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 $12$ and the height of the cylinder is $18$ . If the volume of the solid is $72 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-09-07T14:02:58

The area of the base is $36/11pi$ .

Start by drawing a diagram.

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The formula for volume of a cone is $V= 1/3r^2hpi$ and the formula for volume of a cylinder is $V = pir^2h$ .

Let $V_t$ denote the total volume.

$V_t = V_"cylinder" + V_"cone"$

$72pi = pir^2h + 1/3r^2hpi$

$72pi = 18pir^2 + 12(1/3r^2pi)$

$72pi = 18pir^2 + 4r^2pi$

$72pi = 22pir^2$

$36/11 = r^2$

$r = sqrt(36/11)$

$r = 6/sqrt(11)$

The formula for area of a circle is $A = pir^2$ , so the area of the base is $A = (6/sqrt(11))^2pi = 36/11pi$ .

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 12 12 and the height of the cylinder is 18 18 . If the volume of the solid is 72 pi72 pi, what is the area of the base of the cylinder?