Class 10 Mathematics · Chapter 12

12. Surface Areas and Volumes

Step-by-step solutions to Exercise 12.1 & 12.2. NCERT Mathematics Textbook for Class X (CBSE).

Exercise 12.1

12 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Edge of each cube = ∛64 = 4 cm

The resulting cuboid has l = 8 cm, b = 4 cm and h = 4 cm.

Surface area = 2(lb + bh + hl) = 2(32 + 16 + 32) = 2 × 80 = 160

Surface area of the cuboid 160 cm²

2A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

Radius r = 7 cm ; height of cylinder h = 13 – 7 = 6 cm

Inner surface area = CSA of hemisphere + CSA of cylinder = 2πr² + 2πrh = 2πr(r + h)

= 2 × 227 × 7 × (7 + 6) = 44 × 13 = 572

Inner surface area 572 cm²

3A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

r = 3.5 cm ; height of cone h = 15.5 – 3.5 = 12 cm

Slant height l = √h² + r² = √144 + 12.25 = √156.25 = 12.5 cm

TSA = CSA of cone + CSA of hemisphere = πrl + 2πr² = πr(l + 2r)

= 227 × 3.5 × (12.5 + 7) = 11 × 19.5 = 214.5

Total surface area of the toy 214.5 cm²

4A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

The hemisphere stands on a face of the cube. So, its greatest diameter = side of the cube = 7 cm.

Greatest diameter 7 cm

r = 3.5 cm

Surface area = 6 × (side)² – πr² + 2πr² = 6 × 49 + πr²

= 294 + 227 × 3.5 × 3.5 = 294 + 38.5 = 332.5

Surface area of the solid 332.5 cm²

5A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

Edge of cube = l ; radius of hemisphere = l2

Surface area = 6l² – (area of circular top) + (CSA of hemisphere)

= 6l² – π(l2)² + 2π(l2)² = 6l² + πl²4

= l²4(24 + π)

Surface area of the remaining solid l²4(24 + π)

6A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 12.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

r = 2.5 mm ; length of cylinder h = 14 – 5 = 9 mm

Surface area = 2 × CSA of hemisphere + CSA of cylinder = 4πr² + 2πrh = 2πr(2r + h)

= 2 × 227 × 2.5 × (5 + 9) = 2 × 227 × 2.5 × 14 = 220

Surface area of the capsule 220 mm²

7A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)

r = 2 m ; height of cylinder h = 2.1 m ; slant height of cone l = 2.8 m

Canvas area = CSA of cylinder + CSA of cone = 2πrh + πrl = πr(2h + l)

= 227 × 2 × (4.2 + 2.8) = 227 × 2 × 7 = 44 m²

Cost = 44 × 500 = ₹ 22000

Canvas = 44 m²; cost of canvas = ₹ 22000

8From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

r = 0.7 cm ; h = 2.4 cm ; slant height l = √2.4² + 0.7² = √5.76 + 0.49 = √6.25 = 2.5 cm

TSA = CSA of cylinder + area of base + CSA of cone = 2πrh + πr² + πrl = πr(2h + r + l)

= 227 × 0.7 × (4.8 + 0.7 + 2.5) = 2.2 × 8 = 17.6 cm²

Total surface area 18 cm² (to the nearest cm²)

9A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 12.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

r = 3.5 cm ; h = 10 cm

TSA = CSA of cylinder + 2 × CSA of hemisphere = 2πrh + 4πr² = 2πr(h + 2r)

= 2 × 227 × 3.5 × (10 + 7) = 22 × 17 = 374

Total surface area of the article 374 cm²

Exercise 12.2

1A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

r = 1 cm ; height of cone h = 1 cm

Volume = Volume of cone + Volume of hemisphere = 13πr²h + 23πr³

= 13π(1)(1) + 23π(1) = π

Volume of the solid π cm³

2Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

r = 1.5 cm ; height of each cone = 2 cm ; height of cylinder = 12 – 4 = 8 cm

Volume = Volume of cylinder + 2 × Volume of cone = πr²(8) + 2 × 13πr²(2) = πr²(8 + 43)

= 227 × 2.25 × 283 = 66

Volume of air in the model 66 cm³

3A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 12.15).

r = 1.4 cm ; length of cylindrical part h = 5 – 2.8 = 2.2 cm

Volume of one gulab jamun = πr²h + 2 × 23πr³ = πr²(h + 43r)

= 227 × 1.96 × (2.2 + 1.867) = 6.16 × 4.067 ≈ 25.05 cm³

Volume of 45 gulab jamuns = 45 × 25.05 ≈ 1127.3 cm³

Syrup = 30% of 1127.3 = 30100 × 1127.3 ≈ 338.2 cm³

Syrup in 45 gulab jamuns about 338 cm³

4A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see Fig. 12.16).

Volume of cuboid = 15 × 10 × 3.5 = 525 cm³

Volume of one depression = 13 × 227 × 0.5 × 0.5 × 1.4 = 1130 cm³ ≈ 0.367 cm³

Volume of four depressions = 4 × 1130 = 2215 ≈ 1.47 cm³

Volume of wood = 525 – 1.47 = 523.53

Volume of wood 523.53 cm³ (approx.)

5A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Volume of water in cone = 13πr²h = 13 × 227 × 25 × 8 = 440021 cm³

Water that flows out = 14 × 440021 = 110021 cm³

Volume of one lead shot = 43πr³ = 43 × 227 × 18 = 1121 cm³

Number of lead shots = 1100/2111/21 = 100

Number of lead shots 100

6A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

Volume of lower cylinder = π × 12² × 220 = 31680π cm³

Volume of upper cylinder = π × 8² × 60 = 3840π cm³

Total volume = 35520π = 35520 × 3.14 = 111532.8 cm³

Mass = 111532.8 × 8 = 892262.4 g ≈ 892.26 kg

Mass of the pole 892.26 kg (approx.)

7A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Volume of cylinder = π × 60² × 180 = 648000π cm³

Volume of cone = 13π × 60² × 120 = 144000π cm³

Volume of hemisphere = 23π × 60³ = 144000π cm³

Water left = 648000π – (144000π + 144000π) = 360000π cm³

= 360000 × 227 = 1131428.57 cm³ ≈ 1.131 m³

Volume of water left 113142847 cm³ (≈ 1.131 m³)

8A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Volume of spherical part = 43πr³ = 43 × 3.14 × (4.25)³ = 43 × 3.14 × 76.765625 ≈ 321.39 cm³

Volume of neck = πr²h = 3.14 × 1 × 1 × 8 = 25.12 cm³

Total volume = 321.39 + 25.12 = 346.51 cm³

She is not correct. The correct volume is 346.51 cm³