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Question
The perimeters of the two circular ends of a frustum of a cone are 48 cm and 36 cm. If the height of the frustum is 11 cm, then find its volume and curved surface area.
Solution
We have,
Perimeter of upper end, C = 48 cm,
Perimeter of lower end, c = 36 cm and
Height, h = 11 cm
Let radius of upper end be R and the radius of lower end be r.
As, C = 48 cm
⇒ 2πR = 48
`rArr R = 48/(2pi)`
`rArr R = 24/pi "cm"`
Similarly, c = 36 cm
`rArr r = 36/(2pi)`
`rArr r = 18/(pi) "cm"`
And, `l = sqrt((R - r)^2 + h^2)`
`=sqrt((24/pi-18/pi)^2)+ 11^2`
`=sqrt((6/pi)^2 + 11^2)`
`=sqrt(((6xx7)/22)^2 + 11^2)`
`=sqrt((21/11)^2 + 11^2)`
`=sqrt(441+14641)/121`
`= sqrt(15082)/11 "cm"`
Now,
Volume of the frustum` = 1/3 pi"h"("R"^2 + "r"^2 + "Rr")`
`=1/3xxpixx11xx[(24/pi)^2 + (18/pi)^2+(24/pi)xx(18/pi)]`
`= (11pi)/3xx[576/pi^2 + 324/pi^2 + 432/pi^2]`
`= (11pi)/3xx1332/pi^2`
`= 11/3xx(1332xx7)/22`
= 1554 cm3
Also,
Curved surface area of the frustum = π (R + r)l
`= 22/7xx(24/pi+18/pi)xxsqrt(15082)/11`
`= 22/7 xx 24/pixxsqrt(15082)/11`
≈ 42 × 11.164436
≈ 468.91 cm2
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