Question
Find the volume of a cone with diameter of
18 feet and a height of
24 feet. Use 3.14 for π. C
Answers
GPT-4o mini
To find the volume of a cone, you can use the formula:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( V \) is the volume,
- \( r \) is the radius of the base,
- \( h \) is the height of the cone.
Given that the diameter of the cone is 18 feet, the radius \( r \) is half of the diameter:
\[
r = \frac{\text{diameter}}{2} = \frac{18}{2} = 9 \text{ feet}
\]
The height \( h \) is given as 24 feet. Now, substitute \( r \) and \( h \) into the volume formula:
\[
V = \frac{1}{3} \times 3.14 \times (9)^2 \times 24
\]
Calculate \( (9)^2 \):
\[
(9)^2 = 81
\]
Now plug in the values:
\[
V = \frac{1}{3} \times 3.14 \times 81 \times 24
\]
Multiply \( 81 \) and \( 24 \):
\[
81 \times 24 = 1944
\]
Now substitute back into the equation:
\[
V = \frac{1}{3} \times 3.14 \times 1944
\]
Now calculate \( 3.14 \times 1944 \):
\[
3.14 \times 1944 \approx 6109.76
\]
Now take one-third of that value:
\[
V = \frac{6109.76}{3} \approx 2036.59
\]
Therefore, the volume of the cone is approximately:
\[
\boxed{2036.59} \text{ cubic feet}
\]
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( V \) is the volume,
- \( r \) is the radius of the base,
- \( h \) is the height of the cone.
Given that the diameter of the cone is 18 feet, the radius \( r \) is half of the diameter:
\[
r = \frac{\text{diameter}}{2} = \frac{18}{2} = 9 \text{ feet}
\]
The height \( h \) is given as 24 feet. Now, substitute \( r \) and \( h \) into the volume formula:
\[
V = \frac{1}{3} \times 3.14 \times (9)^2 \times 24
\]
Calculate \( (9)^2 \):
\[
(9)^2 = 81
\]
Now plug in the values:
\[
V = \frac{1}{3} \times 3.14 \times 81 \times 24
\]
Multiply \( 81 \) and \( 24 \):
\[
81 \times 24 = 1944
\]
Now substitute back into the equation:
\[
V = \frac{1}{3} \times 3.14 \times 1944
\]
Now calculate \( 3.14 \times 1944 \):
\[
3.14 \times 1944 \approx 6109.76
\]
Now take one-third of that value:
\[
V = \frac{6109.76}{3} \approx 2036.59
\]
Therefore, the volume of the cone is approximately:
\[
\boxed{2036.59} \text{ cubic feet}
\]