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!