# 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 12  and the height of the cylinder is 4 . If the volume of the solid is 16 pi, what is the area of the base of the cylinder?

$A = 2 \pi$

#### Explanation:

We have a solid consisting of a cone on top of a cylinder. We're given the total volume of the solid, the height of the cone and the height of the cylinder. What is the area of the base of the cylinder?

First, let's express the volume like this:

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

substituting in the volume equations for cone and cylinder:

${V}_{\text{solid}} = \pi {r}^{2} \frac{h}{3} + \pi {r}^{2} h$

now let's drop in what we know:

$16 \pi = \pi {r}^{2} \left(\frac{12}{3}\right) + \pi {r}^{2} \left(4\right)$

We're looking for the area of the base of the cylinder. That equation is:

$A = \pi {r}^{2}$

Since we're given that the radii of the cylinder and the cone at the base are equal, we can therefore substitute:

$16 \pi = A \left(\frac{12}{3}\right) + A \left(4\right)$

Now let's solve for A:

$16 \pi = A \left(4\right) + A \left(4\right)$

$8 A = 16 \pi$

$A = 2 \pi$