Surface Area

Exercise 13.3

Part 1

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

Answer: d = 10.5 so, r = 5.25 cm, l = 10 cm

Curved surface area of cone `=π\rl`

`=(22)/7xx5.25xx10``=22xx7.5=165` sq cm


Question 2: Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.

Answer: d = 24 m so, r = 12 m, l = 21 m

CSA of cone `=π\rl`

`=(22)/7xx12xx21`

`=66xx12=792` sq m

Area of base `=πr^2`

`=(22)/7xx12xx12=452.16` sq m

Total surface area = 792 + 452.16 = 1244.16 sq m


Question 3: Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find (i) radius of the base and (ii) total surface area of the cone.

Answer: CSA = 308 sq cm, l = 14 cm, r = ?

CSA `=π\rl`

Or, `308=(22)/7xx\r\xx14`

Or, `r=(308)/(44)=7` cm

Question 4: A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.

Answer: Slant height can be calculated by using Pythagoras Theorem

`l^2=h^2+r^2`

`=10^2+24^2``=100+576=676`

Or, `l=sqrt(676)=26` cm

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

Answer: Answer: CSA of cone `=π\rl`

`=(22)/7xx24xx26`

Hence, Cost = Area × Rate

`=(22)/7xx24xx26xx70` = Rs. 137280



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