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 #9 # and the height of the cylinder is #6 #. If the volume of the solid is #45 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 #9  # and the height of the cylinder is #6   #.  If the volume of the solid is #45 pi#, what is the area of the base of the cylinder?

Aurora Ayala · Accepted Answer

Volume of a cone is given by #V = 1/3pir^2h# Volume of a cylinder is given by #V = pir^2h# #1/3pi(r)^2(9) + r^2(6)(pi) = 45pi# #3r^2pi + 6r^2pi = 45pi# #9r^2pi = 45pi# #9r^2 = (45pi)/pi# #9r^2 = 45# #r^2 = 5# #r = sqrt(5)# The area of a circle is given by #r^2pi# #(sqrt(5))^2pi# =#5pi# The area of the base of the cylinder is #5pi# I hope this is useful!

Ella Armstrong · Answer

undefined Let's denote the radius of the cone $ r $ and the height of the cone $ h_c $. Since the radius of the cylinder is equal to that of the cone, the radius of the cylinder is also $ r $, and the height of the cylinder is denoted as $ h_{cy} $. Given that the height of the cone is 9 units and the height of the cylinder is 6 units, and the volume of the solid is $ 45\pi $, we can set up the equation for the volume of the solid:

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

$$ 45\pi = \frac{1}{3}\pi r^2 h_c + \pi r^2 h_{cy} $$

Given that $ h_c = 9 $ and $ h_{cy} = 6 $, we can substitute these values into the equation:

$$ 45\pi = \frac{1}{3}\pi r^2 \times 9 + \pi r^2 \times 6 $$

$$ 45\pi = 3\pi r^2 + 6\pi r^2 $$

$$ 45\pi = 9\pi r^2 $$

$$ r^2 = \frac{45\pi}{9\pi} $$

$$ r^2 = 5 $$

$$ r = \sqrt{5} $$

Now, we can find the area of the base of the cylinder:

$$ A_{\text{cylinder}} = \pi r^2 $$

$$ A_{\text{cylinder}} = \pi (\sqrt{5})^2 $$

$$ A_{\text{cylinder}} = \pi \times 5 $$

$$ A_{\text{cylinder}} = 5\pi $$

So, the area of the base of the cylinder is $ 5\pi $.