The ratio of the base area and the curved surface area of a conical tent is 40:41. If its height is 18 m, find the air capacity of the tent in terms of `pi.`

Answer

To find the air capacity of a conical tent given the ratio of the base area to the curved surface area and its height, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Ratio**:
The ratio of the base area (A) to the curved surface area (CSA) of the conical tent is given as:
\[ \frac{A}{CSA} = \frac{40}{41} \]

2. **Formulas for Base Area and Curved Surface Area**:
- The base area \( A \) of a cone is given by:
\[ A = \pi r^2 \]
- The curved surface area \( CSA \) of a cone is given by:
\[ CSA = \pi r l \]
where \( l \) is the slant height.

3. **Set Up the Equation**:
Substitute the formulas into the ratio:
\[ \frac{\pi r^2}{\pi r l} = \frac{40}{41} \]
Simplifying gives:
\[ \frac{r}{l} = \frac{40}{41} \]

4. **Express Slant Height**:
From the ratio, we can express \( l \) in terms of \( r \):
\[ l = \frac{41}{40} r \]

5. **Use the Pythagorean Theorem**:
The slant height \( l \) can also be expressed using the height \( h \) and radius \( r \):
\[ l = \sqrt{h^2 + r^2} \]
Given \( h = 18 \) m, we can set up the equation:
\[ \sqrt{h^2 + r^2} = \frac{41}{40} r \]

6. **Square Both Sides**:
Squaring both sides gives:
\[ h^2 + r^2 = \left(\frac{41}{40} r\right)^2 \]
\[ h^2 + r^2 = \frac{1681}{1600} r^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ h^2 = \frac{1681}{1600} r^2 - r^2 \]
\[ h^2 = \left(\frac{1681 - 1600}{1600}\right) r^2 \]
\[ h^2 = \frac{81}{1600} r^2 \]

8. **Substituting the Height**:
Substitute \( h = 18 \):
\[ 18^2 = \frac{81}{1600} r^2 \]
\[ 324 = \frac{81}{1600} r^2 \]

9. **Solve for \( r^2 \)**:
Multiply both sides by \( 1600 \):
\[ 324 \times 1600 = 81 r^2 \]
\[ 518400 = 81 r^2 \]
\[ r^2 = \frac{518400}{81} = 6400 \]

10. **Find \( r \)**:
Taking the square root gives:
\[ r = 80 \text{ m} \]

11. **Calculate the Volume of the Cone**:
The volume \( V \) of the cone is given by:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r = 80 \) m and \( h = 18 \) m:
\[ V = \frac{1}{3} \pi (80^2)(18) \]
\[ V = \frac{1}{3} \pi (6400)(18) \]
\[ V = \frac{1}{3} \pi (115200) \]
\[ V = 38400 \pi \text{ m}^3 \]

### Final Answer:
The air capacity of the conical tent is:
\[ \boxed{38400 \pi \text{ m}^3} \]