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 #39 # and the height of the cylinder is #17 #. If the volume of the solid is #60 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 #39  # and the height of the cylinder is #17   #.  If the volume of the solid is #60 pi#, what is the area of the base of the cylinder?

Sadie Allen · Accepted Answer

#2\pi \ ext{unit}^2# Ler #r# be the radius of common base of cylinder of height #17# & cone of height #39# then the total volume of the composite solid #\pi r^2(17)+1/3\pir^2(39)=60 \pi# #\pir^2(17+13)=60\pi# #\pi r^2=\frac{60\pi}{30}# #\pi r^2=2\pi# # ext{Area of base of cylinder}=2\pi \ ext{unit}^2#

Maverick Smith · Answer

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

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

Given that the radius of the cylinder is equal to the radius of the cone, let's denote it as $ r $. We have:

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

Given:
$$ h_{\text{cone}} = 39 $$
$$ h_{\text{cylinder}} = 17 $$
$$ V = 60 \pi $$

Substituting the given values into the formulas for the volumes of the cone and cylinder, we get:

$$ 60 \pi = \frac{1}{3} \pi r^2 \times 39 + \pi r^2 \times 17 $$

Solve this equation for $ r $, then use the value of $ r $ to find the area of the base of the cylinder using the formula:

$$ \text{Area of base of cylinder} = \pi r^2 $$