# Surface Area

## Exercise 13.2 Part 2

Question 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 m, find its height.

Answer: CSA = 4.4 sq m, r = 0.7 m, h = ?
CSA = 2 π rh
Or, 2 xx (22/7) xx 0.7 xx h = 4.4
Or, h = (4.4)/(4.4) = 1 m

Question 7: The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find
(i) its inner curved surface area,
(ii) the cost of plastering this curved surface at the rate of Rs 40 per m2.

Answer: d = 3.5 m. h = 10 m
CSA of cylinder = π dh
= (22/7) xx 3.5 xx10 = 110 sq m
Cost = Ar\eaxxRa\te
=110 xx 40 = Rs. 4400

Question 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.

Answer: h = 28 m, d = 5 cm = 0.05 m
CSA of cylinder = π dh
= (22/7) xx 0.05 xx 28 = 4.4 sq m

Question 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 1/12 of the steel actually used was wasted in making the tank.

Answer: d = 4.2 m so, r = 2.1 m, h = 4.5 m
CSA of cylinder = π dh
=(22)/(7) xx 4.2 xx 4.5=59.40 sq m

Total surface area of cylinder = 2 π r(r + h)
= 2 xx (22/7) xx 2.1 (2.1 + 4.5)
=2xx(22/7)xx2.1xx6.6=87.12 sq m

As 1/12 of steel is wasted so 11/12 of steel is used in making the cylinder
Hence, actual amount of used steel = 87.12 xx (12/11) = 95.04 sq m

Question 10: In the given figure, you see the frame of a lampshade. It is to be 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.

Answer: d = 20 cm, h = 30 cm, margin = 2.5 cm
Effective h = 30 + 2.5 + 2.5 = 35 cm
CSA of cylinder = 2π rh
= 2 xx (22/7) xx 20 xx 35 = 2200 sq cm

Question 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?

Answer: r = 3 cm, h = 10.5 cm, no. of penholders = 35
CSA of cylinder = 2 π rh
= 2 xx (22/7) xx 3 xx 10.5 = 198 sq cm
Hence, CSA of 35 cylinders = 198 xx 35 = 6930 sq cm

Area of base = π r^2= (22/7) xx 3^2=(22/7)xx9
Hence, area of 35 base = 35 xx (22/7) xx 9 = 990 sq cm
Total area of required cardboard = 6930 + 990 = 7920 sq cm`

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