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 $18$ and the height of the cylinder is $36$ . If the volume of the solid is $520 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-06-28T15:16:51

Here is a diagram.

enter image source here

First, let's identify the formulae for volume of a cone and volume of a cylinder.

$color(blue)(V_"(cone)" = 1/3pir^2h"$

$color(red)(V_"(cylinder)" = r^2pi xx h$

The total volume of the solid is given by adding the volumes of these two solids together. Therefore, we can state that:

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

Let's now identify the elements that we know:

•Total volume
•Height of the cylinder
•Height of the Cone

All that is left for us to find is the radius.

Hence,

$520pi = 36pir^2 + 1/3(18)pir^2$

$520pi= 36pir^2 + 6pir^2$

$520pi = 42pir^2$

$(520pi)/(42pi) = r^2$

$260/21 = r^2$

$sqrt(260/21) = r$

We can now find the area of the base of the cylinder, which because it's a circle, is given by $a = r^2 xx pi$ .

$a = (sqrt(260/21))^2 xx pi$

$a = (260pi)/21$

The area of the base of the cylinder is $(260pi)/21" u^2$ .

Hopefully this helps!

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 18 18 and the height of the cylinder is 36 36 . If the volume of the solid is 520 pi520 pi, what is the area of the base of the cylinder?