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 $1$ $1$ . If the volume of the solid is $480 π$ $480 pi$ , what is the area of the base of the cylinder?

Question

1162 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-20T12:36:41

The area of the base $= 32 π = 100.5 u^{2}$ $=32pi=100.5u^2$

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

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

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

$480 π = a (\frac{1}{3} \cdot 42 + 1)$ $480pi=a(1/3*42+1)$

$480 π = a \cdot 15$ $480pi=a*15$

$a = \frac{480}{15} π = 32 π$ $a=480/15pi=32pi$

$a = 100.5$ $a=100.5$

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 1 11 . If the volume of the solid is 480 pi480π480 pi, what is the area of the base of the cylinder?