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 $36 π$ $36 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2018-04-20T12:13:42

$\frac{9 π}{8}$ $color(blue)((9pi)/8)$

Volume of a cone is given by:

$V = \frac{1}{3} π r^{2} h$ $V=1/3pir^2h$

Volume of a cylinder is given by:

$V = π r^{2} h$ $V=pir^2h$

We know height of cone and cylinder:

$V = \frac{1}{3} π r^{2} (12)$ $V=1/3pir^2(12)$

$V = π r^{2} (28)$ $V=pir^2(28)$

The sum of these two volumes is $36 π$ $36pi$ given:

$:.$

$pir^2(28)+1/3pir^2(12)=36pi$

$28pir^2+4pir^2=36pi$

Factor out $r^2$ :

$r^2(28pi+4pi)=36pi$

$r^2(32pi)=36pi$

$r^2=(36pi)/(32pi)=9/8$

Area of cylinder base:

$A=pir^2=pi(9/8)=(9pi)/8$

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 1212 and the height of the cylinder is 28 2828 . If the volume of the solid is 36 pi36π36 pi, what is the area of the base of the cylinder?