# A solid consists of a cone on top of a cylinder with a radius equal to the cone. The height of the cone is 6  and the height of the cylinder is 8 . If the volume of the solid is 20 pi, what is the area of the base of cylinder?

$\pi {r}^{2} = S A = \frac{30 \pi}{13}$

#### Explanation:

We have a Cone and a Cylinder. For ease of keeping track, I'm going to use N to indicate Cone and L to indicate Cylinder.

We know:

${h}_{N} = 6$

${h}_{L} = 8$

${r}_{L} = {r}_{N}$ and so I can refer to both as simply $r$

${V}_{N} + {V}_{L} = 20 \pi$

What is the area of the base of the cylinder?

Let's first substitute in the formulas for the volumes of the two shapes:

${V}_{N} = \pi {r}^{2} h$

${V}_{L} = \frac{\pi {r}^{2} h}{3}$

$\pi {r}^{2} {h}_{N} + \frac{\pi {r}^{2} {h}_{L}}{3} = 20 \pi$ and now substitute in values:

$\pi {r}^{2} \left(6\right) + \frac{\pi {r}^{2} \left(8\right)}{3} = 20 \pi$

$\frac{\pi {r}^{2} \left(18\right)}{3} + \frac{\pi {r}^{2} \left(8\right)}{3} = 20 \pi$

$\frac{\pi {r}^{2} \left(26\right)}{3} = 20 \pi$

$\frac{{r}^{2} \left(26\right)}{3} = 20$

${r}^{2} \left(26\right) = 60$

${r}^{2} = \frac{60}{26} = \frac{30}{13}$

The area of the base is $S A = \pi {r}^{2}$ and so:

$\pi {r}^{2} = S A = \frac{30 \pi}{13}$