Question

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 `5` and the height of the cylinder is `12`. If the volume of the solid is `36 pi`, what is the area of the base of the cylinder?

Answer 1

I found `8.3` units of area.

Explanation:

The volume of the cone is:
`V_"cone"=1/3pir^2h_1`
The volume of th cylinder is:
`V_"cylimder"=pir^2h_2`
where the "hs" are the heights of the two solids:
We can write:
`36pi=V_"cone"+V_"cylimder"`
i.e.
`36pi=1/3pir^2h_1+pir^2h_2`
`36=1/3r^2*5+r^2*12`
`r^2[5/3+12]=36`
`r^2[41/3]=36`
`r=sqrt(3/41*36)=1.6`
and the area of the base will be:
`A=pir^2=8.3` u.a.