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 #21 # and the height of the cylinder is #2 #. If the volume of the solid is #72 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 #21  # and the height of the cylinder is #2   #.  If the volume of the solid is #72 pi#, what is the area of the base of the cylinder?

Aaliyah Anderson · Accepted Answer

The area of the base #=8pi#

Let #a=# area of the base

Volume of cone is #V_(co)=1/3*a*h_(co)#

Volume of cylinder is #V_(cy)=a*h_(cy)#

Total amount

#V=V_(co)+V_(cy)#

#V=1/3ah_(co)+ah_(cy)#

#72pi=a(1/3*21+2)#

#72pi=a*9#

#a=72/9pi=8pi#

Ella Armstrong · Answer

undefined The volume $ V $ of a cone is given by the formula $ V = \frac{1}{3}\pi r^2h $, where $ r $ is the radius and $ h $ is the height.

The volume $ V_c $ of the cone with radius $ r $ and height $ 21 $ is $ V_c = \frac{1}{3}\pi r^2 \times 21 $.

The volume $ V_{cy} $ of the cylinder with radius $ r $ and height $ 2 $ is $ V_{cy} = \pi r^2 \times 2 $.

Given that the total volume of the solid is $ 72\pi $, we can set up the equation:

$$
V_c + V_{cy} = 72\pi
$$

$$
\frac{1}{3}\pi r^2 \times 21 + \pi r^2 \times 2 = 72\pi
$$

$$
7\pi r^2 + 2\pi r^2 = 72\pi
$$

$$
9\pi r^2 = 72\pi
$$

$$
r^2 = \frac{72\pi}{9\pi}
$$

$$
r^2 = 8
$$

$$
r = \sqrt{8}
$$

The area $ A $ of the base of the cylinder is given by $ A = \pi r^2 $.

Substituting $ r = \sqrt{8} $, we get:

$$
A = \pi (\sqrt{8})^2
$$

$$
A = 8\pi
$$

So, the area of the base of the cylinder is $ 8\pi $.