Question

A tent is made in the form of a frustum of a cone surmounted by another cone. The diameters of the base and the top of the frustum are 20 m and 6 m, respectively, and the height is 24 m. If the height of the tent is 28 m and the radius of the conical part is equal to the radius of the top of the frustum, find the quantity of canvas required.

Answer 1

For the lower portion of the tent:

Diameter of the base= 20 m

Radius, R, of the base = 10 m

Diameter of the top end of the frustum = 6 m

Radius of the top end of the frustum = r = 3 m

Height of the frustum = h = 24 m

Slant height = l 

`= sqrt("h"^2 + ("R" - r)^2)`

`= sqrt(24^2 + (10-3)^2)`

`=sqrt(576+49)`

`=sqrt(625) = 25 "m"` 

For the conical part:
Radius of the cone's base = r = 3 m
Height of the cone = Total height - Height of the frustum = 28-24 = 4 m 

slant height, L, of the cone =`sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5  "m"`

Total quantity of canvas = Curved surface area of the frustum + curved surface area of the conical top

=(πl(R+r)) + πLr

= π(l(R + r)) + Lr)

`=22/7 (25xx13+5xx3)`

`=22/7 (325 + 15)= 1068.57  "m"^2`