A tent is made in the form of a frustum of a cone surmounted by another cone.
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.

Solution:

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 = = 3 m

Height of the frustum = h = 24 m

Slant height = l

$=\sqrt{h^{2}+(R-r)^{2}}$

$=\sqrt{24^{2}+(10-3)^{2}}$

$=\sqrt{576+49}$

$=\sqrt{625}=25 \mathrm{~m}$

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 $=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5 \mathrm{~m}$

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

$=(\pi l(R+r))+\pi L r$

$=\pi(l(R+r)+L r)$

$=\frac{22}{7}(25 \times 13+5 \times 3)$

$=\frac{22}{7}(325+15)=1068.57 \mathrm{~m}^{2}$