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 #12 # and the height of the cylinder is #28 #. If the volume of the solid is #42 pi#, what is the area of the base of the cylinder?

Question

1222 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-01T12:31:30

#(21pi)/16#

Assume the radius of the cylinder/cone as r, height of cone as #h_1#, height of cylinder as #h_2#

Volume of the cone part of solid = #(pi*r^2*h_1)/3#

Volume of the cylinder part of solid = # pi*r^2 * h_2#

What we have is:

#h_1# = 12,#h_2# = 28

#(pi*r^2*h_1)/3# + # pi*r^2 * h_2# = #42*pi#

#(pi*r^2*12)/3# + # pi*r^2 * 28# = #42*pi#

# pi*r^2 * 4# + # pi*r^2 * 28# = #42*pi#

# 32pi*r^2 # = #42*pi#

# r^2 # = #42/32# = #21/16#

Area of the base of the cylinder = #pi*r^2# = #pi*21/16# = #(21pi)/16#