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#