Question
How many cubic meters of dirt are there in a pile, conical in shape, 9 m in diameter, and 4 m high?
\
There are approximately
cubic meters of dirt in the pile.
\
There are approximately
cubic meters of dirt in the pile.
Answers
GPT-4o mini
To calculate the volume of a conical pile of dirt, we can use the formula for the volume \( V \) of a cone:
\[
V = \frac{1}{3} \pi r^2 h
\]
Where:
- \( r \) is the radius of the base of the cone,
- \( h \) is the height of the cone.
Given:
- The diameter of the pile is 9 m, so the radius \( r \) is half of the diameter:
\[
r = \frac{9 \, \text{m}}{2} = 4.5 \, \text{m}
\]
- The height \( h \) of the pile is 4 m.
Now we can substitute the values into the volume formula:
\[
V = \frac{1}{3} \pi (4.5 \, \text{m})^2 (4 \, \text{m})
\]
Calculating \( (4.5)^2 \):
\[
(4.5)^2 = 20.25
\]
Now substituting into the volume formula:
\[
V = \frac{1}{3} \pi (20.25) (4)
\]
Calculating \( 20.25 \times 4 \):
\[
20.25 \times 4 = 81
\]
Now substituting back into the volume formula:
\[
V = \frac{1}{3} \pi (81)
\]
And calculating \( \frac{1}{3} \times 81 \):
\[
\frac{81}{3} = 27
\]
Therefore:
\[
V = 27\pi
\]
Now, using \( \pi \approx 3.14 \):
\[
V \approx 27 \times 3.14 \approx 84.78 \, \text{cubic meters}
\]
Thus, there are approximately **84.78 cubic meters** of dirt in the pile.
\[
V = \frac{1}{3} \pi r^2 h
\]
Where:
- \( r \) is the radius of the base of the cone,
- \( h \) is the height of the cone.
Given:
- The diameter of the pile is 9 m, so the radius \( r \) is half of the diameter:
\[
r = \frac{9 \, \text{m}}{2} = 4.5 \, \text{m}
\]
- The height \( h \) of the pile is 4 m.
Now we can substitute the values into the volume formula:
\[
V = \frac{1}{3} \pi (4.5 \, \text{m})^2 (4 \, \text{m})
\]
Calculating \( (4.5)^2 \):
\[
(4.5)^2 = 20.25
\]
Now substituting into the volume formula:
\[
V = \frac{1}{3} \pi (20.25) (4)
\]
Calculating \( 20.25 \times 4 \):
\[
20.25 \times 4 = 81
\]
Now substituting back into the volume formula:
\[
V = \frac{1}{3} \pi (81)
\]
And calculating \( \frac{1}{3} \times 81 \):
\[
\frac{81}{3} = 27
\]
Therefore:
\[
V = 27\pi
\]
Now, using \( \pi \approx 3.14 \):
\[
V \approx 27 \times 3.14 \approx 84.78 \, \text{cubic meters}
\]
Thus, there are approximately **84.78 cubic meters** of dirt in the pile.