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 $9$ and the height of the cylinder is $6$ . If the volume of the solid is $135 pi$ , what is the area of the base of the cylinder?

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Answer 1 · 2016-09-16T08:58:13

$15pi$

Cone volume $(1/3)pir^2(h_1) = (1/3)pir^2(9)=3pir^2$

Cylinder volume $=pir^2(h_2)=pir^2xx6=6pir^2$

solid volume = cone volume + cylinder volume

$=> 135pi = 3pir^2+6pir^2=9pir^2$

$=> r^2=135/9 =15$

cylinder base area $= pir^2 = 15pi$

Solution 2)

As the cone and the cylinder have the same radius, they have the same base area.

Let $A_(base)$ be the base area

Cone volume $1/3A_(base)h_1=1/3A_(base)xx9=3A_(base)$

Cylinder volume $A_(base)h_2=6A_(base)$

solid volume = cone volume +cylinder volume

$=> 135pi=3A_(base)+6A_(base)=9A_(base)$
$=> A_(base) = (135pi)/9=15pi$

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 9 9 and the height of the cylinder is 6 6 . If the volume of the solid is 135 pi135 pi, what is the area of the base of the cylinder?