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.

Answer 1

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