Question

The ratio of the base area and the curved surface area of a conical tent is $40:41$. If its height is $18\text{m}$, find the air capacity of the tent in terms of $\pi .$

Answer 1

Let radius of the cone be $r$ and its slant height is equal to $l$.

Given $\Rightarrow$ $\frac{\text{base area}}{C.S.A}=\frac{40}{41}$

To Find $\Rightarrow$ Air capacity of the tent.

$\frac{\text{base area}}{C.S.A}=\frac{40}{41}$

$\Rightarrow \frac{\pi r^2}{\pi rl}=\frac{40}{41}$

$\Rightarrow \frac{r}{l}=\frac{40}{41}$

$\Rightarrow l=\frac{41r}{40}$

Also in a cone,

$\begin{array}{cc}{l}^2& =r^2+h^2\\ {\left(\frac{41}{40}\right)}^2{r}^2-r^2& =h^2\\ \frac{81}{1600}{r}^2& =h^2\\ r& =\frac{40}{9}\times h\\ & =\frac{40}{9}\times 18\\ & =80m\end{array}$

Volume of air present in the conical tent is equal to volume of the tent so,

$\begin{array}{cc}\text{Volume of air}& =\frac{1}{3}\pi r^{2}h\\ & =\frac{1}{3}\pi (80{)}^2\cdot 18\\ & =38400\pi {\text{m}}^3\end{array}$