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 #18 # and the height of the cylinder is #36 #. If the volume of the solid is #420 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 #18  # and the height of the cylinder is #36   #.  If the volume of the solid is #420  pi#, what is the area of the base of the cylinder?

Noah Atkinson · Accepted Answer

#color(green)("radius of the cone " = r = sqrt 10, " units"#

#"Given " h_1 = 18, h_2 = 36, V = 420 pi#

#"Volume of the solid " = V = V_(cone) + V_(cyl)#

#V_(cone) = 1/3 pi r^2 h = 1/3 pi r^2 18 = 6 pi r^2#

#V_(cyl) = pi r^2 h = pi r^2 36 = 36 pi r^2#

#V = 6 pi r^2 + 36 pi r^2 = 420 pi#

#42 pi r^2 = 420 pi#

#r^2 = cancel(420 pi)^color(red)(10) / cancel(42 pi) = 10#

#color(green)("radius of the cone " = r = sqrt 10, " units"#

Allison Allen · Answer

undefined The volume of the solid is the sum of the volumes of the cone and the cylinder. The volume of a cone is given by $V_{\text{cone}} = \frac{1}{3}\pi r^2 h$, where $r$ is the radius and $h$ is the height. The volume of a cylinder is given by $V_{\text{cylinder}} = \pi r^2 h$.

Given that the radius of the cone and the cylinder are equal, let's denote the radius as $r$. The height of the cone is 18, and the height of the cylinder is 36. The total volume of the solid is 420$\pi$.

Thus, we have:

$$420\pi = \frac{1}{3}\pi r^2 \times 18 + \pi r^2 \times 36$$

Simplify this equation to solve for $r$:

$$420\pi = 6\pi r^2 + 36\pi r^2$$
$$420\pi = 42\pi r^2$$
$$10 = r^2$$
$$r = \sqrt{10}$$

The area of the base of the cylinder is $\pi r^2$, so substituting the value of $r$ into this formula, we get:

$$A_{\text{base}} = \pi (\sqrt{10})^2$$
$$A_{\text{base}} = \pi \times 10$$
$$A_{\text{base}} = 10\pi$$

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