# 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 66  and the height of the cylinder is 5 . If the volume of the solid is 96 pi, what is the area of the base of the cylinder?

Mar 21, 2016

$\approx 11.2$

#### Explanation:

The volume of the cylinder is given by its height multiplied by the area of its circular base.

${V}_{\text{cylinder" = pi * r^2 * h_"cylinder}}$

• ${h}_{\text{cylinder}} = 5$ in this question.

The volume of a cone is given by a third of its height multiplied by the area of its circular base.

${V}_{\text{cone" = 1/3 * pi * r^2 * h_"cone}}$

• ${h}_{\text{cone}} = 66$ in this question.
• The variable $r$ is reused as the cone has the same radius as the cylinder.

The volume of the entire solid is

${V}_{\text{solid" = V_"cylinder" + V_"cone}}$

$= \pi \cdot {r}^{2} \cdot {h}_{\text{cylinder" + 1/3 * pi * r^2 * h_"cone}}$

$= \pi \cdot {r}^{2} \cdot \left({h}_{\text{cylinder" + 1/3 h_"cone}}\right)$

$= \pi \cdot {r}^{2} \cdot \left(5 + \frac{1}{3} \times 66\right)$

Now it becomes a simple matter to solve for the base area of the cylinder, which is just $\pi {r}^{2}$.

$\pi \cdot {r}^{2} = {V}_{\text{solid}} / \left(5 + \frac{1}{3} \times 66\right)$

$= \frac{96 \pi}{27}$

$= \frac{32 \pi}{9}$

$\approx 11.2$