1. **Problem statement:** We need to find the area of a swimming pool formed by two identical circles of radius 9 m, where each circle passes through the center of the other. 2. **Understanding the problem:** The shape formed is called a lens or vesica piscis, created by the intersection of two circles of equal radius $r=9$ m, with the distance between their centers $d$ equal to the radius (since each passes through the other's center), so $d=9$ m. 3. **Formula for the area of intersection of two circles:** $$ A = 2r^2 \arccos\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2} $$ 4. **Substitute values:** $$ r=9, \quad d=9 $$ 5. Calculate each term: - Calculate $\arccos\left(\frac{9}{2 \times 9}\right) = \arccos\left(\frac{9}{18}\right) = \arccos\left(\frac{1}{2}\right) = \frac{\pi}{3}$ - Calculate $\sqrt{4 \times 9^2 - 9^2} = \sqrt{4 \times 81 - 81} = \sqrt{324 - 81} = \sqrt{243} = 9\sqrt{3}$ 6. Plug back into the formula: $$ A = 2 \times 9^2 \times \frac{\pi}{3} - \frac{9}{2} \times 9\sqrt{3} = 2 \times 81 \times \frac{\pi}{3} - \frac{81}{2} \sqrt{3} $$ 7. Simplify: $$ 2 \times 81 \times \frac{\pi}{3} = 162 \times \frac{\pi}{3} = 54\pi $$ $$ \frac{81}{2} \sqrt{3} = 40.5 \sqrt{3} $$ 8. Final area: $$ A = 54\pi - 40.5 \sqrt{3} \text{ square meters} $$ This is the area of the swimming pool formed by the overlapping circles.