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 #1 #. If the volume of the solid is #320 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 #42  # and the height of the cylinder is #1   #.  If the volume of the solid is #320 pi#, what is the area of the base of the cylinder?

Theodore Abad · Accepted Answer

#{64 pi}/3#

We use the formulas #V = 1/3 pi r^2 h# and #V = pi r^2 h# for the volumes of a cone and a cylinder, respectively. This gives us

#1/3 pi r^2 42 + pi r^2 1 = 320 pi#

or

#15 pi r^2= 320 pi#

So,

#r^2 = {320}/15 = 64/3#

and the area of the base of the cylinder is given by #pi r^2#

Aaliyah Anderson · Answer

undefined Let the radius of the cone and cylinder be $ r $.

The volume of the solid is the sum of the volumes of the cone and cylinder:

$$ V = V_{\text{cone}} + V_{\text{cylinder}} $$

Given that the height of the cone is 42 and its radius is $ r $, the volume of the cone is:

$$ V_{\text{cone}} = \frac{1}{3} \pi r^2 h_{\text{cone}} = \frac{1}{3} \pi r^2 \times 42 $$

Given that the height of the cylinder is 1, the volume of the cylinder is:

$$ V_{\text{cylinder}} = \pi r^2 h_{\text{cylinder}} = \pi r^2 \times 1 $$

Given that the total volume of the solid is 320π, we can write the equation as:

$$ 320\pi = \frac{1}{3} \pi r^2 \times 42 + \pi r^2 $$

Solving for $ r $ gives:

$$ 320\pi = 14\pi r^2 + \pi r^2 $$

$$ 320\pi = 15\pi r^2 $$

$$ r^2 = \frac{320\pi}{15\pi} $$

$$ r^2 = \frac{64}{3} $$

$$ r = \sqrt{\frac{64}{3}} $$

$$ r = \frac{8}{\sqrt{3}} $$

The area of the base of the cylinder is $ \pi r^2 = \pi \left(\frac{8}{\sqrt{3}}\right)^2 $.