# 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 42  and the height of the cylinder is 10 . If the volume of the solid is 200 pi, what is the area of the base of the cylinder?

Jul 10, 2016

The area of the base of the cylinder (which is a circle with radius $r$) $= \pi {r}^{2} = 25 \frac{\pi}{3} s q . u n i t$.

#### Explanation:

Suppose that the radius of cone $= r =$ that of cylinder.

Therefore, volume $V$ of the solid=volume of cone + that of cylinder

$= \frac{1}{3} \cdot \pi \cdot {r}^{2} \cdot$(height of cone)+$\pi \cdot {r}^{2} \cdot$(height of cyl.)

$= \frac{1}{3} \cdot \pi \cdot {r}^{2} \cdot 42 + \pi \cdot {r}^{2} \cdot \left(10\right)$

$= 14 \pi {r}^{2} + 10 \pi {r}^{2}$

$\therefore V = 24 \pi {r}^{2}$

But we are given that $V = 200 \pi .$

Therefore, $24 \pi {r}^{2} = 200 \pi \Rightarrow \pi {r}^{2} = 200 \frac{\pi}{24} = 25 \frac{\pi}{3.} \ldots \ldots \ldots \ldots \left(1\right)$

Hence, the area of the base of the cylinder (which is a circle with radius $r$) $= \pi {r}^{2} = 25 \frac{\pi}{3}$ sq.unit, by $\left(1\right) .$