# 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 24 . If the volume of the solid is 42 pi, what is the area of the base of the cylinder?

##### 1 Answer
Aug 4, 2018

color(maroon)("Cylinder base area " = A = 3/2 pi " sq units"

#### Explanation:

$\text{Volume of cone } {V}_{c o \ne} = \frac{1}{3} \pi {r}^{2} {h}_{1}$

$\text{Volume of cylinder } {V}_{c y l} = \pi {r}^{2} {h}_{2}$

$\text{Volume of solid } V = \frac{1}{3} \pi {r}^{2} {h}_{1} + \pi {r}^{2} {h}_{2} = \pi {r}^{2} \left(\frac{1}{3} {h}_{1} + {h}_{2}\right)$

$\text{Area of cylinder base } = A = \pi {r}^{2}$

"Given " V = 42 pi, h_1 = 12, h_2 = 24, pi r^2 = ?

$\pi {r}^{2} \left(\frac{1}{3} {h}_{1} + {h}_{2}\right) = 42 \pi$

$\pi {r}^{2} = \frac{42 \pi}{\frac{1}{3} {h}_{1} + {h}_{2}}$

$A = \pi {r}^{2} = \frac{42 \pi}{\frac{1}{3} \cdot 12 + 24} = \frac{42 \pi}{28}$

color(maroon)(A = 3/2 pi " sq units"