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

Apr 3, 2016

$\frac{8}{3} \pi$

#### Explanation:

Consider the solid

We can say that the volume of the solid equals the sum of the volume of the cone and the cylinder

color(blue)(V_(cy)+V_(co)=24pi

Volume of cylinder=color(purple)(pir^2h

Volume of cone=color(indigo)(1/3pir^2h
($\frac{1}{3}$ of the volume of cylinder)

Area of the base (the base is a circle)=color(orange)(pir^2

$n$$o t$$e$$: \pi = \frac{22}{7}$

If you take a closer look at the formula, you could see $\pi {r}^{2}$ appearing in both. So,let $\pi {r}^{2}$ be $w$

color(orange)(pir^2=w

$\rightarrow w h + \frac{1}{3} w h = 24 \pi$

$\rightarrow w \cdot 12 + \frac{1}{3} \cdot w \cdot 9 = 24 \pi$

$\rightarrow w \cdot 12 + \frac{1}{\cancel{3}} ^ 1 \cdot w \cdot {\cancel{9}}^{3} = 24 \pi$

$\rightarrow 12 w + 3 w = 25 \pi$

$\rightarrow 15 w = 24 \pi$

color(green)(rArrw=(24pi)/15=8/3pi

(area of the base)