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 $39$ $39$ and the height of the cylinder is $17$ $17$ . If the volume of the solid is $60 π$ $60 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2018-07-28T02:30:28

$2\pi \ \text{unit}^2$

Ler $r$ be the radius of common base of cylinder of height $17$ & cone of height $39$ then the total volume of the composite solid

$\pi r^2(17)+1/3\pir^2(39)=60 \pi$

$\pir^2(17+13)=60\pi$

$\pi r^2=\frac{60\pi}{30}$

$\pi r^2=2\pi$

$\text{Area of base of cylinder}=2\pi \ \text{unit}^2$

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 39 3939 and the height of the cylinder is 17 1717 . If the volume of the solid is 60 pi60π60 pi, what is the area of the base of the cylinder?