Std 8 Mathematics · Chapter 16

16. Surface Area and Volume

Solutions to Practice Sets 16.1, 16.2 and 16.3 — volume of a cuboid, and surface area and volume of a cylinder, with every calculation shown. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

Practice Set 16.1

Formula: Volume of a cuboid = length × breadth × height

1Find the volume of a box if its length, breadth and height are 20 cm, 10.5 cm and 8 cm respectively.

Volume = 20 × 10.5 × 8 = 160 × 10.5 = 1680

Answer: 1680 cubic cm

2A cuboid shaped soap bar has volume 150 cc. Find its thickness if its length is 10 cm and breadth is 5 cm.

Volume = length × breadth × thickness
150 = 10 × 5 × thickness = 50 × thickness
∴ thickness = 150 ÷ 50 = 3

Answer: 3 cm

3How many bricks of length 25 cm, breadth 15 cm and height 10 cm are required to build a wall of length 6 m, height 2.5 m and breadth 0.5 m?

Convert the wall measures to cm: 6 m = 600 cm, 2.5 m = 250 cm, 0.5 m = 50 cm.
Volume of the wall = 600 × 250 × 50 = 75,00,000 cubic cm
Volume of one brick = 25 × 15 × 10 = 3750 cubic cm
Number of bricks = 75,00,0003750 = 2000

Answer: 2000 bricks

4For rain water harvesting, a tank of length 10 m, breadth 6 m and depth 3 m is built. What is the capacity of the tank? How many litres of water can it hold?

Capacity = 10 × 6 × 3 = 180 cubic m
1 cubic m = 1000 litres ∴ 180 cubic m = 180 × 1000 = 1,80,000 litres

Answer: 180 cubic m; it can hold 1,80,000 litres

Practice Set 16.2

Formulae (take π = 227 unless told otherwise): Curved surface area = 2πrh Total surface area = 2πr(h + r)

1In each example, the radius of the base of a cylinder and its height are given. Find the curved surface area and total surface area.

(1) r = 7 cm, h = 10 cm

CSA = 2 × 227 × 7 × 10 = 44 × 10 = 440
TSA = 2 × 227 × 7 × (10 + 7) = 44 × 17 = 748

Answer: 440 sq cm, 748 sq cm

(2) r = 1.4 cm, h = 2.1 cm

2 × 227 × 1.4 = 8.8
CSA = 8.8 × 2.1 = 18.48
TSA = 8.8 × (2.1 + 1.4) = 8.8 × 3.5 = 30.80

Answer: 18.48 sq cm, 30.80 sq cm

(3) r = 2.5 cm, h = 7 cm

CSA = 2 × 227 × 2.5 × 7 = 44 × 2.5 = 110
TSA = 2 × 227 × 2.5 × (7 + 2.5) = 1107 × 9.5 = 10457 ≈ 149.29

Answer: 110 sq cm, 149.29 sq cm

(4) r = 70 cm, h = 1.4 cm

2 × 227 × 70 = 440
CSA = 440 × 1.4 = 616
TSA = 440 × (1.4 + 70) = 440 × 71.4 = 31416

Answer: 616 sq cm, 31416 sq cm

(5) r = 4.2 cm, h = 14 cm

2 × 227 × 4.2 = 26.4
CSA = 26.4 × 14 = 369.60
TSA = 26.4 × (14 + 4.2) = 26.4 × 18.2 = 480.48

Answer: 369.60 sq cm, 480.48 sq cm

CSA = curved surface area, TSA = total surface area. Working out 2πr first saves time, because it appears in both formulae.

2Find the total surface area of a closed cylindrical drum if its diameter is 50 cm and height is 45 cm. (π = 3.14)

r = 50 ÷ 2 = 25 cm, h = 45 cm
TSA = 2πr(h + r) = 2 × 3.14 × 25 × (45 + 25) = 157 × 70 = 10990

Answer: 10,990 sq cm

3Find the area of base and radius of a cylinder if its curved surface area is 660 sq cm and height is 21 cm.

CSA = 2πrh ∴ 660 = 2 × 227 × r × 21 = 132r ∴ r = 5 cm
Area of base = πr² = 3.14 × 5 × 5 = 78.50

Answer: Radius 5 cm, area of base 78.50 sq cm

The key uses π = 3.14 for the area of the base. With π = 22/7 it is 78.57 sq cm; both are acceptable.

4Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.

r = 28 ÷ 2 = 14 cm. The container is open at the top, so the sheet covers the curved surface and one base.
Curved surface = 2 × 227 × 14 × 20 = 88 × 20 = 1760 sq cm
Base = 227 × 14 × 14 = 616 sq cm
Sheet for the container = 1760 + 616 = 2376 sq cm
Lid: a top of radius 14 cm and a rim of height 2 cm.
Lid = 616 + 2 × 227 × 14 × 2 = 616 + 176 = 792 sq cm

Answer: Container 2376 sq cm; lid approximately 792 sq cm

The lid area is approximate because a lid must be very slightly wider than the container to fit over it.

Practice Set 16.3

Formula: Volume of a cylinder = πr²h

1Find the volume of the cylinder if the height (h) and radius of the base (r) are as given below.

(1) r = 10.5 cm, h = 8 cm

V = 227 × 10.5 × 10.5 × 8 = 22 × 1.5 × 10.5 × 8 = 2772

Answer: 2772 cubic cm

(2) r = 2.5 m, h = 7 m

V = 227 × 2.5 × 2.5 × 7 = 22 × 6.25 = 137.5

Answer: 137.5 cubic m

(3) r = 4.2 cm, h = 5 cm

V = 227 × 4.2 × 4.2 × 5 = 22 × 0.6 × 4.2 × 5 = 277.2

Answer: 277.2 cubic cm

(4) r = 5.6 cm, h = 5 cm

V = 227 × 5.6 × 5.6 × 5 = 22 × 0.8 × 5.6 × 5 = 492.8

Answer: 492.8 cubic cm

In (2) the measures are in metres, so the volume is 137.5 cubic metres (the textbook key prints the unit as “c m”).

2How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?

The rod is a cylinder with r = 0.7 cm and h = 90 cm.
V = 227 × 0.7 × 0.7 × 90 = 22 × 0.1 × 0.7 × 90 = 138.6

Answer: 138.6 cubic cm of iron

3How much water will a tank hold if the interior diameter of the tank is 1.6 m and its depth is 0.7 m?

r = 0.8 m, h = 0.7 m
V = 227 × 0.8 × 0.8 × 0.7 = 22 × 0.64 × 0.1 = 1.408 cubic m
1 cubic m = 1000 litres ∴ 1.408 × 1000 = 1408 litres

Answer: 1408 litres

4Find the volume of the cylinder if the circumference of its base is 132 cm and its height is 25 cm.

2πr = 132 ∴ 2 × 227 × r = 132 ∴ r = 132 × 744 = 21 cm
V = 227 × 21 × 21 × 25 = 22 × 3 × 21 × 25 = 34650

Answer: 34,650 cubic cm

For More Information: Euler’s Formula

The completed table from the textbook. In every row F + V = E + 2.

Solid	Faces (F)	Vertices (V)	Edges (E)	F + V	E + 2
Cube	6	8	12	14	14
Cuboid	6	8	12	14	14
Triangular prism	5	6	9	11	11
Triangular pyramid	4	4	6	8	8
Pentagonal pyramid	6	6	10	12	12
Hexagonal prism	8	12	18	20	20

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