Class 10 Maths

NCERT Exemplar Solutions For Class 10 Maths Chapter 6  Triangles

Introduction:

• To see the above pairs of figures, what do you think?
• Figures in a pair are alike, i.e., their shapes are the same.
• Shapes are the same but do not mean that they are necessarily equal in size.
• Shapes the same mean that their curvature is the same, i.e., figures are proportionately (not necessarily exactly) the same insides as well as angles.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6  Difference Between Proportional And Equal

1. In Terms of Sides :

⇒ \(\frac{A B}{P Q}=\frac{2 x}{2 y}=\frac{x}{y}\)

⇒ \(\frac{B C}{Q R}=\frac{3 x}{3 y}=\frac{x}{y}\)

⇒ \(\frac{A C}{P R}=\frac{4 x}{4 y}=\frac{x}{y}\)

⇒  \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\), i.e, corresponding angles are proportional.

Read and Learn More Class 10 Maths Solutions Exemplar

Suppose, the value of x = 2 sandy = 3.5 then the sides of the first triangle are 2 x 2, 3 x 2, 4 x 2, i.e., 4 cm, 6 cm, and 8 cm, and the sides of the second triangle are 2 x 3.5, 3 x 3.5, 4 x 3.5, i.e., 7 cm, 10.5 cm, and 14 cm. So, we can say that the corresponding sides are proportional, but that does not mean that their corresponding sides are equal.

2. In terms of Angles :

Here,\(\frac{\angle A}{\angle P}=\frac{2 x}{2 y}=\frac{x}{y}\)

⇒ \(\frac{\angle B}{\angle Q}=\frac{3 x}{3 y}=\frac{x}{y}\)

⇒ \(\frac{\angle C}{\angle R}=\frac{4 x}{4 y}=\frac{x}{y}\)

⇒ \(\frac{\angle A}{\angle P}=\frac{\angle B}{\angle Q}=\frac{\angle C}{\angle R}\)

i.e., corresponding angles are proportional.

Now, see whether x and y are equal or they will be different.

We know that,

⇒ \(\angle A+\angle B+\angle C =180^{\circ}\)

⇒ \(2 r+3 x+4 r =180^{\circ}\)

⇒ \(9 x =180^{\circ} \Rightarrow x=20^{\circ}\)

⇒ \(\angle A=2 \times 20^{\circ}=40^{\circ}, \angle B =3 \times 20^{\circ}=60^{\circ}, \angle C=4 \times 20^{\circ}=80^{\circ}\)

⇒ \(\angle P+\angle Q+\angle R =180^{\circ}\)

⇒ \(2 y+3 y+4 y =180^{\circ}\)

⇒ \(9 y =180^{\circ} \Rightarrow y=20^{\circ}\)

P=2 \(\times 20^{\circ}=40^{\circ}, \angle Q =3 \times 20^{\circ}=60^{\circ}, \angle R=4 \times 20^{\circ}=80^{\circ}\)

⇒ \(\angle A=\angle P, \angle B=\angle Q\) and \(\angle C=\angle R\) .

From (1) and (2) we conclude that:

• Corresponding sides are proportional does not mean that their corresponding sides are equal.
• Corresponding angles are proportional means that their corresponding angles are necessarily equal.
• So, corresponding parts are proportional means
• Two triangles are similar if their shapes are the same.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6  Definition Of Similar Triangles

Two triangles are similar if their corresponding parts are proportional.

1. their corresponding sides are proportional.
2.  their corresponding angles are equal.

From the above figures, two triangles ABC and PQR are Bi similar, i.e.\(\triangle\)ABC ~ \(\triangle\)PQR, if any one of the following properties occurs :

(1) Their corresponding sides are proportional (or the ratio of their corresponding sides are equal)

⇒ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\) (SSS similarity)

(2) Their corresponding angles are equal

⇒ \(\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R\) (AAA similarity)

If any two corresponding angles are equal

⇒ \(\left.\begin{array}{ll}\text { i.e., } & \angle A=\angle P \text { and } \angle B=\angle Q \\ \text { or } & \angle A=\angle P \text { and } \angle C=\angle R \\ \text { or } & \angle B=\angle Q \text { and } \angle C=\angle R\end{array}\right\}\) (AA similarity)

⇒ Any two corresponding sides are proportional and their included angles are the same i.e., or

⇒ \(\left.\begin{array}{l}
\frac{A B}{P Q}=\frac{B C}{Q R} \text { and } \angle B=\angle Q \\
\frac{A B}{P Q}=\frac{A C}{P R} \text { and } \angle A=\angle P \\
\frac{B C}{Q R}=\frac{A C}{P R} \text { and } \angle C=\angle R
\end{array}\right\}\)

The above three postulates are the postulates of the similarity of triangles

Remark:

1. In an obtuse-angled triangle,

⇒  \(ar(\triangle A B C) =\frac{1}{2} \times \text { Base } \times \text { Corresponding altitude }\)[/latex]

= \(\frac{1}{2} \times B C \times A D\)

In an obtuse triangle, the perpendicular will be inside the triangle.

2. In an obtuse-angled triangle,

⇒ \(ar(\triangle A B C) =\frac{1}{2} \times \text { Base } \times \text { Corresponding altitude }\)

= \(\frac{1}{2} \times B C \times A D\)

In an obtuse triangle, the perpendicular will be outside the triangle.

Note: In \(\triangle\)ABC, when the base is taken as BC, then perpendicular will be drawn from the angle opposite to the base, i.e. \(\angle\)A on the base BC (whether it lies inside the BC or outside the BC).

NCERT Exemplar Solutions For Class 10 Maths Chapter 6  Difference Between Congruency And Similarity

• Two things are said to be congruent if they have the same shape as well as the same size, i.e., the two figures coincide with each other.
• Two things are said to be similar to each other if they have the same shapes but their sizes may be different.

Remark:

1. Congruent triangles are necessarily similar triangles but the converse is not always true.
2. If two triangles are similar to the third triangle, then they are similar to each other.

Basic Proportionality Theorem Or Thale’s Theorem

Theorem 1: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then it divides the other two sides in the same ratio.

Given: \(\triangle\)ABC in which DE || BC and DE intersects AB and AC at D and E respectively.

To Prove: \(\frac{A D}{D B}=\frac{A E}{E C}\)

Construction: Join BE and CD. Draw EL \(\perp AB\) and \(DM \perp AC\)

Proof:

Area of \((\triangle A D E) =\frac{1}{2} \times A D \times E L\)

Area of \((\triangle D B E) =\frac{1}{2} \times D B \times E L \)

⇒ \(\frac{{ar}(\triangle A D E)}{{ar}(\triangle D B E)} =\frac{\frac{1}{2} \times A D \times E L}{\frac{1}{2} \times D B \times E L}=\frac{A D}{D B}\)

Again Area of \(\triangle M D E =\frac{1}{2} \times A E \times D M\)

Area of \(\triangle E C D =\frac{1}{2} \times E C \times D M\)

⇒\(\frac{{ar}(\triangle A D E)}{{ar}(\triangle E C D)} =\frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M}=\frac{A E}{E C}\)

But since \(\triangle\)DBE and \(\triangle\)ECD are on the same base DE and between the same parallels DE and BC, we have

⇒ \({ar}(\triangle D B E)={ar}(\triangle E C D)\)

From (1), (2), and (3), we get \(\frac{A D}{D B}=\frac{A E}{E C}\)

Corollary: In a \(\triangle\)ABC, a line DE || BC intersects AB in D and AC in E, then prove that:

⇒ \(\frac{A B}{D B}=\frac{A C}{E C}\)

⇒ \(\frac{A D}{A B}=\frac{A E}{A C}\)

Proof: (1) From basic proportionality theorem

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}\)

Adding 1 on both sides, we have

⇒ \(\frac{A D}{D B}+1 =\frac{A E}{E C}+1\)

⇒ \(\frac{A D+D B}{D B} =\frac{A E+E C}{E C} \Rightarrow \quad \frac{A B}{D B}=\frac{A C}{E C}\)

(2) From the basic proportionality theorem, we have

⇒ \(\frac{A D}{D B}=\frac{A E}{E C} \quad \Rightarrow \quad \frac{D B}{A D}=\frac{E C}{A E}\)

Adding 1 on both sides, we have

⇒ \(\frac{D B}{A D}+1 =\frac{E C}{A E}+1\)

⇒ \(\frac{D B+A D}{A D}=\frac{E C+A E}{A E}\)

⇒ \(\frac{A B}{A D}=\frac{A C}{A E} \quad \Rightarrow \quad \frac{A D}{A B}=\frac{A E}{A C}\)

Now, we can conclude

In \(\triangle A B C\) if D E || BC

1.  \(\quad \frac{A D}{D B}=\frac{A E}{E C}\) (B.P.T.)
2. \(\frac{A B}{D B}=\frac{A C}{E C}\)
3.  \(\frac{A D}{A B}=\frac{A E}{A C}\)

These Are the Same operations in R.H.S, as in the L.H.S.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6  Converse of Basic Proportionality Theorem

Theorem 2: If a line divides any two sides of a triangle in the same ratio then the line must be parallel to the third side.

Given: \(\triangle\)ABC in which a line l intersects AB at D and AC at E, such that 12

To Prove: DE \\BC

Proof: If possible, let DE is not parallel to BC, then there must be another line through D, which is parallel to BC. Let DE || BC.

Then by the basic proportionality theorem, we have

⇒ \(\frac{A D}{D B} =\frac{A F}{F C}\)

But \(\frac{A D}{D B}=\frac{A E}{E C}\)

From (1) and (2) we have \(\frac{A F}{F C}=\frac{A E}{E C}\)

Now, adding 1 on both sides

⇒ \(\frac{A F}{F C}+1=\frac{A E}{E C}+1 \quad \Rightarrow \quad \frac{A F+F C}{F C}=\frac{A E+E C}{E C}\)

⇒ \(\frac{A C}{F C}=\frac{A C}{E C} \quad \Rightarrow \quad F C=E C\)

This is possible only, when E and F coincide.

Hence D E || B C (D F || BC, by construction)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Solved Examples

Example 1. In the adjoining figure DE || BC and D divide AB in the ratio 2 : 3. Find.

1. \(\frac{A E}{E C}\)
2. \(\frac{A E}{A C}\)

Solution:

Given

In the adjoining figure DE || BC and D divide AB in the ratio 2 : 3.

(1) Since D E || BC, hence by Thale’s theorem

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}\)

Also, \(\frac{A D}{D B}=\frac{2}{3}\) (given)

Therefore, from (1) and (2), \(\frac{A D}{D B}=\frac{A E}{E C}=\frac{2}{3}\)

(2) \(\frac{AE}{AC}\)=\(\frac{A E}{A E+E C}\)

= \(\frac{A E / E C}{A E / E C+E C / E C}\)

= \(\frac{\frac{2}{3}}{\frac{2}{3}+1}=\frac{\frac{2}{3}}{\frac{5}{3}}=\frac{2}{5}\)

Example 2. In the figure, PQ is parallel to MN, if \(\frac{K P}{P M}=\frac{4}{13}\) and KN = 20.4. Find KQ

Solution:

Given In \(\triangle\)KMN, PQ || MN

⇒ \(\frac{K P}{P M}=\frac{K Q}{Q N}\) (by B.P.T. or Thale’s theorem)

Let K Q=x, then Q N=20.4-x

Now, from (1) \(\frac{4}{13} =\frac{x}{20.4-x}\)

81.6-4 x =13 x

17 x =81.6

x =4.8 cm

Therefore K Q=4.8 cm

Example 3. In the figure, DE || BC. If DB = 10.8 cm, AE = 2.7 cm and£C =8.1 cm, find AD. DE || BC

Solution:

Given

In the figure, DE || BC. If DB = 10.8 cm, AE = 2.7 cm and£C =8.1 cm

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}\)

⇒ \(\frac{A D}{10.8}=\frac{2.7}{8.1}\)

AD=\(\frac{1}{3} \times 10.8 \mathrm{~cm}=3.6 \mathrm{~cm}\)

Example 4. In the given figure, in \(\triangle\)ABC, DE || BC so that AD = (4x – 3) cm, AE = (8x – 7) cm. BD = (3x – 1) cm and CE = (5x – 3) cm. Find the value of x.

Solution:

Given

In the given figure, in \(\triangle\)ABC, DE || BC so that AD = (4x – 3) cm, AE = (8x – 7) cm. BD = (3x – 1) cm and CE = (5x – 3) cm.

⇒ \(\triangle\)ABC, DE || BC

⇒ \(\frac{A D}{B D}=\frac{A E}{C E}\)

⇒ \(\frac{4 x-3}{3 x-1}=\frac{8 x-7}{5 x-3}\)

(4 x-3)(5 x-3)=(8 x-7)(3 x-1)

20 x²-27 x+9=24 x²-29 x+7

4 x²-2 x-2=0

2 x²-x-1=0

2 x²-2 x+x-1=0

2 x(x-1)+1(x-1)=0

(2 x+1)(x-1)=0

x=1 or x=-\(\frac{1}{2}\) (by Thale’s theorem)

But when x=-\(\frac{1}{2}\)

A D=\(\left[4 \times\left(-\frac{1}{2}\right)-3\right]\)=-5

Since distance cannot be negative so x \(\neq-\frac{1}{2}\)

Here X = 1

The value of x = 1

Example 5. The bisector (internal or external) of an angle of a triangle divides the opposite side (internally or externally) in the ratio of the sides containing the angle.

Solution:

Given: \(\triangle\)ABC in which the bisector AD of \(\angle\)A meets BC (or BC produced) in D.

To Prove : \(\frac{B D}{D C}=\frac{A B}{A C}\)

Construction: Draw CE || DA meeting BA (produced if necessary) in E.

Proof: Since CE || DA

⇒ \(\angle 1=\angle 2\) (alternate angles)

and \(\angle 3 =\angle 4\) (corresponding angles) (2)

But \(\angle 1=\angle 3\) (given)…(3)

⇒ \(\angle 2=\angle 4\) [from (1), (2) and (3)]

An E=A C (sides opposite to equal angles are equal) …(4)

Now, since CE || DA (construction)

⇒ \(\frac{B D}{D C}=\frac{B A}{A E}\)

⇒ \(\frac{B D}{D C}=\frac{B A}{A C}\) [from (4) and (5)]

Example 6. Prove that any line parallel to parallel sides of a trapezium divides the non-parallel sides proportionally.

Solution:

Given: A trapezium ABCD in which AB || DC.

Also a line EF || AB || DC.

To Prove: \(\frac{A E}{E D}=\frac{B F}{F C}\)

Construction: Join DB which intersects E F at G,

Proof: In \(\triangle A D B\), since E G || A B. ( E F || A B)

⇒ \(\frac{A E}{E D}=\frac{B G}{G D}\) (by B.P. theorem)

Also, in \(\triangle B D C, since G F || DC (E F || D C)\)

⇒ \(\frac{B F}{F C}=\frac{B G}{G D}\) (by B.P. theorem)

From (1) and (2), we have

⇒ \(\frac{A E}{E D}=\frac{B F}{F C}\)

Hence Proved.

Example 7. The side BC of a triangle ABC is bisected at D. O is any point in Al). BO and CO produced meet AC and AB at E and F respectively and AD is produced to X so that D is the mid-point of OX. Prove that AO: AX = AF: AB and show that FE || BC.

Solution:

Given

The side BC of a triangle ABC is bisected at D. O is any point in Al). BO and CO produced meet AC and AB at E and F respectively and AD is produced to X so that D is the mid-point of OX.

Join BX and CX.

Since D is the mid-point of BC (given)

BD = DC …(1)

Also, D is the mid-point of OX (given)

OD = DX …(2)

From (1) and (2), we have

⇒ [Square OBXC is a parallelogram (diagonals bisect each other) B

OC || BX (opposite sides of a parallelogram)

FC || BX ⇒ OE || XC

In \(\triangle A X C\),

⇒ \(\frac{A O}{O X}=\frac{A E}{E C}\) (by B.P. theorem)

From (3) and (4), we have

⇒ \(\frac{A F}{F B}=\frac{A E}{E C}\)

In \(\triangle A B C\), F E || B C. (converse of B.P. theorem)

Hence Proved.

Example 8. In \(\triangle\)ABC, D, and E are two points on AB such that AD = BE. If DP || BC and EQ || AC, prove that PQ || AB.

Solution:

Given: \(\triangle\)ABC in which D and E are two points such that AD = BE.

Also DP || BC and EQ || AC.

To Prove: PQ || AB

Proof: In ΔABC, Since DP || BC

⇒ \(\frac{A D}{D B}=\frac{A P}{P C}\) (by B.P. theorem) …. 1

In \(\triangle A B C\), since E Q || A C

⇒ \(\frac{B E}{E A}=\frac{B Q}{Q C}\) (by B.P. theorem) …. 2

Now, as A D = B E (given) …. 3

A D+D E=B E+D E (adding DE on both sides)

AE = BD ….. 4

From (1), (2), (3) and (4),

⇒ \(\frac{A P}{P C}=\frac{B Q}{Q C}\)

P Q || A B (by the converse of B.P. theroem)

Hence Proved.

Criteria For Similarity Of Two Triangles

• Two triangles are said to be similar if their corresponding angles are equal and corresponding sides are proportional (i.e., the ratios between the lengths of corresponding sides are equal).
• For example if in \(\triangle\)ABC and \(\triangle\)PQR
•  \(\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R\)
• and, \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
• then \(\triangle\)ABC is similar to \(\triangle\)PQR.
• Symbol ~ is used for “is similar to”

Conversely: If \(\triangle\)ABC is similar to \(\triangle\)PQR then

⇒ \(\angle A=\angle P, \angle B=\angle Q, \angle C=\angle R\)

⇒ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)

Theorem 1 (AAA Similarity): If in two triangles the corresponding angles are equal, then their corresponding sides are proportional and hence the triangles are similar.

Given: \(\triangle\)ABC and \(\triangle\)DEF such that

⇒ \(\angle A=\angle D, \angle B=\angle E \text {, and } \angle C=\angle F\)

To Prove: \(\triangle\)ABC ~ \(\triangle\)DEF

Construction: Cut DP =AB and DQ = AC. Join PQ.

Proof: In \(\triangle\)ABC and \(\triangle\)DPQ, we have

A B =D P

A C =D Q (by construction)

⇒ \(\angle A =\angle D\)

⇒ \(\triangle A B C \cong \triangle D P Q\) (by SAS congruence)

⇒ \(\angle B =\angle P\)

But it is given that \(\angle B=\angle E\)

Therefore, \(\angle E=\angle P\)

P Q || E F

⇒ \(\frac{D P}{D E}=\frac{D Q}{D F}\)

⇒ \(\frac{A B}{D E}=\frac{C A}{F D}\)

Similarly \(\frac{A B}{D E}=\frac{B C}{E F}\)

⇒ \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{C A}{F D}\)

Thus \(\angle A=\angle D, \angle B=\angle E, \angle C=\angle F\) and

⇒ \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{C A}{F D}\)

Hence, \(\triangle A B C \sim \triangle D E F\).

Corollary (AA Similarity): If two angles of a triangle are respectively equal to two angles of another triangle then the two triangles are similar.

Proof: In \(\triangle A B C\) and \(\triangle D E F\) , let \(\angle A=\angle D\) and \(\angle B=\angle E\) , then 3rd \(\angle C= 3rd \angle F\)

Thus, the two triangles are equal.

Hence, the two triangles are similar.

Thus, we can see that two triangles can be proved similar if two angles of 1st triangle are equal to two angles of the other triangle respectively.

Theorem 2 (SSS Similarity): If the corresponding sides of two triangles are proportional then their corresponding angles are equal, and hence the two triangles are similar.

Given: \(\triangle\)ABC and \(\triangle\)DFF in which

To Prove: \(\triangle A B C \sim \triangle D E F\)

Construction: Let us take \(\triangle\) A B C and \(\triangle\) D E F such that

⇒ \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}\)

Cut D P=A B and D Q=A C. Join PQ

Proof: \(\frac{A B}{D E}=\frac{A C}{D F}\)

⇒ \(\frac{D P}{D E}=\frac{D Q}{D F}\)

Thus, the given triangles are equiangular and hence similar.

Theorem 3 (SAS Similarity): If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional then the two triangles are similar.

Given: \(\triangle\)ABC and \(\triangle\)DPP in which

⇒ \(\angle A=\angle D and \frac{A B}{D E}=\frac{A C}{D F}\)

To Prove: \(\triangle \triangle B C \sim \triangle D E F\)

Construction: Let us take \(\triangle A B C\) and \(\triangle D E F\) such that

⇒ \(\frac{A B}{D E}=\frac{A C}{D F}\) and \(\angle A=\angle D\)

Cut D P=A B and D Q=A C. Join P Q.

Proof: In \(\triangle A B C\) and \(\triangle D P Q\)

A B=D P (by construction)

⇒ \(\angle A=\angle D\) (given)

A C=D Q (by construction)

⇒ \(\triangle A B C \cong \triangle D P Q\) (by SAS axiom)

⇒ \(\angle A=\angle D, \angle B=\angle P and \angle C=\angle Q \)

Now \(\frac{A B}{D E}=\frac{A C}{D F}\) (given)

⇒ \(\frac{D P}{D E}=\frac{D Q}{D F}\) (AB = DP and AC = DQ)

P Q || E F (by the converse of Thale’s theorem)

⇒ \(\angle P=\angle E and \angle Q=\angle F )\)

⇒ \(\angle P=\angle E\) and \(\angle Q=\angle F \)(corresponding angles)

⇒ \(\angle A=\angle D, \angle B=\angle P=\angle E \angle C=\angle Q=\angle F\)

Thus, \(\angle A=\angle D, \angle B=\angle E and \angle C=\angle F\)

So, the given triangles are equiangular and hence similar.

Some More Results

Result 1: If two triangles are equiangular, prove that the ratio of their corresponding sides is the same as the ratio of their corresponding altitudes.

Given : \(\triangle M B C\) and \(\triangle D E F \)in which \(\angle A=\angle D\), \(\angle B=\angle E\) and \(\angle C=\angle F\) and \(A L \perp B C\) and \(D M \perp E F\) :

To Prove : \(\frac{B C}{E F}=\frac{A L}{D M}\)

Proof : Since \(\triangle A B C\) and \(\triangle D E F\) are equiangular, therefore, \(\triangle A B C \sim \triangle D E F\)

⇒  \(\frac{A B}{D E}=\frac{B C}{E F}\)

In \(\triangle A L B\) and \(\triangle D M E\), we have

⇒ \(\angle A L B=\angle D M E=90^{\circ}\) and \(\angle B=\angle E\)

⇒ \(\triangle A L B \sim \triangle D M E\)

⇒ \(\frac{A B}{D E}=\frac{A L}{D M}\)

From (1), and (2) we get

⇒ \(\frac{B C}{E F}=\frac{A L}{D M}\)

Result 2: If two triangles are equal, prove that the ratio of their corresponding sides is the same as the ratio of the corresponding medians.

Given : \(\triangle\)ABC and \(\triangle\)DEF in which \(\angle\)A = \(\angle\)D, \(\angle\)B = \(\angle\)E and \(\angle\)C = \(\angle\)F and \(\angle\)L and \(\angle\)M are medians.

To Prove : \(\frac{B C}{E F}=\frac{A L}{D M}\)

Proof : Since \(\triangle A B C\) and \(\triangle D E F\) are equiangular, we have \(\triangle A B C \sim \triangle D E F\)

⇒ \(\frac{A B}{D E}=\frac{B C}{E F}\)

⇒ \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{2 B L}{2 E M}=\frac{B L}{E M}\)

But, \(\frac{A B}{D E} =\frac{B L}{E M}\) and \(\angle B=\angle E\)

Now, in \(\triangle A B L\) and \(\triangle D E M\), we have

⇒ \(\frac{A B}{D E}=\frac{B L}{E M}\) and \(\angle B=\angle E\) (given)

⇒ \(\triangle A B L \sim \triangle D E M\) (by SAS similarity)

⇒ \(\frac{A B}{D E} =\frac{A L}{D M}\) … 2

From 1 And 2, we get

∴ \(\frac{B C}{E F}=\frac{A L}{D M}\)

Result 3: If two triangles are equiangular, show that the ratio of the corresponding sides is the same as the ratio of the corresponding angle bisector segments.

Given : \(\triangle\)ABC and \(\triangle\)DEF in which \(\angle\)A = \(\angle\)D, \(\angle\)B = \(\angle\)E and A

⇒ \(\angle\)C = \(\angle\)F and AX and DY are the bisectors of \(\angle\)A and \(\angle\)D
respectively.

To Prove: \(\frac{B C}{E F}=\frac{A X}{D Y}\)

Proof : Since \(\triangle\)ABC and \(\triangle\)DEF are equiangular, we have \(\triangle\)ABC ~ \(\triangle\)DEF

⇒ \(\frac{A B}{D E}=\frac{B C}{E F}\) …. 1

Now, \(\angle A=\angle D \Rightarrow \frac{1}{2} \angle A=\frac{1}{2} \angle D\)

⇒ \(\angle B A X=\angle E D Y\)

Thus in \(\triangle A B X\) and \(\triangle D E Y\), we have

⇒ \(\angle B A X =\angle E D Y\) (proved)

⇒ \(\angle B =\angle E\) (given)

⇒ \(\triangle A B X \sim \triangle D E Y\) (by AA similarity)

⇒ \(\frac{A B}{D E}=\frac{A X}{D Y}\) ….. 2

From (1) and (2), we get

⇒ \(\frac{B C}{E F}=\frac{A X}{D Y}\)

Remark:

1. If two triangles ABC and DEF are similar, then
2. \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}\)
3. \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{A B+B C+A C}{D E+E F+D F}\) (using ratio and proportion)
4. \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}=\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle D E F}\)
5. Thus, if two triangles are similar, then their corresponding sides are proportional and they are proportional to the corresponding perimeters.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Traingle Importance Of Correct Namings

Then which of the following is correct?

•  \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\).
• \(\frac{A B}{P R}=\frac{B C}{P Q}=\frac{A C}{Q R}\).
• \(\frac{A B}{Q R}=\frac{B C}{P Q}=\frac{A C}{P R}\)

You will say option (a) is correct, but your answer is wrong.

Until we do not know which part of one triangle is equal to which part of another triangle, how can we give the correct answer? We should give the name of the other triangle according to the first triangle. We must see the corresponding parts.

For Example,

Here , \(\angle A=\angle Q\)

⇒ \(\angle C=\angle P\)

⇒ \(\angle B=\angle R\)

Since two angles of \(\triangle\)ABC, are equal to two angles of \(\triangle\)PQR, therefore two triangles are similar to each other. For this, write the name of the first triangle say \(\triangle\)CAB. Now we have to write the name of the other triangle correspondingly, (according to \(\triangle\)CAB)

As C comes in 1st place and C = P, So  P will come in 1st place also

As A comes in 2nd place and A = Q, So Q will come in 2nd place

As the remaining place of first, A is for B so the remaining place of second A will be for R.

Corresponding sides are proportional.

⇒ \(\frac{\text { first two of } \triangle C A B}{\text { first two of } \Delta P Q R} =\frac{\text { last two of } \triangle C A B}{\text { last two of } \triangle P Q R}=\frac{\text { first and last of } \triangle C A B}{\text { first and last of } \triangle P Q R}\)

⇒ \(\frac{C A}{P Q} =\frac{A B}{Q R}=\frac{C B}{P R}\) (no need to see the figures, you can confuse there) …(1)

Also, you can write

1st angle of first \(\Delta\)=I angle of other \(\Delta\) i.e., \(\angle C=\angle P\)

2nd angle of first \(\Delta\)= 2nd angle of other \(\Delta\) i.e., \(\angle A=\angle Q \)

and 3rd angle of first \(\Delta\)= 3rd angle of other \(\Delta\) i.e. \(\angle B=\angle R\)

If we write the name of first \(\triangle\) as BCA then the name of other \(\triangle\) will be as follows (As \(\angle\)A = \(\angle\)Q => if A comes in 3rd place, so Q will also come in 3rd place and as \(\angle\)C = \(\angle\)P if C comes in 2nd places, so P will also come in 2nd place, etc.)

So, now for \(\triangle\)BCA.

(2) Then put

(3) Then finally (remaining place)

Hence, \(\triangle B C A-\triangle R P Q\)

The corresponding sides are proportional

⇒ \(\frac{\text { first two of } \triangle B C A}{\text { first two of } \triangle R P Q} =\frac{\text { last two of } \triangle B C A}{\text { last two of } \triangle R P Q}\)

= \(\frac{\text { first and last of } \triangle B C A}{\text { first and last of } \triangle R P Q}\)

B C =\(\frac{C A}{P Q}=\frac{B A}{R Q}\)

This is also the same as in (1)

Now, we shall see one example based on the above discussion

Example : D is a point on the side BC of a \(\triangle\)ABC, such that \(\angle\)ADC = \(\angle\)BAC.

Prove that: \(\frac{C A}{C D}=\frac{C B}{C A}\)

Solution:

Don’t bother to write the correct namings of two triangles. We know that as \(\angle\)1 = \(\angle\)2 (given)

So we can take the As in which \(\angle\)1 and \(\angle\)2 occur i.e., \(tri\angle\)ADC and \(tri\angle\)ABC

Now, in \(tri\angle\)s ADC and ABC

⇒ \(\angle\)1 = \(\angle\)2

Now From We Can Write,

⇒ \(\triangle \mathrm{A} D \mathrm{~A} \sim \triangle B A C\)

Now, from this, we can write

⇒ \(\frac{\text { first two of } \triangle A D C}{\text { first two of } \triangle B A C}=\frac{\text { last two of } \triangle A D C}{\text { last two of } \triangle B A C}\)

=\(\frac{\text { first and last of } \triangle A D C}{\text { first and last of } \triangle B A C}\)

i.e., \(\frac{A D}{B A}=\underbrace{\frac{D C}{A C}=\frac{A C}{B C}}\)

or \(\frac{C A}{C D}=\frac{C B}{C A}\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Traingle Importance Of Correct Namings Solved Examples

Example 1. In the given figure \(\triangle\)ACB ~ \(\triangle\)APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm, find C/1 and AQ.

Solution:

Given

In the given figure \(\triangle\)ACB ~ \(\triangle\)APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm, AP = 2.8 cm

Since, \(\triangle\)ACB ~ \(\triangle\)APQ (given)

⇒ \(\frac{A C}{A P}=\frac{C B}{P Q}=\frac{A B}{A Q}\)

(don’t see the figure) write \(\frac{\text { first two }}{\text { first two }}=\frac{\text { last two }}{\text { last two }}=\frac{\text { first and last }}{\text { first and last }}\)

⇒ \(\frac{A C}{2.8}=\frac{8}{4}=\frac{6.5}{A Q}\)

Consider, \(\frac{A C}{2.8}=\frac{8}{4}\)

4 A C=8 \(\times 2.8 \quad \Rightarrow \quad A C=5.6 \mathrm{~cm}\)

Consider, \(\frac{8}{4}=\frac{6.5}{A Q}\)

2 A Q=6.5 or A Q=3.25 cm

Example 2. The triangles shown in adjoining figures are similar. Find the values of a and b.

Solution:

Given

The triangles shown in adjoining figures are similar.

⇒ \(\angle A=\angle R, \angle B=\angle P\)and \(\angle C=\angle Q\)

⇒ \(\triangle A B C \sim \triangle R P Q\) (AAA similarity)

⇒ \(\frac{A B}{R P}\) … First Two

= \(\frac{B C}{P Q}\) ……. Last Two

= \(\frac{A C}{R Q}\) ….. First And Last

⇒ \(\frac{a}{8} =\frac{7.5}{6}=\frac{4}{b}\)

⇒ \(\frac{a}{8} =\frac{7.5}{6}\) and \(\frac{4}{b}=\frac{7.5}{6}\)

a =\(\frac{8 \times 7.5}{6}\) and \(b=\frac{4 \times 6}{7.5}\)

a =10 and b=3.2

Example 3. In the given figure if AD || BC, find the value of x.

Solution:

Given: AD || BC

Proof: In \(\triangle\)AOD and \(\triangle\)BOC

⇒ \(\angle 1=\angle 2\) (alternate angles)

⇒ \(\angle 3=\angle 3\) (vertically opposite angles)

i.e., \(\triangle A O D \sim \triangle C O B\) [AA corollary]

⇒ \(\frac{A O}{O C}=\frac{O D}{O B}\)

⇒ \(\frac{3}{x-3}=\frac{x-5}{3 x-19}\)

(x-3)(x-5)=3(3 x-19) \(\quad \Rightarrow \quad x^2-5 x-3 x+15=9 x-57\)

⇒ \(x^2-17 x+72=0 \quad \Rightarrow \quad x^2-9 x-8 x+72=0\)

x(x-9)-8(x-9)=0 or (x-9)(x-8)=0

x-9=0 or x=9

x-8=0 or x=8

Hence x=9,8

Example 4. Find ∠P in the figure below

Solution:

In \(\triangle A B C\) and \(\triangle Q R P\), we have

⇒ \(\frac{A B}{Q R}=\frac{3.6}{7.2}=\frac{1}{2}, \frac{B C}{R P}=\frac{6}{12}=\frac{1}{2}\)

and \(\frac{C A}{P Q}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}\)

Thus,\(\frac{A B}{Q R}=\frac{B C}{R P}=\frac{C A}{P Q}\)

Hence, by SSS criterion,

⇒ \(\triangle A B C \sim \triangle Q R \quad\left(\quad \frac{\text { first two }}{\text { first two }}=\frac{\text { last two }}{\text { last two }}=\frac{\text { first and last }}{\text { first and last }}\right)\)

⇒ \(\angle C =\angle P\) (third angle = third angle)

But \(\angle C=180-(\angle A+\angle B)=180^{\circ}-\left(70^{\circ}+60^{\circ}\right)=50^{\circ}\)

⇒ \(\angle P=50^{\circ}\)

Example 5. In the figure AC || BD, prove that:

1. \(\triangle A C E \sim \triangle B D E\)
2. \(\frac{A E}{C E}=\frac{B E}{D E}\)

Solution:

(1) A C || B D (given)

and, \(\left.\begin{array}{l}\angle 1=\angle 2 \\ \angle 3=\angle 4\end{array}\right\}\)

∴ \(\triangle \widehat{A C E \sim \triangle B D E}\) (by AA criterion)

(2) Now, since \(\triangle A C E \sim \triangle B D E\)

⇒ \(\frac{A E}{B E}=\frac{C E}{D E}\) (corresponding sides of similar \(\Delta\) s are proportional)

∴ \(\frac{A E}{C E}=\frac{B E}{D E}\) (by alternendo)

Example 6. \(\triangle\)ABC is an isosceles triangle with Ab = AC And D is a point on AC such that \(B C^2=A C \times C D\), Prove that BD = BC

Solution:

Given

\(\triangle\)ABC is an isosceles triangle with Ab = AC And D is a point on AC such that \(B C^2=A C \times C D\),

We have

⇒ \(B C^2 =A C \times C D\) and A B=A C

⇒ \(B C \times B C =A C \times C D\) and \(\angle B =\angle C\)

There may be 3 possibilities:

Possibility 1

⇒ \(\frac{B C}{A C}=\frac{C D}{B C}\)

and \(\angle C=\angle B\)

⇒ \(\triangle B \overline{D C} \sim \triangle A \overline{B C}\)

But in this case, \(\angle C \neq \angle B\)

So, the position will not suit.

Therefore, not possible.

Possibility 2.

⇒ \(\frac{B C}{C D}=\frac{A C}{B C}\)

⇒ \(\angle B=\angle C\)

⇒ \(\triangle \widehat{A B C} \sim \triangle \widehat{B D C}\)

But in this case, \(\angle B \neq \angle C\)

So, the position will not suit. Therefore, not possible.

Possibility 3.

⇒ \(\frac{B C}{A B} =\frac{C D}{B C}\)( A B=A C)

⇒ \(\angle C =\angle B \)

⇒ \(\triangle \overline{B C} D \sim \Delta \overline{A B} C\)

All positions are satisfied.

So, possible.

⇒ \(\triangle B C D \sim \triangle A B C\) (by SAS criterion)

⇒ \(\frac{B C}{A B} =\frac{C D}{B C}=\frac{B D}{A C}\)

⇒ \(\frac{B C}{A B} =\frac{B D}{A C}\)

BC =B D (A B=A C, given)

Hence Proved.

We could take directly the 3rd possibility and the solution seems to be much easier but our main aim is to make you understand why we take the 3rd one.

Example 7. In the given figure if \(\angle B=90^{\circ}\) and B D is perpendicular to A C then prove that :

1.  \(\triangle A D B \sim \triangle B D C\)
2. \(\triangle A D B \sim \triangle A B C\)
3.  \(\triangle B D C \sim \triangle A B C\)
4. \(B D^2=A D \times D C\)
5.  \(A B^2=A D \times A C\)
6. \(B C^2=C D \times A C\)
7.  \(A B^2+B C^2=A C^2\)

Solution:

(1) In \(\triangle A D B\) and \(\triangle B D C\),

⇒ \(\angle 5=\angle 4 (each 90^{\circ}\) )

Searching for Second Angle :

We know that \(\angle 1+\angle 2=90^{\circ}\)

Also, since \(\angle 4=90^{\circ}\)

⇒ \(\angle 2+\angle 3=90^{\circ}\) (angle sum property)

From (1) and (2), we get

⇒ \(\angle 1+\angle 2=\angle 2+\angle 3 \quad\left(\text { each } 90^{\circ}\right)\)

⇒ \(\angle 1=\angle 3\)

Now, in \(\triangle A D B\) and \(\triangle B D C\),

⇒ \(\angle 5=\angle 4\) (each \(90^{\circ}\) )

⇒ \(\angle 1=\angle 3\) (just proved)

⇒ \(\triangle A D B \sim \triangle B D C\) (AA corollary)

Hence Proved.

(2) In \(\triangle A D B\) and \(\triangle A B C\),

⇒\(\angle 5=\angle 1+\angle 2\)

⇒ \(\left(\right. each \left.90^{\circ}\right)\)

∴ \(\triangle \widehat{D B \sim \triangle A B C}\) (common)

(3) In \(\triangle B D C\) and \(\triangle A B C\),

⇒ \(\angle 4 =\angle 1+\angle 2 (each 90^{\circ} )\)

⇒ \(\angle 3 =\angle 3\) (common)

⇒ \(\triangle B D C \sim \triangle A B C\) (AA corollary)

Hence Proved.

(4) For \(B D^2=A D \times D C\), we need to prove two \(\Delta s\) similar which contain \(B D, A D\) and D C as sides.

Obviously, these are \(\triangle A B D\) and \(\triangle B D C\), we have proved these \(\triangle s\) as similar in part (1).

So, \(\Delta \underline{A \overline{D B}} \sim \Delta \underline{B} \bar{D} C\) [from part (1)]

⇒ \(\underbrace{\frac{A D}{B D}=\frac{D B}{D C}}_{\text {Taking first two }}=\frac{A B}{B C}\) (corresponding sides of similar triangles are proportional)

⇒ \(B D^2=A D \times D C\)

Hence Proved.

(5) For \(A B^2=A D \times A C\), we need to prove two triangles similar which contain AB, AD and AC as sides. Obviously, these are AABD and AABC. We have already proved these As as similar in part (2)

⇒ \(\triangle A D B \sim \triangle A B C\) [from part (2)]

⇒ \(\frac{A D}{A B}=\frac{D B}{B C}=\frac{A B}{A C}\) (corresponding sides of similar }

⇒ \(A B^2=A D \times A C\) …(1) Hence Proved.

So,\(\Delta \widehat{B D C} \sim \triangle \widehat{A B C}\) [from part (3)]

⇒ \(\frac{B D}{A B}=\frac{D C}{B C}=\frac{B C}{A C}\) (corresponding sides of similar triangles are proportional)

⇒ \(B C^2=C D \times\) A C … 2

Hence Proved.

(7) Adding results (1) and (2), we get

⇒ \(A B^2+B C^2 =A D \times A C+C D \times A C\)

= A C(A D+C D)

= \(A C \times A C=A C^2\)

Hence Proved.

Example 8. In the given figure, DEFG is a square, and ABAC = 90°. Prove that

1. \(\triangle\)AGF ~ \(\triangle\)DBG
2.  \(\triangle\)AGF-\(\triangle\)EFC
3. \(\triangle\)DBG – \(\triangle\)EFC
4. \(DE^2\) = BD x EC

Solution:

(1) Since Square DEFG is a square

GF || BC

⇒ \(\angle 2=\angle 4\) and \(\angle 6=\angle 9\) (corresponding \(\angle s\) )

Now, in \(\triangle A G F\) and \(\triangle D B G\),

⇒ \(\angle 5\) = \(\angle 1\)

⇒ \(\angle 4 =\angle 2\)

⇒ \(\triangle A G F \sim \triangle D B G\) (each \(90^{\circ}\) )

⇒ \(\angle 5=\angle 8\)

⇒ \(\angle 4=\angle 2\) (corresponding \(\angle \mathrm{s}\) )

⇒ \(\triangle A G F \sim \triangle D B G\) (AA corollary)

Hence Proved.

(2) In \(\triangle A G F\) and \(\triangle E F C\),

⇒ \(\angle 5=\angle 8\) (each \(90^{\circ}\) )

⇒ \(\angle 6=\angle 9\) (corresponding \(\angle s\) )

⇒ \(\triangle A G F \sim \triangle E F C\)(AA corollary)

Since \(\triangle A G F \sim \triangle D B G\) [proved in (1)]

and \(\triangle A G F \sim \triangle E F C\) [proved in (2)]

⇒ \(\triangle A G F \sim \triangle D B G \sim \triangle E F C\)

Hence Proved.

(4) Now, since \(\triangle \widehat{D B G} \sim \triangle \overparen{E F C}\)

⇒ \(\frac{D B}{E F}=\frac{D G}{E C}\) (corresponding \(\Delta s\) are proportional)

DG \(\times E F=B D \times E C\)

D E \(\times D E=B D \times E C\)( DG = FE = DE being the sides of square)

⇒ \(D E^2=B D \times E C\)

If we have to prove only the fourth part i.e., prove that \(DE^2\) = BD x EC, then no need to prove the first two parts.

For \(DE^2\) = BD x EC, we need to prove two As similar which contain DE, BD, and EC.

Obviously, these are \(\triangle G B D\) and \(\triangle F E C\).

So, \(\angle I=\angle 8 (each .90^{\circ})\)

Searching for Second Angle :

As we know that \(\angle 5=90^{\circ}\)

⇒ \(\angle 2+\angle 9=90^{\circ}\) (angle sum property)

Also since \(\angle 1=90^{\circ}\)

⇒ \(\angle 2+\angle 3=90^{\circ}\) (angle sum property)

From (1) and (2),

⇒ \(\angle 2+\angle 3=\angle 2+\angle 9\)

⇒ \(\angle 3=\angle 9\)

Now, in \(\triangle G B\) and \(\triangle F E C\),

⇒ \(\angle 1=\angle 8\) (each \(90^{\circ}\) )

⇒ \(\angle 3=\angle 9\) (just proved)

⇒ \(\frac{D B}{E F}=\frac{B G}{F C}=\frac{D G}{E C}\)

Taking the first and last

⇒ \(\triangle D B G \sim \triangle E F C\) (AA corollary)

⇒ \(\frac{D B}{E F}=\frac{B G}{F C}=\frac{D G}{E C}\) (corresponding sides of similar)

⇒ \(D G \times E F=B D \times E C\)

⇒ \(D E \times D E=B D \times E C\) ( D G=E F=D E, sides of a square)

⇒ \(D E^2=B D \times E C\)

Hence Proved.

Example 9. In the adjoining Figure, if a = 18,b = 12, c = 14, and d = 8, what is the measure of x?

Solution:

Given

In the adjoining Figure, if a = 18,b = 12, c = 14, and d = 8

We have, \(\frac{P R}{Q R}=\frac{12}{18}=\frac{2}{3}\) …… 1

⇒ \(\frac{S R}{P R}=\frac{8}{12}=\frac{2}{3}\) …… 2

Now in \(\triangle P R S\) and \(\triangle Q R P\),

Since, \(\frac{P R}{Q R}=\frac{S R}{P R}\) [from (1) and (2)]

⇒ \(\angle R=\angle R\)

⇒ \(\Delta P \widehat{R S}-\triangle Q \overparen{R P}\) (common)

⇒ \(\frac{P R}{Q R}=\frac{S R}{P R}=\frac{P S}{Q P}\) (SAS similarity)

⇒ \(\frac{2}{3}=\frac{2}{3}=\frac{x}{14}\)

∴\(\frac{x}{14}=\frac{2}{3} \quad \Rightarrow \quad x=\frac{28}{3}\)

Example 10. In the adjoining figure, AB || CD || EF. Prove that \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\).

Solution:

In \(triangle\) BFE and \(\triangle\) BDC

⇒ \(\angle 1=\angle 2\) (corresponding \(\angle\) s as E F || C D )

⇒ \(\angle 3=\angle 3\) (common)

⇒ \(\underline{B F E} \sim \triangle B C\) (AA corollary)

⇒ \(\frac{B F}{B D}=\frac{F E}{D C}\)

⇒ \(\frac{B F}{B D}=\frac{z}{y}\) (corresponding sides of similar triangles are proportional)

In \(\triangle D F E\) and \(\triangle D B A\),

⇒ \(\angle 4=\angle 5\) (corresponding \(\angle \)s as E F || A B )

⇒ \(\angle 6=\angle 6\) (common)

⇒ \(\triangle D F E \sim \triangle D B A\) (AA corollary)

⇒ \(\frac{D F}{D D}=\frac{F E}{D A}\) (Corresponding Sides Of Similar Triangles Are Proportional)

⇒ \(\frac{D F}{D B}=\frac{z}{x}\)

Remember, the common part in (1) and (2) should be in the denominator so that we can add easily) Adding (1) and (2), we get

⇒ \(\frac{B F}{B D}+\frac{D F}{D B} =\frac{z}{y}+\frac{z}{x}\)

⇒ \(\frac{B F+D F}{B D} =z\left(\frac{1}{y}+\frac{1}{x}\right)\)

⇒ \(\frac{B D}{B D} =z\left(\frac{1}{x}+\frac{1}{y}\right) \quad \Rightarrow \quad 1=z\left(\frac{1}{x}+\frac{1}{y}\right)\)

∴  \(\frac{1}{z} =\frac{1}{x}+\frac{1}{y}\)

Example 11. Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn, intersecting AC in L and AD produced in E. Prove that EL = 2BL.

Solution:

Given

Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn, intersecting AC in L and AD produced in E.

In \(\triangle\)s BMC and DME,

⇒ \(\angle 1 =\angle 2\) (alternate \(\angle\)s as B C || D E)

C M =D M (M is the mid-point of DC)

⇒ \(\angle 3 =\angle 4\) (vertically opposite angles)

⇒ \(\triangle B M C \simeq \triangle E M D)\)

BC = E D (corresponding parts of the congruent triangle)….. 1

But, BC = A D (opposite sides of a parallelogram)…… 2

Adding (1) and (2), we get

2 BC = D E+A D \(\Rightarrow 2 B C = A E\)

⇒ \(\frac{B C}{A E} =\frac{1}{2}\)

Now in \(\triangle B C L\) and \(\triangle E A L\)

⇒\(\angle 5 =\angle 6\)

⇒ \(\angle 7 =\angle 8\) (vertically opposite angles)

⇒ (\(\angle\)s as B C || A E )

⇒ \(\triangle B C L \sim \triangle E A L\) (AA Collary)

⇒ \(\frac{B C}{E A}=\frac{B L}{E L}\) (corresponding sides of similar triangles are proportional) ….(4)

⇒ \(\frac{1}{2}=\frac{B L}{E L}\) [ from (3) and (4)]

EL = 2BL

Hence Proved.

Example 12. The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm. Find the corresponding side of the second triangle.

Solution:

Given

The perimeters of two similar triangles are 25 cm and 15 cm respectively. If one side of the first triangle is 9 cm.

Let AB = 9cm

Since \(\triangle A B C \sim \triangle P Q R\) (given)

⇒ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=k\) corresponding sides of similar , triangles are proportional

AB= k. PQ, BC = k. QR, AC = k. PR

⇒ \(\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R} =\frac{A B+B C+C A}{P Q+Q R+R P}\)

=\(\frac{k \cdot P Q+k \cdot Q R+k \cdot P R}{P Q+Q R+R P}\)

=\(\frac{k(P Q+Q R+R P)}{(P Q+Q R+R P)}\) = k

From (1) and (2), we get

⇒ \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}=\frac{\text { Perimeter }(\triangle A B C)}{\text { Perimeter }(\triangle P Q R)}\)

⇒ \(\frac{9}{P Q}=\frac{25}{15} \Rightarrow P Q=\frac{9 \times 15}{25}\)=5.4

Hence, the corresponding side of the second triangle is 5.4 cm.

Example 13. A lamp is 3.3 m above ground. A boy I Hi em tall walks away from (be base of (bis lamp post at a speed of 0.8 m/s. Find (be the length of (be a shadow of the boy after 4 seconds.

Solution:

Given

A lamp is 3.3 m above ground. A boy I Hi em tall walks away from (be base of (bis lamp post at a speed of 0.8 m/s.

Let AB = 3.3 m be the lamp post and CD = 1.1 m be the position of the boy after 4 seconds. Also, let the shadow of the boy after 4 see = x m

Distance travelled by boy in 4 sec =y = 0.8 x 4 = 3.2 m

Now, in \(\triangle\)AEB and \(\triangle\)CED

⇒\(\angle E A B=\angle E C D=90^{\circ}\) (obvious)

⇒ \(\angle 1=\angle 1\) (common)

⇒ \(\triangle A E B \sim \triangle C E D\) (AA corollary)

⇒ \(\triangle A E B \sim \triangle C E D\)

⇒  –\(\frac{A E}{C E}=\frac{A B}{C D}\) (corresponding sides of similar triangles are proportional)

⇒ \(\frac{x+y}{x}=\frac{3.3}{1.1}\) (110 cm =1.1 m)

x+y=3 x \(\quad \Rightarrow y=2 x\)

x=\(\frac{y}{2}=\frac{3.2}{2}=1.6\)

Hence, the length of the shadow of the boy after 4 seconds is 1.6 m

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Ratio Of The Areas Of Two Similar Triangles

Theorem: The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

Given : \(\triangle A B C \sim \triangle D E F\)

To Prove : \(\frac{{ar}(\triangle A B C)}{{ar}(\triangle D E F)}=\frac{A B^2}{D E^2}=\frac{A C^2}{\dot{D F}}=\frac{B C^2}{E F^2}\)

Construction : Draw A L \(\perp B C\) and D M \(\perp E F\)

Proof : Consider,\(\frac{{ar}(\triangle A B C)}{{ar}(\triangle D E F)}=\frac{\frac{1}{2} \times B C \times A L}{\frac{1}{2} \times E F \times D M}=\left(\frac{B C}{E F}\right) \cdot\left(\frac{A L}{D M}\right)\) Equation 1

Now, since \(\triangle A B C \sim \triangle D E F\)

⇒ \(\angle B=\angle E\)(corresponding angles of similar triangles are equal)

In \(\triangle A L B\) and \(\triangle D M E\),

⇒ \(\angle B=\angle E\) [ from (2)]

⇒ \(\angle 1=\angle 2\) (each \(90^{\circ}\), by construction)

⇒ \(\triangle A L B \sim \triangle D M E\)(AA corollary)

⇒ \(\frac{A B}{D E}=\frac{A L}{D M}=\frac{B L}{E M}\) (corresponding sides of similar triangles are proportional)

But \(\frac{A B}{D E}=\frac{B C}{E F}=\frac{A C}{D F}(\triangle A B C \sim \triangle D E F)\)

From (3) and (4), we have

⇒ \(\frac{A L}{D M}=\frac{B C}{E F}\)

From (1) and (5), we have

⇒ \(\frac{{ar}(\triangle A B C)}{{ar}(\triangle D E F)}=\frac{B C}{E F} \times \frac{B C}{E F}=\frac{B C^2}{E F^2}\)

=\(\frac{A B^2}{D E^2}\)

=\(\frac{A C^2}{D F^2}\) From 4

Hence, \(\frac{{ar}(\triangle A B C)}{{ar}(\triangle D E F)}=\frac{A B^2}{D E^2}=\frac{B C^2}{E F^2}=\frac{A C^2}{D F^2}\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Ratio Of The Areas Of Two Similar Triangles Solved Examples

Example 1. In figures \(\triangle\)ABC and \(\triangle\)DEF are similar, the areas of 4 MBC is 9 sq. cm and that of ADEF is 16 sq. cm. If EF = 4.2 cm, find BC.

Solution:

Given

In figures \(\triangle\)ABC and \(\triangle\)DEF are similar, the areas of 4 MBC is 9 sq. cm and that of ADEF is 16 sq. cm. If EF = 4.2 cm

ar(\(\triangle\)ABC) = 9 sq. cm

ar(\(\triangle\)DEF) = 16 sq. cm

EF= 4.2 cm

BC= ?

Since the ratio of areas of two similar triangles is equal to the ratio of squares on their corresponding sides

⇒ \(\frac{{ar}(\triangle A B C)}{{ar}(\triangle D E F)}=\frac{B C^2}{E F^2} \quad \Rightarrow \quad \frac{9}{16}=\left(\frac{B C}{E F}\right)^2 \)

⇒ \(\frac{B C}{4.2}=\frac{3}{4} \quad \Rightarrow \quad B C=\frac{3 \times 4.2}{4}=3.15 \mathrm{~cm}\)

Example 2. In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD, and AB = 2 x CD. If the area of \(\triangle\)AOB = 84 cm², find the area of \(\triangle\)COD.

Solution:

Given

In a trapezium ABCD, O is the point of intersection of AC and BD, AB || CD, and AB = 2 x CD. If the area of \(\triangle\)AOB = 84 cm²,

In \(\triangle AOB\) and \(\triangle COD\),

⇒ \(\angle 1 =\angle 2\) (alternate angles)

⇒ \(\angle 3 =\angle 4\) (alternate angles)

⇒ \(\Delta A O B \sim \triangle C O D\) (AA corollary)

⇒ \(\frac{{ar}(\triangle A O B)}{{ar}(\triangle C O D)} =\frac{A B^2}{C D^2}\)

(The ratio of the area of two similar triangles is equal to the ratio of the square of the corresponding sides)

= \(\frac{(2 C D)^2}{C D^2}\)

= \(\frac{4}{1}\) ( A B=2 C D, given )

∴ \(\frac{84}{{ar}(\triangle C O D)}=\frac{4}{1} \quad \Rightarrow \quad {ar}(\triangle C O D)=\frac{84}{4} \mathrm{~cm}^2=21 \mathrm{~cm}^2\)

Example 3. In the figure, DE || BC and the ratio of the areas of \(\triangle\) and trapezium BDEC is 4: 5. Find the ratio of DE: BC. If BD=2cm, Then find AD

Solution:

Given

In the figure, DE || BC and the ratio of the areas of \(\triangle\) and trapezium BDEC is 4: 5. Find the ratio of DE: BC. If BD=2cm

In \(\triangle A D E\) and \(\triangle A B C\).

⇒ \(\angle 1=\angle 2\) (corresponding angles as } \(D E \mid B C \)

⇒ \(\angle A=\angle A\) (common)

⇒ \(\triangle A D E-\triangle A B C\)

⇒ \(\frac{a r \Delta A D E}{{ar}(A B C}=\frac{A D^2}{A B^2}=\frac{D E^2}{B C^2}=\frac{A E^2}{A C^2}\) (AA corollary)(ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides)

Now;\(\frac{{ar}(\text { trapezium } B D E C)}{{ar}(\Delta+D E)}=\frac{5}{4}\)

⇒ \(\frac{{ar}(\Delta+B C)-{ar}(\Delta+D E)}{{ar}(\Delta+D E)}=\frac{5}{4}\)

⇒ \(\frac{{ar}(\Delta+B C)}{{ar}(\Delta+D E)}-1=\frac{5}{4} \quad \Rightarrow \quad \frac{{ar}(\Delta A B C)}{{ar}(\Delta A D E)}=\frac{9}{4}\)

⇒ \(\frac{{ar}(\Delta+D E)}{{ar}(\Delta+B C)}=\frac{4}{9}\)

⇒ \(\frac{D E^2}{B C^2} =\frac{4}{9}\)

⇒ \(\frac{D E}{B C} =\frac{2}{3}\)

D E: B C =2: 3 from (1) and (2)

Now, from (1) and (2), we get

⇒ \(\frac{{ar}(\Delta A D E)}{{ar}(\Delta A B C)} =\frac{A D^2}{A B^2} =\frac{4}{9}\)

⇒ \(\frac{A D}{A B} =\frac{2}{3}\Rightarrow \frac{A D}{A D+B D}=\frac{2}{3}\)

⇒ \(\frac{A D+B D}{A D} =\frac{3}{2} \Rightarrow 1+\frac{B D}{A D}=\frac{3}{2}\)

⇒ \(\frac{B D}{A D} =\frac{3}{2}-1 =\frac{1}{2}\)

∴ \(\frac{2}{A D} =\frac{1}{2} \Rightarrow A D=4 \mathrm{~cm}\)

Example 4. X and Y are points on the sides AB and BC respectively of \(\triangle\) ABC such that XY || AC and XY divides AABC into two parts equal in area. Find \(\frac{A X}{A B}\)

Solution:

Given

X and Y are points on the sides AB and BC respectively of \(\triangle\) ABC such that XY || AC and XY divides AABC into two parts equal in area.

Since X Y || A C

In \(\triangle\) BXY and \(\triangle BAC\),

⇒ \(\angle 1=\angle 2\) (corresponding angles)

⇒ \(\angle B=\angle B\) (common)

⇒ \(\triangle B X Y \sim \triangle B A C\) (AA corollary)

⇒ \(\frac{{ar}(\Delta B X Y)}{{ar}(\triangle B A C)}=\frac{B X^2}{B A^2}\) (area ratio theorem)

But \({ar}(\triangle B X Y)=\frac{1}{2} {ar}(\triangle A B C)\) (given)

⇒ \(\frac{{ar}(\triangle B X Y)}{{ar}(\triangle A B C)}=\frac{1}{2}\)

From (1) and (2), we get

⇒ \(\frac{B X^2}{A B^2}=\frac{1}{2} \Rightarrow \frac{B X}{A B}=\frac{1}{\sqrt{2}}\)

⇒ \(\frac{A B-A X}{A B}=\frac{1}{\sqrt{2}} \Rightarrow \frac{A B}{A B}-\frac{A X}{A B}=\frac{1}{\sqrt{2}}\)

⇒ \(1-\frac{A X}{A B}=\frac{1}{\sqrt{2}} \Rightarrow \frac{A X}{A B}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}\)

Hence, \(\frac{A X}{A B}=\frac{2-\sqrt{2}}{2}\)

Example 5. CE and DE are equal chords of a circle with centre O. If \(\angle A O B\) = 90°, find \(\text{ar}(\triangle C E D): \text{ar}(\triangle A O B)\).

Solution:

Given

CE and DE are equal chords of a circle with centre O. If \(\angle A O B\) = 90°

Since O is the centre of the circle.

⇒ \(\angle\)E = 90° (angle in a semicircle is right angle)

\(\angle\)I = \(\angle\)2 = 45°

(angles opposite to equal sides are equal as CE = DE)

Also, since Z\(\angle\)AOB = 90° (given)

⇒ \(\angle\)3 = \(\angle\)4 = 45°

(angles opposite to equal sides are equal as OA = OB, each radius) Now, in \(\triangle\)CED and \(\triangle\)AOB

⇒ \(\angle E =\angle A O B\)

⇒ \(\angle \mathrm{l}=\angle 3\)

⇒ \(\triangle C E D \sim \triangle A O B\)

⇒ \(\frac{\text{ar}(\triangle C E D)}{\text{ar}(\triangle A O B)}=\frac{C D^2}{A B^2}\)

= \(\frac{(2 O B)^2}{O A^2+O B^2}\)

(CD = diameter and OB = radius so, CD = 2 • OB and using Pythagoras theorem)

= \(\frac{4 O B^2}{O B^2+O B^2}\) (OA = OB, each radius)

= \(\frac{4 O B^2}{2 O B^2}=\frac{2}{1}\)

Hence, \(\frac{\text{ar}(\triangle C E D)}{\text{ar}(\triangle \triangle O B)}=\frac{2}{1}\)

Example 6. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding altitudes.

Solution:

Given: \(\triangle A B C \sim \triangle D E F, A L \perp B C\) and \(D M \perp E F\)

To Prove: \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle D E F)}=\frac{A L^2}{D M^2}\)

Proof: \(\ln \triangle A L B\) and \(\triangle D M E\),

(because the ratio of areas of two similar triangles is equal to the ratio of squares of the corresponding sides

From (1) and (2) we have \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle D E F)}=\frac{A L^2}{D M^2}\)

Hence Proved.

Example 7. Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding angle bisector segments.
Solution:

Given: \(\triangle\)ABC ~ \(\triangle\)DEF in which AX and DY are the bisectors of \(\angle\)A and \(\angle\)D respectively.

To Prove: \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle D E F)}=\frac{A B^2}{D E^2}\)

Proof: Given \(\triangle A B C \sim \triangle D E F\)

Therefore, \(\angle A =\angle D\)

⇒ \(\frac{1}{2} \angle A=\frac{1}{2} \angle D\)

⇒ \(\angle B A X=\angle E D Y\)

Now, in \(\triangle A B X\) and \(\triangle D E Y\), we have

⇒ \(\angle B A X=\angle E D Y\)

and \(\angle B =\angle E\)

⇒ \(\triangle A B X \sim \triangle D E Y\)

Hence \(\frac{A B}{D E}=\frac{A X}{D Y}\)…..(1) given

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Pythagoras Theorem

Theorem 1: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right-angled triangle ABC in which \(\angle\)B = 90°.

To Prove: \((Hypotenuse^2=(\text { Base })^2+ Perpendicular)^2\)

i.e., \(A C^2=A B^2+B C^2\)

Construction: From B, draw \(B D \perp A C\)

Proof: In \(\triangle A D B\) and \(\triangle A B C\)

⇒ \(\angle 1=\angle 2\)

and \(\angle A =\angle A\) (common)

Hence, \(\triangle M D B \sim \triangle A B C\) (by AA corollary)

⇒ \(\frac{A D}{A B} =\frac{A B}{A C}\) (corresponding sides of similar triangle are proportional)

⇒ \(A B^2=A D \times A C\)……(1)

Now in \(\triangle B D C\) and \(\triangle A B C\)

⇒ \(\angle 5=\angle 2\) (each 90^{\circ}[/latex])

and \(\angle C=\angle C\) (common)

Hence, \(\triangle B D C \sim \triangle A B C\) (by AA corollary)

⇒ \(\frac{D C}{B C}=\frac{B C}{A C}\) (corresponding sides of similar triangles are proportional)

⇒ \(B C^2=A C \times D C\)…… (2)

Adding (1) and (2), we get

⇒ \(A B^2+B C^2=A D \times A C+A C \times D C\)

⇒ \(A B^2+B C^2=A C(A D+D C)\)

⇒ \(A B^2+B C^2=A C \times A C\)(because A D+D C=A C)

⇒ \(A B^2+B C^2=A C^2\)

Theorem 2 (Converse of Pythagoras Theorem): In a triangle, if the square of one side is equal to the sum of squares of the other two sides, then the angle opposite to the longest side is a right angle.

Given: A triangle ABC such that \(A C^2=A B^2+B C^2\)

To Prove: \(\angle\)B = 90°

Construction: Construct a triangle DEF such that DE =AB, EF = BC and \(\angle\)E = 90°. Join DF.

Proof: In right \(\triangle\)DEF

⇒ \(D F^2=D E^2+E F^2\) (by Pythagoras theorem)

⇒ \(D F^2=A B^2+B C^2\) (given DE = AB and Ef = BC)

⇒ \(D F^2=A C^2\) (given \(A C^2=A B^2+B C^2\))

DF = AC

Now, in \(\triangle A B C\) and \(\triangle D E F\)

AB = DE (by construction)

BC=EF (by construction)

AC = DF from (1)

⇒ \(\triangle A B C \cong \triangle D E F\)

Hence \(\angle E=\angle B=90^{\circ}\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Some Important Results Based Uponpythagoras Theorem

Result 1: In an obtuse-angled triangle ABC, obtuse-angled at B, if \(A D \perp C B\) (produced), prove that \(A C^2=A B^2+B C^2\) + 2BC x BD.

Given: An obtuse triangle ABC, obtuse-angled at B and AD is perpendicular to CB produced.

To Prove: \(A C^2=A B^2+B C^2\) + 2BC x BD

Proof: Since \(\triangle\)ADB is right angled triangle therefore by Pythagoras’ theorem in \(\triangle\)ADC

⇒ \(A C^2=A D^2+D C^2\)

⇒ \(A C^2=A D^2+(D B+B C)^2\) (because DC=D B+BC)

⇒ \(A C^2=A D^2+D B^2+B C^2+2 D B \cdot B C\)

⇒ \(A C^2=\left(A D^2+D B^2\right)+B C^2+2 B C \cdot B D\)

⇒ \(A C^2=A B^2+B C^2+2 B C \cdot B D\)

Result 2: In an acute-angled triangle ABC, if \(A D \perp B C\), prove that: \(A C^2=A B^2+B C^2-2 B C \times B D\)

Given: \(\triangle A B C\) which is an acute-angled triangle and \(A D \perp B C\)

To Prove: \(A C^2=A B^2+B C^2-2 B C \times B D\)

Proof: In \(\triangle A D C, \angle D=90^{\circ}\), hence by Pythagoras theorem

⇒ \(A C^2=A D^2+D C^2\)

⇒ \(A C^2=A D^2+(B C-B D)^2\) ( D C=B C-B D)

⇒ \(A C^2=A D^2+B C^2+B D^2-2 B C \times B D\)

⇒ \(A C^2=\left(A D^2+B D^2\right)+B C^2-2 B C \times B D\)

⇒ \(A C^2=A B^2+B C^2-2 B C \times B D\) (because \(A D^2+B D^2=A B^2\))

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Pythagoras Theorem Solved Examples

Example 1. The sides of a triangle are 5 cm, 8 cm and 11 cm respectively. Determine whether it is a right-angled triangle or not.
Solution:

Given

The sides of a triangle are 5 cm, 8 cm and 11 cm respectively.

5 = 25

8 = 64

and 11 = 121

We find that 1 = 5 + 8

Hence, it is not a right-angled triangle.

Example 2. A ladder 25 m long reaches a window of a building 20 m above the ground. Determine the distance of the foot of the ladder from the building.
Solution:

Given

A ladder 25 m long reaches a window of a building 20 m above the ground.

Let AC = 20 m be the wall of a building. A is the window and AB = 25 m be the ladder.

In right \(\triangle\)ABC

⇒ \(A B^2=B C^2+A C^2\) (by Pythagoras theorem)

⇒ \(25^2=B C^2+20^2\)

625 = \(B C^2+400\)

⇒ \(B C^2\) = 225

BC = 15

The distance of the foot of the ladder from the building = 15 cm

Example 3. P and Q are the mid-points of the sides C A and CB respectively of a \(\triangle\)ABC, oght anglers at C, prove that:

1. \(4 A Q^2=4 A C^2+B C^2\)
2. \(4 B P^2=4 B C^2+A C^2\)
3. \(4\left(A Q^2+B P^2\right)=5 A B^2\)

Solution:

Given

P and Q are the mid-points of the sides C A and CB respectively of a \(\triangle\)ABC, oght anglers at C,

1. In \(\triangle A Q C, \angle C=90^{\circ}\)

Using Pythagoras theorem. \(A Q^2=A C^2+Q C^2\)

⇒ \(A Q^2=A C^2+\left(\frac{B C}{2}\right)^2\) (because Q is the mid-point of BC)

⇒ \(A Q^2=A C^2+\frac{B C^2}{4}\)

⇒ \(4 A Q^2=4 A C^2+B C^2\)……(1) Hence proved.

2. ln \(\triangle B P C, \angle C=90^{\circ}\)

Using Pythagoras theorem, we have \(B P^2=B C^2+P C^2\)

⇒ \(B P^2=B C^2+\left(\frac{A C}{2}\right)^2\) (since P is the mid-point of AC)

⇒ \(B P^2=B C^2+\frac{A C^2}{4}\)

∴ \(4 B P^2=4 B C^2+A C^2\)……(2) Hence Proved.

3. Adding (1) and (2), we get

⇒ \(4 A Q^2+4 B P^2=4 A C^2+B C^2+4 B C^2+A C^2\)

⇒ \(4\left(A Q^2+B P^2\right)=5 A C^2+5 B C^2\)

⇒ \(4\left(A Q^2+B P^2\right)=5\left(A C^2+B C^2\right)\)

⇒ \(4\left(A Q^2+B P^2\right)=5 A B^2\) (because \(A C^2+B C^2=A B^2\))

Hence Proved.

Example 4. ABC is a right triangle, right angled at C. If p is the length of the perpendicular from C to AB and AB = c, BC = a and CA = b, then prove that

1. pc = ab
2. \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\)

Solution:

Given: ABC is a right triangle, right angled at C, and p is the length of the perpendicular from C to AB.

Proof:

1. \(\text{ar}(\triangle A B C)=\frac{1}{2} \times A B \times C D\)

= \(\frac{1}{2} c p\)

Also, \(\text{ar}(\triangle M B C)=\frac{1}{2} \times \Lambda C \times B C\)

= \(\frac{1}{2} b a\)……(2)

From (1) and (2), we get, \(\frac{1}{2} p c=\frac{1}{2} a b\)

pc = ab……(3)

Hence Proved.

2. From (3) we get c = \(\frac{a b}{p}\)

In \(\triangle A B C\) \(c^2=a^2+b^2\) (by Pythagoras theorem)

⇒ \(\left(\frac{a b}{p}\right)^2=a^2+b^2 \quad \Rightarrow \quad \frac{a^2 b^2}{p^2}=a^2+b^2 \)

⇒ \(\frac{1}{p^2}=\frac{a^2+b^2}{a^2 b^2} \quad \Rightarrow \quad \frac{1}{p^2}=\frac{a^2}{a^2 b^2}+\frac{b^2}{a^2 b^2}\)

⇒ \(\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{a^2}\)

Hence Proved.

Alternative proof (Using trigonometry)

Let \(\angle C A B=\theta\)

⇒ \(\angle C B A=90^{\circ}-\theta\) (\(\angle ACB\) = 90°)

In right \(\triangle C D A\), \(\sin \theta=\frac{p}{b}\)…….(1)

In right \(\triangle C D B\), \(\sin (90-\theta)=\frac{p}{a}\)

⇒ \(\sin (90-\theta)=\frac{p}{a}\) (because \(\sin (90-\theta)=\cos \theta\))……(2)

Squaring and adding equations (1) and (2), we get \(\sin ^2 \theta+\cos ^2 \theta=\frac{p^2}{b^2}+\frac{p^2}{a^2}\)

1 = \(p^2\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)

∴ \(\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}\) (using identity \(\sin ^2 \theta+\cos ^2 \theta=1\))

Hence Proved.

Example 5. O is any point in the Interior of rectangle ABCD. Prove that \(O B^2+O D^2=O C^2+O A^2\).
Solution:

ABCD is a rectangle and O is a point interior to the rectangle. Through O, draw a line parallel to BC meeting AB and DC at F and E respectively.

Then EC = PB and DE = AF

L.H.S. = \(O B^2+O I^2\)

= \(\left(O F^2+F B^2\right)+\left(O E^2+D E^2\right)\)

= \(O F^2+E C^2+O E^2+A F^2\) (by Pythagoras theorem)

= \(\left(O R^2+E C^2\right)+\left(O F^2+A F^2\right)\)

= \(O C^2+O A^2\) FB and \(D E=A F\) (by Pythagoras theorem)

= R.H.S.

Hence, \(O B^2+O D^2=O C^2+O \Lambda^2\)

Hence Proved.

Example 6. In the given figure, S and T trisect QR of a right-angled triangle PQR, right-angled at Q. Prove that 8 PT2 = 3PR² + 5PS².
Solution:

Given

In the given figure, S and T trisect QR of a right-angled triangle PQR, right-angled at Q.

Let QS = ST=TR =x

Now by Pythagoras theorem Prove that \(8 P T^2=3 P R^2+5 P S^2\).

Let QS = ST = TR = x

Now by Pythagoras theorem \(P R^2=P Q^2+Q R^2=P Q^2+(3 x)^2\)

= \(P Q^2+9 x^2\)

⇒ \(3 P R^2=3 P Q^2+27 x^2\) ……(1)

Also, \(P S^2=P Q^2+Q S^2\) (by Pythagoras theorem)

= \(P Q^2+x^2\)

⇒ \(5 P S^2=5 P Q^2+5 x^2\)…..(2)

Adding (1) and (2), we get

R.H.S. = \(3 P R^2+5 P S^2=8 P Q^2+32 x^2=8(P Q^2+ Q T^2)\)

= \(8\left(P Q^2+Q T^2\right)=8 P T^2\)

= L.H.S (by Pythagoras theorem)

Hence Proved.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Exercise 6.1

Question 1. Fill in the blanks using the correct word given in brackets:

1. All circles are……… (congruent, similar)
Solution: Similar

2. All squares are……… (similar, congruent)
Solution: Similar

3. All……..triangles are similar. (isosceles, equilateral)
Solution:  equilateral

4. Two polygons of the same number of sides are similar, if (1) their corresponding angles are … (2) their corresponding sides are
Solution: equal, proportional

Question 2. Give two different examples of

1. Similar
2. Non-similar

Solution:

1. The two circles are similar to each other.

Two squares are similar to each other.

2. A circle and a square are not similar.

A parallelogram and a rhombus are not similar.

Question 3. State whether the following quadrilaterals are similar or not
Answer:

Here the sides are proportional but the corresponding angles are not equal.

Figures are not similar

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Exercise 6.2

Question 1. In figures, (1) and (2), DE || BC. Find EC in (1) and AD in (2).
Solution:

(1) In \(\triangle A B C: D E || B C\)

⇒ \(\frac{A D}{B D}=\frac{A E}{E C}\)

⇒ \(\frac{1.5}{3}=\frac{1}{E C}\)

EC = \(\frac{1 \times 3}{1.5}=2 \mathrm{~cm}\)

(2) ln MBC: DE || BC

⇒ \(\frac{A D}{B D}=\frac{A E}{E C} \Rightarrow \frac{A D}{7.2}=\frac{1.8}{5.4}\)

⇒ \(\frac{A D}{7.2}=\frac{1}{3} \Rightarrow A D=\frac{7.2}{3}=2.4 \mathrm{~cm}\)

Question 2. E and F are points on the sides of PQ and PR respectively of an \(\triangle\)PQR. For each of the following cases, state whether EF || QR:

1. PE = 3.9 cm. EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
2. PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
3. PQ = 1 .28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution:

Given

E and F are points on the sides of PQ and PR respectively of an \(\triangle\)PQR.

1. Here PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

⇒ \(\frac{P E}{E Q}=\frac{3.9}{3}=\frac{1.3}{1}\)

and \(\frac{P F}{F R}=\frac{3.6}{2.4}=\frac{1.5}{1}\)

because \(\frac{P E}{E Q} \neq \frac{P F}{F R}\)

EF is not parallel to QR.

2. Here PE = 4 cm, QE = 4.5 cm, PF = 8 cm, FR = 9 cm

EF || QR

3. Here, PE = 0.18 cm, PQ – 1.28 cm, PF = 0.36 cm, PR = 2.56 cm

⇒ \(\frac{P E}{P Q}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}\)

and \(\frac{P F}{P R}=\frac{0.36}{2.56}=\frac{36}{256}=\frac{9}{64}\)

⇒ \(\frac{P E}{P Q}=\frac{P F}{P R}\)

EF || Q R

Question 3. in Figure, if LM || CB and LN || CD, prove that = \(\frac{A M}{A B}=\frac{A N}{A D}\)

Solution:

In \(\triangle A B C, L M \| C B\)

⇒ \(\frac{A M}{A B}=\frac{A L}{A C}\)……..(1)

In \(\triangle A C D, L N \| C D\)

⇒ \(\frac{A L}{A C}=\frac{A N}{A D}\)….(2)

From equations (1) and (2), \(\frac{A M}{A B}=\frac{A N}{A D}\)

Hence Proved.

Question 4. In the figure, DE || AC and DF || AE. Prove that \(\frac{B F}{F E}=\frac{B E}{E C}\).

Solution:

Here D E || A C

In \(\triangle A B C, \quad \frac{B E}{E C}=\frac{B D}{D A}\)……(1)

Again D F }| AE,

In \(\triangle A B E, \quad \frac{B D}{D A}=\frac{B F}{F E}\)……(2)

From equations (1) and (2), \(\frac{3 F}{E E}=\frac{B E}{E C}\)

Hence proved.

Question 5. In the figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:

In the figure, DE || OQ and DF || OR.

Now, from B.P.T.,

In \(\triangle P Q O\), \(\frac{P E}{E Q}=\frac{P D}{D O}\)….(1)

and, in \(\triangle P O R\), \(\frac{P F}{F R}=\frac{P D}{D O}\)…..(2)

From equations (1) and (2), \(\frac{P E}{E Q}=\frac{P F}{F R}\)

Now, in \(\triangle P Q R, \quad \frac{P E}{E Q}=\frac{P F}{F R}\)

E F || QR (from the converse of B.P.T)

Hence Proved.

Question 6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

In the figure, AB || PQ (given)

⇒ \(\frac{O A}{A P}=\frac{O B}{B Q}\) (from B.PT.)…..(1)

In the figure, \(A C \| P R\)(given)

⇒ \(\frac{O A}{A P}=\frac{O C}{C R}\) from B.P.T. …(2)

From equations (1) and (2), \(\frac{O B}{B Q}=\frac{O C}{C R} \Rightarrow B C \| Q R\) (from the converse of B.P.T.)

Hence Proved.

Question 7. Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Solution:

Given: \(\triangle\)ABC in which D is the mid-point of AB and DE is parallel to BC, meets AC at E

To prove: AE = EC

Proof: DE || BC

From B.P.T., \(\frac{A D}{D B}=\frac{A E}{E C}\)

but AD = D B (because D is midpoint)

⇒ \(\frac{A D}{D B}=\frac{A D}{A D}=1\)

From equations (1) and (2), \(\frac{A E}{E C}=1 \Rightarrow A E=E C\)

Hence Proved.

Question 8. Using Theorem 2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Solution:

Given: In \(\triangle\)ABC, D is the midpoint ofAB and E is the mid-point of AC.

To prove: DE || BC

Proof: D and E are the mid-points of AB and AC respectively.

AD = DB

and AE = EC

⇒ \(\frac{A D}{D B}=1\) and \(\frac{A E}{E C}=1\)

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}\)

From the converse of B.P.T., DE || BC.

Hence Proved

Question 9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\)

Solution:

ABCD is a trapezium whose diagonals intersect each other at point O.

Draw EO || AB || DC.

In \(\triangle\) ADC, EO || DC

⇒ \(\frac{A E}{E D}=\frac{A O}{O C}\)…..(1)

In \(\triangle D A B, E O \| A B\)

⇒ \(\frac{D E}{E A}=\frac{D O}{O B} \Rightarrow \frac{A E}{E D}=\frac{O B}{O D}\)……(2)

From equations (1) and (2), \(\frac{A O}{O C}=\frac{O B}{O D}\)

Hence Proved.

Question10. The diagonals of a quadrilateral ABCD intersect each other at point O such that \(\frac{A O}{B O}=\frac{C O}{D O}\). Show that ABCD is a trapezium.
Solution:

Given

The diagonals of a quadrilateral ABCD intersect each other at point O such that \(\frac{A O}{B O}=\frac{C O}{D O}\).

Draw EO || DC which meets AD at E

Now, in \(\triangle A D C, E O \| D C\)

⇒ \(\frac{A E}{E D}=\frac{A O}{O C}\)…..(1)

Given, \(\frac{A O}{O C}=\frac{B O}{O D}\)…..(2)

From equations (1) and (2) \(\frac{A E}{E D}=\frac{B O}{O D} \Rightarrow E O \| A B(\ln \triangle D A B)\)

but E O || DC

AB || DC

⇒ square A B C Dis a trapezium.

Hence Proved.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Exercise 6.3

Question 1. State which pairs of triangles in figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution:

1. In \(\triangle M B C\) and \(\triangle P Q R\),

⇒ \(\angle A=\angle P=60^{\circ}, \angle B=\angle Q=80^{\circ},\), \(\angle C=\angle R=40^{\circ}\)

⇒ \(\triangle A B C-\triangle P Q R\)

⇒ \(\triangle A B C-\triangle P Q R\) (from A.A.A. similarity)

2. Here \(\frac{A B}{Q R}=\frac{2}{4}=\frac{1}{2}, \frac{B C}{R P}=\frac{2.5}{5}=\frac{1}{2}\)

and \(\frac{A C}{Q P}=\frac{3}{6}=\frac{1}{2}\)

⇒ \(\frac{A B}{Q R}=\frac{B C}{R P}=\frac{A C}{Q P}\)

⇒ \(\triangle M B C-\triangle Q R P\) (from S.S.S. similarity)

3. Here \(\frac{M P}{E D}=\frac{2}{4}=\frac{1}{2}, \frac{P L}{D F}=\frac{3}{6}=\frac{1}{2}\),

⇒ \(\frac{L M}{F E}=\frac{2.7}{5}=\frac{27}{50}\)

⇒ \(\frac{M P}{E D}=\frac{P L}{D F} \neq \frac{L M}{F E}\)

∴ \(\triangle L M P\) and \(\triangle D E F\) are not similar.

4. Here \(\frac{M P}{Q P}=\frac{2.5}{5}=\frac{1}{2}, \frac{M L}{Q R}=\frac{5}{10}=\frac{1}{2}\)

⇒ \(\frac{M P}{Q P}=\frac{M L}{Q R}\) and \(\angle M=\angle Q\)

⇒ \(\triangle M N L-\triangle Q P R\) (from S.S.A. similarity)

5.Here \(\frac{A B}{F D}=\frac{2.5}{5}=\frac{1}{2}, \angle A=\angle F=80^{\circ}\)

but AC and DE are unknown.

∴ \(\triangle A B C\) and \(\triangle D E F\) are not similar.

6. ln \(\triangle D E F\), \(\angle F=180^{\circ}-\angle D-\angle E\)

= \(180^{\circ}-70^{\circ}-80^{\circ}=30^{\circ}\)

In \(\triangle P Q R\),

⇒ \(\angle P =180^{\circ}-\angle Q-\angle R\)

= \(180^{\circ}-80^{\circ}-30^{\circ}=70^{\circ}\)

Now, \(\angle D=\angle P, \angle E=\angle Q, \angle F=\angle R\)

⇒ \(\triangle D E F \sim \triangle P Q R\) (from AAA similarity)

Question 2. In figure, \(\triangle O D C \sim \triangle O B A\), \(\angle B O C=125^{\circ}\) and \(\angle C D O=70^{\circ}\). Find \(\angle D O C\), \(\angle D C O\) and \(\angle O A B\).

Solution:

Here \(\angle B O C=125^{\circ}\)

and \(\angle C D O=70^{\circ}\)

⇒ \(\angle D O C+\angle B O C=180^{\circ}\) (linear pair)

⇒ \(\angle D O C=180^{\circ}-\angle B O C\)

= \(180^{\circ}-125^{\circ}=55^{\circ}\)

In \(\triangle D O C\),

⇒ \(\angle D C O=180^{\circ}-\angle C O D-\angle O D C\)

= \(180^{\circ}-55^{\circ}-70^{\circ}=55^{\circ}\)

⇒ \(\triangle O D C-\triangle O B A\)

⇒ \(\angle O C D=\angle O A B\)

∴ \(\angle O A B=55^{\circ}\)

Question 3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{O A}{O C}=\frac{O B}{O D}\)
Solution:

Given

Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O.

Draw EO || AB || CD

Now, in \(\triangle\)ADC,

EO || DC

⇒ \(\frac{A E}{E D}=\frac{A O}{O C}\) (from B.P.T.)…(1)

ln \(\triangle A B D\)

EO || AB

⇒ \(\frac{A E}{E D}=\frac{B O}{O D}\) (from B.P.T.)…(2)

From equations (1) and (2) \(\frac{A O}{O C}=\frac{B O}{O D}\)

Hence proved.

Question 4. In figure, \(\frac{Q R}{Q S}=\frac{Q T}{P R}\) and \(\angle 1=\angle 2\). Show that \(\triangle P Q S \sim \triangle T Q R\).

Solution:

Given

In figure, \(\frac{Q R}{Q S}=\frac{Q T}{P R}\) and \(\angle 1=\angle 2\).

In \(\triangle P Q R, \angle 1=\angle 2\) (given)

PR = PQ….(1) (sides opposite to equal angles in \(\Delta\) are equal)

Given, \(\frac{Q R}{Q S}=\frac{Q T}{P R} \Rightarrow \frac{Q R}{Q S}=\frac{Q T}{Q P}\) [from equation (1)]

⇒ \(\frac{Q S}{Q R}=\frac{Q P}{Q T}\)

In \(\triangle P Q S\) and \(\triangle T Q R\),

⇒ \(\frac{Q S}{Q R}=\frac{Q P}{Q T}\) and \(\angle S Q P=\angle R Q T\) (each \(\angle 1\))

⇒ \(\triangle P Q S \sim \triangle T Q R\) (from S.A.S. similarity)

Hence Proved.

Question 5. S and T are points on sides PR and QR of \(\triangle P Q R\) such that \(\angle P=\angle R T S\). Show that \(\triangle R P Q \sim \triangle R T S\).
Solution:

Given: \(\triangle R P Q\) and \(\triangle R T S\) in which \(\angle P=\angle R T S\)

To show: \(\triangle R P Q \sim \triangle R T S\)

Proof: \(\ln \triangle R P Q\) and \(\triangle R T S\), \(\angle 1=\angle 2\) (given)

⇒ \(\angle 3=\angle 3\) (common)

⇒ \(\triangle R P Q \sim \Delta R T S\) (from A.A. corollary)

Hence Proved.

Question 6. In the figure, if \(\triangle A B E \cong \triangle A C D\) shows that \(\triangle A D E \sim \triangle A B C\).
Solution:

Given, \(\triangle A B E \cong \triangle A C D\)

AE=AD….(1)

and AB=AC….(2)

⇒ \(\frac{A D}{A B}=\frac{A I}{A C}\) and \(\angle A=\angle A\) (common)

⇒ \(\triangle\) ADE \(\triangle\) ABC (from S.A.S. similarity)

Hence Proved.

Question 7. In the figure, altitudes AD and CE of \(\triangle\) ABC Intersect each other at the point Show that:

1. \(\triangle A E P-\triangle C D P\)
2. \(\triangle A B D-\triangle C B E\)
3. \(\triangle A E P \sim \triangle A D B\)
4. \(\triangle P D C-D B E C\)

Solution:

1. In the figure, \(\angle A E P=\angle C D P\) (each \(90^{\circ}\))

and \(\angle A P E=\angle C P D\) (vertically opposite angles)

⇒ \(\triangle A E P \sim \triangle C D P\) (from A.A.A. similarity)

Hence Proved.

2. In figure, \(\angle A D B=\angle C E B\) (each \(90^{\circ}\))

and \(\angle A B D=\angle C B E\) (each = \(\angle B\))

⇒ \(\triangle A B D \sim \triangle C B E\) (from A.A.A. similarity)

Hence Proved.

3.  In figure, \(\angle A E P=\angle A D B\) (each \(90^{\circ}\))

and \(\angle P A E=\angle B A D\) (common angle)

\(\triangle A E P \sim \triangle A D B\) (from A.A.A. similarity)

Hence Proved.

4. In figure, \(\angle P D C=\angle B E C\) (each \(90^{\circ}\))

and \(\angle P C D=\angle B C E\) (common angle)

⇒ \(\triangle P D C \sim \triangle B E C\) (from A.A.A. similarity)

Hence Proved.

Question 8. E is a point on the side AD produced of parallelogramABCD and BE intersects CD at F. Show that \(\triangle A B E \sim \triangle C F B\)
Solution:

Given

E is a point on the side AD produced of parallelogramABCD and BE intersects CD at F.

AE || BC and BE is a transversal.

⇒ \(\angle\) 1 = \(\angle\) 2(alternate angles)

⇒ \(\angle\) 3 = \(\angle\) 4(opposite angles of parallelogram)

⇒ \(\triangle A B E \sim \triangle C F B\) (A. A similarly)

Question 9. In the figure, ABC and AMP are two right. triangles, right-angled at B and M respectively. Prove that

1. \(\triangle A B C \sim \triangle A M P\)
2. \(\frac{C A}{P A}=\frac{B C}{M P}\)

Solution:

Given

In the figure, ABC and AMP are two right. triangles, right-angled at B and M respectively.

1. In \(\triangle A B C\) and \(\triangle A M P\)

⇒ \(\angle A B C=\angle A M P\) (each \(90^{\circ}\))

⇒ \(\angle B A C=\angle M A P\) (common)

⇒ \(\triangle A B C \sim \triangle A M P\) (from A.A. similarity)

2. Divide (2) by (1)

⇒ \(\triangle A B C \sim \triangle A M P\)

⇒ \(\frac{C A}{P A}=\frac{B C}{M P}\) (ratio of corresponding sides)

Hence Proved.

Question 10. CD and GH are respectively the bisectors of \(\angle\) ACB and \(\angle\) EGF such that D and H lie on sides AB and FE of \(\triangle\) ABC and \(\triangle\) EFG respectively. If latex]\triangle[/latex] ABC ~ \(\triangle\) FEG.

1. \(\frac{C D}{G H}=\frac{A C}{F G}\)
2. \(\triangle D C B \sim \triangle H G E\)
3. \(\triangle D C A \sim \triangle H G F\)

Solution:

1. In \(\triangle A C D\) and \(\triangle F G H\)

⇒ \(\angle C A D=\angle G F H\)

(because \(\triangle A B C \sim \triangle F E G\))

but \(\triangle A B C \sim \triangle F E G\) (given)

⇒ \(\angle C=\angle G\)

⇒ \(\angle A C D=\angle F G H\)

From equations (1) and (2)

⇒ \(\triangle A C D \sim \triangle F G H\) (from A.A. similarity)

⇒ \(\frac{C D}{G H}=\frac{A C}{F G}\)

2. \(\triangle A B C \sim \triangle F E G\)

⇒ \(\angle A C B=\angle F G E\)

⇒ \(\frac{1}{2} \angle A C B=\frac{1}{2} \angle F G E\)

⇒ \(\angle D C B=\angle H G E\) and \(\angle D B C=\angle H E G\)

⇒ (because \(\triangle A B C \sim \triangle F E G \Rightarrow \angle B=\angle E\))

⇒ \(\triangle D C B \sim \triangle H G E\) (A.A. similarity)

3. \(\triangle A B C \sim \triangle F E G\)

⇒ \(\angle C A B=\angle G F E\)

⇒ \(\angle C A D=\angle G F H\)

⇒ \(\angle D A C=\angle H F G\)

and \(\triangle A B C \sim \triangle F E G\)

⇒ \(\angle A C B=\angle F G E\)

⇒ \(\frac{1}{2} \angle A C B=\frac{1}{2} \angle F G E\)

⇒ \(\angle D C A=\angle H G F\)

Now \(\angle D A C=\angle H F G\)

and \(\angle D C A=\angle H G F\)

⇒ \(\triangle D C A \sim \triangle H G F\) (A.A. similarity)

Hence Proved.

Question 11. In figure, E is a point on side CB produced of an isosceles triangle ABC with AB =AC. If \(A D \perp B C\) and \(E F \perp A C\), prove that \(\triangle A B D \sim \triangle E C F\).

Solution:

Given

In figure, E is a point on side CB produced of an isosceles triangle ABC with AB =AC. If \(A D \perp B C\) and \(E F \perp A C\)

⇒ \(\triangle A B C\) is isosceles in which,

AB = AC

⇒ \(\angle A C B=\angle A B C\) (angles opposite to equal sides)

⇒ \(\angle E C F=\angle A B D\)

and \(\angle E F C=\angle A D B\) (each \(90^{\circ}\))

⇒ \(\triangle E C F \sim \triangle A B D\) (from A.A. similarity)

Hence Proved.

Question 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \(\triangle\)PQR. Show that \(\triangle A B C \sim \triangle P Q R \text {. }\).

Solution:

Given: \(\triangle A B C\) and \(\triangle P Q R\), in AD \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\)

To show: \(\triangle A B C \sim \triangle P Q R\)

Proof: \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\) (given)

⇒ \(\frac{A B}{P Q}=\frac{2 B D}{2 Q M}=\frac{A D}{P M}\) (AD and PM are medians)

⇒ \(\frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}\)

⇒ \(\triangle A B D \sim \triangle P Q M\) (from S.S.S. similarity)

⇒ \(\angle B=\angle Q\) (corresponding angles are equal of similar triangles)

Now, in \(\triangle A B C\) and \(\triangle P Q R\), \(\frac{A B}{P Q}=\frac{B C}{Q R}\) (given)

and \(\angle B=\angle Q\) (proved above)

⇒ \(\triangle A B C \sim \triangle P Q R\) (from S.A.S. similarity)

Hence Proved.

Question 13. D is a point on the side BC of a triangle ABC such that triangle ABC such that \(\angle A D C=\angle B A C\). Show that \(C A^2=C B \cdot C D\).
Solution:

Given

D is a point on the side BC of a triangle ABC such that triangle ABC such that \(\angle A D C=\angle B A C\).

In \(\triangle A B C\) and \(\triangle D A C\),

⇒ \(\angle B A C=\angle A D C\) (given)

⇒ \(\angle A C B=\angle D C A\) (common)

⇒ \(\triangle A B C \sim \triangle D A C\)

⇒ \(\frac{C D}{C A}=\frac{C A}{C B}\) (A.A. similarity)

⇒ \(C A^2=C B \times C D\)

Hence Proved.

Question 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional
to sides PQ and PR and median PM of another triangle PQR. Show that triangle PQR. Show that \(\triangle A B C \sim \triangle P Q R\).
Solution:

Given: \(\triangle A B C\) and \(\triangle P Q R\) in which AD and PS are the median such that

⇒ \(\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P S}\)

To show: \(\triangle A B C \sim \triangle P Q R\)

Construction: Produce AD to E such that AD=DE. Join EC, produce PS to T such that PS=ST.

Join TR.

Proof: In \(\triangle A B D\) and \(\triangle E C D\),

BD = DC (Dis mid-point of BC)

⇒ \(\angle\)5 = \(\angle\)6 (vertically opposite angle)

AD = DE (from construction)

⇒ \(\triangle A B D \cong \triangle E C D\) (S.A.S. congruency)

AB=EC (congruent parts of congruent triangles are equal)…(1)

Similarly, we can prove that \(\triangle P Q S \cong \triangle T R S\)

PQ=TR…..(2)

Now, \(\frac{A B}{P Q}=\frac{A C}{P R}=\frac{A D}{P S}\)

⇒ \(\frac{E C}{T R}=\frac{A C}{P R}=\frac{2 A D}{2 P S}\) (from equations (1) and (2))

⇒ \(\frac{E C}{T R}=\frac{A C}{P R}=\frac{A E}{P T}\)

⇒ \(\triangle A E C \sim \triangle P T R\) (S.S.S. similarity)

⇒ \(\angle 1=\angle 2\) (corresponding angles of similar triangles are equal)…..(3)

Similarly, we can prove that \(\angle 3=\angle 4\)…..(4)

Adding equations (3) and (4).

⇒ \(\angle 1+\angle 3=\angle 2+\angle 4 \Rightarrow \angle A=\angle P\)

Now, in \(\triangle A B C\) and \(\triangle P Q R\).

⇒ \(\frac{A B}{P Q}=\frac{A C}{P R}\) (given)

and \(\angle A=\angle P\)(proved above)

∴ \(\Delta B C-\triangle P Q R\) (S.A.S. similarity)

Question 15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time, a tower casts a shadow 28 m long. Find the height of the tower.
Solution:

Given

A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time, a tower casts a shadow 28 m long.

Let MC be a pole of 6m in height and its shadow is DC = 4m. At the same time, tower AB, whose height is h metre, has its shadow BD = 28 m.

Now, \(\triangle D C M \sim \triangle D B A\)

⇒ \(\frac{D C}{D B}=\frac{C M}{B A}\)

⇒ \(\frac{4}{28}=\frac{6}{h} \quad \Rightarrow \quad h=\frac{28 \times 6}{4}=42\)

Height of tower = 42 ~m

Question 16. If AD and PM are medians of triangles ABC and PQR respectively where \(\triangle A B C – [latex]\triangle P Q R\) prove that \(\frac{A B}{P Q}=\frac{A D}{P M} \text {. }\)
Solution:

Given: AD and PM are the medians of \(\triangle A B C\) and \(\triangle P Q R\) respectively.

and \(\triangle A B C \sim \triangle P Q R\)

To prove: \(\frac{A B}{P Q}=\frac{A D}{P M}\)

Proof: In \(\triangle A B D\) and \(\triangle P Q M\),

⇒ \(\angle B=\angle Q\) (because \(\triangle A B C \sim \triangle P Q R\))

⇒ \(\frac{A B}{P Q}=\frac{\frac{1}{2} B C}{\frac{1}{2} Q R}\) (because \(\frac{A B}{P Q}=\frac{B C}{Q R}\))

⇒ \(\frac{A B}{P Q}=\frac{B D}{Q M}\)

⇒ \(\triangle A B D \sim \triangle P Q M\) (from S.A.S. similarity)

∴ \(\frac{A B}{P Q}=\frac{A D}{P M}\) Hence Proved.

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Exercise 6.4

Question 1. Let \(\triangle\)ABC ~ \(\triangle\)DEF areas be, respectively, 64 cm EF = 15.4 cm, find BC.
Solution:

Given

Let \(\triangle\)ABC ~ \(\triangle\)DEF areas be, respectively, 64 cm EF = 15.4 cm

⇒ \(\triangle A B C \sim \triangle D E F\)

⇒ \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle D E F)}=\frac{B C^2}{E F^2} \Rightarrow \frac{64}{121}=\frac{B C^2}{E F^2}\)

⇒ \(\frac{8}{11}=\frac{B C}{E F}\)

∴ \(BC=\frac{8}{11} \times E F=\frac{8}{11} \times 15.4=11.2 \mathrm{~cm}\)

Question 2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Solution:

Given

Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD

In \(\triangle\)AOB and \(\triangle\)COD, \(\angle\)1 = \(\angle\)2 (vertically opposite angles)

⇒  \(\angle\)3 = \(\angle\)4 (alternate angles, AB || DC)

⇒ \(\triangle A O B \sim \triangle C O D\) (A.A. corollary)

⇒ \(\frac{\text{ar}(\triangle A O B)}{\text{ar}(\triangle C O D)}=\frac{A B^2}{C D^2}\)

(because the ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides)

= \(\frac{(2 C D)^2}{C D^2}\) (because AB=2CD, given)

= \(\frac{4 C D^2}{C D^2}=\frac{4}{1}\)

∴ \(\text{ar}(\triangle A O B):\text{ar}(\triangle C O D)=4: 1\)

Question 3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that \(\frac{\text{ar}(A B C)}{\text{ar}(D B C)}=\frac{A O}{D O}\).

Solution:

Given

In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O

Draw \(A L \perp B C\) and \(D M \perp B C\)

In \(\triangle O L A\) and \(\triangle O M D\),

⇒ \(\angle A L O=\angle D M O=90^{\circ}\)

and \(\angle A O L=\angle D O M\) (vertically opposite angles)

⇒ \(\triangle O L A \sim \triangle O M D\) (from A.A.A. similarity)

⇒ \(\frac{A L}{D M}=\frac{A O}{D O}\)

= \(\frac{A L}{D M}=\frac{A O}{D O}\)

Now \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle D B C)}=\frac{\frac{1}{2} \times(B C) \times(A L)}{\frac{1}{2} \times(B C) \times(D M)}\) [from equation (1)]

Therefore, \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle D B C)}=\frac{A O}{D O}\) Hence Proved.

Question 4. If the areas of two similar triangles are equal, prove that they are congruent.
Solution:

Given, \(\triangle A B C \sim \triangle P Q R\)

⇒ \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle P Q R)}=\frac{A B^2}{P Q^2}=\frac{B C^2}{Q R^2}=\frac{C A^2}{R P^2}\)

But \(\text{ar}(\triangle A B C)=\text{ar}(\triangle P Q R)\)

⇒ \(\frac{A B^2}{P Q^2}=\frac{B C^2}{Q R^2}=\frac{C A^2}{R P^2}=1\)

AB = PQ, BC = QR

CA = RP

⇒ \(\triangle A B C \cong \triangle P Q R\)(from S.S.S.)

Hence Proved.

Question 5. D, E and F are respectively the mid-points of sides AB, BC and CA of \(\triangle\) ABC. Find the ratio of the areas of \(\triangle\)DEF and \(\triangle\)ABC.
Solution:

Given: \(\triangle\)ABC in which D, E and F are the mid-points of sides BC, CA and AB respectively.

To find: The ratio of the areas of \(\triangle\)DEF and \(\triangle\)ABC.

Proof: D, E, and F are mid-points.

⇒  \(E F\| B C\) and \(E F=\frac{1}{2} B C\)

EF = BD

Similarly, \(D E=B F\) (because \(D E \| A B\) and \(\frac{1}{2} A B\))

So, \(E F B D\) is a parallelogram.

⇒ \(\angle B=\angle 1\)

Similarly, AFDE and EFDC are parallelograms.

⇒ \(\angle A=\angle 2\) and \(\angle C=\angle 3\)

i.e., \(\triangle A B C\) and \(\triangle D E F\) are equiangular.

⇒ \(\triangle D E F \sim \triangle A B C\)

⇒ \(\frac{\text{ar}(\triangle D E F)}{\text{ar}(\triangle A B C)}=\frac{D E^2}{A B^2}=\frac{F B^2}{A B^2}=\frac{(A B / 2)^2}{A B^2}=\frac{1}{4}\)

Therefore, \(\frac{\text{ar}(\triangle D E F)}{\text{ar}(\triangle A B C)}=\frac{1}{4}\)

Question 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
Solution:

Given: \(\triangle\)ABC ~ \(\triangle\)DEF, AP and DQ are the medians of AABC and ADEF respectively.

To prove: \(\frac{\text{ar}(\triangle M B C)}{\text{ar}(\triangle D E F)}=\frac{A P^2}{D Q^2}\)

Proof: AP and DQ are medians.

So, BP=PC and EQ=QF

and given that \(\triangle A B C \sim \triangle D E F\)

So, \(\frac{A B}{D E}=\frac{A C}{D F}=\frac{B C}{E F}\) and \(\angle A=\angle D, \angle B=\angle E\) and \(\angle C=\angle F\)

Now, \(\frac{A B}{D E}=\frac{B C}{E F}\)

⇒ \(\frac{A B}{D E}=\frac{2 B P}{2 E Q} \Rightarrow \frac{A B}{D E}=\frac{B P}{E Q}\)

and \(\angle B=\angle E\)

⇒ \(\triangle A B P \sim \triangle D E Q\)(S.A.S. similarity)

Now, in \(\triangle A B C\) and \(\triangle D E F\), \(\frac{\text{ar}(\triangle \triangle B C)}{\text{ar}(\triangle D E F)}=\frac{A B^2}{D E^2}\)

(because the ratio of the areas of two similar triangles is equal to the ratio of the squares of its corresponding sides)

From equations (1) and (2), \(\frac{\text{ar}(\triangle A B C)}{text{ar}(\triangle D E F)}=\frac{A P^2}{D Q^2}\)

Hence Proved.

Question 7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Solution:

Given: A square ABCD and an equilateral triangle BCE, formed on side BC and \(\triangle\) ACF formed on side AC.

To prove: \(\text{ar}(\triangle B C E)=\frac{1}{2} \text{ar}(\triangle A C F)\)

Proof: \(\triangle B C E\) and \(\triangle A C F\) are equilateral triangles, so each angle of both is \(60^{\circ}\).

So, \(\triangle B C E \sim \triangle A C F\)

⇒ \(\frac{\text{ar}(\triangle B C E)}{\text{ar}(\triangle A C F)}=\frac{B C^2}{A C^2}=\frac{B C^2}{2(B C)^2}=\frac{1}{2}\)

Therefore, \(\text{ar}(\triangle B C E)=\frac{1}{2} \times \text{ar}(\triangle A C F)\)

Hence Proved.

Question 8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is:

1. 2: 1
2. 1: 2
3. 4: 1
4. 1: 4

Solution: 3. 4: 1

⇒ \(\triangle A B C\) and \(\triangle B D E\) are two equilateral triangles such that

BD = \(\frac{1}{2} B C\)

BD = \(\frac{1}{2} A B\) (because BC=A B)

⇒ \(\frac{A B}{B D}=\frac{2}{1}\)

⇒ \(\triangle A B C \sim \triangle B D E\)(triangles are equilateral)

⇒ \(\frac{\text{ar}(\triangle A B C)}{\text{ar}(\triangle B D E)}=\frac{A B^2}{B D^2}=\left(\frac{2}{1}\right)^2=4: 1\)

Question 9. The sides of two similar triangles are in

1. 2 : 3
2. 4: 9
3. 81: 16
4. 16: 81

Solution: 4. 16: 81

The ratio of the areas of two similar triangles = (ratio of the sides of triangles)2

= \(\left(\frac{4}{9}\right)^2=\frac{16}{81}=16: 81\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Exercise 6.5

Question 1. The sides of the triangles are given below. Determine which of them are right triangles. In the case of a right triangle, write the length of its hypotenuse.

1. 7 cm, 24 cm, 25 cm
2. 3 cm, 8 cm, 6 cm
3. 50 cm, 80 cm, 100 cm
4. 13 cm, 12 cm, 5 cm

Solution:

Here

1. 7² + 24² = 49 + 576 = 625 = 252

The triangle is a right-angled hypotenuse = 25 cm

2. 3² + 6² = 9 + 36 = 45 82

The triangle is not right-angled.

3. 50² + 80² = 2500 + 6400 = 8900* 1002

The triangle is not right-angled.

4. 5²+ 12² = 25 + 144= 169 = 132

∴  Triangle is right-angled

hypotenuse = 13 cm

Question 2. PQR is a triangle right angled at P and M is a point on QR such that \(P M \perp Q R\). Show that \(P M^2=Q M \cdot M R\).
Solution:

Given

PQR is a triangle right angled at P and M is a point on QR such that \(P M \perp Q R\).

In \(\triangle P Q R, \angle P=90^{\circ}\)

and \(\angle 2+\angle 4=90^{\circ}\)

⇒ \(\angle 1+\angle 2=\angle 2+\angle 4 \Rightarrow \angle 1=\angle 4\)

Similarly, \(\angle 3=\angle 2\)

⇒ \(\triangle P Q M \sim \triangle R P M\) (from A.A. similarity)

⇒ \(\frac{Q M}{P M}=\frac{P M}{R M}\)

∴ \(P M^2=Q M \cdot M R\) Hence Proved

Question 3. In the figure, ABD is a triangle right-angled at A and AC¹ BD. Show that

1. AB² = BC • BD
2. AC² =BC- DC
3. AD = BD CD

Solution:

1. In \(\triangle A B C\) and \(\triangle D B A\),

⇒ \(\angle A B C=\angle D B A\) (common)

⇒ \(\angle A C B=\angle D A B\) (each \(90^{\circ}\))

⇒ \(\triangle A B C \sim \triangle D B A\)(A.A. similarity)

⇒ \(\frac{A B}{D B}=\frac{B C}{B A} \Rightarrow A B^2=B C \cdot B D\)

Hence Proved.

2. In \(\triangle A B C\) and \(\triangle D A C\),

⇒ \(\angle A B C=180^{\circ}-\angle A C B-\angle B A C\)

= \(180^{\circ}-90^{\circ}-\angle B A C\)

= \(90^{\circ}-\angle B A C=\angle D A C\)

and \(\angle A C B=\angle D C A\) (each \(90^{\circ}\))

⇒ \(\triangle A B C \sim \triangle D A C\) (A.A. similarity)

⇒ \(\frac{A C}{D C}=\frac{B C}{A C} \Rightarrow A C^2=B C \cdot D C\)

Hence Proved.

3.In \(\triangle D A C\) and \(\triangle D B A\),

⇒ \(\angle A D C=\angle B D A\) (common)

⇒ \(\angle D C A=\angle D A B\) (each \(90^{\circ}\))

⇒ \(\triangle D A C \sim \triangle D B A\) (A.A. similarity)

⇒ \(\frac{D A}{D B}=\frac{D C}{D A} \Rightarrow A D^2=B D \cdot C D\)

Hence Proved.

Question 4.ABC is an isosceles triangle right angled at C. Prove that \(A B^2=2 A C^2\).
Solution:

Given

ABC is an isosceles triangle right angled at C.

In \(\triangle A B C\),

AC = BC

and \(\angle C=90^{\circ}\)

⇒ \(A B^2=A C^2+B C^2\)

⇒ \(A B^2=A C^2+A C^2\) [from equation (1)]

⇒ \(A B^2=2 A C^2\)

Hence Proved.

Question 5. ABC is an isosceles triangle with AC = BC. If \(A B^2=2 A C^2\), prove that ABC is a right triangle.
Solution:

Given

ABC is an isosceles triangle with AC = BC. If \(A B^2=2 A C^2\)

In \(\triangle A B C\),

and AC = BC

⇒ \(A B^2\) =2 A C^2[/latex]

⇒ \(A B^2\) = \(A C^2+A C^2\)

⇒ \(A B^2\) and \(=A C^2+B C^2\)

⇒ \(\angle C\) and \(=90^{\circ}\) (from equation (1))

⇒ \(\angle C=90^{\circ}\)

⇒ \(\triangle A B C\) is a right-angled triangle.

Hence Proved.

Question 6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Solution:

Given

ABC is an equilateral triangle of side 2a.

Let ABC be an equilateral triangle in which,

AB = BC = CA = 2a

AB = BC = CA = 2a

AP \(\perp\) BC

BP = \(\frac{1}{2} B C\)

= \(\frac{1}{2}(2 a)=a\)

In \(\triangle A B P\),

⇒ \(A B^2=A P^2+B P^2\)

⇒ \(A P^2=A B^2-B P^2=(2 a)^2-a^2=3 a^2\)

⇒ \(A P=a \sqrt{3}\)

Length of each altitude = \(a \sqrt{3}\) units.

Question 7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
Solution:

Let ABCD be a rhombus whose side is ‘a’

The diagonals of the rhombus intersect each other at point O at right angle.

In right \(\triangle\) A O B,

⇒ \(O A^2+O B^2=a^2\) (from Pythagoras theorem)

⇒ \((\frac{1}{2} A C)^2+\left(\frac{1}{2} B D\right)^2=a^2\) (diagonals bisect each other)

⇒ \(\frac{1}{4} A C^2+\frac{1}{4} B D^2=a^2\)

⇒ \(A C^2+B D^2=4 a^2\)

⇒ \(A C^2+B D^2=a^2+a^2+a^2+a^2\)

⇒ \(A C^2+B D^2=A B^2+B C^2+C D^2+D A^2\)

Hence Proved

Question 8. In the figure, O is A a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB. Show that :

1.  \(O A^2+O B^2+O C^2= O D^2-O E^2-O F^2=A F^2 +B D^2+C E^2 \)
2.  \(A F^2+B D^2+C E^2=A E^2+C D^2+B F^2\)

Solution:

Given

In the figure, O is A a point in the interior of a triangle ABC, OD \(\perp\) BC, OE \(\perp\) AC and OF \(\perp\) AB.

(1) In \(\triangle A O F\),

⇒ \(A F^2=O A^2-O F^2\)

In \(\triangle B O D\),

⇒ \(B D^2=O B^2-O D^2\)

In \(\triangle C O E\),

⇒ \(C E^2=O C^2-O E^2\)

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

⇒ \(A F^2 +B D^2+C E^2\)

=\(O A^2+O B^2+O C^2-O F^2-O D^2-O E^2\)

Hence Proved.

(2) From part 1

= \(O A^2+O B^2+O C^2-O F^2-O D^2-O E^2\)

= \(\left(O A^2-O E^2\right)+\left(O B^2-O F^2\right)+\left(O C^2-O D^2\right)\)

=\(A E^2+B F^2+C D^2\)

Hence Proved.

Question 9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

Solution:

Given

A ladder 10 m long reaches a window 8 m above the ground.

Let PR is a ladder of length 10 m which reaches at a window of height 8 m from the ground. The distance of the lower end of the ladder from the base of the wall is QR.

Now, \(\angle P Q R=90^{\circ}\)

In \(\triangle P Q R\),

⇒ \(P R^2=P Q^2+Q R^2\)

⇒ \(Q R^2=P R^2-P Q^2\)

=\(10^2-8^2\)

=100-64=36

Distance of the lower end of the ladder from the base of the wall = 6 m

Question 10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Solution:

Given

A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.

Let PQ be a vertical pole of height 18 m. A wire whose length is 24 m, is tied at the upper-end P of the pole. Its other end is tied at point R on the ground.

In \(\triangle P Q R\),

⇒ \(Q R^2+P Q^2 =P R^2\)

⇒ \(Q R^2 =P R^2-P Q^2=24^2-18^2\)

=576-324=252

Q R =\(\sqrt{252}=6 \sqrt{7} \mathrm{~m}\)

Question 11. An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after \(1 \frac{1}{2}\) hours?

Solution:

Given

An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 same airport and flies due west at a speed of 1200 km per hour.

Distance covered by aeroplane in North in 1 \(\frac{1}{2}=\frac{3}{2}\) hrs is A B=1000 \(\times \frac{3}{2}=1500 \mathrm{~km}\).

Similarly, the distance covered by aeroplane in the west in \(\frac{3}{2} \mathrm{hrs}\).

⇒ \(B C=1200 \times \frac{3}{2}=1800 \mathrm{~km}\)

In \(\triangle A B C, A C^2=A B^2+B C^2\)

=\((1500)^2+(1800)^2\)

= 2250000+3240000 = 5490000

∴ \(A C=\sqrt{5490000}=300 \sqrt{61} \mathrm{~km}\)

Question 12. Two poles of heights 6 m and 1 1 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Solution :

Given

Two poles of heights 6 m and 1 1 m stand on a plane ground. If the distance between the feet of the poles is 12 m,

Let AB and CD be two vertical poles, and then

AB = 6 m, CD = 11m

and AC = 12m

Draw BE || AC, then

CE = AB = 6 m,

BE = AC = 12 m

DE = CD- CE = 11m – 6m = 5m

Now in right \(\triangle B E D\),

⇒ \(B D^2 =B E^2+D E^2\)

=\(12^2+5^2\)

=144+25

=169 \(\mathrm{~m}^2 \Rightarrow B D=13 m\)

Therefore, the distance between poles =13 m.

Question 13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C . Prove that \( A E^2+B D^2=A B^2+D E^2\).

Solution :

Given

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C .

In \(\triangle A C E, A E^2=A C^2+C E^2\) …..(1)

In \(\triangle A B C, \quad A B^2=A C^2+B C^2\) …. (2)

In \(\triangle C D E, \quad D E^2=C D^2+C E^2\) ….(3)

In \(\triangle D C B, \quad B D^2=C D^2+B C^2\) …..(4)

Adding equations (1) and (4),

⇒ \(A E^2+B D^2= \left(A C^2+C E^2\right)+\left(C D^2+B C^2\right)\)

= \(\left(A C^2+B C^2\right)+\left(C D^2+C E^2\right)\)

= \(A B^2+D E^2\) [from equations (2) and (3)]

Hence Proved.

Question 14. The perpendicular from A on side BC of a \(\triangle\)ABC intersects BC at D such that DB = 3 CD. Prove that 2AB² = 2AC² + BC²

Solution:

Given, In \(\triangle A B C\), A D \(\perp B C\) and D B=3 C D

To prove : \(2 A B^2=2 A C^2+B C^2\)

Proof: \(D B=3 C D \Rightarrow \frac{D B}{C D}=\frac{3}{1}\)

Let \(D B=3 x \quad \Rightarrow \quad C D\)=x

⇒ \(\frac{D B}{B C}=\frac{3 x}{4 x}=\frac{3}{4} \quad \Rightarrow \quad D B=\frac{3}{4} B C\)

and \(\frac{D C}{B C}=\frac{x}{4 x}=\frac{1}{4} \quad \Rightarrow \quad D C=\frac{1}{4} B C\)…

Now, from Pythagoras theorem,

⇒ \(A B^2=A D^2+B D^2=\left(A C^2-D C^2\right)+B D^2\) (from Pythagoras theorem)

=\(A C^2-\frac{1}{16} B C^2+\frac{9}{16} B C^2\)

=\(A C^2+\frac{8}{16} B C^2=A C^2+\frac{1}{2} B C^2\)

⇒ \(2 A B^2=2 A C^2+B C^2\)

Hence Proved.

Question 15. In an equilateral triangle ABC, D is a point on side BC Such That B D=\(\frac{1}{3} B C\). Prove that \(9 A D^2=7 A B^2\).

Solution :

Given

In an equilateral triangle ABC, D is a point on side BC Such That B D=\(\frac{1}{3} B C\).

Let \(A B=B C=A C=6 x\)

and \(B D=\frac{1}{3} B C=\frac{1}{3} \times 6 x=2 x\)

and B E=E C=\(\frac{B C}{2}\)=3 x ( In equilateral triangle, perpendicular bisects the base)

D E=B E-B D=3 x-2 x=x

Now, from Pythagoras theorem,

⇒ \(A B^2= A E^2+B E^2\)

= \(A D^2-D E^2+B E^2\) (from Pythagoras theorem)

⇒ \((6 x)^2 =A D^2-x^2+(3 x)^2\)

⇒ \(A D^2 =36 x^2+x^2-9 x^2=28 x^2\)

⇒ \(9 A D^2= 9 \times 28 x^2=9 \times 7 \times 4 x^2\)

=\(v7 \times 36 x^2=7(A B)^2\)

⇒ \(9 A D^2 =7 A B^2\)

Hence Proved.

Question 16. In an equilateral triangle, prove that three times the square of one, side is equal to four times the square of one of its altitudes.

Solution:

Let AB=BC = CA=a

In \(\triangle\)ADB and \(\triangle\)ADC,

⇒ \({\begin{array}{lr}
A B=A C \text { (sides of equilateral triangle) } \\
\angle 1=\angle 2 & \text { (each } 90^{\circ} \text { ) } \\
A D=A D & \text { (common) }
\end{array}}. \)

⇒ \(\triangle A D B \cong \triangle A D C\) (R.H.S.)

BD = DC =\(\frac{B C}{2}=\frac{a}{2}\) (c.p.c.t.)

Now, in \(\triangle A D B\) , from Pythagoras theorem

⇒ \(A B^2=A D^2+B D^2\)

⇒ \(a^2=A D^2+\left(\frac{a}{2}\right)^2\)

⇒ \(A D^2=a^2-\frac{a^2}{4} \Rightarrow A D^2=\frac{3 a^2}{4}\)

⇒ \(4 A D^2=3 a^2\)

⇒ \(4 A D^2=3 A B^2\)

Therefore, three times the square of the side of an equilateral triangle is equal to 4 times the square of its altitude.

Hence Proved

Question 17. Tick the correct answer and justify: In \(\triangle A B C\), A B=6 \(\sqrt{3} \mathrm{~cm}\), A C=12 cm and B C=6 \cm. The angle B is :

1. \(120^{\circ}\)
2. \(60^{\circ}\)
3. \(90^{\circ}\)
4. \(45^{\circ}\)

Solution :

⇒ \(A B^2+B C^2 =(6 \sqrt{3})^2+6^2\)

=\(108+36=144=12^2=A C^2\)

⇒ \(\angle B =90^{\circ}\)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Exercise 6.6

Question 1. n figure, PS is the bisector of \(\angle\)QPR of \(\triangle\)PQR. Prove that \(\frac{Q S}{S R}=\frac{P Q}{P R}\)

Solution:

Solution: Draw R T \\ S P which intersects the produced QP at T.

Now, R T \\ S P and R P is a transversal.

⇒ \(\angle S P R=\angle P R T\) (alternate angles) …(1)

Again, SP \\ RT and QT is a transversal.

⇒ \(\angle Q P S=\angle P T R\)

Given, \(\angle Q P S=\angle S P R \Rightarrow \angle P T R=\angle P R T\) [from equations (1) and (2)]

P R=P T

In \(\triangle Q R T\), S P \\ R T

⇒ \(\frac{Q S}{S R}=\frac{Q P}{P T} \quad \Rightarrow \quad \frac{Q S}{S R}=\frac{P Q}{P R}\) [from equation (3)]

Hence proved

Question 2. In the figure, D is a point on hypotenuse AC of \(\triangle\) ABC, such that BD \( \perp\) AC, DM \( \perp\) BC and DN \( \perp\) AB. Prove that :

1. \(D M^2=D N \cdot M C\)
2. \(D N^2=D M \cdot A N\)

Solution :

In right \(\triangle A B C\), BD is perpendicular to A-C.

⇒ \(\triangle B D C \sim \triangle A B C \sim \triangle A D B\)

(1) In \(\triangle B D C\)

⇒ \(\triangle D M \perp \sim B C\)

⇒ \(\frac{M C}{D M}=\frac{M D}{B M}\)

⇒ \(D M^2=B M \cdot M C\)

In square B M D N, \(\quad \angle D=\angle M=\angle N=90^{\circ}\)

⇒ In square B M D N is a rectangle. From equation (1),

⇒ \(D M^2=D N \cdot M C\)

Hence Proved.

(2) In \(\triangle A D B, D N \perp A B\)

In \(\triangle A N D \sim \triangle D N B, \frac{D N}{B N}=\frac{A N}{D N}\)

⇒ \(D N^2=B N \cdot A N\)

but B N=D M,

⇒ \(D N^2=D M \cdot A N\)

Hence Proved.

Question 3. Figure A B C is a triangle in which \(\angle A B C>90^{\circ}\) and \(A D \perp C B\) are produced. Prove that \(A C^2=A B^2+B C^2+2 B C \cdot B D\).

Solution :

An obtuse-angled triangle A B C in which \(\angle B\) is an obtuse angle and AD is perpendicular to produce C B,

To prove : \(A C^2=A B^2+B C^2+2 B C \times B D\)

Proof : \(\triangle A D C\) is the right triangle.

In \(\triangle A D C\), from Pythagoras theorem,

⇒ \(A C^2=A D^2+D C^2\)

⇒ \(A C^2=A D^2+(D B+B C)^2\) (D C=D B+B C)

⇒ \(A C^2=A D^2+D B^2+B C^2+2 D B \cdot B C\)

⇒ \(A C^2=\left(A D^2+D B^2\right)+B C^2+2 B C \cdot B D\)

⇒ \(A C^2=A B^2+B C^2+2 B C \cdot B D\)

Hence Proved.

Question 4. In figure, \(\triangle\)ABC is a triangle in which \(\angle\)ABC < 90° and AD \(\perp\) BC. Prove that AC² = AB² + BC² – 2BC • BD.

Solution:

Given: \(\triangle A B C\) is an acute-angled triangle and \(A D \perp B C\).

To prove: \(A C^2=A B^2+B C^2-2 B C \times B D\)

Proof: In \(\triangle A D C, \angle D=90^{\circ}\). So from Pythagoras theorem,

⇒ \(A C^2=\left(A D^2+B D^2\right)+B C^2-2 B C \times B D\)

⇒\(A C^2=A B^2+B C^2-2 B C \times B D\) (\(A D^2+B D^2=A B^2\))

Hence Proved.

Question 5. Figure AD is a median of a triangle ABC and AM \(\perp\) BC. Prove that

1. \(A C^2=A D^2+B C \cdot D M+\left(\frac{B C}{2}\right)^2 \)
2. \(A B^2=A D^2-B C \cdot D M+\left(\frac{B C}{2}\right)^2\)
3. \(A C^2+A B^2=2 A D^2+\frac{1}{2} B C^2\)

Solution:

⇒\(\triangle A B C\) is acute angled.

AD is a median.

B D=C D=\(\frac{1}{2} B C\)

AM is the altitude.

(1) In \(\triangle A C M\) ,

⇒ \(A C^2=A M^2+M C^2\)

⇒ \(A C^2=A M^2+(M D+C D)^2\)

= \(A M^2+M D^2+2 M D \cdot C D+C D^2\)

= \(\left(A M^2+D M^2\right)+(2 \cdot C D) \cdot D M+C D^2\)

= \(A D^2+(B C) \cdot D M+\left(\frac{1}{2} B C\right)^2\)

∴ \(A C^2=A D^2+B C \times D M+\frac{1}{4} B C^2\)

(2) In \(\triangle A B M\),

⇒ \(A B^2=A M^2+B M^2\)

= \(A M^2+(B D-D M)^2\)

= \(A M^2+B D^2+D M^2-2 B D \cdot D M\)

= \(A M^2+D M^2+\left(\frac{1}{2} B C\right)^2-B C \cdot D M\)

∴ \(A B^2 =A D^2-B C \times D M+\frac{1}{4} B C^2\)

Hence Proved.

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

⇒ \(A C^2+A B^2=2 A D^2+\frac{1}{2} B C^2\)

Hence Proved.

Question 6. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides

Solution:

Draw DE \(\perp\) AB and CF perpendicular to produce AB.

In \(\triangle A E D\) and \(\triangle B F C\),

AD = B C (opposite sides of the parallelogram)

⇒ \(\angle D E A =\angle C F B\) (each \(90^{\circ}\))

D E =C F (perpendicular distance between two parallel lines)

⇒ \(\triangle A E D \cong \triangle B F C\)

An E=B F (corresponding parts of congruent triangles)

= \((A F^2+C F^2)+\left(D E^2+B E^2\right)\) (from Pythagoras theorem)

= \((A B+B F)^2+\left(B C^2-B F^2\right)+\left(A D^2-A E^2\right)+(A B-A E)\) (from Pythagoras theorem)

= \(A B^2+B F^2+2 A B \cdot B F+B C^2 -B F^2+A D^2-A E^2+A B^2+A E^2 -2 A B \cdot A E\)

= \(A B^2+2 A B \cdot A E+B C^2 +A D^2+C D^2-2 A B \cdot A E\)(A E=B F and A B=C D)

= \(A B^2+B C^2+C D^2+D A^2\)

Now, L.H.S. = \(A C^2+B D^2\)

Hence Proved.

Question 7. In the figure, two chords AB and CD intersect each other at point p. Prove that :

1. \(\triangle A P C=\triangle D P B\)
2. \(A P \cdot P B=C P \cdot D P\)

Solution:

(1) In \(\triangle P^P C\) and \(\triangle D P B\),

⇒ \(\angle A P C=\angle D P B\) (vertically opposite angles)

⇒ \(\angle C A P=\angle B D P\) (angles of same segment)

⇒ \(\triangle A P C \sim \triangle D P B\) (from A.A. similarity)

Hence Proved.

(2) From part (1),

⇒ \(\triangle A P C \sim \triangle D P B \Rightarrow \frac{A P}{D P}=\frac{C P}{P B}\)

⇒ \(A P \cdot P B=C P \cdot D P\)

Hence Proved.

Question 8. In the figure, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

1. \(\triangle\)PAC – \(\triangle\)PDB
2. PA. PB = PC. PD

Solution:

(1) In \(\triangle P A C\) and \(\triangle P D B\),

⇒ \(\angle A P C =\angle D P B\) (common)

⇒ \(\angle P A C =180^{\circ}-\angle B A C\)

=\(\angle P D B\)

⇒ \(\Delta P A C \sim \triangle P D B\) (from A.A. similarity)

(2) \(\Delta P A C \sim \triangle P D B\)

⇒ \(\frac{A P}{D P} =\frac{P C}{P B}\)

⇒ \(P A \cdot P B =P C \cdot P D \)

Hence Proved.

Question 9. In the figure, D is a point on side BC of \(\triangle\) ABC such that \(\frac{B D}{C D}=\frac{A B}{A C}\). Prove that AD is the bisector of \(\angle\)BAC

Solution:

Produce BA to M such that

A M=A C

Join CM.

In \(\triangle M C\),

A M =A C

A C M =\(\angle A M C\) .

and \(\frac{B D}{C D}=\frac{A B}{A C}\)

⇒ \(\frac{B D}{C D}=\frac{B A}{A M}\) (given) from equation ( 1 ]

D A \\ C M

Now, \(\angle B A D=\angle A M C\) (corresponding angles)

and \(\angle D A C=\angle A C M\) (alternate angles)

From equations (2), (3) and (4)

⇒ \(\angle D A C=\angle B A D\)

A D bisects \(\angle B A C\).

Hence Proved.

Question 10. Nazima is fly Fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (common) (see figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Solution:

Description of figure :

C → tip of fishing rod DC

AC string, AF → fly

The initial length of the string

A C=\(\sqrt{(180)^2+(240)^2}\) (from Pythagoras theorem)

A C=\(\sqrt{32400+57600}\)

=300 cm

But, When Nazima pulls the string at the rate of 5 cm/sec, she pulls it 5 x 12 cm in 12 seconds.

So she pulls a 60 cm string from C to D. The remaining string CA will now CG – 300 – 60 = 240 cm because when the string pulls then its tip and fly both will move from A to G. So the horizontal distance of fly G from Nazima E will be EG.

Now, \([latex]G B^2= C G^2-C B^2\) (from Pythagoras theorem)

=\((240)^2-(180)^2\)

=57600-32400=25200

60 \(\times 2.64575\)=158.745 cm

Required distance =E G=G B+B E

=158.745+120

= 278.745 cm =2.79 m (approximately)

NCERT Exemplar Solutions For Class 10 Maths Chapter 6 Triangle Multiple Choice Question And Answers

Question 1. The length of the diagonals of a rhombus is 16 cm and 12 cm. The length of its side is :

1. 9 cm
2. 10 cm
3. 12 cm
4. 8 cm

Answer: 2. 10cm

Question 2. In \(\triangle A B C\) and \(\triangle P Q R\), \(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}\), then

1. \(\triangle P Q R \sim \triangle C A B\)
2. \(\triangle P Q R \sim \triangle A B C\)
3. \(\triangle P Q R \sim \triangle C B A\)
4. \(\triangle P Q R \sim \triangle B C A\)

Answer: 1. \(\triangle P Q R \sim \triangle C A B\)

Question 3. Points D and E are on the sides AB and AC respectively of an \(\triangle\)ABC such that AD = 2 cm, BD = 3 cm, BC = 7.5 cm and DE || BC, then the length of DE is :

1. 2.5 cm
2. 3 cm
3. 5 cm
4. 6 cm

Answer: 2. 3cm

Question 4. In \(\triangle\)ABC and \(\triangle\)DEF, \(\angle\)A = \(\angle\)D, \(\angle\)F = \(\angle\)C and AB = 3DE then two triangles are :

1. congruent but not similar
2. similar but not congruent
3. neither congruent nor similar
4. congruent and similar both

Answer: 2. Similar but not congruent

Question 5. In \(\triangle\)ABC and \(\triangle\)DEF, \(\frac{A B}{D E}=\frac{B C}{F D}\)then these triangles will be similar if:

1. \(\angle B=\angle E\)
2. \(\angle A=\angle D\)
3. \(\angle B=\angle D\)
4. \(\angle A=\angle\)

Answer: 3. \(\angle B=\angle D\)

Question 6. S is a point on side PQ of \(\triangle P Q R\) such that P S=Q S=R S, then :

1. \(P Q \cdot Q R=R S^2\)
2. \(P R^2+Q R^2=P Q^2\)
3. \(Q S^2+R S^2=Q R^2\)
4. \(P S^2+R S^2=P R^2\)

Answer: 3. \(Q S^2+R S^2=Q R^2\)

Question 7. If \(\triangle\)ABC ~ \(\triangle\)PQR and BC = 5QR then ar(\(\triangle\)ABC) : ar(\(\triangle\)PQR) is :

1. 1: 25
2. 25: 1
3. 1: 5
4. 5: 1

Answer: 1. 1: 25

Question 8. The areas of two similar triangles are in the ratio 4: 9. The ratio on the sides of the triangle will be :

1. 2 : 3
2. 4: 9
3. 81: 16
4. 16: 81

Answer: 1. 2 : 3

Question 9. The side of an equilateral triangle is 2a. The length of each altitude will be :

1. \(a \sqrt{2}\)
2. \(2 a \sqrt{3}\)
3. \(a \sqrt{3}\)
4. 3a

Answer: 3. \(a \sqrt{3}\)

Question 10. The corresponding sides of two similar triangles are in the ratio 9:4. The ratio of the areas of these triangles will be :

1. 2 : 3
2. 4: 9
3. 81: 16
4. 16:81

Answer: 3. 81: 16

Question 11. the following figure, a line segment PQ is drawn parallel to base BC of \(\triangle\)ABC. If PQ: BC = 1 : 3 then the ratio ofAP and PB will be :

1. 1: 4
2. 1 : 3
3. 1: 2
4. 2 : 3

Answer: 3. 1: 2

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Introduction

An average is which represents the large number of observations in a concise or single numerical data. It is a representative value around which all the values of the variable concentrate. It is called the measure of central tendency.

The word ‘average’ has been defined differently by various authors. According to Dr. Bowley, “Statistics is the science of averages.”

According to Gorton and Cowden, “An average is a single value within the range of the data used to represent all the values in the series.”

It is clear from the above definitions that an average is a single value that represents a group of values.

Read and Learn More Class 10 Maths Solutions Exemplar

Objectives Of Statistical Averages

The statistical average has the following objectives:

1. Provide a Brief Picture of Data: We can present complex data in a simple manner and concise form with the help of averages. An ordinary person can easily remember these averages.
2. Comparative Study: Measures of central value make easier the comparison of two or more than two groups. For example, it is impossible to conclude any result on the basis of marks obtained by all the students of two colleges but we can easily conclude by comparing the average marks obtained by both colleges.
3. Representation of the Group: Averages represent the picture. the whole group and that value enables Us to gel an idea of the entire data.

Properties Of Statistical Averages

Good averages should possess the following properties:

1. An average should be easy to understand and simple to compute.
2. It should be properly defined.
3. It should be based on all the observations.
4. It should be used for further statistical computation.

Types Of Averages

There are different types:

1. Mathematical Averages:
   • Arithmetic mean,
   • Geometric mean,
   • Harmonic mean.
2. Average Related with Position:
   • Median,
   • Mode

Statistics Mean

Mean is that value that can be calculated by dividing the sum of all terms of the series by the number of terms of the series.

Characteristics of Mean

1. It depends on all the terms of the group.
2. It can be calculated easily.
3. Its value is definite and based on calculations.
4. We can use algebraic methods on it.
5. The algebraic sum of deviations measured from the mean is always zero.
6. It can be determined in every condition.
7. Mean can be a value that does not exist in the series.

For example, the average of 5, 9, and 10 is \(\frac{5+9+10}{3}=8\) which is not an observation in the series. Now, we shall discuss the mean of grouped data.

Direct Method for the Mean of a Grouped Frequency Distribution

Step 1: Find the class mark x_i for each class: \(x_i=\frac{\text { lower limit }+ \text { upper limit }}{2}\)

Step 2: Find x_i for each i.

Step 3 : Find the mean using the formula, mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)

Direct Method For The Mean Of A Grouped Frequency Distribution Solved Examples

Question 1. Find the mean of the following frequency distribution by direct method:

Solution:

Now, \(\text { mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{2170}{74}=29.32\)

Question 2. If the mean of the following data is 26, then find the value of p:

Solution:

Now, arithmetic mean \(\bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}\)

⇒ \(26=\frac{890+15 p}{30+p}\)

⇒ 780 + 26p = 890 + 15p

⇒ 11p = 110

⇒ p = 10

The value of p = 10

Question 3. Find the mean from the following data:

Solution:

First, we will convert the given table into a table containing class intervals and corresponding frequencies:

Now, \(\text { mean } \bar{x}=\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{1485}{80}=18.5625\)

Assumed-Mean Method for the Mean of Grouped Frequency Distribution

Step 1: Find the class marks x_i for each class interval: \(x_i=\frac{\text { lower limit }+ \text { upper limit }}{2}\)

Step 2: Choose a suitable value of x_i from the middle values as the assumed mean and represent it by ‘a’.

Step 3: Find the deviations d_i =X_i – a for each i.

Step 4: Fin d_id_i for each i.

Step 5: Find the mean using the formula: \(\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}\)

Question 4. Find the mean for the following distribution table by shortcut method:

Solution:

Let assumed mean a = 35

⇒ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=35+\frac{160}{50}=35+3.2=38.2\)

Question 5. The height of 84 trees is given in the following table. Find their arithmetic mean by shortcut method:

Solution:

Let assumed mean a = 55

∴ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}=55+\frac{450}{84}=55+5.36=60.36 \mathrm{~cm}\)

Step Deviation Method for the Mean of Grouped Frequency Distribution

Step 1: Find the class marks x_i for each class. \(x_i=\frac{\text { lower limit }+ \text { upper limit }}{2}\)

Step 2: Choose a suitable value of x_i from the middle values as the assumed mean and represent it by ‘a’.

Step 3: Find h = upper limit – a lower limit which is the same for all the classes.

Step 4: Calculate \(u_i=\frac{x_i-A}{h}\) for each i.

Step 5: Calculate f_iμ_i for each i.

Step 6: Find the mean using the formula \(\bar{x}=a+\frac{\Sigma f_i u_i}{\sum f_i} \times h\).

An Important Result

First, we should know the meanings of step and deviation (used in the step-deviation method for finding the mean).

Deviation

All xi’s are deviated (displaced) from their places by adding or subtracting the same non-zero quantity from each value. This is called the deviation.

Step

Big jump or fall in all xi’s in multiples i.e., all xi’s are deviated from their places by multiplying or by dividing with the same number.

But if we divide by only 1, then the step-deviation method and assumed method are the same.

So, if a student is asked to find the mean by step-deviation method and he solves this by assumed mean method then there is no harm.

Question 6. Find the mean of the following table by step deviation method:

Solution:

Here, h = 3

Let assumed mean a = 13

∴ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i \cdot u_i}{\Sigma f_i} \times h=13+\frac{0}{174} \times 3=13\)

Question 7. The marks obtained by 30 students are given in the following table. Find their mean-by-step deviation method:

Solution:

Let assumed mean a = 25

Here, h = 10

∴ \(\text { mean } \bar{x}=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h=25+\frac{2}{30} \times 10=25.67\)

Question 8. Find the mean for the following data by step-deviation method:

Solution:

The given data can be written in the following form:

Now, \(\text { mean }=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h=35+\frac{-48 \times 10}{75}=28.6\)

Question 9. Find the mean for the following distribution:

Solution:

First, we will convert the given table into the exclusive form:

Here, h = 5

Let assumed mean a = 42

Now, \(\text { meàn }=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h=42+\frac{(-79) \times 5}{70}=36.36\)

Question 10. Find the class limits corresponding to each frequency if the mean of the following distribution is 33 and the assumed mean is 35:

Solution:

Here, we are given \(\bar{x}=33 \text { and } a=35\)

Now, \(\text { mean } \bar{x}=a+\frac{\Sigma f_i u_i}{\Sigma f_i} \times h\)

⇒ \(33=35+\frac{-20}{100} \times h\)

⇒ -200 = -20 h ⇒ h = 10

Now, if \(u_i=-3 \Rightarrow \frac{x_i-35}{10}=-3 \Rightarrow x_i-35=-30 \Rightarrow x_i=5\) and width i.e., h = 10, So, class limits of the first group is 0-10. (subtract and add \(\frac{h}{2}\) from lower and upper boundaries) if \(u_i=-2 ⇒ \frac{x_i-35}{10}=-2 ⇒ x_i-35=-20 x_i=15\) and width i.e., h = 10. So class limits of the second group are 10-20 (subtract and add \(\frac{h}{2}\) from lower and upper boundaries). Similarly, the class limits of other groups are 20-30, 30-40, 40-50 and 50-60.

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Median

The median is the value of the variable which divides the group into two equal parts, one part comprising all values greater and the other all values less than the median.

or

The median of a series is the value of that actual or estimated when a series is arranged in ascending or descending order which divides the distribution into two equal parts.

Properties acre Merits of Median

1. It is easy to calculate and simple to understand. In some cases, it can be located simply by inspection.
2. It is a definite average.
3. It eliminates the effect of extreme values of variable by which it is not affected.
4. It is not capable of further algebraic treatment.
5. It can be determined graphically.
6. Median can be defined for qualitative data where it is possible to rank the observations in some order.
7. Median is especially useful in the case of open-end distributions.

Demerits of Median

1. It requires the data to be arrayed which is difficult if a number of data are large.
2. Our determination is not based on all the observations.
3. The median multiplied by the number of items does arithmetic means.

Median for Ungrouped Data

Method: Arrange the data in ascending or descending order of their magnitudes. Let the total number of observations be N.

1. If N is odd, then median = \(\left(\frac{N+1}{2}\right) \text { th term }\)
2. If N is even, then median = \(\frac{\frac{N}{2} \text { th term }+\left(\frac{N}{2}+1\right) \text { th term }}{2}\)

Median for Grouped Data

Method: We make the cumulative frequency table for the grouped data. Now we will find that class interval by dividing the total number of frequencies by 2 in which the median lies. This class interval is called the median class. Now we will use the following result to find the median:

Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

where l₁ = lower limit of the median class

l₂ = upper limit of median class

i = l₂ – l₁

f = frequency of median class

N = sum of frequencies

C = cumulative frequency of the class preceding (just before) the median class

Remark:

If an inclusive (discontinuous) series is given then first of all we will prepare an exclusive series from the given inclusive series. To make it we subtract and add the same number i.e.,

⇒ \(\frac{\text { Lower limit of a class – Upper limit of previous class }}{2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Median Solved Examples

Question 1. Find the median from the following table:

Solution:

Here, N = 65

∴ For median class \(\frac{N}{2}=\frac{65}{2}=32.5\)

∴ Median class = 20-30

Here l₁=20, l₂ = 30

⇒ i = 30-20= 10,

f= 18, C= 19

∴ Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i=20+\frac{(32.5-19)}{18} \times 10=20+\frac{135}{18}=27.5\)

Question 2. Find the median from the following table:

Solution:

Here, N = 100

⇒ \(\frac{N}{2}=\frac{100}{2}=50\)

Median class = 40-55

Here l₁ = 40, l₂ = 55

⇒ i = 55 – 40 = 15, f = 44, C = 26

Now, Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i=40+\frac{(50-26)}{44} \times 15\)

= \(40+\frac{15 \times 24}{44}=40+8.18=48.18\)

Question 3. Find the median from the following data:

Solution:

Converting into a simple frequency table :

Here, N = 600

⇒ \(\frac{N}{2}=\frac{600}{2}=300\)

∴ Median class = 40-50

and l₁ = 40, l₂ = 50, i = 50 – 40 = 10, f = 200, C = 236

Now, median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i=40+\frac{(300-236)}{200} \times 10\)

= \(40+\frac{10 \times 64}{200}=40+3.2=43.2\)

Question 4. Find the median for the following frequency distribution:

Solution:

First, we will convert the given data into exclusive form:

Here, N = 400

⇒ \(\frac{N}{2}=\frac{400}{2}=200\)

Median class is 165.5 – 168.5

Now, l₁ = 165.5, l₂ = 168.5, i = 1 68.5 – 165.5 = 3, f= 136, C = 132

and median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(165.5+\frac{(200-132)}{136} \times 3\)

= \(=165.5+\frac{3 \times 68}{136}=167\)

Question 5. Find the median for the following data:

Solution:

Construct the following table from the given data:

Here, N = 50

⇒ \(\frac{N}{2}=\frac{50}{2}=25\)

The median class is 30 – 40

Now, l₁= 30, l₂ = 40, i = 40 – 30 = 10, f = 13, C = 22

and median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(30+\frac{(25-22)}{13} \times 10=32.31\)

Question 6. Find the missing frequency if the median for the given distribution is 24:

Solution:

Here, N = 62 + p

⇒ Median = 24

⇒ The median class is 20-30

∴ l₁ = 20, l₂ = 30 ⇒ i = 30- 20 = 10

f = 25, C = 30

Now, Median \(M=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

⇒ \(24=20+\frac{\left(\frac{62+p}{2}-30\right)}{25} \times 10\)

⇒ \(4=\frac{10}{25}\left(\frac{62+p-60}{2}\right)\)

20 = p + 2 ⇒ p = 18

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Mode

Mode is the value of a variable that occurs most often, i.e., the value of the observation having the maximum frequency (most frequent item of the group).

Modal Class

The class having maximum frequency is called the modal class.

Computation of Mode for a Continuous Frequency Distribution

The formula for Calculating Mode:

Mode = \(l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)

Where, l = lower limit of modal class

l = frequency of modal class

h = width of the modal class

f₀ = frequency of the class preceding the modal class

f₂ = frequency of the class following the modal class

Step 1: Determine the class of maximum frequency.

Step 2: Obtain all values required in the formula.

Step 3: Substitute values in the formula and solve.

Mode Solved Examples

Question 1. Compute the mode for the following frequency distribution:

Solution:

The modal class is 20-30 as it has the maximum frequency.

∴ I = 20, f₁ = 28, f₀ = 16, f₂ = 20

and h = 10

Mode \(M=l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

Mode \(M=20+\frac{28-16}{2 \times 28-16-20} \times 10\)

= \(20+\frac{12}{56-16-20} \times 10=20+\frac{12}{20} \times 10=26 \)

Hence, mode = 26

Question 2. Calculate the value of mode for the following frequency distribution:

Solution:

The given data is in inclusive form. So, we convert it into exclusive form, as given below:

The modal class is 12.5 – 16.5 as it has the maximum frequency.

l =12.5, f₁ = 15, f₀ = 12, f₂ = 14, h = 4

Mode = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

Mode = \(12.5+\frac{15-12}{2 \times 15-12-14} \times 4\)

= \(12.5+\frac{3}{4} \times 4=12.5+3=15.5\)

Hence, mode = 15.5

Question 3. The mode of the following series is 17.3. Find the missing frequency:

Solution:

15-20 is the modal class as mode 17.3 lies in this class.

Here, l = 15, f₁ = 24, f₀=x (say), f₂ = 17 and h = 5 and mode = 17.3

∴ Mode(M) = \(l+\frac{f_1-f_0}{2 f_1-f_0-f_2} \times h\)

⇒ \(17.3=15+\frac{24-x}{2 \times 24-x-17} \times 5\)

⇒ \(17.3=15+\frac{24-x}{31-x} \times 5\)

⇒ 2.3 (31 – x) = 120-5x

⇒ 2.3 x 31 – 2.3 x = 120 – 5x

⇒ 5x – 2.3 x = 120 – 71.3

⇒ 2.7 x = 48.7

∴ \(x=\frac{48.7}{2.7}=18.03\)

Hence, the missing frequency is 18.

Question 4. A survey regarding the heights (in cm) of 50 girls in class 10 of a school was conducted and the following data was obtained:

Find the mean, median, and mode of the above data.

Solution:

1. Mean: Mean = \(\frac{\Sigma f_i x_i}{\Sigma f_i}=\frac{7490}{50}=149.80\)

2. Median: Here, \(\frac{N}{2}=\frac{50}{2}=25\)

∴ The median class is 150 – 160

∴ Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(150+\frac{25-22}{20} \times 10\)

= 150 + 1.5 = 151.5

3. Mode: The modal class is 150-160 as it has the maximum frequency.

∴ Mode = \(\text { Mode }=l+\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right) \times h\)

= \(150+\left(\frac{20-12}{40-12-8}\right) \times 10\)

= \(150+\frac{8}{20} \times 10\)

= 150 + 4 = 154.

Hence, the mean height of the girls = 149.80 cm

the median height = 15 1.5 cm

Empirical Relationship Between Three Measures of Central Tendency

The relationship between mean, median, and mode is.

Mode = 3(Median) – 2(Mean)

and the modal height = 154 cm

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 – Empirical Relationship between Three Measures of Central Tendency Solved Examples

Question 1. Find the mean and median of the following frequency distribution:

Also, find the mode of the following data.

Solution:

We have

Let assumed mean A = 55, h= 10, Σf= 50 and Σfu = -27

Mean \(\bar{x}=A+\left[h \times \frac{\Sigma(f u)}{\Sigma f}\right]\)

⇒ \(\bar{x}=55+\left(10 \times \frac{-27}{50}\right)\)

⇒ \(\bar{x}=55-5.4\)

⇒ \(\bar{x}=49.6\)

Here, N = 50 \(\frac{N}{2}=\frac{50}{2}=25\)

Cumulative frequency greater than 25 is 36 and the corresponding class is 50-60.

∴ l₁=50,/= 12, l₂ = 60, C = 24, i = 60-50= 10

Now, \(\text { median }(M)=l_1+\frac{\left(\frac{N}{2}-C\right)}{f} \times i\)

= \(50+\frac{10}{12}(25-24)=50+\frac{10}{12}\)

= 50 + 0.83 = 50.83

Mode = 3(Median) – 2(Mean)

Mode = 3 x 50.83- 2 x 49.60

= 152.49-99.20 = 53.29

Mode = 53.29

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Cumulative Frequency Curve (Or Ogive)

An ogive is a freehold graph showing the curve of a cumulative frequency distribution.

If we plot the points talcing the upper limit of the class intervals as x coordinates and their corresponding cumulative frequencies asy coordinates and then join these points by a free hand curve, the curve so obtained is called the cumulative frequency curve.

We may follow the following steps:

1. Step 1: Construct a cumulative frequency table.
2. Step 2: Mark the actual class limits along the X-axis.
3. Step 3: Mark the cumulative frequency of respective classes along the Y-axis.
4. Step 4: Plot the points corresponding to cumulative frequency at each upper limit point.
5. Step 5: Join the points plotted by a free-hand curve.

Less Than Series

When we mark the upper-class limits along the X-axis and corresponding cumulative frequency polygon along the 7-axis, then the obtained curve is called cumulative frequency curve or ogive for less than series.

Greater Than Series

When we mark the lower class limits along the X-axis and corresponding cumulative frequency polygon along the Y-axis, then the curve so obtained is called cumulative frequency curve or ogive for greater than series.

To Obtain Median From Cumulative Frequency Curve

Method-1

Step 1: Draw a cumulative frequency curve for the given frequency distribution.

Step 2: Locate \(\frac{N}{2}\) on the Y-axis. Let it be M.

Step 3: From point M draw a line parallel to X-axis cutting the curve at point N.

Step 4: From N, draw NP perpendicular to the X-axis cutting the X-axis at P.

The value of P is the median of the data.

Method-2

Step 1: Draw both ogives (less than series and more than series) on the same axis.

Step 2: From the point of intersection of these curves draw perpendicular X-axis. Let it meet the X-axis at point M. Value of M is the required median.

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 – To Obtain Median From Cumulative Frequency Curve Solved Examples

Question 1. Draw a less than cumulative frequency curve (ogive) for the following distribution:

Solution:

Taking upper-class limits along the X-axis and corresponding cumulative frequencies along the Y-axis mark the points (10, 7) (15, 16), (20, 28), (25, 36), and (30, 42).

Join the points marked by a free-hand curve.

Remark:

Students are advised to leave some points (one on two) without joining, which do not lie on the curve. It is not necessary to join all the points.

This will not affect the median. You can imagine that the given information is not correct, but you have to make a free-hand curve.

Question 2. In a study of the cases of diabetes the following data was obtained:

Draw a less than ogive for the above data.

Solution:

The given frequency distribution is discontinuous, to convert it into a continuous distribution, we subtract 0.5 from the lower limit and add 0.5 to the upper limit respectively. So, the continuous cumulative frequency distribution table is as follows:

Required ogive is given below:

Question 3. Construct a less than ogive and a more than ogive from the following data:

Solution:

Cumulative frequency distribution table according to ages (in years) less than and more than are:

Question 4. During the medical checkup of 35 students of a class, their weights were recorded as follows:

Draw a less than type and a more than type ogive from the given data. Hence obtain the median weight from the graph.

Solution:

1. Less than Series:

We may prepare the less-than series as follows:

Scale:\(\left\{\begin{array}{l}
\text { Along the } X \text {-axis, } 5 \text { small div. }=1 \mathrm{~kg} . \\
\text { Along the } Y \text {-axis, } 10 \text { small div. }=5 \mathrm{~kg} .
\end{array}\right.\)

We plot the points A(40, 3), B(42, 5), C(44, 9), D(46, 14), E(48, 28), F(50, 32) and G(52, 35).

Join AB, BC, CD, DE, EF, and FG with a free hand to get the curve representing ‘Less than series’.

2. More than Series:

We may prepare more than one series as follows:

On the same graph-paper as above we plot the points E(38, 35), Q(40, 32), R(42, 30), S(44, 26), 7(46, 21); (7(48, 7) and 7(50, 3). Join EQ, QR, RS, ST, TV, and UV with a free hand to get the curve representing ‘More than series’.

Scale:\(\left\{\begin{array}{l}
\text { Along the } X \text {-axis, } 5 \text { small div. }=1 \mathrm{~kg} \\
\text { Along the } Y \text {-axis, } 10 \text { small div. }=5 \text { students }
\end{array}\right.\)

The two curves intersect at point L. Draw LM ⊥ OX.

∴ Median weight = OM = 46.5 kg

Question 5. Find the median for the following distribution:

Solution:

Using the given data, first of all, construct a cumulative frequency table and then draw ogive (cumulative frequency curve).

Since, the no. of terms, n = 35

∴ Median = \(\left(\frac{n+1}{2}\right) \text { th term }\)

= \(\left(\frac{35+1}{2}\right) \text { th term }=18 \text { th term }\)

Through mark 18 on the Y-axis, draw a horizontal line that meets the curve at point A.

Through point A, on the curve, the draw meets the X-axis at point B. The value of point B on the X-axis is the median.

It is clear from the cumulative frequency curve drawn that the die median is 26.

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Excercise 14.1

Question 1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Which method did you use for finding the mean, and why?

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{162}{20}=8.1\)

∴ The mean number of plants per house = 8.1.

The direct method is used here because the values of X_i and f_i are very small.

Question 2. Consider the following distribution of daily wages of 50 workers of a factory.

Find the mean daily wages of the workers of the factory by using an appropriate method.

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{27260}{50}=545.20\)

∴ Mean wages of workers = ₹ 545.20

Question 3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is ₹ 18. Find the missing frequency f:

Solution:

Now, mean pocket allowance = \(\frac{\sum f_i \cdot x_i}{\sum f_i}\)

⇒ \(18=\frac{752+20 f}{44+f}\)

⇒ 792 + 18f = 752 + 20f

⇒ 792 – 752 = 20f – 18f

⇒ 40 = 2f ⇒ f = 20

The missing frequency f = 20

Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Solution:

Let assumed mean A = 75.6

∴ Mean heartbeats per minute

⇒ \(\bar{x}=A+\frac{\sum f_i \cdot d_i}{\sum f_i}=75.5+\frac{12}{30}\)

= 75.5 + 0.4 = 75.9

The mean heartbeats per minute for these women = 75.9

Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes:

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Solution:

Here, h = 3

Let assumed mean A = 57

∴ Mean \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot u_i}{\sum f_i}\)

= \(57+\frac{3 \times 25}{400}=57+0.1875\)

= 57.1875

∴ Mean number of mangoes = 57.1875

≈ 57.19

Here, the assumed mean method is used.

Question 6. The table below shows the daily expenditure on food of 25 households in a locality.

Find the mean daily expenditure on food by a suitable method.

Solution:

Here, h = 50

Let assumed mean A = ₹ 225

∴ Mean \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot u_i}{\sum f_i}\)

= \(225+50 \times \frac{(-7)}{25}\)

= 225 – 14 = ₹ 221

Therefore, the mean daily expenditure on food per family = ₹ 211.

Question 7. To find out the concentration of SO₂ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

Find the mean concentration of SO₂ in the air.

Solution:

∴ Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{2}{30}\)

= 0.0986 parts per million.

The mean concentration of SO₂ in the air = 0.0986 parts per million.

Question 8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{499}{40}=12.475\)

∴ The mean number of absences of students = 12.475 days per student

Question 9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Solution:

Mean \(\bar{x}=\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{2430}{35}=69.43\)

∴ The mean percentage rate of literacy = 69.43

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.2

Question 1. The following table shows the ages of the patients admitted to a hospital during a year:

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

Lot assumed mean A = 40

∴ Mean \(\bar{x}=A+\frac{\sum f_1 \cdot d_i}{\sum f_i}=40+\frac{(-370)}{80}\)

= 40 – 4.625

= 35.375 years

For more,

Highest frequency = 23

∴ Modal class: 35 – 45

Now, l₁= 35, l₂ = 45

f = 23, f₁ =21, f₂ = 14

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(35+\frac{(23-21)(45-35)}{46-21-14}\)

= \(35+\frac{2 \times 10}{11}\)

= 35. 1.8 = 36.8 years.

Mode = 36.8 years.

Question 2. The following data gives information on the observed lifetimes (in hours) of 225 electrical components:

Determine the modal lifetime of the components.

Solution:

Highest frequency = 61

∴ Modal class: 60 – 80

Now, l₁ = 60, l₂ = 80

f = 61, f₁ = 52, f₂ = 38

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(60+\frac{(61-52)(80-60)}{2 \times 61-52-38}\)

= \(\begin{equation}
60+\frac{9 \times 20}{32}=60+5.625
\end{equation}\)

= 65.625 hours

Question 3. The following data gives the distribution of the total monthly household expenditure of 200 families in a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Solution:

Highest frequency = 40

∴ Modal class: 1500-2000

Now, l₁ = 1500, l₂ = 2000

f = 40, f₁ = 24, f₂= 33

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(1500+\frac{(40-24)(2000-1500)}{2 \times 40-24-33}\)

= \(1500+\frac{16 \times 500}{23}\)

= 1500 + 347.83 = ₹ 1847.83

Let assumed A = ₹ 2750

Here, h = 500

Mean Monthly expenditure

⇒ \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot u_i}{\sum f_i}\)

⇒ \(2750+\frac{500 \times(-35)}{200}\)

= 2750 – 87.50 = ₹ 2662.50

Mean Monthly expenditure = ₹ 2662.50

Question 4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools in India. Find the mode and mean of this data. Interpret the two measures.

Solution:

Here, maximum frequency = 10

∴ Model class: 30-35

Now, l₁ = 30, l₂ = 35

f = 10, f₁ = 9, f_i2 = 3

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(30+\frac{(10-9)(35-30)}{2 \times 10-9-3}\)

= \(30+\frac{5}{8}=30+0.625\)

= 30.625 students per teacher

For mean,

let assumed mean A = 27.5

Now, mean = \(A+\frac{\sum f_i \cdot d_i}{\sum f_i}=27.5+\frac{60}{35}\)

= 27.5 + 1.7

= 29.2 students per teacher

So, in the higher secondary schools of India, the mean number of students per teacher is 29.2 states while the number of students per teacher in maximum states is 30.625.

Question 5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Find the mode of the data.

Solution:

Here, maximum frequency = 18

∴ Modal class: 4000 – 5000

Now, l₁ = 4000, l₂ = 5000

f = 18, f₁ = 4, f₂ = 9

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(4000+\frac{(18-4)(5000-4000)}{2 \times 18-4-9}\)

= \(4000+\frac{14 \times 1000}{23}\)

= 4000 + 608.7 = 4608.7

≈ 4608 run (approximately)

The mode of the data = 4608 run (approximately)

Question 6. A student noted the number of cars passing through a spot on the road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:

Solution:

Here, maximum frequency = 20

∴ Model class: 40-50

Now, l₁ = 40, l₂ = 50

f = 20, f₁ = 12, f₂ = 11

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(40+\frac{(20-12)(50-40)}{2 \times 20-12-11}\)

= \(40+\frac{8 \times 10}{17}=40+4.7\)

= 44.7

The mode of the data = 44.7

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.3

Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean, and mode of the data and compare them.

Solution:

Let assumed mean A = 115

Here, h = 20

Here, N = 68

⇒ \(\frac{N+1}{2}=\frac{68+1}{2}=34.5\)

∴ Median class = 125 – 145

Now, l₁ = 125, l₂ = 145

f = 20, C = 22

and median (M) = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(125+\frac{\left(\frac{68}{2}-22\right)(145-125)}{20}\)

= \(125+\frac{12 \times 20}{20}\)

= 125 + 12 = 137

Mean \(\bar{x}=A+h \cdot \frac{\sum f_i \cdot d_i}{\sum f_i}\)

= \(115+\frac{20 \times 75}{68}\)

= 115 + 22.06 = 137.06

Again, maximum frequency = 20

∴ Modal class: 125-145

Now, l₁ = 125, l₂= 145

f = 20, f₁ = 13, f₂ = 14

∴ Mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(125+\frac{(20-13)(145-125)}{2 \times 20-13-14}\)

= \(125+\frac{7 \times 20}{13}\)

= 125 + 10.77 = 135.77

The values of median, mean and mode are equal approximately.

Question 2. If the median of the distribution given below is 28.5. find the value of x and y:

Solution:

Now, 45 + x + y = 60

⇒ x + y = 15 → (1)

Given that, median = 28.5

∴ Median class: 20 – 30

Now, l₁ = 20, l₂ = 30

f = 20, c = 5 + x, N = 60

Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

⇒ \(28.5=20+\frac{\left(\frac{60}{2}-5-x\right)(30-20)}{20}\)

⇒ \(28.5-20=\frac{(25-x)(10)}{20}\)

⇒ \(8.5=\frac{25-x}{2} \quad ⇒ \quad 17=25-x\)

⇒ x = 25 – 17 = 8

From equation (1),

8 + y = 15

⇒ y = 15 – 8 = 7

∴ x = 8, y = 7

The value of x and y is 8 and 7.

Question 3. A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons aged 18 years onwards but less than 60 years.

Solution:

N = 100

∴ \(\frac{N+1}{2}=\frac{100+1}{2}=50.5\)

⇒ Median class: 35 – 40

∴ l₁ = 35, l₂ = 40

f = 33, C = 45

Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(35+\frac{\left(\frac{100}{2}-45\right)(40-35)}{33}\)

= \(35+\frac{5 \times 5}{33}=35+0.76\)

= 35.76 years.

Median = 35.76 years.

Question 4. The lengths of40 the leaves of a plant are measured correctly to the nearest millimeter and the data obtained is represented in the following table:

Find the median length of the leaves.

[Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to 117.5-126.5, 126.5-135.5, …, 171.5 – 180.5.]

Solution:

∴ \(N=40 \Rightarrow \frac{N+1}{2}=\frac{40+1}{2}=20.5\)

⇒ Median class: 144.5 – 153.5

l₁ = 144.5, l₂ = 153.5

f = 12, C = 17

and median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(144.5+\frac{\left(\frac{40}{2}-17\right)(153.5-144.5)}{12}\)

= \(144.5+\frac{3 \times 9}{12}=144.5+2.25\)

= 146.75 mm

Therefore, the median length of leaves = 146.75 mm

Question 5. The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median lifetime of a lamp.

Solution:

Here, N = 400

∴ \(\frac{N+1}{2}=\frac{400+1}{2}=200.5\)

Median class: 3000 – 3500

Now, l₁ = 3000, l₂ = 3500

f = 86, C = 130

and median = \(=l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(=3000+\frac{\left(\frac{400}{2}-130\right)(3500-3000)}{86}\)

= \(3000+\frac{70 \times 500}{86}\)

= 3000 + 406.98 = 3406.98 hours

Therefore, the mean lifetime of lamps = 3406.98 hours.

Question 6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

Solution:

N = 100

∴ \(\frac{N+1}{2}=\frac{100+1}{2}=50.5\)

⇒ Median class: 7 – 10

Now, l₁ = 7, l₂ = 10

f = 40, C = 36

∴ Median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(7+\frac{\left(\frac{100}{2}-36\right)(10-7)}{40}\)

= \(7+\frac{14 \times 3}{40}\)

= 7 + 1.05 = 8.05

Mean = \(\frac{\sum f_i \cdot x_i}{\sum f_i}=\frac{832}{100}=8.32\)

Fir mode,

Maximum frequency = 40

∴ Modal class 7 – 10

Now, l₁ = 7, l₂ = 10

f = 40, f₁ = 30, f₂ = 16

and mode = \(l_1+\frac{\left(f-f_1\right)\left(l_2-l_1\right)}{2 f-f_1-f_2}\)

= \(7+\frac{(40-30)(10-7)}{2 \times 40-30-16}\)

= \(7+\frac{10 \times 3}{34}\)

= 7 + 0.88 = 7.88

Mode = 7.88

Question 7. The distribution below gives the weights of 30 students in a class. Find the median weight of the students.

Solution:

Here, N = 30

∴ \(\frac{N+1}{2}=\frac{30+1}{2}=15.5\)

⇒ Median class: 55 – 60

Now, l₁ = 55, l₂ = 60, f = 6, C= 13

and median = \(l_1+\frac{\left(\frac{N}{2}-C\right)\left(l_2-l_1\right)}{f}\)

= \(55+\frac{\left(\frac{30}{2}-13\right)(60-55)}{6}\)

= \(55+\frac{2 \times 5}{6}\)

= 55 + 1.67

= 56.67 kg

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Exercise 14.4

Question 1. The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less type cumulative frequency distribution, and draw its ogive.

Solution:

Cumulative frequency curve (ogive)

Question 2. During the medical check-up of 35 students of a class, their weights were recorded as follows:

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Solution:

To find the median:

1. Take the point \(\left(0, \frac{N}{2}\right) \equiv\left(0, \frac{35}{2}\right)\) on Y-axis.
2. Draw a line parallel to the X-axis from points \(\left(0, \frac{35}{2}\right)\) which is the curve at point P.
3. Find the abscissa of point P (read). It is 46.8.
4. Therefore, the required median = 46.8 kg.

Question 3. The following table gives the production yield per hectare of the heat of wheat of 100 farms in a village.

Change the distribution to a more than type distribution, and draw its ogive.

Solution:

Convert the given distribution to the ‘more than’ type distribution.

Distribution of “More than” type

NCERT Exemplar Solutions for Class 10 Maths Chapter 14 Statistics Multiple Choice Questions And Answers

Question 1. In the formula \(\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i}\) to determine the mean of grouped data, di is the deviation from and the following:

1. The lower limit of classes
2. The upper limit of classes
3. Mid-point of classes
4. Frequencies of classes

Answer: 3. Mid-point of classes

Question 2. If x_i is the mid-points of class intervals of grouped data and f_i are their corresponding frequencies, then the value of Σ(f_i – x_i – \(\bar{x}\)) is:

1. -1
2. 0
3. 1
4. None of these

Answer: 2. 0

Question 3. For the distribution

the sum of the upper limits of the median class and modal class is:

1. 70
2. 50
3. 60
4. 30

Answer: 1. 70

Question 4. The abscissa of the point of intersection of cumulative frequency curves of ‘less than type’ and ‘more than type’ gives the ______ of the data:

1. Mean
2. Median
3. Mode
4. All of these

Answer: 2. Median

Question 5. A cumulative frequency curve is necessary in

1. Mean
2. Median
3. Mode
4. All of these

Answer: 2. Median

Question 6. For the distribution

the upper limit of the median class is:

1. 19
2. 19.5
3. 29
4. 29.5

Answer: 4. 29.5

Question 7. In the distribution

the modal class is:

1. 20 – 30
2. 30 – 40
3. 40 – 50
4. 50 – 60

Answer: 3. 40 – 50

Question 8. The modal class of the following frequency distribution will be

1. 1 – 3
2. 3 – 5
3. 5 – 7
4. 7 – 9

Answer: 2. 3 – 5

Question 9. The arithmetic mean of 3, 4, 6, and x is 5. The value of x will be:

1. 5
2. 2
3. 7
4. 3

Answer: 3. 7

Question 10. The arithmetic mean of positive even numbers from 1 to 10 will be:

1. 2
2. 4
3. 6
4. 5

Solution: 3. 6

Question 11. 1 hejnean ol n observations is \(\bar{x}\). If each observation is increased by a, then the new mean will be:

1. \(\bar{x}\)+ a
2. \(\bar{x}\)+ a/2
3. \(\bar{x}\)– a
4. \(\bar{x}\)– a/2

Answer: 1. \(\bar{x}\)+ a

Question 12. For a frequency distribution, the relation between mean, median, and mode is:

1. Mode = 3 mean – 2 median
2. Mode = 2 median – 3 mean
3. Mode = 3 median – 2 mean
4. Mode = 3 median + 2 mean

Answer: 3. Mode = 3 median – 2 mean

Question 13. The mean of natural number from 1 to 9 will be:

1. 3
2. 5
3. 8
4. 9

Answer: 2. 5

Question 14. For a frequency distribution, the mean is 20.5 and the median is 20. Its mode will be:

1. 21.5
2. 17
3. 19
4. 20.5

Answer: 3. 19

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Introduction

In our daily lives, we come across some situations, for which we do not have any definite answer, although we know what the possible results (outcomes) are.

Probability For Example :

1. When a coin is tossed, it is certain that it will come down but it may turn up a head or may not turn up a head.
2. When a die is thrown, it may turn up a particular number 3 or not.
3. It may rain today.
4. The above situations predict die uncertainty (chance), therefore all are referred to a phenomenon which may or may not occur. The probability estimates the degree of uncertainty regarding the happening of a given phenomenon.

Read and Learn More Class 10 Maths Solutions Exemplar

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Experiment

An operation which gives some outcomes (results) under some given conditions is known as the ‘experiment’.

For Example:

1. “Tossing a coin” is an experiment with two outcomes: head and tail.
2. “Throwing a die” is an experiment with six outcomes: 1,2, 3, 4, 5 and 6. The plural of die is dice.

Probability Trial

Experimenting repeatedly under similar conditions is called a trial.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Sample Space

The set (collection) of all possible outcomes of usually denoted by S.

For Example:

(1). In tossing a coin ← experiment

Sample space denoted by S=\(\left\{\begin{array}{cc}
H & T \\
(\text { Head ) } & \text { (Tail) }
\end{array}\right\}\)

(2). In throwing a die ← experiment

S = {1,2, 3, 4, 5, 6}

(3). In tossing a coin 3 times ← experiment

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

(4). In throwing two dice simultaneously ← experiment

S=\(\left\{\begin{array}{l}
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6) \\
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6) \\
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6) \\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6) \\
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6) \\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Event

Any outcome or combination of some outcomes constitutes an event. Each event is a subset (part) of the sample space.

For Example:

(1). Let S = {1, 2, 3, 4, 5, 6}, when a die is thrown and let E₁ is the event of “getting an odd number”.

∴ E₁ = {1, 3, 5}

(2). Let E₂ be the event of “getting an even number”.

∴ E₂ = {2, 4, 6}

(3) Let E₃ be the event of “getting a prime number”.

∴ E₃ = {2,3, 5}

All above events are part of a subset of a sample space.

Remark: Total number of elements in a sample space = (Number of outcomes of an event)No‘of times

For Example:

(1). In tossing a coin 3 times,

The total number of elements in a sample space denoted by

n(S) = (2)^{3 → No.of times}

Possible outcomes of a coin = 8

(2). In throwing 2 dice,

n(S) = (6)^{2 → No.of times}

Possible outcomes of an experiment = 36

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Simple Event Or Elementary Event

An event is called a simple event if it contains only one element. In terms of set, it is a singleton set.

For Example:

(1) In “tossing a coin”, sample space S = {H, T}

Let A = event of getting a head = {H} and B = event of getting a tail = {T}

Here A and B are simple events.

(2). In “tossing a coin 2 times”, S = {(H, H), (H, T), (T, H), (T, T)}

Let A = event of getting a head on the first coin and getting a tail on the second coin = {(H, T)}

B = event of getting a head on the first coin = {(H, H), (H, T)}

Here, A is a simple event but B is not a simple event as B has two sample points.

To understand the experiment, sample space and event, are given below.

It is necessary that event is a part of the sample space

About the Playing Cards

1. A pack of playing cards has 52 cards in all.

2. It has 2 Colours – [Red (26 cards); Black (26 cards)]

3. It has 4 suites

Numbering on each card of each suit as :

Actually, A, J, Q and IC are written in place of 1, 11, 12 and 13.

So, in a pack of 52 cards, 4 cards of the same number are there.

Note: Kings, Queens and Jacks are called the face cards while the aces together with face cards are called the honour cards.

For Example: Number 2 will be in spade, club, heart as well as in diamond. Similarly, the number IC will be in spade, dub, heart and diamond also.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Equally Likely Events

The events are said to be equally likely when we have no reason to believe that one is more likely to occur than the other. Each of the possible outcomes has an equal chance of occurring.

For Example:

1. In “rolling a die”. There are six possible outcomes in the sample space, each is equally likely to occur.
2. In “tossing a coin”, getting a tail or a head are equally likely events.

Now, let us try to understand the meaning of probability.

(Probability: probable, perhaps, possibly, chances etc.)

For this, first, we see some experiments.

(1 ). In tossing a coin, S = {H, T}

What are the chances of getting a head—only {H}?

You will say 50% or fifty-fifty—Correct.

∴ \(50 \% \text { means } \frac{50}{100}, \text { i.e., } \frac{1}{2}=\frac{n(\text { favourable cases })}{n(S)}\)

Our favourable case is getting a head ⇒ n(E) = 1

and the total number of outcomes in a sample space ⇒ n(S)= 2

(2). In throwing a die, S = {l, 2, 3, 4, 5, 6}

What are the chances of getting an even number, i.e., {2,4, 6}?

Again, you will say 50% or fifty-fifty—Correct.

∴ Probability of an event =\(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}\)

Now, tell me O’ girl!

What is the probability of getting a head exactly 2 times when we toss a coin 3 times simultaneously?

Now, you see example (1) of sample space and give the correct answer.

Yes! Your answer is correct. It is \(\frac{3}{8}\)

Now, you can make understand to your friend easily that
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

In these 8 outcomes, only 3 underlined contain exactly two heads.

∴ Our favourable cases = 3 and total number of outcomes = 8

∴ Required probability = \(\frac{3}{8}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Minimum And Maximum Values Of Probability Of An Event

The value of the probability of any event always lies between 0 and 1. When the probability of an event is 0, it is said to be an impossible event and when the probability of an event is l, it is said to be a sure event.

Explanation: We know that a set (collection) of favourable outcomes is a part of the whole sample space.

∴ Number of favourable cases \(\leq\) Total number of cases.

⇒ \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }} \leq 1\) → (1)

Now, see the following example :

“In throwing a die numbered l to 6”, find the probability of getting a number greater than 6”.

Here there is no such possibility to get a number which is greater than 6.

Favourable number of cases ≥ 0

So. always a number of favourable cases \(\leq\) 0

∴ \(\frac{\text { Number of favourable cases }}{\text { Total number of cases }} \leq 0\)

Combining Eqs. (1) and (2), we get

∴0 \(\leq\) Probability of an event \(\leq\) 1

Remember: In each experiment, we assume that the chance of each outcome is the same.

• Die or die must be fair.
• The coin must be unbiased (We do not consider the possibility that the coin can land on its side). We will consider that coin will land on any one of its faces, i.e., either head or tail.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Theorems Of Addition Of Probabilities

If an event £ can happen in any one of the n different ways with the probabilities P₁, P₂, P₃ …. respectively, then the probability for the event £ is equal to the sum of the individual probabilities P₁, P₂, P₃, …Pn

For Example:

When a die is rolled, then any one of the numbers,1,2, 3, 4, 5 or 6 can be shown up.

Total number of outcomes = 6

Number of even numbers (2, 4 or 6)= 3

∴ Probability of getting an even number = \(\frac{\text { Favourable outcomes }}{\text { Total number of outcomes }}=\frac{3}{6}=\frac{1}{2}\)

Also, Probability of getting 2 = \(\frac{1}{6}\)

Probability of getting 4 = \(\frac{1}{6}\)

Probability of getting 6 = \(\frac{1}{6}\)

∴ Probability of getting an even number = \(\frac{1}{6}\) + \(\frac{1}{6}\) + \(\frac{1}{6}\) = \(\frac{1}{2}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Complementary Events

If E is an event, then the event ‘not £’ is a complementary event of E.

For Example:

When we throw a die, let E be the event getting a number less than or equal to 2, then the event ‘not E’, i.e., getting a number greater than 2 is a complementary event of E. Complementary of E is denoted by E or \((\bar{E})\)

∴ \(P(E)+P(\bar{E})\)=1

Let E be an event, then the number of outcomes favourable to E is greater than or equal to zero and is less than or equal to the total number of outcomes. It follows than

∴ \(0 \leq P(E) \leq 1\).

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Exhaustive Events

All the possible outcomes of any trial taken together are called exhaustive events.

For Example:

From a group of 2 boys and 3 girls, two children are selected at random. Describe the events:

S = {B₁B₂, B₁G₁, B₁G₂, B₂G₂, B₂G₁,B₂G₂, B₂G₂, G₁G₂, G₂G₃, G₁G₃}

E₁ = event that both the selected children are girls

= {G₁G₂, G₂G₃, G₁G₃}

E₂ = event that the selected children consist of 1 boy and 1 girl

= {B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₂}

E₃ = event that at least one boy is selected

= {B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃}

Here E₁ and E₂ together do not constitute S.

∴ E₁ and E₂ are not exhaustive events,

and E₂ and E₃ also do not form the whole sample space S.

∴ E₂ and E₃ are not exhaustive events.

But E₁ and E₃ together form a whole sample space.

∴ E₁ and E₃ are exhaustive events.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Solved Examples

Question 1. In any situation that has only two possible outcomes, each outcome will have probability \(\frac{1}{2}\). Find whether it is true or false.

Solution:

False,

The probability of each outcome will be \(\frac{1}{2}\) only when the two outcomes are equally likely.

For Example:

On the eve of Deepawali, we are lighting the candles in night on the roof.

The candles may be “lightening off due to heavy rain” or “they may not be lightening off” are not equally likely events because they do have not equal chances of occurring.

The possibility of falling rain at the time of the festival Deepawali is very less.

Question 2. A marble is chosen at random from 6 marbles numbered 1 to 6. Find the probability of getting a marble having number 2 and 6 on it.

Solution:

Given

A marble is chosen at random from 6 marbles numbered 1 to 6.

The favourable case is to get a marble on which both numbers 2 and 6 are written. But there is no such marble.

So, n(E) = 0 and n(S) = 6

∴ Required probability =\(\frac{n(E)}{n(S)}=\frac{0}{6}\)=0

Question 3. A marble is chosen at random from 6 marbles numbered 1 to 6. Find the probability of getting a marble having the number 2 or 6 on it.

Solution:

Given

A marble is chosen at random from 6 marbles numbered 1 to 6.

Here n(E) = 2   [either 2 or 6]

and n(S) = 6

∴ Required probability = \(\frac{2}{6}=\frac{1}{3}\)

Question 4. A fair coin is tossed

1. List the sample space.
2. What is P(tail)?

Solution:

(1). When a fair coin is tossed, there are two possible outcomes. Either the head appears on the uppermost face or the tail appears on the uppermost face.

S = {H, T}; n {S} = 2

(2). \(P(\text { tail })=\frac{\text { No. of favourable outcomes }}{\text { No. of possible outcomes }}=\frac{1}{2}\)

Question 5. A fair die is tossed. List the sample space. State the probability of the event :

1. A number less than 3 appears
2. A number greater than 3 appears
3. A number greater than or equal to 3 appears

Solution:

When a fair die is tossed, the number of possible outcomes is 1, 2, 3, 4, 5 and 6.

Sample space S = {1,2, 3,.4, 5, 6); n (S) = 6

(1). Let A be the event which denotes “a number less than 3 appears”.

A= {1,2}, n(A) = 2;

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(2). Let E be the event which denotes “a number greater than 3” appears.

E= {4,5,6}; n{E) =3

⇒ P(E)=\(\frac{n(E)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

(3). Let B be the event in which “a number greater than or equal to 3” appears.

B = {3, 4, 5, 6}; n(B) = 4

⇒ P(B)=\(\frac{n(B)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

Question 6. Two dice are thrown simultaneously. Find the probability of getting:

1. An even number is the sum
2. An even number as the product
3. The sum as a prime number
4. A total of at least 10
5. A doublet.

Solution:

Elementary events associated with the random experiment of throwing two dice are :

{1,1} {1,2} {1,3} {1,4} {1,5} {1,6}

{2,1} {2,2} {2,3} {2,4} {2,5} {2,6} .

{3,1} {3,2} {3,3} {3,4} {3,5} {3,6}

{4,1} {4,2} {4,3} {4,4} {4,5} {4,6}

{5,1} {5,2} {5,3} {5,4} {5,5} {5,6}

{6,1} {6,2} {6,3} {6,4} ‘{6,5} {6,6}

∴ Total number of elementary events = 6 x 6 = 36.

(1). Let A be the event of getting an even number as the sum, i.e., 2, 4, 6, 8, 10, 12.

The sum is even.

∴ Sum may be 2 or 4 or 6 or 8 10 or 12.

So, elementary events favourable to events are (1,1), (1,3), (3,1),(2,2), (1,5), (5, 1), (2, 4), (4, 2),'(3, 3), (2, 6), (6, 2), (4, 4), (5, 3), (3, 5), (5, 5), (6, 4), (4, 6) and (6, 6). C

clearly, a favourable number of outcomes =18

Hence, required probability = \(\frac{18}{36}=\frac{1}{2}\)

(2). Let B be the event of getting an even number as the product, i.e., 2, 4, 6, 8, 10, 12, 16, 18,20,24,30,36.

∴ Elementary events favourable to event £ are : (1, 2), (2, 1), (1, 4), (4, 1), (2, 2), (1, 6), (6, 1), (2, 3), (3, 2), (2, 4), (4, 2), (2, 5), (5, 2), (2, 6), (6, 2), (3, 4), (4, 3), (4, 4), (3, 6), (6, 3), (4, 5), (5, 4), (4, 6), (6, 4), (5, 6), (6, 5), (6, 6).

∴ Favourable number of outcomes = 27

Hence, required probability = \(\frac{27}{36}=\frac{3}{4}\)

(3). Let C be the event of getting the sum as a prime number, i.e., 2, 3, 5, 7, 11.

Elementary events favourable to event C are :(1, 1), (1 2), (2, 1),(1, 4), (4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (2, 5), (5, 2), (3, 4), (4, 3), (6, 5), (5, 6)

∴ Favourable number of outcomes =15

Hence, required probability = \(\frac{15}{36}=\frac{5}{12}\)

(4). Let D be the event of getting a total of at least 10, i.e., 10, 11, 12.

Then, the elementary events favourable to D are (6, 4), (4, 6), (5, 5), (6, 5), (5, 6) and (6, 6).

∴ Favourable number of outcomes = 6

Hence, required probability = \(\frac{6}{36}=\frac{1}{6}\)

(5). Let E be the event of getting a Doublet, i.e., (1,1), (2,2), (3,3), (4, 4), (5,5), (6,6).

Clearly, a favourable number of outcomes = 6

∴ \(P(E)=\frac{6}{36}=\frac{1}{6}\)

Question 7.

• A die is thrown once. Find the probability of getting :
  1. A prime number
  2. A number between 3 and 6
  3. A number greater than 4
  4. A number at most 4
  5. A factor of 6.
• When two dice are thrown find the probability of obtaining :
  1. A total of 6
  2. A total of 10
  3. Same number on both dice
  4. A total of 9.
• Two different dice are thrown together. Find the probability that the numbers obtained :
  1. Have a sum of less than 7
  2. Have a product of less than 16
  3. Is a doublet of odd numbers.

Solution:

1. As a die is rolled once, therefore there are six possible outcomes, i.c., 1,2,3, 4, 5, 6.

(1). Let A be an event “getting a prime number”.

Favourable cases for a prime number are 2, 3, 5,

i.e., n(A) = 3

Hence, P(A)=\(\frac{n(A)}{n(S)}=\frac{3}{6}=\frac{1}{2}\)

(2). Let A be an event “getting a number between 3 and 6”.
Favourable cases for events A are 4 or 5.

i.e., n(A) = 2

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(3). Let A be an event “a number greater than 4”.

Favourable cases for events A are 5 and 6.

i.e., n(A) = 2

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)

(4). Let A be the event of getting a number at most 4.

∴ A = {1,2, 3, 4} ⇒ n(A) = 4, n(S) = 6

∴ Required probability = \(\frac{n(A)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

(5). Let A be the event of getting a factor of 6.

∴ A = {1,2, 3, 6} r ⇒ n(A) = 4, n(S) = 6

∴ Required probability = \(\frac{n(A)}{n(S)}=\frac{4}{6}=\frac{2}{3}\)

2. Since a pair of dice is thrown once, so there are 36 possible outcomes.

i.e., n(S) = 36

(1). Let A be an event “a total 6”. Favourable cases for a total of 6 are (2, 4), (4, 2), (3, 3),(5, 1), (1,5).

i.e., n(A) = 5

Hence, P(A)=\(\frac{n(A)}{n(S)}=\frac{5}{36}\)

(2). Let A be an event “a total of 10”. Favourable cases for a total of 10 are (6, 4), (4, 6),(5,5).

i.e., n(A) = 3

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

(3). Let A be an event “the same number of both the dice”. Favourable cases for the same number on both the dice are (1, 2), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6).

i.e.. n(A)= 6

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{6}{36}=\frac{1}{6}\)

(4). Let A be an event “of getting a total of 9”. Favourable cases for a total of 9 are (3,6), (6,3), (4,5), (5,4). ”

i.e., n(A) = 4

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{4}{36}=\frac{1}{9}\)

3. We have, n(S) = 36

(1). Let A be an event “a sum less than 7” i.e., 2, 3, 4, 5, 6.

Favourable cases for a sum less than 7 are : (1, 1), (1, 2), (2, l),(l,3), (3, 1), (2, 2), (1,4), (4, 1), (2, 3), (3, 2), (1,5), (5, 1), (2, 4), (4, 2), (3, 3)

i.e., n(A) = 15

⇒ P(A)=\(\frac{n(A)}{n(S)}=\frac{15}{36}=\frac{5}{12}\)

(2). Let/1 be an event “product less than 16” i.e., 1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15.

Favourable cases for a product less than 16 are : (1, 1), (1, 2), (2, 1),(1, 3), (3, 1), (1, 4), (4, 1), (2, 2), (1, 5), (5, 1), (1, 6), (6, 1), (2, 3), (3, 2), (2, 4), (4, 2), (3, 3), (2, 5),(5, 2), (2, 6), (6, 2), (3, 4), (4, 3), (3, 5), (5, 3).

n(A) = 25

Hence, P(A)=\(\frac{n(A)}{n(S)}=\frac{25}{36}\)

(3). Let A be an event “a doublet of odd numbers”.

Favourable cases for a doublet of odd numbers are (1, 1), (3, 3), (5, 5)

i.e., n(A) = 3

Hence P(A)=\(\frac{n(A)}{n(S)}=\frac{3}{36}=\frac{1}{12}\)

Question 8. A standard deck of 52 cards is shuffled. Ritu draws a single card from the deck at random. What is the probability that the card is a jack?

Solution:

Given

A standard deck of 52 cards is shuffled. Ritu draws a single card from the deck at random.

S = Sample space of all possible outcomes, or 52 cards.

Thus, n(S) = 52

E = Event of selecting a jack. There are four jacks in the deck; jack of hearts, of diamonds, of spades, and of clubs.

Thus n(E) = 4 .

∴ P(E)=\(\frac{\text { Number of possible jacks }}{\text { Total number of possible cards }}=\frac{4}{52}=\frac{1}{13}\)

Hence, the probability that the card is jack is \(\frac{1}{13}\)

Question 9. A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is:

1. Red
2. Black or white
3. Not black.

Solution:

Given

A bag contains 5 black, 7 red and 3 white balls. A ball is drawn from the bag at random.

Total number of balls = 5 + 7 + 3= 15

1. Number of red balls = 7

∴ P(drawing a red ball) = \(\frac{7}{15}\)

2. Number of black or white balls = 5 + 3 = 8

∴ P(drawing a black or white ball) =\(\frac{8}{15}\)

3. Number of balls which are not black =15-5 = 10

∴ P(drawing a ball that is not black) = \(\frac{10}{15}\) = \(\frac{2}{3}\)

Hence, the probability of getting a red ball, a black or white ball and not a black ball are \(\frac{7}{15}, \frac{8}{15}\) and \(\frac{2}{3}\) respectively.

Question 10. The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18. What is the number of rotten apples in the heap?

Solution:

Given

The probability of selecting a rotten apple randomly from a heap of 900 apples is 0.18

Total number of apples in a heap = 900

Let the number of rotten apples in a heap = x

Number of rotten apples = \(\frac{\text { Number of rotten apples }}{\text { Total number of apples in a heap }}\)

Now, probability (rotten apple) = \(0.18=\frac{x}{900}\)

∴ x = 900 x 0.18 = 162

∴ Number of rotten apples in a heap of apples =162

Question 11. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of drawing:

1. An ace
2. ‘2’ of spades
3. ‘10’ of a black suit.

Solution:

1. The number of possible outcomes of drawing an ace is 4 as the number of aces in the deck is 4.

The total number of outcomes relating to drawing a card from a deck of 52 cards is 52.

∴ P(drawing an ace) = \(\frac{4}{52}=\frac{1}{13}\)

2. As there is only one card ‘2’ of spades, there is one favourable outcome only.

∴ Probability of drawing ‘2’ of spades = \(\frac{1}{52}\)

3. Since, there are two cards of TO’ of a black suit, one 10 of spade and the other 10 of the club, the probability of drawing 10 of a black suit = \(\frac{2}{52}=\frac{1}{26}\)

Question 12. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

1. A two-digit number,
2. A number divisible by 5.

Solution:

Given

A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box

We have, n(S) = 90

(1) Let A be the event of getting “a two-digit number”.

∴ Favourable cases are 10, 11, 12, 13, 14, …, 90.

∴ n(A)=81

∴ P(A)=\(\frac{n(A)}{n(S)}=\frac{81}{90}=\frac{9}{10}\)

(2) Let B be the event of getting “a number divisible by 5 “.

∴ Favourable cases are 10,15,20,25,30, …, 90.

Let there be n in numbers.

T_n =90

∴ 10+(n-1) 5 =90 [\(a_n\) of A.P.=a+(n-1)d]

⇒ (n-1) 5 =80 ⇒ n-1 = 16 ⇒ n=17

∴ n(B) =17

∴ P(B) =\(\frac{n(B)}{n(S)}=\frac{17}{90}\)

Question 13. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to all of the numbers 1,2, 3, …,12. What is the probability that it will point to

1. 10
2. An odd number
3. A number which is a multiple of 3?

Solution:

Given

A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to all of the numbers 1,2, 3, …,12.

n(S) = Total number of elementary events = 12.

1. Favourable number of elementary events that point to 10 = 1

∴ Required probability = \(\frac{1}{12}\)

2. Let A be the event “point of the arrowhead towards an odd number”

∴ n(A) = {1,3, 5, 7, 9, 11} = 6

∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{6}{12}=\frac{1}{2}\)

3. Let E be the event “pointing towards a number which is a multiple of 3”

∴ n(E) = {3, 6, 9, 12} = 4

∴ P(E) = \(\frac{n(A)}{n(S)}=\frac{4}{12}=\frac{1}{3}\)

Question 14. A bag contains 15 white balls and some black balls. If the probability of drawing a black ball is thrice that of a white ball, find the number of black balls in the bag.

Solution:

Given

A bag contains 15 white balls and some black balls. If the probability of drawing a black ball is thrice that of a white ball

Number of white balls in a bag =15

Suppose the number of black balls in the bag = x

Total number of balls in the bag = 15 + x

P(white ball) =\(\frac{15}{15+x}\)

P(black ball) =\(\frac{x}{15+x}\)

Now, P(black ball ) =3 P( white ball )

⇒ \(\frac{x}{15+x} =3 \times \frac{15}{15+x} \quad \Rightarrow \quad x=45\)

Therefore, the number of black balls in the bag is 45

Question 15. Find the probability of having 53 Tuesdays in a:

1. Non-leap year
2. Leap year.

Solution:

1. A non-leap year contains 365 days. So, by dividing it by 7, we get 52 weeks and 1 more day.

So, since 52 weeks are there, it means 52 Tuesdays will also be there necessarily with probability 1 and 1 more day may be either Sun. or Mon. or Tue. or Wed. or Thur. or Fri. or Sat.

So, to get 53 Tuesdays, we have to select one more Tuesday from these 7 possibilities with probability \({1}{7}\)

Therefore, probability of having 53 Tuesdays in a non-leap year = \(1 \times \frac{1}{7}=\frac{1}{7}\)

2. In a leap year, there are 366 days and 364 days make 52 weeks and therefore 52 Tuesdays. So, the probability of getting 52 Tuesdays till now is l(sure event). The remaining two days can be

Sunday – Monday
Monday – Tuesday
Tuesday – Wednesday
Wednesday – Thursday
Thursday – Friday
Friday – Saturday
Saturday – Sunday

Hence, favourable outcomes = 2 and total outcomes = 7.

Therefore, the probability of having 53 Tuesdays in a leap year is \(1 \times \frac{2}{7}=\frac{2}{7}\)

Question16. It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. What is the probability that it is a non-defective bulb?

Solution:

Given

It is known that a box of 600 electric bulbs contains 12 defective bulbs. One bulb is taken out at random from this box.

The number of non-defective bulbs in the box = 600 – 12 = 588

So, probability of taking out a non-defective bulb = \(\frac{588}{600}=\frac{49}{50}\)=0.98

Question 17. Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

1. The same day
2. Different days
3. Consecutive days?

Solution:

Given

Two customers are visiting a particular shop in the same week (Monday to Saturday). Each is equally likely to visit the shop on any day as on another day.

1. Both the customers visit the shop on the same day, i.e., there are 6 days (Monday to Saturday) and each is equally likely.

∴ Probability of both will visit on the same day

= First will go on Monday and the other will also go on Monday or the first will go on Tuesday and the other will go on Tuesday:

or first will go on Saturday and other will also go on Saturday

=\(\frac{1}{6} \times \frac{1}{6}+\frac{1}{6} \times \frac{1}{6}+\ldots+\frac{1}{6} \times \frac{1}{6}=6 \times \frac{1}{6} \times \frac{1}{6}=\frac{1}{6}\)

Alternative Method:

The first person will definitely go to the market on any of the six days with probability 1 (sure event). If both go to the market the same day then the second person has only 1 choice (only that 1 day when the first goes). So, its probability of going to the market = \(\frac{1}{6}\)

∴ Required probability = \(1 \times \frac{1}{6}=\frac{1}{6}\)

2. The first person will definitely go to the market on any of the six days with probability 1 (sure event). Since both will not go on the same day, so second person has now only 5 choices (except this 1 day when first go to the market). So, the probability of the second man go to the other day = \(\frac{5}{6}\)

∴ Required probability = \(1 \times \frac{5}{6}=\frac{5}{6}\)

3. There are 5 consecutive days, i.e., Monday Tuesday, Tuesday Wednesday, Wednesday Thursday, Thursday Friday, Friday Saturday.

∴ Required probability = 5 \times \frac{1}{6} \times \frac{1}{6}=\frac{5}{36}

Question 18. Find the probability of getting 52 Sundays in a leap year.

Solution:

A leap year has 366 days, so by dividing it by 7, we get 52 weeks and 2 days more. 52 weeks means 52 Sundays surely. Now, what will you say?

Perhaps you will say that the probability of getting 52 Sundays in a leap year is 1. Your answer is not correct. Why?

Think about the two remaining days. If from the remaining 2 days, 1 day is the Sunday, then there are 53 Sundays in a leap year.

So, the question does not end at the stage that probability is 1. We have to consider necessarily the two remaining days.

The remaining 2 days may be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).

For only 52 Sundays we want the combination of (Sunday, and Monday) or (Saturday, and Sunday not to occur. So, to get the required probability.

= (52 Sundays with probability 1) and (5 other possibilities out of 7 with probability \(\left(\frac{5}{7}\right)\))

= \(1 \times \frac{5}{7}=\frac{5}{7}\)

Question 19. If a number x is chosen at random from the numbers -2, -1, 0, 1, 2. What is the probability that \(x^2<2\)?

Solution:

We are presenting a table for x and x.

Clearly, x can take any one of the five values.

∴ Total number of cases = 5

Now, x³ < 2 when x = -1 or 0 or 1

∴ Favourable cases = 3

∴ Required probability =\(\frac{\text { Number of favourable cases }}{\text { Total number of cases }}=\frac{3}{5}\)

Question 20. A circle with a radius of 10 cm is drawn somewhere on a rectangular piece of paper of dimensions 10 cm x 30 cm, This paper is kept horizontally on the tabletop and a point is marked on the rectangular paper without seeing It. bind the probability that it will marked outside the circle. {Take \(\pi=\frac{22}{7}\)

Solution:

Given

A circle with a radius of 10 cm is drawn somewhere on a rectangular piece of paper of dimensions 10 cm x 30 cm, This paper is kept horizontally on the tabletop and a point is marked on the rectangular paper without seeing It.

Area of rectangle =40 \(\times 30 \mathrm{~cm}^2=1200 \mathrm{~cm}^2 \)

Area of circle =\(\pi(10)^2=100 \pi \mathrm{cm}^2\)

∴ Area of remaining portion =(1200-100 \(\pi) \mathrm{cm}^2\)

∴ Required probability =\(\frac{\text { Favourable portion }}{\text { Total portion }}=\frac{100(12-\pi)}{1200}\)

= \(\frac{12-\frac{22}{7}}{12}=\frac{84-22}{7 \times 12}=\frac{62}{7 \times 12}=\frac{31}{42}\)

Question 21. A square dart board is placed in the first quadrant from x = 0 to,x = 6 and = 0 to x = 6. A triangular region on the dartboard is enclosed y = 2, x = 6 and y = x. Find the probability that a dart that randomly hits the dartboard will land in the triangular region formed by the three lines.

Solution:

Given

A square dart board is placed in the first quadrant from x = 0 to,x = 6 and = 0 to x = 6. A triangular region on the dartboard is enclosed y = 2, x = 6 and y = x.

Area of square board =\((6)^2=36 \mathrm{sq}\). units

Now, DE = 6 – 2 = 4 units

BD = 6 – 2 = 4 units

∴ Area of \(\triangle B D E =\frac{1}{2} \times D E \times B D=\frac{1}{2} \times 4 \times 4\)

= 8 sq units

Hence, required probability =\(\frac{Area of \triangle B D E}{{ Area of square } O A B C}\)

= \(\frac{8}{6 \times 6}=\frac{2}{9}\)

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.1

Question 1. Complete the following statements:

1. Probability of an event E + Probability of the event ‘not E’ = ________.
2. The probability of an event that cannot happen is ________ Such an event is called ___________.
3. The probability of an event that is certain to happen is _______ Such an event is called _______.
4. The sum of the probabilities of all the elementary events of an experiment is _________.
5. The probability of an event is greater than or equal to _______ and less than or equal to ________.

Solution:

1. Probability of an event E + Probability of the event ‘not F = 1.
2. The probability of an event that cannot happen is zero. Such an event is called an impossible event.
3. The probability of an event that is certain to happen is 1. Such an event is called a certain event.
4. The sum of the probabilities of all the elementary events of an experiment is 1.
5. The probability of an event is greater than or equal to zero and less than or equal to one.

Question 2. Which of the following experiments have equally likely outcomes? Explain.

1. A driver attempts to start a car. The car starts or does not start.
2. A player attempts to shoot a basketball. She/he shoots or misses the shot.
3. A trial is made to answer a true-false question. The answer is right or wrong.
4. A baby is born. It is a boy or a girl?

Solution:

1. A driver attempts to start a car. Here the possibility of starting a car is more than that of not starting the car. So this experiment is not equally likely.
2. A player attempts to shoot a basketball.
   • In the same condition, the possibilities of shoots or misses the shoots are not the same. So this experiment is not equally likely.
3. A trial is made to answer a true-false question. The possibilities of right or wrong answers are the same. So this experiment is equally likely.
4. A baby is born. The possibilities of this baby being a boy or a girl are equal. So this experiment is equally likely.

Question 3. Why is tossing a coin considered to be a fair way of deciding which team should get the ball at the beginning of a football game?

Solution:

Tossing a coin is considered to be a fair way because the coin is symmetric and the possibilities of getting head or tail are equal.

Question 4. Which of the following cannot be the probability of an event?

1. \(frac{2}{3}\)
2. -1.5
3. 15%
4. 0.7

Solution: 2. -1.5

Because the probability of an event cannot be smaller than zero.

Question 5. If P(E) = 0.05, what is the probability of ‘not E’?

Solution :

P(E) = 0.05

Probability of ‘E-not’ = 1 – P(E)

= 1 – 0.05 = 0.95

Question 6. A bag contains lemon-flavoured candies only. Malini takes out one candy without looking into the bag. What is the probability that she takes out:

1. An orange-flavoured candy?
2. A lemon-flavoured candy?

Solution :

There are only lemon-flavoured candies in the bag.

If a candy is drawn at random from the bag, then

1. The probability of drawing an orange-flavoured candy = 0.

So. the probability of drawing an orange-flavoured candy = 0.

2. The event of drawing a lemon-flavoured candle is a certain event.

So. the probability of drawing a lemon-flavoured candy = 1

Question 7. It is given that in a group of 3 students, the probability of 2 students not having the same birthday is 0.992. What is the probability that the 2 students have the same birthday?

Solution:

The probability of 2 students not having the same birthday = 0.992

∴ The probability of 2 students having the same birthday = 1 – 0.992 = 0.008

Question 8. A bag contains 3 red balls and 5 black balls. A ball is drawn at random from the bag. What is the probability that the ball drawn is

1. Red?
2. Not red?

Solution:

Total balls 3 + 5 = 8

Total possible outcomes of drawing a ball at random from the bag = 8

1. Favourable outcomes of drawing a red ball = 3

∴ The probability of drawing a red ball

Favourable outcomes of = \(\frac{\text { drawing a red ball }}{\text { Total possible outcomes }}\)

=\(\frac{3}{8}\)

2. Probability that the ball drawn is not red = 1 – probability that the ball drawn is red.

Question 9. A box contains 5 red marbles, 8 white marbles and 4 green marbles. One marble is taken out of the box at random. What is the probability that the marble taken out will be

1. Red?
2. White?
3. Not green?

Solution:

Total marbles = 5 + 8 + 4=17 Total possible outcomes of drawing a marble at random from the box = 17

1. Favourable outcomes of drawing a red marble = 5.

∴ The probability that the marble drawn is red

Favourable outcomes of = \(\frac{\text { drawing a red ball }}{\text { Total possible outcomes }}\)

=\(\frac{5}{17}\)

2. Favourable outcomes of drawing a white marble = S

∴ Possibility of drawing a white marble

Favourable outcomes of \(\frac{\text { drawing a red ball }}{\text { Total possible outcomes }}\)

= \(\frac{8}{17}\)

3. Favourable outcomes of drawing a green marble = 4

⇒ Favourable outcomes that the marble drawn is not green = 17 – 4 =13

∴ The probability that the marble drawn is not green

Favourable outcomes = \(\frac{\text { marble drawn is not green }}{\text { Total possible outcomes }}\)

= \(\frac{13}{17}\)

Question 10. A piggy bank contains a hundred 50 p coins, fifty ₹ 1 coins, twenty ₹ 2 coins and ten ₹ 5 coins. If it is equally likely that one of the tyre coins will fall out when the bank is turned upside down, what is the probability that the coin?

1. Will be a 50 p coin?
2. Will not be a ₹ 5 coin?

Solution:

No. of coins of 50 paise =100

No. of coins of ₹ 1 =50

No. of coins of ₹ 2 =20

No. of coins of ₹ 5 = 10

Total coins = 100 + 50 + 20 + 10 = 180

Total outcomes that one coin will fall out = 180

1. The favourable outcomes that the coin fall out is 50 paise = 100

∴ Probability that coin fall out is of 50 paise =\(\frac{100}{180}=\frac{5}{9}\)

2. The favourable outcome that the coin falls out is ₹ 5 = 10

∴ Probability that coin fall out is of Rs 5 = =\(\frac{10}{180}=\frac{1}{18}\)

Therefore, the probability that the coin falls out is not ₹ 5

= \(1-\frac{1}{18}=\frac{17}{18}\)

Question 11. Gopi buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 5 male fish and 8 female fish. What is the probability that the fish taken out is a male fish?

Solution :

Total no. of fish = 5 + 8 = 13

Total possible outcomes of taking a fish = 13

Favourable outcomes of taking a male fish = 5

Probability of a male fish

Favourable outcomes

= \(\frac{\text { drawing a male fish }}{\text { Total no. of possible outcomes of drawing a fish }}\)

= \(\frac{5}{3}\)

Question 12. A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1,2,3,4,5,6,7,8. and these are equally likely outcomes.

What is the probability that it will point at:

1. 8?
2. An odd number?
3. A number greater than 2?
4. A number less than 9?

Solution :

Total no. of outcomes of pointing by an arrow =8

1. Favourable outcomes that the arrow indicates the number 8 = 1

∴ Probability of this event = \(\frac{1}{8}\)

Favourable outcomes that the arrow indicates an odd number = {1,3,5, 7} = 4

∴ Probability of this event = \(\frac{4}{8}=\frac{1}{2}\)

3. Favourable outcomes that the arrow indicates a number greater than 2 = {3,4,5,6,7,8} = 6

∴ Probability of this event = \(\frac{6}{8}\) =\(\frac{3}{4}\)

4. Favourable outcomes that the arrow indicates a number less than 9. = {1,2, 3, 4, 5, 6, 7, 8} = 8

∴ Probability of this event = \(\frac{8}{8}\)=1

Question 13. A die is thrown once. Find the probability of getting:

1. A prime number
2. A number lying between 2 and 6
3. An odd number.

Solution:

Possible outcomes in one throw of a die = {1,2,3,4,5,6}

Total possible outcomes = 6

1. Prime numbers = {2, 3, 5} = 3

∴ Probability of getting a prime number = \(\frac{3}{6}=\frac{1}{2}\)

2. Numbers lying between 2 to 6 = {3,4,5} = 3

∴ Probability of this event = \(\frac{3}{6}=\frac{1}{2}\)

3. Odd numbers = {1, 3, 5} = 3

∴ Probability of this event =\(\frac{3}{6}=\frac{1}{2}\)

Question 14. One card is drawn from a well-shuffled deck of 52 cards. Find the probability of getting:

1. A king of red colour
2. A face card
3. A red-face card
4. The jack of hearts
5. A spade
6. The Queen of diamonds

Solution:

Total possible outcomes of drawing a card from the pack of cards = 52.

1. Favourable outcomes of getting a king of red colour = 2.

∴ Probability of drawing a icing of red colour = =\(\frac{2}{52}=\frac{1}{26}\) .

2. Favourable outcomes of getting a face card = 12

∴ Probability of drawing a face card = \(\frac{12}{52}=\frac{3}{13}\)

3. Favourable outcomes of getting a face card of red colour = 6

∴ Probability of drawing a face card of red colour =\(\frac{6}{52}=\frac{3}{26}\)

4. Favourable outcomes of drawing a jack of heart = 1

∴ Probability of drawing a jack of heart = \(\frac{1}{52}\)

5. Favourable outcomes of drawing a card of spade = 13

∴ Probability of drawing a card of spade = \(\frac{13}{52}={1}{4}\)

6. Favourable outcomes of drawing a queen of diamonds = 1

∴ Probability of drawing a queen of diamond = \(\frac{1}{52}\)

Question 15. Five cards—the ten, jack, queen, king and ace of diamonds, are well-shuffled with their face downwards. One card is then picked up at random.

1. What is the probability that the card is the queen?
2. If the queen is drawn and put aside, what is the probability that the second card picked up is
   1. An ace?
   2. A queen?

Solution:

Total favourable outcomes of drawing a card from the five playing cards = 5

1. Favourable outcomes of drawing a queen = 1

∴ Probability of drawing a queen = \(\frac{1}{5}\)

2. Remaining cards when ‘queen’ is drawn and put aside = 4

(1). Favourable outcomes of drawing a card of ace = 1

∴ Probability of drawing the second card ace = \(\frac{1}{4}\)

(2). Favourable outcomes of drawing a queen = 0

∴ Probability of drawing the second card of queen = \(\frac{0}{4}\)=0

Question 16. 12 defective pens are accidentally mixed with 132 good ones. It is not possible to just look at a pen and tell whether or not it is defective. One pen is taken out at random from this lot. Determine the probability that the pen taken out is a good one.

Solution :

No. of good pens = 132

No. of defective pens = 12

Total pens = 132 + 12 = 144

Total favourable outcomes of drawing a pen = 144

Favourable outcomes of drawing a good pen = 132

∴ Probability of drawing a good pen = \(\frac{132}{144}\) = \(\frac{11}{12}\)

Question 17.

1. A lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective?
2. Suppose the bulb drawn in (1) is not defective and is not replaced. Now one bulb is drawn at random from the rest. What is the probability that this bulb is not defective?

Solution:

Total bulbs = 20

Defective bulbs = 4

1. If a bulb is drawn at random then total possible outcomes = 20

Favourable outcomes of a defective bulb = 4.

∴ Probability of a defective bulb = \(\frac{4}{20}=\frac{1}{5}\)

2. If the bulb drawn is not defective and it is not replaced. Now a bulb is drawn then

Total possible outcomes =19

Favourable outcomes of a defective bulb = 4

∴ Probability of a defective bulb = \(\frac{4}{19}\)

⇒ Probability of a non-defective bulb =\(1-\frac{4}{19}=\frac{15}{19}\)

Question 18. A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears

1. A two-digit number
2. A perfect square number
3. A number divisible by 5.

Solution:

Total no. of discs = 90

Possible outcomes of drawing a disc = 90

1. Two digit numbers = {10, 11, 12, …90}

Favourable outcomes of drawing a two’ digit number = 81

Now, the probability of getting a two-digit number on the disc =\(\frac{81}{90}=\frac{9}{10}\)

2. Perfect square numbers in given numbers
= {1,4, 9, …,81}

⇒ Favourable outcomes of drawing a perfect square number = 9

Now, the probability that the number marked on the disc is a perfect square =\(\frac{9}{90}=\frac{1}{10}\)

3. Numbers divisible by 5 in the given numbers = (5, 10, 15,…, 90}

⇒ Favourable outcomes of a number drawn divisible by 5 = 18

Now, the probability that the number marked on the disc is divisible by 5 = \(\frac{18}{90}=\frac{1}{5}\)

Question 19. A child has a die whose six faces show the letters as given below:

A B C D E A

The die is thrown once. What is the probability of getting

1. A?
2. D?

Solution:

Total possible outcomes in a throw of die = 6

1. Favourable outcome of getting A = 2

∴ Probability of getting A = \(\frac{2}{6}=\frac{1}{3}\)

2. Favourable outcomes of getting D = 1

∴ Probability of getting D = \(\frac{1}{6}\)

Question 20. Suppose you drop a die at random on the rectangular region shown. What is the probability that it will land inside the circle with a diameter of 1 m?

Solution:

Area of rectangle = \(3 \times 2=6 \mathrm{~m}^2\)

Diameter of circle =1 m

⇒ Radius of circle =\(\frac{1}{2} \mathrm{~m}\)

⇒ Area of circle =\(\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4} \mathrm{~m}^2\)

Now, the probability that in one throw of a die, the die lands inside the circle

= \(\frac{\text { Area of circle }}{\text { Area of rectangle }}\)

= \(\frac{\pi / 4}{6}=\frac{\pi}{4 \times 6}=\frac{\pi}{24}\)

Question 21. A lot consists of 144 ball pens of which 20 are defective and the others are good. Nuri will buy a pen if it is good, but will not buy if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that:

1. She will buy it?
2. She will not buy it?

Solution:

Total number of ball pens = 144

No. of defective ball pens = 20

∴ No. of good ball pens = 144 – 20 = 124

No. of favourable outcomes of drawing a pen randomly = 144

1. To buy the pen, it must be a good one.

∴ Favourable outcomes of drawing a good ball pen=124

⇒ Probability of buying a good ball pen =\(\frac{124}{144}=\frac{31}{36}\)

2. If the ball pen is defective then she will not buy it.

∴ Favourable outcomes of drawing a defective ball pen = 20

⇒ Probability of drawing a defective ball pen =\(\frac{20}{144}=\frac{5}{36}\)

Question 22. Two dice, one blue and one grey, are thrown at the same time,

1. Complete the following table:

2. A student argues that ‘there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability \(\frac{1}{11}\). Do you agree with this argument? Justify your answer.

Solution :

1. In a throw of two dice, possible outcomes are

⇒ \(\left\{\begin{array}{llllll}
(1,1), & (1,2), & (1,3), & (1,4), & (1,5), & (1,6) \\
(2,1), & (2,2), & (2,3), & (2,4), & (2,5), & (2,6) \\
(3,1), & (3,2), & (3,3), & (3,4), & (3,5), & (3,6) \\
(4,1), & (4,2), & (4,3), & (4,4), & (4,5), & (4,6) \\
(5,1), & (5,2), & (5,3), & (5,4), & (5,5), & (5,6) \\
(6,1), & (6,2), & (6,3), & (6,4), & (6,5), & (6,6)
\end{array}\right\}\)

Total possible outcomes = 36

Outcome of getting the sum 2 = (1, 1)

Favourable outcomes of getting a sum 2 = 1

∴ Probability of getting the sum 2 = \(\frac{1}{36}\)

Outcomes of getting the sum 3 = (1, 2), (2, 1)

Favourable outcomes of getting the sum 3 = 2

∴ Probability of getting the sum 3 = \(\frac{2}{36}={1}{18}\)

Outcomes of getting the sum 4 = (1,3), (2, 2), (3, 1)

Favourable outcomes of getting the sum 4 = 3

∴ Probability of getting the sum 4 = \(\frac{3}{36}={1}{12}\)

Outcomes of getting the sum 5 = (1,4), (2, 3), (3, 2), (4, 1)

Favourable outcomes of getting the sum 5=4

∴ Probability of getting the sum 5 = \(\frac{4}{36}={1}{9}\)

Outcomes of getting the sum 6 = (1,5), (2, 4), (3, 3), (4, 2), (5, 1)

Favourable outcomes of getting the sum 6 = 5

∴ Probability of getting the sum 6 = \(\frac{5}{36}\)

Outcomes of getting the sum 7 = (1,6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)

Favourable outcomes of getting the sum 7 = 6

∴ Probability of getting the sum 7 = \(\frac{6}{36}={1}{6}\)

Outcomes of getting the sum 8 = (2, 6), (3, 5), (4, 4), (5, 3), (6, 2)

Favourable outcomes of getting the sum 8 = 5

∴ Probability of getting the sum 8 = \(\frac{5}{36}\)

Outcomes of getting the sum 9 = (3,6), (4, 5), (5, 4), (6, 3)

Favourable outcomes of getting the sum 9 = 4

∴ Probability of getting the sum 9 = \(\frac{4}{36}={1}{9}\)

Outcomes of getting the sum 10 = (4, 6),(5, 5), (6, 4)

Favourable outcomes of getting the sum 10 = 3

∴ Probability of getting the sum 10 = \(\frac{3}{36}={1}{12}\)

Outcomes of getting the sum 11 = (5, 6), (6, 5)

Favourable outcomes of getting the sum 11 = 2

∴ Probability of getting the sum 11= \(\frac{2}{36}={1}{18}\)

Outcomes of getting the sum 12 = (6, 6)

Favourable outcomes of getting the sum 12 = 1

∴ Probability of getting the sum 12 = \(\frac{1}{36}\)

Now, the given table will be as follows:

2. The student’s argument is wrong because the favourable outcomes of every event are different.

Question 23. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanifwins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise. Calculate the probability that Hanif will lose the game.

Solution :

Given

A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Hanifwins if all the tosses give the same result i.e., three heads or three tails, and loses otherwise.

All possible outcomes in three throws of a coin of ₹ 1.

HHH, HHT, HTH, THH, HTT, THT, TTH,

TTT → Total possible outcomes = 8

Favourable outcomes of 3 heads = 1

Favourable outcomes of 3 tails = 1

∴ Favourable outcomes to win = 2

Now, favourable outcomes to lose = 8 – 2 = 6

Therefore, the probability to lose

= \(\frac{\text { Favourable outcomes to lose }}{\text { Total possible outcomes }}\)

= \(\frac{6}{8}=\frac{3}{4}\)

Question 24. A die is thrown twice. What is the probability that:

1. 5 will not come up either time?
2. 5 will come up at least once?

[Hint: Throwing a die twice and throwing two dice simultaneously are treated as the same experiment]

Solution:

In two throws of a die, the all possible outcomes are :

(1.1), (1,2), (1,3), (1,4), (1,5), (1,6)

(2.1), (2,2), (2,3), (2,4), (2,5), (2,6)

(3.1), (3,2), (3,3), (3,4), (3,5), (3,6)

(4.1), (4,2), (4,3), (4,4), (4,5), (4,6)

(5.1), (5,2), (5,3), (5,4), (5,5), (5,6)

(6.1), (6,2), (6,3), (6,4), (6,5), (6,6)

Total possible outcomes = 36

Outcomes in which 5 comes up = 11

Outcomes in which 5 does not come up = 36 – 11 = 25

1. Favourable outcomes that 5 does not come up = 25

∴ Probability of not getting 5 = \(\frac{25}{36}\)

2. Favourable outcomes that 5 will come up at least once = 11

∴ Probability that 5 will come up at least once = \(\frac{11}{36}\)

Question 25. Which of the following arguments are correct and which are not correct? Clive reasons for your answer.

1. If two coins are tossed simultaneously there are three possible outcomes—two heads, two tails or one of each. Therefore, for each of these outcomes, the probability is \(\frac{1}{3}\)
2. If a die is thrown, there are two possible outcomes—an odd number or an even number. Therefore, the probability of getting an odd number is \(\frac{1}{2}\)

Solution:

1. When two coins are tossed together, then there are not three outcomes, but the following four outcomes are obtained:

HH, HT, TH, TT

So, the .ugumom of the student is false.

2. In one throw of a die

All possible outcomes {1, 2, 3, 4, 5, 6} 6

Outcomes of getting an even number = {2. 4, 6} = 3

Outcomes of gelling an odd number = (1,3,5} =3

∴ Possibility of getting an odd number= \(\frac{3}{6}=\frac{1}{2}\)

Therefore, the argument of the student is true.

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Exercise 15.2

Question 1. Two customers Shvam and Ehta visit a particular shop in the same week (Tuesday to Saturday), Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on

1. The same day?
2. Consecutive days?
3. Different days?

Solution:

Hence, the total possible outcomes are as follows:

Toral possible outcomes = 25

1. Outcomes that both customers visit on the same day

= (T. T), (W, W), (Th. Th), (F, F), (S, S)

Total favourable outcomes = 5

So. is probability Drat Bodi will visit on the same day

= \(\frac{5}{25}=\frac{1}{5}\)

2. Outcomes that both customers visit on consecutive days

= (T, W), (W; Th), (Th, F), (F, S), (W, T), (Th, W), (F, Th), (S, F)

Total favourable outcomes = 8

So, the probability that both visit on consecutive days the shop

= \(\frac{8}{25}\)

3. Probability that both visit the shop on the same day = \(\frac{1}{5}\) [from part (1)]

∴ The probability that both visit the shop on different days

1= \(\frac{1}{5}= {4}{5}\)

Question 2. A die is numbered in such a way that its faces show the numbers 1,2,2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a tew values of the total score on the two throws:

Number in the first throw What is the probability that the total score is

1. Even?
2. 6?
3. At least 6?

Solution:

Total possible outcomes = 36

1. Favourable outcomes of getting the sum as an even number = 18

∴ Probability of getting the sum as even number = \(\frac{18}{36}=\frac{1}{2}\)

2. Favourable outcomes of getting a sum 6 = 4

∴ Probability of getting a sum 6= \(\frac{4}{36}=\frac{1}{9}\)

3. Favourable outcomes of getting a sum less than 6=15.

∴ Probability of getting a sum less than 6 =\(\frac{15}{36}=\frac{5}{12}\)

Question 3. A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.

Solution:

Given

A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball

Let the number of blue balls =x

Given, the number of red balls = 5

Total balls = x + 5

∴ The probability of drawing a blue ball

⇒ P(B)=\(\frac{x}{x+5}\)

and the probability of drawing a red ball

⇒ P(R)=\(\frac{5}{x+5}\)

Given that, P(B) =2 \(\times P(R)\)

⇒ \(\frac{x}{x+5} =\frac{2 \times 5}{x+5}\)

⇒ x =10

⇒ Number of blue balls =10

Question 4. A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball?

If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.

Solution:

Given

A box contains 12 balls out of which x are black. If one ball is drawn at random from the box,

Total balls in the box = 12

Black balls = x

∴ Probability of drawing a black ball = \(\frac{x}{12}\)

On putting 6 more black balls in the box, Total balls = 12 + 6= 18 Black balls = x + 6

Now, probability of drawing a black ball = \(\frac{x+6}{12}\)

Given that, \(\frac{x+6}{18} =2 \times \frac{x}{12}\)

⇒ \(\frac{x+6}{18} =\frac{x}{6}\)

⇒ x + 6 = 3 x

⇒ 2 x = 6 ⇒ x = 3

Question 5. A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green 2 is \(\frac{2}{3}\). Find the number of blue balls in the jar.

Solution:

Given

A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green 2 is \(\frac{2}{3}\).

Total marbles in jar = 24

Let green marbles in jar = x

∴ Probability of drawing 1 green marble from jar = \(\frac{x}{24}\)

Given that, \(\frac{x}{24}=\frac{2}{3}\)

x = \(\frac{2}{3} \times 24\)

⇒ x = 16

∴ Green marbles = 16

⇒ Blue marbles = 24 – 16 = 8

NCERT Exemplar Solutions for Class 10 Maths Chapter 15 Probability Multiple Choice Question And Answers

Question 1. The probability of a sure event is:

1. 0
2. 1
3. -l
4. \(\frac{1}{2}\)

Answer: 2. 1

Question 2. The sum of the probabilities ofan event and its complementary event is:

1. -1
2. 0
3. 1
4. None of these

Answer: 3. 1

Question 3. The probability of winning a match by Ravi is \(\frac{2}{5}\). The probability of his losing the match is

1. \(\frac{3}{5}\)
2. \(\frac{2}{5}\)
3. 1
4. 0

Answer: 1. \(\frac{3}{5}\)

Question 4. If P(A) represents the probability of an event, A, then:

1. P(A)<0
2. P(A)>1
3. 0 ≤ P(A) ≤1
4. -1 ≤ P(A)≤ 1

Answer: 3. 0 ≤ P(A) ≤1

Question 5. When a dice is thrown, the probability of getting an even number less than 3, is:

1. 0
2. \(\frac{1}{2}\)
3. \(\frac{1}{3}\)
4. \(\frac{1}{6}\)

Answer: 4. \(\frac{1}{6}\)

Question 6. The probability of selecting a prime number from the numbers 1 to 20, is:

1. \(\frac{7}{20}\)
2. \(\frac{2}{5}\)
3. \(\frac{9}{20}\)
4. \(\frac{1}{2}\)

Answer: 2. \(\frac{2}{5}\)

Question 7. In a year, which is not a leap year, the probability of 53 Monday is:

1. \(\frac{6}{7}\)
2. \(\frac{3}{7}\)
3. \(\frac{2}{7}\)
4. \(\frac{1}{7}\)

Answer: 4. \(\frac{1}{7}\)

Linear Equations In Two Variables Introduction

First of all we shall understand the meaning of terms used in the name of the chapter “Linear equations in two variables”.

Consider the following equations:

⇒ \(3 x-8 y=21,2 x+5 y=14, \frac{7}{4} x+\frac{9}{5} y=\frac{5}{6}\)

Read and Learn More Class 10 Maths Solutions Exemplar

Each of these equations contains two variables over R (real numbers). An equation of the form an ax + by + c = 0, where a, b, c are real numbers (a ≠ 0, b ≠ 0) is called a linear equation in two variables x and b, where a is the coefficient of x, b is the coefficient of y and c is a constant term.

Solution Of The Simultaneous Linear Equations

Any pair of values of x and y which satisfies the equation is called the solution of the equation. It means a solution of two equations is the point (x, y) of the intersection of two straight lines or any two curves. This point will lie on both the equations (curves).

If the equation contains only one variable, the graph of such type of equation will be a straight line parallel to the x-axis or y-axis according to a form of equation as ay = b or ax = b, where a ≠ 0. The graph of a linear equation in two variables is also a straight line.

The general form of a pair of linear equations in two variables x and y is

a₁x +b₁y +c₁ = 0

a₂x + b₂y + c₂ = 0

where a₁,b₁,C₁,a₂,b₂ and C₂ are all real numbers.

A system of linear equations will have either a unique solution or infinitely many solutions or no solution. If a system of simultaneous linear equations has a solution (either unique or infinitely many), then the system is said to be consistent otherwise it is said to be an inconsistent system.

Consistent System: A system of simultaneous linear equations is said to be consistent if it has at least one solution.

In-consistent System: A system of simultaneous linear equations is said to be inconsistent if it has no solution.

Graphical Method To Solve The Simultaneous Linear Equations

Let the given system of linear equations be

a₁x +b₁y +c₁ = 0 ……(1)

a₂x + b₂y + c₂ = 0 ……(2)

A pair of linear equations in two variables will be represented by two straight lines, both to be considered together. The following three possibilities can happen:

1. The two lines intersect at exactly one point. Let the coordinates of a point of intersection be (a, b).
Solution:

Given

The two lines intersect at exactly one point. Let the coordinates of a point of intersection be (a, b)

Then, x = a and y = b are the unique solutions of the given system of equations.

2. The two lines are parallel to each other.
Solution:

Then there is no common solution to the given system of equations. Thus, in this case, the given system is inconsistent.

3. One line overlaps the other i.e., the two lines are coincident.
Solution:

Then the given system of equations has infinitely many solutions.

Simultaneous Linear Equations Solved Examples

Question 1. Solve graphically the system of linear equations x +y =10 and x-y = 4.
Solution:

Given

x +y =10 and x-y = 4

For x +y= 10, it can be written as y = 10- x

For equation x-y = 4, it can be written as y = x- 4

Now we plot the points A(10, 0), B(0, 10) and P(4, 0), Q(0, -4)

We observe that the two lines representing the two equations intersect at points (7, and 3). Hence, \(\left.\begin{array}{l}
x=7 \\
y=3
\end{array}\right\}\) is a solution of given equation.

Question 2. Show graphically that the system of equations 2x + 4y = 10 and 3x + 6y = 12 has no solution.
Solution:

The given equation is

2x + 4y = 1 0

4y = 1 0 – 2x

⇒ \(y=\frac{5-x}{2}\)

Now for equation 3x + 6y – 12

6y = 12-3x

⇒ \(y=\frac{4-x}{2}\)

Now plot the points 2l(1, 2), B(3, 1), C(5, 0) and P(2, 1), Q(0, 2), R(4, 0) on the same graph paper.

We observe from the graph that the two lines are parallel to each other i.e., they do not intersect each other at any point. Hence, the given system of equations has no solution.

Question 3. Show graphically that the system of equations 3x -y=2, 9x-3y = 6 has an infinite = number of solutions. When this system has infinitely many solutions, then find any two solutions. Also, find the y-intercept made by this line.
Solution:

For equation 3x -y = 2

y = 3x-2 ……..(1)

and for equation 9x- 3y = 6

3y = 9x- 6

y = \(\frac{9 x-6}{3}\)

Now, plot the points A(-1, -5), B(0, -2), C(2, 4) and P(-2, -8), Q(1, 1), R(2, 4) on the graph paper. We observe from the graph that the two lines coincide. Hence, a given system of equations has an infinite number of solutions.

For any two solutions, Put x = 0 in equation (1), we get = -2 If we put x = 1 in the equation (1), we get = 1

So, two solutions are \(\left.\left.\begin{array}{l}
x=0 \\
y=-2
\end{array}\right\} \text { and } \begin{array}{l}
x=1 \\
y=1
\end{array}\right\}\)

This line (both lines are the same) cuts the y-axis at 2 units below the origin. So, its y-intercept (part) is -2

Algebraic Methods Of Solving Simultaneous Linear Equations In Two Variables

Elimination by “Substitution” Method (or Simply “Substitution Method”)

Step 1: From any of the given two equations, find the value of one variable in terms of other.

Step 2: Substitute the value of the variable obtained in step 1 in the other equation. In this way, have eliminated one variable with the help of substitution. So the name fell as “elimination by substitution”. Now, solve it to get the value of one variable.]

Step 3: Substituting the value of the variable obtained in step 2 in the result of step 1 get the value of the remaining unknown variable.

Algebraic Methods Solved Examples

Question 1. Solve x +y = 7 and 3x- 2y = 11.
Solution:

Given x +y = 7 …(1)

and 3x-2y= 11 …(2)

From equation (1) we get y = 7 -x …(3)

Substituting the value of y from equation (3) in (2), we get

3x-2 (7-x) = 11

3x- 14 + 2r= 11

5x- 14= 11

5x = 25 or x = 5

Substituting the value of x in equation (3), we get

y = 7 – 5 ⇒ y = 2

∴ Solution is \(\left.\begin{array}{l}
x=5 \\
y=2
\end{array}\right\}\)

Question 2. Solve 15x- 8y = 29 and 17x + 12y = 75.
Solution:

Given equations are 15x- 8y = 29 ….(1)

and 17x + 12y = 75 …….(2)

From equation (1), we get

⇒ \(x=\frac{29+8 y}{15}\) …..(3)

Substituting the value of x from equation (3) in (2), we get

⇒ \(17\left(\frac{29+8 y}{15}\right)+12 y=75\)

⇒ 493 + 136y + 180y = 1125

⇒ 316y= 1125-493

⇒ 316y = 632

⇒ y = 2

Substituting the value of y in equation (3) we get

⇒ \(x=\frac{29+8 \times 2}{15}\)

⇒ \(x=\frac{45}{15}\)

⇒ x = 3

Hence, the solution is \(\left.\begin{array}{l}
x=3 \\
y=2
\end{array}\right\}\)

Elimination by Equating the Coefficients (Elimination Method)

Step 1: Multiply both the equations by a suitable number, so that the coefficients of x or the coefficients of y in both the equations become equal.

Step 2: If these coefficients are of the same sign then subtract the new equations otherwise add them.

Step 3: Solve the resulting equation to find the value of unknown. In this process, you will see that we have eliminated one variable. So, the name fell as “elimination by equating the coefficients”.

Step 4: Substitute this value in any of the two equations given and find the value of the other unknown.

Example 3. Solve 3x- 4y = 20 and x + 2y = 5.
Solution:

The given equations are 3x- 4y = 20 ….(1)

and x+2y=5 …..(2)

Multiplying equation (1) by 1 and (2) by 2, we get

3x- 4y = 20 ……(3)

2x+4y = 10 ……(4)

Adding these equations, we get

5x = 30

x=6

Substituting x = 6 in equation (1), we get

3 × 6- 4y = 20

18 -4y = 20

-4y = 2

y = \(-\frac{1}{2}\)

Hence, the solution is \(\left.\begin{array}{l}
x=6 \\
y=-\frac{1}{2}
\end{array}\right\}\)

Question 4. Solve 3x – 17y = 23 and \(\frac{x}{3}+\frac{y}{4}=4 .\)
Solution:

The given equations are 3x-y = 23 ….(1)

and \(\frac{x}{3}+\frac{y}{4}=4\)

4x + 3y = 48 ….(2)

Multiplying equation (1) by 3 and (2) by 1, we get

9x-3y = 69 …..(3)

4x + 3y = 48 …..(4)

Adding equations (3) and (4), we get

13x = 117

⇒ x = 9

Substituting x = 9 in equation (1), we get

3 x 9-y = 23

⇒ 27 -y = 23

⇒ -y =-4

⇒ y = 4

Hence, the solution is \(\left.\begin{array}{l}
x=9 \\
y=4
\end{array}\right\}\)

Question 5. Solve \(4 x+\frac{6}{y}=15 \text { and } 6 x-\frac{8}{y}=14 \text {. }\)
Solution:

The given equation are \(4 x+\frac{6}{y}=15\)…(1)

and \(6 x-\frac{8}{y}=14\)

Multiplying equation (1) by 4 and (2) by 3, we get

⇒ \(16 x+\frac{24}{y}=60\) …..(3)

⇒ \(18 x-\frac{24}{y}=42\) …..(4)

Adding equations (3) and (4), we get

34x = 102 ⇒ x = 3

Substituting x = 3 in equation (1), we get

⇒ \(4 \times 3+\frac{6}{y}=15\)

⇒ \(\frac{6}{y}=15-12\)

⇒ \(\frac{6}{y}=3\)

⇒ 3y = 6

or y = 2

Hence, the solution is \(\left.\begin{array}{l}
x=3 \\
y=2
\end{array}\right\}\)

Question 6. Solve for x and y, \(\frac{1}{2 x}-\frac{1}{y}=-1 \text { and } \frac{1}{x}+\frac{1}{2 y}=8 \text {. }\)
Solution:

The given equations are \(\frac{1}{2 x}-\frac{1}{y}=-1\) ….(1)

and \(\frac{1}{x}+\frac{1}{2 y}=8\) …..(2)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\) then equations (1) and (2) can be written as

⇒ \(\frac{n}{2}-v=-1\)

⇒ u – 2v = -2 …….(3)

and \(u+\frac{v}{2}=8\)

⇒ 2u + v = 16 …….(4)

Multiplying equation (3) by 1 and (4) by 2, we get

⇒ u – 2v = -2 ….(5)

⇒ 4u + 2v = 32 ….(6)

Adding equations (5) and (6), we get

⇒ 5u = 30

⇒ u = 6

Substituting u = 6 in equation (3), we get

⇒ 6-2v = -2

⇒ -2v = -8

⇒ v = 4

Now, \(\frac{1}{x}=u\)

⇒ \(\frac{1}{x}=6\)

⇒ \(x=\frac{1}{6}\)

and \(\frac{1}{y}=v\)

⇒\(\frac{1}{y}=4\)

⇒\(x=\frac{1}{4}\)

Hence, the solution is \(\left.\begin{array}{l}
x=\frac{1}{6} \\
y=\frac{1}{6}
\end{array}\right\}\)

Question 7. Solve \(\frac{20}{x+y}+\frac{3}{x-y}=7 \text { and } \frac{8}{x-y}-\frac{15}{x+y}=5\)
Solution:

Given equations are

⇒ \(\frac{20}{x+y}+\frac{3}{x-y}=7\) ……..(1)

and \(\frac{8}{x-y}-\frac{15}{x+y}=5\) ……(2)

Let \(\) then equation (1) and (2) can be written as

20a + 3b = 7 ……(3)

and -15a + 8b = 5 …….(4)

Multiplying equation (3) by 8 and (4) by 3, we get

160a + 24b = 56 ……(5)

-45a + 24b = 15 ……(6)

Subtracting equation (6) from equation (5), we get

205a = 41

⇒ \(a=\frac{1}{5}\)

Substituting \(a=\frac{1}{5}\) in equation (3), we get

⇒ \(20 \times \frac{1}{5}+3 b=7\)

⇒ 32b = 3

⇒ b = 1

Now, \(\frac{1}{x+y}=a \quad \Rightarrow \quad \frac{1}{x+y}=\frac{1}{5} \quad \Rightarrow x+y=5\) ……(7)

and \(\frac{1}{x-y}=b \quad \Rightarrow \quad \frac{1}{x-y}=1 \quad \Rightarrow x-y=1\) …..(8)

Adding equations (7) and (8), we get

2x= 6

x = 3

Substituting x = 3 in equation (7), we get

3 +y = 5

⇒ y = 2

Hence, the solution is \(\left.\begin{array}{l}
x=3 \\
y=2
\end{array}\right\}\)

Question 8.

1. Solve 8x- 3y = 5xy and 6x-5y = -2xy. How many solutions, does this system of equations have?
2. Solve for x and y, by reducing the following equations in a pair of linear equations: 2x + 3y = 5xy and 3x-2y=xy.

Solution:

Given equations are 8x-3y = 5xy

and 6x- 5y- -2xy

The given equations are not linear in the variables x and y. These can be reduced into linear equations.

If we put x = 0 in either of the equations, we get y = 0. Hence, x = 0 and y = 0 is a solution of these equations.

To find another solution, divide each of the given equations by xy

So, \(\frac{8}{y}-\frac{3}{x}=5\) ……(1)

and \(\frac{6}{y}-\frac{5}{x}=-2\) …..(2)

Let \(\frac{1}{y}=a \text { and } \frac{1}{x}=b\) then from equations (1) and (2), we get

⇒ \(\left.\begin{array}{l}
8 a-3 b=5 \\
6 a-5 b=-2
\end{array}\right\} \text { linear equations }\) …….(3)and (4)

Multiplying equation (3) by 5 and (4) by 3, we get

40a -15b = 25 ……(5)

18a – 15b = -6 ……(6)

Subtracting equation (6) from (5), we get

22a = 31

⇒ \(a=\frac{31}{22}\)

Putting a = \(\frac{31}{22}\) in equation (3), we get

⇒ \(8 \times \frac{31}{22}-3 b=5\)

⇒ \(\frac{124}{11}-3 b=5\Rightarrow-3 b=5-\frac{124}{11}\)

⇒ \(-3 b=-\frac{69}{11}\Rightarrow=\frac{23}{11}\)

Now, \(\frac{1}{y}=a \quad \Rightarrow \quad \frac{1}{y}=\frac{31}{22} \quad \Rightarrow \quad y=\frac{22}{31}\)

and \(\frac{1}{x}=b \quad \Rightarrow \quad \frac{1}{x}=\frac{23}{11} \quad \Rightarrow \quad x=\frac{11}{23}\)

Hence, the solution of the given two equations are \(\left.\left.\begin{array}{l}
x=0 \\
y=0
\end{array}\right\} \text { and } \quad \begin{array}{l}
x=\frac{11}{23} \\
y=\frac{22}{31}
\end{array}\right\}\)

So, the given system of equations has two solutions.

Note:

In the above example 10,

1. If it is given that x ≠ 0, y ≠ 0 or reduce the given equations in a pair of linear equations, then we have a right to divide both the equations by xy. In this way, we get only one solution i.e., \(x=\frac{11}{23}, y=\frac{22}{31}\) only. When no condition is given to the equations then we get the two solutions of this system of equations i.e.,

⇒ \(\left.\left.\begin{array}{l}
x=0 \\
y=0
\end{array}\right\} \text { and } \begin{array}{r}
x=\frac{11}{23} \\
y=\frac{22}{31}
\end{array}\right\}\) (for more details, see “Gold coins” at the end of this chapter).

2. We have,

2x + 3y = 5xy ……(1)

3x-2y=xy …..(2)

Dividing equations (1) and (2) by xy, we get

⇒ \(\frac{2 x}{x y}+\frac{3 y}{x y}=\frac{5 x y}{x y}\Rightarrow \frac{2}{y}+\frac{3}{x}=5\) …..(3)

and \(\frac{3 x}{x y}-\frac{2 y}{x y}=\frac{x y}{x y}\Rightarrow \frac{3}{y}-\frac{2}{x}=1\) …….(4)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\)

∴ Equations (3) and (4) become,

2v + 3u = 5 ….(5)

3v- 2u = 1 ….(6)

Now this is a pair of linear equations, multiplying equation (5) by 2 and equation (6) by 3, we get

4v + 6u = 10 …(7)

9v – 6u = 3 …..(8)

On adding equations (7) and (8), we get

13v = 13

⇒ v = 1

Putting v = 1 in equation (5), we get

2(1) + 3u = 5

3u = 3

u=1

Now, u = 1 ⇒ \(\frac{1}{x}=1\) => x =1

and v = 1 ⇒ \(\frac{1}{y}=1\) => y =l

Hence, the required solution is \(\left.\begin{array}{l}
x=1 \\
y=1
\end{array}\right\}\)

Question 9. Solve \(\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2} \text { and } \frac{7}{2 x+3 y}+\frac{4}{3 x-2 y}=2,\) by reducing them into a pair of linear equations.
Solution:

The given equations

⇒ \(\frac{1}{2(2 x+3 y)}+\frac{12}{7(3 x-2 y)}=\frac{1}{2}\) ….(1)

and \(\frac{7}{2 x+3 y}+\frac{4}{3 x-2 y}=2\) ….(2)

are not linear equations. So, first of all, we shall make these as a pair of linear equations, by putting

⇒ \(\frac{1}{2 x+3 y}=a \text { and } \frac{1}{3 x-2 y}=b\)

Therefore, equations (1) and (2) become,

⇒ \(\frac{1}{2} a+\frac{12}{7} b=\frac{1}{2}\) ….(3)

and 7a + 4b = 2 …….(4)

Multiplying equation (3) by 14, we get

7a + 24b = 7 ……(5)

Subtracting equation (5) from (4), we get

b = \(\frac{1}{4}\)

Putting b = \(\frac{1}{4}\) in equation (4), we get

⇒ \(7 a+4\left(\frac{1}{4}\right)=2 \quad \Rightarrow 7 a=2-1 \quad \Rightarrow a=\frac{1}{7}\)

since, \(a=\frac{1}{7} \quad \Rightarrow \frac{1}{2 x+3 y}=\frac{1}{7} \quad \Rightarrow 2 x+3 y=7\) ……..(6)

and \(b=\frac{1}{4} \quad \Rightarrow \frac{1}{3 x-2 y}=\frac{1}{4} \quad \Rightarrow 3 x-2 y=4\) ……….(7)

Multiplying equation (6) by 2 and equation (7) by 3, we get

⇒ x = 2

On adding, we get

Putting x = 2 in equation (8), we get,

4(2) + 6y= 14

6y = 14- 8

y = 1

Hence, the solution is \(\left.\begin{array}{r}
x=2 \\
y=1
\end{array}\right\}\)

Question 10. Solve 41x + 53y = 135 and 53x + 41y = 147
Solution:

Given equations are 41x + 53y = 135 ……..(1)

53x + 41y = 147 ……..(2)

On adding equations (1) and (2) we get

94x + 94y = 282

⇒ x+y = 3 ………(3)

On subtracting equation (2) from (1), we get

-12x+ 12y = -12

⇒ x-y- 1 …………(4)

Adding equations (3) and (4), we get

2x= 4  ⇒ x = 2

Putting x = 2 in equation (3)

2 + y = 3 ⇒ y = 1

Hence, the solution is \(\left.\begin{array}{r}
x=2 \\
y=1
\end{array}\right\}\)

Question 11. Solve (a – b)x + (a+b)y = a²– 2ab – b² and (a + b)(x +y) = a² + b².
Solution:

Given equations are (a- b)x + (a + b]y = a² – 2ab – b² ….(1)

and (a + b)(x+y) – a² +b²

or (a + b)x + (a +b)y = a² + b² …….(2)

On subtracting equation (2) from (1), we get

(a – b)k- [a + b)x = a² – 2ab- b² – a² – b²

⇒  x(a -b -a -b) = -2b² – 2ab

⇒ -2bx = -2b(b + a)

⇒ x = a +b

Putting x= (a + b) in equation (1), we get

(a – b)(a + b) + (a + b)y = a²– 2ab- b²

a² -b² + (a +b)y = a² – 2ab – b²

(a +b)y = -2ab

⇒ \(y=\frac{-2 a b}{a+b}\)

Hence, the solution is \(\left.\begin{array}{l}
x=a+b \\
y=\frac{-2 a b}{a+b}
\end{array}\right\}\)

Method of Cross Multiplication

Theorem : If a₁x+b₁y +c₁ =0 and a₂x +b₂y + c₂ = 0 be a system of simultaneous linear equations in two variables x and y such that \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} \text { i.e., } a_1 b_2-a_2 b_1 \neq 0\).

Then the system has a unique solution given by

⇒ \(x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1} \text { and } y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\)

Proof: The given equations are

a₁x +b₁y + c₁ = 0

a₂x +b₂y + c₂ = 0

Multiplying equation (1) by b₂ and (2) by b₁ and subtracting we get

⇒ \(\left(a_1 b_2-a_2 b_1\right) x=\left(b_1 c_2-b_2 c_1\right)\)

⇒ \(x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}\)

Again multiplying equation (1) by a2, (2) by A J and subtracting we get

⇒ \(y\left(a_2 b_1-a_1 b_2\right)=c_2 a_1-c_1 a_2\)

⇒ \(y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}\)

Hence, x = \(\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1} \text { and } y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1} \text {. }\)

⇒ \(y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-c_2 b_1}\)

Solved Examples

Question 1. Solve the following system of equations by using the method of cross multiplication 2x+3y = 7 and 6x + 5y = 11.
Solution:

Given

2x+3y = 7 and 6x + 5y = 11

The given equations can be written as 2x + 3y- 7 = 0

⇒ \(\frac{x}{3 \times(-11)-5 \times(-7)}=\frac{y}{-7 \times 6-(-11) \times 2}=\frac{1}{2 \times 5-6 \times 3}\)

⇒ \(\frac{x}{-33+35}=\frac{y}{-42+22}=\frac{1}{10-18}\)

⇒ \(\frac{x}{2}=\frac{y}{-20}=\frac{1}{-8}\)

when \(\frac{x}{2}=-\frac{1}{8} \quad \text { or } \quad x=-\frac{1}{4}\)

and \(\frac{y}{-20}=-\frac{1}{8} \quad \text { or } \quad y=\frac{20}{8}=\frac{5}{2}\)

Hence, \(\left.\begin{array}{l}
x=-\frac{1}{4} \\
y=\frac{5}{2}
\end{array}\right\}\) is the required solution.

Question 2. Solve the following system of equations bx + ay = 2ab and ax -by- a² – b².

Solution:

Given

bx + ay = 2ab and ax -by- a² – b²

Given equations can be written as

bx + ay- 2ab = 0 and ax- by – a² + b² = 0

By cross multiplication method, we have

⇒ \(\frac{x}{a \times\left[-\left(a^2-b^2\right)\right]-(-b) \times(-2 a b)}=\frac{y}{(-2 a b) \times a-\left[-\left(a^2-b^2\right)\right] \times b}=\frac{1}{b \times(-b)-a \times a}\)

⇒ \(\frac{x}{a\left(-a^2+b^2-2 b^2\right)}=\frac{y}{-2 a^2 b+a^2 b-b^3}=\frac{1}{-b^2-a^2}\)

⇒ \(\frac{x}{a\left(-a^2-b^2\right)}=\frac{y}{-a^2 b-b^3}=\frac{1}{-b^2-a^2}\)

⇒ \(\frac{x}{a\left(-a^2-b^2\right)}=\frac{1}{-b^2-a^2}\)

⇒ \(x=\frac{-a\left(a^2+b^2\right)}{-1\left(b^2+a^2\right)}\) or x = a

⇒ \(\frac{y}{-a^2 b-b^3}=\frac{1}{-b^2-a^2}\)

⇒ \(y=\frac{-b\left(a^2+b^2\right)}{-1\left(b^2+a^2\right)}\) or y = b

Hence, \(\left.\begin{array}{l}
x=a \\
y=b
\end{array}\right\}\) is the required solution.

Question 3. Solve the following system of equations for x and y, a(x +y) + b(x -y) = a² – ab + b² and a(x +y) -b(x -y) = a² + ab + b².
Solution:

Given

a(x +y) + b(x -y) = a² – ab + b² and a(x +y) -b(x -y) = a² + ab + b²

Given equations can be written as

⇒ ax + ay +bx -by = a² – ab + b²

⇒ (a + b)x + (a -b)y = a² – ab + b²

and ax+ ay -bx + by =a² +ab +b² ….(1)

⇒ (a – b)x + (a + b)y = a² + ab + b² …..(2)

For equations (1) and (2) by cross multiplication method, we have

⇒ \(\frac{y}{\left[-\left(a^2-a b+b^2\right)\right] \times(a-b)-\left[-\left(a^2+a b+b^2\right) \times(a+b)\right]}=\frac{1}{(a+b)(a+b)-(a-b)(a-b)}\)

⇒ \(\frac{x}{-\left(a^3-b^3\right)+\left(a^3+b^3\right)}\)

⇒ \(\frac{y}{-\left(a^3-a^2 b+a b^2-a^2 b+a b^2-b^3\right)+\left(a^3+a^2 b+a b^2+a^2 b+a b^2+b^3\right)}=\frac{1}{(a+b)^2-(a-b)^2}\)

⇒ \(\frac{x}{2 b^3}=\frac{y}{2 a^2 b-2 a b^2+b^3+2 a^2 b+2 a b^2+b^3}=\frac{1}{4 a b}\)

⇒ \(\frac{x}{2 b^3}=\frac{y}{4 a^2 b+2 b^3}=\frac{1}{4 a b}\)

⇒ \(\frac{x}{2 b^3}=\frac{y}{2 b\left(2 a^2+b^2\right)}=\frac{1}{4 a b}\)

when \(\frac{x}{2 b^3}=\frac{1}{4 a b} \quad \Rightarrow \quad x=\frac{2 b^3}{4 a b} \quad \Rightarrow \quad x=\frac{b^2}{2 a}\)

and \(\frac{y}{2 b\left(2 a^2+b^2\right)}=\frac{1}{4 a b} \Rightarrow y=\frac{2 b\left(2 a^2+b^2\right)}{4 a b} \Rightarrow y=\frac{2 a^2+b^2}{2 a}\)

Hence, \(\left.\begin{array}{rl}
x=\frac{b^2}{2 a} \\
y=\frac{2 a^2+b^2}{2 a}
\end{array}\right\}\) is the required solution.

Condition For Solvability Of Linear Equations

Consistent and In-consistent System: When a system of equations a₁ x + b₁ y = c₁ and a₂x + b₂y = c₂ has a solution, the system is said to be consistent. A consistent system has one or more solutions. When a system has no solution, the system is said to be inconsistent.

Conditions for Solvability

1. If \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2},\) then the given system of equations has a unique solution and thus it is consistent.
Solution:

In this case, the two lines intersect at exactly one point.

2. If \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) then there is no solution and system is in-consistent.
Solution:

In this case, the two lines are parallel to each other. They will not meet at any point.

3. If \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\) then the system has infinitely many solutions; s and thus the system is consistent and the given pair of linear equal.
Solution:

In this case, the two lines coincide each other, i.e., the two lines are the same. So, we say such lines as coincident lines.

Remark:

If c₁ and c₂ are both equal to zero, the solution can be found more easily as follows:

1. When \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2} ，\) then only solution is x = y = 0
2. When \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\) then there are infinite number of non-zero solution.

Linear Equations Solved Examples

Question 1. Show that the following system of equations has a unique solution:

1. 7x – 2y = 3
   22x – 3y = 16
2. 3x + y = 17
   8x + 11y = 37

and also solve the system of equations in each case.

Solution:

Given equations are 7x- 2y = 3 …….(1)

22x- 3y = 16 …….(2)

Here, a₁ = 7, b₁ = -2 and c₁ = 3

a₂ = 22, b₂ = -3 and c₂= 16

Now, \(\frac{a_1}{a_2}=\frac{7}{22} \text { and } \frac{b_1}{b_2}=\frac{-2}{-3}=\frac{2}{3}\)

Since, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\) Hence, the given system has a unique solution.

By cross multiplication method, we have

⇒ \(\frac{x}{(-2) \times(-16)-(-3) \times(-3)}=\frac{y}{(-3) \times 22-(-16) \times 7}=\frac{1}{7 \times(-3)-22 \times(-2)}\)

⇒ \(\frac{x}{32-9}=\frac{y}{-66+112}=\frac{1}{-21+44}\)

⇒ \(\frac{x}{23}=\frac{y}{46}=\frac{1}{23}\)

when \(\frac{x}{23}=\frac{1}{23}\)

x = 1 and \(\frac{y}{46}=\frac{1}{23}\) y = 2

Hence, \(\left.\begin{array}{l}
x=1 \\
y=2
\end{array}\right\}\) is the required solution.

The given system of equations is 3x + y = 17 ……..(1)

8x + 11y = 37 ……..(2)

Here a₁ = 3, b₁ =1 and c₁ = 17

a₂ = 8, b₂ = 11 and c₂ = 37

Now, \(\frac{a_1}{a_2}=\frac{3}{8}, \frac{b_1}{b_2}=\frac{1}{11}\)

Since \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). Hence, the given system has a unique solution.

We can write the equations as

3x+y- 17 = 0 and 8r + 1ly- 37 = 0

By cross multiplication method, we have

⇒ \(\frac{x}{1 \times(-37)-11 \times(-17)}=\frac{y}{(-17) \times 8-(-37) \times 3}=\frac{1}{3 \times 11-8 \times 1}\)

⇒ \(\frac{x}{-37+187}=\frac{y}{-136+111}=\frac{1}{33-8}\)

⇒ \(\frac{x}{150}=\frac{y}{-25}=\frac{1}{25}\)

When \(\frac{x}{150}=\frac{1}{25}\) = x = 6 and \(\frac{y}{-25}=\frac{1}{25}\) = y = -1

Hence, \(\left.\begin{array}{l}
x=6 \\
y=-1
\end{array}\right\}\) is the required solution

Question 2. Find the value of k for which the system of equations 2x + ky = 1 and 3x- 5y = 7 has a unique solution.
Solution:

The given system of equations is

2x +ky = 1 ……..(1)

and 3x- 5y = 7 ……..(2)

Here, a₁ =2, b₁=k and c₁ = 1

a₂ = 3, b₂ = -5 and c₂ = 7

The system has unique solution if \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{2}{3} \neq \frac{k}{-5} \quad \text { or } \quad k \neq \frac{-10}{3}\)

So, k can take any real value except \(\frac{-10}{3}\)

Question 3. Find the value of k for which the system of equations x + 2y = 5, 3x + ky – 1 5 = 0 has no solution.
Solution:

The given system of equations is

x + 2y- 5 = 0 ………(1)

3x + ky- 15 = 0 ……..(2)

Here, a₁ = 1, b₁ = 2, c₁ =-5

a₂ = 3, b₂ = k, c₂ = -15

If the equations have no solution then

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

⇒ \(\frac{1}{3}=\frac{2}{k} \neq \frac{-5}{-15} \Rightarrow \frac{1}{3}=\frac{2}{k} \quad \text { and } \quad \frac{2}{k} \neq \frac{-5}{-15}\)

k = 6 and k ≠ 6 which is impossible.

Hence, there is no such value of k for which the given system has no solution.

Question 4. For what value(s) of a will the system of linear equations αx + 3y = α – 3 and 12x + αy = a have a unique solution?
Solution:

The given equations are

αx + 3y = α – 3

12x + αy = α

Here a₁ = α, b₁ = 3 and c₁ = α- 3

a₂ = 12, b₂ = α and c₂ = α

The system has a unique solution if \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

⇒ \(\frac{\alpha}{12} \neq \frac{3}{\alpha}\)

⇒ a² ≠  36

⇒ a ≠ ± 6

i.e., a ≠ 6 and a ≠ -6

So, α can take any real value except -6 and 6

Question 5. Find the value(s) of k for which the system of equations 5x + 2y = k and 10x + 4y = 3 has infinitely many solutions.
Solution:

The given system of equations is

5x + 2y – k = 0 and 10k + 4y- 3 = 0

Here a₁ = 5, b₁ = 2 and c₁ = k

a₂= 10, b₂ = 4 and c₂ = -3

The system has infinitely many solutions if

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{5}{10}=\frac{2}{4}=\frac{-k}{-3}\)

Hence, for k = \(\frac{3}{2}\), the given system of equations will have infinitely many solutions.

Question 6. Find the values of p and q for which the following system of equations has an infinite number of solutions: 2x + 3y = 7 and (p+q)x+(2p-q)y = 21.
Solution:

The given system of equations is

2x + 3y = 7

(p + q)x + (2p- q)y = 21

Here a₁ = 2, b₁ = 3,

a₂ =P + q, b₂ = q and c₂ = 21

The given system will have an infinite number of solutions if

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{2}{p+q}=\frac{3}{2 p-q}=\frac{7}{21}\)

Taking the first two and last two expressions

⇒ \(\frac{2}{p+q}=\frac{1}{3} \quad \text { and } \quad \frac{3}{2 p-q}=\frac{1}{3}\)

p + q = 6 and 2p-q = 9

Solving p + q = 6 and 2p- q = 9, we get

p = 5 and q = 1

Hence, the given system of equations will have infinitely many solutions, if p = 5 and q = 1.

Question 7. For which value(s) of A, will the pair of equations represent coincident lines k x + 3y – (k + 3) = 0 and 12x + ky + k = 0?
Solution:

Comparing the given equations with

a₁x + b₁y + c₁ = 0

a₂x+ b₂y + c₂ = 0

We have

⇒ \(\frac{a_1}{a_2}=\frac{k}{12}, \frac{b_1}{b_2}=\frac{3}{k} \text { and } \frac{c_1}{c_2}=\frac{-(k+3)}{k}\)

Now, for coincident lines

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{k}{12}=\frac{3}{k}=-\frac{(k+3)}{k}\)

Taking the first two expressions, we get

⇒ \(\frac{k}{12}=\frac{3}{k}\)….(1)

k² = 36

k = +6

Taking the last two expressions, we get

⇒ \(\frac{3}{k}=-\frac{(k+3)}{k}\) …..(2)

3k = -k² – 3k

k² + 6k = 0

k(k + 6) = 0

∴ k = 0 or k = -6

But k = -6 is a value which satisfies both the equations (1) and (2).

∴ k = -6

Linear Equations Solved Examples

Question 1. The sum of the two numbers is 85. If the larger number exceeds four times the smaller one by 5. Find the numbers.
Solution:

Given

The sum of the two numbers is 85. If the larger number exceeds four times the smaller one by 5.

Let the two numbers x andy, and x > y.

According to question, x + y = 85 ……(1)

and x – 4y = 5 …..(2)

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

5y = 80

⇒ y = 16

Putting, y = 16 in equation (1), we get

x+ 16 = 85

⇒ x = 69

Hence, the required numbers are 69 and 16.

Question 2. The sum of the two numbers is 18. The sum of their reciprocals is \(\frac{1}{4}\) Find the numbers.
Solution:

Given

The sum of the two numbers is 18. The sum of their reciprocals is \(\frac{1}{4}\)

Let the two numbers be x and y.

∴ According to the condition,

x + y = 18 …….(1)

According to the 2 condition, \(\frac{1}{x}+\frac{1}{y}=\frac{1}{4}\) ……(2)

⇒ \(\frac{y+x}{x y}=\frac{1}{4}\)

⇒ xy = 4(x+y)

⇒ xy = 4 × 18

⇒ xy = 72

Now, we know that

(x-y)² = (x+y)²-4xy

⇒ (x -y)² = (18)² – 4(72) = 324 – 288 = 36

∴ x-y = ± 6

If x-y = 6 ….(3)

Now, from equations (3) and (1),

∴ x= 12

∴ y = 18-12

= 6

∴ Two numbers are 12 and 6 in both cases.

If x – y = -6 …….(4)

Now, from equations (4) and (1),

∴ x= 6

∴ y = 18-6

= 12

Question 3. Five years hence father’s age will he three times the age of his son. Five years ago, the father was seven times as old as his son. Find their present ages.
Solution:

Given

Five years hence father’s age will he three times the age of his son. Five years ago, the father was seven times as old as his son.

Let the present age of the son be x years and the present age of the father by years.

Five years hence, the age of son = x + 5

and the age of father = y + 5

Using given information

3(x + 5)=y + 5

⇒ 3x-y = -10 …….(1)

Five years ago, age of son = (x- 5) years

age of father = (y- 5) years

∴ 7(x-5)=y-5

7x -y = 30 ………(2)

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

4x = 40

⇒ x = 10

On substituting x = 10 in equation (1)

3 × 10 – y = -10

⇒ 30 -y =-10

⇒ -y = -40

⇒ y = 40

Hence, the present age of the son is 10 years and the present age of a father is 40 years.

Question 4. The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of the digits of the given number exceeds the given number by 27. Find the given number.
Solution:

Given

The sum of the digits of a two-digit number is 9. The number obtained by reversing the order of the digits of the given number exceeds the given number by 27.

Let the digit at the unit’s place be x and the ten’s place be y.

∴ Number = 10y +x

On reversing the digits, the new number = 10x+y

According to the given conditions,

x+y = 9 …..(1)

10x+y = 21 + (10y +x)

9x-9y = 21

x-y = 3 …….(2)

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

2x= 12

x = 6

Putting x= 6 in equation (1), we get

6+y = 9

y = 3

Hence, the required number = 10y +x= 10 x 3 + 6 = 36

Question 5. If 2 is added to the numerator of a fraction it reduces to \(\frac{1}{2}\) and if 1 is subtracted from the denominator, it reduces to \(\frac{1}{3}\) Find the fraction.
Solution:

Given

If 2 is added to the numerator of a fraction it reduces to \(\frac{1}{2}\) and if 1 is subtracted from the denominator, it reduces to \(\frac{1}{3}\)

Let the fraction be \(\frac{x}{y}\), where the numerator is x and the denominator is y.

According to 1 condition, \(\frac{x+2}{y}=\frac{1}{2}\)

2x+ 4 = y

⇒ 2x-y = -4 ……(1)

According to 2 condition, \(\frac{x}{y-1}=\frac{1}{3}\)

⇒ 3x=y- 1

⇒ 3x -y = -1 …..(2)

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

-x = -3

⇒ x = 3

Substituting x = 3 in equation (1), we get

2 x 3 -y = -4

⇒ -y = -10

⇒ y = 10

Hence, the required fraction is \(\frac{x}{y}=\frac{3}{10}\)

Question 6. A man sold a chair and table together for ₹760 thereby making a profit of 25% on chairs and 10% on tables. By selling them together for ₹767.50, he would have made a profit of 1 0% on a chair and 25% on the table. Find the cost price of each.
Solution:

Given

A man sold a chair and table together for ₹760 thereby making a profit of 25% on chairs and 10% on tables. By selling them together for ₹767.50, he would have made a profit of 1 0% on a chair and 25% on the table.

Let the cost price of the chair = ₹x and the cost price of the table = ₹y

Now, two cases arise:

Case 1: 25% profit on chair and 10% profit on table

Selling price of chair = \(₹\left(x+\frac{25 x}{100}\right)=₹ \frac{125 x}{100}\)

Selling price of table = \(₹\left(y+\frac{10 y}{100}\right)=₹ \frac{110 y}{100}\)

But total S.P = ₹760

⇒ \(\frac{125 x}{100}+\frac{110 y}{100}=760 \quad \Rightarrow 125 x+110 y=76000\)

∴ 25x + 22y = 15200 ……(1)

Case 2: 10% profit on chair and 25% profit on table

Selling price of chair = \(₹\left(x+\frac{10 x}{100}\right)=₹ \frac{110 x}{100}\)

Selling price of table = \(₹\left(y+\frac{25 y}{100}\right)=₹ \frac{125 y}{100}\)

But total S.P. = ₹767.50

∴ \(\frac{110 x}{100}+\frac{125 y}{100}=767.50\)

⇒ 110x + 125y = 76750

⇒ 22x + 25y = 15350 ……(2)

To solve equations (1) and (2), adding equations (1) and (2), we get

47x + 47y = 30550

⇒ x + y = 650 …….(3)

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

3x- 3y = -150

⇒ x-y = -50 ….(4)

On adding equations (3) and (4), we get

2x = 600

⇒ x = 300

Putting x = 300 in equation (3), we get

300 +y = 650

⇒ y = 650- 300 = 350

Cost of chair = ₹300 and cost of table = ₹350

Question 7. A lady has 25 paise and 50 paise coins in her purse. If in all she has 40 coins and the value of her money is ₹12.75, how many coins of each type does she have?
Solution:

Given

A lady has 25 paise and 50 paise coins in her purse. If in all she has 40 coins and the value of her money is ₹12.75,

Let a number of 25 paise coins be x and a number of 50 paise coins be y.

According to the given conditions, we have

x + y = 40 ….(1)

and 25x + 50y= 1275 …..(2)

Multiplying equation (1) by 50, we get

50x+50y = 2000

Subtracting equation (3) from (2), we get

-25x = -725

⇒ x = 29

Substituting x = 29 in equation (1), we get

29 + y = 40

⇒ y = 11

Hence, the number of 25 paise coins = 29 and the number of 50 paise coins = 11.

Question 8. 2 men and 7 boys can finish a work in 4 days, while 4 men and 4 boys can finish the same work in 3 days. How long would it take one man or one boy to do it?
Solution:

Given

2 men and 7 boys can finish a work in 4 days, while 4 men and 4 boys can finish the same work in 3 days.

Let 1 man can do a piece of work in* days while a boy can do a piece of work in y ways.

∴ One man’s one day work = \(\frac{1}{x}\)

and one boy’s one day work = \(\frac{1}{y}\)

According to the given conditions,

⇒ \(\frac{2}{x}+\frac{7}{y}=\frac{1}{4}\) ……(1)

and \(\frac{4}{x}+\frac{4}{y}=\frac{1}{3}\) …….(2)

Multiplying equation (1) by 2 and (2) by 1, we get

⇒ \(\frac{4}{x}+\frac{14}{y}=\frac{1}{2}\) ……..(3)

⇒ \(\frac{4}{x}+\frac{4}{y}=\frac{1}{3}\) ……..(4)

Subtracting equation (4) from (3), we get

⇒ \(\frac{14}{y}-\frac{4}{y}=\frac{1}{2}-\frac{1}{3} \Rightarrow \frac{10}{y}=\frac{1}{6}\)

⇒ y = 60

Substituting y = 60 in equation (1), we get

⇒ \(\frac{2}{x}+\frac{7}{60}=\frac{1}{4} \quad \Rightarrow \quad \frac{2}{x}=\frac{1}{4}-\frac{7}{60}\)

⇒ \(\frac{2}{x}=\frac{8}{60}\)

⇒ 8x= 120 or x = 15

Hence, one man would do the work in 15 days and one boy in 60 days.

Question 9. A boat goes 301km upstream and 44 km downstream in 10 hours. In 1 3 hours it can go 40 1cm upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Solution:

Given

A boat goes 301km upstream and 44 km downstream in 10 hours. In 1 3 hours it can go 40 1cm upstream and 55 km downstream.

Let the speed of the boat in still water is x km/hr and the speed of the stream is y km/hr.

∴ speed of boat upstream = (x-y) km/hr

and speed of boat downstream = (x +y) km/hr

We know that, speed = \(\frac{\text { distance }}{\text { time }}\)

According to the given conditions,

Time taken in going 30 km upstream = \(\frac{30}{x-y} \mathrm{hr}\)

Time taken in 44 lem downstream = \(\frac{40}{x+y} \mathrm{hr}\)

∴ \(\frac{30}{x-y}+\frac{44}{x+y}=10\)

Similarly in the second condition,

⇒ \(\frac{40}{x-y}+\frac{55}{x+y}=13\)

Let \(\frac{1}{x-y}=a \text { and } \frac{1}{x+y}=b,\), we get

30a + 44b = 10

and 40a + 55b =13

Multiplying equation (3) by 4 and (4) by 3, we get

120a+ 1766 = 40

120a+ 1656 = 39

Subtracting equation (6) from (5), we get

11b = 1

⇒ b = \(\frac{1}{11}\)

Putting b = \(\frac{1}{11}\) inequation (3), we get

⇒ \(30 a+44 \times \frac{1}{11}=10\)

30a = 6

⇒ \(a=\frac{1}{5}\)

⇒ \(\frac{1}{x-y}=\frac{1}{5}\text { or }x-y=5\) ……(7)

and \(\frac{1}{x+y}=11\text { or }x+y=11\) …….(8)

Solving equations (7) and (8), we get

x = 8 and y = 3

Hence, the speed of a boat in still water = 8 km/hr and the speed of the water stream = 3 km/hr.

Question 10. A boat which travels at the rate of 10.5 km/hr downstream takes 3 times as long time to go to a certain distance up a river as to go the same distance down. Find the rate at which the stream flows.
Solution:

Given

A boat which travels at the rate of 10.5 km/hr downstream takes 3 times as long time to go to a certain distance up a river as to go the same distance down.

Let the speed of the stream = x km/hr

and the speed of the boat in still water = y km/hr

∴ Speed of boat downstream = (x +y) km/hr

and speed of boat upstream = (y- x) km/hr

Let the distance be D km.

∴ According to the problem,

x+y = 10.5 ……(1)

Also, time taken by boat upstream = 3 x time taken by boat downstream

⇒ \(\frac{D}{y-x}=3 \times \frac{D}{10.5}\)

y – x = 3.5 ……(2)

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

2y= 14 ⇒ y = 7

Putting y = 7 in equation (1), we get

x+7 = 10.5 ⇒ x= 3.5

Hence, speed of stream = 3.5 km/hr

Linear Equations In Two Variables Exercise 3.1

Question 1. Aftab tells his daughter, “Seven years ago, 1 was seven times as old as you were then. Also, three years from now, 1 shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Solution:

Given

Aftab tells his daughter, “Seven years ago, 1 was seven times as old as you were then. Also, three years from now, 1 shall be three times as old as you will be.”

Let the present age of Aftab be x years and the present age of his daughter be x years.

7 years ago,

age of Aftab = (x- 7) years

age of his daughter = (y- 7) years

According to the problem,

x-7 = 7(y -7)

⇒ x-7 = 7y- 49

⇒ x-7y = -42

3 years later,

age of Aftab = (x + 3) years

age of his daughter = (y + 3) years

According to the problem,

x + 3 = 3(y + 3)

⇒ x + 3 = 3y + 9

⇒ x- 3y = 6

∴ The algebraic form of the given problem is as follows:

x- 7y = -42 and x- 3y = 6

Now, x-7y = -42

x = 7y – 42

y = 5 then x=7 × 5-42=-7

y = 6 then x-7 × 6 – 42 = 0

y = 7 then A=7 × 7-42 = 7

and x- 3y = 6 ⇒ x = 3y + 6

y = -1 then x = 3 × -1 +6 = 3

y = 0 then x = 3 × 0 + 6 = 6

y = -2 then x = 3 × -2 + 6 = 0

Mark and join the given points in the two tables. It is the graphical form of the given problem.

Question 2. The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.
Solution:

Given

The coach of a cricket team buys 3 bats and 6 balls for ₹3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300.

Let the cost of a bat = ₹x

and cost of a ball = ₹y

According to the problem,

3x + 6y = 3900

x + 2y= 1300 ……….(1)

x + 3y = 1300 ……….(2)

Equations (1 ) and (2) represent the algebraic form of the given problem.

From equation (1),

x = 1300 -2y

Mark the points A(700, 300), B(300, 500), C(100, 600) and join them.

From equation (2), x- 1300- 3y

Mark the points D = (100, 400), E = (400, 300), E(700, 200) and join them.

It is the geometric form of the given problem.

Question 3. The cost of 2 kg of apples and I kg of grapes in a day was found to be ‘ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ‘ 300. Represent the situation algebraically and geometrically.
Solution:

Given

The cost of 2 kg of apples and I kg of grapes in a day was found to be ‘ 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ‘ 300.

Let the cost of 1 kg apple = ₹x

and cost of 1kg grapes = ₹y

According to the problem,

2x+y = 160 ……(1)

4x + 2y = 300

2x+y = 150 ……(2)

Equation (1) and equation (2) is the algebraic form of the given problem.

From equation (1),y = 160- 2x

Mark the points A(50, 60), B(60, 40) and C(70, 20) and join them.

From equation (2), y = 150-2x

Mark the points D(50, 50), E(60, 30) and F(70, 10) and join them.

The lines shown in the graph represent the geometric form of the given problem.

Linear Equations In Two Variables Exercise 3.2

Question 1. Form the pair of linear equations in the following problems and find their solutions graphically.

1. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
2. 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹ 46. Find the cost of one pencil and that of one pen.

Solution:

1. Let many boys = x and a number of girls =y.

According to the problem,

x +y = 10 ……(1)

y – x + 4 …….(2)

From equation (1), x +y = 10 ⇒ y = 10 – x

Mark the points A(5, 5), B(4, 6) and C(3, 7) and join them.

From equation (2), y = x + 4

Mark the points D(1, 5), E(2, 6) and C(3, 7) and join them.

Both lines intersect at points C(3, 7).

∴ x = 3, y = 7

⇒ Number of boys = 3 and number of girls = 7

2. Let the cost of 1 pencil = ₹x

and the cost of 1 pen = ₹y

According to the problem,

5x + 7y = 50

⇒ \(x=\frac{50-7 y}{5}\)

Mark the points A(10, 0), B(3, 5) and C(-4, 10) and join them

And 7x + 5y – 46 ⇒ \(x=\frac{46-5 y}{7}\)

Mark the points B(3, 5) and E(8, -3) and join them.

Two lines intersect at point (3,5).

∴ x = 3,y = 5

∴ Cost of 1 pencil= ₹3

and cost of 1 pen = ₹5.

Question 2. On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2} \text { and } \frac{c_1}{c_2}\), find representing the following pairs of linear equations intersect at a point, are parallel or coincident :

1. 5x- 4y + 8 = 0
   7x + by- 9 = 0
2.  9x + 3y+12 = 0
   18r + 6y + 24 = 0
3. 6x-3y+ 10 = 0
   2x-y + 9 = 0

Solution:

1. Given equations, 5x- 4y + 8 = 0 and 7x + 6y – 9 = 0

Here, a₁ = 5, b₁ = -4, c₁ = 8

a₂ = 7, b₂ = 6, c₂ = -9

Now, \(\frac{a_1}{a_2}=\frac{5}{7}, \frac{b_1}{b_2}=\frac{-4}{6}=\frac{-2}{3}, \frac{c_1}{c_2}=\frac{8}{-9}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Two lines intersect at a point.

2. Given equations, 9x + 3y+ 12 = 0 and 18x + 6y + 24 =0

Here,

a₁ = 9, b₁ = 3, c₁ = 12

a₂ = 18, b₂ = 6, c₂ = 24

Now \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Two lines are coincident

3. Given equations, 6x- 3y + 10 = 0 and 2x-y + 9 =0

Here, a₁ = 6, b₁ = -3, c₁ = 10

a₂ = 2, b₂=-1, c₂ = 9

Now \(\frac{a_1}{a_2}=\frac{6}{2}=\frac{3}{1}, \frac{b_1}{b_2}=\frac{-3}{-1}=\frac{3}{1}\)

and \(\frac{c_1}{c_2}=\frac{10}{9}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Two lines are parallel.

Question 3. On comparing the ratios, \(\frac{a_1}{a_2}, \frac{b_1}{b_2} \text { and } \frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent, or inconsistent.

1. 3x + 2y = 5; 2x- 3y = 7
2. 2x- 3y = 8; 4x- 6y = 9
3. \(\frac{3}{2} x+\frac{5}{3} y=7 ; 9 x-10 y=14\)
4. 5x-3y = 11;-10x+6y = -22
5. \(\frac{4}{3} x+2 y=8 ; 2 x+3 y=12\)

Solution:

1. Given equations,

3x + 2y = 5 ⇒ 3x + 2y – 5 = 0

and 2x- 3y = 7 ⇒ 2x- 3y- 7 = 0

Here, a₁ =3, b₁ = 2, c₁ = -5

a₂ = 2, b₂ =-3, c₂ = -7

Now, \(\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{-3} \text { and } \frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given equations have a unique solution.

⇒ Given pair of equations is consistent.

2. Given equations,

2x-3y = 8 ⇒ 2x-3y-8 = 0

and 4x- 6y= 9 ⇒ 4r-6y-9 = 0

Here, a₁= 2, b₁ = -3, c₁ = -8

a₂ = 4, b₂ = -6, c₂ = -9

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given equations have no solution.

⇒  Given pair of equations is inconsistent.

3. Given equations, \(\frac{3}{2} x+\frac{5}{3} y=7 \Rightarrow \frac{3}{2} x+\frac{5}{3} y-7=0\)

and 9x- 10y = 14 ⇒ 9x- 10y- 14 = 0

Here, \(a_1=\frac{3}{2}, b_1=\frac{5}{3}, c_1=-7\)

and \(a_2=9, b_2=-10, c_2=-14\)

Now, \(\frac{a_1}{a_2}=\frac{3 / 2}{9}=\frac{1}{6}, \frac{b_1}{b_2}=\frac{5 / 3}{-10}=-\frac{1}{6}\)

and \(\frac{c_1}{c_2}=\frac{-7}{-14}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given equations have a unique solution.

⇒ Given pair of equations is consistent.

4. Given equations,

5x-3y =11 ⇒ 5x- 3y – 11 = 0

and -10x + 6y = -22 ⇒ -10x + 6y + 22 = 0

Here, a₁ = 5, b₁ -3, c₁ = -11

a₂ = -10, b₂ = 6, c₂ = 22

Now, \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-11}{22}=\frac{-1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Two lines are coincident i.e., they will have infinitely many solutions.

⇒ Given pair of equations is consistent.

5. Given equations, \(\frac{4}{3} x+2 y=8\)

⇒ \(\frac{4}{3} x+2 y-8=0\)

and 2x + 3y = 12

⇒ 2x + 3y – 12 = 0

Here, \(a_1=\frac{4}{3}, b_1=2, c_1=-8\)

a₂ = 2, b₂ = 3, c₂ =-12

Now, \(\frac{a_1}{a_2}=\frac{4 / 3}{2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{2}{3}, \frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Two lines are coincident i.e., they will have infinitely many solutions.

⇒ Given pair of equations is consistent.

Question 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:

1. x + y = 5, 2x + 2y = 10
2. x-y = 8, 3x- 3y = 16
3. 2x+y-6 = 0, 4x- 2y – 4 = 0
4. 2x-2y-2 = 0, 4x-4y-5 = 0

Solution:

1. Given equations,

x+y = 5 ⇒ x+y-5 = 0

and 2x + 2y = 10 ⇒ 2x + 2y – 10 = 0

Here, a₁ = 1, b₁ = 1, c₁ = -5

a₂ = 2, b₂ = 2, c₂ = -10

Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{1}{2} \text { and } \frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Given equations have infinitely many solutions.

⇒ Given pair of equations is consistent.

Now, x + y = 5 ⇒ y = 5-x

Mark the points (5, 0), (3, 2) and (0, 5) and join them.

It is the required solution of given equations.

2. Given equations,

x-y = 8 ⇒ x -y- 8 = 0

3x-3y = 16 ⇒ 3x- 3y – 16 = 0

a₁ = 1, b₁ =-1, c₁ =-8

a₂= 3, b₂ = -3, c₂= -16

Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3} \text { and } \frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given system of equations has no solution

⇒ Given pair of equations is inconsistent.

3. Given equations,

2x+y – 6 =0 and 4x-2y -4 = 0

Here, a₁ = 2, b₁ = 1, c₁ = -6

a₂ = 4, b₂= -2, c₂ = -4

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given equations have a unique solution.

⇒ Given pair of equations is consistent.

From the first equation,

2x +y- 6 =0 ⇒ y = 6- 2x

Mark the points (3, 0), (2, 2) and (.0, 6) and join them.

From the second equation,

4x- 2y- 4 = 0 ⇒ 2x -y – 2 = 0

y = 2x-2

Mark the points (3, 4), (2, 2) and (1,0) and join them.

Two lines intersect at point (2, 2).

∴ Solution of given pair of equations: x = 2, y = 2

4. Given equations,

2x- 2y – 2 = 0 and 4x- 4y- 5 = 0

Here, a₁ = 2, b₁ = -2, c₁ = -2

a₂ = 4, b₂ = -4, c₂ = -5

Now \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given system of equations has no solution.

⇒ Given pair of equations is inconsistent.

Question5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:

Given

Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m.

Let the length of the garden = x metre

and breadth =y metre.

∴ The perimeter of a rectangular garden

= 2(length + breadth)

= 2(x+y)

So, half the perimeter = x+y

but it is given that, half the perimeter = 36 m

∴ x+y = 36 ….(1)

and x=y + 4 i.e., x-y = 4 …(2)

To solve the equations (1) and (2) we get x=20, y = 16

Therefore, length = 20 m and breadth = 16 m.

Question 6. Given the linear equation 2x + 3y- 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

1. Intersecting lines
2. Parallel lines
3. Coincident lines

Solution: Given a linear equation,

2x+ 3y- 8 = 0

Here, a₁ = 2, b₁ = 3, c₁ =-8

1. We know that, intersecting lines

⇒ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Taking a₂ = 3, b₂ = 4 and c₂ = 2

Required equation, 3x + 4y + 2 = 0

2. We know that, for parallel lines,

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Taking = 4, = 6 and = 5

Required equation, 4x + 6y + 5 = 0

3. We know that, for coincident lines,

⇒ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Taking a₂ = 4, b₂ = 6, c₂ = -16

Required equation, 4x+ 6y- 16 = 0

Question 7. Draw the graphs of the equations x -y +1=0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Solution:

x-y =-1 ⇒ x=y – 1

3x + 2y =12 ⇒ \(x=\frac{12-2 y}{3}\)

Mark the points Q(0, 1),B(-1, 0), P(0, 6) and C(4, 0).

Obtain the graph of the given lines on the graph paper by joining BQ and PC.

Two lines intersect at point A(2, 3).

∴ x = 2, y = 3

Lines intersect x-axis at points B(-l, 0) and C(4, 0).

Draw the perpendicular AM from A to BC

∴ Area of ΔABC = \(\frac{1}{2} \times B C \times A M\)

⇒ \(\frac{1}{2} \times 5 \times 3\)

= 7.5 square units.

Linear Equations In Two Variables Exercise 3.3

Question 1. Solve the following pair of linear equations by the substitution method :

1. x+y= 14, x-y = 4
2. s-t = 3, \(\frac{s}{3}+\frac{t}{2}=6\)
3. 3x-y = 3, 9x-3y = 9
4. 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3
5. \(\sqrt{2} x+\sqrt{3} y=0\), \(\sqrt{3} x-\sqrt{8} y=0\)
6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\), \(\frac{1}{2 x}+\frac{1}{3 y}=2\)

Solution:

1. x+y = 14 …..(1)

x-y = 4 ……(2)

From equation (1),

x+y =14 ⇒ x= 14 -y

Put this value of x in equation (2),

14 -y -y = 4

⇒ -2y = 4-14 =-10

⇒ y = 5

Put the value of y in equation (3),

x = 14-5 = 9

∴ x = 9, y = 5

2. s-t = 3 …..(1)

⇒ \(\frac{s}{3}+\frac{t}{2}=6\) …….(2)

From equation (1)

s-t =3 ⇒ s = 3 + t ……(3)

Put the value of s from equation (3) to equation (2),

⇒ \(\frac{3+t}{3}+\frac{t}{2}=6 \Rightarrow \frac{6+2 t+3 t}{6}=6\)

⇒ 6+5t = 36

⇒ 5t = 36-6 = 30

⇒ t = 6

Put this value of t in equation (3),

s = 3 + t = 3 + 6 = 9

s =9, t = 6

3. 3x -y = 3 …….(1)

9x- 3y = 9 ……(2)

From equation (1),

3x-y=3 ⇒ y = 3x-3 ,..(3)

Put the value of y from equation (3) to equation (2),

9x- 3(3x- 3) = 9

⇒ 9x- 9x + 9 =9

⇒ 9 = 9

From here, we do not obtain the values of x and y.

∴ Given equations have infinitely many solutions.

4. 0.2x + 0.3y = 1.3 + 3y = 13 …….(1)

and 0.4r + 0.5y = 2.3 => 4x + 5y = 23 …..(2)

From equation (1),

2x + 3y = 13 => 2x = 13-3y-

⇒ \(x=\frac{13-3 y}{2}\)…….(3)

Put the value of x from equation (3) to equation (2)

⇒ \(\frac{4(13-3 y)}{2}+5 y=23\)

⇒ 2(13 -3y) + 5y = 23

⇒ 26 – 6y + 5y = 23

⇒ -y = 23 – 26 =-3

⇒ y = 3

Put the value of y in equation (3)

⇒ \(x=\frac{13-3 \times 3}{2}\)

⇒ \(x=\frac{13-9}{2}=2\)

∴ x = 2,y = 3

5. \(\sqrt{2} x+\sqrt{3} y=0\) …….(1)

⇒ \(\sqrt{3} x-\sqrt{8} y=0\) ……(2)

From equation (2),

⇒ \(\sqrt{3} x=\sqrt{8} y\)

⇒ \(x=\frac{\sqrt{8}}{\sqrt{3}} y\) ……..(3)

Put the value of x from equation (3) to equation (1),

⇒ \(\sqrt{2} \cdot \frac{\sqrt{8}}{\sqrt{3}} y+\sqrt{3} y=0 \Rightarrow \frac{4 y+3 y}{\sqrt{3}}=0\)

⇒ 7y = 0

⇒ y = 0

Put this value of y in equation (3),

⇒ \(x=\frac{\sqrt{8}}{\sqrt{3}} \times 0=0\)

∴ x = 0,y = 0

6. \(\frac{3 x}{2}-\frac{5 y}{3}=-2\) ……(1)

⇒ \(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) …..(2)

From equation (1)

⇒ \(\frac{3 x}{2}=-2+\frac{5 y}{3}=\frac{-6+5 y}{3}\)

⇒ \(x=\frac{10 y-12}{9}\) ………(3)

Put the value of x from equation (3) to equation (2),

⇒ \(\frac{1}{3} \cdot\left(\frac{10 y-12}{9}\right)+\frac{y}{2}=\frac{13}{6}\)

⇒ \(\frac{20 y-24+27 y}{-54}=\frac{13}{6}\)

⇒ \(47 y-24=\frac{13}{6} \times 54=117\)

⇒ 47y = 117 + 24= 141

⇒ y = 3

Put this value of/ in equation (3),

⇒ x = \(\frac{10 \times 3-12}{9}=2\)

∴ x = 2, y = 3

Question 2. Solve 2x + 3y = 11 and 2x -4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:

Here, 2x + 3y= 11 ……(1)

2x- 4y = -24 …….(2)

From equation (2), 4y = 2x+ 24

y = \(\frac{x+12}{2}\) …….(3)

Put the value of y from equation (3) to equation (1),

⇒ \(2 x+3\left(\frac{x+12}{2}\right)=11\)

⇒ 4x + 3(x + 12)= 11 x 2

⇒ 4x + 3x + 36 = 22

⇒ 4x = 22- 36

⇒ 7x = -14

⇒ x = -2

Put x = -2 in equation (3),

⇒ \(y=\frac{-2+12}{2} \Rightarrow y=\frac{10}{2}=5\)

Put x =-2 andy = 5 in the equation, y = mx + 3

⇒ 5 = m x (-2) + 3

⇒ 5 = -2m + 3

⇒ 2m = 3-5=-2

⇒ m = -1

The value of ‘m’ = -1

Question 3. Form the pair of linear equations for the following problems and find their solution by substitution method :

1. The difference between two numbers is 26 and one number is three times the other. Find them.
2. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
3. The coach of a cricket team buys 7 bats and 6 balls for ₹3800. Later, she buys 3 bats and 5 balls for ₹1750. Find the cost of each bat and each ball.
4. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ?₹105 and for a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
5. A fraction becomes \(\frac{9}{11}\) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\) Find the fraction.
6. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

1. Let two numbers be x and y. According to the problem,

x-y = 26

⇒ x = 26 +y ……(1)

and x = 3y …..(2)

⇒ 26 +y = 3y

Put the value of x from equation (1),

⇒ 3y -y= 26

⇒ 2y = 26

⇒ y= 13

From equation (2),

x = 3 x 13 = 39

Therefore numbers =39 and 13

2. Let two angles are x and y According to the problem,

x+y = 180°

⇒ x = 180° -y ……(1)

and x =y+ 18° …..(2)

⇒ 180° -y =y+ 18°

Put the value of x from equation (1)

⇒ -y – y = 18°- 180°

⇒ -2y = -162°

⇒ y = 81°

From equation (2),

x = 81°+ 18° = 99°

∴ Required angle= 99° and 81°

3. Let the cost of one bat = ₹x

and the cost of one ball = ₹y

According to the question,

⇒ 7x + 6y = 3800

⇒ 7x = 3800 – 6y

⇒ \(x=\frac{3800-6 y}{7}\) …….(1)

and 3x+5y = 1750 …….(2)

Put the value of x from equation (1),

⇒ \(\frac{11400-18 y+35 y}{7}=1750\)

11400+ 17y = 12250

⇒ 17y= 12250- 11400 = 850

⇒ y = 50

Put the value of y in equation (1),

⇒ \(x=\frac{3800-6 \times 50}{7}=\frac{3500}{7}=500\)

∴ Cost of 1 bat = ₹500

and Cost of 1 ball = ₹50

4. Let fixed charge = ₹x

and the fare for each lem = ₹y

According to the problem,

x+10y = 105

⇒ x = 105 – 10y …..(1)

and x+15y = 155

⇒ 105 – 10y + 15y = 155

Put the value of x from equation (1),

⇒ 5y = 155- 105 = 50

⇒ y = 10

Put the value of y in equation (1),

x = 105- 10 x 10 = 5

∴ Fixed charge = ₹5

and the fare of each km = ₹10

Fare for travelling 25 km

= x + 25y = 5 + 25 x 10

= 5 + 250 = ₹255

5. Let fraction \(\frac{x}{y}\)

According to the problem,

⇒ \(\frac{x+2}{y+2}=\frac{9}{11}\)

⇒ 11x + 22 = 9y + 18

⇒ 11x = 9y-4

⇒ x = \(\frac{9 y-4}{11}\) ……(1)

and \(\frac{x+3}{y+3}=\frac{5}{6}\)

6x + 18 = 5y + 15

6x = 5y- 3

⇒ \(\frac{6(9 y-4)}{11}=5 y-3\)

Put the value of* from equation (1),

⇒ 54y – 24 = 55y- 33

⇒ 54y- 55y = -33 + 24

⇒ -y = -9 ⇒ y = 9

Put the value of y in equation (1)

⇒ \(x=\frac{9 \times 9-4}{11}=7\)

∴ Fraction = \(\frac{x}{y}=\frac{7}{9}\)

6. Let the present age of Jacob = x years

and present age of son = y years

After 5 years,

Jacob’s age = (x + 5) years

son’s age = (y + 5) years

According to the problem,

x + 5 = 3(y + 5)

⇒ x = 3y+ 10 …….(1)

Five years ago,

Jacob’s age = (x- 5) years

son’s age = (y- 5) years

According to the problem, x- 5 = 7(y-5)

⇒ 3y+10-5 = 7y – 35

Put the value of x from equation (1),

⇒ 3y-7y = -35-5 ⇒ -4y =-40

⇒ y = 10

Put the value of y in equation (1),

⇒ x = 3 x 10+10 = 40

∴ The present age of Jacob = 40 years

Present age of son = 10 years

Linear Equations In Two Variables Exercise 3.4

Question 1. Solve the following pair of linear equations bv the elimination method and the substitution method:

1. x +y = 5 and 2x- 3y = 4
2. 3a + 4y = 10 and 2x- 2y = 2
3. 3x- 5y -4 = 0 and 9x = 2y + 7
4. \(\frac{x}{2}+\frac{2 y}{3}=-1 \text { and } x-\frac{y}{3}=3\)

Solution:

1. x +y = 5 ……(1)

2x-3y = 4 …….(2)

Elimination method

Multiply equation (1) by 3 and adding in equation (2),

⇒ \(x=\frac{19}{5}\)

Put this value of or in equation (1),

⇒ \(\frac{19}{5}+y=5\)

⇒ \(y=5-\frac{19}{5}=\frac{6}{5}\)

∴ \(x=\frac{19}{5}, y=\frac{6}{5}\)

Substitution method

From equation (1),

⇒ x + y = 5 ⇒ x = 5 -y ……(3)

Put the value of x from equation (3) to equation (2)

⇒ 2(5 -y) -3y = 4 ⇒ 10 -2y- 3y = 4

⇒ -5y = 4-10 = -6

⇒ \(y=\frac{6}{5}\)

Put this value of y in equation (3),

⇒ \(x=5-\frac{6}{5}=\frac{25-6}{5}=\frac{19}{5}\)

∴ \(x=\frac{19}{5}, y=\frac{6}{5}\)

Here both methods are appropriate.

2. 3x + 4y = 10 …….(1)

2x-2y = 2 ……(2)

Elimination method

Multiply equation (2) by 2 and adding in equation (1),

x = 2

Put the value of x in equation (1)

3×2+4y = 10

⇒ 4y = 10-6 = 4

⇒ y = 1

∴ x = 2, y = 1

Substitution method

From equation (2),

2x- 2y = 2 ⇒ x-y = 1

⇒ x = 1+y ……(3)

Put the value of x from equation (3) to equation (1),

3(1+y) + 4y = 10

⇒ 3 + 3y + 4y = 10

⇒ 7y = 10- 3 = 7

⇒ y = 1

Put this value of y in equation (1),

x = 1+y

⇒ x = 1+1=2

∴ x = 2, y = 1

Here both methods are appropriate.

3. 3x- 5y- 4 = 0

⇒ 3x- 5y = 4 …(1)

9a = 2y + 7

⇒ 9x-2y= 7 …(2)

Elimination method

Multiply equation (1) by 3 and subtracting from equation (2),

⇒ \(y=\frac{-5}{13}\)

Put this value of y in equation (1),

⇒ \(3 x-5\left(\frac{-5}{13}\right)=4\)

⇒ \(3 x=4-\frac{25}{13}=\frac{52-25}{13}=\frac{27}{13}\)

⇒ \(x=\frac{9}{13}\)

⇒ \(x=\frac{9}{13}\), \(x=\frac{-5}{13}\)

Substitution method

From equation (1),

3x- 5y = 4 ⇒ 3x = 4 + 5y

⇒ \(x=\frac{4+5 y}{3}\) ….(3)

Put the value of x from equation (3) to equation (2),

⇒ \(\frac{9(4+5 y)}{3}-2 y=7\)

3(4 + 5y) – 2y = 7k

⇒ 12 + 15y- 2y = 7

⇒ 13y = 7 – 12= -5 ⇒ \(y=\frac{-5}{13}\)

Put this value of y in equation (3),

⇒ \(x=\frac{4+5\left(\frac{-5}{13}\right)}{3}=\frac{52-25}{3 \times 13}=\frac{27}{3 \times 13}=\frac{9}{13}\)

⇒ \(x=\frac{9}{13}, y=\frac{-5}{13}\)

Here both methods are appropriate.

4. \(\frac{x}{2}+\frac{2 y}{3}=-1 \Rightarrow 3 x+4 y=-6\) …..(1)

⇒ \(x-\frac{y}{3}=3 \quad \Rightarrow 3 x-y=9\) => 3x-y = 9 …..(2)

Elimination method

Subtracting equation (2) from (1),

y = -3

Put the value of y in equation (2),

3x- (-3) = 9

⇒ 3x = 9-3 = 6 ⇒ x = 2

x = 2, y- -3

Substitution method

From equation (2),

3x-y= 9

y = 3x- 9 …..(3)

Put this value of y in equation (1)

⇒ 3x + 4(3x- 9) = -6

⇒ 3x + 12x- 36 = -6

⇒ 15x = -6 + 36 = 30

⇒ x = 2

Put this value of x in equation (3),

y = 3×2-9 = -3

∴ x = 2, y = -3

Here both methods are appropriate.

Question 2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

1. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?
2. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
3. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
4. Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 she received.
5. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.

Solution:

1. Let fraction = \(\frac{x}{y}\)

According to the problem,

⇒ \(\frac{x+1}{y-1}=1\)

⇒ x + 1 = y- 1

⇒ x-y = -2 …..(1)

and \(\frac{x}{y+1}=\frac{1}{2}\)

⇒ 2x = y + 1

⇒ 2x-y = 1 ……(2)

Subtracting equation (1) from equation (2),

Put this value of x in equation (1),

3 -y = -2 ⇒ y = 3 + 2 = 5

∴ Fraction = \(\frac{x}{y}=\frac{3}{5}\)

2. Let present age of Noori = A years and present age of Sonu =y years 5 years ago,

the age of Noori = (x- 5) years

the age of Sonu = (y – 5) years

According to the problem,

(x-5) =3(y-5)

⇒ x- 5 = 3y – 15

⇒ x-3y = -10 ……(1)

After 10 years,

the age of Noori = (x + 10) years

the age of Sonu = (y + 10) years

According to the problem,

(x + 10) = 2(y + 10)

⇒ x+ 10 = 2y + 20

⇒ x- 2y = 10 …..(2)

Subtracting equation (2) from equation (1),

⇒ y = 20

Put the value of y in equation (2),

⇒ x- 2 × 20 = 10

⇒ x = 10 + 40 = 50

∴ The present age of Noori = 50 years

and present age of Sonu = 20 years

3. Let two digit number = 10x +y

According to the problem,

x+y = 9 ….(1)

and 9(10x +y) = 2 x number formed by reversing the digits.

= 2(10 +x)

⇒ 90x+ 9y = 20y + 2x

⇒ 88x- 11y = 0

⇒ 8x -y = 0 …..(2)

Adding equations (1) and (2),

⇒ x = 1

Put the value of A in equation (2),

8×1 -y – 0 ⇒ y = 8

∴ Number = 10A +y = 10 x 1 + 8 = 18

4. Let ₹50 note = x

and ₹100 note = y

According to the problem,

x+y = 25

and ₹100 note =y

According to the problem,

⇒ x+y = 25 …..(1)

and 50x + 100y = 2000

⇒ x + 2y= 40 …..(2)

Subtracting equation (1) from equation (2),

Put this value of y in equation (1),

x + 15 = 25

x = 10

∴ ₹50 note =10

and ₹100 note = 15 …(2)

5. Let the fixed charges for the first 3 days = ₹x

and the charge per day after it = ₹y

According to the problem,

charges for 7 days = ? 27

⇒ x+ (7 – 3)y = 27 ⇒ x + 4y = 27 …(1)

and charges for 5 days = ₹21

⇒ x+ (5 – 2)y = 21

⇒ x + 2y = 21 …..(2)

Subtracting equation (2) from equation (1)

⇒ y = 3

Put this value of y in equation (l),

⇒ x + 4×3 = 27 ⇒  x = 27 – 12 = 15

Fixed charges = ₹15

and charges per day after the first 3 days = ₹3

Linear Equations In Two Variables Exercise 3.5

Question 1. Which of the following pairs of linear equations has a unique solution, no solution, or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.

1. x- 3y – 3 = 0, 3x- 9y -2 = 0
2. 2x + y = 5, 3x + 2y = 8
3. 3x- 5y = 20, 6x- 10y = 40
4. x- 3y -7 = 0, 3x-3y- 15 = 0

Solution:

1. x-3y-3 = 0

3x- 9y- 2 = 0

Here, a₁ = 1, b₁ = -3, c₁ =-3

a₂ = 3, b₂ = -9, c₂ = -2

∴ \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3} \text { and } \frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

∴ Given pair of equations has no solution.

2. 2x +y = 5 ⇒ 2x +y – 5 = 0

3x + 2y =8 ⇒ 3x + 2y – 8 = 0

Here, a₁ = 2, b₁ = 1, c₁ = -5

a₂ = 3, b₂ = 2, c₂ = -8

∴ \(\frac{a_1}{a_2}=\frac{2}{3}, \frac{b_1}{b_2}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given pairofequations has a unique solution.

From the cross-multiplication method,

⇒ \(\frac{x}{-8-(-10)}=\frac{y}{-15-(-16)}=\frac{1}{4-3}\)

⇒ \(\frac{x}{2}=\frac{y}{1}=\frac{1}{1} \quad \Rightarrow x=2, y=1\)

3. 3x- 5y = 20 ⇒ 3x-5y- 20 = 0

6x- 10y = 40 ⇒ 6x- 10y- 40 = 0

Here, a₁ =3, b₁, =-5, c₁, =-20

a₂ = 6, b₂ =-10 c₂ = -40

∴ \(\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-5}{-10}=\frac{1}{2}, \frac{c_1}{c_2}=\frac{-20}{-40}=\frac{1}{2}\)

∵ \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

∴ Given pair of equations has infinitely many solutions.

4. x-3y – 7 = 0

3x-3y- 15 = 0

Here, a₁ = 1, b₁ = -3, c₁ = -7

a₂ = 3, b₂ = -3, c₂ = -15

∴ \(\frac{a_1}{a_2}=\frac{1}{3}, \frac{b_1}{b_2}=\frac{-3}{-3}=1\)

∵ \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

∴ Given pair ofequations has aunique solution.

Now, from the cross-multiplication method,

⇒ \(\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}\)

⇒ \(\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}\)

⇒ \(\frac{x}{4}=\frac{y}{-1}=\frac{1}{1}\)

⇒ x = 4, y = -1

Question 2.

1. For which values of a and b does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7, (a-b)x+(a+b)y = 3a+b-2
2. For which value of k will the following pair of linear equations have no solution? 3x- +y = 1, (2k-1)x+(k-1)y = 2k+1

Solution:

1. Here, 2x + 3y – 7 = 0

and (a – b)x + (a + b)y – (3a + b – 2) = 0

Now, a₁ = 2, b₁ = 3, c₁ = -7,

a₂= a – b, b₂ = a + b, c₂ =- (3a +b- 2)

For infinitely many solutions,

⇒ \(\frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-(3 a+b-2)}\)

⇒ \(\left(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\right)\)

⇒ \(\frac{a-b}{2}=\frac{a+b}{3}=\frac{3 a+b-2}{7}\)

From the first and second expressions,

⇒ \(\frac{a-b}{2}=\frac{a+b}{3}\)

⇒ 3a-3b = 2a + 2b

⇒ a = 5b …(1)

From the second and third expressions,

⇒ \(\frac{a+b}{3}=\frac{3 a+b-2}{7}\)

⇒ 7a + 7b = 9a + 3b – 6

⇒ 4b = 2a -6

⇒ 2b = a-3 …(2)

From equations (1) and (2),

2b = 5b -3 ⇒ 6 = 1

Put 6= 1 in equation (1),

a = 5 x 1 ⇒ a = 5

2. Here, 3x+y- 1 =0 …..(1)

and (2k- 1)x + (k- l)y- (2k + 1) = 0 …(2)

Here, a₁ =3, b₁ = 1, c₁ = -1,

a₂ =2k-1, b₂ = k-1, c₂ = -(2k+1)

⇒ \(\frac{a_1}{a_2}=\frac{3}{2 k-1}, \frac{b_1}{b_2}=\frac{1}{k-1}, \frac{c_1}{c_2}=\frac{1}{2 k+1}\)

For no solution, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\) i.e., \(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{1}{2 k+1}\)

⇒ \(\frac{3}{2 k-1}=\frac{1}{k-1} \neq \frac{1}{2 k+1}\)

⇒ 3(k-1) = 2k-1 and k-1 = 2k+1

⇒ 3k-3 = 2k-1 and k = -2

⇒ k = 2 and k = -2

Therefore, the given system will have no solution when k = 2

Question 3. Solve the following pair of linear equations by the substitution and cross-multiplication methods. 8x + 5y = 9, 3x + 2y = 4
Solution:

Given equations

8x + 5y = 9 …….(1)

3x- + 2y = 4 ……(2)

Substitution method

From equation (1),

⇒ 8x = 9 – 5y

⇒ \(x=\frac{9-3 y}{8}\) ……(3)

Put this value of x in equation (2),

⇒ \(\frac{3(9-5 y)}{8}+2 y=4\)

⇒ \(\frac{27-15 y+16 y}{8}=4\)

⇒ 27 +y = 32 = 5

Put the value of/ in equation (3)

⇒ \(x=\frac{9-5 \times 5}{8}=\frac{9-25}{8}=-2\)

⇒ x = -2, y = 5

From the cross-multiplication method,

8x + 5y- 9 = 0

3x + 2y- 4 = 0

⇒ \(\frac{x}{-20+18}=\frac{y}{-27+32}=\frac{1}{16-15}\)

⇒ \(\frac{x}{-2}=\frac{y}{5}=\frac{1}{1}\)

⇒ x = -2, y = 5

Here, both methods are appropriate.

Question 4. Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method :

1. A part of monthly hostel charges are fixed and the remaining depends on the number of days one has taken food in the mess. When student A takes food for 20 days she has to pay ₹1000 as hostel charges whereas student B, who takes food for 26 days, pays ₹1180 as hostel charges. Find the fixed charges and the cost of food per day.
2. A fraction becomes when \(\frac{1}{3}\) is subtracted from the numerator and it becomes \(\frac{1}{4}\) when 8 is added to its denominator. Find the fraction.
3. Yash scored 40 marks on a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
4. Places’ A’ and B are 100 1cm apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
5. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Solution:

Let fixed charge = ₹x

and the cost of food per day = ₹y

∴ x + 20y = 1000 ………(1)

and x + 26y = 1180 ………(2)

Subtracting equation (1) from equation (2),

⇒ y = 30

From equation (1),

x + 20×30 = 1000

⇒ x = 1000-600 = 400

∴ Fixed charge = ₹400

and cost of food per day = ₹30

2. Let, fraction = \(\frac{x}{y}\)

According to the problem,

⇒ \(\frac{x-1}{y}=\frac{1}{3}\)

⇒ 3x-3=y

⇒ 3x-y = 3 …….(1)

and \(\frac{x}{y+8}=\frac{1}{4}\)

⇒ 4x =y + 8 ….(2)

⇒ 4x -y = 8

Subtracting equation (2) from equation (1),

Put the value of x in equation (2),

4×5-y = 8 ⇒ y = 20-8 = 12

∴ Fraction = \(\frac{x}{y}=\frac{5}{12}\)

3. Let number of correct answers = x and number of incorrect answers =7.

According to the first condition,

3x -7 = 40 ……..(1)

According to the second condition,

4x-2y =50 ⇒ 2x -7 = 25 ……..(2)

Subtracting equation (2) from equation (1),

Put the value of x in equation (2),

⇒ 2(15) -7 = 25

⇒ 30 -7 = 25 ⇒ y = 30 – 25 = 5

Total number of questions in a test

= x +7 =15 + 5= 20

4. Let the speed of the car starting from A =xkm/hr and the speed of the car starting from B =                       7 km/hr

Case 1: When they move in the same direction:

Distance covered by car A = AP= x × 5 Km ( ∵ distance = speed x time)

Distance covered by carB = BP = y × 5 km

⇒ but AB = 100 km

⇒ AP-BP = 100

⇒ 5x -5y = 100 ⇒ x-y=20 …….(1)

Case 2: When they move in opposite directions :

Distance covered by car A = AQ=x × 1 km

Distance covered by carB = BQ =y × 1 km

AB = 100 km

AQ + BQ = 100 ⇒ x +y = 100 …(2)

Adding equations (1) and (2),

2x = 1 20 ⇒ x = 60 km/hr

Put x = 60 in equation (2)

y = 100 – 60 = 40 km/hr

∴ Speed Of car A = 60 km/hr and speed of cal- B = 40 km/hr.

5. Let the length of the rectangle be x and breadth y units.

∴ Area of rectangle = xy square units

According to the given conditions,

⇒ (x-5)(y + 3) =xy-9

⇒ xy+ 3x-5y- 15 = xy- 9

⇒ 3x- 5y = 6 …….(1)

and (x + 3)(y + 2) = xy + 67

⇒ xy + 2x + 3y + 6= xy + 67

⇒ 2x + 3y =61 ……..(2)

Multiplying equation (1) by 3 and equation (2) by 5

9x- 15y = 18 …(3)

10x + 15y = 305

Adding equations (3) and (4)

19x = 323 ⇒ x = 17

Put x = 17 in equations (1)

3 x 17 – 5y = 6

⇒ 51-5y = 6

⇒ -5y = -45 ⇒ y = 9

Therefore, the length of the rectangle = 17 units and the breadth = 9 units.

Linear Equations In Two Variables Exercise 3.6

Question 1. Solve the following pairs of equations by reducing them to a pair of linear equations :

1. \(\frac{1}{2 x}+\frac{1}{3 y}=2 \quad \frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)
2. \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2 \quad \frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\)
3. \(\frac{4}{x}+3 y=14 \quad \frac{3}{x}-4 y=23\)
4. \(\frac{5}{x-1}+\frac{1}{y-2}=2 \quad \frac{6}{x-1}-\frac{3}{y-2}=1\)
5. \(\frac{7 x-2 y}{x y}=5 \quad \cdot \quad \frac{8 x+7 y}{x y}=15\)
6. 6x + 3y = 6xy, 2x + 4y = 5xy
7. \(\frac{10}{x+y}+\frac{2}{x-y}=4\),\(\frac{15}{x+y}-\frac{5}{x-y}=-2\)
8. \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\), \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\)

Solution:

1. Given equations, \(\frac{1}{2 x}+\frac{1}{3 y}=2,\)

⇒ \(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

Let, \(\frac{1}{x}=a \text { and } \frac{1}{y}=b\)

∴ Given equations, \(\frac{a}{2}+\frac{b}{3}=2\)

⇒ 3a + 2b = 12 …….(1)

and \(\frac{a}{3}+\frac{b}{2}=\frac{13}{6}\)

⇒ 2a+3b = 13 ……….(2)

Multiplying equation (1) by 2 and equation (2) by 3 and subtracting,

⇒ b = 3

Put the value of b in equation (1),

⇒ 3a + 2 × 3 = 12

⇒ 3a = 12-6 = 6

⇒ a = 2

Now, \(\frac{1}{x}=a \quad \Rightarrow \quad x=\frac{1}{a}=\frac{1}{2}\)

and \(\frac{1}{y}=b \quad \Rightarrow \quad y=\frac{1}{b}=\frac{1}{3}\)

∴ \(x=\frac{1}{2}, y=\frac{1}{3}\)

2. Given equations, \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\)

⇒ \(\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1\)

Let \(\frac{1}{\sqrt{x}}=A \text { and } \frac{1}{\sqrt{y}}=B\)

∴ Given equations, 2A + 3B =2 …(1)

4A-9B = -1 ……(2)

Multiplying equation (1) by 3 and adding in equation (2),

⇒ \(A=\frac{1}{2}\)

Put the value of A in equation (1),

⇒ \(2 \times \frac{1}{2}+3 B=2\)

35 = 2-1 = 1 = B = \(\frac{1}{3}\)

Now, \(\frac{1}{\sqrt{x}}=A \quad \Rightarrow \sqrt{x}=\frac{1}{A}\)

⇒ \(x=\frac{1}{A^2}=\frac{1}{\left(\frac{1}{2}\right)^2}=4\)

and \(\frac{1}{\sqrt{y}}=B \Rightarrow \sqrt{y}=\frac{1}{B}\)

⇒ \(y=\frac{1}{B^2}=\frac{1}{\left(\frac{1}{3}\right)^2}=9\)

∴ x = 4, y = 9

3. Given equations, \(\frac{4}{x}+3 y=14, \frac{3}{x}-4 y=23\)

Let \(\frac{1}{x}=A\)

∴ Given equations, 4A + 3y = 14 ………(1)

3A – 4y = 23 ………(2)

Multiplying equation (1) by 4 and equation (2) by 3 and adding

⇒ A = 5

Put the value of A in equation (1),

4 × 5 + 3y = 14

⇒ 3y = 14 – 20 = -6

⇒ y = -2

Now, \(\frac{1}{x}=A \Rightarrow x=\frac{1}{A}=\frac{1}{5}\)

⇒ \(x=\frac{1}{5}, y=-2\)

4. Given equations, \(\frac{5}{x-1}+\frac{1}{y-2}=2\)

and \(\frac{6}{x-1}-\frac{3}{y-2}=1\)

Let \(\frac{1}{x-1}=A \text { and } \frac{1}{y-2}=B\)

∴ Given equations, 5A + B = 2 …..(1)

6A-3B = 1 ……(2)

Multiplying equation (1) by 3 and adding in equation (2),

⇒ \(A=\frac{7}{21}=\frac{1}{3}\)

Put the value of A in equation (1),

⇒ \(5 \times \frac{1}{3}+B=2 \quad \Rightarrow B=2-\frac{5}{3}=\frac{1}{3}\)

Now, \(\frac{1}{x-1}=A\)

⇒ \(x-1=\frac{1}{A}=\frac{1}{1 / 3}=3\)

⇒ x = 4

and \(\frac{1}{y-2}=B\)

⇒ \(y-2=\frac{1}{B}=\frac{1}{1 / 3}=3\)

⇒ y = 5

∴ x = 4, y = 5

5. Given equations, \(\frac{7 x-2 y}{x y}=5\)

⇒ \(\frac{7 x}{x y}-\frac{2 y}{x y}=5\)

⇒ \(\frac{7}{y}-\frac{2}{x}=5\) …..(1)

and \(\frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15\)

⇒ \(\frac{8}{y}+\frac{7}{x}=15\) ……(2)

Let \(\frac{1}{y}=A \quad \text { and } \quad \frac{1}{x}=B\)

∴ From equation (1), 7A-2B =5 …(3

From equation (2), 8A + 7B = 15 …(4)

Multiplying equation (3) by 7 and equation (4) by 2 and adding,

⇒ A = 1

Put the value of A in equation (4),

8 × 1 + 7B = 15

⇒ 7B = 15 – 8 = 7

⇒ B = 1

Now, \(\frac{1}{x}=B \Rightarrow x=\frac{1}{B}=\frac{1}{1}=1\)

and \(\frac{1}{y}=A \Rightarrow y=\frac{1}{A}=\frac{1}{1}=1\)

∴ x = 1, y =1

6. Given equations, 6x+3y = 6xy

⇒ \(\frac{6 x}{3 x y}+\frac{3 y}{3 x y}=\frac{6 x y}{3 x y} \Rightarrow \frac{2}{y}+\frac{1}{x}=2\) …….(1)

and 2x + 4y = 5xy

⇒ \(\frac{2 x}{x y}+\frac{4 y}{x y}=\frac{5 x y}{x y} \Rightarrow \frac{2}{y}+\frac{4}{x}=5\) ……..(2)

Let \(\frac{1}{y}=A \quad \text { and } \quad \frac{1}{x}=B\)

∴ From equation (1), 2A + B = 2 …..(3)

From equation (2), 2A + 4B = 5 …..(4)

Subtracting equation (3) from equation (4),

B=1

Put the value of B in equation’ (3),

2A + 1 = 2 ⇒ 2A = 2-1 = 1

⇒ \(A=\frac{1}{2}\)

Now, \(\frac{1}{y}=A\)

⇒ \(y=\frac{1}{A}=\frac{1}{1 / 2}=2\)

⇒ \(\frac{1}{x}=B \quad \Rightarrow x=\frac{1}{B}=\frac{1}{1}=1\)

∴ x = 1, y = 2

7. Given equations,

⇒ \(\frac{10}{x+y}+\frac{2}{x-y}=4\) …..(1)

⇒ \(\frac{15}{x+y}-\frac{5}{x-y}=-2\) ….(2)

Let \(\frac{1}{x+y}=A \text { and } \frac{1}{x-y}=B\)

∴ From equation (1), 10A + 25 = 4

From equation (2), 15A – 55 = -2

Multiplying equations (3) by 5 and equation (4) by 2 and adding,

⇒ \(A=\frac{16}{80}=\frac{1}{5}\)

Put the value of A in equation (3),

⇒ \(10 \times \frac{1}{5}+2 B=4 \Rightarrow 2+2 B=4\)

2B = 4-2 = 2 ⇒ B = 1

Now \(\frac{1}{x+y}=A=\frac{1}{5}\)

⇒ x+y = 5 …..(5)

and \(\frac{1}{x-y}=B=1\)

⇒ x-y = 1 …..(6)

Adding equation (5) and equation (6)

⇒ x= 3

Put the value of x in equation (5)

3 + = 5 ⇒ y = 2

∴ x = 3, y = 2

8. Given equations, \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) …….(1)

and \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) ….(2)

Let \(\frac{1}{3 x+y}=A \text { and } \frac{1}{3 x-y}=B\)

From equation (1)

⇒ \(\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4}\) ……(1)

and \(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=-\frac{1}{8}\) …..(2)

Let \(\frac{1}{3 x+y}=A \text { and } \frac{1}{3 x-y}=B\)

From equation (1), \(A+B=\frac{3}{4}\) ……(3)

From equation (2), \(\frac{A}{2}-\frac{B}{2}=-\frac{1}{8}\)

⇒ \(A-B=-\frac{2}{8} \Rightarrow A-B=-\frac{1}{4}\) ……(4)

Adding equations (3) and (4)

A = \(\frac{1}{4}\)

Put the value of A in equation (3),

⇒ \(\frac{1}{4}+B=\frac{3}{4}\)

⇒ \(B=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}\)

Now \(\frac{1}{3 x+y}=A=\frac{1}{4}\)

⇒ 3x-y= 4 ….(5)

and \(\frac{1}{3 x-y}=B=\frac{1}{2}\)

⇒ 3x-y= 2 ….(6)

Adding equations (5) and (6),

⇒ x= 1

Put the value of x in equation (5),

⇒ 3 × 1 +y = 4

⇒ y = 4-3=1

∴ x = 1, y = 1

Question 2. Formulate the following problems as a pair of equations, and hence find their solutions:

1. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
2. 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
3. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

1. Let the speed of Ritu in still water = x km/hr

and speed of current =y km/hr

∴ speed in downstream = (x +y) km/hr

and speed in upstream = (x -y) km/hr

Now from \(\text { time }=\frac{\text { distance }}{\text { speed }}\)

⇒ \(\frac{20}{x+y}=2 \text { and } \frac{4}{x-y}=2\)

⇒ x + y = 10 …..(1)

and x-y = 2 ….(2)

Adding equations (1) and (2),

2x = 12 ⇒ x= 6

Put x = 6 in equation (1),

6 +y = 10 ⇒ y = 4

∴ Speed of Ritu in still water = 6 km/hr

and speed of current = 4 km/hr

2. Let 1 woman can finish the work in x days and 1 man can finish the work in y days.

∴ 1 day work of 1 woman = \(\frac{1}{x}\)

1 day work of 1 man = \(\frac{1}{y}\)

1-day work of 2 women and 5 men \(\frac{2}{x}+\frac{5}{y}=\frac{5 x+2 y}{x y}\)

∴ Time taken to finish the work = \(\frac{x y}{5 x+2 y}\)

Given that, \(\frac{x y}{5 x+2 y}=4\)

Similarly in second case, \(\frac{x y}{6 x+3 y}=3\)

Then, \(\frac{5 x+2 y}{x y}=\frac{1}{4} \text { and } \frac{6 x+3 y}{x y}=\frac{1}{3}\)

⇒ \(\frac{20}{y}+\frac{8}{x}=1 \text { and } \frac{18}{y}+\frac{9}{x}=1\)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\)

∴ 20v + 8u = 1 …..(1)

and 18v+9u = 1 ….(2)

Multiplying equation (1) by 9 and equation (2) by 8,

180v-144v = 9-8

⇒ 36v = 1

⇒ v = \(\frac{1}{36}\)

Put v = \(\frac{1}{36}\) in equation (2),

⇒ \(18 \times \frac{1}{36}+9 u=1\)

⇒ 9u = \(1-\frac{1}{2}\)

⇒ 9u = \(\frac{1}{2}\)

⇒ u = \(\frac{1}{18}\)

Now, \(u=\frac{1}{18} \quad \text { and } \quad v=\frac{1}{36}\)

⇒ \(\frac{1}{x}=\frac{1}{18} \text { and } \frac{1}{y}=\frac{1}{36}\)

⇒ x= 18 and y = 36

Therefore, 1 woman can finish the work alone in 18 days and 1 man can finish the work alone in 36 days.

3. Let the speed of train = x km/hr and speed of bus =y km/hr

In the first case, Ruhi goes 60 km from the train and 240 km from a bus in 4 hours.

⇒ \(\frac{60}{x}+\frac{240}{y}=4\)

⇒ \(\frac{15}{x}+\frac{60}{y}=1\) …..(1)

Similarly in the second case, she goes 100 km from the train and 200 km from a bus in \(4 \frac{1}{6}\) hours.

⇒ \(\frac{100}{x}+\frac{200}{y}=4+\frac{1}{6}\) ,\(\left(10 \text { minutes }=\frac{1}{6} \mathrm{hr}\right)\)

⇒ \(\frac{100}{x}+\frac{200}{y}=\frac{25}{6}\)

⇒ \(\frac{24}{x}+\frac{48}{y}=1\) ….(2)

Let \(\frac{1}{x}=u \text { and } \frac{1}{y}=v\)

⇒ 15u + 60v = 1 ……(3)

and 24u + 48v = 1 …..(4)

Multiplying equation (3) by 24 and equation (4) by 15 and subtracting,

24(15u + 60v)- 15(24u + 48v) =24-15

⇒ 440v – 720v = 9

⇒ 20v = 9

⇒ v = \(\frac{9}{720}=\frac{1}{80}\)

Put v = \(\frac{1}{80}\)

⇒ \(15 u+60 \times \frac{1}{80}=1\)

⇒ \(15 u=1-\frac{3}{4}=\frac{1}{4}\)

⇒ u = \(\frac{1}{60}\)

Now, u = \(\frac{1}{60}\) and v = \(\frac{1}{80}\)

⇒ \(\frac{1}{x}=\frac{1}{60} \text { and } \frac{1}{y}=\frac{1}{80}\)

⇒ x = 60 and y = 80

Therefore, the speed of the train = 60 km/hr and speed of bus = 80 km/hr.

Linear Equations In Two Variables Exercise 3.7 (Optional)

Question 1. The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Solution:

Given

The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years.

For convenience, let their ages be according to their names.

Now, A – B = ± 3 ……(1) (As we don’t know who is older)

and \(\left.\begin{array}{rl}
D=2 A \\
B=2 C \Rightarrow C=\frac{B}{2}
\end{array}\right\}\)

and D – C = 30

But we don’t know who is older, so D-C= ±30

⇒ \(2 A-\frac{B}{2}= \pm 30 \Rightarrow 4 A-B= \pm 60\) ……..(2)

Now from equations (1) and (2), four cases are possible:

Question 2. One says, “Give me a hundred, friend! f shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara 2]

[Hint : x + 100 = 2(y- 100),y + 10 = 6(x- 10)].

Solution:

Let them have ₹x and ₹y respectively.

∴ According to the given condition,

⇒ x+ 100 = 2 (y- 100)

⇒ x – 2y = -300 …..(1)

and 6(x- 10) = + 10 ⇒ 6x-y = 70 …(2)

Multiplying equation (2) by 2 and subtracting from equation (1),

x- 12x = -300 – 140 ⇒ -11x = -440

x = ₹40

Put in c(|untion (1),

40 – 2y = -300

2y – 340 ⇒ y = ₹170

So, they have ₹40 and ₹170 respectively.

Question 8. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Solution:

Let speed of train = x km/hr and distance covered = y km

Time taken at this speed = \(\frac{y}{x} \text { hours }\)

If speed = (x + 10) km/hr

then time taken = \(\frac{y}{x+10} \text { hours }\)

∴ \(\frac{y}{x+10}=\frac{y}{x}-2 \Rightarrow \frac{y}{x+10}=\frac{y-2 x}{x}\)

⇒ xy = (x+10)(y-2x)

⇒ xy = xy-2x²+10y-20x

⇒ -2x²+ 10y-20x = 0 …(1)

If speed = (x- 10) km/hr

then time taken = \(\frac{y}{x-10} \mathrm{hr}\)

∴\(\frac{y}{x-10}=\frac{y}{x}+3\)

⇒ \(\frac{y}{x-10}=\frac{y+3 x}{x}\)

⇒ xy = (x-10)(y+3x)

⇒ xy = xy + 3x²-10y-30x

⇒ 3x²– 10y-30x = 0 ……(2)

Adding equations (1) and (2),

x² – 50x = 0 ⇒ x(x-50) = 0

⇒ x = 0 or x-50 = 0

⇒ x = 0 or x = 50

x = 0 is not possible

∴ x = 50 km/hr

Put the value of x in equation (1),

-2(50)² + 10y- 20×50 = 0

⇒ -5000+ 10y- 1000 = 0

⇒ 10y = 6000

⇒ y = 6000km

∴ Distance covered = 600 km

Question 4. ‘The students of a class are made to stand in rows. If 3 students arc extra in a row, there would be 1 row. If 3 students are less in a and there would be 2 rows more. Find the number of students in the class.
Solution:

Given

‘The students of a class are made to stand in rows. If 3 students arc extra in a row, there would be 1 row. If 3 students are less in a and there would be 2 rows more.

Let the number of rows = x

and number of students in each row = y

Total number of students in the class = xy

According to the first condition,

(y + 3) (x- 1) = xy

⇒ xy-y + 3xr3 = xy

⇒ 3x-y = 3 …(1)

According to the second condition,

(y-3)(x + 2)=xy

⇒ xy + 2y-3x- 6 = xy

⇒ 3x-2y = -6 ……(2)

Subtracting equation (2) from equation (1),

y = 9

Put y = 9 in equation (1),

3x- 9 =3

⇒ 3x= 12

⇒ x = 4

∴ Total number of students in the class = xy = 4 x 9 = 36

Question 5. In ΔABC, ∠C = 3∠B = 2(∠A+∠B). Find the three angles.
Solution:

Given

In ΔABC, ∠C = 3∠B = 2(∠A+∠B).

Let ∠A=x° and ∠B=y°

∴ ∠C = 3∠B ⇒ ∠C = 3y°

and ∠C = 2(∠A + ∠B)

⇒ 3y° = 2(x° +y°)

⇒ 2x + 2y-3y =0

⇒ 2x-y = 0 ……(1)

∵ ∠A + ∠B + ∠C = 180° (sum of three angles of Δ)

⇒ x+y + 3y = 180

⇒ x + 4y = 180 …..(2)

Multiplying equation (2) by 2

2x+ 8y = 360 …….(3)

Subtracting equation (3) from (1),

-9y = -360 ⇒ y = 40°

Put y = 40° in equation (1),

2x- 40 = 0

⇒ 2x = 40 ⇒ x = 20°

So, ∠A = 20°, ∠B = 40°

and ∠C = 3 × 40° = 120°

Question 6. Draw the graphs of the equations 5x-y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis.
Solution:

5x-y = 5 y ⇒ 5x-5

Mark the points A(0, -5), B( 1, 0) and C(2, 5) and join them.

Again, 3x -y = 3 ⇒ y= 3x-3

Mark the points D(0, -3), B( 1, 0) and E(2, 3) and join them.

The coordinates of the vertices of ΔBDA formed by two lines and the y-axis are as follows :

B( 1, 0), A(0, -5) and D(0, -3)

Now area of ΔBAD = \(\frac{1}{2} \times A D \times B O\)

⇒ \(\frac{1}{2} \times 2 \times 1\)

= 1 sq. units

Question 7. Solve the following pair of linear equations :

1. px + qy -p – q , qx-py = p + q
2. ax + by = c, bx + ay = 1 + c
3. \(\), ax + by = a2 +b2
4. (a – b)x + (a + b)y = a2 – 2ab – b2, (a + b)(x +y) =a2 +b2
5. 152x – 378y = -74, -378x + 152y = -604

Solution:

1. Given equation,

px + qy = p-q ….(1)

and qx – py = p+q

Multiplying equation (1) by p and equation (2) by q and adding,

⇒ \(\left(p^2+q^2\right) x=p^2+q^2\)

⇒ x = 1

Put the value of x in equation (1),

p×1+qy = p-q

⇒ qy = -q

⇒ y = -1

∴ x = 1, y = -1

2. ax + by = c

⇒ ax + by -c = 0

⇒ bx + ay = 1 +c ⇒ bx + ay- (1+ c) = 0

Now,

∴ \(\frac{x}{-b(1+c)+a c}=\frac{y}{-b c+a(1+c)}=\frac{1}{a^2-b^2}\)