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Question
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. find the curved surface area of the frustum.
Solution 1
Perimeter of upper circular end of frustum = 18 cm
2πr1 =18
r1 = 9/π
Perimeter of lower end of frustum = 6 cm
2πr2 = 6
r2 = 3/π
Slant height (l) of frustum = 4 cm
CSA of frustum = π (r1 + r2) l
`=pi(9/pi+3/pi)4`
= 12x4
= 48 cm2
Therefore, the curved surface area of the frustum is 48 cm2.
Solution 2
We have,
Perimeter of upper end, C=18cm,
Perimeter of lower end, c=6 cm and
Slant Height, l=4 cm
Let the radius of upper end be R and the radius of lower end be r.
As, C = 18 cm
⇒2πR = 18
`=> R = 18/(2pi)`
`=> R = 9/pi`cm
Similary c = 6 cm
`=> r = 6/(2pi)`
`=> r = 3/pi` cm
Curved surface area of the frustum=π(R + r)l
`=pi(9/pi + 3/pi) xx 4`
`= pi xx 12/pi xx 4`
`= 48 cm^2`
Hence, the curved surface area of the frustum is 48 cm2
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