# 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 39  and the height of the cylinder is 17 . If the volume of the solid is 90 pi, what is the area of the base of the cylinder?

##### 1 Answer
May 3, 2016

This question involves some basic algebra and really simple calculations. It has a very good feel to it as you work through it.

Area of the base = $3 \pi$

#### Explanation:

What do we already know?
The cylinder and the cone have the same radius, but we do not have the actual length of the radius.
We have the height of the cone and the cylinder.
We know the total volume of the shape.

Write down a word equation;
vol of cylinder + vol cone = Vol of shape

Now use the formulae: $\pi {r}^{2} h + \frac{\pi {r}^{2} H}{3} = 90 \pi$

Factorise (common factor): $\pi {r}^{2} \left(h + \frac{H}{3}\right) = 90 \pi$

Divide both sides by $\pi$ and substitute: ${r}^{2} \left(17 + \frac{39}{3}\right) = 90$

Simplify: ${r}^{2} \times 30 = 90$

This gives us: ${r}^{2}$ = $\frac{90}{30}$ = 3

We could calculate the radius, but it is not necessary, we need ${r}^{2}$ to find the area of the base which is a circle.

Area = $\pi {r}^{2} = 3 \pi$

Let's check: $\pi {r}^{2} h + \frac{\pi {r}^{2} H}{3}$

= $3 \pi \left(17\right) + \frac{3 \pi \left(39\right)}{3}$

=$51 \pi + 39 \pi$

=$90 \pi$