1. **Problem statement:** We have two circles representing the ranges of transmitters on boats A and B. - Circle A has center A and radius $r_A = 7$ km. - Circle B has center B and radius $r_B = 5$ km. - The distance between centers $d = 10$ km. We need to find the area of the overlapping (intersection) region of these two circles. 2. **Formula for the area of intersection of two circles:** If two circles with radii $r_A$ and $r_B$ have centers separated by distance $d$, the area of their intersection is given by: $$ A = r_A^2 \cos^{-1}\left(\frac{d^2 + r_A^2 - r_B^2}{2 d r_A}\right) + r_B^2 \cos^{-1}\left(\frac{d^2 + r_B^2 - r_A^2}{2 d r_B}\right) - \frac{1}{2} \sqrt{(-d + r_A + r_B)(d + r_A - r_B)(d - r_A + r_B)(d + r_A + r_B)} $$ 3. **Calculate each term:** Calculate the cosine inverse arguments: $$ \alpha = \frac{d^2 + r_A^2 - r_B^2}{2 d r_A} = \frac{10^2 + 7^2 - 5^2}{2 \times 10 \times 7} = \frac{100 + 49 - 25}{140} = \frac{124}{140} = 0.8857 $$ $$ \beta = \frac{d^2 + r_B^2 - r_A^2}{2 d r_B} = \frac{10^2 + 5^2 - 7^2}{2 \times 10 \times 5} = \frac{100 + 25 - 49}{100} = \frac{76}{100} = 0.76 $$ 4. **Evaluate the inverse cosines:** $$ \cos^{-1}(0.8857) \approx 0.485 \text{ radians} $$ $$ \cos^{-1}(0.76) \approx 0.707 \text{ radians} $$ 5. **Calculate the square root term:** $$ \sqrt{(-d + r_A + r_B)(d + r_A - r_B)(d - r_A + r_B)(d + r_A + r_B)} = \sqrt{(-10 + 7 + 5)(10 + 7 - 5)(10 - 7 + 5)(10 + 7 + 5)} $$ $$ = \sqrt{2 \times 12 \times 8 \times 22} = \sqrt{4224} \approx 65.01 $$ 6. **Calculate the area:** $$ A = 7^2 \times 0.485 + 5^2 \times 0.707 - \frac{1}{2} \times 65.01 $$ $$ = 49 \times 0.485 + 25 \times 0.707 - 32.505 $$ $$ = 23.765 + 17.675 - 32.505 = 8.935 \text{ km}^2 $$ **Final answer:** The area of the overlapping region is approximately **8.94 km²**.