**Question:**

A tent consists of a frustum of a cone, surmounted by a cone. If the diameter of the upper and lower circular ends of the frustum be 14 m and 26 m, respectively, the height of the frustum be 8 m and the slant height of the surmounted conical portion be 12 m, find the area of the canvas required to make the tent. (Assume that the radii of the upper circular end of the frustum and the base of the surmounted conical portion are equal.)

**Solution:**

For the frustum:

Upper diameter = 14 m

Upper Radius, *r* = 7 m

Lower diameter = 26 m

Lower Radius, *R* = 13 m

Height of the frustum= *h* = 8 m

Slant height *l *=

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

$=\sqrt{8^{2}+(13-7)^{2}}$

$=\sqrt{64+36}$

$=\sqrt{100}=10 \mathrm{~m}$

For the conical part:

Radius of the base =* r* = 7 m

Slant height = *L* =12 m

Total surface area of the tent = Curved area of frustum + Curved area of the cone

$=\pi \mathrm{l}(\mathrm{R}+\mathrm{r})+\pi \mathrm{rL}$

$=\frac{22}{7} \times 10(13+7)+\frac{22}{7} \times 7 \times 12$

$=\frac{22}{7}(200+84)=892.57 \mathrm{~m}^{2}$