**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 = *r *= 3 m

Height of the frustum = *h* = 24 m

Slant height =* l*

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

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

$=\sqrt{576+49}$

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

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

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

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

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

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