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 #42 # and the height of the cylinder is #10 #. If the volume of the solid is #144 pi#, what is the area of the base of the cylinder?

Question
Answer 1

#A_("base") = 6pi#

#V_("cone") = 1/3pir_("cone")^2h_("cone")#

#V_("cylinder") = pir_("cylinder")^2h_("cylinder")#

We have that #r_("cone") = r_("cylinder")# so we shall just denote these as #r#.

#V_("total") = V_("cone") + V_("cylinder") #

#V_("total") = pir^2(1/3h_("cone") + h_("cylinder"))#

#therefore pir^2 = (V_("total"))/(1/3h_("cone") + h_("cylinder"))#

Notice that #pir^2# is precisely the area of the base of the cylinder, which is what we want to calculate so just plug in the numbers:

#A_(base) = (144pi)/(1/3*42 + 10) = (144pi)/(14+10) = (144pi)/(24) = 6pi#

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