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 $27$ $27$ and the height of the cylinder is $7$ $7$ . If the volume of the solid is $42 π$ $42 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2017-02-03T14:19:19

The area is $= (\frac{21}{8} π) u^{2}$ $=(21/8pi) u^2$

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

Volume of cone is $V_{c o} = \frac{1}{3} \cdot a \cdot h_{c o}$ $V_(co)=1/3*a*h_(co)$

Volume of cylinder is $V_{c y} = a \cdot h_{c y}$ $V_(cy)=a*h_(cy)$

Total volume

$V = V_{c o} + V_{c y}$ $V=V_(co)+V_(cy)$

$V = \frac{1}{3} a h_{c o} + a h_{c y}$ $V=1/3ah_(co)+ah_(cy)$

$42 π = a (\frac{1}{3} \cdot 27 + 7)$ $42pi=a(1/3*27+7)$

$42 π = a \cdot 16$ $42pi=a*16$

$a = \frac{42}{16} π = \frac{21}{8} π$ $a=42/16pi=21/8pi$

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 2727 and the height of the cylinder is 77 . If the volume of the solid is 42π42 pi, what is the area of the base of the cylinder?