Class- VIII-CBSE-Mathematics Mensuration

CBSE NCERT Solutions for Class 8 Mathematics Chapter 11

Back of Chapter Questions

Exercise 11.1
1. A square and a rectangular field with measurements as given in the figure have
the same perimeter. Which field has a larger area?

Solution:
Given, side of square = 60 m and length of rectangle = 80m
We know that, perimeter of square = 4a
And perimeter of rectangle= 2(l + b)
Perimeter of square = 4 × 60 = 240m
Perimeter of rectangle = 2[l + b]
= 2(80 + b)
Given, Perimeter of Square = Perimeter of Rectangle
⇒ 240 = 2(80 + b)
⇒ b = 40m
We know that, Area of square= a2
Area of rectangle = lb
Area of Square = 60 × 60 = 3600m2
Area of rectangle = 80 × 40 = 3200m2
Hence, area of square is greater than area of rectangle.
2. Mrs. Kaushik has a square plot with the measurement as shown in the 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 m2 .

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Solution:
Given, Side of square = 25m
Length of house = 15m
Breadth of house = 20m
We know that, Area of square = a2
Area of rectangle = lb
Area of remaining portion = Area of square plot − Area of house
Now, area of Square = 25 × 25 = 625 m2
And area of house = 15 × 20 = 300 m2
Hence, area of remaining portion = 625 m2 − 300 m2 = 325 m2
Total cost = ₹ (325 × 55) = ₹ 17,875.
Hence, the total cost for developing a garden is ₹17,875.
3. The shape of a garden is rectangular in the middle and semicircular 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) meters].

Solution:
Given total length = 20 m
And diameter of the semicircle = 7 m

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7
⇒ Radius of the semicircle = m = 3.5m
2

Length of rectangular field = 20 − (3.5 + 3.5)

= 20 − 7 = 13 m
Breadth of the rectangular field = 7m
Area of rectangular field = l × b
= 13 × 7
= 91m2
1
Area of two semi circles = 2 × 2 × 𝜋 × r 2
1 22
=2× × × 3.5 × 3.5
2 7
= 38.5m2
Therefore, area of garden = 91 + 38.5 = 129.5 m2
Perimeter of two semi circular arcs = 2 × πr
22
=2× × 3.5
7
= 22 m
Hence, Perimeter of garden = 22 + 13 + 13
= 48 m

4. 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 m2 ? (If required you can split the tiles in whatever way you want to
fill up the corners).
Solution:

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Given, base of a parallelogram = 24 cm.

And height of a parallelogram= 10 cm.
Area of a parallelogram = bh
Area of one tile = 24cm × 10cm = 240 cm2
We know that 1m = 100cm
Total area = 1080 × 10000 = 10800000cm2 .
Hence,
Total area
Total tiles required = Area of one tile
1080 × 10000
=
240
= 45000 Tiles
Therefore, 45000 tiles are required to cover area of 1080 m2
5. 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 = 2πr, Where
r is the radius of the circle.

(a)

(b)

(c)
Solution:
(a) Given figure

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Given that diameter of a circle = 2.8 cm

Diameter
Radius of circle = 2

= 1.4 cm
Circumference of the semi-circle = πr
22
= × 1.4
7
= 4.4
Total distance covered =4.4cm + 2.8cm = 7.2cm

(b)
Given diameter of a semicircle = 2.8 cm
2.8
Radius of semicircle = = 1.4 cm
2

Circumference of the semi-circle = πr

22
= × 1.4
7
= 4.4
Total distance covered = 1.5 + 2.8 + 1.5 + 4.4
= 10.2 cm

(c)
Given diameter of a semicircle = 2.8 cm

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2.8
Radius of semicircle = = 1.4 cm
2

Circumference of the semi-circle = πr

22
= × 1.4
7
= 4.4
Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm
Therefore, for food piece of shape (b), ant have to take a longer round.

Exercise 11.2
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 perpendicular distance between them is 0.8 m.

Solution:
Given, Length of parallel sides of trapezium are 1 m and 1.2 m
Perpendicular distance between parallel sides = 0.8 m
1
We know, Area of Trapezium = 2 h(a + b)
1
Area = × (1 + 1.2) × 0.8
2
= 0.88 m2
Therefore, Area of top surface of table is 0.88 m2
2. The area of a 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 unknown parallel side be 𝑥.
Given, Area of a trapezium is 34 cm2 .
And length of one of parallel sides of trapezium = 10 cm

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1
We know, area of Trapezium = h(a + b)
2

1
⇒ 34 = × 4 × (10 + 𝑥)
2
⇒ 34 = 2 (10 + 𝑥)
⇒ 17 = 10 + 𝑥
⇒ 𝑥 = 17 − 10 = 7 cm
Hence, length of the other parallel side = 7 cm
3. 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:

Given, Length of the fence of a trapezium shaped field ABCD = 120 m

BC = 48 m
CD = 17 m
AD = 40 m
Now, AB= 120 − 48 − 17 − 40
= 15 m
Draw a perpendicular from D on BC
AB = DE
1
Area of Trapezium = 2 h(a + b)
1
= × 15 × (40 + 48)
2
= 660 m2

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Class- VIII-CBSE-Mathematics Mensuration

Therefore, area of field = 660 m2

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:

Given, Length of base = 24 m.

Height of upper triangle = 13 m
Height of lower triangle = 8 m
1
We know, area of triangle = 2 × height × base
1
Hence, Area of upper triangle = 2 × 13 × 24 = 156 m2
1
And area of lower triangle = 2 × 8 × 24 = 96 m2

Therefore, Area of field = 156 + 96 = 252 m2

5. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.
Solution:
1
We Area of rhombus = 2 × Diagonal 1 × Diagonal 2

Given, Length of diagonal 1 = 7.5 cm

Length of diagonal 2 = 12 cm
1
Area of rhombus = × 7.5 × 12
2

= 45 cm2
Therefore, area of given rhombus = 45 cm2

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Class- VIII-CBSE-Mathematics Mensuration

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:
1
We know that area of rhombus = 2 × product of its diagonals

Since, a rhombus is also a parallelogram

∴ Area of rhombus = 5cm × 4.8cm = 24cm2
Let the length of another diagonal be 𝑥.
1
Now, 24cm2 = 2 (8 cm × 𝑥)
24 × 2
⇒𝑥= cm = 6 cm
8
Thus, the length of other diagonal is 6 cm.
7. 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 m2 is ₹ 4.
Solution:
Given length of diagonals are 45 cm and 30 cm.
1
We know that area of rhombus = 2 × product of its diagonals.
1
Area of rhombus (Each tile) = × 45 × 30 = 675 cm2
2
Now, area of 3000 tiles = 3000 × 675 cm2
= 2025000 cm2
= 202.5 m2
Given that the cost of polishing is 4m2 .
Cost of polishing 202.5 m2 area = ₹ (4 × 202.5)
= ₹ 810
8. 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 m2 and the
perpendicular distance between the two parallel sides is 100 m, find the length of
the side along the river.

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Solution:
Given, Perpendicular distance (Height) = 100 m .
Area of field = 10500 m2 .
Let the length of side alongside the road be 𝑥. Given that side along the river is
twice the side along the road.
Hence, length of other side = 2𝑥.
1
We know that, Area of Trapezium = 2 × sum of parallel sides ×
perpendicular distance.
1
⇒ 10500 = × (𝑥 + 2𝑥) × 100
2
2
⇒ 3𝑥 = 10500 ×
100
⇒ 3𝑥 = 210
⇒ 𝑥 = 70
Therefore, length of side along the road = 70 m.
and length of side along the river = 140 m.
Hence, Length of side along the river = 140 m.
9. 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.

Solution:

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From the given figure, Area of trapezium ABCH = Area of trapezium DEFG
1
We know that, Area of Trapezium = 2 × sum of parallel sides ×
perpendicular distance.
1
Hence, area of Trapezium ABCH = [2 (4)(11 + 5)] m2

1
= ( × 4 × 16) m2
2
= 32m2
Area of rectangle HGDC = length × breadth = 11 × 5 = 55 m2
Therefore, Area of octagon = Area of trapezium ABCH + Area of trapezium
DEFG + Area of rectangle HGDC
Area of Octagon = 32 m2 + 32 m2 + 55 m2
= 119 m2
Hence, area of the octagon is 119 m2
10. There is a pentagonal shaped park as shown in the figure. For finding its area
Jyoti and Kavita 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?

Solution:

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Class- VIII-CBSE-Mathematics Mensuration

(a) From the given figure, Area of pentagon = Area of trapezium AEDF +
Area of trapezium ABCF
Area of pentagon = 2 (Area of trapezium ABCF)

1
We know that, Area of Trapezium = 2 × sum of parallel sides ×
perpendicular distance.
1 15
Area of pentagon = [2 × 2 (15 + 30) ( 2 )] m2

= 337.5 m2
(b) From the given diagram, Area of rectangle BCDE = 15 × 15 = 225 m2

1
Area of triangle ABE = 2 × Base × Height
1
= × 15 × 15
2
= 112.5 m2
Area of pentagon ABCDE = Area of rectangle BCDE +Area of triangle
ABE
= 225 + 112.5 = 337.5 m2 .
11. 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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Solution:

Given that width of each section is same.

∴ IB = BJ = CK = CL = DM = DN = AO = AP
And IL = IB + BC + 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 = 4 cm
Area of section BEFC = Area of section DGHA
We know that, Area of Trapezium
1
= × sum of parallel sides × perpendicular distance.
2

1
Area of section BEFC = [ (20 + 28)(4)] cm2 = 96 cm2
2
Area of section ABEH = Area of section CDGF
1
= [ (16 + 24)(4)]
2
= 80cm2

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EXERCISE 11.3
1. There are two cuboidal boxes as shown in the adjoining figure. Which box
requires the lesser amount of material to make?

Solution:
We know, Total surface area of the cuboid =2 × (L × H + B × H + L × B)
And total surface area of the cube = 6 L2
Given, Dimensions of cuboid are 60 cm × 40 cm × 50 cm
And dimensions of cube are 50 cm × 50 cm × 50 cm
Total surface area of cuboid = [2{(60)(40) + (40)(50) + (50)(60)}] cm2
= (2 × 7400)cm2
= 14800 cm2
Total surface area of cube = 6 (50 cm)2
= 15000 cm2
Thus, the cuboidal box will require lesser amount of material than cube.
2. A suitcase with measures 80 cm × 48 cm × 24 cm is to be covered with a
tarpaulin cloth. How m any meters of tarpaulin of width 96 cm is required
to cover 100 such suitcases?
Solution:
Given dimensions of suitcase are 80 cm × 48 cm × 24 cm
We know that total surface area of the cuboid = 2 (L × H + B × H + L × B)
Total surface area of suitcase = 2[(80) (48) + (48) (24) + (24) (80)]
= 2[3840 + 1152 + 1920]
= 13824 cm2
Total surface area of 100 suitcases = (13824 × 100)cm2
= 1382400 cm2

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Area of tarpaulin = Length × Breadth

⇒ 1382400 cm2 = Length × 96 cm
1382400
⇒ Length = cm
96
= 14400 cm
= 144 m
Therefore, 144 m of tarpaulin is required to cover 100 suitcases.
3. Find the side of a cube whose surface area is 600 cm2 .
Solution:
Given surface area of cube = 600 cm2
We know that Surface area of cube = 6 (Side)2
Let side of cube be L cm.
⇒ 600 cm2 = 6L2
⇒ L2 = 100 cm2
⇒ L = 10 cm
Therefore, the side of the cube is 10 cm.
4. 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?

Solution:
Given, Length of the cabinet = 2m
Breadth of the cabinet = 1m
Height of the cabinet = 1.5 m
Area of the cabinet that was painted = 2H (L + B) + L × B
= [2 × 1.5 × (2 + 1) + (2)(1)]m2
= (9 + 2)m2
= 11m2

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Class- VIII-CBSE-Mathematics Mensuration

Therefore, area of the cabinet that was painted is 11 m2 .

5. 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 m2 of
area is painted. How many cans of paint will she need to paint the room?
Solution:
Given, Length of the cabinet= 15 m
Breadth of the cabinet = 10 m
Height of the cabinet = 7 m
Area of the cabinet that was painted = 2H (L + B) + L × B
= [2(7)(15 + 10) + 15 × 10] m2
= 500m2
Given 100 m2 can be painted from 1 can.
500
Number of cans required = 100

=5
Hence, number of cans required is 5.
6. Describe how the two figures at the right are alike and how they are different.
Which box has larger lateral surface area?

Solution:
Given, side length of cube = 7 cm
Radius of cylinder = 3.5 cm
Height of cylinder = 7 cm.
We know, Lateral Surface area of the cube = 4L2
= 4 (7 )2
= 196 cm2
Lateral surface area of the cylinder = 2πrh

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22 7
= (2 × × × 7) cm2
7 2
= 154 cm2
Hence, cube has the larger surface area than cylinder.
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:

Given, Radius of cylinder = 7m

Height of cylinder = 3m
We know surface area of cylinder = 2πr (r + h)
22
Surface area = [2 × × 7(7 + 3)] m2
7

= 440m2
Thus, 440m2 sheet of metal is required.
8. The lateral surface area of a hollow cylinder is 4224 cm2 . It is cut along its height
and formed a rectangular sheet of width 33 cm. Find the perimeter of rectangular
sheet?
Solution:

Given
Lateral surface area of cylinder = 4224 cm2
Clearly, Area of cylinder = Area of rectangular sheet
⇒ 4224 cm2 = 33 cm × Length

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4224cm2
⇒ Length =
33cm
= 128cm
Now, perimeter of rectangular sheet= 2[L + B]
= [2 (128 + 33)] cm
= (2 × 161) cm
= 322 cm
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 length is
1 m.
Solution:

Given, Road roller takes 750 revolutions.

Length of road roller = 1 m
Diameter of a road roller = 84 cm
In 1 revolution, area of the road covered = 2πrh
22
=2× × 42cm × 1m
7
22 42 264 2
=2× × m × 1m = m
7 100 100
264
In 750 revolutions, area covered on the road = 750 × 100 m2 = 1980m2

10. A company packages its milk powder in cylindrical container whose base has a
diameter of 14 cmand 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.

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Solution:

Given, Height of cylinder = 20 cm.

Diameter of cylinder = 14 cm.
Height of the label = 20 cm − 2 cm − 2 cm
= 16 cm
14
Radius of the label = cm
2

= 7 cm
Area of the label = 2π (Radius) (Height)
22
= (2 × × 7 × 16) cm2
7
= 704 cm2
Hence, area of the label = 704 cm2 .

Exercise 11.4
1. 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

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Solution:
In given situation, we will find its volume to know how much it can hold.
In given situation, we will find surface area to find number of cement bags
required to plaster it.
In given situation, we will find volume to find number of smaller tanks that can be
filled with water from it.
2. 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?

Solution:
We know that Volume of cylinder= πr 2 h
If measures of r and h are same, then the cylinder with greater radius will have
greater area.
7
Given, Radius of cylinder A = 2 cm = 3.5 cm
14
Radius of cylinder B = cm = 7cm
2

Height of cylinder A = 14 cm
Height of cylinder B = 7 cm.
Therefore, volume of cylinder B is greater.
Volume of cylinder A = πr 2 h
22
= × 3.5 × 3.5 × 14
7
= 539 cm3
Volume of cylinder B = πr 2 h
22
= ×7×7×7
7

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= 1078 cm3
Surface area of cylinder A = 2πr(r + h)
22
=2× × 3.5(3.5 + 14)
7
= 385 cm2
Surface area of cylinder B = 2πr(r + h)
22
=2× × 7(14)
7
= 616 cm2
The cylinder with higher volume has higher surface area.
3. Find the height of a cuboid whose base area is 180 cm2 and volume is 900 cm3?
Solution:
Given, Base area of a cuboid = 180 cm2
Volume of a cuboid = 900 cm3
We know that, Volume = base area × height
Volume
⇒ Height =
Base area
900cm3
= 180cm2

= 5 cm
4. 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?
Solution:
Given that the dimensions of cuboid are 60 cm × 54 cm × 30 cm
And side length of cube = 6 cm
We know that volume of cuboid = lbh
Volume of cuboid = 60 cm × 54 cm × 30 cm
= 97200 cm3
Side of the cube = 6 cm
Volume of cube = a3
Volume of the cube = 216 cm3

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Volume of cuboid
Number of cubes = Volume of 1 cube

97200cm3
=
216cm3
= 450
Hence, 450 cubes can be placed in cuboid of given dimensions.
5. Find the height of the cylinder whose volume is 1.54 m3 and diameter of the base
is 140 cm?
Solution:
Given that the diameter of the base = 140 cm
Volume of the cylinder = 1.54 m3
140
Also, radius (r) of the base = cm = 70cm
2

70
= m = 0.7m
100
We know, Volume of cylinder = πr 2 h
22
⇒ 1.54m3 = × 0.7m × 0.7m × h
7
⇒ h = 1m
Hence, Height of cylinder = 1m
6. 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?
Solution:
Given, Radius of cylinder = 1.5 m
Length of cylinder = 7 m
Volume of cylinder = πr 2 h
22
=( × 1.5 × 1.5 × 7) m3
7
= 49.5 m3
1 m3 = 1000 L
Required quantity = (49.5 × 1000)L = 49500L
Hence, 49500 L milk can be stored in tank.
7. If each edge of a cube is doubled,

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How many times will its surface area increase?

How many times will its volume increase?
Solution:
Let initially the edge of the cube be L.
We know surface area of cube = 6L2
If each edge of the cube is doubled, then it becomes 2L.
New surface area = 6(2L)2 = 4 × 6L2
Clearly, the surface area will be increased by 4 times
Initial volume of the cube = L3
When each edge of the cube is doubled, it becomes 2L.
New volume = (2L)3 = 8L3 = 8 𝑥 L3
Clearly, the volume of the cube will be increased by 8 times.
8. Water is pouring into a cuboidal reservoir at the rate of 60 liters per minute. If the
volume of reservoir is 108 m3 , find the number of hours it will take to fill the
reservoir.
Solution:
Given, Volume of cuboidal reservoir = 108 m3
= (108 × 1000)L
= 108000 L
Rate at which Water pouring = (60 × 60)L
= 3600 L per hour
108000
Required number of hours = 3600

= 30 hours .
Therefore, 30 hours are required to fill the reservoir.

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