Question

The perimeters of the ends of a frustum of a right circular cone are $44cm$ and $33cm.$ If the height of the frustum is $16cm,$ find its volume, the slant surface, and the total surface.

Answer 1

Let $R_1$ and $R_2$ be the radius of the ends of the frustum of the cone.

So,

$2\pi R_1=44cm$

$\Rightarrow 2\times \frac{22}{7}\times R_1=44$

$\Rightarrow R_1=\frac{44\times 7}{44}$

$\Rightarrow R_1=7cm$

$2\pi R_2=33cm$

$\Rightarrow 2\times \frac{22}{7}\times R_2=33$

$\Rightarrow R_2=\frac{33\times 7}{44}$

$\Rightarrow R_2=5.25cm$

Now,

$\begin{array}{cc}Volume& =\frac{1}{3}\pi h\left[R_1^2+R_2^2+R_1{R}_2\right]\\ & =\frac{1}{3}\times \frac{22}{7}\times 16\left[7^2+{5.25}^2+7\times 5.25\right]\\ & =1899.33c{m}^3\end{array}$

Slant height of the frustum of the cone will be,

$\begin{array}{cc}l& =\sqrt{{\left(R_1-R_2\right)}^2+h^2}\\ & =\sqrt{{\left(7-5.25\right)}^2+{16}^2}\\ & =\sqrt{259.0625}\\ & =16.10cm\end{array}$

$\begin{array}{cc}CSA& =\pi \left(R_1+R_2\right)l\\ & =\frac{22}{7}\left(7+5.25\right)\times 16.10\\ & =619.85c{m}^2\end{array}$

$\begin{array}{cc}TSA& =CSA+\pi \left(R_1^2+R_2^2\right)\\ & =619.85+\frac{22}{7}\left(7^2+{5.25}^2\right)\\ & =860.475c{m}^2\end{array}$