In this question we have to find the curved

surface area of frustum given that the slant

height of the frustum of a cone is 4 cm and

the perimeters of its circular ends are 18 cm and 6 cm.

This question is from r d sharma class 10 maths.

I found this question while doing mensuration of

class 10.

I need help in getting the solution.

Given,

Slant height of frustum of cone (l) = 4 cm

Let ratio of the top and bottom circles be r1 and r2

And given perimeters of its circular ends as 18 cm and 6 cm

âŸ¹ 2Ï€r1 = 18 cm; 2Ï€r2 = 6 cm

âŸ¹ Ï€r1= 9 cm and Ï€r2 = 3 cm

We know that,

Curved surface area of frustum of a cone = Ï€(r1 + r2)l

= Ï€(r1 + r2)l

= (Ï€r1+Ï€r2)l = (9 + 3) Ã— 4 = (12) Ã— 4 = 48 cm2

Therefore, the curved surface area of the frustum = 48 cm2