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 $42$ and the height of the cylinder is $10$ . If the volume of the solid is $144 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-07-28T18:17:09

$A_("base") = 6pi$

The volume of each component is given by

$V_("cone") = 1/3pir_("cone")^2h_("cone")$

$V_("cylinder") = pir_("cylinder")^2h_("cylinder")$

We have that $r_("cone") = r_("cylinder")$ so we shall just denote these as $r$ .

$V_("total") = V_("cone") + V_("cylinder")$

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

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

Notice that $pir^2$ is precisely the area of the base of the cylinder, which is what we want to calculate so just plug in the numbers:

$A_(base) = (144pi)/(1/3*42 + 10) = (144pi)/(14+10) = (144pi)/(24) = 6pi$

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 42 42 and the height of the cylinder is 10 10 . If the volume of the solid is 144 pi144 pi, what is the area of the base of the cylinder?