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 $18$ $18$ and the height of the cylinder is $1$ $1$ . If the volume of the solid is $84 π$ $84 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-06-27T02:59:17

$12 π$ $12pi$

Assume the radius of the cylinder/cone as r, height of cone as $h_{1}$ $h_1$ , height of cylinder as $h_{2}$ $h_2$

Volume of the cone part of solid = $\frac{π \cdot r^{2} \cdot h_{1}}{3}$ $(pi*r^2*h_1)/3$

Volume of the cylinder part of solid = $π \cdot r^{2} \cdot h_{2}$ $pi*r^2 * h_2$

What we have is:

$h_{1}$ $h_1$ = 18, $h_{2}$ $h_2$ = 1

$\frac{π \cdot r^{2} \cdot h_{1}}{3}$ $(pi*r^2*h_1)/3$ + $π \cdot r^{2} \cdot h_{2}$ $pi*r^2 * h_2$ = $84 \cdot π$ $84*pi$

$\frac{π \cdot r^{2} \cdot 18}{3}$ $(pi*r^2*18)/3$ + $π \cdot r^{2} \cdot 1$ $pi*r^2 * 1$ = $84 \cdot π$ $84*pi$

$π \cdot r^{2} \cdot 6$ $pi*r^2 * 6$ + $π \cdot r^{2} \cdot 1$ $pi*r^2 * 1$ = $84 \cdot π$ $84*pi$

$π \cdot r^{2} \cdot 7$ $pi*r^2 * 7$ = $84 \cdot π$ $84*pi$

$r^{2}$ $r^2$ = $\frac{84}{7}$ $84/7$ = 12

Area of the base of the cylinder = $π \cdot r^{2}$ $pi*r^2$ = $π \cdot 12$ $pi*12$ = $12 π$ $12pi$

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 18 1818 and the height of the cylinder is 1 11 . If the volume of the solid is 84 pi84π84 pi, what is the area of the base of the cylinder?