Class 9 Maths

# 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: Curved Surface Area (CSA) = 4.4 sq m, r = 0.7 m, h = ?

CSA =2πrh

Or, 2xx(22)/7xx0.7xxh=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)/7xx3.5xx10=110 sq m

Cost = Area × Rate =110xx40 = 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)/7xx0.05xx28=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)/7xx4.2xx4.5=59.40 sq m

Total surface area of cylinder =2πr(r+h)

=2xx(22)/7xx2.1(2.1+4.5)=2xx(22)/7xx2.1xx6.6=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 =87.12xx(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

=2xx(22)/7xx20xx35=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 =2π\rh

=2xx(22)/7xx3xx10.5=198 sq cm

Hence, CSA of 35 cylinders =198xx35 = 6930 sq cm

Area of base =π\r^2

=(22)/7xx3^2=(22)/7xx9

Hence, area of 35 bases =35xx(22)/7xx9=990 sq cm

Total area of required cardboard = 6930 + 990 = 7920 sq cm