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 $18$ $18$ and the height of the cylinder is $21$ $21$ . If the volume of the solid is $420 π$ $420 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2017-01-12T09:16:02

The area is $= 48.9 u^{2}$ $=48.9u^2$

The volume of the cylinder $=$ $=$ area of base $\cdot$ $*$ height

The volume of the cone $= \frac{1}{3} \cdot$ $=1/3*$ area of base $\cdot$ $*$ height

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

Volume of the cone $= \frac{1}{3} \cdot a \cdot 18 = 6 a$ $=1/3*a*18=6a$

Volume of cylinder $= a \cdot 21 = 21 a$ $=a*21=21a$

Total $= 6 a + 21 a = 27 a = 420 π$ $=6a+21a=27a=420pi$

$a = \frac{420}{27} π = 48.9 u^{2}$ $a=420/27pi=48.9 u^2$

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 18 1818 and the height of the cylinder is 21 2121 . If the volume of the solid is 420 pi420π420 pi, what is the area of the base of the cylinder?