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 `33` and the height of the cylinder is `14`. If the volume of the solid is `70 pi`, what is the area of the base of the cylinder?

Answer 1

Solve for the radius, then find the Area to be `16/5pi`

Explanation:

We're given a description of a solid consisting of a cone sitting on top of a cylinder. We're given the Volume of the solid, the height of the cone, the height of the cylinder, and the radius of the cone and cylinder are equal. We're asked for the Area of the base of the cylinder.

So let's put this altogether in one equation and start substituting in and expanding as we can.

`V_(solid)=V_(co)+V_(cy)` where co is for cone and cy is for cylinder.

Let's substitute in the equations for the volumes of the cube and cylinder:

`V_(solid)=1/3pir^2h+pir^2h` - keeping in mind that the first term is for the cone and the second for the cylinder.

We can substitute in what we know:

`70pi=1/3pir^2(33)+pir^2(14)`

`70pi=pir^2(11)+pir^2(14)`

So at this point, there is no term for the area of the cylinder! But what we can solve for (and incidentally, what we need to solve for) is the radius. So let's do that. We can divide by `pi`:

`70=r^2(11)+r^2(14)=25r^2`

Then we can divide by 25 and take the square root:

`r=sqrt(70/25)=sqrt(70)/5`

Not the easiest radius to deal with, but here goes:

The area of the base of the cylinder, `A_(bcy)=pir^2=pi*(sqrt(70)/5)^2=70/25pi=16/5pi`