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
⟹ 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
= π(r1 + r2)l
= (πr1+πr2)l = (9 + 3) × 4 = (12) × 4 = 48 cm2
Therefore, the curved surface area of the frustum = 48 cm2