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

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Answer 1 · 2016-10-09T10:16:11

$pir^2=SA=(30pi)/13$

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:

$h_N=6$

$h_L=8$

$r_L=r_N$ and so I can refer to both as simply $r$

$V_N+V_L=20pi$

What is the area of the base of the cylinder?

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

$V_N=pir^2h$

$V_L=(pir^2h)/3$

$pir^2h_N+(pir^2h_L)/3=20pi$ and now substitute in values:

$pir^2(6)+(pir^2(8))/3=20pi$

$(pir^2(18))/3+(pir^2(8))/3=20pi$

$(pir^2(26))/3=20pi$

$(r^2(26))/3=20$

$r^2(26)=60$

$r^2=60/26=30/13$

The area of the base is $SA=pir^2$ and so:

$pir^2=SA=(30pi)/13$

A solid consists of a cone on top of a cylinder with a radius equal to the cone. The height of the cone is 6 6 and the height of the cylinder is 8 8 . If the volume of the solid is 20 pi20 pi, what is the area of the base of cylinder?