To find the area of the section of the pool that is blocked off at an angle of 75°, we can use the formula for the area of a sector of a circle. The formula is:
\[ A = \frac{\theta}{360} \times \pi r^2 \]
where:
- \( A \) is the area of the sector,
- \( \theta \) is the angle of the sector in degrees,
- \( r \) is the radius of the circle.
- First, we need to find the radius of the pool. Since the diameter is 18 ft, the radius \( r \) is:
\[ r = \frac{18}{2} = 9 \text{ ft} \]
- Now we can plug the values into the formula:
\[ A = \frac{75}{360} \times \pi (9^2) \]
- Calculating \( 9^2 \):
\[ 9^2 = 81 \]
- Now substituting this into the area formula:
\[ A = \frac{75}{360} \times \pi \times 81 \]
- Simplifying \( \frac{75}{360} \):
\[ \frac{75}{360} = \frac{75 \div 15}{360 \div 15} = \frac{5}{24} \]
- Now, substitute this back into the area formula:
\[ A = \frac{5}{24} \times \pi \times 81 = \frac{405}{24} \pi \]
- Now calculating \( \frac{405}{24} \):
\[ \frac{405 \div 3}{24 \div 3} = \frac{135}{8} = 16.875 \]
- So, the area is:
\[ A = 16.875 \pi , \text{ft}^2 \]
Thus, the area of the section that is blocked off is
\[ \boxed{16.875 \pi} , \text{ft}^2. \]