# NCERT Solutions For Class 8 Maths Chapter 11 Mensuration NCERT Solutions for Class 8 Maths Chapter 11 Mensuration- It is a branch of mathematics in which we deal with different geometric shapes both 2D and 3D shapes. It is a process of measurement. In this chapter, you will study different geometrical shapes like circle, quadrilateral, triangle, cube, cuboid, sphere, cylinder, cone and their area and volume. Also, you will learn to find area of polygon, area of trapezium, area of general quadrilaterals and special quadrilaterals. In NCERT solutions for Class 8 Maths chapter 11 Mensuration, you will find questions related to finding area and volume 3D shapes. A 3D shape can be measured by three dimensions-length, width, and depth/height. As the 3D shape is bounded by a number of planes or surfaces, you can measure the total surface area (TSA), lateral surface area (LSA), curved surface area (CSA) and volume (V). For 2-Dimensional shapes, you can measure the area(A) and perimeter(P).

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There are 4 exercises with 34 questions in this chapter. All these questions are explained in NCERT solutions for Class 8 Maths chapter 11 Mensuration in a detailed manner using diagrams. It is easy for you to visualize and understand the problem. Only using the formulas and finding the answer is not enough, you should know how these formulas are derived so you can find the area or volume of the new shape you may come across. In NCERT solutions for Class 8 Maths chapter 11 Mensuration, you will find some new ways to solve the problem. It will give you more concept clarity and help you in solving a new problem. Check NCERT Solutions from Classes 6 to 12 to learn Science and Maths. Here you will get the detailed NCERT Solutions for Class 8 by clicking on the link.

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## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.2 Let Us Recall

Question:(a) Match the following figures with their respective areas in the box.

1.

Area of above shape = =

Area of above shape =

Area of above shape =

Area of above triangle =

Area of above shape = b*h =

Question:(b) Write the perimeter of each shape.

2.

Perimeter of shape =

perimeter of shape = =

perimeter of shape =

perimeter of shape =

perimeter of this shape cannot be calculated only with height and breadth,we need slant height or angle.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration-Exercise: 11.1

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?

Given:

Perimeter of square = perimeter of rectangle

Area of square

Area of rectangle =

Hence, the area of the square is greater than the area of the rectangle.

Question: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 Rs. 55 per .

Area of plot =

Area of house =

Area of garden around house =

the rate = 55 per .

the total cost of developing a garden around the house=

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 this garden [Length of rectangle is 20 – (3.5 + 3.5) metres]. Answer:

Length of rectangle is 20 – (3.5 + 3.5) metres=13 metres

Breadth of rectangle = 7 metres

diameter of circular side = 7metres

Area of garden = area of rectangle + 2 times area of semi circular part

Area of garden

Perimetre of garden

metres

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 (If required you can split the tiles in whatever way you want to fill up the corners).

Area of parallelogram

a floor of area = =

Number of tiles = = tiles

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 , where r is the radius of the circle.

(a) circumference

(b)circumference

(c) circumference

Hence, the perimeter of (b) is highest so ant have to take a longer round in (b).

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.3 Area Of Trapezium

Question:1 Nazma’s sister also has a trapezium-shaped plot. Divide it into three parts as shown (Fig 11.4). Show that the area of a trapezium

.

Area of trapezium WXYZ = Area of triangle with base 'c' + area of rectangle + area of triangle with base'd'

Taking 'h' common, we get

Replacing

Hence proved that the area of a trapezium

Question:2 If h = 10 cm, c = 6 cm, b = 12 cm, d = 4 cm, find the values of each of its parts separetely and add to find the area WXYZ. Verify it by putting the values of h, a and b in the expression .

Area of trapezium WXYZ = Area of traingle with base'c'+area of rectangle +area of triangle with base 'd'

the area WXYZ by the expression

.

Hence,we can conclude area from given expression and calculated area is equal.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.3 Area of Trapezium

Question:1 Find the area of the following trapeziums (Fig 11.8)

(i) Area of trapezium

(ii)

Area of trapezium

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.4 Area Of A General Quadrilateral

Question:1 We know that parallelogram is also a quadrilateral. Let us also split such a quadrilateral into two triangles, find their areas and hence that of the parallelogram. Does this agree with the formula that you know already? (Fig 11.12)

This agree with the formula that we know already.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.4.1 Area Of A Special Quadrilaterals

Question:1 Find the area of these quadrilaterals (Fig 11.14).

(i) Area=

(ii) Area=

(iii)Area=

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.5 Area Of A Polygon

Question:(i) Divide the following polygons (Fig 11.17) into parts (triangles and trapezium) to find out its area.

FI is a diagonal of polygon EFGHI NQ is a diagonal of polygon MNOPQR

(i)

Area of polygon EFGHI = area of EFI + area of quadrilateral FGHI

Draw a diagonal FH

Area of polygon EFGHI = area of EFI + area of FGH + area of FHI

ii)

Area of polygon MNOPQR = area of quadrilateral NMRQ+area of quadrilateral NOPQ

Draw diagonal NP and NR .

Area of polygon MNOPQR = area of NOP+area of NPQ +area of NMR +area of NRQ

Question:(ii) Polygon ABCDE is divided into parts as shown below (Fig 11.18). Find its area if , , , and perpendiculars , , . Area of Polygon ABCDE = area of

Area of

Area of trapezium

Area of Area of

So, the area of polygon ABCDE = ....

Area of Polygon ABCDE = area of + area of trapezium BCHF + area of +area of

Question:(iii) Find the area of polygon MNOPQR (Fig 11.19) if , , , , NA, and are perpendiculars to diagonal MP.

the area of polygon MNOPQR

= area of MAN + area of trapezium ACON+area of CPO + area of MBR+area of trapezium BDQR+area of DPQ

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration-Exercise: 11.2

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.

Area of table =

Question:2 The area of a trapezium is and the length of one of the parallel sides is and its height is . Find the length of the other parallel side.

Let the length of the other parallel side be x

area of a trapezium =

cm

Hence,the length of the other parallel side is 7cm

Question:3 Length of the fence of a trapezium shaped field is . If , and , find the area of this field. Side is perpendicular to the parallel sides and .

, and ,

Length of the fence of a trapezium shaped field = =

Area of trapezium =

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.

The diagonal of a quadrilateral shaped field is 24 m

the perpendiculars are 8 m and 13 m.

the area of the field =

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

Area of rhombus =

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.

Rhombus is a type of parallelogram and area of parallelogram is product of base and height.

So,Area of rhombus = base height

Let the other diagonal be x

Area of rhombus =

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 is 4.

Area of 1 tile =

Area of 3000 tiles =

Total cost of polishing =

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. If the area of this field is 10500 and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Let the length of the side along road be x m.

Then according to question, lenght of side along river will be 2x m.

So equation becomes :

or

or

So the length of the side along the river is 2x = 140m.

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:

Area of octagonal surface = Area of rectangular surface + 2(Area of trapezium surface)

Area of recatangular surface =

Area of trapezium surface =

So total area of octagonal surface = 55 + 2(32) = 55 + 64 = 119

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?

Area of pentagonal park according to Jyoti's diagram :-

= 2(Area of trapezium) =

Area of pentagonal park according to Kavita's diagram :-

= Area of triangle + Area of square.

Question:11 Diagram of the adjacent picture frame has outer dimensions and inner dimensions . Find the area of each section of the frame, if the width of each section is same.

Area of opposite sections will be same.

So area of horizontal sections,

And area of vertical sections,

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.7.1 Cuboid

Question:1 Find the total surface area of the following cuboids (Fig 11.31):

(i) Total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)

(ii) Total surface area = 2 (h × l + b × h + b × l) = 2(lb + bh + hl)

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.7.2 Cube

Question: Find the surface area of cube A and lateral surface area of cube B (Fig 11.36).

Surface area of cube A =

=

Lateral surface area of cube B =

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.7.3 Cylinders

Question:1 Find the total surface area of the following cylinders following figure

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

(i) Area =

(ii) Area =

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration-Exercise: 11.3

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

Surface area of cuboid (a)

Surface area of cube (b)

So box (a) requires the lesser amount of material to make.

Question:2 A suitcase with measures is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width is required over such suitcases?

Surface area of suitcase .

Area of such 100 suitcase will be

So lenght of tarpaulin cloth

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

Surface area of cube

So,

or

or .

Thus side of cube is 10 cm.

Question:4 Rukhsar painted the outside of the cabinet of measure . How much surface area did she cover if she painted all except the bottom of the cabinet.

Required area = Total area - Area of bottom surface

Total area

Area of bottom surface

So required area =

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 of area is painted. How many cans of paint will she need to paint the room?

Total area painted by Daniel :

( Bottom surface is excluded.)

So, Area

No. of cans of paint required

Thus 5 cans of paint are required.

Question:6 Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

The two figures have same height.The diference between them is one is cylinder and another is cube.

lateral surface area of cylinder =

lateral surface area of cube =

Cube 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?

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

Question:8 The lateral surface area of a hollow cylinder is . It is cut along its height and formed a rectangular sheet of width . Find the perimeter of rectangular sheet?

Lenght of rectangular sheet

So perimeter of rectangular sheet = 2(l + b)

.

Thus perimeter of 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 length is 1 m.

Area in one complete revolution of roller = 2πrh.

Question:10 A company packages its milk powder in a 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:

Area of label = 2πrh

Here height is found out as = 20-2-2 = 16 cm.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.8.1 Cuboid

Question:1 Find the volume of the following cuboids

(i) Volume of cuboid is given as:

,

So, Given that Length = 8cm, Breadth = 3cm, and height = 2 cm so,

its volume will be = .

Aslo for Given Surface area of cuboid and height = 3 cm we can easily calculate the volume:

;

So, Volume =

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.8.2 Cube

Question: Find the volume of the following cubes

(a) with a side 4 cm (b) with a side 1.5 m

(a) Volume of cube having side equal to 4cm will be

or .

(b) When having side length equal to 1.5m then ,

.

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration Topic 11.8.3 Cylinder

Question:1 Find the volume of the following cylinders.

(i) The volume of a cylinder given as = .

or given radius of cylinder = 7cm and length of cylinder = 10cm.

So, we can calculate the volume of the cylinder = (Take the value of )

The volume of cylinder = .

(ii) Given for the Surface area = and height = 2m .

we have .

## NCERT Solutions for Class 8 Maths Chapter 11 Mensuration-Exercise: 11.4

Question:1(a) Given a cylindrical tank, in which situation will you find surface area and in
which situation volume.

To find how much it can hold.

(a) To find out how much the cylindrical tank can hold we will basically find out the volume of the cylinder.

Question:1(b) Given a cylindrical tank, in which situation will you find surface area and in
which situation volume.

Number of cement bags required to plaster it.

(b) if we want to find out the cement bags required to plaster it means the area to be applied, we then calculate the surface area of the bags .

Question:1(c) Given a cylindrical tank, in which situation will you find surface area and in
which situation volume.

To find the number of smaller tanks that can be filled with water from it.

(c) We have to find out the volume.

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?

Given the diameter of cylinder A = 7cm and the height = 14cm.

Also, the diameter of cylinder B = 14cm and height = 7cm.

We can easily suggest whose volume is greater without doing any calculations:

As volume is directly proportional to the square of the radius of cylinder and directly proportional to the height of the cylinder hence

B has more Volume as compared to A because B has a larger diameter.

Verifying:

Volume of A :

and Volume of B : .

Hence clearly we can see that the volume of cylinder B is greater than the volume of cylinder A.

The cylinder B has surface area of = .

and the surface area of cylinder A = .

The cylinder with greater volume also has greater surface area.

Question:3 Find the height of a cuboid whose base area is 180 cm 2 and volume is 900 cm 3?

Given that the height of a cuboid whose base area is and volume is ;

As

So, we have relation:

or, Height = 5cm.

Question:4 A cuboid is of dimensions . How many small cubes with side 6 cm can be placed in the given cuboid?

So given the dimensions of cuboid hence it's the volume is equal to =

We have to make small cubes with side 6cm which occupies the volume =

Hence we have now one cube having side length = 6cm volume = .

So, total numbers of small cubes that can be placed in the given cuboid =

Hence 450 small cubes can be placed in that cuboid.

Question:5 Find the height of the cylinder whose volume is 1.54 m 3 and diameter of the base is 140 cm ?

Given that the volume of the cylinder is and having its diameter of base = 140cm.

So, as ;

hence putting in the relation we get;

The height of the cylinder would be = .

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?

Volume of the cylinder = V =

So the quantity of milk in litres that can be stored in the tank is 49500 litres.

Question:7(i) If each edge of a cube is doubled,

how many times will its surface area increase?

The surface area of cube =

So if we double the edge l becomes 2l.

New surface area =

Thus surface area becomes 4 times.

Question:7(ii) If each edge of a cube is doubled,

how many times will its volume increase?

Volume of cube =

Since l becomes 2l, so new volume is : .

Hence volume becomes 8 times.

Question:8 Water is pouring into a cubiodal reservoir at the rate of 60 litres per minute. If the volume of reservoir is , find the number of hours it will take to fill the reservoir.

Given that the water is pouring into a cuboidal reservoir at the rate of 60 litres per minute.

Volume of the reservoir is , then

The number of hours it will take to fill the reservoir will be:

As we know .

Then = 108,000 Litres ;

Time taken to fill the tank will be:

or .

## Mensuration Class 8 Math Chapter 11-Topics

• Let us Recall

• Area of Trapezium

• Area of a General Quadrilateral

• Area of a Polygon

• Solid Shapes

• Surface Area of Cube, Cuboid, and Cylinder

• The volume of Cube, Cuboid, and Cylinder

• Volume and Capacity

### NCERT Solutions for Class 8 Maths: Chapter-Wise

 Chapter -1 Rational Numbers Chapter -2 Linear Equations in One Variable Chapter-3 Understanding Quadrilaterals Chapter-4 Practical Geometry Chapter-5 Data Handling Chapter-6 Squares and Square Roots Chapter-7 Cubes and Cube Roots Chapter-8 Comparing Quantities Chapter-9 Algebraic Expressions and Identities Chapter-10 Visualizing Solid Shapes Chapter-11 Mensuration Chapter-12 Exponents and Powers Chapter-13 Direct and Inverse Proportions Chapter-14 Factorization Chapter-15 Introduction to Graphs Chapter-16 Playing with Numbers

### NCERT Solutions for Class 8: Subject-Wise

• NCERT Solutions for Class 8 Maths
• NCERT Solutions for Class 8 Science

Some Important Formulas From NCERT Solutions For Class 8 Maths Chapter 11 Mensuration To Remember-

• Area of a trapezium

• Area of a rhombus

• The surface area of a cuboid

2(lb+bh+lh)

l- length of the cuboid

h- height of the cuboid

• The surface area of a cube

• The Surface area of a cylinder

h- height of the cylinder

• The volume of a cuboid

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l- length of the cuboid