# 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 144 pi, what is the area of the base of the cylinder?

Jul 28, 2016

${A}_{\text{base}} = 6 \pi$

#### Explanation:

The volume of each component is given by

${V}_{\text{cone") = 1/3pir_("cone")^2h_("cone}}$

${V}_{\text{cylinder") = pir_("cylinder")^2h_("cylinder}}$

We have that ${r}_{\text{cone") = r_("cylinder}}$ so we shall just denote these as $r$.

${V}_{\text{total") = V_("cone") + V_("cylinder}}$

V_("total") = pir^2(1/3h_("cone") + h_("cylinder"))

$\therefore \pi {r}^{2} = \left({V}_{\text{total"))/(1/3h_("cone") + h_("cylinder}}\right)$

Notice that $\pi {r}^{2}$ is precisely the area of the base of the cylinder, which is what we want to calculate so just plug in the numbers:

${A}_{b a s e} = \frac{144 \pi}{\frac{1}{3} \cdot 42 + 10} = \frac{144 \pi}{14 + 10} = \frac{144 \pi}{24} = 6 \pi$