Question

A solid consists of a cone on top of a cylinder with a radius equal to the cone. The height of the cone is `6` and the height of the cylinder is `8`. If the volume of the solid is `20 pi`, what is the area of the base of cylinder?

Answer 1

`pir^2=SA=(30pi)/13`

Explanation:

We have a Cone and a Cylinder. For ease of keeping track, I'm going to use N to indicate Cone and L to indicate Cylinder.

We know:

`h_N=6`

`h_L=8`

`r_L=r_N` and so I can refer to both as simply `r`

`V_N+V_L=20pi`

What is the area of the base of the cylinder?

Let's first substitute in the formulas for the volumes of the two shapes:

`V_N=pir^2h`

`V_L=(pir^2h)/3`

`pir^2h_N+(pir^2h_L)/3=20pi` and now substitute in values:

`pir^2(6)+(pir^2(8))/3=20pi`

`(pir^2(18))/3+(pir^2(8))/3=20pi`

`(pir^2(26))/3=20pi`

`(r^2(26))/3=20`

`r^2(26)=60`

`r^2=60/26=30/13`

The area of the base is `SA=pir^2` and so:

`pir^2=SA=(30pi)/13`