Question

directions: for each of the following questions, show or explain your work to receive full credit. no work = no credit!
problem 1:
a cylinder and cone have the same height and radius. the height of each is 5 cm, and the radius is 2 cm. calculate the volume of the cylinder and the cone.

problem 2:
the volume of this cone is 36π cubic units.
what is the volume of a cylinder that has the same base area and the same height?

problem 3:
a cone - shaped popcorn cup has a radius of 5 centimeters and a height of 9 centimeters. how many cubic centimeters of popcorn can the cup hold? use 3.14 as an approximation for pi, and give a numerical answer.

Explanation:

Problem 1

Step 1: Recall the formula for the volume of a cylinder

The formula for the volume of a cylinder is \( V_{cylinder} = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. Given \( r = 2 \, \text{cm} \) and \( h = 5 \, \text{cm} \).
\[
V_{cylinder} = \pi \times (2)^2 \times 5
\]
\[
V_{cylinder} = \pi \times 4 \times 5
\]
\[
V_{cylinder} = 20\pi \, \text{cubic centimeters} \approx 20 \times 3.14 = 62.8 \, \text{cubic centimeters}
\]

Step 2: Recall the formula for the volume of a cone

The formula for the volume of a cone is \( V_{cone} = \frac{1}{3} \pi r^2 h \), with the same \( r = 2 \, \text{cm} \) and \( h = 5 \, \text{cm} \).
\[
V_{cone} = \frac{1}{3} \pi \times (2)^2 \times 5
\]
\[
V_{cone} = \frac{1}{3} \pi \times 4 \times 5
\]
\[
V_{cone} = \frac{20}{3}\pi \, \text{cubic centimeters} \approx \frac{20}{3} \times 3.14 \approx 20.93 \, \text{cubic centimeters}
\]

Step 1: Recall the relationship between the volume of a cone and a cylinder with the same base and height

The volume of a cone with the same base area (\( B \)) and height (\( h \)) as a cylinder is \( \frac{1}{3} \) of the volume of the cylinder. That is, \( V_{cone} = \frac{1}{3} V_{cylinder} \), so \( V_{cylinder} = 3 V_{cone} \).

Step 2: Calculate the volume of the cylinder

Given \( V_{cone} = 36\pi \) cubic units.
\[
V_{cylinder} = 3 \times 36\pi
\]
\[
V_{cylinder} = 108\pi \, \text{cubic units}
\]

Step 1: Recall the formula for the volume of a cone

The formula for the volume of a cone is \( V = \frac{1}{3} \pi r^2 h \), where \( r = 5 \, \text{cm} \), \( h = 9 \, \text{cm} \), and \( \pi \approx 3.14 \).
\[
V = \frac{1}{3} \times 3.14 \times (5)^2 \times 9
\]

Step 2: Simplify the expression

First, calculate \( (5)^2 = 25 \).
\[
V = \frac{1}{3} \times 3.14 \times 25 \times 9
\]
The \( \frac{1}{3} \) and \( 9 \) can be simplified: \( \frac{1}{3} \times 9 = 3 \).
\[
V = 3.14 \times 25 \times 3
\]
\[
V = 3.14 \times 75
\]
\[
V = 235.5 \, \text{cubic centimeters}
\]

Answer:

Volume of the cylinder: \( 20\pi \, \text{cm}^3 \) (or \( 62.8 \, \text{cm}^3 \)); Volume of the cone: \( \frac{20}{3}\pi \, \text{cm}^3 \) (or approximately \( 20.93 \, \text{cm}^3 \))

Problem 2