NCERT Class 9 Solutions: Surface Areas and Volumes (Chapter 13) Exercise 13.2 – Part 3

Q-9 Find

  1. The lateral or curved surface area of a closed cylindrical petrol storage tank that is 4.2m in diameter and 4.5m high.

  2. How much steel was actually used, if 112 of the steel actually used was wasted in making the tank.

Solution:

Areas in a cylinder and its volume

Recap of Cylinder Areas and Volume

Areas in a cylinder and its volume

Solution (I)

The tank radius (r) =4.22m=2.1m The tank height (h) =4.5m

Curved surface area = 2πrhm2

  • =(2×227×2.1×4.5)m2

  • =59.4m2

Solution (II)

Total surface area of the tank

  • 2πr(r+h)m2

  • [2×227×2.1(2.1+4.5)]m2

  • 87.12m2

Now assume that xm2 is the actual steel used in making tank. Than we know that 112 of actual steel used was wasted. Therefore the remaining must match the area of the tank.

Therefore, (1112)×x=87.12

  • 12112×x=87.12

  • 1112×x=87.12

  • x=87.12×1211

  • x=95.04m2

Q-10 In the figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20cm and height of 30cm . A margin of 2.5cm 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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the Frame of a Lampshade

The frame of a lampshade, frame diameter20cm, height30cm, margin2.5cm.

Solution: Note that he lampshade is a cylinder. Now,

  • The cloth cylinder radius (r) =202cm=10cm

  • Since, 2.5cm of margin must be added to both side of the height. Therefore, the cloth cylinder height (h) =30cm+2×2.5cm=35cm

  • Therefore, cloth required for covering the lampshade is the curved surface area of our cloth cylinder = 2πrh=(2×227×10×35)cm2 = 2200cm2

Q-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 3cm and height 10.5cm. 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?

Solution:

  • Radius (r) of the penholder =3cm

  • Height (h) of the penholder =10.5cm

Cardboard required by 1 competitor, is found by realizing that the penholder has a base but no top. Therefore, area of cardboard required for 1 penholder = curved surface area + area of the base

  • =2πrh+πr2

  • =[(2×227×3×10.5)+227×32]cm2

  • =(198+1987)cm2

  • =1386+1987cm2

  • =15847cm2

Therefore, cardboard required for 35 competitors (35×15847)cm2=7920cm2

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