NCERT Solutions for Class 8 Math Chapter 11 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among class 8 students for Math Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the NCERT Book of class 8 Math Chapter 11 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s NCERT Solutions. All NCERT Solutions for class 8 Math are prepared by experts and are 100% accurate.

#### Page No 171:

#### Question 1:

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 × Breadth

It is given that the perimeter of the square and the rectangle are the same.

160 m + 2 × Breadth = 240 m

Breadth of the rectangle = = 40 m

Area of square = (Side)^{2} = (60 m)^{2} = 3600 m^{2}

Area of rectangle = Length × Breadth = (80 × 40) m^{2} = 3200 m^{2}

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

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##### Video Solution for Mensuration (Page: 171 , Q.No.: 1)

NCERT Solution for Class 8 math - Mensuration 171 , Question 1

#### Page No 171:

#### Question 2:

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 Rs 55 per m^{2}.

#### Answer:

Area of the square plot = (25 m)^{2} = 625 m^{2}

Area of the house = (15 m) × (20 m) =300 m^{2}

Area of the remaining portion = Area of square plot − Area of the house

= 625 m^{2} − 300 m^{2} = 325 m^{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^{2} = Rs (55 × 325)

= Rs 17,875

#### Page No 171:

#### Question 3:

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 the 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 part = π*r *

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

Perimeter of the garden = AB + Length of both semi-circular regions BC and

DA + CD

= 13 m + 22 m + 13 m = 48 m**

Area of the garden = Area of rectangle + 2 × Area of semi-circular region

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##### Video Solution for Mensuration (Page: 171 , Q.No.: 3)

NCERT Solution for Class 8 math - Mensuration 171 , Question 3

#### Page No 171:

#### Question 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 m^{2}?
(If required you can split the tiles in whatever way you want to fill
up the corners).

#### Answer:

Area of parallelogram = Base × Height

Hence, area of one tile
= 24 cm × 10 cm = 240 cm^{2}

Required number of
tiles =**
**

= 45000 tiles

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

#### Page No 171:

#### Question 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* = 2*πr*, where *r* is the radius of the circle.

#### Answer:

(a)Radius (*r*) of semi-circular part =

Perimeter of the given figure = 2.8 cm + π*r*

(b)Radius (*r*) of semi-circular part =

Perimeter of the given figure = 1.5 cm + 2.8 cm + 1.5 cm +π (1.4 cm)

(c)Radius (*r*) of semi-circular part =

Perimeter of the figure(c) = 2 cm + π*r* + 2 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.

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##### Video Solution for Mensuration (Page: 171 , Q.No.: 5)

NCERT Solution for Class 8 math - Mensuration 171 , Question 5

#### Page No 177:

#### 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 perpendicular distance between them is 0.8 m.

E

#### Answer:

Area of trapezium = (Sum of parallel sides) × (Distances between parallel sides)

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##### Video Solution for Mensuration (Page: 177 , Q.No.: 1)

NCERT Solution for Class 8 math - Mensuration 177 , Question 1

#### Page No 178:

#### Question 2:

The
area of a trapezium is 34 cm^{2}
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.

#### Answer:

It is given that,area of trapezium = 34 cm^{2} and height = 4 cm

Let the length of one
parallel side be *a*. We know that,

Area of trapezium = (Sum of parallel sides) × (Distances between parallel sides)

Thus, the length of the other parallel side is 7 cm.

#### Page No 178:

#### Question 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.

#### Answer:

Length of the fence of trapezium ABCD = AB + BC + CD + DA

120 m = AB + 48 m + 17 m + 40 m

AB = 120 m − 105 m = 15 m

Area of the field ABCD

#### Page No 178:

#### 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.

#### Answer:

It is given that,

Length of the diagonal,
*d *= 24 m

Length of the
perpendiculars, *h*_{1}
and *h*_{2}, from the opposite vertices to the diagonal
are *h*_{1} = 8 m and *h*_{2} = 13 m

Area of the quadrilateral

Thus, the area of the
field is 252 m^{2}.

#### Page No 178:

#### Question 5:

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

#### Answer:

Area of rhombus = (Product of its diagonals)

Therefore, area of the given rhombus

=

= 45 cm^{2}

#### Page No 178:

#### 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.

#### 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 × Height = 5 cm × 4.8 cm = 24 cm^{2}

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

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

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##### Video Solution for Mensuration (Page: 178 , Q.No.: 6)

NCERT Solution for Class 8 math - Mensuration 178 , Question 6

#### Page No 178:

#### Question 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 m^{2}
is Rs 4.

#### Answer:

Area of rhombus = (Product of its diagonals)

Area of each tile

= 675 cm^{2}

Area of 3000 tiles = (675 × 3000) cm^{2} = 2025000
cm^{2} = 202.5 m^{2}

The cost of polishing is Rs 4 per m^{2}.

Cost of polishing 202.5 m^{2} area = Rs (4 × 202.5)
= Rs 810

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

#### Page No 178:

#### Question 8:

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. It the area of this field is 10500 m^{2} and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

#### Answer:

Let the length of the field along the road be *l *m. Hence, the length of the field along the river will be 2*l* m.

Area of trapezium = (Sum of parallel sides) (Distance between the parallel sides)

Thus, length of the field along the river = (2 × 70) m = 140 m

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##### Video Solution for Mensuration (Page: 178 , Q.No.: 8)

NCERT Solution for Class 8 math - Mensuration 178 , Question 8

#### Page No 178:

#### Question 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.

#### Answer:

Side of regular octagon = 5 cm

Area of trapezium ABCH = Area of trapezium DEFG

Area of rectangle HGDC
= 11 × 5 = 55 m^{2}

Area of octagon = Area of trapezium ABCH + Area of trapezium DEFG

+ Area of rectangle HGDC

=
32 m^{2} + 32 m^{2} + 55 m^{2} = 119 m^{2}

#### Page No 178:

#### Question 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?

#### Answer:

Jyoti’s way of finding area is as follows.

Area of pentagon = 2 (Area of trapezium ABCF)

=
337.5 m^{2}

Kavita’s way of finding area is as follows.

Area of pentagon = Area of ΔABE + Area of square BCDE

#### Page No 178:

#### Question 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.

#### Answer:

Given that, the width of each section is same. Therefore,

IB = BJ = CK = CL = DM = DN = AO = AP

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

Area of section ABEH = Area of section CDGF

$\Rightarrow $Area of section ABEH = Area of section CDGF

= $\left[\frac{1}{2}\left(16+24\right)\left(4\right)\right]=80{\mathrm{cm}}^{2}$

#### Page No 186:

#### Question 1:

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 × 7400) cm^{2}

= 14800 cm^{2}

Total surface area of cube (b) = 6 (50 cm)^{2} = 15000 cm^{2}

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

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##### Video Solution for Mensuration (Page: 186 , Q.No.: 1)

NCERT Solution for Class 8 math - Mensuration 186 , Question 1

#### Page No 186:

#### Question 2:

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 × 100) cm^{2} = 1382400 cm^{2}

Required tarpaulin = Length × Breadth

1382400 cm^{2}
= Length × 96 cm

Length = = 14400 cm = 144 m

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

#### Page No 186:

#### Question 3:

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}

600 cm^{2} =
6*l*^{2}

*l*^{2}=
100 cm^{2}

*l* = 10 cm

Thus, the side of the cube is 10 cm.

#### Page No 186:

#### Question 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?

#### 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 = 2*h* (*l* + *b*) + *lb*

=
[2 × 1.5 × (2 + 1) + (2) (1)] m^{2}

=
[3(3) + 2] m^{2}

=
(9 + 2) m^{2}

=
11 m^{2}

#### Page No 186:

#### Question 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 m^{2} of area is painted. How many cans
of paint will 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

= 2*h* (*l* + *b*) + *lb*

= [2(7) (15 + 10) + 15 ×10] m^{2}

= [14(25) + 150] m^{2}

= 500 m^{2}

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

Number of cans required to paint an area of 500 m^{2}

=

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

#### Page No 186:

#### Question 6:

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 = 4*l*^{2} = 4 (7 cm)^{2} = 196 cm^{2}

Lateral surface area of
the cylinder = 2π*rh*
cm^{2} = 154 cm^{2}

Hence, the cube has larger lateral surface area.

#### Page No 186:

#### 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?

#### Answer:

Total surface area of
cylinder = 2π*r* (*r* + *h*)

m^{2}

= 440
m^{2}

Thus, 440 m^{2}
sheet of metal is required.

#### Page No 186:

#### Question 8:

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

4224 cm^{2} =
33 cm × Length

Thus, the length of the rectangular sheet is 128 cm.

Perimeter of the rectangular sheet = 2 (Length + Width)

= [2 (128 + 33)] cm

= (2 × 161) cm

= 322 cm

#### Page No 186:

#### 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 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 = 2π*rh*

In 750 revolutions, area of the road covered

=

= 1980 m^{2}

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##### Video Solution for Mensuration (Page: 186 , Q.No.: 9)

NCERT Solution for Class 8 math - Mensuration 186 , Question 9

#### Page No 186:

#### Question 10:

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

Label is in the form of a cylinder having its radius and height as 7 cm and 16 cm.

Area of the label = 2π (Radius) (Height)

#### Page No 191:

#### Question 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.

#### 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 No 191:

#### Question 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?

#### Answer:

The heights and diameters of these cylinders A and B are interchanged.

We know that,

Volume of cylinder

If measures of *r*
and *h* are same, then the cylinder with greater radius will
have greater area.

Radius of cylinder A = cm

Radius of cylinder B = cm = 7 cm

As the radius of cylinder B is greater, therefore, the volume of cylinder B will be greater.

Let us verify it by calculating the volume of both the cylinders.

Volume of cylinder A

Volume of cylinder B

Volume of cylinder B is greater.

Surface area of cylinder A

Surface area of cylinder B

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

#### Page No 191:

#### Question 3:

Find
the height of a cuboid whose base area is 180 cm^{2}
and volume is 900 cm^{3}?

#### Answer:

Base area of the cuboid
= Length × Breadth = 180 cm^{2}

Volume of cuboid = Length × Breadth × Height

900 cm^{3} =
180 cm^{2} × Height

Thus, the height of the cuboid is 5 cm.

#### Page No 191:

#### Question 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?

#### Answer:

Volume of cuboid = 60
cm × 54 cm ×
30 cm = 97200 cm^{3}

Side of the cube = 6 cm

Volume of the cube =
(6)^{3} cm^{3} = 216 cm^{3}

Required number of cubes =

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

#### Page No 191:

#### Question 5:

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

#### Answer:

Diameter of the base = 140 cm

Radius (*r*) of
the base

Volume of cylinder

Thus, the height of the cylinder is 1 m.

#### Page No 191:

#### Question 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?

#### Answer:

Radius of cylinder = 1.5 m

Length of cylinder = 7 m

Volume of cylinder

1m^{3} = 1000 L

Required quantity = (49.5 × 1000) L = 49500 L

Therefore, 49500 L of milk can be stored in the tank.

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##### Video Solution for Mensuration (Page: 191 , Q.No.: 6)

NCERT Solution for Class 8 math - Mensuration 191 , Question 6

#### Page No 191:

#### Question 7:

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 = 6*l*^{2}

If
each edge of the cube is doubled, then it becomes 2*l*.

New
surface area = 6(2*l*)^{2} = 24*l*^{2} = 4
× 6*l*^{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 2*l*.

New
volume = (2*l*)^{3} = 8*l*^{3 }= 8 ×
*l*^{3}

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

#### Page No 192:

#### Question 8:

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

#### Answer:

Volume of cuboidal reservoir = 108 m^{3} = (108 × 1000) L = 108000 L

It is given that water is being poured at the rate of 60 L per minute.

That is, (60 × 60) L = 3600 L per hour

Required number of hours = 30 hours

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

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##### Video Solution for Mensuration (Page: 192 , Q.No.: 8)

NCERT Solution for Class 8 math - Mensuration 192 , Question 8

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