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 $520 π$ $520 pi$ , 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.

$V_{(cone)} = \frac{1}{3} π r^{2} h$ $color(blue)(V_"(cone)" = 1/3pir^2h"$

$V_{(cylinder)} = r^{2} π \times h$ $color(red)(V_"(cylinder)" = r^2pi xx h$

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

$V_{total} = h_{(cylinder)} r^{2} π + \frac{1}{3} π r^{2} h_{(cone)}$ $V_"total" = h_"(cylinder)"r^2pi + 1/3pir^2h_"(cone)"$

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,

$520 π = 36 π r^{2} + \frac{1}{3} (18) π r^{2}$ $520pi = 36pir^2 + 1/3(18)pir^2$

$520 π = 36 π r^{2} + 6 π r^{2}$ $520pi= 36pir^2 + 6pir^2$

$520 π = 42 π r^{2}$ $520pi = 42pir^2$

$\frac{520 π}{42 π} = r^{2}$ $(520pi)/(42pi) = r^2$

$\frac{260}{21} = r^{2}$ $260/21 = r^2$

$\sqrt{\frac{260}{21}} = r$ $sqrt(260/21) = r$

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

$a = {(\sqrt{\frac{260}{21}})}^{2} \times π$ $a = (sqrt(260/21))^2 xx pi$

$a = \frac{260 π}{21}$ $a = (260pi)/21$

The area of the base of the cylinder is $\frac{260 π}{21} u^{2}$ $(260pi)/21" u^2$ .

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 1818 and the height of the cylinder is 3636 . If the volume of the solid is 520π520 pi, what is the area of the base of the cylinder?