MAK LABS — Std 9 Maths Part 2 Chapter 9 Surface Area and Volume — Practice Sets 9.1 to 9.3 & Problem Set 9 Solutions

1Length, breadth and height of a cuboid shape box of medicine is 20 cm, 12 cm and 10 cm respectively. Find the surface area of vertical faces and total surface area of this box.

Given: l = 20 cm, b = 12 cm, h = 10 cm

Area of vertical faces = 2(l + b) × h = 2(20 + 12) × 10 = 2 × 32 × 10 = 640 cm²

Total surface area = 2(lb + bh + lh) = 2(20 × 12 + 12 × 10 + 20 × 10) = 2(240 + 120 + 200) = 2 × 560 = 1120 cm²

∴ Vertical faces = 640 cm², total surface area = 1120 cm²

2Total surface area of a box of cuboid shape is 500 sq. unit. Its breadth and height is 6 unit and 5 unit respectively. What is the length of that box?

Given: Total surface area = 500, b = 6, h = 5

2(lb + bh + lh) = 500

2(6l + 30 + 5l) = 500

11l + 30 = 250

11l = 220 ∴ l = 20

∴ Length of the box = 20 units

3Side of a cube is 4.5 cm. Find the surface area of all vertical faces and total surface area of the cube.

Given: side l = 4.5 cm

Area of vertical faces = 4l² = 4 × (4.5)² = 4 × 20.25 = 81 cm²

Total surface area = 6l² = 6 × 20.25 = 121.5 cm²

∴ Vertical faces = 81 cm², total surface area = 121.5 cm²

4Total surface area of a cube is 5400 sq. cm. Find the surface area of all vertical faces of the cube.

6l² = 5400 ∴ l² = 900

Area of vertical faces = 4l² = 4 × 900 = 3600 cm²

∴ Area of all vertical faces = 3600 cm²

5Volume of a cuboid is 34.50 cubic metre. Breadth and height of the cuboid is 1.5 m and 1.15 m respectively. Find its length.

Volume = l × b × h

34.50 = l × 1.5 × 1.15 = 1.725 l

l = 34.51.725 = 20

∴ Length of the cuboid = 20 m

6What will be the volume of a cube having length of edge 7.5 cm?

Volume = l³ = (7.5)³ = 7.5 × 7.5 × 7.5 = 421.875 cm³

∴ Volume = 421.875 cm³

7Radius of base of a cylinder is 20 cm and its height is 13 cm, find its curved surface area and total surface area. (π = 3.14)

Given: r = 20 cm, h = 13 cm

Curved surface area = 2πrh = 2 × 3.14 × 20 × 13 = 1632.8 cm²

Total surface area = 2πr(r + h) = 2 × 3.14 × 20 × (20 + 13) = 125.6 × 33 = 4144.8 cm²

∴ Curved surface area = 1632.8 cm², total surface area = 4144.8 cm²

8Curved surface area of a cylinder is 1980 cm² and radius of its base is 15 cm. Find the height of the cylinder. (π = 22/7)

2πrh = 1980

2 × 227 × 15 × h = 1980

6607 × h = 1980

h = 1980 × 7660 = 3 × 7 = 21

∴ Height of the cylinder = 21 cm

1Perpendicular height of a cone is 12 cm and its slant height is 13 cm. Find the radius of the base of the cone.

l² = r² + h²

13² = r² + 12²

r² = 169 − 144 = 25 ∴ r = 5

∴ Radius of the base = 5 cm

2Find the volume of a cone, if its total surface area is 7128 sq. cm and radius of base is 28 cm. (π = 22/7)

Total surface area = πr(l + r)

7128 = 227 × 28 × (l + 28) = 88(l + 28)

l + 28 = 712888 = 81 ∴ l = 53 cm

h² = l² − r² = 53² − 28² = 2809 − 784 = 2025 ∴ h = 45 cm

Volume = 13 πr²h = 13 × 227 × 28 × 28 × 45 = 22 × 4 × 28 × 15 = 36960 cm³

∴ Volume of the cone = 36960 cm³

3Curved surface area of a cone is 251.2 cm² and radius of its base is 8 cm. Find its slant height and perpendicular height. (π = 3.14)

πrl = 251.2

3.14 × 8 × l = 251.2

25.12 l = 251.2 ∴ l = 10 cm

h² = l² − r² = 100 − 64 = 36 ∴ h = 6 cm

∴ Slant height = 10 cm, perpendicular height = 6 cm

4What will be the cost of making a closed cone of tin sheet having radius of base 6 m and slant height 8 m if the rate of making is Rs. 10 per sq. m?

A closed cone needs its total surface area of tin.

Total surface area = πr(l + r) = 227 × 6 × (8 + 6) = 227 × 84 = 264 m²

Cost = 264 × 10 = Rs. 2640

∴ Cost = Rs. 2640

No value of π is given; π = 22/7 is used. With π = 3.14 the area is 263.76 m² and the cost Rs. 2637.60.

5Volume of a cone is 6280 cubic cm and base radius of the cone is 30 cm. Find its perpendicular height. (π = 3.14)

Volume = 13 πr²h

6280 = 13 × 3.14 × 30 × 30 × h = 942 h

h = 6280942 = 203 ≈ 6.67

∴ Perpendicular height = 20/3 cm ≈ 6.67 cm

6Surface area of a cone is 188.4 sq. cm and its slant height is 10 cm. Find its perpendicular height. (π = 3.14)

Taking the given surface area as the curved surface area:

πrl = 188.4 ∴ 3.14 × r × 10 = 188.4 ∴ 31.4 r = 188.4 ∴ r = 6 cm

h² = l² − r² = 100 − 36 = 64 ∴ h = 8 cm

∴ Perpendicular height = 8 cm

If 188.4 cm² is taken as the total surface area, r(10 + r) = 60, which has no whole-number solution. The intended value is the curved surface area.

7Volume of a cone is 1212 cm³ and its height is 24 cm. Find the surface area of the cone. (π = 22/7)

With 1212 cm³ the radius is not exact (r² ≈ 48.2). The intended volume is 1232 cm³, which gives r = 7 cm. The solution below uses 1232 cm³.

13 πr²h = 1232

13 × 227 × r² × 24 = 1232 ∴ 1767 r² = 1232

r² = 1232 × 7176 = 49 ∴ r = 7 cm

l² = r² + h² = 49 + 576 = 625 ∴ l = 25 cm

Total surface area = πr(l + r) = 227 × 7 × (25 + 7) = 22 × 32 = 704 cm²

∴ Surface area of the cone = 704 cm²

8The curved surface area of a cone is 2200 sq. cm and its slant height is 50 cm. Find the total surface area of cone. (π = 22/7)

πrl = 2200 ∴ 227 × r × 50 = 2200 ∴ r = 2200 × 722 × 50 = 14 cm

Area of base = πr² = 227 × 14 × 14 = 616 cm²

Total surface area = 2200 + 616 = 2816 cm²

∴ Total surface area = 2816 cm²

9There are 25 persons in a tent which is conical in shape. Every person needs an area of 4 sq. m of the ground inside the tent. If height of the tent is 18 m, find the volume of the tent.

Area of the base = 25 × 4 = 100 m²

Volume = 13 × area of base × height = 13 × 100 × 18 = 600 m³

∴ Volume of the tent = 600 m³

10In a field, dry fodder for the cattle is heaped in a conical shape. The height of the cone is 2.1 m and diameter of base is 7.2 m. Find the volume of the fodder. If it is to be covered by polythene in rainy season then how much minimum polythene sheet is needed?

Given: h = 2.1 m, r = 7.2 ÷ 2 = 3.6 m; π = 22/7

Volume = 13 πr²h = 13 × 227 × 3.6 × 3.6 × 2.1 = 22 × 0.1 × 12.96 = 28.512 ≈ 28.51 m³

l² = r² + h² = 12.96 + 4.41 = 17.37 ∴ l ≈ 4.17 m

Polythene covers only the curved surface (the base is on the ground).

Curved surface area = πrl = 227 × 3.6 × 4.17 ≈ 47.18 m²

∴ Volume ≈ 28.51 m³; polythene needed ≈ 47.18 m²

1Find the surface areas and volumes of spheres of the following radii. (i) 4 cm (ii) 9 cm (iii) 3.5 cm (π = 3.14)

(i) r = 4 cm
Surface area = 4 × 3.14 × 16 = 200.96 cm²
Volume = ⁴⁄₃ × 3.14 × 64 = 267.95 cm³ (approx.)

(ii) r = 9 cm
Surface area = 4 × 3.14 × 81 = 1017.36 cm²
Volume = ⁴⁄₃ × 3.14 × 729 = 3052.08 cm³

(iii) r = 3.5 cm
Surface area = 4 × 3.14 × 12.25 = 153.86 cm²
Volume = ⁴⁄₃ × 3.14 × 42.875 = 179.50 cm³ (approx.)

200.96 cm², 267.95 cm³
1017.36 cm², 3052.08 cm³
153.86 cm², 179.50 cm³

2If the radius of a solid hemisphere is 5 cm, then find its curved surface area and total surface area. (π = 3.14)

Curved surface area = 2πr² = 2 × 3.14 × 25 = 157 cm²

Total surface area = 3πr² = 3 × 3.14 × 25 = 235.5 cm²

∴ Curved surface area = 157 cm², total surface area = 235.5 cm²

3If the surface area of a sphere is 2826 cm² then find its volume. (π = 3.14)

4πr² = 2826 ∴ 12.56 r² = 2826 ∴ r² = 225 ∴ r = 15 cm

Volume = ⁴⁄₃ × 3.14 × 15 × 15 × 15 = ⁴⁄₃ × 3.14 × 3375 = 14130 cm³

∴ Volume = 14130 cm³

4Find the surface area of a sphere, if its volume is 38808 cubic cm. (π = 22/7)

⁴⁄₃ × 227 × r³ = 38808

r³ = 38808 × 2188 = 9261 ∴ r = 21 cm

Surface area = 4πr² = 4 × 227 × 21 × 21 = 5544 cm²

∴ Surface area = 5544 cm²

5Volume of a hemisphere is 18000π cubic cm. Find its diameter.

⅔πr³ = 18000π

r³ = 18000 × 3 ÷ 2 = 27000 ∴ r = 30 cm

Diameter = 2r = 60 cm

∴ Diameter = 60 cm

1If diameter of a road roller is 0.9 m and its length is 1.4 m, how much area of a field will be pressed in its 500 rotations?

In one rotation the roller presses an area equal to its curved surface area.

Curved surface area = 2πrh = πdh = 227 × 0.9 × 1.4 = 3.96 m²

Area in 500 rotations = 500 × 3.96 = 1980 m²

∴ Area pressed = 1980 m²

2To make an open fish tank, a glass sheet of 2 mm gauge is used. The outer length, breadth and height of the tank are 60.4 cm, 40.4 cm and 40.2 cm respectively. How much maximum volume of water will be contained in it?

Thickness of glass = 2 mm = 0.2 cm.

Inner length = 60.4 − 2 × 0.2 = 60 cm; inner breadth = 40.4 − 2 × 0.2 = 40 cm

The tank is open at the top, so glass is only at the bottom: inner height = 40.2 − 0.2 = 40 cm

Volume of water = 60 × 40 × 40 = 96000 cm³ = 96 litres

∴ Maximum volume of water = 96000 cm³ (96 litres)

3If the ratio of radius of base and height of a cone is 5 : 12 and its volume is 314 cubic metre. Find its perpendicular height and slant height. (π = 3.14)

Let r = 5k and h = 12k.

Volume = 13 × 3.14 × 25k² × 12k = 314k³

314k³ = 314 ∴ k³ = 1 ∴ k = 1

h = 12 m, r = 5 m

l² = r² + h² = 25 + 144 = 169 ∴ l = 13 m

∴ Perpendicular height = 12 m, slant height = 13 m

4Find the radius of a sphere if its volume is 904.32 cubic cm. (π = 3.14)

⁴⁄₃ × 3.14 × r³ = 904.32

r³ = 904.32 × 34 × 3.14 = 2712.9612.56 = 216 ∴ r = 6 cm

∴ Radius = 6 cm

5Total surface area of a cube is 864 sq. cm. Find its volume.

6l² = 864 ∴ l² = 144 ∴ l = 12 cm

Volume = l³ = 12³ = 1728 cm³

∴ Volume = 1728 cm³

6Find the volume of a sphere, if its surface area is 154 sq. cm.

4πr² = 154 ∴ 4 × 227 × r² = 154 ∴ r² = 154 × 788 = 12.25 ∴ r = 3.5 cm

Volume = ⁴⁄₃ × 227 × 3.5 × 3.5 × 3.5 = 5393 ≈ 179.67 cm³

∴ Volume ≈ 179.67 cm³

No value of π is given; π = 22/7 is used because it gives an exact radius.

7Total surface area of a cone is 616 sq. cm. If the slant height of the cone is three times the radius of its base, find its slant height.

Let the radius be r; then l = 3r.

Total surface area = πr(l + r) = πr(3r + r) = 4πr²

4 × 227 × r² = 616 ∴ r² = 616 × 788 = 49 ∴ r = 7 cm

l = 3 × 7 = 21 cm

∴ Slant height = 21 cm

8The inner diameter of a well is 4.20 metre and its depth is 10 metre. Find the inner surface area of the well. Find the cost of plastering it from inside at the rate Rs. 52 per sq. m.

The inner surface of the well is the curved surface of a cylinder with r = 2.1 m, h = 10 m.

Curved surface area = 2πrh = 2 × 227 × 2.1 × 10 = 132 m²

Cost = 132 × 52 = Rs. 6864

∴ Inner surface area = 132 m², cost = Rs. 6864

9The length of a road roller is 2.1 m and its diameter is 1.4 m. For levelling a ground 500 rotations of the road roller were required. How much area of ground was levelled by the road roller? Find the cost of levelling at the rate of Rs. 7 per sq. m.

Area levelled in one rotation = curved surface area = 2πrh = 2 × 227 × 0.7 × 2.1 = 9.24 m²

Area in 500 rotations = 500 × 9.24 = 4620 m²

Cost = 4620 × 7 = Rs. 32340

∴ Area levelled = 4620 m², cost = Rs. 32340

9. Surface Area and Volume

Practice Set 9.1

Practice Set 9.2

Practice Set 9.3

Problem Set 9