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 #18 # and the height of the cylinder is #36 #. If the volume of the solid is #420 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 #18  # and the height of the cylinder is #36   #.  If the volume of the solid is #420  pi#, what is the area of the base of the cylinder?

Evelyn Arboleda · Accepted Answer

#10pi# Volume of cone#=pir^2h/3# Volume of cylinder#=pi r^2h# #:.pir^2h/3+pir^2h=420pi# #:.pir^2*18/3+pir^2*36=420pi# #:.pir^2(6+36)=420pi# #:.42pir^2=420pi# #:.42r^2=420# #:.r^2=10# #:.r=sqrt10# Area of a circle#=pir^2# #:.=pi(sqrt10)^2# #:.=10pi#

Autumn Armstrong · Answer

undefined The volume of the solid is given by the sum of the volumes of the cone and the cylinder. Since the radius of the cylinder is equal to the radius of the cone, let's denote it as $ r $. The volume of the cone is $\frac{1}{3}\pi r^2 h $, where $ h $ is the height of the cone, and the volume of the cylinder is $\pi r^2 h $, where $ h $ is the height of the cylinder. Given that the volume of the solid is $ 420\pi $, we have the equation $\frac{1}{3}\pi r^2 \cdot 18 + \pi r^2 \cdot 36 = 420\pi $. Solving this equation for $ r $, we find $ r = 3 $. Therefore, the area of the base of the cylinder is $ \pi r^2 = \pi \cdot 3^2 = 9\pi $.