Question
A cone-shaped building has a height of 11.4 meters and a base with a diameter of 12 meters. The building will be filled with road salt that costs $20 per cubic meter.
How much will it cost to fill the building with road salt?
Use 3.14 for π.
Enter your answer in the box
$
How much will it cost to fill the building with road salt?
Use 3.14 for π.
Enter your answer in the box
$
Answers
GPT-4o mini
To calculate the cost to fill the cone-shaped building with road salt, we first need to find the volume of the cone. The formula for the volume \( V \) of a cone is given by:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the base and \( h \) is the height of the cone.
1. **Find the radius**:
The diameter of the base is given as 12 meters, so the radius \( r \) is:
\[
r = \frac{\text{diameter}}{2} = \frac{12 \, \text{m}}{2} = 6 \, \text{m}
\]
2. **Use the height**:
The height \( h \) of the cone is given as 11.4 meters.
3. **Calculate the volume**:
Now we substitute the values into the volume formula:
\[
V = \frac{1}{3} \times 3.14 \times (6 \, \text{m})^2 \times (11.4 \, \text{m})
\]
First, calculate \( r^2 \):
\[
r^2 = 6^2 = 36 \, \text{m}^2
\]
Now plug this into the volume formula:
\[
V = \frac{1}{3} \times 3.14 \times 36 \times 11.4
\]
Calculate \( 36 \times 11.4 \):
\[
36 \times 11.4 = 410.4
\]
Now multiply by \( \pi \):
\[
3.14 \times 410.4 = 1287.696
\]
Finally, divide by 3:
\[
V = \frac{1287.696}{3} \approx 429.232
\]
Therefore, the volume of the cone is approximately \( 429.232 \, \text{m}^3 \).
4. **Calculate the cost**:
The road salt costs $20 per cubic meter. Thus, the total cost \( C \) is:
\[
C = 429.232 \, \text{m}^3 \times 20 \, \text{\$}/\text{m}^3 = 8584.64 \, \$
\]
Thus, the cost to fill the building with road salt is approximately **$8584.64**.
\[
V = \frac{1}{3} \pi r^2 h
\]
where \( r \) is the radius of the base and \( h \) is the height of the cone.
1. **Find the radius**:
The diameter of the base is given as 12 meters, so the radius \( r \) is:
\[
r = \frac{\text{diameter}}{2} = \frac{12 \, \text{m}}{2} = 6 \, \text{m}
\]
2. **Use the height**:
The height \( h \) of the cone is given as 11.4 meters.
3. **Calculate the volume**:
Now we substitute the values into the volume formula:
\[
V = \frac{1}{3} \times 3.14 \times (6 \, \text{m})^2 \times (11.4 \, \text{m})
\]
First, calculate \( r^2 \):
\[
r^2 = 6^2 = 36 \, \text{m}^2
\]
Now plug this into the volume formula:
\[
V = \frac{1}{3} \times 3.14 \times 36 \times 11.4
\]
Calculate \( 36 \times 11.4 \):
\[
36 \times 11.4 = 410.4
\]
Now multiply by \( \pi \):
\[
3.14 \times 410.4 = 1287.696
\]
Finally, divide by 3:
\[
V = \frac{1287.696}{3} \approx 429.232
\]
Therefore, the volume of the cone is approximately \( 429.232 \, \text{m}^3 \).
4. **Calculate the cost**:
The road salt costs $20 per cubic meter. Thus, the total cost \( C \) is:
\[
C = 429.232 \, \text{m}^3 \times 20 \, \text{\$}/\text{m}^3 = 8584.64 \, \$
\]
Thus, the cost to fill the building with road salt is approximately **$8584.64**.