Surface Area and Volumes – Exercise 13.2 – Class IX

For previous exercise – Surface Area and Volumes – Exercise 13.2 – Class IX


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

Solution:

Given, h = 14 cm ; surface area = 88 cm2

Curved surface area of the cylinder = 2πrh

88 = 2 x 22/7 x r x 14

r = 88×7/2x22x14 = 1 cm

Hence, base diameter of the cylinder = 2r = 2x 1 = 2 cm


  1. 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 meters of the sheet are required for the same?

Solution:

h = 1 m

d = 140 cm

r = d/2 = 140/2 = 70 cm = 0.7 m

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

= 2 x 22/7 x 0.7 x (1 + 0.7)

= 2 x 22/7 x 0.7 x 1.7

= 7.48 m2


  1. Surface Area and Volumes – Exercise 13.2 – Class IXA metal pipe is 77 cm long. The inner diameter of a cross-section is 4 cm, the outer diameter being 4.4 cm (see Fig. 13.11). Find its

(i) inner curved surface area,

(ii) outer curved surface area,

(iii) total surface area.

Solution:

h = 77 cm

Inner diameter = 4 cm ⇒ r = d/2 = 4/2 = 2 cm

Outer  diameter = 4.4 cm ⇒ r  =  d/2= 4.4/2 = 2.2 cm

(i) inner curved surface area = 2πRh

= 2 x 22/7 x 2 x 77

= 968 cm2

(ii) outer curved surface area = 2πrh

= 2 x 22/7  x 2.2 x 77

= 1064.8 cm2

(iii) total surface area =inner curved surface area + outer curved  surface area + two base rings

=  2πrh + 2π Rh +  2π ( R2 – r2)

=  968  + 1064.8  + 2x22/7x(2.22 – 22)

= 968 + 1064.8 + 5.28

= 2038. 08 cm2


  1. 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

Solution:

d = 84 cm ; r = d/2= 84/2 = 42 cm

h = 120 cm

Surface area of the roller = 2πrh  = 2 x 22/7 x  42 x 120

=31680 cm2

Given, it takes 500 complete revolutions to move once over to level a playground. Then the area of the playground = 31680 x 500 = 15840000 cm2 = 1584 m2


  1. 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

Solution:

d = 50 cm ; r = 50/2 = 25 cm

h =  3.5 m

Curved surface area of  the pillar = 2πrh = 2 x 22/7 x 25 x 3.5 =550m2

The cost of painting the pillar for 1m2 is  Rs. 12.50

Then the cost  of 550m2 at the rate ofRs12.50=12.50*550=Rs. 68.75


  1. 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.

Solution:

Given, curved surface area of  the cylinder 4.4 m2  and radius, r = 0.7 m

surface  area = 2πrh

4.4 = 2x 22/7 x 0.7 x h

h = 4.4/4.4  = 1 m


  1. 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.

Solution:

Inner diameter ; r = 3.5 m and h = 10m

(i) inner curved surface area = 2πrh = 2 x 22/7 x 3.5 x 10 = 220 m2

(ii) The cost of plastering of curved surface for 1m2 is Rs. 40

Then, the cost of plastering for 220m2 = 220 x 40 = Rs. 8800


  1. 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.

Solution:

h = 28 m and d = 5 cm; r = 5/2 = 2.5 cm = 0.025 cm

Total surface area =  2πr (r  + h) = 2 x 22/7 x 0.025 x(0.025 + 28) = 0.1571 x(28.025) = 4.403 m2


  1. 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.

Solution:

d = 4.2 m ; r  = 4.2/2=2.1 m

h = 4.5 m

(i) Curved surface area  of the  cylinder = 2πrh = 2 x 22/7x 2.1 x 4.5 =59.4  m2

(ii) Total surface area =2πr(r+h)

=  2 x 22/7x 2.1 (2.1  +   4.5)

=  87.12m2

Lets total steel area of steel used be x m2

Area of steel being wasted = 1/12  of x m2 = x/12 m2

Area of steel used in the tank = (x – x/12) m2 = 11x/12 m2

87.12 = 11x/12

x = 87.12×12/11 = 95.04 m2

Hence, 95.04 m2 of steel was actually used in tank.


  1. Surface Area and Volumes – Exercise 13.2 – Class IXIn Fig. 13.12, 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.

Solution:

d = 20 cm; r = d/2 = 20/2 = 10 cm

h = 30 cm

Circumference of the base of the frame = 2πr = 2 x 22/7 x 10 = 62. 86 cm

3Height of the cloth required for covering frame (including margin) = (30+2.5+2.5)cm = 35 cm

Also, breadth of the cloth = circumference of the base of the frame

Area of  the cloth required  for covering the lampshade = length x breadth = 35 x 20 π cm2 = 35 x 20 x 22/7 cm2 = 2200 cm2


  1. The students of a Vidyalaya were asked to participate in a competition for making and decorating penholder 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?

Solution:

Here, r = 3 cm, h = 10.5 cm

The penholder have only one base i.e., these are open at one end.

Total surface area of 1 penholder = 2πrh + πr2 = πr (2h + r)

= 22/7 × 3 (2 × 10.5 + 3) cm2

= 22/7 × 3 × 24 cm2

Total surface area of 35 penholder = 22/7 × 3 × 24 × 35 cm2 = 7920 cm2

Hence, 7920 cm2 of cardboard is needed.


For next exercise – Surface Area and Volumes – Exercise 13.3 – Class IX


 

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