Surface Areas and Volumes – Exercise 13.3 – (MATHEMATICS) – 9th Class

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Exercise 13.3 

Assume \pi = {{22} \over 7} unless stated otherwise.

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

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Sol. Curved surface area =  \pi rl= {{22} \over 7} \times {{10.5} \over 2} \times 10c{m^2}

= 165 cm²


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

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Sol. Total surface area = πrl + πr² = πr(l + r)

 = {{22} \over 7} \times 12 \times (21 + 12)c{m^2}

= 1244.57 m2


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.

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Sol. (i) Given: πrl = 308 \Rightarrow {{22} \over 7} \times r \times 14 = 308

 \Rightarrow r = {{308} \over {44}} = 7cm

∴ Radius of the base = 7 cm.

(ii) Total surface area = πrl + πr²
 = 308 + {{22} \over 7} \times {(7)^2} = 462c{m^2}


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 Rs.70.

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Sol. Base radius = 24 m, height = 10 m.

(i) Slant height =  = \sqrt {{{(24)}^2} + {{(10)}^2}m} = \sqrt {576 + 100} m
 = \sqrt {676} m = 26m

(ii) Total canvas required = πrl =  = {{22} \over 7} \times 24 \times 26{m^2}

∴ Cost = Rs.\left( {70 \times {{22} \over 7} \times 24 \times 26} \right) = Rs.137280


5. What 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 n = 3.14).

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Sol. Slant height aof the tent = \sqrt {{{(6)}^2} + {{(8)}^2}} m = \sqrt {36 + 64} m  = 10 m.

Area of tarpailin required = πrl = 3.14 × 6 × 10 m²

= 188.4 m².

∴ Length of tarpaulin required = {{188.4} \over 3} = 62.8 m.

Total length required including wastage = (62.8 + 0.2) m = 63 m.


6. The slant height and  base diameter of a conical tomb are 25 in and 14 m respectively. Find the cost of whitewashing its curved surface at the rate of t210 per 100 m2.

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Sol. Curved surface area = πrl = {{22} \over 7} \times 7 \times 25{m^2} = 550{m^2}

Cost of white-washing =  Rs.\left( {210 \times {{550} \over {100}}} \right) = Rs.1155


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.

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Sol. Slant height of joker’s cap

 = \sqrt {{{(24)}^2} + {{(7)}^2}} cm
 = \sqrt {576 + 49} cm
 = \sqrt {625} cm = 25cm

∴ Area of sheet required for 10 caps

 = 10 \times \pi rl = 10 \times {{22} \over 7} \times 7 \times 25c{m^2} = 5500c{m^2}


8. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each one 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 \sqrt {1.04}  = 1.02.)

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Sol. Slant height of each cone (l) = \sqrt {{{(1)}^2} + {{(0.2)}^2}} m
\sqrt {1.04} m = 1.02m
∴ Area to be painted for 50 hollow cones

= 50 × πrl

= (50 × 3.14 × 0.2 × 1.02) m²

= 32.028 m²

Cost of painting = Rs.(12 × 32.028)
= Rs. 384.34 (approx).


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