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 #60 # and the height of the cylinder is #16 #. If the volume of the solid is #520 pi#, what is the area of the base of the cylinder?

Let ( r ) be the radius of both the cone and the cylinder.

The volume of the solid can be expressed as the sum of the volumes of the cone and the cylinder:

[ V = V_{\text{cone}} + V_{\text{cylinder}} ]

Given that the radius ( r ) is equal for both shapes, the volume formulas become:

[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}} ] [ V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} ]

Substituting the given values:

[ V = \frac{1}{3} \pi r^2 (60) + \pi r^2 (16) ]

Given that ( V = 520 \pi ), we can solve for ( r ):

[ 520 \pi = \frac{1}{3} \pi r^2 (60) + \pi r^2 (16) ] [ 520 = \frac{1}{3} r^2 (60) + 16r^2 ] [ 520 = 20r^2 + 16r^2 ] [ 520 = 36r^2 ] [ r^2 = \frac{520}{36} ] [ r^2 = \frac{130}{9} ] [ r = \sqrt{\frac{130}{9}} ]

Now, to find the area of the base of the cylinder, which is a circle, we use the formula for the area of a circle:

[ A_{\text{cylinder base}} = \pi r^2 ]

Substitute the value of ( r ):

[ A_{\text{cylinder base}} = \pi \left(\sqrt{\frac{130}{9}}\right)^2 ] [ A_{\text{cylinder base}} = \pi \left(\frac{130}{9}\right) ] [ A_{\text{cylinder base}} = \frac{130\pi}{9} ]

So, the area of the base of the cylinder is ( \frac{130\pi}{9} ).