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 $9$ and the height of the cylinder is $12$ . If the volume of the solid is $24 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-04-03T08:41:51

$8/3 pi$

Consider the solid

enter image source here

We can say that the volume of the solid equals the sum of the volume of the cone and the cylinder

$color(blue)(V_(cy)+V_(co)=24pi$

Volume of cylinder $=color(purple)(pir^2h$

Volume of cone $=color(indigo)(1/3pir^2h$
( $1/3$ of the volume of cylinder)

Area of the base (the base is a circle) $=color(orange)(pir^2$

$n$ $ot$ $e$ $:pi=22/7$

If you take a closer look at the formula, you could see $pir^2$ appearing in both. So,let $pir^2$ be $w$

$color(orange)(pir^2=w$

$rarrwh+1/3wh=24pi$

$rarrw*12+1/3*w*9=24pi$

$rarrw*12+1/cancel3^1*w*cancel9^3=24pi$

$rarr12w+3w=25pi$

$rarr15w=24pi$

$color(green)(rArrw=(24pi)/15=8/3pi$

(area of the base)

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 9 9 and the height of the cylinder is 12 12 . If the volume of the solid is 24 pi24 pi, what is the area of the base of the cylinder?