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 $12$ $12$ . If the volume of the solid is $45 π$ $45 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2017-05-10T10:18:09

The area is $= 9.4 u^{2}$ $=9.4u^2$

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

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

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

$45 π = a (\frac{1}{3} \cdot 9 + 12)$ $45pi=a(1/3*9+12)$

$45 π = a \cdot 15$ $45pi=a*15$

$a = \frac{45}{15} π$ $a=45/15pi$

$a = 3 π$ $a=3pi$

$a = 9.4 u^{2}$ $a=9.4u^2$

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 99 and the height of the cylinder is 1212 . If the volume of the solid is 45π45 pi, what is the area of the base of the cylinder?