# A solid consists of a cone on top of a cylinder with a radius equal to that of the cone. The height of the cone is 9  and the height of the cylinder is 12 . If the volume of the solid is 48 pi, what is the area of the base of the cylinder?

Apr 15, 2016

$3.2 \pi$

#### Explanation:

Volume of a right circular cone is given by the expression
${V}_{\text{cone}} = \pi {r}^{2} \frac{h}{3}$, where $\pi {r}^{2}$ is the area of the base and $h$ is the vertical height.
Also that Volume of a cylinder ${V}_{\text{cylinder}} = \pi {r}^{2} h$, where $\pi {r}^{2}$ is the area of the base and $h$ is cylinder's height.

Volume of the given solid $= {V}_{\text{cone"+V_"cylinder}}$
$= \frac{1}{3} \text{area of the base of cone"xx"height of cone"+"area of the base"xx"height of cylinder}$
Inserting given values

$48 \pi = \frac{1}{3} \text{area of the base of cone"xx9+"area of the base of cylinder} \times 12$

Since, radius of both cone and cylinder are equal, rearranging and solving for area of the base of cylinder
$15 \times \text{area of the base of cylinder} = 48 \pi$
$\text{Area of the base of cylinder} = {\cancel{48}}^{16} / {\cancel{15}}_{5} \pi$
$= 3.2 \pi$