A solid consists of a cone on top of a cylinder with a radius equal to the cone. The height of the cone is $6$ $6$ and the height of the cylinder is $8$ $8$ . If the volume of the solid is $20 π$ $20 pi$ , what is the area of the base of cylinder?

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Answer 1 · 2016-10-09T10:16:11

$π r^{2} = S A = \frac{30 π}{13}$ $pir^2=SA=(30pi)/13$

We have a Cone and a Cylinder. For ease of keeping track, I'm going to use N to indicate Cone and L to indicate Cylinder.

We know:

$h_{N} = 6$ $h_N=6$

$h_{L} = 8$ $h_L=8$

$r_{L} = r_{N}$ $r_L=r_N$ and so I can refer to both as simply $r$ $r$

$V_{N} + V_{L} = 20 π$ $V_N+V_L=20pi$

What is the area of the base of the cylinder?

Let's first substitute in the formulas for the volumes of the two shapes:

$V_{N} = π r^{2} h$ $V_N=pir^2h$

$V_{L} = \frac{π r^{2} h}{3}$ $V_L=(pir^2h)/3$

$π r^{2} h_{N} + \frac{π r^{2} h_{L}}{3} = 20 π$ $pir^2h_N+(pir^2h_L)/3=20pi$ and now substitute in values:

$π r^{2} (6) + \frac{π r^{2} (8)}{3} = 20 π$ $pir^2(6)+(pir^2(8))/3=20pi$

$\frac{π r^{2} (18)}{3} + \frac{π r^{2} (8)}{3} = 20 π$ $(pir^2(18))/3+(pir^2(8))/3=20pi$

$\frac{π r^{2} (26)}{3} = 20 π$ $(pir^2(26))/3=20pi$

$\frac{r^{2} (26)}{3} = 20$ $(r^2(26))/3=20$

$r^{2} (26) = 60$ $r^2(26)=60$

$r^{2} = \frac{60}{26} = \frac{30}{13}$ $r^2=60/26=30/13$

The area of the base is $S A = π r^{2}$ $SA=pir^2$ and so:

$π r^{2} = S A = \frac{30 π}{13}$ $pir^2=SA=(30pi)/13$

A solid consists of a cone on top of a cylinder with a radius equal to the cone. The height of the cone is 66 and the height of the cylinder is 88 . If the volume of the solid is 20π20 pi, what is the area of the base of cylinder?