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 $6$ $6$ . If the volume of the solid is $130 π$ $130 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-12-15T06:15:57

The answer is $= 45.4$ $=45.4$

Let area of base $= a$ $=a$

The volume of the cone $= \frac{1}{3} \cdot a \cdot h$ $=1/3*a*h$

The volume of the cylinder $= a \cdot H$ $=a*H$

Total volume $V = \frac{1}{3} a h + a H = a (\frac{h}{3} + H)$ $V=1/3ah+aH=a(h/3+H)$

$V = 130 π$ $V=130pi$

height of cone $h = 9$ $h=9$

height of cylinder $H = 6$ $H=6$

So,

$a (\frac{9}{3} + 6) = 130 π$ $a(9/3+6)=130pi$

$9 a = 130 π$ $9a=130pi$

$a = \frac{130}{9} π = 45.4$ $a=130/9pi=45.4$

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 99 and the height of the cylinder is 6 66 . If the volume of the solid is 130 pi130π130 pi, what is the area of the base of the cylinder?