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 #90 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 #90 pi#, what is the area of the base of the cylinder?

Theodore Abad · Accepted Answer

The area of the base is #=31.4u^2#

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)#

#90pi=a(1/3*21+2)#

#90pi=a*9#

#a=90/9pi=10pi#

#a=31.4#

Caroline Aucoin · Answer

undefined Let's denote the radius of both the cone and the cylinder as $ r $. Given that the height of the cone is 21 and the height of the cylinder is 2, we can set up the volume equation for the solid:

$$
\text{Volume of cone} + \text{Volume of cylinder} = 90\pi
$$

The volume of a cone is $\frac{1}{3}\pi r^2 h$, where $ r $ is the radius and $ h $ is the height. Similarly, the volume of a cylinder is $\pi r^2 h$. Substituting the given values:

$$
\frac{1}{3}\pi r^2 \times 21 + \pi r^2 \times 2 = 90\pi
$$

Simplifying and solving for $ r $, we get:

$$
7\pi r^2 + 2\pi r^2 = 90\pi
$$
$$
9\pi r^2 = 90\pi
$$
$$
r^2 = \frac{90\pi}{9\pi}
$$
$$
r^2 = 10
$$

Taking the square root of both sides to find $ r $:

$$
r = \sqrt{10}
$$

The area of the base of the cylinder is given by $ \pi r^2 $, so substituting $ r = \sqrt{10} $:

$$
\text{Area of base} = \pi (\sqrt{10})^2 = 10\pi
$$

Therefore, the area of the base of the cylinder is $ 10\pi $ square units.