Surface Area and Volumes – Exercise 13.3 – Class IX

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


Assume π = 22/7 , unless stated otherwise.

  1. Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

Solution:

r = 10.5/2 cm = 5.25 cm , l = 10 cm

Curved surface area of the cone = πrl = 22/7 x 5.25 x 10 = 165 cm2


  1. Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Solution:

l = 21 m and d = 24 m ; r = d/2 = 24/2  = 12 m

Total surface area of the cone = πr(l  + r) = 22/7 x 12 x (21 + 12) = 1244. 57 m2


  1. Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find

(i) radius of the base  

(ii) total surface area of the cone.

Solution:

Curved surface area of a cone = 308 cm2

l = 14 cm

(i) Curved surface area of a cone = πrl

308 = 22/7 x r x 14

r = 308 x 7/22×14 = 7

(ii) total surface area of the cone = πr(l + r)

= 22/7 x 7 x(14 + 7)

= 462 cm


  1. A conical tent is 10 m high and the radius of its base is 24 m. Find

(i) slant height of the tent.

(ii) cost of the canvas required to make the tent, if the cost of 1 m2  canvas is Rs 70.

Solution:

l = 10 cm ; r = 24 m

(i) l2 = h2 + r2

l2= 102 + 242

l2 = 100 + 576 = 676

l = √676 = 26 m

(ii) curved surface area of the tent = πrl = 22/7 x 24 x 26 = 1961.143 m2

Cost of 1m2 canvas = Rs. 70

Cost of  1961.143m2 of canvas = 70 x 1961.143 = Rs. 137280


  1. What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm (Use π = 3.14).

Solution:

r = 6m ; h = 8m

We have, l2 = r2 + h2

l2 = 62 + 82

l2 = 36 + 64

l2 = 100

l = 10 m

Curved surface area of the tent = πrl = 3.14 x 6 x 10/3 m + 20 m = 62.8 m + 0.2 m = 63 m


  1. The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white-washing its curved surface at the rate of Rs 210 per 100 m2.

Solution:

l = 25 m

d =14 m

r = d/2 = 14/2 = 7 m

Curved surface area = πrl = 22/7 x 7 x 25 = 550 m2

The cost of white-washing its curved surface at the rate for 100 m2 = Rs 210

The cost of white-washing its curved surface at the rate for 550 m2 = 550 x 210/100  = Rs. 115500/100 = Rs. 1155


  1. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.

Solution:

r = 7 cm

h = 24

l2 = h2 + r2

l2 = 242 + 72 = 625

l = 25 cm

Total cured  surface of 1 cap = πrl =22/7 x 7x 25 = 550 cm2

Area of sheet  required to make 10  such caps = 10 x 550 = 5500 cm2


  1. A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2 , what will be the cost of painting all these cones? [Use π = 3.14 and take √(1.04) = 1.02]

Solution:

r = 40/2 = 20 cm = 0.20 m

h = 1 m

l2 = h2 + r2 = 12 + (0.2)2 = 1.04

l = 1.02 m

Curved surface  area  of 1 cone = πrl

Curved surfacearea of 50 cones = 50 x 3.14 x 0.2 x 1.02 m2 = 32.028 m2

Cost of painting an area of 1m2 =Rs. 12

Cost of painting an area of 32.028 m2 = Rs. 12 x 32.028 = Rs. 384.34


For next exercise – Surface Area and Volumes – Exercise 13.4 – Class IX


 

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