ICSE Class 7 Maths Chapter 32 Volume and Surface Area

Chapter 32: Volume And Surface Area For A Regular Shaped Body

32.1 Volume And Surface Area

1. Volume

Volume of a body is the space occupied by it.

Infact:

1. For a solid body:

Its volume = The volume of the material in the body

= Amount of space occupied by the body

2. For a hollow body:

(i) Its external volume = Amount of space occupied by the body.

(ii) Its internal volume = Amount of space inside the body.

(iii) Volume of material used to make the body = External volume of the body - Its internal volume

2. Surface Area

Surface area of a body is the sum of the areas of all its faces.

32.2 Cuboid

In everyday life, we come across many objects like, a match box, a brick, etc. Each of these objects is a cuboid in shape.

A body which has six faces, all rectangles, is called a cuboid or a rectangular solid.

As shown in the adjoining figure:

(i) The cuboid has 12 edges, namely:

AB, BC, CD, DA, PQ, QR, RS, SP, PA, QB, SD and RC such that the opposite edges are equal in length.

\[\Rightarrow \quad PA = QB = SD = RC = \text{length } (l) \text{ of the cuboid,}\]

\[AB = DC = PQ = SR = \text{breadth } (b) \text{ of the cuboid,}\]

and, \[AD = BC = PS = QR = \text{height } (h) \text{ of the cuboid.}\]

(ii) The cuboid has 6 faces, namely:

ABCD, PQRS, ABQP, DCRS, ADSP and BCRQ such that the opposite faces are equal in area.

\[\Rightarrow \quad \text{area of face ABQP} = \text{area of face DCRS} = l \times b,\]

\[\text{area of face ABCD} = \text{area of face PQRS} = b \times h,\]

and, \[\text{area of face ADSP} = \text{area of face BCRQ} = h \times l\]

Total surface area of the cuboid

= sum of the areas of its six faces

= area of (ABQP + DCRS + ABCD + PQRS + ADSP + BCRQ)

\[= l \times b + l \times b + b \times h + b \times h + h \times l + h \times l\]

\[= 2(l \times b + b \times h + h \times l).\]

Volume of the cuboid = Its length × breadth × height

\[= l \times b \times h.\]

Example 1

Find the total surface area and the volume of a cuboid, whose:

(i) length = 25 cm, breadth = 20 cm and height = 12 cm.

(ii) length = 1.8 m, breadth = 1.2 m and height = 60 cm.

Solution

(i) Given: \[l = 25 \text{ cm}, b = 20 \text{ cm and } h = 12 \text{ cm.}\]

\[\therefore \text{Total surface area} = 2(l \times b + b \times h + h \times l)\]

\[= 2(25 \times 20 + 20 \times 12 + 12 \times 25) \text{ cm}^2\]

\[= \mathbf{2080 \text{ cm}^2}\]

And, \[\text{volume} = l \times b \times h\]

\[= 25 \text{ cm} \times 20 \text{ cm} \times 12 \text{ cm}\]

\[= \mathbf{6000 \text{ cm}^3}\]

(ii) Given: \[l = 1.8 \text{ m}, b = 1.2 \text{ m and } h = 60 \text{ cm} = \frac{60}{100} \text{ m} = 0.6 \text{ m}\]

\[\therefore \text{Total surface area} = 2(l \times b + b \times h + h \times l)\]

\[= 2(1.8 \times 1.2 + 1.2 \times 0.6 + 0.6 \times 1.8) \text{ m}^2\]

\[= \mathbf{7.92 \text{ m}^2}\]

And, \[\text{volume} = l \times b \times h\]

\[= 1.8 \text{ m} \times 1.2 \text{ m} \times 0.6 \text{ m}\]

\[= \mathbf{1.296 \text{ m}^3}\]

Teacher's Note

Understanding cuboid dimensions helps us calculate storage space in boxes, containers, and furniture used in everyday life.

32.3 Cube

A cuboid whose length, breadth and height are equal to each other is called a cube.

For a cube: length = breadth = height and each of its six faces is a square.

Since, each face of a cube is a square: area of each face = (side)^2 = l^2

\[\therefore \text{Total surface area of the cube}\]

= Sum of the areas of its six faces

\[= 6 \times (\text{side})^2 = 6 \times l^2\]

Volume of the cube = Its length × breadth × height

\[= l \times l \times l \quad \lbrack \because \text{In a cube, } l = b = h \rbrack\]

\[= l^3\]

Example 2

Find the total surface area and the volume of a cube with each side:

(i) 16 cm

(ii) 1.2 m

Solution

(i) Total surface area of the cube = \[6 \times (\text{side})^2\]

\[= 6 \times (16)^2 \text{ cm}^2 = \mathbf{1536 \text{ cm}^2}\]

And, \[\text{volume} = (\text{side})^3\]

\[= (16)^3 \text{ cm}^3 = \mathbf{4096 \text{ cm}^3}\]

(ii) Total surface area of the cube = \[6 l^2\]

\[= 6 \times (1.2)^2 \text{ m}^2 = \mathbf{8.64 \text{ m}^2}\]

And, \[\text{its volume} = l^3 = (1.2)^3 \text{ m}^3 = \mathbf{1.728 \text{ m}^3}\]

Teacher's Note

Cubes are used in packaging, dice games, and storage systems where equal dimensions simplify calculations and design.

32.4 Area Of The Four Walls Of A Room

A room is in the form of a hollow cuboid with 4 walls, each rectangular in shape. The opposite walls being equal in area.

In one pair of opposite walls, the area of each wall = \[l \times h\]

and in the other pair of opposite walls, the area of each wall = \[b \times h.\]

\[\therefore \text{Total area of 4 walls of a room (including door and windows)}\]

\[= 2 \times l \times h + 2 \times b \times h\]

\[= 2(l + b) \times h\]

Example 3

(i) Find the area of the four walls of a room whose dimensions are 8 m, 4.5 m and 3 m. Find the cost of distempering the walls at the rate of ₹ 20 per m².

(ii) Also, find the cost of white washing its roof at the rate of ₹ 15 per m².

Solution

Given: Length (l) = 8 m, breadth (b) = 4.5 m and height (h) = 3 m.

(i) \[\text{Area of four walls} = 2(l + b) \times h\]

\[= 2(8 + 4.5) \times 3 \text{ m}^2\]

\[= 2 \times 12.5 \times 3 \text{ m}^2 = \mathbf{75 \text{ m}^2}\]

Cost of distempering 1 m² = ₹ 20

\[\therefore \text{Cost of distempering 75 m}^2 = 75 \times \text{₹} 20 = \mathbf{\text{₹ 1,500}}\]

(ii) \[\text{Area of the roof} = l \times b\]

\[= 8 \text{ m} \times 4.5 \text{ m} = 36 \text{ m}^2\]

Cost of white washing 1 m² = ₹ 15

\[\therefore \text{Cost of white washing 36 m}^2 = 36 \times \text{₹} 15 = \mathbf{\text{₹ 540}}\]

Teacher's Note

Calculating wall areas is essential for estimating painting, wallpaper, or plastering costs in home renovation projects.

Example 4

The volume of a tank, which is cuboid in shape, is 6.4 m³. The dimensions of its base are 2 m × 1.6 m. Find the depth of the tank.

Solution

Given: Length of the tank, \[l = 2 \text{ m}\], its breadth, \[b = 1.6 \text{ m and its volume} = 6.4 \text{ m}^3.\]

To find: depth (h) of the tank.

Since, the volume of the tank = \[l \times b \times h\]

\[\Rightarrow \quad 6.4 = 2 \times 1.6 \times h\]

\[\Rightarrow \quad h = \frac{6.4}{2 \times 1.6} \text{ m} = 2 \text{ m}\]

\[\therefore \text{Depth of the tank} = \mathbf{2 \text{ m}}\]

Teacher's Note

Water tank depth calculations are crucial for ensuring adequate water storage capacity in household and agricultural applications.

Example 5

Bricks of size 20 cm, 10 cm and 8 cm are used to build a wall whose length, breadth and height are 12 m, 30 cm and 3.5 m respectively. How many bricks will be required?

Solution

\[\therefore \quad \text{Volume of one brick} = 20 \text{ cm} \times 10 \text{ cm} \times 8 \text{ cm} = 1600 \text{ cm}^3\]

And, \[\text{volume of the wall} = 12 \text{ m} \times 30 \text{ cm} \times 3.5 \text{ m}\]

\[= 1200 \text{ cm} \times 30 \text{ cm} \times 350 \text{ cm}\]

\[= 12600000 \text{ cm}^3\]

\[\text{Number of bricks required} = \frac{\text{Volume of the wall}}{\text{Volume of each brick}}\]

\[= \frac{12600000}{1600} = \mathbf{7875}\]

Teacher's Note

Calculating the number of bricks needed helps in material planning and cost estimation for construction projects.

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