator is \(sin \theta\) then convert the question in cosec \(\theta\) and cot \(\theta\) and if the single term is cos \(\theta\) then convert the question in see \(\theta\) and tan \(\theta\).

As in this question in R.H.S. single term \(sin \theta\) is in the numerator so we will use \(cosec^2 \theta-\cot ^2 \theta\) for 1.

(dividing Nr and Dr by \(\sin \theta\) to convert it in } \(cosec \theta and \cot \theta \text { ) }\)

= \(\frac{{cosec} \theta+\cot \theta-1}{1+\cot \theta-{cosec} \theta}=\frac{{cosec} \theta+\cot \theta-\left({cosec}^2 \theta-\cot ^2 \theta\right)}{1+\cot \theta-{cosec} \theta}\)

= \(\frac{({cosec} \theta+\cot \theta)-({cosec} \theta+\cot \theta)({cosec} \theta-\cot \theta)}{1+\cot \theta-{cosec} \theta} \)

= \(\frac{({cosec} \theta+\cot \theta)[1-{cosec} \theta+\cot \theta]}{1+\cot \theta-{cosec} \theta}={cosec} \theta+\cot \theta \)

= \(\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}=\frac{(1+\cos \theta)}{\sin \theta}=\frac{(1+\cos \theta)(1-\cos \theta)}{\sin \theta(1-\cos \theta)}\)

= \(\frac{1-\cos ^2 \theta}{\sin \theta(1-\cos \theta)}=\frac{\sin ^2 \theta}{\sin \theta(1-\cos \theta)}=\frac{\sin \theta}{1-\cos \theta}=\text { R.H.S. }\)

Hence Proved.

\(\frac{\sec \theta+1-\tan \theta}{\tan \theta+1-\sec \theta}=\frac{\sin \theta}{1-\cos \theta}\)

Example 12. If x=r \(\sin A \cos C\), y=r \(\sin A \sin C\) and z=r \(\cos A\), then prove that :

⇒ \(r^2=x^2+y^2+z^2\)

Solution:

Here, x=r sin A cos C, y=r sin A sin C and z=r cos A

Now, R.H.S. = \(x^2+y^2+z^2\)

= \((r \sin A \cos C)^2+(r \sin A \sin C)^2+(r \cos A)^2\)

= \(r^2 \sin ^2 A \cos ^2 C+r^2 \sin ^2 A \sin ^2 C+r^2 \cos ^2 A\)

= \(r^2 \sin ^2 A\left(\cos ^2 C+\sin ^2 C\right)+r^2 \cos ^2 A \quad\left(\cos ^2 C+\sin ^2 C=1\right)\)

= \(r^2 \sin ^2 A+r^2 \cos ^2 A\)

= \(r^2\left(\sin ^2 A+\cos ^2 A\right)\)

= \(r^2\) =L.H.S.

\(r^2=x^2+y^2+z^2\)

Hence Proved.

Identities And Equations

Identities are special + type of equations that are true for all values of the variable while equations are true for some particular values of the variable.

Solved Examples

Example 1. Check whether the equation \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}=\frac{\sec \phi+1}{\sec \phi-1}\) is an identity or not?

Solution:

L.H.S. = \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}=\frac{\frac{\sin \phi}{\cos \phi}+\sin \phi}{\frac{\sin \phi}{\cos \phi}-\sin \phi}\)

= \(\frac{\sin \phi \sec \phi+\sin \phi}{\sin \phi \sec \phi-\sin \phi}=\frac{\sin \phi(\sec \phi+1)}{\sin \phi(\sec \phi-1)}=\frac{\sec \phi+1}{\sec \phi-1}\)= R.H.S.

L.H.S. = R.H.S.

And this equation is true for all values of \phi.

The given equation is an identity.

Example 2. Check whether the following equation \(\tan ^4 \theta+\tan ^6 \theta=\tan ^3 \theta \sec ^2 \theta\) is an identity or not?

Solution:

Given equation,

⇒ \(\tan ^4 \theta+\tan ^6 \theta =\tan ^3 \theta \sec ^2 \theta\)

⇒ \(\tan ^4 \theta\left(1+\tan ^2 \theta\right) =\tan ^3 \theta \sec ^2 \theta\)

⇒ \(\tan ^4 \theta \sec ^2 \theta =\tan ^3 \theta \sec ^2 \theta\)

L.H.S. \(\neq\) R.H.S.

It is not an identity.

Again, \(\tan ^4 \theta \sec ^2 \theta-\tan ^3 \theta \sec ^2 \theta=0\)

⇒ \(\tan ^3 \theta \sec ^2 \theta(\tan \theta-1)\) =0

⇒ \(\tan ^3 \theta\)=0 or \(\sec ^2 \theta=0\) or \(\tan \theta-1\)=0

Therefore, given equation satisfies only for \(\theta=0^{\circ}\) and \(\theta=45^{\circ}\).

The given equation is not an identity.

Alternate Method : For \(\theta=60^{\circ}\)

L.H.S. =\(\tan ^4 60^{\circ}+\tan ^6 60^{\circ}\)

= \((\sqrt{3})^4+(\sqrt{3})^6=9+27=36\)

and R.H.S. =\(\tan ^3 60^{\circ} \cdot \sec ^2 60^{\circ}\)

= \((\sqrt{3})^3(2)^2=12 \sqrt{3}\)

L.H.S. \(\neq\) R.H.S.

The given equation is not an identity.

Example 3. Solve : \(2 \sin ^2 \theta=\frac{1}{2}, 0^{\circ}<\theta<90^{\circ}\)

Solution:

⇒ \(2 \sin ^2 \theta =\frac{1}{2}\)

⇒ \(\sin ^2 \theta =\frac{1}{2 \times 2}=\frac{1}{4}\)

⇒ \(\sin \theta =\sqrt{\frac{1}{4}}=\frac{1}{2}\)

⇒ \(\sin \theta =\sin 30^{\circ}\)

∴ \(\theta =30^{\circ}\)

Example 4. Find the value of \(\theta if 2 \cos 3 \theta=1\) and \(0^{\circ}<\theta<90^{\circ}\).

Solution:

2 \(\cos 3 \theta\) =1

⇒ \(\cos 3 \theta =\frac{1}{2}\)

⇒ \(\cos 3 \theta =\cos 60^{\circ}\)

3 \(\theta =60^{\circ}\)

∴ \(\theta =20^{\circ}\)

The Value of θ = 20°

Example 5. Find the value of \(\theta\) if \(\sec ^2 \theta+\tan ^2 \theta=\frac{5}{3}\) and θ lies in first quadrant.

Solution:

⇒ \(\sec ^2 \theta+\tan ^2 \theta =\frac{5}{3}\)

⇒ \(1+\tan ^2 \theta+\tan ^2 \theta =\frac{5}{3} \quad \Rightarrow \quad 2 \tan ^2 \theta=\frac{5}{3}-1=\frac{2}{3}\)

⇒ \(\tan ^2 \theta=\frac{1}{3} \quad \Rightarrow \quad \tan ^2 \theta=\left(\frac{1}{\sqrt{3}}\right)^2\)

⇒ \(\tan \theta=\frac{1}{\sqrt{3}}\) (talking positive sign only)

⇒ \(\tan \theta=\tan 30^{\circ} \Rightarrow \theta=30^{\circ}\)

The value of \(\theta\) if \(\sec ^2 \theta+\tan ^2 \theta=\frac{5}{3}\) = 30°

Example 6. If \(0^{\circ}<\alpha<90^{\circ}\), then solve the equation \(\frac{\sin \alpha}{1-\cos \alpha}+\frac{\sin \alpha}{1+\cos \alpha}=4\).

solution:

⇒ \(\frac{\sin \alpha}{1-\cos \alpha}+\frac{\sin \alpha}{1+\cos \alpha} =4\)

⇒ \(\frac{\sin \alpha(1+\cos \alpha)+\sin \alpha(1-\cos \alpha)}{(1-\cos \alpha)(1+\cos \alpha)}=4\)

⇒ \(\frac{\sin \alpha+\sin \alpha \cos \alpha+\sin \alpha-\sin \alpha \cos \alpha}{1-\cos ^2 \alpha}\) =4

⇒ \(\frac{2 \sin \alpha}{\sin ^2 \alpha}\)=4

⇒ \(\frac{2}{\sin \alpha}\)=4

⇒ \(\sin \alpha =\frac{1}{2}=\sin 30^{\circ}\)

⇒ \(\alpha=30^{\circ}\)

Example 7. If \(0^{\circ}<\theta<90^{\circ}\), then find the value of \(\theta\) from the equation \(\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3\)

Solution:

⇒ \(\frac{\cos ^2 \theta}{\cot ^2 \theta-\cos ^2 \theta}=3\)

⇒ \(\frac{\cos ^2 \theta}{\frac{\cos ^2 \theta}{\sin ^2 \theta}-\cos ^2 \theta}=3 \Rightarrow \frac{\cos ^2 \theta}{\cos ^2 \theta\left(\frac{1}{\sin ^2 \theta}-1\right)}=3\)

⇒ \(\frac{1}{\frac{1}{\sin ^2 \theta}-1}=3 \quad \Rightarrow \quad \frac{1}{\frac{1-\sin ^2 \theta}{\sin ^2 \theta}}=3\)

⇒ \(\frac{\sin ^2 \theta}{1-\sin ^2 \theta}=3 \quad \Rightarrow \quad \frac{\sin ^2 \theta}{\cos ^2 \theta}=3\)

⇒ \(\tan ^2 \theta=(\sqrt{3})^2\)

⇒ \(\tan \theta=\sqrt{3}\) (taking positive sign only)

⇒ \(\tan \theta=\tan 60^{\circ}\)

∴ \(\theta=60^{\circ}\)

The Value of θ = 60°

Complementary Angles

Two angles are said to be complementary angles if their sum is 90°.

\(\theta\) and (90°- \(\theta\)) are complementary angles.

Trigonometric Ratios Of Complementary Angles

Let a rotating ray rotate 90° in an anticlockwise direction from the initial position OX and reach OY and after this, it rotates ‘9’ angle in a clockwise direction and reaches the OA position.

⇒ \(\angle\)XOA = 90° – \(\theta\).

Now, P is a point on side OA. PM and PN are perpendiculars from P to OX and OY respectively.

⇒ \(\sin \left(90^{\circ}-\theta\right)=\frac{P M}{O P}=\frac{O N}{O P}=\cos \theta\)

⇒ \(\cos \left(90^{\circ}-\theta\right)=\frac{O M}{O P}=\frac{P N}{O P}=\sin \theta\)

⇒ \(\tan \left(90^{\circ}-\theta\right)=\frac{P M}{O M}=\frac{O N}{P N}=\cot \theta\)

⇒ \(\cot \left(90^{\circ}-\theta\right)=\frac{O M}{P M}=\frac{P N}{O N}=\tan \theta\)

⇒ \(\sec \left(90^{\circ}-\theta\right)=\frac{O P}{O M}=\frac{O P}{P N}= cosec \theta\)

⇒ \({cosec}\left(90^{\circ}-\theta\right)=\frac{O P}{P M}=\frac{O P}{O N}=\sec \theta\)

⇒ \(\sin \left(90^{\circ}-\theta\right)=\cos \theta\)

⇒ \(\cos \left(90^{\circ}-\theta\right)=\sin \theta\)

⇒ \(\tan \left(90^{\circ}-\theta\right)=\cot \theta\)

⇒ \(\cot \left(90^{\circ}-\theta\right)=\tan \theta\)

⇒ \(\sec \left(90^{\circ}-\theta\right)={cosec} \theta\)

⇒ \({cosec}\left(90^{\circ}-\theta\right)=\sec \theta\)

Trigonometry Trigonometric Ratios Of Complementary Angles

Summary :

⇒ \(\sin \left(90^{\circ}-\theta\right)=\cos \theta\)

⇒ \(\cos \left(90^{\circ}-\theta\right)=\sin \theta\)

⇒ \(\tan \left(90^{\circ}-\theta\right)=\cot \theta\)

⇒ \(\cot \left(90^{\circ}-\theta\right)=\tan \theta\)

⇒ \(\sec \left(90^{\circ}-\theta\right)= cosec \theta\)

cosec\(\left(90^{\circ}-\theta\right)=\sec \theta\)

Solved Examples

Example 1. Evaluate the following :

  1. \(\frac{\sin 58^{\circ}}{\cos 32^{\circ}}\)
  2.  \(\frac{\sec 42^{\circ}}{{cosec} 48^{\circ}}\)

Solution:

  1.  \(\frac{\sin 58^{\circ}}{\cos 32^{\circ}}=\frac{\sin \left(90^{\circ}-32^{\circ}\right)}{\cos 32^{\circ}}=\frac{\cos 32^{\circ}}{\cos 32^{\circ}}\)=1
  2. \(\frac{\sec 42^{\circ}}{{cosec} 48^{\circ}}=\frac{\sec \left(90^{\circ}-48^{\circ}\right)}{{cosec} 48^{\circ}}=\frac{{cosec} 48^{\circ}}{{cosec} 48^{\circ}}\)=1

Example 2. Evaluate:

  1. \(\tan 42^{\circ}-\cot 48^{\circ}\)
  2. \(\sec 36^{\circ}- cosec 54^{\circ}\)

Solution:

(1)\(\tan 42^{\circ}-\cot 48^{\circ} =\tan 42^{\circ}-\cot \left(90^{\circ}-42^{\circ}\right)\)

=\(\tan 42^{\circ}-\tan 42^{\circ}\)=0

(2) \(\sec 36^{\circ}- cosec 54^{\circ}=\sec 36^{\circ}- cosec\left(90^{\circ}-36^{\circ}\right)\)

=\(\sec 36^{\circ}-\sec 36^{\circ}\)=0

Example 3. Prove that :

  1. \(\sin 42^{\circ} \cos 48^{\circ}+\sin 48^{\circ} \cos 42^{\circ}=1\)
  2.  \(\cos 70^{\circ} \cos 20^{\circ}-\sin 70^{\circ} \sin 20^{\circ}=0\)

Solution:

(1) L.H.S. =\(\sin 42^{\circ} \cos 48^{\circ}+\sin 48^{\circ} \cos 42^{\circ}\)

=\(\sin 42^{\circ} \cos \left(90^{\circ}-42^{\circ}\right)+\sin \left(90^{\circ}-42^{\circ}\right) \cos 42^{\circ}\)

= \(\sin 42^{\circ} \sin 42^{\circ}+\cos 42^{\circ} \cos 42^{\circ}\)

= \(\sin ^2 42^{\circ}+\cos ^2 42^{\circ}\)

= 1 = R.H.S.

Hence Proved.

(2) L.H.S. = cos 70° cos 20° – sin 70° sin 20°

= cos 70° cos 20° – sin (90° – 20°) sin(90° – 70°)

= cos 70° cos 20° – cos 20° cos 70°

= 0 = R.H.S.

Example 4. Without using trigonometric tables, evaluate :

⇒ \(\left(\frac{\tan 20^{\circ}}{{cosec} 70^{\circ}}\right)^2+\left(\frac{\cot 20^{\circ}}{\sec 70^{\circ}}\right)^2+2 \tan 15^{\circ} \tan 37^{\circ} \tan 53^{\circ} \tan 60^{\circ} \tan 75^{\circ}\)

Solutions:

⇒ \(\left(\frac{\tan 20^{\circ}}{{cosec} 70^{\circ}}\right)^2+\left(\frac{\cot 20^{\circ}}{\sec 70^{\circ}}\right)^2+2 \tan 15^{\circ} \tan 37^{\circ} \tan 53^{\circ} \tan 60^{\circ} \tan 75^{\circ}\)

= \(\left\{\frac{\tan 20^{\circ}}{{cosec}\left(90^{\circ}-20^{\circ}\right)}\right\}^2+\left\{\frac{\cot 20^{\circ}}{\sec \left(90^{\circ}-20^{\circ}\right)}\right\}^2\)

+2 \(\tan 15^{\circ} \tan 37^{\circ} \tan \left(90^{\circ}-37^{\circ}\right) \cdot(\sqrt{3}) \tan \left(90^{\circ}-15^{\circ}\right)\)

⇒ \(+2 \tan 15^{\circ} \tan 37^{\circ} \tan \left(90^{\circ}-37^{\circ}\right) \cdot(\sqrt{3}) \tan \left(90^{\circ}-15^{\circ}\right)\)

= \(\left(\frac{\tan 20^{\circ}}{\sec 20^{\circ}}\right)^2+\left(\frac{\cot 20^{\circ}}{{cosec} 20^{\circ}}\right)^2+2 \tan 15^{\circ} \tan 37^{\circ} \cot 37^{\circ} \cdot(\sqrt{3}) \cot 15^{\circ}\)

= \(\left(\frac{\sin 20^{\circ} / \cos 20^{\circ}}{1 / \cos 20^{\circ}}\right)^2+\left(\frac{\cos 20^{\circ} / \sin 20^{\circ}}{1 / \sin 20^{\circ}}\right)^2+2 \sqrt{3} \tan 15^{\circ} \tan 37^{\circ} \cdot \frac{1}{\tan 37^{\circ}} \cdot \frac{1}{\tan 15^{\circ}}\)

= \(\sin ^2 20^{\circ}+\cos ^2 20^{\circ}+2 \sqrt{3}=1+2 \sqrt{3}\)

Example 5. Without using trigonometric tables, evaluate the following :

⇒ \(\frac{{cosec}^2\left(90^{\circ}-\theta\right)-\tan ^2 \theta}{4\left(\cos ^2 48^{\circ}+\cos ^2 42^{\circ}\right)}-\frac{2 \tan ^2 30^{\circ} \sec ^2 52^{\circ} \sin ^2 38^{\circ}}{\left({cosec}^2 70^{\circ}-\tan ^2 20^{\circ}\right)}\)

Solution:

⇒ \(\frac{{cosec}^2\left(90^{\circ}-\theta\right)-\tan ^2 \theta}{4\left(\cos ^2 48^{\circ}+\cos ^2 42^{\circ}\right)}-\frac{2 \tan ^2 30^{\circ} \sec ^2 52^{\circ} \sin ^2 38^{\circ}}{\left({cosec}^2 70^{\circ}-\tan ^2 20^{\circ}\right)}\)

= \(\frac{\sec ^2 \theta-\tan ^2 \theta}{4\left\{\cos ^2\left(90^{\circ}-42^{\circ}\right)+\cos ^2 42^{\circ}\right)}-\frac{2\left(\frac{1}{\sqrt{3}}\right)^2 \sec ^2\left(90^{\circ}-38^{\circ}\right) \sin ^2 38^{\circ}}{{cosec}^2\left(90^{\circ}-20^{\circ}\right)-\tan ^2 20^{\circ}}\)

= \(\frac{1}{4\left(\sin ^2 42^{\circ}+\cos ^2 42^{\circ}\right)}-\frac{2{cosec}^2 38^{\circ} \cdot \sin ^2 38^{\circ}}{3\left(\sec ^2 20^{\circ}-\tan ^2 20^{\circ}\right)}\)

= \(\frac{1}{4}-\frac{2{cosec}^2 38^{\circ} \times \frac{1}{{cosec}^2 38^{\circ}}}{3}=\frac{1}{4}-\frac{2}{3}=\frac{-5}{12}\)

Example 6. Prove that :

  1. \(\sin \left(40^{\circ}-\theta\right)-\cos \left(50^{\circ}+\theta\right)\)=0
  2. \(\sec \left(65^{\circ}+\theta\right)-{cosec}\left(25^{\circ}-\theta\right)\)=0

Solution:

(1) L.H.S. =\(\sin \left(40^{\circ}-\theta\right)-\cos \left(50^{\circ}+\theta\right)\)

=\(\sin \left\{90^{\circ}-\left(50^{\circ}+\theta\right)\right\}-\cos \left(50^{\circ}+\theta\right)\)

=\(\cos \left(50^{\circ}+\theta\right)-\cos \left(50^{\circ}+\theta\right)\)=0= R.H.S.

(2) L.H.S. =\(\sec \left(65^{\circ}+0\right)- {cosec}\left(25^{\circ}-\theta\right)\)

=\(\sec \left\{90^{\circ}-\left(25^{\circ}-0\right)\right\}-{cosec}\left(25^{\circ}-\theta\right)\)

=\({cosec}\left(25^{\circ}-\theta\right)- {cosec}\left(25^{\circ}-\theta\right)\)=0= R.H.S.

Example 7. Express each of the following in terms of trigonometric ratios of angles between 0° and 45°.

  1. sin 70° + sec 70°
  2.  tan 65° + cosec 65°
  3. cos 81° + cot 80°

Solution:

  1. sin 70° + sec 70° = sin (90° – 20°) + sec (90° – 20°) = cos 20° + cosec 20°
  2. tan 65° + cosec 65° = tan (90° – 25°) + cosec (90° – 25°) = cot 25° + sec 25°
  3. cos 81° + cot 80° = cos (90° – 9°) + cot (90° -10°) = sin 9° + tan 10°

Example 8. If sin 3A = cos (A – 26°) where 3A is an acute angle, then find the value of A

Solution:

Given that,

sin 3A = cos (A – 26°).

cos (90° – 3A) = cos (A – 26°) ⇒ 90° – 3A =A – 26°

⇒ -471=-116° ⇒ A =29°

The value of A =29°

Example 9. If sin (\(\theta\)+ 24°) = cos \(\theta\) and \(\theta\) + 24° is an acute angle, then find the value of \(\theta\).

Solution:

Given that,

sin (\(\theta\)+ 24°) = cos \(\theta\)

⇒ sin (\(\theta\) + 24°) = sin (90° – \(\theta\))

⇒ \(\theta\) + 24° = 90° – \(\theta\)

⇒ 2\(\theta\) = 66°

∴ \(\theta\)= 33°

The value of θ = 33°

Example 10. If A, B, C are the angles of \(\triangle\) M B C, show that \(\sin \frac{B+C}{2}=\cos \frac{A}{2}\).

Solution:

In \(\triangle\) A B C,

A+B+C=\(180^{\circ}\)

B+C=\(180^{\circ}-A\)

⇒ \(\frac{B+C}{2}=90^{\circ}-\frac{A}{2}\)

⇒ \(\sin \frac{B+C}{2}=\sin \left(90^{\circ}-\frac{A}{2}\right)\)

⇒ \(\sin \frac{B+C}{2}=\cos \frac{A}{2}\)

Example 11. If \(\sin 36^{\circ}\)=p, then find \(\sin 54^{\circ}\) in terms of p.

Solution: 

We have, \(\sin 36^{\circ}\)=p

⇒ \(\sin ^2 36^{\circ}=p^2 \quad \Rightarrow \quad 1-\cos ^2 36^{\circ}=p^2\)

⇒ \(\cos ^2 36^{\circ}=1-p^2 \quad \Rightarrow \quad \cos ^2\left(90^{\circ}-54^{\circ}\right)=1-p^2\)

⇒ \(\sin ^2 54^{\circ}=1-p^2\)

∴ \(\sin 54^{\circ}=\sqrt{1-p^2}\)(taking only positive sign as \(54^{\circ}\) lies in 1 quadrant)

Example 12. If \(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \tan 4^{\circ} \ldots \tan 89^{\circ}=x^2-8\), then find the value of x .

Solution:

⇒ \(x^2-8=\left(\tan 1^{\prime \prime} \tan 89^{\circ}\right)\left(\tan 2^{\prime \prime} \tan 88^{\circ}\right)\left(\tan 3^{\circ} \tan 87^{\circ}\right) \ldots\left(\tan 44^{\circ} \tan 46^{\circ}\right) \tan 45^{\circ}\)

⇒ \(\left.=\left[\tan 1^{\circ} \tan \left(90^{\circ}-1^{\circ}\right)|| \tan 2^{\circ \prime} \tan \left(90^{\circ}-2^{\circ}\right)\right] \mid \tan 3^{\circ} \tan \left(90^{\circ}-3^{\circ}\right)\right]\) … \(\left|\tan 44^{\circ} \tan \left(90^{\circ}-44^{\circ}\right)\right| \times 1\)

=\(\left(\tan 1^{\prime \prime} \cot 1^{\circ}\right)\left(\tan 2^{\circ} \cot 2^{\circ}\right)\left(\tan 3^{\circ} \cot 3^{\circ}\right) \ldots\left(\tan 44^{\circ} \cot 44^{\circ}\right)\)

=\(1 \times 1 \times 1 \times \ldots \times 1 \quad(\tan x \cdot \cot x=\tan x \cdot \frac{1}{\tan x}=1). \)

⇒ \(x^2-8\)=1

⇒ \(x^2=9 \quad \Rightarrow \quad x= \pm 3\)

The value of  \( \quad x= \pm 3\)

Introduction Of Trigonometry Exercise 8.1

Question 1. In \(\triangle A B C\), right-angled at B, AB = 24 cm, BC = 7 cm. Determine :

  1. sin A, cos A
  2. sin C, cos C

Solution :

In \(\triangle A B C\),

Trigonometry In Triangle ABC, Right Angled At B

AB = 24 cm, BC = 7 cm and \(\angle B=90^{\circ}\)

From Pythagoras theorem,

⇒ \(A C^2 =A B^2+B C^2=24^2+7^2\)

=576+49=625

AC = 25 cm

(1) \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{7}{25}\)

⇒ \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{24}{25}\)

(2) \(\sin C=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{24}{25}\).

⇒ \(\cos C=\frac{\text { base }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{7}{25}\).

Question 2. In figure, find \(\tan P-\cot R\),

Solution :

In \(\triangle\) PQR,

Trigonometry The Value Of Tan P - Cot R

⇒ \(P Q^2+Q R^2 =P R^2\)

⇒ \(Q R^2 =P R^2-P Q^2\)

⇒ \(Q R^2 =(13)^2-(12)^2\)

=169-144=25 \(\Rightarrow Q R=5 \mathrm{~cm}\)

Now, \(\tan P=\frac{\text { perpendicular }}{\text { base }}=\frac{5}{12}\)

⇒ \(\cot R=\frac{\text { base }}{\text { perpendicular }}=\frac{5}{12}\)

∴ \(\tan P-\cot R=\frac{5}{12}-\frac{5}{12}=0\)

\(\tan P-\cot R\) = 0

Question 3. If \(\sin A=\frac{3}{4}\), calculate cos A and tan A.

Solution :

⇒ \(\sin A=\frac{3}{4}\)

In \(\triangle A B C\),

Trigonometry The Value Of Cos A And Tan A

Let BC = 3k

and AC = 4k

⇒ \(A B^2+B C^2 =A C^2\)

⇒ \(A B^2 =A C^2-B C^2\)

⇒ \(A C^2 =(4 k)^2-(3 k)^2\)

= \(16 k^2-9 k^2=7 k^2\)

A B = \(\sqrt{7} k\)

Now, \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{\sqrt{7} k}{4 k}=\frac{\sqrt{7}}{4}\)and \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{3 k}{\sqrt{7} k}=\frac{3}{\sqrt{7}}\)

Question 4. Given 15 cotA = 8, find sin A and Sec A.

Solution :

⇒ \(15 \cot A=8 \Rightarrow \cot A=\frac{8}{15}\)

Trigonometry The Value Of Sin A And Sec A

Let base =8 k=A B

and Perpendicular =15 k=B C

In \(\triangle A B C\),

⇒ \(A C^2=A B^2+B C^2\)

= \((8 k)^2+(15 k)^2\)

= \(64 k^2+225 k^2=289 k_A^2\)

AC = 17 k

Now, \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}\)

= \(\frac{B C}{A C}=\frac{15 k}{17 k}=\frac{15}{17}\)

and \(\sec A=\frac{\text { hypotenuse }}{\text { base }}\)

= \(\frac{A C}{A B}=\frac{17 k}{8 k}=\frac{17}{8}\)

Question 5. Given \(\sec \theta=\frac{13}{12}\), calculate all other trigonometric ratios.

Solution :

⇒ \(\sec \theta=\frac{13}{12}\)

Introduction to Trigonometry

⇒ \(\sec \theta=\frac{\text { hypotenuse }}{\text { base }}=\frac{13}{12}\)

Let, in \(\triangle A B C\), \(\angle B=90^{\circ}\) and \(\angle A=\theta\)

Let, A C=13 k and A B=12 k

Now, \(A B^2+B C^2=A C^2\)

⇒ \(B C^2 =A C^2-A B^2=(13 k)^2-(12 k)^2\)

= \(169 k^2-144 k^2=25 k^2\)

BC = 5 k

Now, \(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{5 k}{13 k}=\frac{5}{13}\)

⇒ \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{12 k}{13 k}=\frac{12}{13}\)

⇒ \(\tan \theta=\frac{\text { perpendicular }}{\text { base }}=\frac{5 k}{12 k}=\frac{5}{12}\)

cosec \(\theta=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{13 k}{5 k}=\frac{13}{5}\)

∴ \(\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{12 k}{5 k}=\frac{12}{5}\)

Question 6. If \(\angle A\) and \(\angle B\) are acute angles such that \(\cos A=\cos B\), then show that \(\angle A=\angle B\).

Solution :

Let, in \(\triangle A B C, \angle C=90^{\circ}\)

Trigonometry The Acute Angles Of Triangle ABC

⇒ \(\angle A\) and \(\angle B\) are acute angles.

Given,\(\cos A=\cos B\)

Question 7. If \(\cot \theta=\frac{7}{8}\), evaluate :

  1.  \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\),
  2. \(\cot ^2 \theta\)

Solution :

⇒ \(\cot \theta=\frac{7}{8}\)

Trigonometry In Triangle ABC The Value Of Cot

⇒ \(\cot \theta=\frac{\text { base }}{\text { perpendicular }}=\frac{7}{8}\)

Let, in \(\triangle A B C\), \(\angle B=90^{\circ}\) and \(\angle A=\theta\)

Let base AB = 7k

and perpendicular BC = 8k

Now, \(A C^2=A B^2+B C^2\)

=\((7 k)^2+(8 k)^2\)

=\(49 k^2+64 k^2=113 k^2\)

A C =\(\sqrt{113} k\)

Now, \(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}\)

=\(\frac{8 k}{\sqrt{113 k}}=\frac{8}{\sqrt{113}}\)

and \(\cos \theta=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}\)

= \(\frac{7 k}{\sqrt{113 k}}=\frac{7}{\sqrt{113}}\)

(1) \(\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}\)

= \(\frac{\left(1+\frac{8}{\sqrt{113}}\right)\left(1-\frac{8}{\sqrt{113}}\right)}{\left(1+\frac{7}{\sqrt{113}}\right)\left(1-\frac{7}{\sqrt{113}}\right)}\)

= \(\frac{(1)^2-\left(\frac{8}{\sqrt{113}}\right)^2}{(1)^2-\left(\frac{7}{\sqrt{113}}\right)^2}=\frac{1-\frac{64}{113}}{1-\frac{49}{113}}\)

= \(\frac{113-64}{113-49}=\frac{49}{64}\)

(2) \(\cot ^2 \theta=\left(\frac{7}{8}\right)^2=\frac{49}{64}\)

Question 8. If \(3 \cot A=4\), check whether \(\frac{\left(1-\tan ^2 A\right)}{\left(1+\tan ^2 A\right)}=\cos ^2 A-\sin ^2 A\) or not.

Solution :

3 cot A=4

⇒ \(\cot A=\frac{4}{3}\)

Now, \(\cot A=\frac{\text { base }}{\text { perpendicular }}=\frac{4}{3}\)

Trigonometry In Triangle ABC The Value Of Cos

In \(\triangle A B C, \angle B =90^{\circ}\)

base A B =4 k, perpendicular B C = 3 k

⇒ \(A C^2 =A B^2+B C^2=(4 k)^2+(3 k)^2\)

= \(16 k^2+9 k^2=25 k^2\)

A C = 5 k

Now, \(\tan A=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{3 k}{4 k}=\frac{3}{4}\)

⇒ \(\cos A=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}\)

⇒ \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{3 k}{5 k}=\frac{3}{5}\)

Now, \( \frac{1-\tan ^2 A}{1+\tan ^2 A}=\frac{1-\left(\frac{3}{4}\right)^2}{1+\left(\frac{3}{4}\right)^2}\)

= \(\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{16-9}{16+9}=\frac{7}{25}\)

and \(\cos ^2 A-\sin ^2 A=\left(\frac{4}{5}\right)^2-\left(\frac{3}{5}\right)^2\)

=\(\frac{16}{25}-\frac{9}{25}=\frac{16-9}{25}=\frac{7}{25}\)

∴ \(\frac{1-\tan ^2 A}{1+\tan ^2 A}=\cos ^2 A-\sin ^2 A\)

Question 9. In triangle ABC, right-angled at B, if \(\tan A=\frac{1}{\sqrt{3}}\), find the value of:

  1. \(\sin A \cos C+\cos A \sin C\)
  2.  \(\cos A \cos C-\sin A \sin C\)

Solution :

In \(\triangle A B C, \angle B=90^{\circ}\)

Trigonometry In Triangle ABC, Right Angled At B Of Tan A

⇒ \(\tan A=\frac{1}{\sqrt{3}}\)

⇒ \(\tan A =\frac{\text { perpendicular }}{\text { base }}\)

= \(\frac{1}{\sqrt{3}}\)

Let perpendicular B C = k and base \(A B=k \sqrt{3}\)

Now, \(A C^2 =A B^2+B C^2=(k \sqrt{3})^2+k^2\)

= \(3 k^2+k^2=4 k^2\)

A C = 2 k

Now, \(\sin A=\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}\)

⇒ \(\cos A =\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}\)

⇒ \(\sin C =\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{A B}{A C}\)

= \(\frac{k \sqrt{3}}{2 k}=\frac{\sqrt{3}}{2}\)

⇒ \(\cos C=\frac{\text { base }}{\text { hypotenuse }}=\frac{B C}{A C}=\frac{k}{2 k}=\frac{1}{2}\)

(1) \(\sin A \cos C+\cos A \sin C =\frac{1}{2} \times \frac{1}{2}+\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}\)

= \(\frac{1}{4}+\frac{3}{4}=1\)

(2) \(\cos A \cos C-\sin A \sin C=\frac{\sqrt{3}}{2} \times \frac{1}{2}-\frac{1}{2} \times \frac{\sqrt{3}}{2}=0\)

Question 10. In \(\triangle\) PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P, and tan P.

Solution :

In \(\triangle P Q R\), \(\angle Q=90^{\circ}\)

and P Q =5 cm

P R+Q R =25  → Equation 1

Now, \(P R^2=P Q^2+Q R^2\)

Trigonometry The Values OF Sin P, Cos P, Tan P

⇒ \(P R^2-Q R^2=P Q^2\)

⇒ \((P R-Q R)(P R+Q R) =5^2\)

⇒ \((P R-Q R) \times 25 =2\)

P R-Q R =1  → Equation 2

Adding equations (1) and (2),

2 \(\cdot P R=26 \quad \Rightarrow \quad P R=13\)

From equation (1)

Now, Q R =25-P R=25-13=12

⇒ \(\sin P =\frac{Q R}{P R}=\frac{12}{13}\)

⇒ \(\cos P =\frac{P Q}{P R}=\frac{5}{13}\)

∴ \(\tan P =\frac{Q R}{P Q}=\frac{12}{5}\)

Question 11. State whether the following are true or false. Justify your answer.

  1. The value of tan A is always less than 1.
  2.  sec A = \(\frac{12}{5}\) for some value of angle A.
  3. cos A is the abbreviation used for the cosecant of angle A.
  4. cot A is the product of cot and A.
  5. \(\sin \theta=\frac{4}{3}\) for some angle \(\theta\).

Solution :

(1) False, \(\tan A=\frac{\text { perpendicular }}{\text { base }}\)

tan A < l is possible only when the perpendicular is smaller the base but it is not always necessary, hypotenuse

(2) True, sec A=\(\frac{\text { hypotenuse }}{\text { base }}\)

Hypotenuse is always greater than the base.

Therefore, sec A = \(\frac{12}{5}\), is true tor some angle A

(3) False, cos A, is the brief form of the cosine of \(\angle\)A

Introduction To Trigonometry Exercise 8.2

Question 1. Evaluate the following :

  1.  \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)
  2.  \(2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}\)
  3. \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+{cosec} 30^{\circ}}\)
  4.  \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)
  5. \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)

Solution :

(1) \(\sin 60^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 60^{\circ}\)

=\(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{1}{2}\)

=\(\frac{3}{4}+\frac{1}{4}=\frac{4}{4}=1\)

(2) \(2 \tan ^2 45^{\circ}+\cos ^2 30^{\circ}-\sin ^2 60^{\circ}\)

=\(2(1)^2+\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2\)

=\(2+\frac{3}{4}-\frac{3}{4}=2\)

(3) \(\frac{\cos 45^{\circ}}{\sec 30^{\circ}+{cosec} 30^{\circ}}\)

=\(\frac{1 / \sqrt{2}}{\frac{2}{\sqrt{3}}+\frac{2}{1}}\)

⇒ \(=\frac{1}{\sqrt{2}\left(\frac{2+2 \sqrt{3}}{\sqrt{3}}\right)}\)

⇒ \(=\frac{\sqrt{3}}{2 \sqrt{2}(\sqrt{3}+1)}\)

= \(\frac{\sqrt{3} \cdot \sqrt{2}(\sqrt{3}-1)}{2 \sqrt{2}(\sqrt{3}+1) \cdot \sqrt{2}(\sqrt{3}-1)}\)

= \(\frac{\sqrt{2}(3-\sqrt{3})}{2 \cdot 2(3-1)}=\frac{3 \sqrt{2}-\sqrt{6}}{8}\)

(4) \(\frac{\sin 30^{\circ}+\tan 45^{\circ}-{cosec} 60^{\circ}}{\sec 30^{\circ}+\cos 60^{\circ}+\cot 45^{\circ}}\)

= \(\frac{\frac{1}{2}+1-\frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}}+\frac{1}{2}+1}\)

= \(\frac{\frac{\sqrt{3}+2 \sqrt{3}-4}{2 \sqrt{3}}}{\frac{4+\sqrt{3}+2 \sqrt{3}}{2 \sqrt{3}}}=\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4}\)

= \(\frac{3 \sqrt{3}-4}{3 \sqrt{3}+4} \times \frac{3 \sqrt{3}-4}{3 \sqrt{3}-4}\)

= \(\frac{27+16-24 \sqrt{3}}{(3 \sqrt{3})^2-(4)^2}\)

= \(\frac{43-24 \sqrt{3}}{27-16}\)

= \(\frac{43-24 \sqrt{3}}{11}\)

(5) \(\frac{5 \cos ^2 60^{\circ}+4 \sec ^2 30^{\circ}-\tan ^2 45^{\circ}}{\sin ^2 30^{\circ}+\cos ^2 30^{\circ}}\)

=\(\frac{5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-(1)^2}{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}\)

=\(\frac{\frac{5}{4}+4 \times \frac{4}{3}-1}{\frac{1}{4}+\frac{3}{4}}=\frac{\frac{15+64-12}{12}}{1}=\frac{67}{12}\)

Question 2. Choose the correct option and justify your choice :

(1) \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)= ?

  1. \(\sin 60^{\circ}\)
  2. \(\cos 60^{\circ}\)
  3. \(\tan 60^{\circ}\)
  4. \(\sin 30^{\circ}\)

(2) \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\) ?

  1. \(\tan 90^{\circ}\)
  2. 1
  3. \(\sin 45^{\circ}\)
  4. 0

(3) \(\sin 2 A=2 \sin A\) is true when A=

  1. \(0^{\circ}\)
  2. \(30^{\circ}\)
  3. \(45^{\circ}\)
  4. \(60^{\circ}\)

(4) \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)=

  1. \(\cos 60^{\circ}\)
  2. \(\sin 60^{\circ}\)
  3. \(\tan 60^{\circ}\)
  4. \(\sin 30^{\circ}\)

Solution:

(1) 1

⇒ \(\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}=\frac{2 \times \frac{1}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)

= \(\frac{\frac{2}{\sqrt{3}}}{1+\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{4}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}\)

(2) 4

⇒ \(\frac{1-\tan ^2 45^{\circ}}{1+\tan ^2 45^{\circ}}=\frac{1-(1)^2}{1+(1)^2}=\frac{1-1}{1+1}=\frac{0}{2}=0\)

(3) 1

If A =\(0^{\circ}\)then \(2 A=0^{\circ}\)

⇒ \(\sin 2 A =\sin 0^{\circ}=0\)

and \(2 \sin A =2 \sin 0^{\circ}\)

=2 \(\times 0=0\)

So, for \(A=0^{\circ}, \sin 2 A=2 \sin A\)

(4) 3

⇒ \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}} =\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}=\frac{2 / \sqrt{3}}{2 / 3}\)

= \(\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}=\tan 60^{\circ}\)

Question 3. If \(\tan (A+B)=\sqrt{3}\) and \(\tan (A-B)=\frac{1}{\sqrt{3}}\) ; \(0^{\circ}<A+B \leq 90^{\circ}\) ; A>B, find A and B.

Solution :

⇒ \(\tan (A+B)=\sqrt{3}\)

⇒ \(\tan (A+B)=\tan 60^{\circ}\)

⇒ \(A+B=60^{\circ}\)

and \(tan (A-B)=\frac{1}{\sqrt{3}}\)

⇒ \(\tan (A-B)=\tan 30^{\circ} \Rightarrow A-B=30^{\circ}\)

Adding equations (1) and (2)

⇒ \(A+B=60^{\circ}\)

⇒ \(A-B=30^{\circ}\)

⇒ \( 2 A=90^{\circ}\)

⇒ \(A \quad A=45^{\circ}\)

Put the value of A in equation (1),

⇒ \(45^{\circ}+B=60^{\circ} \Rightarrow \quad B\)

A=\(45^{\circ}\) and B=\(15^{\circ} \quad 45^{\circ}=15^{\circ}\)

Question 4. State whether the following are true or false. Justify your answer.

  1. \(\sin (A+B)=\sin A+\sin B\)
  2. The value of \(\sin \theta\) increases as \(\theta\) increases.
  3.  The value of \(\cos \theta\) increases as \(\theta\) increases.
  4. \(\sin \theta=\cos \theta\) for all values of θ.
  5.  cot A is not defined for A=\(0^{\circ}\)

Solution :

(1) False,

Let A=\(30^{\circ}\) and B=\(60^{\circ}\)

⇒ \(\sin (A+B)=\sin \left(30^{\circ}+60^{\circ}\right)=\sin 90^{\circ}=1\)

and \(\sin A+\sin B=\sin 30^{\circ}+\sin 60^{\circ}\)

= \(\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}+1}{2} \neq 1 \)

⇒ \(\sin (A+B) \neq \sin A+\sin B\)

(2) True, as the value of \(\theta\) varies from \(\theta^{\circ}\) to \(90^{\circ}\) then the value of \(\sin \theta\) varies from 0 to 1 .

(3) False, as the value of \(\theta\) varies front 0° to 90° then the value of cos \(\theta\) varies from 1 to 0,

i.e, the value of cos \(\theta\) decreases.

(4) False,

⇒ \(\theta\) = 0° then sin \(\theta\) = sin 0° = 0

and \(\cos 0=\cos \theta^{\circ}=1\)

⇒ \(\sin 0 \times \cos 0\), if \(\theta=0^{\circ}\)

(5) True,

⇒ \(A=0^{\circ}\) than \(\cot A=\cot 0^{\circ}\) which is not defined.

Exercise 8.3

Question 1. Evaluate :

  1. \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
  2. \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
  3. \(\cos 48^{\circ}-\sin 42^{\circ}\)
  4. \(cosec 31^{\circ}-\sec 59^{\circ}\)

Solution :

(1) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}} =\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)

=\(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}=1\)

\(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) = 1

(2)\(\frac{\tan 26^{\circ}}{\cot 64^{\circ}} =\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}\)

=\(\frac{\cot 64^{\circ}}{\cot 64^{\circ}}=1\)

\(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) =1

(3)\(\cos 48^{\circ}-\sin 42^{\circ} =\cos \left(90^{\circ}-42^{\circ}\right)-\sin 42^{\circ}\)

=\(\sin 42^{\circ}-\sin 42^{\circ}=0\)

\(\cos 48^{\circ}-\sin 42^{\circ}\) =0

(4) \({cosec} 31^{\circ}-\sec 59^{\circ}\)

= \({cosec}\left(90^{\circ}-59^{\circ}\right)-\sec 59^{\circ}\)

=\(\sec 59^{\circ}-\sec 59^{\circ}=0\)

\(cosec 31^{\circ}-\sec 59^{\circ}\)= 0

Question 2. Show that:

  1. \(\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1\)
  2. \(\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0\)

Solution :

(1) L.H.S. =\(\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}\)

=\(\tan 48^{\circ} \cdot \tan 23^{\circ} \cdot \tan \left(90^{\circ}-48^{\circ}\right)\)

⇒ \(\cdot \tan \left(90^{\circ}-23^{\circ}\right)\)

= \(\tan 48^{\circ} \cdot \tan 23^{\circ} \cdot \cot 48^{\circ} \cdot \cot 23^{\circ}\)

= \(\tan 48^{\circ} \cdot \tan 23^{\circ} \cdot \frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)

= 1 = R.H.S.

Hence Proved.

(2) L.H.S.=\(\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ}+\sin 52^{\circ}\)

=\(\cos \left(90^{\circ}-52^{\circ}\right) \cdot \cos \left(90^{\circ}-38^{\circ}\right)\)

⇒ \(-\sin 38^{\circ} \cdot \sin 52^{\circ}\)

=\(\sin 52^{\circ} \cdot \sin 38^{\circ}-\sin 38^{\circ} \cdot \sin 52^{\circ}\)

= 0 = R.H.S

Hence Proved.

Question 3. If \(\tan 2 A=\cot \left(A-18^{\circ}\right)\), where 2 A is an acute angle, find the value of A.

Solution:

⇒ \(\tan 2 A =\cot \left(A-18^{\circ}\right)\)

⇒ \(\cot \left(90^{\circ}-2 A\right) =\cot \left(A-18^{\circ}\right)\)

⇒ \(90^{\circ}-2 A =A-18^{\circ}\)

⇒ \(90^{\circ}+18^{\circ} =A+2 A\)

⇒ \(3 A =108^{\circ}\)

A =\(36^{\circ}\)

The value of A =\(36^{\circ}\)

Question 4. If \(\tan A=\cot B\), prove that

A+B=\(90^{\circ}\).

Solution :

⇒ \(\tan A=\cot B\)

⇒ \(\tan A=\tan \left(90^{\circ}-B\right)\)

⇒ \(A=90^{\circ}-B\)

⇒ \(A+B=90^{\circ}\)

Hence Proved.

Question 5. If \(\sec 4 A={cosec}\left(A-20^{\circ}\right)\), where 4 A is an acute angle, find the value of A.

Solution :

⇒ \(\sec 4 A = cosec \left(A-20^{\circ}\right)\)

cosec\(\left(90^{\circ}-4 A\right) = cosec\left(A-20^{\circ}\right)\)

⇒ \(90^{\circ}-4 A =A-20^{\circ}\)

⇒ \(90^{\circ}+20^{\circ} =A+4 A\)

⇒ \(5 A =110^{\circ}\)

A =\(22^{\circ}\)

The value of A =\(22^{\circ}\)

Question 6. If A, B and C are interior angles of a triangle A B C, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)

Solution :

In \(\triangle A B C\)

⇒ \(A+B+C=180^{\circ}\)

⇒ \(B+C=180^{\circ}-A\)

⇒ \(\frac{B+C}{2}=\frac{180^{\circ}-A}{2}\)

= \(\frac{180^{\circ}}{2}-\frac{A}{2}=90^{\circ}-\frac{A}{2}\)

⇒ \(\sin \left(\frac{B+C}{2}\right)=\sin \left(90^{\circ}-\frac{A}{2}\right)\)

⇒ \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)

Hence Proved.

Exercise 8.4

Question 1. Express the trigonometric ratios \(\sin A, \sec A\) and \(\tan A\) in terms of \(\cot A\).

Solution :

⇒ \(\sin A=\frac{1}{cosec^2 A}\)

⇒ \(\sin A=\frac{1}{\sqrt{{cosec}^2 A}}\)

⇒ \(\sin A=\frac{1}{\sqrt{1+\cot ^2 A}}\)

⇒ \(\sec A=\sqrt{\sec ^2 A}\)

⇒ \(\sec A=\sqrt{1+\tan ^2 A}\)

⇒ \(\sec A=\sqrt{1+\left(\frac{1}{\cot A}\right)^2}\)

⇒ \(\sec A=\sqrt{1+\frac{1}{\cot ^2 A}}=\sqrt{\frac{\cot ^2 A+1}{\cot ^2 A}}\)

and \(\tan A=\frac{1}{\cot A}\)

Question 2. Write all the other trigonometric ratios of \(\angle A\) in terms of \sec A.

Solution :

⇒ \(\sin A=\sqrt{\sin ^2 A} =\sqrt{1-\cos ^2 A}\)

= \(\sqrt{1-\frac{1}{\sec ^2 A}}=\sqrt{\frac{\sec ^2 A-1}{\sec ^2 A}}\)

⇒ \(\sin A =\frac{\sqrt{\sec ^2 A-1}}{\sec A}\)

cos A =\(\frac{1}{\sec A}\)

tan A =\(\sqrt{\tan ^2 A}\)

⇒ \(\tan A =\sqrt{\sec ^2 A-1}\)

⇒ \(\cot A =\frac{1}{\tan A}=\frac{1}{\sqrt{\tan ^2 A}}\)

⇒ \(\cot A=\frac{1}{\sqrt{\sec ^2 A-1}}\)

⇒ \(cosec A=\sqrt{{cosec}^2 A}=\sqrt{1+\cot ^2 A}\)

= \(\sqrt{1+\frac{1}{\tan ^2 A}}\)

= \(\sqrt{\frac{1+\tan ^2 A}{\tan ^2 A}}=\sqrt{\frac{\sec ^2 A}{\sec ^2 A-1}}\)

cosec A=\(\frac{\sec A}{\sqrt{\sec ^2 A-1}}\)

Question 3. Evaluate:

  1. \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)
  2. \(\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\)

Solution:

(1) \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)

=\(\frac{\sin ^2\left(90^{\circ}-27^{\circ}\right)+\sin ^2 27^{\circ}}{\cos ^2\left(90^{\circ}-73^{\circ}\right)+\cos ^2 73^{\circ}}\)

=\(\frac{\cos ^2 27^{\circ}+\sin ^2 27^{\circ}}{\sin ^2 73^{\circ}+\cos ^2 73^{\circ}}=\frac{1}{1}=1\)

  1. \(\frac{\sin ^2 63^{\circ}+\sin ^2 27^{\circ}}{\cos ^2 17^{\circ}+\cos ^2 73^{\circ}}\)=1

(2) \(\sin 25^{\circ} \cdot \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\)

= \(\sin 25^{\circ} \cdot \cos \left(90^{\circ}-25^{\circ}\right)\)

⇒ \(+\cos 25 \sin \left(90^{\circ}-25^{\circ}\right)\)

= \(\sin 25^{\circ} \cdot \sin 25^{\circ}+\cos 25^{\circ} \cdot \cos 25^{\circ}\)

= \(\sin ^2 25^{\circ}+\cos ^2 25^{\circ}=1\)

\(\sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}\) = 1

Question 4. Choose the correct option. Justify your choice.

(1) 9 \(\sec ^2 A-9 \tan ^2 A\)=

  1. 1
  2. 9
  3. 8
  4. 0

(2) \((1+\tan \theta+\sec \theta)\)

  1. \((1+\cot \theta- cosec \theta)\)=
  2. 0
  3. 1
  4. 2

(3) \((\sec A+\tan A)(1-\sin A)=\)

  1. sec A
  2. sin A
  3. cosec A
  4. cos A

(4) \(\frac{1+\tan ^2 A}{1+\cot ^2 A}\)=

  1. \(\sec ^2 A\)
  2. -1
  3. \(\cot ^2 A\)
  4. \(\tan ^2 A\)

Solution :

(1) Answer. (2)

⇒ \(9 \sec ^2 A-9 \tan ^2 A=9\left(\sec ^2 A-\tan ^2 A\right)\)

=9 \(\times \)1=9

9 \(\sec ^2 A-9 \tan ^2 A\)= 9

(2) Answer. (3)

⇒ \((1+\tan \theta+\sec \theta)(1+\cot \theta-{cosec} \theta)\)

=\(1+\cot \theta-{cosec} \theta+\tan \theta+\tan \theta \cdot \cot \theta\)

⇒ –\(\tan \theta \cdot {cosec} \theta+\sec \theta+\sec \theta \cdot \cot \theta-\sec \theta \cdot{cosec} \theta\)

= \(1+\cot \theta-{cosec} \theta+\tan \theta +\frac{\sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}-\frac{\sin \theta}{\cos \theta} \cdot \frac{1}{\sin \theta}+\sec \theta\)

⇒ \(+\frac{1}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta}-\sec \theta {cosec} \theta\)

=\(1+(\cot \theta+\tan \theta)-{cosec} \theta\)

⇒ \(+1-\sec \theta+\sec \theta+ cosec \theta\)

⇒ \(-\sec \theta cose \theta\)

= \(2+\left(\frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}\right)-\sec \theta {cosec} \theta\)

= \(2+\frac{\cos { }^2 \theta+\sin ^2 \theta}{\sin \theta \cdot \cos \theta}-\sec \theta{cosec} \theta\)

= \(2+\frac{1}{\sin \theta \cos \theta}-\frac{1}{\cos \theta \sin \theta}=2\) .

\((1+\tan \theta+\sec \theta)\) =1

(3) Answer. (4)

⇒ \((\sec A+\tan A)(1-\sin A)\)

= \(\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)\)

= \(\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)\)

= \(\frac{1-\sin ^2 A}{\cos A}=\frac{\cos ^2 A}{\cos A}=\cos A \)

\((\sec A+\tan A)(1-\sin A)=\) =cos A

(4) Answer. (4)

⇒ \(\frac{1+\tan ^2 A}{1+\cot ^2 A} =\frac{1+\tan ^2 A}{1+\frac{1}{\tan ^2 A}}=\frac{1+\tan ^2 A}{\frac{\tan ^2 A+1}{\tan ^2 A}}\)

=\(\tan ^2 A\)

\(\frac{1+\tan ^2 A}{1+\cot ^2 A}\)= \(\tan ^2 A\)

Trigonometry Multiple-Choice Questions

Question 1. The value of \(\sin 45^{\circ}+\cos 45^{\circ}\) is :

  1. 2
  2. \(\sqrt{2}\)
  3. \(\frac{\sqrt{3}}{2}\)
  4. 1

Answer: 2. \(\sqrt{2}\)

The value of \(\sin 45^{\circ}+\cos 45^{\circ}\) is \(\sqrt{2}\)

Question 2. If \(\sin A=\frac{3}{2}\) then the value of tan A is:

  1. \(\frac{5}{3}\)
  2. \(\frac{4}{3}\)
  3. \(\frac{3}{4}\)
  4. \(\frac{3}{5}\)

Answer: 3. \(\frac{3}{4}\)

The value of tan A is \(\frac{3}{4}\)

Question 3. If \(\sin A+\sin ^2 A=1\) then the value of \(\left(\cos ^2 A+\cos ^4 A\right)\) is :

  1. 1
  2. \(\frac{1}{2}\)
  3. 2
  4. \(\frac{1}{3}\)

Answer: 1. 1

The value of \(\left(\cos ^2 A+\cos ^4 A\right)\) is 1.

Question 4. If \(\cos 8 \alpha=\sin \alpha\) and \(8 \alpha<90^{\circ}\) then the value of \(\tan 3 \alpha\) is :

  1. 0
  2. 1
  3. \(\sqrt{3}\)
  4. \(\frac{1}{\sqrt{3}}\)

Answer: 4. \(\frac{1}{\sqrt{3}}\)

The value of \(\tan 3 \alpha\) is  \(\frac{1}{\sqrt{3}}\)

Question 5. In \(\triangle A B C\), \(\angle C=90^{\circ}\). The value of \(\cos (A+B)\) is :

  1. 1
  2. 0
  3. -1
  4. \(\frac{1}{2}\)

Answer: 2.  0

The value of \(\cos (A+B)\) is 0

Question 6. If \(4 \tan \theta=3\) then the value of \(\frac{4 \sin \theta-3 \cos \theta}{\sin \theta+\cos \theta}\) is :

  1. \(-\frac{16}{25}\)
  2. \(\frac{16}{25}\)
  3. 0
  4. 4

Answer: 3.  0

Question 7. The value of \(\sin \left(60^{\circ}+\theta\right)-\cot \left(30^{\circ}-\theta\right)\) is :

  1. 0
  2. 2 \(\tan \theta\)
  3. \(2 \cot \theta\)
  4. \(2 \sqrt{3}\)

Answer: 1.  0

Question 8. If \(\cos \theta=\sin \theta, 0 \leq \theta<90^{\circ}\), then angle \(\theta\) is equal to :

  1. \(0^{\circ}\)
  2. \(30^{\circ}\)
  3. \(45^{\circ}\)
  4. \(60^{\circ}\)

Answer: 3. \(45^{\circ}\)

\(\theta\) is equal to \(45^{\circ}\)

Question 9. If \(\sin A=\frac{3}{5}\) then the value of \(\cos A\) is:

  1. \(\frac{5}{4}\)
  2. \(\frac{4}{5}\)
  3. \(\frac{5}{3}\)
  4. \(\frac{4}{3}\)

Answer: 2. \(\frac{4}{5}\)

The value of \(\cos A\) is \(\frac{4}{5}\)

Question 10. If \(\sec \theta=2\) then the value of \(\theta\) is:

  1. \(30^{\circ}\)
  2. \(45^{\circ}\)
  3. \(60^{\circ}\)
  4. \(90^{\circ}\)

Answer: 3. \(60^{\circ}\)

The value of \(\theta\) is \(60^{\circ}\)

Question 11. If \(\cos ^2 \theta=\frac{1}{2}\) then the value of \(\sin ^2 \theta\) is:

  1. \(\frac{1}{4}\)
  2. \(\frac{\sqrt{3}}{2}\)
  3. \(\frac{1}{\sqrt{2}}\)
  4. \(\frac{1}{2}\)

Answer: 4. \(\frac{1}{2}\)

The value of \(\sin ^2 \theta\) is \(\frac{1}{2}\)

Question 12. If \(\tan \theta=\frac{2 a b}{a^2-b^2}\) then the value of \(\cos \theta\) is:

  1. 1
  2. \(\frac{a^2-b^2}{a^2+b^2}\)
  3. \(\frac{a^2+b^2}{a^2-b^2}\)
  4. \(\frac{2 a b}{a^2+b^2}\)

Answer: 2.  \(\frac{a^2-b^2}{a^2+b^2}\)

The value of \(\cos \theta\) is \(\frac{a^2-b^2}{a^2+b^2}\)

Question 13. If cosec A= A, \(0^{\circ} \leq A \leq 90^{\circ}\) then \(\angle A\) is equal to :

  1. \(120^{\circ}\)
  2. \(60^{\circ}\)
  3. \(45^{\circ}\)
  4. \(30^{\circ}\)

Answer: 3. \(45^{\circ}\)

Question 14. If \(3 \cot A=4\) then the value of \(\sec A\) is :

  1. \(\frac{3}{4}\)
  2. \(\frac{5}{4}\)
  3. \(\frac{2}{3}\)
  4. \(\frac{3}{2}\)

Answer: 2. \(\frac{5}{4}\)

The value of \(\sec A\) is \(\frac{5}{4}\)