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 $28$ . If the volume of the solid is $42 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-07-01T12:31:30

$(21pi)/16$

Assume the radius of the cylinder/cone as r, height of cone as $h_1$ , height of cylinder as $h_2$

Volume of the cone part of solid = $(pi*r^2*h_1)/3$

Volume of the cylinder part of solid = $pi*r^2 * h_2$

What we have is:

$h_1$ = 12, $h_2$ = 28

$(pi*r^2*h_1)/3$ + $pi*r^2 * h_2$ = $42*pi$

$(pi*r^2*12)/3$ + $pi*r^2 * 28$ = $42*pi$

$pi*r^2 * 4$ + $pi*r^2 * 28$ = $42*pi$

$32pi*r^2$ = $42*pi$

$r^2$ = $42/32$ = $21/16$

Area of the base of the cylinder = $pi*r^2$ = $pi*21/16$ = $(21pi)/16$

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 28 28 . If the volume of the solid is 42 pi42 pi, what is the area of the base of the cylinder?