Question

The slant height of the frustum of a cone is 4 cm and the perimeters (i.e., circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.​

Answer 1

Given slant height(l)=4cm

Circumference1(C1)=18cm

2 pie r1 = 18cm

2*22/7*r1 = 18cm

therefore, r1 = 63/22cm

similarily C2 = 6cm

so, r2 = 21/22cm

now CSA of frustum of cone = pie l (r1+r2)

22/7*4(63/22+21/22)

84/22 * 4 * 22/7

=48 sq. cm.------------Ans.

Answer 2

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

⟹ 2πr1 = 18 cm; 2πr2 = 6 cm

⟹ πr1= 9 cm and πr2 = 3 cm

We know that,

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

= (πr1+πr2)l = (9 + 3) × 4

 = (12) × 4 = 48 cm2

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