NCERT Solutions for Class 8 Maths Chapter 9 Mensuration

NCERT
SOLUTIONS
CLASS - 8TH

aglase .co

Book : Mathematics Ncert Solutions | Chapter-11 Maths

Class : 8th
Subject : Maths
Chapter : 11
Chapter Name : Mensuration

Exercise 11.1

Q1 A square and a rectangular field with measurements as given in the figure have the same perimeter.
Which field has a larger area?

Answer. Perimeter of square = 4 (Side of the square) = 4 (60 m) = 240 m.
Perimeter of rectangle = 2 (Length + Breadth)
= 2 (80 m + Breadth)
= 160 m + 2 x Breadth
It is given that the perimeter of the square and the rectangle are the same.
160 m + 2 x Breadth = 240 m
80
Breadth of the rectangle = ( ) m = 40 m
2

Area of square = ( Side ) = (60m) = 3600m
2 2 2

Area of rectangle = Length x Breadth = (80 x 40) m = 3200 m
2 2

Thus, the area of the square field is larger than the area of the rectangular field.

Page : 171 , Block Name : Exercise 11.1

Q2 Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to
construct a house in the middle of the plot. A garden is developed around the house. Find the total cost
of developing a garden around the house at the rate of ₹ 55 per m 2

Answer. Area of the square plot = (25m) 2
= 625m
2

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Area of the house = (15 m) x (20 m) = 300 m 2

Area of the remaining portion = Area of square plot - Area of the house
= 625 m - 300 m = 325 m .
2 2 2

The cost of developing the garden around the house is Rs 55 per m . 2

Total cost of developing the garden of area 325 m = Rs (55 x 325) 2

= Rs 17,875

Page : 171 , Block Name : Exercise 11.1

Q3 The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the
diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].

Answer. Length of the rectangle = [20 - (3.5 + 3.5)] metres = 13 m
Circumference of 1 semi-circular parts = nr = ( × 3.5) = 11 m 22

7

Circumference of both semi-circular parts = ( 2 x 11 ) m = 22 m

Perimeter of the garden = AB + Length of both semi-circular regions BC and DA + CD
= 13m + 22m + 13m = 48 m
Area of the garden = Area of rectangle + 2 x Area of two semi-circular regions
1 22 2 2
= [(13 × 7) + 2 × × × (3.5) ] m
2 7

2
= (91 + 38.5)m
2
= 129.5m

Page : 171 , Block Name : Exercise 11.1

Q4 A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is
10 cm. How many such tiles are required to cover a floor of area 1080 m ? (If required you can split the
2

tiles in whatever way you want to fill up the corners).

Answer. Area of parallelogram Base x Height
Hence, area of one tile = 24 cm x 10 cm = 240 cm 2

Required number of tiles =
Area of the floor

Area of each tile
2
2 (1080×10000)cm
=
1080m

240cm
2
=
240cm
2
tiles
(∵ 1m = 100cm) = 45000

Thus, 45000 tiles are required to cover a floor of area 1080 m . 2

Page : 171 , Block Name : Exercise 11.1

Q5 An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-
piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained
by using the expression c = 2nr, where r is the radius of the circle.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. (a) Radius (r) of semi-circular part = ( 2.8

2
) cm = 1.4cm

Perimeter of the given figure = 2.8 cm + nr
= 2.8 cm + ( cm
22
× 1.4)
7

= 2.8 cm = 4.4 cm
= 7.2 cm
(b) Radius (r) of semi-circular part = ( 2.8

2
) cm = 1.4cm

Perimeter of the given figure = 1.5 + 2.8 cm + 1.5 cm + n (1.4 cm)
22
= = 5.8cm + (1.4cm) 7

= = 4cm + 22
× (1.4cm)
7

= 4 cm + 4.4 cm
= 8.4 cm
(c) Radius (r) of semi-circular part = (
2.8
) cm = 1.4cm
2

Perimeter of the figure(c) = 2 cm + nr + 2 cm
= 4cm +
22
× (1.4cm)
7

= 4 cm + 4.4 cm
= 8.4 cm
Thus, the ant will have to take a longer round for the food-piece (b), because the perimeter of the figure
given in alternative (b) is the greatest among all.

Page : 171 , Block Name : Exercise 11.1

Exercise 11.2

Q1 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 perpendicular distance between them is 0.8 m.

Answer. Area of trapezium = 1

2
(Sum of parallel sides) x (Distances between parallel sides)
1
2 2
= [ (1 + 1.2)(0.8)] m = 0.88m
2

Page : 177 , Block Name : Exercise 11.2

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q2 The area of a trapezium is 34 cm and the length of one of the parallel sides is 10 cm and its height is
2

4 cm. Find the length of the other parallel side.

Answer. It is given that, area of trapezium = 34cm and height = 4 cm
2

Let the length of one parallel side be a. We know that,
Area of trapezium = (Sum of parallel sides) x (Distances between parallel sides)
1

2
1
2
34cm = (10cm + a) × (4cm)
2

34 cm = 2(10 cm + a)
17 cm = 10 cm + a
a = 17 cm - 10 cm = 7 cm
Thus, the length of the other parallel side is 7 cm.

Page : 178 , Block Name : Exercise 11.2

Q3 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.

Answer. Length of the fence of trapezium ABCD = AB + BC + CD + DA
120 m = AB + 48 m + 17 m + 40m
AB = 120 m - 105 m = 15 m
Area of the field ABCD = = (AD + BC) × AB
1

2

1
2
= [ (40 + 48) × (15)] m
2

1
2
= ( × 88 × 15) m
2

2
= 660m

Page : 178 , Block Name : Exercise 11.2

Q4 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.

Answer. It is given that,
Length of the diagonal, d = 24 m
Length of the perpendiculars, h and h , from the opposite vertices to the diagonal are h = 8m and h
1 2 1 2

= 13 m
Area of the quadrilateral = = d (h + h )1

2
1 2

1
= (24m) × (13m + 8cm)
2
1
= (24m)(21m)
2
2
= 252m

Thus, the area of the field is 252m . 2

Page : 178 , Block Name : Exercise 11.2

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q5 The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Answer. Area of rhombus = (Product of its diagonals)
1

2

Therefore, area of the given rhombus
1
= × 7.5cm × 12cm
2
2
= 45cm

Page : 178 , Block Name : Exercise 11.2

Q6 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.

Answer. Let the length of the other diagonal of the rhombus be x.
A rhombus is a special case of a parallelogram.
The area of a parallelogram is given by the product of its base and height.
Thus, area of the given rhombus = Base x Height = 6 cm x 4 cm = 24cm 2

Also, area of rhombus = ( Product of its diagonals )
1

2

2 1
⇒ 24cm = (8cm × x)
2

24×2
⇒ x = ( ) cm = 6cm
8

Thus, the length of the other diagonal of the rhombus is 6 cm.

Page : 178 , Block Name : Exercise 11.2

Q7 The floor of a building consists of 3000 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 m is ₹ 4.
2

Answer. Area of rhombus = 1

2
(Product of diagonals)
Area of each tile
1 2
= ( × 45 × 30) cm
2

2
= 675cm

Area of 3000 tiles = (675 x 3000) cm = 2025000 cm = 202.5 m
2 2 2

The cost of polishing is Rs 4 per m . 2

Cost of polishing 202.5 m area = Rs (4 x 202.5) = Rs 810
2

Thus, the cost of polishing the floor is Rs 810.

Page : 178 , Block Name : Exercise 11.2

Q8 Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side
along the road. If the area of this field is 10500 m and the perpendicular distance between the two
2

parallel sides is 100 m, find the length of the side along the river.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. Let the length of the field along the road be I m. Hence, the length of the field along the river
will be 2/ m.
Area of trapezium = (Sum of parallel sides) (Distance between the parallel sides)
1

2

2 1
⇒ 10500m = (l + 2l) × (100m)
2

2×10500
3l = ( ) m = 210m
100

l = 70 m
Thus, length of the field along the river = (2 x 70) m - 140 m

Page : 178 , Block Name : Exercise 11.2

Q9 Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the
area of the octagonal surface.

Answer.
Side of regular octagon = 5 cm
Area of trapezium ABCH = Area of trapezium DEFG
Area of trapezium ABCH = [ 1

2
2
(4)(11 + 5)] m = (
1

2
2
× 4 × 16) m = 32m
2

Area of rectangle HGFC = 11 x 5 = 55 m 2

Area of octagon = Area of trapezium ABCH + Area of trapezium DEEG + Area of rectangle HGDC
2 2 2 2
= 32m + 32m + 55m = 119m

Page : 178 , Block Name : Exercise 11.2

Q10 There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

divided it in two different ways.

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

Answer. Jyoti's way of finding area is as follows.

Area of pentagon = 2 (Area of trapezium ABCF)
1 15 2
= [2 × (15 + 30) ( )] m
2 2

2
337.5m

Kavita's way of finding area is as follows.

Area of pentagon = Area of ΔABE + Area of square BCDE
1 2 2
= [ × 15 × (30 − 15) + (15) ] m
2

1 2
= ( × 15 × 15 + 225) m
2

2
= (112.5 + 225)m
2
= 337.5m

Page : 178 , Block Name : Exercise 11.2

Q11 Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions
16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer.
Given that, the width of each section is same. Therefore,
IB = BJ = CK = CL = DM = DN = AO = AP
IL = IB + 20 + CL
28 = IB + 20 + CL
IB + CL = 28 cm - 20 cm = 8 cm
IB = CL = 4 cm
Hence, IB = BJ = CK = CL = DM = DN = AO = AP = 4cm
Area of section BEFC = Area of section DGHA
1 2 2
= [ (20 + 28)(4)] cm = 96cm
2

Area of section ABEH = Area of section CDGF

Page : 178 , Block Name : Exercise 11.2

Q1 There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount
of material to make?

Answer. We know that,
Total surface area of the cuboid = 2 (lh + bh + lb)
Total surface area of the cube — 6(l) 2

Total surface area of cuboid (a) = [2{(60) (40) + (40) (50) + (50) (60)}] cm
2

= [2(2400 + 2000 + 3000)] cm 2

= (2 x 7400) cm 2

= 14800 cm 2

Total surface area of cube (b) = 6 (50cm) = 15000 cm .
2 2

Thus, the cuboidal box (a) will require lesser amount of material.

Exercise 11.3

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 186 , Block Name : Exercise 11.3

Q2 A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a tarpaulin cloth. How many
metres of tarpaulin of width 96 cm is required to cover 100 such suitcases?

Answer. Total surface area of suitcase = 2[(80) (48) + (48) (24) + (24) (80) ]
= 2[3840 + 1152 + 1920]
= 13824 cm 2

Total surface area of 100 suitcases = (13824 x 100) cm = 1382400 cm
2 2

Required tarpaulin = Length x Breadth
1382400 cm = Length x 96 cm
2

1382400
Length = ( ) cm = 14400 cm = 144 cm
96

Thus, 144 m of tarpaulin is required to cover 100 suitcases.

Page : 186 , Block Name : Exercise 11.3

Q3 Find the side of a cube whose surface area is 600 cm ? 2

Answer. Given that, surface area of cube = 600 cm 2

Let the length of each side of cube be l.
Surface area of cube = 6( Side ) 2

2 2
600cm = 6l
2 2
l = 100cm

L = 10 cm.

Page : 186 , Block Name : Exercise 11.3

Q4 Rukhsar painted the outside of the cabinet of measure 1 m × 2 m × 1.5 m. How much surface area
did she cover if she painted all except the bottom of the cabinet.

Answer. Length (l) of the cabinet = 2 m
Breadth (b) of the cabinet = 1 m
Height (h) of the cabinet = 1.5 m
Area of the cabinet that was painted = 2h (l+b) + lb
= [2 x 1.5 x (2 + 1) + (2) (1) ] m 2

= [3(3) + 2] m 2

=(9+2)m 2

= 11 m 2

Page : 186 , Block Name : Exercise 11.3

Q5 Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of 15 m, 10
m and 7 m respectively. From each can of paint 100 m of area is painted. How many cans of paint will
2

she need to paint the room?

Answer. Given that,
Length (l) — 15 m, breadth (b) = 10 m, height (h) = 7 m
Area of the hall to be painted = Area of the wall + Area of the ceiling
= 2h (1 + b) + 1b
= [2(7) (15 + 10) + 15 x 10]m 2

= [14(25) + 150] m 2

= 500 m 2

It is given that 100 m area can be painted from each can.
2

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Number of cans required to paint an area of 500 m . 2

= 500/100 = 5
Hence, 5 cans are required to paint the walls and the ceiling of the cuboidal hall.

Page : 186 , Block Name : Exercise 11.3

Q6 Describe how the two figures at the right are alike and how they are different. Which box has larger
lateral surface area?

Answer. Similarity between both the figures is that both have the same heights.
The difference between the two figures is that one is a cylinder and the other is a cube.
Lateral surface area of the cube = 4l = 4(7cm) = 196cm
2 2 2

22 7
Lateral surface area of the cylinder = 2rh = = (2 × 7
×
2
× 7)
2 2
cm =154cm

Hence, the cube has larger lateral surface area.

Page : 186 , Block Name : Exercise 11.3

Q7 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?

Answer. Total surface area of cylinder = 2r (r+h)
22
= [2 × × 7(7 + 3)]
7 2
m

= 440 m 2

Thus, 440 m sheet of metal is required.
2

Page : 186 , Block Name : Exercise 11.3

Q8 The lateral surface area of a hollow cylinder is 4224 cm . It is cut along its height and formed a
2

rectangular sheet of width 33 cm. Find the perimeter of rectangular sheet?

Answer. A hollow cylinder is cut along its height to form a rectangular sheet.
Area of cylinder = Area of rectangular sheet
4224cm = 33 cm x Length
2

2

Length = 4224cm
= 128cm
33cm

Thus, the length of the rectangular sheet is 128 cm.
Perimeter of the rectangular sheet — 2 (Length + Width)
= [2 (128 + 33)] cm
= (2 x 161) cm
= 322 cm

Page : 186 , Block Name : Exercise 11.3

Q9 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 length is 1 m.

Answer. In one revolution, the roller will cover an area equal to its lateral surface area.
Thus, in 1 revolution, area of the road covered = 2rh
22
= 2 × × 42cm × 1m
7
22 42
= 2 × × m × 1m
7 100
264 2
= m
100

In 750 revolutions, area of the road covered

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

264
2
= (750 × )m
100

2
= 1980m

Page : 186 , Block Name : Exercise 11.3

Q10 A company packages its milk powder in cylindrical container whose base has a diameter of 14 cm
and height 20 cm. 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.

Answer. Height of the label = 20 cm - 2 cm - 2 cm = 16 cm
Radius of the label = = ( ) cm = 7cm
14

2

Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm.
Area of the label = 2 (Radius) (Height)
22 2 2
= (2 × × 7 × 16) cm = 704cm
7

Page : 186 , Block Name : Exercise 11.3

Exercise 11.4

Q1 Given a cylindrical tank, in which situation will you find surface area and in which situation volume.
(a) To find how much it can hold.
(b) Number of cement bags required to plaster it.
(c) To find the number of smaller tanks that can be filled with water from it.

Answer. (a) In this situation, we will find the volume.
(b) In this situation, we will find the surface area.
(c) In this situation, we will find the volume.

Page : 191 , Block Name : Exercise 11.4

Q2 Diameter of cylinder A is 7 cm, and the height is 14 cm. Diameter of cylinder B is 14 cm and height
is 7 cm. Without doing any calculations can you suggest whose volume is greater? Verify it by finding
the volume of both the cylinders. Check whether the cylinder with greater volume also has greater
surface area?

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Answer. The heights and diameters of these cylinders A and 3 are interchanged.
We know that,
Volume of cylinder = πr h 2

If measures of r and h are same, then the cylinder with greater radius will have greater area.
Radius of cylinder A = 7

2 cm

Radius of cylinder B = = ( 14

2
) = 7cm
cm

As the radius of cylinder 3 is greater, therefore, the volume of cylinder 3 will be greater.
Let us verify it by calculating the volume of both the cylinders.
Volume of cylinder A = πr h 2

22 7 7 3
= ( × × × 14) cm
7 2 2

3
= 539cm

Volume of cylinder B = πr h 2

22 3
= ( × 7 × 7 × 7) cm
7

3
= 1078cm

Volume of cylinder B is greater.
Surface area of cylinder A = = 2πr(r + h)
22 7 7 2
= [2 × × ( + 14)] cm
7 2 2

7+28
2
= [22 × ( )] cm
2

35 2
= (22 × ) cm
2

2
= 385cm

Surface area of cylinder B = 2πr(r + h)
22 2
= [2 × × 7 × (7 + 7)] cm
7

2
= (44 × 14)cm
2
= 616cm

Thus, the surface area of cylinder a is also greater than the surface area of cylinder A.

Page : 191 , Block Name : Exercise 11.4

Q3 Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3 ?

Answer. Base area of the cuboid = Length x Breadth = 180 cm 2

Volume of cuboid = Length x Breadth x Height
900 cm = 180 cm x Height
3 2

Height (900/180)cm = 5 cm
Thus, the height of the cuboid is 5 cm.

Page : 191 , Block Name : Exercise 11.4

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Q4 A cuboid is of dimensions 60 cm × 54 cm × 30 cm. How many small cubes with side 6 cm can be
placed in the given cuboid?

Answer. Volume of cuboid = 60 cm x 54 cm x 30 cm = 97200 cm 3

Side of the cube = 6 cm
Volume of the cube = (6) cm = 216cm 3 3 3

Required number of cubes = Volume of the cuboid

Volume of the cube
97200
= = 450
216

Thus, 450 cubes can be placed in the given cuboid.

Page : 191 , Block Name : Exercise 11.4

Q5 Find the height of the cylinder whose volume is 1.54 m and diameter of the base is 140 cm ?
3

Answer. Diameter of the base = 140 cm
Radius (r) of the base = ( ) cm = 70cm = 140

2
70

100
m

Volume of cylinder = πr h 2

3 22 70 70
1.54m = × m × m × h
7 100 100

1.54×100
h = ( ) m = 1m
22×7

Thus, the height of the cylinder is 1 m.

Page : 191 , Block Name : Exercise 11.4

Q6 A milk tank is in the form of cylinder whose radius is 1.5 m and length is 7 m. Find the quantity of
milk in litres that can be stored in the tank?

Answer. Radius of cylinder = 1.5 m
Length of cylinder = 7 m
Volume of cylinder = πr h 2

22 3
= ( × 1.5 × 1.5 × 7) m
7

3
= 49.5m
3
1m = 1000L

Required quantity = (49.5 x 1000) L 49500 L
Therefore, 49500 L of milk can be stored in the tank.

Page : 191 , Block Name : Exercise 11.4

Q7 If each edge of a cube is doubled,
(i) how many times will its surface area increase?
(ii) how many times will its volume increase?

Answer. (i) Let initially the edge of the cube be l.
Initial surface area = 6l 2

If each edge of the cube is doubled, then it becomes 2l.
New surface area = 6(2t) = 24l = 4 × 6l 2 2 2

Clearly, the surface area will be increased by 4 times.
(ii) Initial volume of the cube = l 3

When each edge of the cube is doubled, it becomes 2l.
New volume (2l) = 8l = 8 × l 3 3 3

Clearly, the volume of the cube will be increased by 8 times.

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Book : Mathematics Ncert Solutions | Chapter-11 Maths

Page : 191 , Block Name : Exercise 11.4

Q8 Water is pouring into a cuboidal reservoir at the rate of 60 litres per minute. If the volume of
reservoir is 108 m , find the number of hours it will take to fill the reservoir.
3

Answer. Volume of cuboidal reservoir = 108 m = (108 x 1000) L = 108000 L
3

It is given that water is being poured at the rate of 60 L per minute.
That is, (60 x 60) L = 3600 L per hour
Required number of hours = 108000

3600
= 30 hours

Thus, it will take 30 hours to fill the reservoir.

Page : 192 , Block Name : Exercise 11.4

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