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 $5$ $5$ and the height of the cylinder is $12$ $12$ . If the volume of the solid is $36 π$ $36 pi$ , what is the area of the base of the cylinder?

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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 $5$ $5$ and the height of the cylinder is $12$ $12$ . If the volume of the solid is $36 π$ $36 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-04-23T18:01:35

I found $8.3$ $8.3$ units of area.

Explanation:

The volume of the cone is:
$V_{cone} = \frac{1}{3} π r^{2} h_{1}$ $V_"cone"=1/3pir^2h_1$
The volume of th cylinder is:
$V_{cylimder} = π r^{2} h_{2}$ $V_"cylimder"=pir^2h_2$
where the "hs" are the heights of the two solids:
We can write:
$36 π = V_{cone} + V_{cylimder}$ $36pi=V_"cone"+V_"cylimder"$
i.e.
$36 π = \frac{1}{3} π r^{2} h_{1} + π r^{2} h_{2}$ $36pi=1/3pir^2h_1+pir^2h_2$
$36 = \frac{1}{3} r^{2} \cdot 5 + r^{2} \cdot 12$ $36=1/3r^2*5+r^2*12$
$r^{2} [\frac{5}{3} + 12] = 36$ $r^2[5/3+12]=36$
$r^{2} [\frac{41}{3}] = 36$ $r^2[41/3]=36$
$r = \sqrt{\frac{3}{41} \cdot 36} = 1.6$ $r=sqrt(3/41*36)=1.6$
and the area of the base will be:
$A = π r^{2} = 8.3$ $A=pir^2=8.3$ u.a.

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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 $5$ $5$ and the height of the cylinder is $12$ $12$ . If the volume of the solid is $36 π$ $36 pi$ , what is the area of the base of the cylinder?

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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 $5$ $5$ and the height of the cylinder is $12$ $12$ . If the volume of the solid is $36 π$ $36 pi$ , what is the area of the base of the cylinder?