Class 8 Maths

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Introduction

A paper is a very common example of a plane surface. The curve obtained by joining several points consecutively without lifting the pencil from the paper is called a plane curve or simple curve or simply curve.

Some examples of simple curves are as follows

A circle is a very common example of a plane curve.

Some types of plane curves are as follows :

1. Open curve: A curve that does not cut itself is called an open curve.

Read and Learn More NCERT Solutions For Class 8 Maths

For example:

   is an open curve.

2. Closed curve:  A curve that cuts itself is called a closed curve.

For example:

 is a closed curve.

3. Simple closed curve: A closed curve that does not pass through a point more than once is called a simple closed
curve.

For example: 

is a simple closed curve.

Note: A curve can be straight also, i.e., A straight line is also a curve. It is an open

Question 1. Match the following: (Caution A figure may match more than one type)
Solution:

Compare your matchings with those of your friends. Do they agree?
Solution:

1-C

2-B

3-A

4-D

Please compare yourself your matchings with those of friends and find out whether they agree.

Polygons

A polygon is a simple closed-curve formed of only line segments. A triangle is a very common example of a polygon

Question 1. Try to give a few more examples and non-examples for a polygon.
Solution:

The following curves are examples of a polygon:

The following curves are non-examples of a polygon:

Question 2. Draw a rough figure of a polygon and identify its sides and vertices.
Solution:

Sides : AB, BC, CD, DE, EF, FG, GH, HI, IJ, JK, KL and LA.

Vertices: A, B, C, D, E, F, G, H, I, J, KandL.

Convex And Concave Polygons

A polygon is said to be convex if it has no portion of its diagonals in its exterior otherwise it is said to be a concave polygon.

Question 1. Can you find how these types of polygons differ from one another

Solution:

Yes, some of these are convex polygons while the others are concave polygons.

These polygons differ in the sense that in a convex polygon, the line segment joining any two points inside it lies completely inside the polygon whereas this property does not hold well in a concave polygon.

Question 2. Polygons that are convex have no portions of their diagonals in their exteriors. Is it true with concave polygons?
Solution:  No, it is not true with concave poly polygons

Question 3. Study the figures given below

Then try to describe in your own words what we mean by a convex polygon and what we mean by a concave polygon. Give two rough sketches of each kind
Solution:

By a convex polygon, we mean a polygon that has no portion of its diagonals in its exterior. On the contrary, by a concave polygon, we mean a polygon that has some portion of its diagonals in its exterior

1. Two rough sketches of a convex polygon

2. Two rough sketches of a concave polygon

Regular And Irregular Polygons

A polygon that is both ‘equiangular’ (has all angles of equal measure) and ‘equilateral’ (has all sides of equal measure) is called a regular polygon,

For example: A square, an equilateral triangle.

A polygon which is equiangular but not equilateral is called an irregular polygon. For example: a rectangle

Question 1. Is a rectangle a regular polygon? my?
Solution: No, A rectangle is not a regular polygon because it is equiangular but not equilateral

Question 2. Is an equilateral triangle a regular polygon? Why?
Solution: Yes, An equilateral triangle is a regular polygon because it is both equiangular and equilateral.

Question 3. Is there a triangle that is equilateral but not equiangular?
Solution:

No, there is no triangle which is equilateral but not equiangular

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.1

Question 1.  Given Here are some figures

Classify each of them based on the following;

1. Simple curve
2. Simple closed curve
3. Polygon
4. Convex polygon
5. Concave polygon

Solution:

1. Simple curve-1,2,5,6,7
2. Simple closed curve-1,2,5,6,7
3. Polygon – 1,2
4. Convex polygon – 2
5. Concave polygon – 1

Question 2. What is a regular polygon? State the name of a regular polygon of

1. 3 sides
2. 4 sides
3. 6 sides.

Solution:

A polygon, both ‘equilateral’ and ‘equiangular’, is called a regular polygon.

1. 3 sides: The name of the regular polygon of 3 sides is an equilateral triangle.
2. 4 sides: The name of the regular polygon of 4 sides is square.
3. 6 sides: The name of the regular polygon of 6 sides is a regular hexagon.

Sum Of The Measures Of The Exterior Angles Of A Polygon

The sum of the measures of the exterior angles of a polygon is 360°

Draw a polygon on the floor, using a piece of chalk. (In the figure, a pentagon ABCDE is shown)

We want to know the total measure of angles, i.e.

m∠1 + m∠2 + m∠3 + m∠4 + m∠5

Start at A. Walk along \(\overline{\mathrm{AB}} \text {. }\). On reaching B, we need to turn through an angle of m∠1, to walk along \(\overline{\mathrm{BC}} \text {. }\)

When we reach C, we need to turn E D through an angle of m∠2 to walk along CD.

We continue to move in this manner until we return to side AB. We would have made one complete turn.

∴  m∠1 + m∠2 + m∠3 + m∠4 + m∠5 = 360°

This is true whatever the number of E sides ofthe polygon.

Therefore, the sum of the measures of y the external angles of any polygon is 360°

Take a regular hexagon

Question 1. What is the sum of the measures of its exterior angles r, y, z, p, q, r?
Solution:

The sum of the measures of its exterior angles x, y, z, p , q, r is 360°.

Question 2. Is x = y = z = p = q = r? Why?
Solution:

Yes, x = y = z = p = q = r because

x + a = y + a = z + a = p + a = q + a = r + a

= 180°

Linear pair property

Question 3. What is the measure of each?

1. Exterior angle
2. Interior angle

Solution:

1. The measure of each exterior angle is \(\frac{360^{\circ}}{6}=60^{\circ}.\) = 60°

The sum of the measures of the exterior angles of a polygon is 360°. Also, all interior angles are of equal Consequently, all exterior angles are of equal measure (by Unear pair property).

2. The measure of each interior angle is 180°-60° = 120°

Note:

1. Each exterior angle of an n–sided regular polygon

⇒ \(\frac{360^{\circ}}{6}=60^{\circ} .\)

2. Number of sides of a regular polygon

= \(\frac{360^{\circ}}{n} .\)

3. Each interior angle of an n-sided regular polygon

= \(\frac{(n-2) 180^{\circ}}{n} \text { Or } \frac{(2 n-4) 90^{\circ}}{n}\)

=  \(\frac{(2 n-4)}{n}\)right angles.

Question 4. Repeal, this activity in the cases of

1. A regular octagon
2. A regular 20-gon.

Solution:

1. A regular octagon

1. The measure of each exterior angle = \(\frac{360^{\circ}}{8}=45^{\circ} .\)
2. The measure of each interior angle is 180° -45° =135°

2. A regular 20-gon

1. The measure of each exterior angle \(\frac{360^{\circ}}{20}=18^{\circ} .\)
2. The measure of each interior angle is 180° -18° =162°

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.2

Question 1. Find x in the following figures

Solution:

1.  x°+ 125° + 125° = 360°.

∵ The sum of the measures of the exterior angles of a polygon is 360°

x + 250° = 360°

x = 360° -250°

x  = 110°

2. x + 70° + 90° + 60° + (180°- 90°)

= 360°

x + 160° + 60° + 90° = 360°

x + 220° + 90° = 360°.

The sum of the measures of the exterior angles of a polygon is 360° and the linear pair property.

x+ 310° = 360°

x = 360° – 310°

x = 50°.

Question 2. Find the measure ofeach exterior angle of a regular polygon of

1. 9 sides
2. 15sides

Solution:

1. 9 sides:

Here, n = 9

Measure of each exterior angle

= \(\frac{360^{\circ}}{n}=\frac{360^{\circ}}{9}\)

= \(40^{\circ}\)

2. 15 sides:

No, since 22 is not a factor of

Hero, n = 15

Measure of each exterior angle

= \(\frac{360^{\circ}}{n}=\frac{360^{\circ}}{15}\)

= \(24^{\circ}\)

Question 3. How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution:

Let the number of sides be n. Then

n = \(\frac{360^{\circ}}{\text { Each exterior angle }} \text {. }\)

n = \(\frac{360^{\circ}}{24^{\circ}}\)

= 15

Hence, the number of sides is 15.

Question 4. How many sides does a regular polygon have if each of its interior angles is 165°?
Solution: 

Each interior angle = 165°

∴  Each exterior angle

= 180° – 165° = 15°

∴ Linear pair property

Let the number of sides be n. Then

n = \(\frac{360^{\circ}}{\text { Each exterior angle }}\)

n = \(\frac{360^{\circ}}{15^{\circ}}\)

= 24

Hence, the number of sides is 24.

Question 5.

1. Is it possible to have a regular polygon with a measure of each exterior angle as 22°?
2. Can it be an interior angle of a regular polygon? Why?

Solution:

1. No, since 22 is not a factor of m 360°

2. No, because by linear pair property, each exterior angle is 180° -22°= 158°

Which 360°n – 360°15 = 24°. is not. a factor of 360

Question 6.

1. What is the minimum interior angle possible for a regular polygon? Why ?
2. What is the maximum exterior angle possible for a regular polygon

Solution:

1. Minimum number of sides of a regular polygon = 3.

• A regular polygon of 3 sides is an equiangular triangle.
• Each interior angle of an equilateral triangle

n = \(\left(\frac{n-2}{n}\right) 180^{\circ}\)

= \(\left(\frac{3-2}{3}\right) 180^{\circ}\)

= 60°

Hence, the minimum interior angle possible for a regular polygon is 60°.

2. We know that the measure of an interior angle + the measure of a corresponding exterior angle

= 180°

By linear pair property

∵ Minimum interior angle possible for a regular polygon = 60°

∴ Maximum exterior angle possible for a regular polygon = 180°- 60° = 120°

Kinds Of Quadrilaterals

The important types of quadrilaterals are as follows :

• Trapezium
• Kite
• Parallelogram
• Rhombus
• Rectangle
• Square

Trapezium

A quadrilateral, which has only one pair of parallel sides, is called a trapezium.

Question 1. Study the above, [inures and discuss with your friends why some of them are trapeziums while some are not,(Note: The arrow marks indicate parallel lines).
Solution:

Some of them are trapeziums because they have only one pair of parallel sides. Some of them are not trapeziums because they have no pair of parallel sides.

Note: One of these figures is not a trapezium although it has only one pair of parallel  sides because a trapezium is essentially a quadrilateral whereas this figure is a pentagon

Question 2. Take identical cut-outs of congruent triangles of sides 3 cm, 4 cm, and 5 cm. Arrange them as shown

You get a trapezium. (Check it!) Which are the parallel sides here? Should the non-parallel sides be equal? You can get two more trapeziums using the same set of triangles. Find them out and discuss their shapes
Solution:

DE = CB (= 3 cm)

DC = EB (= 5 cm)

∴ Quadrilateral DCBE parallelogram

∴ DC || EB => DC|| AB

∴ ABCD is a trapezium. Its parallel sides are AB and DC.

Its non-parallel sides DA and CB are D 5 cm c not equal.

Thus, the non-parallel sides need not to be equal.

We get two more trapeziums using the same set of triangles. These are shown below

Question 3. Take four sol squares from your and your friend’s instrument boxes. Use different numbers of them to place side-by-side and obtain different trapeziums.

If the non-parallel sides of a trapezium arc are of equal length, we call it an isosceles trapezium. Did you get an isosceles trapezium in any of your investigations?

Solution:

Taking the different numbers of set-squares and pinning (hem side-by-side, we obtain the following trapeziums

No, we did not get an isosceles trapezium in any of our investigations.

Kite

A quadrilateral, which has exactly two pairs of equal consecutive sides, is called a kite.

Question 4. Take a thick white sheet. Fold the paper once. Draw two line segments of different lengths as shown in the figure

Cut along the line segments and open up. You have the shape of a kite. Has the kite any line symmetry?

Fold both the diagonals of the kite. Use the set square to check if they cut at right angles. Are the diagonals equal in length?

Verify by (paper-folding or measurement) if the diagonals bisect each other, i j By folding an angle of the kite on its opposite, check for angles of equal measure.

Observe the diagonal folds; do they indicate any diagonal being an angle bisector? Share your findings with others and list them.

Solution:

The kite has only one line of symmetry shown by a dotted line segment (a diagonal) AC and ∠B = ∠D

The diagonal AC is an angle bisector as it bisects ∠A and ∠C both.

Yes, the diagonals are cut at right angles.

No, the diagonals are not equal in length.

Yes, one of the diagonals bisects the other.

Question 5. Show that ΔABC mid ΔADC congruent, What do we infer from this?
Solution:

In ΔABC and  ADC

AB = AD

One pair of consecutive sides

BC = DC

Another pair of consecutive sides common

AC = AC

Δ ABC = Δ ADC

SSS Congruence Axiom

∴ ∠AC = ∠DAC

∠BCA = ∠DCA

i-e., diagonal AC bisects ∠BAD and ∠BCD both.

Parallelogram

A quadrilateral whose opposite sides are parallel is called a parallelogram

Question 1.  Study these figures and try to describe in your own words what we mean by a parallelogram. Share your observations with your friends
Solution:

By a parallelogram, we mean a quadrilateral whose opposite sides are parallel.

We observe that in a parallelogram,

Opposite sides are parallel and equal

Opposite angles are equal

Adjacent angles are supplementary

Diagonals bisect each other but the not equal

Elements Of A Parallelogram

The elements of a parallelogram are as follows :

1. Two pairs of opposite sides
2. Four pairs of adjacent sides
3. Two pairs of opposite equal angles
4. Four pairs of adjacent angles.

Question 2.  Are \(\overline{B C} \text { and } \overline{C D}\) adjacent sides too? Try to find two more pairs of adjacent sides.

Solution:

Yes, \(\overline{B C} \text { and } \overline{C D}\) are also adjacent sides. Two more pairs of adjacent sides are CD and DA; DA and AB.

Question 3. Identify other pairs of adjacent angles of the parallelogram
Solution:

The other pairs of adjacent angles of the parallelogram are∠C, ∠D, and ∠D, ∠A.

Question 4.

1.  Place \(\overline{A^{\prime} B^{\prime}}\) over \( \overline{D C}\). Do they coincide?
Solution:

Yes, they coincide.

2. What can you say about the lengths \( \overline{A B}\) and \( \overline{D C}\)
Solution:  These lengths are equal

3. Examine the lengths \( \overline{A D}\) and \( \overline{B C}\) What do you find?
Solution: These lengths are equal

Question 5.  Take two identical set squares with angles 30°- 60°- 90°andplace them adjacently to form a parallelogram as shown in the figure. Does this help you to verify the above property?

Solution: 

Yes, this helps us to verify the above property. We can further strengthen this idea through a logical argument.

Consider a parallelogram ABCD. Draw any one diagonal, say \(\overline{\mathrm{AC}}\). Looking at the angles,

∠1 =∠2 and ∠3 = ∠4

Alternate interior angles

Since in triangles ABC and ADC,

∠1 = ∠2, ∠3 =∠4 (alternate interior angles), and \(\overline{\mathrm{AC}}\) is common;

So, by ASA congruency condition,

Δ ABC≅ Δ CDA

This gives AB = DC and BC = AD

Question 6. Does this tell you anything about the measures of angles A and C ? Examine the same for angles B and D. State your findings.
Solution:

Findings: Angles A and C are equal.

Similarly, angles B and D are equal.

Also, pairs of consecutive angles (adjacent angles) ∠A and ∠B, ∠B and ∠C, ∠C and ∠D, ∠D and ∠A are supplementary.

Angles Of A Parallelogram

Note point

Property :

The opposite angles of a parallelogram are of equal measure

Question 1.  Take two identical 30°- 60°- 90° set squares and form a parallelogram as before. Does the figure obtained help us to confirm the above property

Solution:

Yes, the figure obtained helps us to confirm the above property. We can further justify this idea through logical arguments.

If \( \overline{A C}\) and \( \overline{B D}\) are the diagonals of the parallelogram, we find that

∠ 1 = ∠2 and ∠3 = ∠4

Alternate interior angles

Studying Δ ABC and Δ ADC separately, helps us to see that, by ASA-congruency condition,

Δ ABC ≅  Δ CDA

1 ∠1 = ∠2, ∠3 = ∠4, \( \overline{A C}\) and \( \overline{B D}\). This shows that ∠B and ∠D have the same measure. In the same way, we can get m ∠A = m∠C.

Question 2. Identify two more pairs of supplementary angles from the figure

Solution:

Two more pairs of supplementary angles from the figure are ∠B, ∠C, and ∠C, ∠D

Note Point

Property: The adjacent angles in a parallelogram are supplementary.

Question 3.  After showing m ∠R = m ∠N = 70°, can you find m ∠I and m∠G by any other method?
Solution:

RG || IN and RI is a transversal

∴ ∠R + ∠ I = 180°

Consecutive interior angles

70° + ∠ I = 180°

∠l  = 180° – 70° = 110°

Again,  RI || GN and RG is a transversal

∴∠R + ∠ G = 180°

Consecutive interior angles

70° +  ∠G = 180°

∠G = 180° -70° = 110°

∠I = ∠G = 110°.

Question 4. Is the mid-point, the same as O?
Solution: Yes

Question 5. Does this show that diagonal \(\overline{D B}\) bisects the diagonal \(\overline{A C}\)at point O ? Discuss it with your friends
Solution: Yes

Question 6. Repeat this activity to find where the I mid-point of \(\overline{D B}\) could lie.
Solution:

The midpoint of \(\overline{D B}\) will also lie at

Diagonals Of A Parallelogram

Note Point

Property:

The diagonals of a parallelogram bisect each other at the point. of their intersection.

Proof: Consider the parallelogram given below:

In Δ AOB and Δ COD

∠1 = ∠2

∠3 = ∠ 4

Alternate interior ∠ s

AB = DC

Opp. sides of parallelogram

∴  AOB ≅ COD

ASA congruence criterion

This gives AO = CO and BO = DO

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.3

Question 1. Given a parallelogram ABCD. x + 100° = 180° Complete each statement along with the definition or property used

AD  = _______________

∠DCB =_______________

OC =_______________

m∠DAB + m∠CDA = _______________

Solution:

1. AD = BC

∵ Opposite sides of a supplementary parallelogram are equal

2. ∠DCB =∠DAB

∵ Opposite angles of a parallelogram are equal

3. OC = OA

∵ Diagonals of a parallelogram bisect each other

m∠DAB + m∠CDA = 180°

∵ Adjacent angles of a parallelogram are supplementary.

Question 2. Consider the following parallelo¬ grams. Find the values ofthe unknowns x, y, z.

Solution:

1. y = 100° 

∵ Opposite angles of a parallelogram are equal

x + 100° = 180°

Adjacent angles in a parallelogram are supplementary

x = 180° – 100°

x = 80°

z = x = 80°

∵ Opposite angles of a parallelogram are of equal measure

Thus, x = 80°, y = 100° and z = 80°

2. x + 50° = 180°

Adjacent angles in a parallelogram are supplementary

x= 180°- 50° = 130°

y = x = 130°

∵ Opposite angles of a parallelogram are of equal measure

180° – z = 50°

∵ Linear pair property and opposite angles of a parallelogram are of equal measure

z = 180° – 50° = 130°

Thus, x = 130°, y = 130° and z = 130°

3. x = 90°

x + y 4- 30° = 180°

Vertically opposite angles are equal

X + y + 30° = 180°

By angle sum property of a triangle

90° + y + 30° = 180°

120° + y = 180°

y = 180°- 120° = 60°

∵ AD|| BC and AC is a transversal

∴ y = z

Alternate interior angles

But y = 60°

z = 60°

Thus, .x = 90°, y = 60° and z = 60°

4. y = 80°

∵ Opposite angles of a parallelogram are of equal measure

x + 80° = 180°

Adjacent angles in a parallelogram are supplementary

x = 180° – 80°

x = 100°

180° – z + 80° = 180°

∵ Linear pair property and adjacent angles in a parallelogram are supplementary. Solution:

z = 80°

Thus, x = 100°, y = 80° and z = 80°

5. y = 112°

∵ Opposite angles of a parallelogram are equal

x+ y + 40° = 180°

∵ By angle sum property of a triangle

⇒  x + 112° + 40° = 180°

⇒  x + 152° = 180°

x = 180° – 152°

x = 28°

Z = X

Alternate interior angles

But x = 28°

∴ z = 28°

Thus, x= 28°, y = 112° and z = 28°

Question 3. Can a quadrilateral ABCD be a parallelogram if

1. ∠D + ∠B = 180°?
2. AB = DC = 8 cm, AD = 4 cm, and BC = 4.4 cm?
3. ∠A = 70° and ZC = 65° ?

Solution:

1. Can be, but needs not be.

2. No, in a parallelogram, opposite sides are equal; but here, AD ≠ BC.

3. No, in a parallelogram, opposite angles are of equal measure; but here ∠A ≠∠ C.

Question 4. Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure
Solution:

 

ABCD is a quadrilateral that is not a x parallelogram. Here,  ∠A = ∠ C. It is a kite. °

Question 5. The measures of two adjacent angles of a parallelogram are in the ratio 3: 2. Find the measure of each of the angles of the parallelogram.
Solution:

Let the two adjacent angles be 3x° and 2x

Then,

3x° + 2x° = 180°

∵ The sum of the two adjacent angles of a parallelogram is 180°

5x° = 180°

x° = \(\frac{180^{\circ}}{5}\)

x° = 36°

3x° = 3 × 36° = 108°

2x° = 2 × 36° = 72°

Since the opposite angles of a parallelogram are of equal measure, therefore the measures of the angles of the parallelogram are 72°, 108°; 72°, and 108°.

Question 6. Two adjacent angles of a parallelogram have equal measure. Find the immure of each of the angles of the parallelogram.
Solution:

Let the two adjacent, angles of a parallelogram be .v° each.

Then, x° + x° = 180°

The sum of the two adjacent angles of a parallelogram is 180°

2x° = 180°

x° =  \(\frac{180^{\circ}}{2}\)

x° = 90°

Since the opposite angles of a parallelogram are of equal measure, therefore, the measure of each of the angles of the parallelogram is 90°, i.e., each angle of the parallelogram is a right angle

Question 7. The adjacent figure HOPE is a parallelogram. Find the angle measures x, y, and z. State the properties you use to find them
Solution:

x = ∠HOP = 180° – 70° = 110°

The opposite angles of a parallelogram are of equal measure

∵ HOPE is a ||gm

∴ HE || OP

And HP is a transversal

∴ y = 40°

An alternate interior angles

40° + z + x = 180°

The adjacent angles in a parallelo¬ gram are supplementary

40° + z + 110° = 180°

z + 150° = 180°

z = 180° – 150°

z = 30°.

Thus, x =110°, y = 40°  and z = 30°

Question 8. The following figures GUNS mid RUNS are parallelograms. Find x and y. (Lengths are in cm)

Solution:

1. For GUNS:

Since the opposite sides of a parallelogram are of equal length, therefore, 3x = 18

x = \(\frac{18}{3}\) = 6

And, – 1 = 26

3y = 26 + 1

3y = 27

x = \(\frac{27}{9}\)

Hence, x = 6; y = 9.

2. For RUNS:

Since the diagonals of a parallelogram bisect each other,

∴ x + y = 16______________ (1)

y + 7 = 20 ______________ (2)

From (2), y = 20 – 7 = 13

Putting y = 13 in (1), we get

x + 13 = 16 ⇒  x = 16 – 13

= 3.

Hence, x = 3; y = 13

Question 9. In the below figure, both RISK and CLUE are parallelograms. Find the value of x

Solution:  ∠RISK is a parallelogram

∴ ∠RIS = ∠RKS = 120°

The opposite angles of a parallelo¬ gram are of equal measure

Also, ∠RIS + ∠ISK = 180°.

The adjacent angles in a parallelo¬ gram are supplementary

120° + ∠ISK = 180°

∠ISK = 180° – 120°

∠ISK = 60°_____________ (1)

CLUE is a parallelogram

∠CES = ∠CLU = 70°_____________ (2)

The opposite angles of a parallelo¬ gram are of equal measure

In triangle EST,

x° + ∠TSE + ∠TES = 180°

By angle sum property of a triangle

⇒ x° + ∠ISK + ∠CES = 180°

⇒ x° + 60° + 70° = 180°

From (1) and (2)

x° + 130° = 180°

x° = 180° – 130°

x° = 50°.

Question 10. Explain how this figure is a trapezium. Which of its two sides is parallel

Solution:

∠KLM + ∠NML = 80°+ 100°

= 180°

∴ KL || NM

∵ The sum of consecutive interior angles is 180°

∴ KLMN is a trapezium.

Its two sides \(\overline{\mathrm{KL}} \text { and } \overline{\mathrm{NM}}\) are parallel.

Question 11. Find m∠C in the figure, if \(\overline{A B} \| \overline{D C}\)

Solution:

∴ \(\overline{\mathrm{AB}} \| \overline{\mathrm{DC}}\)

m∠C + m∠B = 180°

The sum of consecutive interior angles is 180

m∠C + 120° = 180°

m∠C = 180° – 120° = 60

Question 12. Find the measure of ∠P and ∠S, if \(\overline{S P} \| \overline{R Q}\) in the figure. (If you find m∠R, is there more than one method to find m∠P
Solution: 

⇒ \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) and \(\overline{\mathrm{SR}}\)isatransversal

∴m ∠P + m ∠Q = 180°

∴ The sum of consecutive interior angles is 180°

130° = 180°

m ∠P = 180° – 130°

m∠P = 50°

Again,

∵ \(\overline{\mathrm{SP}} \| \overline{\mathrm{RQ}}\) and \(\overline{\mathrm{SR}}\) and SR is a transversal.

m∠R + m ∠S = 180°

The sum of consecutive interior angles is 180°

90° + m∠S = 180°

m∠S = = 180° – 90°

= 90°

Yes; there is one more method of finding m ∠P if m ∠R is given and that is by using the angle sum property of a quadrilateral.

We have,

m∠P+m∠Q+m∠R+m∠S = 360°

m∠P+ 130° + 90° + 90° = 360°

∠S = 90° as above

m∠P + 310° = 360°

m∠P = 360°-310°

= 50°

Some Special Parallelograms

Rhombus

A quadrilateral whose all four sides are of equal length is called a rhombus.

Note Point

Property: The diagonals of a rhombus are perpendicular bisectors of one other.

A Rectangle

A rectangle is a parallelogram with equal angles.

Question 1. What Is The Full Meaning Of This Definition? Discuss With Your Friends.
Solution:

A rectangle is a parallelogram in which every angle is a right angle.

Note point:

Property: The diagonals of a rectangle are of equal length.

A Square

Property: A square is a rectangle whose all four sides have equal length.

The diagonals of a square are perpendicular bisectors of each other

Question 2. Take a square sheet, say PQRS. Fold along both the diagonals. Are their mid-points the same?

Check if the angle at 0 is 90° by using a set square. see that This verifies the property stated above.

Solution:

Yes, their mid-points are the same.

Yes, the angle at 0 is 90°.

We can justify this also by arguing logically

ABCD is a square whose diagonals meet at 0.

(Since the square is a parallelogram)

By the SSS congruency condition, we now see that

Δ AOD =Δ  COD

OA = OC, OD = OD, AD = CD

m ∠AOD = m∠COD

These angles being a linear pair, each is a right angle

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Exercise 3.4

Question 1. State whether True or False :

1. All rectangles are squares
2. All rhombuses are parallelograms
3. All squares are rhombuses and also rectangles
4. All squares are not parallelograms
5. All kites are rhombuses
6. All rhombuses are kites
7. Allparallelograms are trapeziums
8. All squares are trapeziums.

Solution:

(2), (3), (6), (7), (8) are true

(1), (4), (5) are false

Question 2. Identify all the quadrilaterals that have.

1. Four sides of equal length
2. Four right angles

Solution:

The quadrilaterals that have four sides of equal length are square and rhombus.

The quadrilaterals that have right angles are square and rectangle

Question 3. Explain how a square is

1. A quadrilateral
2. A parallelogram
3. A rhombus
4. A rectangle.

Solution:

1.  A quadrilateral: A square is 4 sides, so it is a quadrilateral.
2. A parallelogram: A square has its opposite sides parallel; so it is a parallelogram.
3. A rhombus: A square is a parallelogram with all 4 sides equal, so it is a rhombus.
4. A rectangle: A square is a parallelogram with each angle a right angle, so it is a rectangle

Question 4. Name the quadrilaterals whose diagonals:

1. Bisect each other
2. Are perpendicular bisectors of each other
3. Are equal.

Solution:

1. Bisect each other: The names of the quadrilaterals whose diagonals bisect each other are parallelogram; rhombus, square, and rectangle.
2. Are perpendicular bisectors of each other: The names of the quadrilaterals whose diagonals are perpendicular bisectors of each other are rhombus and square.
3. Are equal: The names of the quadrilaterals whose diagonals are equal are square and rectangle

Question 5. Explain why a rectangle is a convex quadrilateral.
Solution:

A rectangle is a convex quadrilateral because both of its diagonals he wholly in its interior.

Question 6. ABC is a right-angled triangle and AC = BD 0 is the mid-point of the side opposite to the right angle. Explain why is equidistant from A, B, and C. (The dotted lines are drawn addi¬ tionally to help you).

Solution:

Construction: Produce 130 to D such that BO = OD. Join AD and CD.

Proof: AO = OC

∵ O is the mid-point of AC by construction

BO=OD

∴ Diagonals of quadrilateral ABCD bisect each other.

∴  Quadrilateral ABCD isparallelogram.

Now, ∠ ABC = 90°

ABCD is a rectangle.

Since the diagonals of a rectangle bisect each other, therefore,

O is the mid-point of AC and BD both.

But AC = BD

The diagonals of a rectangle are equal

∴  \(\mathrm{OA}=\mathrm{OC}=\frac{1}{2} \mathrm{AC}=\frac{1}{2} \mathrm{BD}=\mathrm{OB}\)

OA = OB = OC

Question 7. A mason has made a concrete slab. He needs it to be rectangular. In what different ways can he make sure that it is rectangular
Solution:

Different ways:

1. If each angle at the corner is 90° in
2. If the diagonals are equal in length.

Question 8. A square was defined as a rectangle with all sides equal. Can we define it as a rhombus with equal angles? Explore this idea
Solution:

Yes, a square can be defined as a rhombus with equal angles. If a rhombus has equal angles

Then the measure o each angle of the rhombus will be \(\frac{360^{\circ}}{4^{\circ}}\) = 90°

Thus, then it becomes a quadrilateral with all sides equal and each angle of measure 90° and hence a square.

Question 9. Can a trapezium have all angles equal? Can it have all sides equal? Explain.
Solution:

If a trapezium has all angles equal, then either it becomes a rectangle or a square.

If a trapezium has all sides equal, then either it becomes a rhombus or a square

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Multiple Choice Questions

Question 1. What, is the number of sides of a triangle

1. 1
2. 2
3. 3
4. 4

Solution: 3. 3

Question 2. What is the number of vertices of a triangle?

1. 1
2. 2
3. 3
4. 4

Solution: 3. 3

Question 3. What is the number of sides of a quadrilateral?

1. 1
2. 2
3. 3
4. 4

Solution: 4. 4

Question 4. What is the number of vertices of a quadrilateral?

1. 1
2. 2
3. 3
4. 4

Solution: 4. 4

Question 5. How many diagonals does a quadrilateral have?

1. 1
2. 2
3. 3
4. 4

Solution: 2. 2

Question 6. How many diagonals does a triangle have?

1. 0
2. 1
3. 2
4. 4

Solution: 1. 0

Question 7. How many diagonals does a regular hexagon have?

1. 2
2. 0
3. 4
4. 9

Solution: 4. 9

Question 8. What is the name of a regular polygon of 3 sides?

1. Equilateral triangle
2. Square
3. Regular hexagon
4. Regular octagon.

Solution: 1. Equilateral triangle

Question 9. What is the name of a regular polygon of 6 sides?

1. Square
2. Equilateral triangle
3. Regular hexagon
4. Regular octagon.

Solution: 3. Regular hexagon

Question10. What is the name of a regular polygon of 4 sides?

1. Regular hexagon
2. Regular octagon
3. Square
4. Equilateral triangle.

Solution: 3. Square

Question 11. The sum of the measures of the exterior angles of any polygon is

1. 90°
2. 180°
3. 360°
4. 720°

Solution: 3. 360°

Question 12. The number of sides of a regular polygon, whose exterior angle has a measure of 45°. is

1. 4°
2. 6°
3. 8°
4. 10°

Solution: 3. 8°

Number of sides =\(\frac{360^{\circ}}{45^{\circ}}\)

= 8°

Question 13. The measure of each exterior angle of a regular polygon of 9 sides is

1. 30°
2. 40°
3. 60°
4. 45°

Solution: 2. 40°

Required measure =\(\frac{360^{\circ}}{9^{\circ}}\)

= 40°

Question 14. The measure of each exterior angle of a regular polygon of 15 sides is

1. 30°
2. 45°
3. 60°
4. 24°

Solution: 4. 24°

Required measure =\(\frac{360^{\circ}}{24^{\circ}}\)

= 24°

Question 15. How many sides does a regular polygon have if the measure of an exterior angle is 24°?

1. 6°
2. 9°
3. 15°
4. 12°

Solution: 3. 15°

Number of sides =\(\frac{360^{\circ}}{15^{\circ}}\)

Question 16. How many sides does a regular polygon have if each of its interior angles is 165°

1. 12°
2. 24°
3. 9°
4. 6°

Solution: 2. 24°

Each exterior angle = 180° -165°= 15°

Number of sides = \(\frac{360^{\circ}}{15^{\circ}}\)

= 24°

Question 17. Which of (ho following statements is false?

1. All the angles of a rectangle are equal
2. No angle of a rectangle can be obtuse
3. The diagonals of a rectangle bisect each other
4. The opposite sides of a rectangle are not equal.

Solution: 4. The opposite sides of a rectangle are not equal.

Question 18. Which of the following statements is false?

1. Asquare is a rectangle whose adjacent sides are equal
2. A square is a rhombus whose one angle is a right angle
3. The diagonals of a square bisect each other at right angles
4. The diagonals of a square do not divide the whole square into four equal parts.

Solution: 4. The diagonals of a square do not divide the whole square into four equal parts.

Question 19. Which of the following statements is false?

1. All the rectangles are parallelograms
2. All the squares are rectangles
3. All the parallelograms are rectangles
4. All the rhombuses are parallelo¬ grams.

Solution: 3. All the parallelograms are rectangles

Question 20. Which ofthe following statements is true?

1. All the rectangles are squares
2. All the parallelograms are rhombuses
3. All the squares are rhombuses
4. Each parallelogram is a trapezium.

Solution: 3. All the squares are rhombuses

Question 21. Which ofthe following statements is true?

1. All the rhombuses are squares
2. Each square is a parallelogram
3. Each parallelogram is a square
4. Each trapezium is a parallelogram.

Solution: 2. Each square is a parallelogram

Question 22. One angle of a parallelogram is a right angle. The name ofthe quadrilateral is

1. Square
2. Rectangle
3. Rhombus
4. Kite

Solution: 2. Rectangle

Question 23. Two adjacent sides of a rod angle are equal. The name of the quadrilateral is kite

1. Square
2. Kite
3. Rhombus
4. None of these.

Solution: 1. Square

Question 24. Which of the following statements is false?

1. All the four sides of a parallelogram are equal.
2. The opposite angles of a parallelogram are equal
3. The diagonals of a parallelogram bisect each other
4. All the four sides of a rhombus are equal

Solution: 1. All the four sides of a parallelogram are equal.

Question 25. Which of the following statements is false?

1. All the four angles of a rhombus are equal
2. The diagonals of a rhombus bisect each other at right angles
3. A rectangle is a parallelogram
4. All squares are rectangles.

Solution: 1. All the four angles of a rhombus are equal

Question 26. If one angle of a parallelogram is of 65°, then the measure of the adjacent angle is

1. 65°
2. 115°
3. 25°
4. 90°.

Solution: 2. 115°

The measure of the adjacent angle

= 180° – 65° = 115°

Question 27. If ∠A of a parallelogram ABCD is of60°, then the measure of the opposite angle ∠C is

1. 60°
2. 120°
3. 30°
4. None of these.

Solution: 1. 60°

∠C =∠A = 60°.

Question 28. If all four sides of a parallelogram are equal and the adjacent angles are 120° and 60°, then the name of the quadrilateral is

1. Rectangle
2. Square
3. Rhombus
4. Kite

Solution: 3. Rhombus

Question 29. If the length of a side of a rhombus is 6 cm, (when the perimeter of the rhombus 

1. 6 cm
2. 12 cm
3. 24 cm
4. 3 cm.

Solution: 3. 24 cm

Perimeter = 4 side = 4× 6 = 24 cm

Question 30. In a kite, what is false?

1. The diagonals are perpendicular to each other
2. One ofthe diagonals bisects the other
3. Both the pairs of opposing angles are equal
4. All the four sides are equal.

Solution: 4. All the four sides are equal.

Question 31. ABCD is a rectangle. Its diagonals meet at O.

OA = 2x-1, OD = 3x-2. Find x 

1. 1
2. 2
3. 3
4. -1

Solution: 1. 1

3x – 2 = 3x – 1

3x -2x = -1+ 2

x = 1

Question 32. Find the perimeter of the rectangle ABCD

1. 6 cm
2. 12 cm
3. 3 cm
4. 24 cm.

Solution: 2. 12 cm

Perimeter = 2 (4 + 2) cm = 12 cm

Question 33. The four angles of a quadrilateral are in the ratio 1: 2 : 3: 4. The measure of its smallest angle is

1. 120°
2. 36°
3. 18°
4. 10°.

Solution: 2. 36°

Sum of the ratios =1 + 2+3 + 4=10

∴ Smallest angle =\(\frac{1}{10} \times 360^{\circ}\)

Question 34. In a parallelogram ABCD, ∠A : ∠B = 1: 2. Then, ∠A =

1. 30°
2. 60°
3. 45°
4. 90°.

Solution: 2. 60°

∠A + ∠B = 180°

∠A : ∠B = 1:2

Sum of the ratios =1 + 2 = 3

∴ \(\angle \mathrm{A}=\frac{1}{3} \times 180^{\circ}=60^{\circ}\)

Question 35. Two adjacent angles of a parallelogram are of equal measure. The measure of each angle of the parallelogram is

1. 45°
2. 30°
3. 60°
4. 90°.

Solution: 4. 90°.

x° + x° = 180°

2x° = 180°.

x° = \(\frac{180^{\circ}}{2^{\circ}}\)

x° = 90°

Question 36. ABCD is a parallelogram as shown. Find x and y

1. 1,7
2. 2,6
3. 3,5
4. 4,4

Solution: 3. 3,5

x + y = 8

y + 5 = 10

y=5

x + 5 = 8

x = 3.

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals True Or False

1. The number of diagonals in a decagon is 35 – True

2. The measure of each interior angle of a regular pentagon is 108° – True

3. The minimum number of sides a polygon can have is 4 – False

4. A regular polygon is both ‘equiangular’ and ‘equilateral’ – True

NCERT Solutions For Class 8 Maths Chapter 3 Understanding Quadrilaterals Fill In The Blanks

1. The name of a three-sided regular polygon is → An equilateral triangle

2. The sum of the adjacent angles of a parallelogram is  →  180°

3. The angle between the diagonals of a rhombus is → 90°

4. The maximum number of obtuse angles that a quadrilateral can have is → 3

5. Apala has a beautiful garden in her flat. The shape ofthe garden is triangular. The sum of all the exterior angles taken in order of the triangular garden is → 360°

6. The sides of a rectangle are 24 cm and 10 cm. Find the length of one of the diagonals → 26 cm

7. ABCD is a parallelogram. Find ∠A – ∠C →0

8. Two adjacent angles of a parallelogram are (3x + 3)° and (lx + 7)°. Find x → 17

9. In the following figure, find the value of x → 70°

10. In the following figure, find the measure of ∠ABC – 130°

Mensuration

Mensuration Introduction

We know that the perimeter of a closed figure is the distance around its boundary. Also, the area of a closed figure is the measurement of the region covered by it.

• Moreover, the volume of a solid is the amount of space occupied by it. We know how to find the areas and perimeters of various plane figures such as triangles, parallelograms, rectangles, rhombuses, squares, circles, pathways, and borders in rectangular shapes, etc.
• Here, we shall learn to solve the problems related to perimeters and areas of general quadrilaterals and trapeziums.
• We shall also learn to solve the problems related to areas of polygons (regular and irregular) by using the formula for the area of a triangle and that for the area of a trapezium.
• Moreover, we shall also learn to find out the surface areas, and volumes of cubes, cuboids, and cylinders.

Area Of A Polygon

We use the method of triangulation which means splitting into triangles.

Read and Learn More NCERT Solutions For Class 8 Maths

Question 1.  Divide the following polygon (Figures) into parts (triangles and trapezium) to find out its area.

Solution:

2. Polygon ABODE is divided into parts as shown below. Find its area ifAD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm, and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

Solution:

Area of polygon ABCDE = Area of A AFB + Area of trapezium FBCH + Area of A CHD + Area of A ADE …..(1)

Area of Δ AFB = (I) =\(\frac{1}{2}\) x AF x BF = \(\frac{1}{2}\) x 3 x 2 = 3 cml²

Area of trapezium FBCH = (2) = FH x \(\frac{(\mathrm{BF}+\mathrm{CH})}{2}=3 \times \frac{(2+3)}{2}=7.5 \mathrm{~cm}^2\)

FH = AH-AF = 6- 3 = 3

Area of Δ CHD = (3) = \(\frac{1}{2} \times \mathrm{HD} \times \mathrm{CH}=\frac{2 \times 3}{2}=3 \mathrm{~cm}^2\)

HD = AD -AH = 8 – 6 = 2cm

Area of Δ ADE = (4) \(\frac{1}{2} \times \mathrm{AD} \times \mathrm{GE}=\frac{1}{2}\) x 8 x 2.5 = 10 cm2

From (1),

Area of polygon ABCDE = Area [(1) + (2) + (3) + (4)]

= 3 + 7.5 + 3 + 10 = 23.5 cml²

3. Find the area of polygon MNOPQR (Figure), if

MP = 9 cm, MD = 7 cm, MC = 6 cm, MB – 4 cm, MA = 2 cm, NA, OC, QD, and RB are perpendicular to diagonal MP.

Solution:

Area of polygon MNOPQR

= Area of A MAN + Area of trapezium ACON + Area of A OOP + AreaofAPDQ + Area of trapezium DBRQ + AreaofARBM …(1)

Given:

MP = 9 cm, MD = 7 cm, MC = 6 cm, MB = 4 cm MA = 2 cm

Area of A MAN = (1) = \(\frac{\mathrm{MA} \times \mathrm{AN}}{2}=\frac{2 \times 2.5}{2}=2.5 \mathrm{~cm}^2\)

Area of trapezium ACON = (2) = \(\frac{(\mathrm{AN}+\mathrm{CO}) \times \mathrm{AC}}{2}=\frac{(\mathrm{AN}+\mathrm{CO}) \times(\mathrm{MC}-\mathrm{MA})}{2}\)

\(=\frac{(2.5+3) \times(6-2)}{2}=11 \mathrm{~cm}^2\)

Area of A 0CP = (3) = \(\frac{\mathrm{CP} \times \mathrm{OC}}{2}=\frac{(\mathrm{MP}-\mathrm{MC}) \times \mathrm{OC}}{2}=\frac{(9-6) \times 3}{2}\) = 4.5 cml²

AreaofAPDQ = (4) = \(\frac{\mathrm{PD} \times \mathrm{DQ}}{2}=\frac{(\mathrm{MP}-\mathrm{MD}) \times \mathrm{DQ}}{2}=\frac{(9-7) \times 2}{2}\) = 2cm2

Area of trapezium DBRQ = (5) \(\frac{(\mathrm{DQ}+\mathrm{BR}) \times \mathrm{BD}}{2}=\frac{(2+2.5) \times(\mathrm{MD}-\mathrm{MB})}{2}\)

⇒  \(\frac{(2+2.5) \times(7-4)}{2}=6.75 \mathrm{~cm}^2\)

Area of ARBM= (6) =\(\frac{\mathrm{MB} \times \mathrm{RB}}{2}=\frac{4 \times 2.5}{2}=5 \mathrm{~cm}^2\)

From (1),

Area of polygon MNOPQR

= Area [(1) + (2) + (3)+ (4)+ (5) + (6)]

= 2.5 cml² + 11cml² + 4.5 cml² + 2 cml² + 6.75 cml² + 5 cml²

= 31.75 cml²

Mensuration Exercise 9.1

Question 1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and the perpendicular distance between them is 0.8 m.

Solution:

Area of the top surface of the table

⇒  \(\frac{1}{2} h(a+b)\)

⇒  \(\frac{1}{2} \times 0.8 \times(1.2+1)\)

⇒  \(0.88 \mathrm{~m}^2\)

Question 2. The area of trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Solution:

Let the length of the other parallel side be b cm

Area of trapezium = \(\frac{1}{2} h(a+b)\)

⇒  \(34 =\frac{1}{2} \times 4 \times(10+b)\)

⇒  \(34 =2 \times(10+b)\)

⇒  \(10+b =\frac{34}{2}\)

10 + b = 17

b = 17 – 10

b = 7cm

Hence, the length of another parallel side is 7cm

Question 3. The length of the fence of a trapezium-shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m, and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Solution:

Fence of the trapezium-shaped field

ABCD = 120 m

AB + BC + CD + DA = 120

AB + 48 + 17 + 40 = 120

AB + 105 = 120

AB = 120 – 105

AB = 15 m

Area of the field

⇒  \(\frac{(B C+A D) \times A B}{2}\)

⇒  \(\frac{(48+40) \times 15}{2}=660 \mathrm{~m}^2 .\)

Hence, the area of the trapezium-shaped field is 660 ml²

Question 4. The diagonal of a quadrilateral-shaped field is 24 m and the perpendiculars dropped on it. from the remaining opposite vertices are 8 m and 13 m. Find the area, of the field

Solution: Area of the field

⇒  \(\frac{1}{2} d\left(h_1+h_2\right)\)

⇒  \(\frac{24 \times(8+13)}{2}=\frac{24 \times 21}{2}\)

= 12 x 21 = 252 ml².

Hence, the area of the quadrilateral-shaped field is 252 ml².

Question 5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:

Area of the rhombus

⇒  \(\frac{1}{2} \times d_1 \times d_2=\frac{1}{2} \times 7.5 \times 12\)

= 45cml²

Hence, the area of the rhombus is 45cml²

Question 6. Find the area of a rhombus whose side is 5 cm. and whose altitude is 4.8 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.
Solution:

Area of the rhombus

= base (b) x altitude (h)

= 5 x 4.8 = 24 cml²

Area of the rhombus

⇒  \(\frac{1}{2} \times d_1 \times d_2\)

⇒  \(24 =\frac{1}{2} \times 8 \times d_2\)

⇒  \(24 =4 d_2\)

⇒  \(d_2 =\frac{24}{4}=6 \mathrm{~cm}\)

Hence, the length of the other diagonal of a rhombus is 6 cm.

Question 7. The floor of a building consists of 3,000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m2 is 4.
Solution:

Area of a tile = \(=\frac{1}{2} \times d_1 \times d_2=\frac{1}{2} \times 45 \times 30\)

= 675 cm2

Area of the floor = 675 x 3,000 cm2 = 20,25,000 cm²

⇒  \(\frac{20,25,000}{100 \times 100} \mathrm{~m}^2\)

lm² = 100 X 100 cml²

202.50 ml²

The cost of polishing per ml² = 4

Total cost of polishing the floor

= 202.50 x 4

= $ 810.

Hence, the total cost of polishing the floor is? 810

Question 8. Mohan wants to buy a trapezium field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10,500 m2 and the perpendicular distance between the two parallel sides is 100 m., find the length of the side along the river.

Solution:

Let the length of the side along the road be m. Then, the length of the side along the river is 2x m.

Area of the field = 10,500square meters

⇒ \(\frac{1}{2} h(a+b)=10500\)

⇒ \(\frac{100 \times(2 x+x)}{2}=10,500\)

⇒ \(150 x =10,500\)

⇒ \(x =\frac{10,500}{150}\)

x= 70

2x = 2 x 70 = 140m

Hence, the length of the side along the river is 140 m.

Question 9. The top surface of the raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.
Solution:

Area of the octagonal surface

= Area of rectangular surface + 2(Area of trapezoidal surface)

Since the octagon is regular, therefore, the two trapeziums will be congruent.

Hence, their surface area of 111 will be equal.

⇒ \(11 \times 5+2 \times\left[\frac{(5+11) \times 4}{2}\right] \mathrm{m}^2\)

55 + 64 m2 = 119 ml²

Question 10. There isapentagonalshaped’park as shown in the figure. ForfindingitsareaJyoti and Kavita divided it into two different ways

Find the area of this park using both ways. Can you suggest some other way offind¬ ing its area?

Solution: Jyoti’s diagram

Here, the pentagonal shape is split into two congruent trapeziums.

Area of the park

⇒ \(2 \times\left[\frac{(15+30)}{2} \times \frac{15}{2}\right]\)

⇒ \(\frac{675}{2}\)

= 337.5 m2

Kavita’s diagram:

Here, the pentagonal shape is split into a square and a triangle.

Area of the park

= Area of square + Area of triangle

⇒ \(=15 \times 15 \mathrm{~m}^2+\frac{15 \times(30-15)}{2} \mathrm{~m}^2\)

⇒ \(=225 \mathrm{~m}^2+\frac{225}{2} \mathrm{~m}^2\)

= 225 m2 + 112.5 ml²

= 337.5 ml²

Another way of finding the area

Here, the pentagonal shape is split into three triangles out of which two triangles are congruent.

Area of the park \(=\frac{15 \times(30-15)}{2} \mathrm{~m}^2 +2 \times\left[\frac{15 \times 15}{2}\right] \mathrm{m}^2 \)

Question 11. The diagram of the adjacent picture frame has outer dimensions = 24 cm x 28 cm and inner dimensions 16 cm x 20 cm. Find the area of each section of the frame, if the width of each section is the same

Solution:

Area of the right section of the frame

⇒ \(\frac{28+20}{2} \times \frac{24-16}{2} \mathrm{~cm}^2\)

⇒ \(\frac{48}{2} \times \frac{8}{2}=24 \times 4=96 \mathrm{~cm}^2\)

The section is a trapezium

Similarly, the area of the left section of the frame = 96 cml²

The area of the upper section of the frame

⇒ \(\frac{24+16}{2} \times \frac{28-20}{2} \mathrm{~cm}^2\)

⇒ \(\frac{40}{2} \times \frac{8}{2}=20 \times 4 \mathrm{~cm}^2\)

= 80 cml²

Similarly, the area of the lower section of the frame

= 80 cml²

Solid Shapes

Two-dimensional figures are the faces of three-dimensional shapes. If two faces of a shape are identical, then they are called congruent faces.

Surface Area Of Cube, Cuboid, And Cylinder

The surface area of a solid is the sum of the areas of its faces.

Mensuration Cuboid

Total surface area of a cuboid = 2 (lb + bh + hl)

where l, b, and h are the length, width, and height of the cuboid respectively

Question Find the total surface area of the following cuboids (Figure):

Solution:

The total surface area of the first cuboid

= 2(lb + bh + hi)

= 2(6 x 4 + 4 x 2 + 2 x 6)

= 2(24 + 8 + 12) = 88 cml²

The total surface area of the second cuboid

= 2(lb+ bh + hi)

= 2(4 x 4 + 4 x 10 + 10 x 4)

= 2(16 + 40 + 40) = 192 cml²

Question 1 . Cover the lateral surface of a cuboidal duster (which your teacher uses in the classroom) using a strip of brown sheet of paper, such that it just fits around the surface. Remove the paper. Measure the area of the paper. Is it the lateral surface area of the duster?
Solution:

(i) Yes, we can say that the area of this strip of brown sheet paper is equal to the lateral surface area of the duster.

Question 2. Measure the length, width, and height of your classroom and find

1. The total surface area of the room, ignoring the area of windows and doors.
2. The lateral surface area of this room.
3. The total area of the room which is to be white-washed.

Solution:

2.  Please find yourself

Question 3. Can we say that the total surface area of a cuboid

= lateral surface area + 2 x area of base?

Question 4. If we interchange the lengths of the base and the height of a cuboid to get another cuboid, will its lateral surface area change?
Solution:

1. Yes, we can say that the total surface area of a cuboid

= lateral surface area + 2 x area of base.

2. Lateral surface area of cuboid (i) = 2(1 + b) h

The lateral surface area of the cuboid (ii)

= 2(h + b) l. These results are different.

Hence, yes; the lateral surface area will change if we interchange the length of the base and the height of a cuboid

Mensuration Cube

The total surface area of a cube = 6l², where l is the side of the cube. The lateral surface area of a cube = 4l², where l is the side of the cube.

Question: Draw the pattern shown on a squared paper and cut it out. (You know that this pattern is a net of a cube). Fold it along the lines and tape the edges to form a cube.

1. What is the length, width, and height of the cube? Observe that all the faces of a cube are square. This makes the length, height, and width of a cube equal.
2. Write the area of each of the face. Are they equal?
3. Write the total surface area of this cube
4. If each side of the cube is l, what will be the area of each face?
5. Can we say that the total surface area of a cube of side l is 6l²?

Solution:

(a) Length of the cube

= Width of the cube

= Length of the cube of side s

Area of each face = l x l = l².

Yes, they are equal.

(c) Total surface area of this cube = 6l².

Area of each face = l².

Yes, we can say that the total surface area of a cube of side l is 6l²

10. Find the surface area of Cube A and the lateral surface area of Cube B.

Solution:

The surface area of the cube (A)

= 6l²

= 6(10)2cm²

= 600 cm²

The lateral surface area of the cube (B)

= 4l²

= 4 x 8 x 8cm²

= 256 cm²

Question .1 Two cubes each with side b are joined to form a cuboid (Figure). What is the surface area, of this cuboid?

How will you arrange 12 cubes of equal length to form a cuboid of the smallest surface area. ?

After the surface area of a cube is painted, the cube is cut into 64 smaller cubes of the same dimensions (Figure).

How many have no face painted? 1 face painted? 2faces painted? 3faces painted?
Solution:

By joining two cubes on each side b,

we get a cuboid whose

length (L) = b + b = 2b units

breadth (B) = b units

and height (H) = b units

The surface area of this cuboid

= 2(L x B + B x H + H x L)

= 2[2b x b + b x b + b x 2b] sq. units

= 2[2b² + b² + 2b² sq. units

= 10b² sq. units

≠ 12b². (No, it is not equal to 1262)

Again, by joining three cubes each with ® side b, we get a cuboid whose

length (L) = 6 + 6 + 6 = 36 units

breadth (B) = 6 units

height (H) = 6 units

The surface area of the cuboid formed by joining the three cubes

= 2 (L x B + B x H + H x L)

= 2 x [3b x b + b x b + b x 3b] sq. units

= 2 x [3b² + b² + 3b²] sq. units

= 14b² sq. units

≠186²(No, it is not equal to 18b²)

(ii) In the first arrangement,

L = 66, B = 26, H = 26

Surface area

= 2 x (6b x 2b + 2b x 2b + 2b x 6b)

= 2 x (12b² + 4b² + 12b²)

= 56b²

In the second arrangement,

L = 4b

B = 6

H = 3b

Surface area

= 2 x (4b x b + b x 3b + 3b x 4b)

= 2 x (4b² + 3b² + 12b²)

= 38b²

Hence, for the smallest surface area, the second arrangement must be made.

(Hi) 16 cubes have no faces painted.

24 cubes have 1 face painted.

16 cubes have 2 faces painted.

8 cubes have 3 faces painted.

Mensuration Cylinders

Formulae :

The lateral (curved) surface area of a cylinder = 2kvh

Total surface area of a cylinder = 2nr (h + r)

where r is the radius of the base and h is the height of the cylinder.

We take n to be \(\frac{22}{7}\) unless otherwise stated

Question .1  Find the total surface area of the following cylinders.

Solution:

For First Cylinder

r = 14 cm

h = 8 cm

Total surface area of the cylinder = 2nr(r + h)

\(=2 \times \frac{22}{7} \times 14 \times(14+8) \mathrm{cm}^2\)

= 2 x 22 x 2 x 22 = 88 x 22

= 1936 cm²

For Second Cylinder

\(r=\frac{2}{2} \mathrm{~m}=1 \mathrm{~m}\)

h = 2 m

The total surface area of the cylinder

= 2πr (r + h)

⇒ \(2 \times \frac{22}{7} \times 1 \times(1+2) \mathrm{m}^2\)

⇒ \(\frac{132}{7} \mathrm{~m}^2=18 \frac{6}{7} \mathrm{~m}^2\)

Question 2.  Note that the lateral surface area of a cylinder is the circumference of the base x height of the cylinder. Can. we write the lateral surface area of the cuboid as the perimeter of the base x height of the cuboid.

Solution:

Lateral surface area

= 2 x [(l x h) + (b x h)]

= 2 (l + b) x h

= Perimeter of base b x height of the cuboid

Yes, we can write the surface area of a cuboid as

perimeter of base x height of the cuboid

Mensuration Exercise 9.2

Question 1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires a lesser amount of material to make?

Solution:

First Cuboidal Box

l = 60 cm, b = 40 cm, h = 50 cm

Total surface area

= 2(lb + bh + hl)

= 2(60 x 40 + 40 x 50 + 50 x 60) cm²

= 2(2400 + 2000 + 3000) cm²

= 2(7400) cm²

= 14800 cm²

Second Cuboidal Box

l = 50 cm, b = 50 cm, h = 50 cm

Total surface area

= 2(lb + bh + hl)

= 2(50 x 50 + 50 x 50 + 50×50) cm²

= 2(2500 + 2500 + 2500) cm²

= 2(7500) cm²

= 15000 cm²

Since the total surface area of the first cuboidal box is less than the total surface area of the second cuboidal box, therefore, box (a) requires a lesser amount of material to make.

Question 2. A suitcase with measures 80 cm x 48 cm x 24 cm is to be covered with a tarpaulin cloth. How many meters of tarpaulin of width 96 cm is required to cover 100 such suitcases?
Solution:

Here,

Length of the suitcase (l) = 80 cm

The breadth of the suitcase (b) = 48 cm

Height of the suitcase (h) = 24 cm

The total surface area of the suitcase

= 2(lb + bh + hl)

= 2(80 x 48 + 48 x 24 + 24 x 80) cm2

= 2(3840 + 1152 + 1920) cm²

= 2(6912) cm²

= 13824 cm²

Width of tarpaulin = 96 cm

Length of tarpaulin required to cover 1 suitcase

⇒ \(\frac{\text { Total surface area of the suitcase }}{\text { Width of tarpaulin }}\)

⇒ \(=\frac{13824}{96}=144 \mathrm{~cm}\)

Length of tarpaulin required to cover 100 such suitcases

= 144 x 100 cm = 14400 cm

⇒ \(\frac{14400}{100} \mathrm{~m}=144 \mathrm{~m}\)

Hence, 144 m of tarpaulin is required.

Question 3. Find the side of a cube whose surface area is 600 cm2.
Solution:

Let the side of the cube be a cm. Then, the total surface area of the cube

= 6a² cm²

According to the question,6a² = 600

⇒ \(a^2=\frac{600}{6}\)

a²= 100

a = √1oo

a = 10 cm

Hence, the side of the cube is 10 cm.

Question 4. Rukhsar painted the outside of the cabinet measuring 1 m x 2 m x 1.5 m. How much surface area did she cover if she painted all except the bottom of the cabinet?

Solution:

I = 2m 7 cm

b = 1 m

h = 1.5 m

Required area

= 2[2 x 1 + 1 x 1.6 + 1.5 x 2] -2xl

= 2[2 +1.5 + 3] – 2

= 13 m²-2 m²

= 11 m²

Hence, she covered 11 m² of surface area.

Question 5. Daniel is painting the walls and ceiling of the cuboidal hall with length, breadth, and height of 15 m, 10 m, and 1 m respectively. From each can ofpaint100 m² of area is painted, How many cans of paint will she need, to paint the room?
Solution:

Z = 15 m

b = 10 m

h = 7 m

Surface area to be painted

= 2(1 x 6 + b x h + h x l) – l x b

= 2 (15 x 10 + 10 x 7 + 7 x 15) m² – (15 x 10) m²

= 2(150 + 70 + 105) m² – 150 m²

= 2(325) m² – 150 m²

= 650 m² – 150 m²

= 500 m²

Number of cans needed \(=\frac{\text { Surface area to be painted }}{\text { Area painted by } 1 \text { can }}\)

⇒ \(=\frac{500}{100}=5\)

Hence, she will need 5 cans of paint to paint the room.

Question 6. Describe how the two figures (below) are alike and how they are different. Which box has a larger lateral surface area?

Solution:

Likeness→ Both have the same heights.

Difference → One is a cylinder, the other is a cube;

The cylinder is a solid obtained by revolving a rectangular area about one side.

A cube is a solid enclosed by six square faces.

A cylinder has two circular faces whereas a cube has six square faces

For First Box

Diameter = 7 cm

Radius (r) =\(\frac{\text { Diameter }}{2}=\frac{7}{2} \mathrm{~cm}\)

Height (h) = 7 cm

Lateral surface area = 2nrh

⇒ \(2 \times \frac{22}{7} \times \frac{7}{2} \times 7\) = 154 cm²

For Second Box

l = 7 cm

b = 7 cm

h = 7 cm

Lateral surface area

= 4 = 4 x (7)2

= 196cm2

Hence, the second box has a larger lateral surface area

Question 7. A closed cylindrical tank of radius 7 m and height 3 m is made from a sheet of metal. How much sheet of metal is required?

Solution:

r = 7 m

h = 3 m

Total surface area

= 2πr(r + h)

⇒ \(2 \times \frac{22}{7} \times 7 \times(7+3)\) = 440 m²

Hence, 440 m² of metal sheet is required.

Question  8. The lateral surface area of a hollow cylinder is 4224 cm². It is cut along its height and forms a rectangular sheet of width 33 cm. Find the perimeter of the rectangular sheet.
Solution:

The lateral surface area of the hollow

cylinder = 4224 cm²

The area of the rectangular sheet = 4224 cm2

Length x width = 4224 cm²

Length x 33 = 4224

Length = \(\frac{4224}{33}\)

Length = 128 cm

The perimeter of the rectangular sheet

= 2(Length + Breadtn)

= 2(128 + 33) cm

= 2(161) cm

= 322 cm

Hence, the perimeter of the rectangular sheet is 322 cm.

Question 9. A road roller takes 750 complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is 84 cm and the length is 1 m.

Solution:

The diameter of the road roller = 84 cm

Radius (r) of the road roller \(\frac{84}{2} \mathrm{~cm}=42 \mathrm{~cm}\)

Length (h) of the road roller = 1 m = 100 cm

The lateral surface area of the road roller = 2πrh

⇒ \(2 \times \frac{22}{7} \times 42 \times 100\)

= 26,400 cm2

The area of the road is covered in 1 complete revolution

= 26,400 cm²

The area of the road covered 750 complete revolutions

= 26,400 x 750 cm²

= 1,98,00,000 cm²

⇒ \(\frac{1,98,00,000}{100 \times 100} \mathrm{~m}^2\)

= 1,980 m²

Question 10. A company packages its milk ponder in a cylindrical container whose base has a diameter of 14 cm and a height of 20 cm. The company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area, of the label

Solution:

For a cylindrical container

The diameter of the base = 14 cm

Radius of the base (r) =\(\frac{14}{2} \mathrm{~cm}\)

= 7 cm

Height (h) = 20 cm

The curved surface area of the container = 2πrh

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

= 880 cm²

Surface area of the label

= 880 cm²

⇒ \(-2\left(2 \times \frac{22}{7} \times 7 \times 2\right) \mathrm{cm}^2\)

= 880 cm2 – 176 cm²

= 704 cm²

Hence, the surface area of the label is 704 cm².

Aliter:

The label is in the form of a cylinder for which

radius (R) =\(\frac{14}{2}=7 \mathrm{~cm}\)

height (H) = 20 – (2 + 2) = 16 cm

Area of the label = 2πRH

⇒ \(=2 \cdot \frac{22}{7} \cdot 7 \cdot 16=704 \mathrm{~cm}^2\)

Hence, the surface area of the label is 1704 cm².

Volume Of Cube, Cuboid And Cylinder

The volume of a three-dimensional object is the amount of space occupied by it. Volume is measured in cubic units.

Question 1 cubic cm

= 1 cm x 1 cm x 1 cm

= 1 cm³

= 10 mm x 10 mm x 10 mm

= mm³

1 cubic m

= lm x lm x lm = lm3

1 cubic mm

= 1 mm x l mm x 1 mm 1 cubic mm

= 1 mm³

= 0.1 cm x 0.1 cm x 0.1 cm

cm³

1 cubic mm

= 1 mm x l mm x 1 mm 1 cubic mm

= 1 mm³

= 0.1 cm x 0.1 cm x 0.1 cm

=100cm³

1 cubic mm

= 1 mm x l mm x 1 mm 1 cubic mm

= 1 mm³

= 0.1 cm x 0.1 cm x 0.1 cm

cm³

Cuboid Formula

Volume of cuboid = Ibh

where l, b, and h are the length, width, and height of the cuboid respectively.
OR

The volume of the cuboid = area of the base x height cuboid.

YouTakecan36arrangecubes of the qualm size many(ie., ways.lengthObserveofeachthecubefollowingis same).tableArrangeandfillthemin the to blank torn a

We have the following observations:

Since then, we have used 36 cubes of equal size to form these 36 cubic units. Also, the volume of each cuboid is the height of the cuboid. From the above example ese cuboids, the volume of each cuboid = l x b x h

Since l x b is the area of its b

The volume of cuboid = Area of base x Therefore we can also say that,

Question 1.  Can you think of such objects whose

Solution: The volume of godow reservoirs, etc., can be found by this method.

Question Find the volume of the following cuboids (Figure).

Solution:

(i) Volume of the cuboid

= l x b x h

= (8x3x2) cm³

= 48 cm³

(ii) Volume of the cuboid

= Area of Base x Height

⇒ \(24 \times \frac{3}{100}\)

⇒ \(\frac{72}{100}=0.72 \mathrm{~m}^3\)

Cube

The volume of the cube = 1³, where l is the side of the cube.

Question 1.  Find the volume of the following cubes.

1. with a side of 4 cm
2. with a side of 1.5 m

Solution:

Volume of the cube = (Side)³

= 4 x 4 x 4 cm³

= 64 cm³

(b) Volume of the cube = (Side)³

= 1.5 x 1.5 x 1.5 m³

= 3.375 m³

Question 2.  Arrange 64 cubes of equal size in as many ways as you can to form a cuboid. Find the surface area of each arrangement. Can solid shapes of the same volume have the same surface area?
Solution:

Some arrangements are as follows :

The surface area in the arrangement

(1) = 2 x (64 x l + i x l + l x 64)

258 square units

2 x (32 x 2 + 2 x l + i x 32)

= 196 square units

= 2 x (16 x 2 + 2 x 2 + 2 x 16)

= 136 square units

= 2 x (16 x 4 + 4 x 1 + 1x 16)

= 168 square units

=2 x (8 x 4 + 4 x 2 + 2 x 8) = 112 square units

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

= 96 square units, etc.

Also, the volume of the cuboid obtained in each case is 64 cubic units. So, No, we cannot say that solid shapes of the same volume need to have the same surface area.

Question 3. The company sells biscuits. For packing purposes, they are using cuboidal boxes :

box A → 3 cm x 8 cm x 20 cm,

box B → 4 cm x 12 cm x 10 cm.

1. What size of the box will be economical for the company? Why?
2. Can you suggest any other size (dimensions) that has the same volume but is more economical than these?

Solution:

Box A: Total surface area

= 2 x (3 x 8 + 8 x 20 + 20 x 3) cm²

= 2 x (24 +160 + 60) cm²

= 2 x 244 cm² = 488 cm²

Box B: Total surface area

= 2 x (4 x 12 + 12 x 10 + 10 x 4) cm²

= 2 x (48 + 120 + 40) cm²

= 2 x208 cm2 = 416 cm².

-Since the total surface area of box B is lesser than the total surface area of box A, therefore, the size of 4 cm x 12 cm x 10 cm of the

The box will be economical for the company to use for packing purposes.

(The volume of both the boxes are same as given below)

Again, the volume of the cuboidal box A

= 3 x 8 x 20

= 480 cm³ and, volume of the cuboidal box B

= 4 x 12 x 10

= 480 cm³

Yes, we can suggest another size (dimensions) which has the same volume but is more economical than these. It is 10 m x 6 m x 8 m.

Volume = 10 x 6 x 8 = 480 cm³

Total surface area = 2 x (10 x 6 + 6 x 8 + 8 x 10) cm2 = 376 cm².

This size (dimensions) has the same volume but it is more economical than box A and box B.

Cylinder Formula:

Volume of cylinder = πr²h

where r is the radius of the base and h is the height of the cylinder

Question 1.   Find the volume of the following cylinders :

Solution:

r = 7 cm, h = 10 cm

The volume of the cylinder = πr²h

⇒ \(=\frac{22}{7} \times 7^2 \times 10\)

⇒ \(=\frac{22}{7} \times 7 \times 7 \times 10\)

= 1540 cm³

The Volume of the cylinder

= Area of base x height

= (250 x 2) m³

= 500 m³

Volume And Capacity

There is not much difference between these two words.

Volume refers to the amount of space occupied by an object.

Capacity refers to the quantity that a container holds.

Note: If a water tin holds 100 cm³ of water, then the capacity of the water tin is 100 cm³.

Capacity is also measured in terms of liters.

The relation between liter and cm³ is,

1 mL = 1 cm³, 1 L = 1000 cm³.

lm3 = 10,00,000 cm³= 1000 L.

Mensuration Exercise 9.3

Question 1. Given a cylindrical tank, in which situation will you find surface area, and in which situation is volume?

1. To find how much it can hold,
2. Number of cement bags required to plaster it.
3. To find the number of smaller tanks that can be filled with water from it.

Solution:

Volume

Surface area

Volume.

Question 2. The diameter of cylinder A is 7 cm, and the height is 14 cm. The diameter of cylinder B is 14 cm and the height is 7cm. Without doing any calculations canyon suggest whose volume is greater? Verify it by finding the volume of both cylinders. Check whether the cylinder with greater volume also has a greater surface area.

Solution:

The volume of cylinder B is greater.

For Cylinder A

⇒ \(r=\frac{7}{2} \mathrm{~cm}\)

h = 14 cm

Volume = nr²h

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

= 539 cm³

For Cylinder B

⇒ \(r=\frac{14}{2} \mathrm{~cm}=7 \mathrm{~cm}\)

h = 7 cm

Volume = nr2h

⇒ \(\frac{22}{7} \times 7 \times 7 \times 7\)

= 1078 cm³.

By actual calculation of volumes of both the cylinders, it is verified that the volume of cylinder B is greater

Surface Area

For Cylinder A

Surface area = 2jir (r + h)

⇒ \(2 \times \frac{22}{7} \times \frac{7}{2} \times\left(\frac{7}{2}+14\right)\)

⇒ \(2 \times \frac{22}{7} \times \frac{7}{2} \times \frac{35}{2}\)

= 385 cm²

For Cylinder B

Surface area

=27tr(r + h)

⇒ \(2 \times \frac{22}{7} \times 7 \times(7+7)\)

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

By actual calculation of the surface area of both the cylinders, we observe that the cylinder with greater volume (Cylinder B) has a greater surface area.

Question 3. Find the height of a cuboid whose base area is 180 cm² and volume is 900 cm³.
Solution:

Height of the cuboid = \(=\frac{\text { Volume of the cuboid }}{\text { Base area of the cuboid }}\)

⇒ \(=\frac{900}{180}\) = 5 cm

Question 4. A cuboid is of dimensions 60 cm x 54 cm x 30 cm. How many small cubes with sides 6 cm can be placed in the given cuboid?
Solution:

Volume of the cuboid = 60 x 54 x 30 cm³

= 97200 cm³

The volume of a small cube

= 6 x 6 x 6 cm³

= 216 cm³

Number of small cubes that can be placed in the given cuboid

⇒ \(\frac{\text { Volume of the cuboid }}{\text { Volume of a small cube }}\)

⇒ \(\frac{97200}{216}=450\)

Hence, 450 small cubes can be placed in the given cuboid.

Question 5. Find the height of the cylinder whose volume is 1.54 m² and the diameter of the base is 140 cm.
Solution:

The diameter of the base = 140 cm

Radius of the base (r)= \(=\frac{140}{2} \mathrm{~cm}=70 \mathrm{~cm}\)

Area of the base = nr²

⇒ \(=\frac{22}{7} \times 70 \times 70\)

= 15400 cm²

The volume of the cylinder

= 1.54 m³

= 1.54 x 100 x 100 x 100 cm³

= 15,40,000 cm³

Height of the cylinder

⇒ \(\frac{\text { Volume of the cylinder }}{\text { Area of the base of the cylinder }}\)

⇒ \(\frac{(2)}{(1)}\)

⇒ \(=\frac{15,40,000}{15400}\)

= 100 cm

= 1 m

Hence, the height of the cylinder is 1 m.

Question 6. A milk tank is in the form of a cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of milk in liters that can be stored in the tank

solution:

For milk tank

r = 1.5 m

h = 7 m

Capacity = nFh

⇒ \(=\frac{22}{7} \times 1.5 \times 1.5 \times 7\)

⇒ \(=\frac{22}{7} \times \frac{15}{10} \times \frac{15}{10} \times 7\)

= 49.5 m³ I

= 49.5 x 1000 L I

[lm³ = 1000 L]

= 49500 L.

Hence, the quantity of milk that can be storedin the tank is 49500 litres.

Question 7. If each edge of a cube is doubled,

1. how many times will its surface area increase?
2. how many times will its volume increase?

Solution:

Let the original edge of the cube be a cm.

Then, its new edge = 2a cm

The original surface area of the cube = 6a² cm²

The new surface area of the cube

= 6(2a)² cm²

= 24a² cm²

= 4 (6a² cm²)

= 4 original surface area,

Hence, its surface area will increase 4 times.

Originalvolume ofthe cube = a³ cm³

A new volume of the cube

= (2a)³ cm³

= 8a³ cm³

= 8 x the original volume of the cube.

Hence, its volume will increase 8 times

Question 8. Water is poured into a cuboidal g reservoir at the rate of 60 liters per minute. If the volume of the reservoir is 108 m3, find the BL number of hours it will take to fill the reservoir

Solution:

Volume of reservoir

= 108 m³

= 108 x 1000 L

[1 m³ = 1000 L]

= 108000 L

Water poured per minute = 60 L

Time taken to fill the reservoir

⇒ \(\frac{\text { Volume of the reservoir }}{\text { Water poured per minute }}\)

⇒ \(\frac{108000}{60} \mathrm{~m}\)

⇒ \(\frac{108000}{60 \times 60} \text { hours }\)

= 30 hours

Hence, the number of hours it will take to fill the reservoir is 30

Mensuration Multiple-Choice Question And Solutions

Question 1. 1 cm³ =

1. 1000 mm³
2. 100 mm³
3. 10 mm³
4. \(\frac{1}{1000} \mathrm{~mm}^3 .\)

Solution: 1. 1000 mm³

Question 2. 1 m³ =

1. 1000000 cm³
2. 100 cm³
3. 10 cm³
4. \(\frac{1}{1000} \mathrm{~cm}^3 .\)

Solution: 1. 1000000 cm³

Question 3. 1 mm³ =

1. 0.001 cm³
2. 0.01cm³
3. 0.1 cm³
4. 100 cm³

Solution: 1. 0.001 cm³

Question 4. 1 cm³ =

1. 0.000001m³
2. 0.01m³
3. 0.1m³
4. 1000m³

Solution: 1. 0.000001m³

Question 5. The surface area of a cuboid of length l, breadth b, and height h is

1. lbs
2. lb + bh + hl
3. 2 (lb + bh + hi)
4. 2 (l+ b)h.

Solution: 3. 2 (lb + bh + hi)

Question 6. The surface area of a cube of edge a is

1. 4a²
2. 6a²
3. 3a²
4. a²

Solution: 2. 6a²

Question 7. The total surface area of a cylinder of base radius r and height h is

1. 2πr (r + h)
2. πr (r + h)
3. 2πrh
4. 2πr²

Solution: 1. 2πr (r + h)

Question 8. The volume of a cuboid of length l, breadth b, and height h is

1. lbh
2. lb + bh + hl
3. 2 (lb + bh + hl)
4. 2(l + b)h.

Solution: 1. lb

Question 9. The volume of a cube of edge a is

1. a²
2. a³
3. a⁴
4. 6a²

Solution: 2. a³

Question 10. The volume of a cylinder of base radius and height his

1. 2πrh
2. πr²h
3. 2πr (r + h)
4. \(\frac{1}{3} \pi r^2 h .\)

Solution: 2. nr2h

Question 11. 1 L =

1. 10 cm³
2. 100 cm³
3. 1000 cm³
4. 10000 cm³

Solution: 3. 1000 cm³

Question 12. 1 m³ =

1. 1L
2. 10 L
3. 100L
4. 1000 L

Solution: 4. 1000 L

Question 13. The area ofthe trapezium

1. 9 cm²
2. 6 cm²
3. 7 cm²
4. 24 cm²

Solution: 1. 9 cm²

Area \(=\frac{(4+2) 3}{2}=9 \mathrm{~cm}^2\)

Question 14. The area of the trapezium

1. 6 cm²
2. 4 cm²
3. 3 cm²
4. 9 cm²

Solution: 1. 6 cm²

Area \(=\frac{(3+1) 3}{2}=6 \mathrm{~cm}^2\)

Question 15. The perimeter of the trapezium

1. 12 cm
2. 24 cm
3. 6 cm
4. 18 cm.

Solution: 1. 12 cm

Perimeter = 3+ 3 + 2 + 4 = 12 cm

Question 16. The area of a rhombus is 60 cm². One diagonal is 10 cm. The other diagonal is

1. 6 cm
2. 12 cm
3. 3 cm
4. 24 cm.

Solution: 2. 12 cm

⇒ \(\frac{1}{2}\)x 10 X = d260 => d2 = 12cm

Question 17. The area of a trapezium is 40 cm². Its parallel sides are 12 cm and 8 cm. The distance between the parallel sides is

1. 1cm
2. 2cm
3. 3cm
4. 4cm

Solution: 4. 4cm

⇒ \(\frac{(12+8) d}{2}=\) – 40 =» d = 4cm

Question 18. 8 persons can stay in a cubical room. Each person requires 27 m³ of air. The side of the cube is

1. 6m
2. 4m
3. 3m
4. 2m

Solution: 1. 6m

Volume = 8 x 27 = 216 m3.

Side = 216 = 6 m

Question 19. If the height of the cuboid becomes zero, it will take the shape of a

1. cube
2. parallelogram
3. circle
4. rectangle

Solution: 4. rectangle

Height = \(\frac{80}{20}=4 \mathrm{~m} .\)

Question 20. The volume of a room is 80 m³. The area of the floor is 20 m². The height of the 1 room is

1. 1m
2. 2m
3. 3m
4. 4m

Solution: 4. 4m

Question 21. The floor of a room is a square of side 6 m. Its height is 4 m. The volume of the room is

1. 140 m³
2. 142 m³
3. 144 m³
4. 145 m³

Solution: 3. 144 m³

Volume = 6x6x4 = 144 m³

Question 22. The base radius and height of a right circular cylinder are 14 cm and 5 cm respectively. Its curved surface is

1. 220 cm²
2. 440 cm²
3. 1232 cm²
4. 2π x 14 x (14 + 5) cm²

Solution: 2. 440 cm²

Curved surface =\(2 \times \frac{22}{7} \times 14 \times 5\)

= 440 cm².

Question 23. The heights of the two right circular cylinders are the same. Their volumes are respectively 16πm³  and 8lπm³. The ratio of their base radius

1. 16:81
2. 4:9
3. 2: 3
4. 9: 4.

Solution: 2. 4:9

⇒ \(\frac{\pi r_1^2 h}{\pi \pi_2^2 h}=\frac{16 \pi}{81 \pi} \Rightarrow \frac{r_1}{r_2}=\frac{4}{9}\)

Question 24. The ratio of the radii of two right circular o cylinders is 1: 2 and the ratio of their heights is 4: 1. The ratio of their volumes is

1. 1:1
2. 1:2
3. 2:1
4. 4:1

Solution: 1. 1:1

⇒ \(\frac{r_1}{r_2}=\frac{\pi(1)^2 4}{\pi(2)^2 1}=1: 1 \text {. }\)

Question 25. A glass in the form of a right circular cylinder is half full of water. Its base radius is 3 cm and its height is 8 cm. The volume of water is

1. 18π cm³
2. 36π cm³
3. 9π cm³
4. 36π cm³

Solution: 2. 36πcm³

Volume = \(\frac{1}{2} \pi \times 3 \times 3 \times 8=36 \pi \mathrm{cm}^3 .\)

Question 26. The base area of the right circular cylinder is 16ft cm³. Its height is 5 cm. Its curved surface area is

1. 40π cm²
2. 30π cm²
3. 20π cm²
4. 10π cm²

Solution: 1. 40π cm²

πr² = 16π => r = 4 cm

Curved surface area = 2xπX4x5 = 40π cm²

Question 27. The base radius and height of a right circular cylinder are 5 cm and 10 Its total surface area is

1. 150ft cm²
2. 150 cm²
3. 300ft cm²
4. 300n cm²

Solution: 1.150π cm²

Total surface area = 2πr (h + r)

= 2π5 (10 + 5) = 150π cm².

Mensuration True-False

Write whether the following statements are True or False:

1. The volume of a solid is the measurement of the space occupied by it: True

2. The areas of any two faces of a cuboid are equal:  False

3. Two cuboids with equal volumes essentially have equal surface areas: False

4. The radii of two cylinders having the same volume are in the ratio 1 : 3. Then, the ratio of their heights is 9: 1: True

5. The volume of a cube is 216 cm³. Its surface area is 216 cm3: True

Mensuration Fill In The Blanks

1. The areas of any two faces of a cube are_______: Equal

2. The total surface area of a cube, whose volume is 1 cm³, is_____:   1cm²

3. 1 litre = _______cm³:  1000

4. 1 mm =_______ m : \(\frac{1}{1000} \mathrm{~m}\)

5. The ratio of radii of two cylinders is 1: 2 and heights are in the ratio 4: 5. Find the ratio of their volumes:  1:5

6. A metal sheet 64 cm long, 27 cm broad, and 8 cm thick is melted into a cube. Find the side of the cube:  24 cm

7. A cube of side 2 cm is cut into 1 cm cubes. What is the percentage increase in the volume after such a cutting:   0%

8. Find the area of the following figure: 10 cm²

Algebraic Expressions And Identities Addition And Subtraction Of Algebraic Expressions

While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them; then handle the unlike terms. Note that the sum of x number of like terms is another like term whose coefficient is the sum of the coefficients of the like terms being added.

Addition And Subtraction Of Algebraic Expressions Exercise 8.1

Question 1. Add the following.

xb – bc, bc – cx, cx – xb

x – b + xb, b – c + be, c – x + xc

2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

l² + m², m² + n² n² + l², 2lm + 2mn + 2nl

Read and Learn More NCERT Solutions For Class 8 Maths

Solution: 1. xb – bc, bc – cx, cx – xb

Thus, the sum of the given expressions is 0.

2. x-b + xb, b-c + bc, c-x+xc

Thus, the sum of the given expressions is ab + bc + cx

3. 2p²q² – 3pq + 4, 5 + 7pq – 3p²q²

4. l² + m², m² + n², n² + l², 2lm + 2mn + 2nl

Thus, the sum of the given expressions is 2(l² + m² + n² + Im + mn + nl).

Question 2.

1. Subtract 4x – 7xb + 3b + 12 from 12x – 9xb + 5b – 3
2. Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + lOxyz
3. Subtract 4p²q – 3pq + 5pq² – 8p + 7q – 10 from IS – 3p – 11q + 5pq – 2pq² + 5p²q

Solution:

Thus, the required answer is 8a – 2ab + 26- 15

Thus, the required answer is 28 + bp- 18q + 8pq- Ipq² + p²q

8.2 Multiplication Of Algebraic Expressions: Introduction

There exist x number of situations when we need to multiply algebraic expressions. For example, in finding the xerox of x rectangle whose sides are given xs expressions

Question 1. Can you think of two more such situations, where we may need to multiply algebraic expressions?

[Hint: Think of speed and time; Think of interest to be paid, the principal and the rate of simple interest; etc.].

Solution:

Distance = Speed x Time

Simple Interest

Principal x Rate of simple interest

⇒ \(=\frac{\text { per annum } \times \text { Time in years }}{100}\)

Multiplying A Monomial By A Monomial

A monomial multiplied by an x monomial always gives an x monomial.

Multiplying Two Monomials

In the product of two monomials

Coefficient = coefficient of first monomial x coefficient of second monomial

Algebraic factor – algebraic factor of first monomial x algebraic factor of second monomial

Multiplying Three Or More Monomixls

We first multiply the first two monomials and then multiply the resulting monomial by the third monomial. This method can be extended to the product of any number of monomials.

Rules Of Signs

(1) The product of two factors is positive or negative accordingly xs the two factors have like signs or unlike signs. Note that

1. (+) x (+) = +
2. (+) X (-) = –
3. (-) x (+) = –
4. (-) x (-) = +

(2) If x is x vxrixble xnd p, q xre positive inteqers, then

x^p x x^q = x^p+q

Question 1. Find 4x x 5y x 7z. First, find 4x x 5y and. multiply it by 7z; or first, find 5y x 7z and multiply it by 4x. Is the result the same? What do you observe? Does the order in which you carry out the multiplication matter?
Solution:

4x x 5y x 7z

4x x 5y = (4 x 5) x (x x y)

= 20 x (xy) = 20 xy

(4x x 5y) x 7z = 20xy x lz

= (20 x 7) x (xy x z)

= 140 x (XYZ)

= 140xyz …(1)

OR

5y x 7z = (5 x 7) x (y x z)

– 35 X (yz) = 35yz

4x X (5y X 7z) = 4x X 35yz

= (4 x 35) x (x x yz)

= 140 x (xyz)

= I40xyz…..(2)

We observe from Eqns. (1) and (2) that the result is the same. It shows that the product of monomials is associative, i.e., the order in which we multiply monomials does not matter.

Addition And Subtraction Of Algebraic Expressions Exercise 8.2

Question 1. Find the product of the following pairs of monomials:

1. 4, 7p
2. -4p, 7p
3. -4p, 7pq
4. 4p³, -3p
5. 4p, 0.

Solution:

(1)  4, 7p

4 x 7p = (4 x 7) x p

= 28 x p

= 28p

(2)  – 4p, Ip

(-4p)x(7p) = {(-4) x 7} x (P x P)

= (-28) x p2

= -28p2

(3)  -4p, 7pq

(- 4p) x (Ipq) = {(- 4) x 7} x (p x (pq)}

= (-28) x (p x p x q)

= (-28) x p²

= -28p²q

(4)  4p², – 3p

(4p³) x ( -3p) = {4 x (_ 3)} x (p³ Xp)

= (-12) X p⁴

= – 12p⁴

(5)  4p, 0

(4p) x 0 = (4×0) x p

0 x p = 0

Question 2. Find the press of rectangles with the following pairs of monomials xs their lengths and breadths respectively:

(p, q) ; (10m, 5n); (20x² 5y²); (4x, 3.x²); (3mn, 4np).

Solution:

1.  (p, q)

Length = p

Brexdth = q

Area of the rectangle

= Length x Brexdth

= P x q

= pq

2.  (10m, 5n)

Length = 10 m

Brexdth = 5 n

Area of the rectangle

= Length x Brexdth

= (10m) x (5n)

= (10 x 5) x (m x n)

= 50 x (mn)

= 50 mn

3. (2Ox² , 5y²)

Length = 20X²

Brexdth = 5y²

Area of the rectangle

= Length x Brexdth

= (20x²) x (5y²)

= (20 x 5) x (x² x y²)

= 100 x (x2/)

= 100x²y²

4.  (4x,3x²)

Length = 4x

Brexdth = 3X²

Area of the rectangle

= Length x Brexdth

= (4x) x (3x²)

= (4 x 3) x (x x x²)

= 12 x x³ = 12X³

5.  (3mn, 4np)

Length = 3mn

Brexdth = 4np

Area of the rectangle

= Length x Brexdth

= (Smn) x (4np)

= (3×4) x (mn) x (np)

= 12 x m x (n x n)xp

= 12 mn²p.

Question 3. Complete the table of products.

Solution:

We have.

1. 2x x -5y = {2 x (- 5)} x (x x y) = – lOxy

2x x 3X² = (2 x 3) x (x x x²) = 6X³

2x x (- 4xy) = {2 x (-4)} x x x (xy)

= -8 X (x X x) X y

= – 8 x²y

2x X 7x²y = (2 X 7) X (x X x²) x y = 14x²y

2x x (-9X²y²) = {2 x (-9)> x (x x x²) Xy²

= – 18x³y²

2. (-5y)x(2x)

= {(-5) x 2} X (y X x)

= – 10 30

(_5y) x (- 5p)

= {(-5) x (-5)} x y xy

= 25 y²

(~5y) x 3X2

= {(-5) x 3} x yx2

= — 15.x²y

(-5y) x (— 430’)

= {(-5) X (- 4)} x XX (y X y)

= 20 xy²

(-5y) x (7x2y)

= {(-5) X 7} X X2 X (yXy)

= -35 x²y²

(~5y) x (-qx2ÿ)

= {(-5) x (- 9)} x x2x(yxy2)

= 45 x²y³

3. 3x² x 2x = (3 x 2) x (x2 x x) = 6X²

3X² x 3x² = (3 x 3) x (x² x x²) = 9 x⁴

3x⁴ x (- 4 xy) = 3 x (- 4)} x (x2 x x) x y = – 12x³y

3x² x 7x²y

= (3 x 7) x (x2 x x1) xy

= 21×4;

3x² x (- 9X² y²)

= {3 x (- 9)} x (x² x x²)xy²

= – 27xy

4. (-4xy) x 2x

= {(_4) x 2} x {x x x) xy

= -8x²y

(-4xy) x (-5y)

= {(- 4) x (- 5)} X X x (y x y)

= 20xy²

(-4x²y) x 3X²

= {(- 4) x 3} x (ix x²) x y

= – 12x²y

(-430) x (- 4ry)

= {(-4) X (_4)}x (ixi)x(y Xy)

= 16x²y²

(-4xy) X Ixÿy

= {(-4) X 7} X (x X X2) X (y X y)

= -28x³y²

(- 4xy) X (- 9X2/)

={(-4) x (-9)} X (x X x2) X (y X y2)

= 36x³y³

5. 7x²y x 2x

= (7 x 2) x (x2 x x) x y

= 14x²y

7x²y x (- 5y)

= {7 x (-5)} x x² X (y Xy)

= – 35x²y²

7x²y X(3x²)

– (7 x 3) X (*2 x x2) Xy = 21x4y

7x²y x (- 4 xy)

= {7 X (-4)} X (x² x x) X (y X y)

= -28x³y²

7x2y x 7x2p

= (7 x 7) X (x² X X²) x (y X y)

= 49x⁴y²

7-X²y X (_ 9x²ÿ2)

= {7X(-9)} X (x²X X²) X (y x y²)

= -63x⁴y³

(- 9x²,y²) x 2x

= {(-9) x 2} x x( x x) x y2

= -18x³y²

(-9xV) x (-5y)

= {(-9) x (-5)} x x² x x² (y²x y)

– 45x²y³

(- 9x2y²) x (3X²)

= {(- 9) x 3} x (x² x x²) x y2

= – 27x⁴ y²

(- 9x²y) x (- 4xy)

= {(-9) x (-4)} x (x² x X) x (y2 x y)

= 36 x⁴

(~9x2y2) x (7x2y)

= {(-9)x7}x(x2Xx2)X(y2xy)

= – 63 xy

(-9x2y²) x (- 9xY)

= {(-9) x (-9)}x ft? xÿ x (y2 x y2)

= 81 x⁴y⁴

Question 4. Obtain the volume of rectangular boxes with the following length, breadth, and height respectively:

1. 5x, 3x², 7x.4
2. 2p, 4q, 8r
3. xy, 2x²y, 2xy²
4. x, 2b, 3c.

Solution:

1.  5x, 3x², 7×4

Length = 5x

Brexdth = 3x²

Height = 7×4

Volume of the rectxnqulxr box

= Lenqth x Brexdth x Heiqht

= (5x) x (3x²) x (7×4)

= (5 x 3 x 7) x (x x x² x x4)

= 105×7

2.   2p, xq, 8r

Lenqth = 2p

Brexdth = xq

Heiqht = 8r

Volume of the rectxnqulxr box

= Lenqth x Brexdth x Heiqht

= (2p) x (xq) x (8r)

= (2 x 4 x 8) x (p x q x r)

= 64 pqr

3.  xy, 2x²y, 2xy²

Length = xy

Brexdth = 2x²y

Height = 2xy²

The volume of the rectangular box

= Length x Breadth x Height

= (xy) x (2x²y) x (2xy²)

= (2 x 2) x (x x x² x x) x (yxyxy²)

= 4x4y4

4.  x, 2b, 3c

Length = x

Brexdth = 2b

Height = 3c

The volume of the rectangular box

= Length x Breadth x Height

= (x) x (2b) x (3c)

= (2 x 3) x (x x b x c)

= 6xbc

Question 5. Obtain the product of

1. xy, yz, zx
2. x, -x², x3
3. 2, 4y, 8y2, 16y³
4. x, 2b, 3c,6xbc
5. m, -mn, mn

Solution:

1.  xy, yz, zx

Required product

= (xy) x (yz) x (zx)

= (x x X) X (y X y) X (Z X z)

= x² x x ²z²

= x2²y²z²

2.  x, -x², x³

Required product

= (x) x (-x²) x (x²)

= – (x x x² x x³)

= -x⁶

3.  2, 4y, 8y² 16y³

Required product

= (2) x (4y) x (8y²) x (16y³)

= (2 x 4 x 8 x 16) x (y x y2 x y3)

= 1024y⁶

4.  x, 2b, 3c, 6xbc

Required product

= (x) x (26) x (3c) x (6xbc)

= (2 x 3 x 6) x (x x x) x (6 x 6) x(c x c)

= 36 x²b²c²

5.  m, – mn, mn

Required product

= (m) x (-mn) x (mnp)

= (- 1) x (m x m x m) x (n xn) xp

= -m²n²p

Multiplying A Monomial By A Polynomial

While multiplying x polynomial by x monomial, we multiply every term in the polynomial by the monomial.

Using the distributive law, we carry out the multiplication term by term.
It states that if P, Q, and R are three monomials, then

1. P x (Q + R) = (P x Q) + (p x R)
2.  (Q + R) x p = (Q x P) + (R x P)

Question. Find the product:

1.  2x(3x + 5xy) (H)x,2 (2xb – 5c).

Solution:

1.  2x (3x + 5ry)

2x (3x + 5xy) = (2x) x (3x) + (2x) x (5xy)

= (2 x 3) x(xxx) + (2 x 5) x (x x x) x y

= 6x² + 10 x2y

2.  a² (2xb-5c)

x2(2xb – 5c) = (x2) x (2xb) – (x2) x (5C)

= (1 x 2) x (x2 x x) X b -(1 X 5) X x2 X c

= 2a³b – 5a²c

8.4.2 Multiplying A Monomixl By A Trinomial

By using the distributive law, we carry out the multiplication term by term

Question. Find the product: (4p² + 5p + 7) x 3p.
Solution:

(4p² + 5p + 7) x 3p.

= (4p²x 3p) + (5p x 3p)+ (7 x 3p)

= (4 x 3) x (p² x P) + (5 x 3) x (p x p) + (7 x 3) x p

= 12p3+ 15p² + 21p

Addition And Subtraction Of Algebraic Expressions Exercise 8.3

Question 1. Carry out the. multiplication of the expressions in exch for the following pairs:

1. 4p, q + r
2. ab, a – b
3. a + b, 7a²b²
4. a²-9, 4a
5. PQ + qr + rp, 0

Solution:

1.   4p, q + r

(4p) x (q+ r) = (dp) x (q) + (4 p) x (r)

By distributive law

= (4 x 1) x p x q + (4 x 1) x p x r

= 4pq + 4pr

(2) xb, x-b

(ab) x (a-b) = (ab) x (a) – (ab) x (b)

By distributive law

= (1 x 1) x (a x a) xb

– (1 x 1) a x (b x b)

= a²b – ab²

3.  a+b, 7a²b²

(x +b) x (7a²b²)

= x x 7a² b² + 5 x 7×2 b2

By distributive law

= (1 x 7) x (a x x²) x b² + (1 x 7) x a² x (a x b2)

= 7a³b² + 7a²b³

a²– 9, 4a

(a²– 9) x (4a)

= a²x 4a – 9 x 4a

By distributive law

= ( 1 x 4) x (a² x a) – (9 x 4) x a

= 4a³-36a

PQ + qr + rp, 0

(PQ + qr+ rp) x (0)

= (pq) x 0 + (qr) x 0 + (rp) x 0.

By distributive law

= 0 + 0 + 0 = 0.

Question 2. Complete the table:

Solution:

We have,

(1) a x (b + c + d)

= a x b + a x c + a x d

= ab + ac + ad

(2) (X + y – 5) x 5xy

= pC x 5pc_y + y x 5*y + (-5) x 5xy

= (1 X 5) X (x x X) X y

+ (1 x 5) X X X (y X y)

+ {(-5) x 5}x xy

= 5x²y + 5xy² – 25xy

(3) p x (6p² – 7p + 5)

= p x 6p² + p x (- 7p) + p x 5

= (1 x 6) x (p xp²)

+ {1 x (- 7)} x (p x p) + 5 x p

= 6p³ – 7p² + 5p

(4) 4p²q² x (p² – q²)

= 4p²q² x p² + 4p²q² x (-q²)

= (4 x 1) x (p² x p²) x q²

+ {4 X (- 1)} X p² X (q² X q²)

= 4p⁴ q² – 4p² q⁴.

(5)(a + b + c) x abc

= a x abc + b x abc + c x abc

= (ax a) x b x c

+ a x (b x b) x c

+ a x b x (c x c)

= a² bc + ab² c + abc²

Question 3. Find the product:

(a²) x(2a²²) x (4a²⁶)

⇒ \(\left(\frac{2}{3} x y\right) \times\left(\frac{-9}{10} x^2 y^2\right)\)

⇒ \(\left(-\frac{10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

x x x² x x³x x⁴

Sol. (i) (x2) x (2×22) x (4×26)

= (1 x 2 x 4) x (x2 x x22 x x26)

= 8 X a^2+22+26

Bylaws of exponents

= 8 x a⁵⁰ = 8×50

⇒ \(\left(\frac{2}{3} x y\right) \times\left(-\frac{9}{10} x^2 y^2\right)\)

⇒ \(\left\{\frac{2}{3} \times\left(-\frac{9}{10}\right)\right\} \times\left(x \times x^2\right) \times\left(y \times y^2\right)\)

⇒ \(-\frac{3}{5} x^3 y^3\) I by laws of exponents

3.   \(\left(-\frac{10}{3} p q^3\right) \times\left(\frac{6}{5} p^3 q\right)\)

⇒ \(=\left\{\left(-\frac{10}{3}\right) \times \frac{6}{5}\right\} \times\left(p \times p^3\right) \times\left(q^3 \times q\right)\)

= -4p⁴q⁴.    Bylaws of exponents

(4) X x x² X x³ X x⁴

X x x2 x x3 X x4 = (1 X 1 X 1 X 1) X X¹ X x² X x³ X x⁴

= (1) X x ^1+2+3+4

= X10.

Bylaws of exponents

Question 4. (x) Simplify: 3x (4x -5) +3 and find Us values for (i) x = 3, (ii) x =\(\frac{1}{2}\)

b) Simplify: x(x2 + x + 1) + 5 xnd find its vxlue for

(i) x = 0, (ii) x = l xnd (Hi) x=-l.

Solution :

1.  3x (4x-5) + 3

= (3x) (xx) – (3x) (5) + 3

= (3 x 4) x (x x x) – (3 x 5) x x + 3

= 12x² – 15x + 3

(i) Whenx = 3,

12x² – 15.x + 3

= 12(3)² – 15(3) + 3

= 12 x 9 – 45 + 3

= 108 – 45 + 3 = 66

2.  When x =\(\frac{1}{2}\)

12X² – 15.x + 3

⇒ \(12\left(\frac{1}{2}\right)^2-15\left(\frac{1}{2}\right)+3\)

⇒ \(12\left(\frac{1}{4}\right)-\frac{15}{2}+3\)

⇒ \(3-\frac{15}{2}+3\)

⇒ \(6-\frac{15}{2}\)

⇒ \(\frac{12-15}{2}\)

⇒ \(-\frac{3}{2}\)

(b) a(a² + x + 1) + 5

a x a² + a x a + (a x 1 + 5 )

= a³ + a² + a+ 5

1.  When a = 0

a³ + a²+ a + 5 = (0)3 + (0)2 + (0) + 5

=0+0+0+5=5

2.  When a =1

a³ + a² + a+ 5 = (l)3 + (l)2 + (1) + 5

=l+l+l+5=8

3.  When a=-l

a³ +a² + a+ 5

= (- 1)3 + (- l)2 + (- 1) + 5

= —1 + 1 — 1 + 5 = 4

Question  5.

1. add : p(p – q), q(q – r) and r(r-p)
2. add: 2x(z – x – y)and 2y (z -y- x)
3. Subtract: 31 (l – 4m. + 5n) from 41 (10n-3m + 21)
4. Subtract: 3x(x +b+c)-2b(x-b+c) from 4c(- x + b+ c).

Solution:

(1) First expression

= p(p-q)=pxp-pxq

= p² – pq

Second expression

= q(q-r)

= q x q – q x r

= q² – qr

Third expression

= r(r – p)

=r x r – r x p

= r² – rp

adding the three expressions,

image

First expression

= 2x (z – x – y)

= (2x) X (2)-(2x) X (x) – (2x) X (y)

= 2xz -2x² – 2xy

Second expression

= 2y(z-y-x)

= (2y) x (z) – (2y) x (y) – (2y)x(x)

= 2yz – 2y² – 2yx

adding the two expressions,

image

First expression

= 31(1 – 4m + 5n)

= (31) x (l) – (31) x (4m) + (31) x (bn)

= 312 – 12lm + 15In

Second expression

= 41 (10n – 3m + 2l)

= (4l) x (lOn) – (4l) x (3m) + (4l) x (2l)

= 40ln – 12lm + 8l²

Subtrading,

image

First expression

= 3x (x + 6 + c) – 26 (x – 6 + c)

= [(3x) x (x) + (3x) x (6) + (3x) x (c)]

-[(26) x (x) + (26) x (6) – (26) x (c)]

= [3a² + 3ab + 3ac]- [2ab- 2b² + 2bc]

= 3a² + 2b² + 3ab – 2ab- 2bc + 3ac

– 3a² + 2b² + ab- 2bc + 3ac

Second expression

= 4c (- x + 6 + c)

= 4c x (-x) + 4c x 6 + 4c x c

= -4xc + 46c + 4c2

Subtracting

image

8.5 Multiplying A Polynomial By A Polynomial

We multiply the term of one polynomial by each term of the other polynomial. Also, we combine the like terms in the product.

Multiplying A Binomial By A Binomial

We use distributive law and multiply each of the two terms of one binomial by exchanging the two terms of the other binomial and combining like terms in the product.

Thus, if P, Q, R, and S are four monomials, then

(P + Q) x (R + S) = P x (R + S) +Qx(R + S)

= (P x R + P x S) + (QXR + QXS)

= PR + PS + QR + QS.

8.5.2 Multiplying A Binomial By A Trinomial

We use distributive law and multiply each of the three terms in the trinomial by exchanging the two terms in the binomial and combining like terms in the product.

Addition And Subtraction Of Algebraic Expressions Exercise 8.4

Question 1. Multiply the binomial:

1. (2x + 5) and (4x – 3)
2. (y – 8) and (3y – 4)
3. (2.51 – 0.5 m) and (2.51 + 0.5m)
4. (x + 3b) and (x + 5)
5. (2pq + 3q²) and (3pq – 2q²)
6. \(\left(\frac{3}{4} x^2+3 b^2\right) \text { and } 4\left(x^2-\frac{2}{3} b^2\right)\)

Solution:

(i) (2x + 5) xnd (4x – 3)

(2x + 5) x (4x – 3)

= (2x) x (4x – 3) + 5 x (4x – 3) by distributive law

= (2x) x (4x) – (2x) x (3) + (5) x (4x) – (5) x (3)

= 8x² – 6x + 20x – 15

= 8x² + (20x – 6x) – 15 Combining like terms

= 8x² + 14x – 15

2. (y – 8) and (3y – 4) by distributive law

(y – 8) x (3y – 4)

= y x (3y – 4) – 8 x (3y – 4) – (8) x (3y) + 8 x 4

= (y) x (3y) – (y) x (4)

= 3y² – 4y -24y + 32 Combining like terms

3. (2.5 l – 0.5 m) xnd (2.5 l + 0.5 m)

(2.5 l – 0.5 m) x (2.5 l + 0.5 m)

= (2.5/) x (2.5/ + 0.5 m) – (0.5 m) x (2.5/ + 0.5 m) by distributive law

= (2.5 Z) x (2.5 Z) + (2.5 /) x (0.5 m) – (0.5 m) x (2.5 /) – (0.5 m) x (0.5 m)

= 6.25 12 + 1.25 Zm – 1.25 m/- 0.25 m² Combining like terms

= 6.25 l² + (1.25 lm – 1.25 lm) – 0.25 m²

= 6.25 l² – 0.25 m²

4. (x + 3 b) and (x + 5)

(a + 35) x (x + 5)

= a x (c +5) + (3b) x (x + 5) by distributive law

= (a) x (x) + (a) x (5) + (3b) x (x) + (3b) x (5)

= ax + 5a + 3bx + 15b

5.  (2pq + 3q²) xnd. (3pq – 2q²)

(2pq + 3q²) x (3pq – 2q²)

= (2pq) x (3pq – 2q²) + (3q²) x (3pq – 2q²) by distributive law

= (2pq) x (3pq) – (2pq) x (2q²) + (3q²) x (3pq) – (3q²) x (2q²)

= 6p2q2 – 4pq³ + 9pq³ – Qqi

= 6p²q² + (9pq3 – 4pq3) – 6q4 Combining like terms

= 6p²q² + 5pqz – 6q⁴

⇒ \(\left(\frac{3}{4} x^2+3 b^2\right) \text { and } 4\left(x^2-\frac{2}{3} b^2\right)\)

⇒ \(\left(\frac{3}{4} x^2+3 b^2\right) \times 4\left(x^2-\frac{2}{3} b^2\right)=\left(\frac{3}{4} x^2+3 b^2\right) \times\left(4 x^2-\frac{8}{3} b^2\right)\)

⇒ \(\frac{3}{4} x^2 \times\left(4 x^2-\frac{8}{3} b^2\right)+3 b^2 \times\left(4 x^2-\frac{8}{3} b^2\right)\) by distributive law

⇒ \(\left(\frac{3}{4} x^2\right) \times\left(4 x^2\right)-\left(\frac{3}{4} x^2\right) \times\left(\frxc{8}{3} b^2\right)+\left(3 b^2\right) \times\left(4 x^2\right)-\left(3 b^2\right) \times\left(\frac{8}{3} b^2\right)\)

= 3a⁴ – 2a²b² + 12b²a² – 8b⁴

= 3a⁴ + (12a²b² -2a²b²) -8b⁴

= 3a⁴ + 10a²b² – 8b⁴Combining like terms

Question 2. Find the product:

1. (5 – 2x) (3 + x)
2. (x + 7y) (7x-y)
3. (x² + b) (x + b²)
4. (p² – q²) (2p + q)

Solution:

(i) (5 – 2x) (3 + x)

= (5) x (3 + x) – (2x) x (3 + x)

= (5) x (3) + (5) x (x) – (2x) X (3) – (2x) X (X)

= 15 + 5x – 6x – 2x²

= 15 – x – 2x²

2.  (x + 7y) (7x – y)

= (x) x (7x – y) + (7y) x (7x – y)

= (x) x (7x) – (x) x (y) + (7y) x (7x) – (7y) x (y)

= 7x² – xy + 49yx – 7y²

= 7x² + 48xy – 7y²

3.  (a² + b) (a + b²)

= a² x (a+ b²) + b x (a + b²)

= (a²) x (a) + (a²) x (b²) + (b) x (a) + (b) x (b²)

= a³ + a²b² + ab + b³

= a³ + a²b² + ab + b³

4.  (p² – q²) (2p + q)

= p² x (2p + q) – q² x (2p + q)

= (p²) x (2p) + (p²) x (q) – (q²) x (2p)- (q²) x (q)

= 2p³+ p²q – 2q²p – q³

= 2p³ + p²q – 2pq² – q³

Question 3. Simplify :

1. (x² – 5) (x+ 5) + 25
2. (x² + 5) (b³ + 3) + 5
3. (l + s²) (l² – s)
4. (x + b)(c – d) + (x – 6) (c + d)+ 2 (xc + bd)
5. (x + y) (2x + y) + (x + 2y)(x – y)
6.  (x + y) (x² – xy + y²)
7. (1.5x – 4y)(1.5x + 4y + 3)- 4.5x + I2y
8. (x + b + c) (x. + b – c).

Solution:

1.   (x² – 5) (x + 5) + 25

= x²(x + 5) – 5(x + 5) + 25

= (x² x X) + (x² x 5) – (5 x X) – (5 x 5) + 25

= x³ + 5X² – 5x – 25 + 25

= x³ + 5×2 – 5x

2.  (a² + 5) (b³ + 3) + 5

= a²(b³ + 3) + 5(b³ + 3) + 5

= (a² x b³) + (a² x 3) + (5 x b³) +(5a³) + 5

= a²b³ + 3a² + 5b³ + 15 + 5

= a²b³ + 3a² + 5b³ + 20

3.  (t + s²) (t² – s)

= t(t² – s) + s²(t² – s)

= (tx t²) – (t X s) + (s² X t²) – (s² X s)

= t² -ts + s²t² – s³

4.  (a + b) (c – d) + (a – b) (c + d) + 2(ac + bd)

= a(c – d) + b(c – d) + a(c + d) – b(c + d) + 2(oc + bd)

= ac – ad + be – bd + ac + ad – be – bd + 2ac + 2bd by distributive law

= (ac + ac + 2ac) + (ad- ad) + (be- be) + (2bd – bd – bd)

= 4ac combines the like terms

5.  (x + y) (2x + y) + (x + 2y) (x – y)

= x(2x + y) + y(2x + y) + x(x – y) + 2y(x – y)

= (X X 2x) + (x X y) + (y X 2x) + (y x y) + (x x x) – (x X y) (2y x x)- (2y x y

By distributive law

– 2×2 + xy + 2xy + y² + x2 – xy + 2xy – 2y²

= (2x² + x²) + (xy + 2xy – xy + 2xy) + (y2 – 2y²)

= 3x² + 4xy – y²

6.  (x + y) (x² – xy + y²)

x(x² – xy + y²)= y(x² – xy + y²)

(x x x²) – (x x xy) + (x x y²) + (y x x²) – (y x xy) + (y x y² ) | by distributive law

x³ – x²y + xy² + x²y – xy² + y3

= x^-3 + x²y – x²y) + (xy² – xy²) + y3   combining the like terms

= x^–3 + y²

7.  (1.5x – 43-) (1.5x + 4y + 3) – 4.5x + 12y

= 1.5x (1.5x + 4y + 3)- 4y (1.5x + 4y + 3) – 4.5x + 12y | by distributive law

= (1.5x x 1.5x) + (1.5x x 4y) + (1.5x X 3) – (4y x 1.5x) – (4y x 4y) – (4y x 3) – 4.5x + 12y | by distributive law

2.25x² + 6xy + 4.5x – 6xy – 16y2 – 12y – 4.5x + 12y

= 2.25x² + (6xy – 6xy) – 16y² + (4.5x – 4.5x) + (12y – 12y) I combining the like terms

= 2.25x² – 16y²

8.  (a + b + c) (a + b – c)

= a(a + b-c) + b(a + b-c) + c(a + 6-c) I by distributive law

= a2 + ab – ac + ab + b² – be + ae + be – c²    I by distributive law

= a²+ (ab + ab) + (ac – ac) + b² + (be – be) – c² combining the like terms

= a² + 2ab + b2 – c²

Addition And Subtraction Of Algebraic Expressions Multiple-Choice Questions And Solutions

Question 1. The expression x + 3 is in

1. one variable
2. two variables
3. no variable
4. none of these.

Solution: 1. one variable

Question 2. The expression 4xy + 7 is in

1. one variable
2. two variables
3. no variable
4. none of these.

Solution: 2. two variables

Question 3. The expression x + y + z is in

1. one variable
2. no variable
3. three variables
4. two variables.

Solution: 3. three variables

Question 4. The value of 5x when x = 5 is

1. 5
2. 10
3. 25
4. -5.

Solution: 3. 25

Question 5. The value of x² – 2x + 1 when x = 1 is

1. l
2. 2
3. -2
4. 0.

Solution: 4. 0

Question 6. The value of x² + y2 when x = 1, y = 2 is

1. l
2. 2
3. 4
4. 5.

Solution: 4. 5

Question 7. The value of x² – 2yx + y2 when x = 1, y = 2 is

1. l
2. -l
3. 2
4. -2.

Solution: 1. 1

Question 8. The value of x² – xy + y2 when x = 0, y = 1 is

1. 0
2. -l
3. l
4. none of these.

Solution: 3. 1

Question 9. The sum of lx, lOx, and 12x is

1. 17x
2. 22x
3. 19x
4. 29x.

Solution: 4. 29x.

Question 10. The sum of 8pq and -17pq is

1. pg
2. 9pqr
3. -9pq
4. -pg.

Solution: 3.-9pq

Question 11. The sum of 5x², -7X², 8x², 11x² and -9x² is

1. 2x²
2. 4x²
3. 6x²
4. 8x²

Solution: 4. 8x².

Question 12. The sum ofx² y²-z² and z²-x² is

1. 0
2. 3x²
3. 3y²
4. 3z²

Solution: 1. 0

Question 13. What do you get when you subtract – 3xy from 5xy?

1. 3xy
2. 5xy
3. 8xy
4. xy.

Solution: 3. 8xy

Question 14. The result of the subtraction of law from 0 is 4

1. 0
2. 7x
3. -7x
4. x

Solution: 3. -7x

Question 15. The result of subtraction of 3x from – 4x is

1. -7x
2. 7x
3. x
4. -x.

Solution: 1. -7x

Question 16. The product of 4mn and 0 is

1. 0
2. 1
3. mn
4. 4mn.

Solution: 1. 0

Question 17. The product of 5x and 2y is

1. xy
2. 2×7
3. 5xy
4. 10xy.

Solution: 4. 10xy.

Question 18. The product of lx and -12x is

1. 84x²
2. -84x²
3. x²
4. -x².

Solution: 2. -84x²

Question 19. The area of a rectangle whose length and breadth are 9 and 4y respectively is

1. 4y³
2. 9y³
3. 36y³
4. 13y³

Solution: 3. 36y³

Question 20. The area of a rectangle with length 2l2m and breadth 3Im? is

1. 6l3m³
2. l3m³
3. 2l3m³
4. 4l3m³

Solution: 1. 6l3m³

Question 21. The volume of a cube of side 2a is

1. 4a³
2. 2a
3. 8a³
4. 8.

Solution: 3. 8a³

Question 22. The volume of a cuboid of dimensions a, 6, c is

1. abc
2. a2b2c2
3. a3b3c3
4. none of these.

Solution: 1. abc

Question 23. The product of x², – x³, – x⁴ is

1. x⁹
2. x⁵
3. x⁷
4. x⁶

Solution: 1. x⁹

Addition And Subtraction Of Algebraic Expressions True-False

Write whether the following statements are True or False:

1. In a polynomial, the exponents of the variables are always non-negative integers:   True

2. a (b + c) = ab + ae is called the distributive property : True

3. a² – b² is the product of (a – b) and (a + b): True

Addition And Subtraction Of Algebraic Expressions Fill in the Blanks

1. The value of, when 92 -72 = 8p, is___________ : 4

2. On subtracting -a²b² from 2ab², we get___________:  2a²b²

3. The difference between the squares of two consecutive natural numbers is their_______: Sum

4. Area of a rectangle with length 3 ab² and breadth 4 ac² is_______:  12a²b²c²

5. Write the product of the monomials -12a, -15a², a²b:  180 a³b³

6. Add a² + b² – c²  b²  + c² – a²  and c²  + a² – b² :  a² + b² + c²

7. Find the value of, if lOOOy = (981)² – (19)²:  962

NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Introduction

In the preceding class, we have learned about algebraic expressions and equations. Here, we shall confine ourselves to the study of linear equations in one variable.

An equation essentially contains a sign of equality (=). It is a statement of equality involving an unknown quantity.

→ Also, the expressions we use to form a linear equation are linear only, i.e., the highest power of the variable occurring in the expression is 1, and that too only in one variable. A linear equation may have linear expressions on both sides of the sign of equality.

The expression on the left of the sign of equality is called the Left Hand Side whereas the expression on the right of the sign of equality is called the Right Hand Side.

Read and Learn More NCERT Solutions For Class 8 Maths

→  The value of the variable for which LHS = RHS is called a solution of the linear equation.

To find the solution of a linear equation in one variable, we assume that the two sides of the equation are balanced.

→  We are free to perform the same operation (suitable) on both sides of the equation such as we can add to or subtract from both sides of the equation the same quantity (number).

→  Also, we can multiply or divide both sides of the equation by the non-zero quantity (number).

Solving Equations Having The Variable On Both Sides

We transpose the terms so that the terms containing the variables are on the LHS and constant numbers on the RHS.

Then, we can solve the equation by simplifying both sides and dividing by a suitable number (if required).

NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Exercise 2.1

Solve the following equations and check your results

Question 1. 3x = 2x + 18
Solution:

3x = 2x + 18

3x- 2x = 18

Transposing 2x from RHS to LHS

x = 18

This is the required solution

Check:

LHS = 3x = 3 (18)

= 54

RHS = 2x + 18 = 2 (18)+18

= 36 + 18

= 54

∴ x = 18

∴ LHS = RHS

2. 5t – 3 = 3t – 5
Solution:

5t – 3 = 3t – 5

5t – 3t, = -5 + 3

Transposing 3t from RHS to LHS- 3 from LHS to RHS

2t =-2

t = \(\frac{2}{2}\)

t = -1

Dividing both sides by 2

This is the required solution
.
Check:

LHS = 5t – 3 = 5(- 1)- 3

= -5 – 3 = -8

∴ t = -1

RHS = 3t – 5 = 3(- 1)- 5

= -3 – 5

= -8

∴ LHS = RHS

Question 3. 5x + 9 = 5 + 3x
Solution:

5x + 9 = 5 + 3x

5x – 3x = 5 – 9

Transposing 3 from RMS to LHS and 9 from LHS to RMS

2 x =-4

x = \(\frac{-4}{2}\)

x = – 2

Dividing both sides by 2

This is the required solution

Check:

LHS =  5x + 9

= 5(-2)+9 = -10+9

= -1

RHS =  5 + 3x

= 5+3(-2) = 5 -6

= -1

∴ LHS = RHS

Question 4. 4z + 3 = 6 + 2z
Solution:

4z + 3 = 6 + 2z

4z-2z = 6-3

Transposing 2z from RHS to LHS 3 from LHS to RHS

2z = 3

z= \(\frac{3}{2}\)

Dividing both sides by 2

This is the required solution

Check:

LHS = 4z + 3

= 4 (\(\frac{3}{2}\)) +3

= (\(\frac{12}{2}\)) +3 = 6+3

= 9

RHS = 6 + 2z

= 6+ 2(\(\frac{3}{2}\))

=  6+ \(\frac{6}{2}\) = 6+3

= 9

∴ LHS = RHS

Question 5. 2x – 1 = 14 – x
Solution:

2x – 1 = 14 – x

2x + x = 14+ 1

Transposing- x from RHS to LHS and -1 from LHS to RHS

3x = 15

x= \(\frac{15}{3}\)

x = 5

Dividing both sides by 3

This is the required solution

Check:

LHS = 2x – 1 = 2(5) – 1

= 10 – 1

= 9

RHS = 14 – x

=14 – 5

= 9

∴ LHS = RHS

6. 8x + 4 = 3(x – 1) + 7
Solution:

8x + 4 = 3(x – 1) + 7

8x + 4 = 3x – 3 + 7

8x+4 = 3x + 4

8x- 3x = 4-4

Transposing 3x from RHS to LHS and 4 from LHS to RHS

5x = 0

x = \(\frac{0}{5}\)

= 0

Dividing both sides by 5

This is the required solution

Check:

LHS =  8x + 4 = 8(0)+ 4

= 4

RHS =  3(x – 1) + 7= 3(0-1)+7

= -3+7

= 4

∴ LHS = RHS

Question 7. \(x=\frac{4}{5}(x+10)\)
Solution:

x = \(\frac{4}{5}(x+10)\)

5x = 4(x + 10)

Multiplying both sides by 5

5x = 4x + 40

5x-4x = 40

Transposing 4x from RHS to LHS

x = 40

This is the required solution

Check:

LHS = 5x   = 5(40)

=  200

RHS = 4(x + 10) = 4 (40+10)

= 4(50)

= 200

∴ LHS = RHS

8. \(\frac{2 x}{3}+1=\frac{7 x}{15}+3 \)
Solution: 

⇒ \(\frac{2 x}{3}+1=\frac{7 x}{15}+3 \)

⇒ \(\frac{2 x}{3}-\frac{7 x}{15}=3-1\)

Transposing \(\frac{7 x}{15}\) from RHS to LHS and 1 from LHS to RHS

⇒  \(\frac{2 x}{3}-\frac{7 x}{15}=2\)

⇒  \(15\left(\frac{2 x}{3}-\frac{7 x}{15}\right)=2 \times 15\)

Multiplying both sides by 15

∴ LCM (3, 15) = 15

10x – 7x = 30

3x = 30

x = \(\frac{30}{3}\)

x = 10

This is the required solution

∴ LHS = RHS

Question 9. \(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Solution:

⇒ \(2 y+\frac{5}{3}=\frac{26}{3}-y\)

⇒ \(2 y+y=\frac{26}{3}-\frac{5}{3}\)

Transposing – y from RHS to LHS \(\frac{5}{3}\) and from LHS to RHS

3 y= \(\frac{26-5}{3}\)

3 y= \(\frac{21}{3}\)

3y = 7

y = \(\frac{7}{3}\)

∴ LHS = RHS

10. 3m-5m-\(\frac{8}{5}\)
Solution:

3m-5m= –\(\frac{8}{5}\)

3m-5m-\(\frac{8}{5}\)

Transposing 5m from RHS to LHS

= -2m = –\(\frac{8}{5}\)

m = \(\frac{-8}{5 \times(-2)}=\frac{4}{5}\)

This is the required solution

∴ LHS = RHS

Reducing Equations To Simpler Form

We multiply both sides of the equation by the LCM of the denominators of the terms in the expressions occurring in the given equation.

→ We transpose properly so that all the variable terms come on LHS and constant terms on RHS.

→ Then, combining like terms on both sides of the equation and dividing both sides by a suitable number (if required), we can find out the required solution.

Finally, we check this solution for its validity. Reject if the solution is invalid

NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Exercise 2.2

Solve the following linear equations:

Question 1. \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:

⇒ \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)

We have, \(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)

It is a linear equation since it involves linear expressions only

⇒ \(\frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}\)

Transposing \(\frac{x}{3}\) from RHS to LHS and \(\frac{-1}{5}\) from LHS to RHS

⇒  \( \frac{3 x-2 x}{6}= \frac{5+4}{20}\)

⇒  \(\frac{x}{6}= \frac{9}{20}\)

x = \(\frac{9}{20} \times 6=\frac{9}{10} \times 3\)

Multiplying both sides by 6

x = \(\frac{27}{10}\)

This is the required solution

Check:

LHS = \(\frac{x}{2}-\frac{1}{5}=\frac{1}{2} \times \frac{27}{10}-\frac{1}{5} \)

= \(\frac{27}{20}-\frac{1}{5}=\frac{27-4}{20} \)

= \(\frac{23}{20}\)

RHS = \(\frac{x}{3}+\frac{1}{4}=\frac{1}{3} \times \frac{27}{10}+\frac{1}{4} \)

=\(\frac{9}{10}+\frac{1}{4}=\frac{18+5}{20}\)

∴ LCM(10,4)= 20

= \(\frac{23}{20}\)

∴ LHS = RHS

Question 2. \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)
Solution:

⇒ \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21\)

It is a linear equation since it involves linear expressions only

⇒ \(\frac{6 n-9 n+10 n}{12}=21\)

⇒ \(\frac{7 n}{12}\) =21

n = \(21 \times \frac{12}{7}=36\)

Multiplying both sides by \(\frac{12}{7}\)

This is the required solution

Check:

LHS =  \(\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}\)

= \(\frac{36}{2}-\frac{3}{4} \times 36+\frac{5}{6} \times 36\)

18 – 27 + 30 = 21

RHS = 21

∴ LHS = RHS

Question 3. \(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:

⇒ \(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)

It is linear equation since it involves linear expressions only.

x- \(\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-7\)

Transposing \(\frac{5 x}{2}\) from RHS to LHS and 7 from LHS to RHS

⇒ \(\frac{6 x-16 x+15 x}{6}=\frac{17-42}{6}\)

∴  LCM (3,2) = 6

⇒ \(\frac{5 x}{6}=\frac{-25}{6} \)

⇒ \(\frac{-25}{6} \times \frac{6}{5}\)

Multiplying both sides by \(\frac{6}{5}\)

x =-5

This is the required solution

∴ LHS = RHS

Question 4. \(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:

⇒ \(\frac{x-5}{3}=\frac{x-3}{5}\)

It is a linear equation since it involves linear expressions only.

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

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

Transposing \(\frac{x}{5}\) From RHS to LHS and-\(\frac{x}{5}\)from LHS to RHs

⇒ \(\frac{5 x-3 x}{15}=\frac{25-9}{15}\)

⇒ \(\frac{2 x}{15}=\frac{16}{15} \)

=\(\frac{16}{15} \times \frac{15}{2}=8\)

Multiply both sides by \(\frac{15}{2}\)

This is the required solution

∴ LHS = RHS

Question 5. \(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)
Solution:

⇒ \(\frac{3 t-2}{4}-\frac{2 t+3}{3}=\frac{2}{3}-t\)

It is a linear equation since it involves linear expressions only.

⇒ \(\frac{3}{4} t-\frac{2}{4}-\frac{2}{3} t-\frac{3}{3}=\frac{2}{3}-t\)

⇒  \(\frac{3}{4} t-\frac{1}{2}-\frac{2}{3} t-1=\frac{2}{3}-t\)

⇒  \(\frac{3}{4} t-\frac{2}{3} t+t=\frac{2}{3}+\frac{1}{2}+1\)

Transposing – t from RHS to LHS and – \(\frac{1}{2}\) and -1 from LHs to RHS

⇒  \(\frac{9 t-8 t+12 t}{12}\)

= \(\frac{4+3+6}{6}\)

LCM (4, 3) = 12; LCM (3, 2) = 6

⇒  \(\frac{13 t}{12}=\frac{13}{6} \)

= \(\frac{13}{6} \times \frac{12}{13}=2\)

Multiplying both sides by \(\frac{12}{3}\)

This is the required solution

∴ LHS = RHS

Question 6. \(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)

It is a linear equation since it involves linear expressions only

m- \(\frac{m}{2}+\frac{1}{2}=1-\frac{m}{3}+\frac{2}{3} \)

m- \(\frac{m}{2}+\frac{m}{3}=1+\frac{2}{3}-\frac{1}{2}\)

Transposing – \(\frac{m}{3}\) from RHS to LHS and \(\frac{1}{2}\) from LHS to RHS

⇒ \(\frac{6 m-3 m+2 m}{6}=\frac{6+4-3}{6}\)

⇒  \(\frac{5 m}{6}=\frac{7}{6} \)

m = \(\frac{7}{6} \times \frac{6}{5}=\frac{7}{5}\)

Multiplying both sides by \(\frac{6}{5}\)

This is the required solution

∴ LHS = RHS

Question 7. 3(t – 3) = 5 (2t + 1)
Solution:

3(t – 3) = 5(2t + 1)

3t – 9 = 10t + 5

3t – 10t = 5 + 9

-7t = 14

t =- \( \frac{14}{7}\) = -2

t =-2

Dividing both sides by  – 7

This is the required solution.

∴ LHS = RHS

Question 8. 15 (y -4)- 2(y – 9) + 5(y + 6) = 0
Solution:

15(y- 4)- 2(y- 9) + 5(y + 6) = 0

15y – 60 – 2y + 18 + 5y + 30 = 0

15y – 2y- 5y -60 +18 +30 = 0

18y – 12 = 0

18y = 12

Transposing- 12 from LHS to RHS

y =\(\frac{12}{18} \)

y =\(\frac{12 \div 6}{18 \div 6}\)

=\(\frac{2}{3}\)

This is the required solution.

Check:

LHS = \(15(y-4)-2(y-9)+5(y+6) \)

= \(15\left(\frac{2}{3}-4\right)-2\left(\frac{2}{3}-9\right)+5\left(\frac{2}{3}+6\right) \)

= \(15\left(\frac{2-12}{3}\right)-2\left(\frac{2-27}{3}\right)+5\left(\frac{2+18}{3}\right) \)

= \(15\left(-\frac{10}{3}\right)-2\left(-\frac{25}{3}\right)+5\left(\frac{20}{3}\right) \)

= \(-50+\frac{50}{3}+\frac{100}{3} \)

= \(\frac{-150+50+100}{3}=\frac{0}{3}=0\)

RHS = 0

∴ LHS = RHS

Question 9. 3(5z – 7) -2 (9z – 11) = 4 (8z – 13) – 17
Solution:

3(5z – 7) -2 (9z – 11) = 4 (8z – 13) – 17

15 z-21-18z+22 = 32z -52-17

-3z+1 = 32z- 69

– 3z- 32z = – 69-1

35z = – 70

z = \(\frac{-70}{-35}=\frac{70}{35}\)

= 2

This is the required solution.

∴ LHS = RHS

Question 10. 0.25 (4f – 3) = 0.05(10f – 9)
Solution:

0.25(4f- 3) = 0.05(10f- 9)

f- 0.75 = 0.5f- 0.45

f- 0.5f = – 0.45 + 0.75

Transposing 0.5f from RHS to LHS and – 0.75 from LHS to RHS

0.5f = 0.30

f = \(\frac{0.30}{0.5}\)

= 0.6

Dividing both sides by 0.5

This is the required solution

∴ LHS = RHS

NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Multiple-Choice Questions

Question 1. The standard form of a linear equation is 7. The root, of the equation 3y + 4 = by – 4 in one variable x is

1. ax + 6=0
2. ax² + bx + c = 0
3. ox³ + bx² + cx + d = 0
4. ax⁴ + bx³ + cx² + dx + e = 0.

Solution: 1. ax + 6=0

Question 2. Of the following, the linear equation in one variable x, is

1. \(\frac{4}{x}=\frac{x}{4}\)
2. \(\frac{1}{x}+\frac{1}{x-1}=1\)
3. \(\frac{x}{2}+\frac{x}{3}=\frac{1}{4}\)
4. 2 + 2x + 3 = 0

Solution: 3. \(\frac{x}{2}+\frac{x}{3}=\frac{1}{4}\)

Question 3. The degree of the equation x² – 2x + 1 = x² – 3 is

1. 1
2. 2
3. 0
4. 3

Solution: 1. 1

x² – 2x + 1  = x² – 3

x² – 2x + 1 – x² + 3= 0

– 2x + 4 = 0

-2x = – 4

2x =  4

Question 4. The root of the equation 2x+3 = 2(x- 4) is

1. 2
2. 4
3. 0
4. Does not exist

Solution:  4. Does not exist

2x +3 = 2 (x- 4)

2x + 3 = 2x – 8

3 = – 8 which is impossible.

Question 5.The value of x in \(\frac{3}{4x}\) = 7-x is

1. 4
2. 3
3. \(\frac{7}{3}\)
4. 7

Solution: 1. 4

⇒ \(\frac{3}{4} x=7-x \)

⇒ \(\frac{3}{4} x+x=7\)

⇒ \(\frac{7}{4} x=7 \)

x = 4

Question 6. The root of the equation 2y = 5(3 + y) is

1. 5
2. \(\frac{-1}{5}\)
3. -5
4. -1/5

Solution: 3. -5

2y = 5 (3 + y)

⇒   2y= 15 + 5y

⇒  by – 2y = – 15

⇒  3 y =-15

y =-\(\frac{15}{3}\)

= -5

Question 7. The root of the equation 3y + 4 = 5y – 4

1. 1
2. 3
3. 4
4. 2

Solution: 4. 2

3y+ 4 = 5y- 4

5y – by = 4 + 4

y= \(\frac{8}{2}\)

= 4

Question 8. The root of the equation 9z- 15 – 9- 3z is

1. 1
2. 2
3. 3
4. 4

Solution: 2. 2

2y = 8

92-15 = 9 – 32

⇒  92 + 32 = 9+15

12 z = 24

z = \(\frac{24}{12}\)

= 2

Question 9. The root of the equation 3x= \(\frac{20}{7}-x\) is

1. \(\frac{7}{5}\)
2. \(\frac{5}{7}\)
3. \(-\frac{7}{5}\)
4. \(-\frac{5}{7}\)

Solution: 2. \(\frac{5}{7}\)

3 x= \(\frac{20}{7}-x \)

3 x+x= \(\frac{20}{7}\)

4 x= \(\frac{20}{7}\)

x = \(\frac{20}{7 \times 4}=\frac{5}{7}\)

Question 10. The root of the equation 2y = 5 (7- y) is

1. 5
2. -5
3. 3
4. – 3.

Solution: 1. 5

2y = 5 (7 – y) => 2y = 35 – 5y 2y + 5y = 35

7y = 35

y = \(\frac{35}{7}\)

= 5

Question 11. The root of the equation (2x – 1) + (x – 1) = x + 2 is

1. 1
2. 2
3. -1
4. -2

Solution: 2. 2

(2x – 1) +(x – 1) = x + 2

2x – 2 + x-1 = x+2

2x + x -x = 2+2

2x = 4

x= \(\frac{4}{2}\)

= 2

Question 12. The root of the equation, 13x – 14 = 9x + 10 is

1. 1
2. 2
3. 3
4. 6

Solution:  4. 6

13x – 14 = 9x + 10

13x- 9x = 10+14

4x = 24

x = \(\frac{24}{4}\)

= 6

Question 13. The root of the equation,11x – 5- x + 6 = 2x + 17 is

1. 1
2. 2
3. 3
4. 4

Solution: 2. 2

11x – 5 – x + 6 = 2x+ 17

8x = 16

x = 2

NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable True-False

State whether the following are True or False statements

1. A polynomial of degree 1 is called a linear polynomial – True

2. The equations 2.v + 1 = 5 and 4* + 2 = 10 are identical – True

3. If x is an odd number, then the next even number is x + 2 –False

4. Ifa and b are positive integers, then the solution of the linear equation ax = b is always positive – True

5. Two different equations can never have the same solution – False

NCERT Solutions For Class 8 Maths Chapter 2 Linear One Equations In Variable Fill In The Blanks

1. In a linear equation, the highest power ofthe variable appearing in the equation is → One

2. The other name of a solution of an equation is → Root

3. If x is an even number, then the next even number is → x+2

4. A term of the equation can be transposed from LHS to RHS by changing its→  Sign

5.  Find the value of x for which the expressions 3x+ 1 and 2x + 9 become equal →  8

6. Is ax²+ bx + c = 0 a linear equation → No