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 #12 #. If the volume of the solid is #15 pi#, what is the area of the base of the cylinder?

Question

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 #12   #.  If the volume of the solid is #15 pi#, what is the area of the base of the cylinder?

James Atkins · Accepted Answer

#pi#

The base of a cylinder is a circle. The area of a circle is #pir^2#; however, we do not know the radius. To find #r#, we can use the information given in the problem.

The volume of a cone is #1/3pir^2h#, where #r# is the radius and #h# is the height. We know the height is #9#, so we can say

#V_"cone" = 1/3pir^2*9 = 3pir^2#

The volume of a cylinder is #pir^2h#, and we know the height is #12#.

#V_"cylinder" = pir^2*12 = 12pir^2#

We also know that the volume of the cone plus that of the cylinder is equal to #15pi#.

#V_"cone" + V_"cylinder" = 15pi#

#3pir^2 + 12pir^2 = 15pi#

#15pir^2 = 15pi#

Dividing by #15pi# on both sides, we get

#r^2 = 1#

#r=1#

Consequently, the cylinder's base area is

#pir^2 = pi* 1^2 = pi#

Caroline Aucoin · Answer

undefined Let $ r $ be the radius of the base of the cylinder (and also the radius of the cone). The volume of the cone is $ \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 \cdot 9 = 3 \pi r^2 $. The volume of the cylinder is $ \pi r^2 h = \pi r^2 \cdot 12 = 12 \pi r^2 $. The total volume of the solid is the sum of the volumes of the cone and the cylinder, which is $ 3 \pi r^2 + 12 \pi r^2 = 15 \pi r^2 $. Given that this is equal to $ 15 \pi $, we can solve for $ r^2 $.

$ 15 \pi r^2 = 15 \pi $ implies $ r^2 = 1 $. Thus, the area of the base of the cylinder is $ \pi r^2 = \pi \cdot 1 = \pi $.