Surface Areas and Volumes - Class 9 NCERT Solutions-10.1, 10.2, 10.3, 10.4

Surface Areas and Volumes

EXERCISE 11.1

Assume π = ²²⁄₇, unless stated otherwise.

Question 1: Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Solution:
Curved Surface Area = π × r × l = ²²⁄₇ × 5.25 × 10 = 165 cm²

Question 2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:
Radius (r) = 12 m
Curved Surface Area = π × r × l = ²²⁄₇ × 12 × 21 = 5544 m²
Base Area = π × r² = ²²⁄₇ × 144 = 2016 m²
Total Surface Area = Curved Surface Area + Base Area = 5544 + 2016 = 7560 m²

Question 3: Curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.

Solution:
(i) Radius (r) = Curved Surface Area / (π × l) = 308 / (²²⁄₇ × 14) = 7 cm
(ii) Base Area = π × r² = ²²⁄₇ × 49 = 154 cm²
Total Surface Area = Curved Surface Area + Base Area = 308 + 154 = 462 cm²

Question 4: A conical tent is 10 m high and the radius of its base is 24 m. Find (i) slant height of the tent. (ii) Cost of the canvas required to make the tent, if the cost of 1 m² canvas is ₹70.

Solution:
(i) Slant Height (l) = √(r² + h²) = √(24² + 10²) = 26.83 m
Curved Surface Area = π × r × l = ²²⁄₇ × 24 × 26.83 = 5076.24 m²
Cost = 5076.24 × 70 = ₹355,337.00

Question 5: What length of tarpaulin 3 m wide will be required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

Solution:
Slant Height (l) = √(r² + h²) = √(6² + 8²) = 10 m
Curved Surface Area = π × r × l = 3.14 × 6 × 10 = 188.4 m²
Length of Tarpaulin = Area / Width = 188.4 / 3 = 62.8 m
Total Length = 62.8 + 0.2 = 63 m

Question 6: The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of ₹210 per 100 m².

Solution:
Radius (r) = 7 m
Curved Surface Area = π × r × l = ²²⁄₇ × 7 × 25 = 550 m²
Cost = (550 / 100) × 210 = ₹1155.00

Question 7: A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:
Slant Height (l) = √(r² + h²) = √(7² + 24²) = 25 cm
Curved Surface Area = π × r × l = ²²⁄₇ × 7 × 25 = 550 cm²
Area for 10 Caps = 550 × 10 = 5500 cm² = 0.55 m²

Question 8: A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is ₹12 per m², what will be the cost of painting all these cones? (Use π = 3.14).

Solution:
Radius (r) = 20 cm
Slant Height (l) = √(r² + h²) = √(20² + 100) = 20.98 cm
Curved Surface Area per Cone = π × r × l = 3.14 × 20 × 20.98 = 1318.52 cm² = 1.318 m²
Total Area = 1.318 × 50 = 65.9 m²
Cost = 65.9 × 12 = ₹790.80

EXERCISE 11.2

Assume π = ²²⁄₇, unless stated otherwise.

Question 1: Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm

Solution:
(i) Surface Area = 4 × π × r² = 4 × ²²⁄₇ × 10.5² = 1386 cm²
(ii) Surface Area = 4 × π × r² = 4 × ²²⁄₇ × 5.6² = 396.8 cm²
(iii) Surface Area = 4 × π × r² = 4 × ²²⁄₇ × 14² = 2464 cm²

Question 2: Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m

Solution:
(i) Radius = 7 cm
Surface Area = 4 × π × r² = 4 × ²²⁄₇ × 7² = 1232 cm²
(ii) Radius = 10.5 cm
Surface Area = 4 × π × r² = 4 × ²²⁄₇ × 10.5² = 1386 cm²
(iii) Radius = 1.75 m
Surface Area = 4 × π × r² = 4 × ²²⁄₇ × 1.75² = 38.5 m²

Question 3: Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)

Solution:
Total Surface Area = 3 × π × r² = 3 × 3.14 × 10² = 3140 cm²

Question 4: The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution:
Surface Area Ratio = (4 × π × (14)²) / (4 × π × (7)²) = 2² = 4

Question 5: A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin-plating it on the inside at the rate of ₹16 per 100 cm².

Solution:
Radius = 5.25 cm
Surface Area = 3 × π × r² = 3 × ²²⁄₇ × 5.25² = 410.0625 cm²
Cost = (410.0625 / 100) × 16 = ₹65.61

Question 6: Find the radius of a sphere whose surface area is 154 cm².

Solution:
Surface Area = 4 × π × r² = 154 cm²
Radius (r) = √(154 / (4 × π)) = 3.5 cm

Question 7: The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution:
Surface Area Ratio = (1/4)² = 1/16

Question 8: A hemispherical bowl is made of steel, 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.

Solution:
Outer Radius = 5.25 cm
Outer Surface Area = 2 × π × Outer Radius × Height = 2 × 3.14 × 5.25 × 5 = 164.79 cm²

Question 9: A right circular cylinder just encloses a sphere of radius r. Find (i) surface area of the sphere, (ii) curved surface area of the cylinder, (iii) ratio of the areas obtained in (i) and (ii).

Solution:
(i) Surface Area of Sphere = 4 × π × r² = 4 × ²²⁄₇ × r²
(ii) Curved Surface Area of Cylinder = 2 × π × r × (2r) = 4 × π × r²
(iii) Ratio = 1:1

EXERCISE 11.3

Assume π = ²²⁄₇, unless stated otherwise.

Question 1: Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm

Solution:
(i) Volume = (1/3) × π × r² × h = (1/3) × ²²⁄₇ × 6² × 7 = 792 cm³
(ii) Volume = (1/3) × π × r² × h = (1/3) × ²²⁄₇ × 3.5² × 12 = 462 cm³

Question 2: Find the capacity in litres of a conical vessel with (i) radius 7 cm, slant height 25 cm (ii) height 12 cm, slant height 13 cm

Solution:
(i) Height = √(25² - 7²) = 24 cm
Volume = (1/3) × π × r² × h = (1/3) × ²²⁄₇ × 7² × 24 = 1232 cm³ = 1.232 L
(ii) Volume = (1/3) × π × r² × h = (1/3) × ²²⁄₇ × (12²) × 13 = 1884 cm³ = 1.884 L

Question 3: The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π = 3.14)

Solution:
Volume = (1/3) × π × r² × h = 1570 cm³
Radius (r) = √((1570 × 3) / (3.14 × 15)) = 10 cm

Question 4: If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.

Solution:
Volume = (1/3) × π × r² × h = 48π cm³
Radius (r) = √((48π × 3) / (π × 9)) = 4 cm
Diameter = 8 cm

Question 5: A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?

Solution:
Radius (r) = 1.75 m
Volume = (1/3) × π × r² × h = (1/3) × 3.14 × 1.75² × 12 = 37.1 m³ = 37.1 kilolitres

Question 6: The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find (i) height of the cone (ii) slant height of the cone (iii) curved surface area of the cone.

Solution:
Radius (r) = 14 cm
Volume = (1/3) × π × r² × h = 9856 cm³
Height (h) = (9856 × 3) / (3.14 × 14²) = 16 cm
Slant Height (l) = √(r² + h²) = √(14² + 16²) = 21.8 cm
Curved Surface Area = π × r × l = 3.14 × 14 × 21.8 = 965.26 cm²

Question 7: A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.

Solution:
Volume = (1/3) × π × (base)² × height = (1/3) × π × 5² × 12 = 300π cm³

Question 8: If the triangle ABC in the Question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

Solution:
Volume = (1/3) × π × (base)² × height = (1/3) × π × 12² × 5 = 720π cm³
Ratio = 720 / 300 = 2.4

Question 9: A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.

Solution:
Radius (r) = 5.25 m
Volume = (1/3) × π × r² × h = (1/3) × 3.14 × 5.25² × 3 = 259.95 m³
Curved Surface Area = π × r × l = 3.14 × 5.25 × √(5.25² + 3²) = 3.14 × 5.25 × 6.05 = 100.45 m²

EXERCISE 11.4

Assume π = ²²⁄₇, unless stated otherwise.

Question 1: Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m

Solution:
(i) Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 7³ = 1436 cm³
(ii) Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 0.63³ = 0.773 m³

Question 2: Find the amount of water displaced by a solid spherical ball of diameter (i) 28 cm (ii) 0.21 m

Solution:
(i) Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 14³ = 11576 cm³
(ii) Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 0.105³ = 0.388 m³

Question 3: The diameter of a metallic ball is 4.2 cm. What is the mass of the ball, if the density of the metal is 8.9 g per cm³?

Solution:
Radius = 2.1 cm
Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 2.1³ = 38.79 cm³
Mass = Volume × Density = 38.79 × 8.9 = 345.7 g

Question 4: The diameter of the moon is approximately one-fourth of the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?

Solution:
Volume Ratio = (1/4)³ = 1/64

Question 5: How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Solution:
Radius = 5.25 cm
Volume = (2/3) × π × r³ = (2/3) × ²²⁄₇ × 5.25³ = 573.95 cm³ = 0.574 L

Question 6: A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.

Solution:
Outer Radius = 1.01 m
Volume of Tank = (2/3) × π × (1.01)³ = 2.732 m³
Volume of Iron = Volume of Outer Sphere - Volume of Inner Sphere = 2.732 - (2/3) × π × (1)³ = 0.335 m³

Question 7: Find the volume of a sphere whose surface area is 154 cm².

Solution:
Radius (r) = √(154 / (4 × π)) = 3.5 cm
Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 3.5³ = 179.59 cm³

Question 8: A dome of a building is in the form of a hemisphere. From inside, it was white-washed at the cost of ₹4989.60. If the cost of white-washing is ₹20 per square metre, find the (i) inside surface area of the dome, (ii) volume of the air inside the dome.

Solution:
(i) Surface Area = 4989.60 / 20 = 249.48 m²
Radius = √(249.48 / (2 × π)) = 9 m
(ii) Volume = (2/3) × π × r³ = (2/3) × 3.14 × 9³ = 3053.6 m³

Question 9: Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the (i) radius r′ of the new sphere, (ii) ratio of S and S′.

Question 10: A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (in mm³) is needed to fill this capsule?

Solution:
Radius = 1.75 mm
Volume = (4/3) × π × r³ = (4/3) × ²²⁄₇ × 1.75³ = 22.6 mm³

Surface Areas and Volumes - Notes CH-11