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NCERT Class 9 Mathematics Chapter 11 Surface Areas and Volumes
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 9 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 9 Mathematics Chapter 11 Surface Areas and Volumes Solutions for All Subject, You can practice these here.
Surface Areas and Volumes
Chapter – 11
| Exercise 11.1 |
Q.1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.
Ans: ∴ Diameter of the base = 10.5cm
Slant height (l) = 10cm
∴ Curved surface area of the cone = πrl
Q.2. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
Ans: Slant height (l) = 21m
Diameter of base = 24m
∴ Total curved surface area of the cone = πr(l + r)
Q.3. Curved surface area of a cone is 308cm² and its slant height is 14 cm. Find
(i) radius of the base and
(ii) total surface area of the cone.
Ans: (i) Slant height (l) = 14cm
Curved surface area = 308cm²
Hence the radius of the base is 7 cm.
(ii) Total surface area of the cone = πr(l + r)
Hence the total surface area of the cone is 462cm²
Q. 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
Ans: (i) h = 10m,
r = 24m
Hence the slant height of the tent is 26 m.
(ii) Curved surface area of the tent = πrl
∴ Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs. 70
Hence the cost of the canvas is Rs. 137280
Q.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 wast- age in cutting is approximately 20 cm (Use π = 3.14 )
Ans: For conical tent
∴ Curved surface area = πrl
= 3.14 x 6 x 10 = 188.4m²
Width of tarpaulin = 3m
Extra length of the material required = 20cm = 0.2m
∴ Actual length of tarpaulin required = 62.8m + 0.2m = 63m
Q.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 Rs. 210 per 100m².
Ans: Slant height (l) = 25 m
Base diameter = 14m
∴ Cost of white-washing the curved surface of the tomb at the rate of Rs. 210 per 100 m².
Q.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.
Ans: Base radius (r) = 7cm
Height (h) = 24cm
∴ Curved surface area of a cap
∴ Curved surface area of 10 caps = 550 x 10 = 5500 cm²
Hence the area of the sheet required to make 10 such caps is 5500cm².
Q.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 Rs. 12 per m² what will be the cost of painting all these cones? (Use π = 3.14 and take
√(1.04) = 1.02 )
Ans: Base diameter = 40cm
∴
Height (h) = 1m
∴ Curved surface area = πrl
= 3.14 x 0.2 x 1.02 = 0.64056m²
∴ Curved surface area of 50 cones
= 0.64056 x 50m² = 32.028m²
∴ Cost of painting all these cones = 32.028 x 12
= 384.336 = Rs. 384.34 (approximately).
| Exercise 11.2 |
Q.1. Find the surface area of a sphere of radius:
(i) 10.5 cm
(ii) 5.6 cm
(iii) 14 cm
Ans: (i) r = 10.5cm
∴ Surface area = 4πr²
(ii) r = 5.6cm
(iii) r = 14cm
Q.2. Find the surface area of a sphere of radius:
(i) 14 cm
(ii) 21 cm
(iii) 3.5 cm
Ans: (i) Diameter = 14cm
(ii) Diameter = 21 cm
(iii) Diameter = 3.5 cm
Q.3. Find the total surface area of a hemisphere of radius 10 cm. (Use π = 3.14
Ans: r = 10cm
∴ Total surface area of the hemisphere
= 3πr² = 3 x 3.14 x (10)² = 942cm²
Q.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.
Ans: Case I. r = 7cm
∴ Case II. r = 14cm
∴ Ratio of surface areas of the balloon
Q.5. A Hemisphere bowl made of brass has an inner diameter 10.5 cm. Find the cost of tin-plating it on the incident the rate of Rs. 16 per 100 cm²
Ans: Inner diameter = 10.5cm
∴ Inner surface area
∴ Cost of tin-plating at the rate of Rs. 16 per 100 cm²
Q.6. Find the radius of a sphere whose surface area is 154 cm².
Ans: Let the radius of the sphere be r cm.
Surface area = 154 cm²
Hence the radius of the sphere is 3.5 cm.
Q.7. The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
Ans: Let the diameter of the earth be 2r.
∴ Surface area of the earth = 4πr²
∴ Ratio of their surface area
Q.8. A Hemisphere 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.
Ans: Inner radius of the bowl = 5 cm
Thickness of steel = 0.25 cm
∴ Outer radius of the bowl = 5 + 0.25 = 5.25cm
∴ Outer curved surface of the bowl
Q.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).
Ans: (i) Surface area of the sphere = 4πr²
(ii) For cylinder
Radius of the base = r
Height = 2r
∴ Curved surface area of the cylinder = 2π(r)(2r) = 4πr²
(iii) Ratio of the areas obtained in (i) and (ii)
