# 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 6 . If the volume of the solid is 135 pi, what is the area of the base of the cylinder?

Sep 16, 2016

$15 \pi$

#### Explanation:

Cone volume $\left(\frac{1}{3}\right) \pi {r}^{2} \left({h}_{1}\right) = \left(\frac{1}{3}\right) \pi {r}^{2} \left(9\right) = 3 \pi {r}^{2}$

Cylinder volume $= \pi {r}^{2} \left({h}_{2}\right) = \pi {r}^{2} \times 6 = 6 \pi {r}^{2}$

solid volume = cone volume + cylinder volume

$\implies 135 \pi = 3 \pi {r}^{2} + 6 \pi {r}^{2} = 9 \pi {r}^{2}$

$\implies {r}^{2} = \frac{135}{9} = 15$

cylinder base area $= \pi {r}^{2} = 15 \pi$

Solution 2)

As the cone and the cylinder have the same radius, they have the same base area.

Let ${A}_{b a s e}$ be the base area

Cone volume $\frac{1}{3} {A}_{b a s e} {h}_{1} = \frac{1}{3} {A}_{b a s e} \times 9 = 3 {A}_{b a s e}$

Cylinder volume ${A}_{b a s e} {h}_{2} = 6 {A}_{b a s e}$

solid volume = cone volume +cylinder volume

$\implies 135 \pi = 3 {A}_{b a s e} + 6 {A}_{b a s e} = 9 {A}_{b a s e}$
$\implies {A}_{b a s e} = \frac{135 \pi}{9} = 15 \pi$