Chapter 11: Surface Areas and Volumes Exercise Questions & Answers
1Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Answer:
Radius \( r = \frac{10.5}{2} = 5.25 \) cm, slant height \( l = 10 \) cm.
Curved Surface Area = \( \pi r l = \frac{22}{7} \times 5.25 \times 10 = 165 \) cm². (Reference: Formula from p. 139, Example 1)
2Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Answer:
Radius \( r = 12 \) m, \( l = 21 \) m.
Total Surface Area = \( \pi r (l + r) = \frac{22}{7} \times 12 \times (21 + 12) = \frac{22}{7} \times 12 \times 33 = 1244.57 \) m². (Reference: p. 139, Total Surface Area formula)
3Curved 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.
Answer:
(i) \( \pi r l = 308 \Rightarrow \frac{22}{7} \times r \times 14 = 308 \Rightarrow r = \frac{308 \times 7}{22 \times 14} = 7 \) cm.
(ii) Total Surface Area = \( \pi r (l + r) = \frac{22}{7} \times 7 \times (14 + 7) = 22 \times 21 = 462 \) cm². (Reference: p. 139)
4A 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.
Answer:
(i) \( l = \sqrt{r^2 + h^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26 \) m.
(ii) Curved Surface Area = \( \pi r l = \frac{22}{7} \times 24 \times 26 = 1961.14 \) m².
Cost = \( 1961.14 \times 70 = ₹137,280 \). (Reference: p. 139, Example 2)
5What length of tarpaulin 3 m wide will be required to make 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).
Answer:
\( l = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \) m.
Curved Surface Area = \( \pi r l = 3.14 \times 6 \times 10 = 188.4 \) m².
Let length of tarpaulin = \( L \) m. Area = \( 3 \times L = 188.4 \Rightarrow L = 62.8 \) m.
Add 0.2 m wastage → Total length = \( 62.8 + 0.2 = 63 \) m. (Reference: p. 140, Example 2 logic)
6The 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.
Answer:
Radius \( r = 7 \) m, \( l = 25 \) m.
Curved Surface Area = \( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \) m².
Cost = \( \frac{550}{100} \times 210 = ₹1155 \). (Reference: p. 139)
7A 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.
Answer:
\( l = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \) cm.
Curved Surface Area of one cap = \( \pi r l = \frac{22}{7} \times 7 \times 25 = 550 \) cm².
For 10 caps = \( 550 \times 10 = 5500 \) cm². (Reference: p. 140, Example 3)
8A 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 and take √1.04 = 1.02)
Answer:
Radius \( r = 20 \) cm = 0.2 m, height \( h = 1 \) m.
\( l = \sqrt{r^2 + h^2} = \sqrt{0.04 + 1} = \sqrt{1.04} = 1.02 \) m.
Curved Surface Area of one cone = \( \pi r l = 3.14 \times 0.2 \times 1.02 = 0.64056 \) m².
For 50 cones = \( 50 \times 0.64056 = 32.028 \) m².
Cost = \( 32.028 \times 12 = ₹384.34 \). (Reference: p. 139, Example 2)
9Find the surface area of a sphere of radius: (i) 10.5 cm (ii) 5.6 cm (iii) 14 cm
10Find the surface area of a sphere of diameter: (i) 14 cm (ii) 21 cm (iii) 3.5 m
Answer:
(i) \( r = 7 \) cm → \( 4\pi r^2 = 616 \) cm²
(ii) \( r = 10.5 \) cm → \( 1386 \) cm²
(iii) \( r = 1.75 \) m → \( 4 \times \frac{22}{7} \times (1.75)^2 = 38.5 \) m² (Reference: p. 142)
11Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14)
Answer:
Total Surface Area = \( 3\pi r^2 = 3 \times 3.14 \times 100 = 942 \) cm². (Reference: p. 143, formula)
12The 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.
Answer:
Surface area ∝ \( r^2 \) → Ratio = \( \frac{7^2}{14^2} = \frac{49}{196} = \frac{1}{4} \). (Reference: p. 142)
13A 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.
Answer:
Radius \( r = 5.25 \) cm.
Inner surface to be plated = Curved Surface Area = \( 2\pi r^2 = 2 \times \frac{22}{7} \times (5.25)^2 = 173.25 \) cm².
Cost = \( \frac{173.25}{100} \times 16 = ₹27.72 \). (Reference: p. 143, Example 5)
14Find the radius of a sphere whose surface area is 154 cm².
Answer:
\( 4\pi r^2 = 154 \Rightarrow r^2 = \frac{154 \times 7}{4 \times 22} = \frac{49}{4} \Rightarrow r = \frac{7}{2} = 3.5 \) cm. (Reference: p. 142)
15The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Answer:
Let Earth’s diameter = \( D \), Moon’s = \( D/4 \) → Radii: \( R \) and \( R/4 \).
Ratio = \( \frac{4\pi (R/4)^2}{4\pi R^2} = \frac{1}{16} \). (Reference: p. 142)
16A 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.
17A 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).
Answer:
Cylinder: radius = \( r \), height = \( 2r \).
(i) Sphere SA = \( 4\pi r^2 \)
(ii) Cylinder CSA = \( 2\pi r h = 2\pi r (2r) = 4\pi r^2 \)
(iii) Ratio = \( 1:1 \). (Reference: p. 145, Fig. 11.10 concept)
18Find the volume of the right circular cone with (i) radius 6 cm, height 7 cm (ii) radius 3.5 cm, height 12 cm
19Find 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
Answer:
(i) \( h = \sqrt{25^2 - 7^2} = \sqrt{576} = 24 \) cm → Volume = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 49 \times 24 = 1232 \) cm³ = 1.232 L
(ii) \( r = \sqrt{13^2 - 12^2} = \sqrt{25} = 5 \) cm → Volume = \( \frac{1}{3} \times \frac{22}{7} \times 25 \times 12 = 314.29 \) cm³ = 0.314 L (Reference: p. 146, Example 8)
27A 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.
Answer:
Radius \( r = 5.25 \) m, \( h = 3 \) m.
Volume = \( \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (5.25)^2 \times 3 = 86.625 \) m³.
\( l = \sqrt{5.25^2 + 3^2} = \sqrt{36.5625} = 6.05 \) m (approx).
Canvas area = Curved Surface Area = \( \pi r l = \frac{22}{7} \times 5.25 \times 6.05 ≈ 99.825 \) m². (Reference: p. 146–147, Example 9)
28Find the volume of a sphere whose radius is (i) 7 cm (ii) 0.63 m