Surface Areas and Volumes – Exercise 13.2 – (MATHEMATICS) – 9th Class

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Exercise 13.2

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

1. The curved surface area of a right circular cylinder of height 14 cm is 88 cm². Find the diameter of the base of the cylinder.

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Sol. Given: 2πrh = 88 \Rightarrow 2 \times {{22} \over 7} \times r \times 14 = 88
⇒             r = 1 cm
∴   diameter of the base = 2 cm.


2. It is required to make a closed cylindrical tank of height 1 m and base diameter 140 cm from a metal sheet. How many square metres of the sheet are required for the same?

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Sol. Base diameter = 140 cm = 1.4 m. Height (h) = 1 m.
∴  Base radius (r) = 0.7 m
Total surface area (required metal sheet)

 = 2\pi r(r + h) = 2 \times {{22} \over 7} \times 0.7(0.7 + 1){m^2}

= 4.4 ×  1.7 m2 = 7.48 m2.


3. A metal pipe is 77 cm long. The inner diameter of a cross section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its
(i) inner curved surface area,
(ii) outer curved surface area,
(iii) total surface area.

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Sol. Length of the pipe (h) = 77 cm
Inner diameter = 4 cm, inner radius (r1) = 2 cm, length = 77 cm,
outer diameter = 4.4 cm, outer radius (r2) = 2.2 cm

(i) Inner curved surface area = 2πr1h
 = 2 \times {{22} \over 7} \times 2 \times 77c{m^2} = 968c{m^2}
(ii) Outer curved surface area = 2πr2h
 = 2 \times {{22} \over 7} \times 2 \times 77c{m^2}
= 1064.8 cm².
(iii) Total surface area = 2π1h + 2πr2h + 2π(r22 – r12)

= 968 + 1064.8 + 2 × {{22} \over 7}(2.2 + 2)(2.2 - 2)

= 968 + 1064.8 + 5.28

= 2038.08 cm².


4. The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m².

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Sol. Area levelled in 1 revolution = 2πrh.

 = 2 \times {{22} \over 7} \times 42 \times 120c{m^2}

= 31680 cm².
in 500 revolutions (area of playground)

 = 31680 \times 500c{m^2} = {{31680 \times 500} \over {10000}}{m^2}

= 1584 m².


5. A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curved surface of the pillar at the rate of  Rs.12.50 per m².

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Sol. Curved surface of pillar =  = 2 \times {{22} \over 7} \times {{25} \over {100}} \times 3.5{m^2}

= 5.5 m2

Cost of painting = Rs.(12.50 × 5.5) = Rs.68.75.


6. Curved surface area of a right circular cylinder is 4.4 m². If the radius of the base of the cylinder is 0.7 in, find its height.

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Sol. 2\pi rh = 4.4 \Rightarrow 2 \times {{22} \over 7} \times 0.7 \times h = 4.4 \Rightarrow h = 1m.


7. The inner diameter of a circular well is 3.5 m. It is 10 in deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of T40 per m2.

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Sol. (i) Inner curved surface area = 2πrh
 = 2 \times {{22} \over 7} \times {{35} \over {20}} \times 10{m^2}
= 110 m2.
(ii) Cost of plastering = Rs.(40 × 110) = Rs.4400.


8. In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.

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Sol. Total radiating surface =  = 2 \times {{22} \over 7} \times {5 \over {200}} \times 28{m^2}

= 4.4 m2


9. Find
(i) the lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.

(ii) how much steel was actually used, if {1 \over {12}} of the steel actually used was wasted in making the tank.

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Sol. (i) Lateral surface area =  2 \times {{22} \over 7} \times 2.1 \times 4.5{m^2}

(ii) Total surface area of the tank = 59.4 + 2 × {{22} \over 7} × 2.1 × 2.1

= 59.4 + 27.72

= 87.12 m2

Let total steel used = x m

∴  \left( {x - {1 \over {12}}x} \right) = 87.12 \Rightarrow x = {{12} \over {11}} \times 87.12{m^2}

= 95.04 m2


10. In the adjoining figure, you see the frame of a lampshade. .It is to he covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find. how much cloth is required for covering the lampshade.

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Sol. Total height including margin = (30 + 2 × 2.5) cm = 35 cm.
Since diameter = 20 cm therefore, radius = 10 cm.

Total cloth required   = 2 \times {{22} \over 7} \times 10 \times 35c{m^2}
= 2200 cm².


11. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition?

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Sol. Total cardboard required
 = 35 \times \left[ {{{22} \over 7} \times 3 \times 10.5 + {{22} \over 7} \times {{(3)}^2}} \right]c{m^2}

= 35 (198 + 28.29) cm² = 7920 cm².


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