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 $42$ $42$ and the height of the cylinder is $6$ $6$ . If the volume of the solid is $320 π$ $320 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2017-01-24T07:09:31

The area is $= 16 π$ $=16pi$

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

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

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

$320 π = a (\frac{1}{3} \cdot 42 + 6)$ $320pi=a(1/3*42+6)$

$320 π = a \cdot 20$ $320pi=a*20$

$a = \frac{320}{20} π = 16 π$ $a=320/20pi=16pi$

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 42 4242 and the height of the cylinder is 6 66 . If the volume of the solid is 320 pi320π320 pi, what is the area of the base of the cylinder?