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 #33 # and the height of the cylinder is #13 #. If the volume of the solid is #256 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 #33  # and the height of the cylinder is #13   #.  If the volume of the solid is #256 pi#, what is the area of the base of the cylinder?

Allison Allen · Accepted Answer

#33.52# A Conical Volume is given by: #V = 1/3 * pi * r^2 * h# A Cylindrical Volume is given by: #V = pi * r^2 * h# Circular Area (base of cylinder) #A = 2*pi * r^2 # Total solid volume = # 256pi = 1/3 * pi * r^2 * h_1 + pi * r^2 * h_2 # Solve for r. # 256 = 1/3 * r^2 * 33 + r^2 * 13# ; #256 = r^2 * 11 + r^2 * 13# #256 = r^2 * 24# ; #256/24 = r^2# ; #r^2 = 10.67 ; # #A = pi * 10.76~~33.52#

Violet Aldrich · Answer

#10.76 pi#

Let's consider the diagram

We need to find the area of the base of the cylinder, which is a circle. The area of a circle is given by

#color(blue)("Area of circle"=pir^2#

Where #r# is the radius of the circle. We need to find #r^2#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

The total volume of the solid is #256pi#

Therefore,

#color(purple)("Volume of cone"+"Volume of cylinder"=256pi #

We use the formulas

#color(orange)("Volume of cone"=1/3pir^2#
#color(orange)("Volume of cylinder"=pir^2h#

Where, #h# is the height and #r# is the radius. Let's put everything in the equation,

#rarr1/3pir^2h+pir^2h=256pi#

#rarr1/3pir^2*33+pir^2* 13=256pi#

#rarr1/(cancel3^1)pir^2*cancel33^11+pir^2* 13=256pi#

#rarrpir^2*11+pir^2*13=256pi#

#rarr24pir^2=256pi#

#rarr24cancelpir^2=256cancelpi#

#rarr24r^2=256#

#rarrr^2=256/24#

#color(violet)(rArrr^2=10.56#

#~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~#

Now, Let's find the area of the base

#color(green)("Area"=pir^2=pi*10.76~~33.52#

Parker Allen · Answer

undefined To find the area of the base of the cylinder, we need to first determine the radius of both the cone and the cylinder.

Let $ r $ be the radius of the cone and the cylinder.

The volume of the solid can be expressed as the sum of the volume of the cone and the volume of the cylinder:

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

Given $ V = 256\pi $, $ h_{\text{cone}} = 33 $, and $ h_{\text{cylinder}} = 13 $, we can solve for $ r $.

$$
256\pi = \frac{1}{3} \pi r^2 \times 33 + \pi r^2 \times 13
$$

$$
256 = \frac{33}{3} r^2 + 13r^2
$$

$$
256 = 11r^2 + 13r^2
$$

$$
256 = 24r^2
$$

$$
r^2 = \frac{256}{24} = \frac{32}{3}
$$

$$
r = \sqrt{\frac{32}{3}} = \frac{4\sqrt{2}}{\sqrt{3}}
$$

The area of the base of the cylinder is given by $ A_{\text{cylinder}} = \pi r^2 $. Substituting the value of $ r $:

$$
A_{\text{cylinder}} = \pi \left(\frac{4\sqrt{2}}{\sqrt{3}}\right)^2
$$

$$
A_{\text{cylinder}} = \pi \times \frac{32}{3}
$$

$$
A_{\text{cylinder}} = \frac{32}{3}\pi
$$

Therefore, the area of the base of the cylinder is $ \frac{32}{3}\pi $.