1. **Problem:** Two circles with radii 6 cm and 8 cm have centers 10 cm apart. Find the area common to both circles. 2. **Formula and explanation:** The area of intersection of two circles with radii $r_1$ and $r_2$ and center distance $d$ is given by: $$\text{Area} = r_1^2 \cos^{-1}\left(\frac{d^2 + r_1^2 - r_2^2}{2 d r_1}\right) + r_2^2 \cos^{-1}\left(\frac{d^2 + r_2^2 - r_1^2}{2 d r_2}\right) - \frac{1}{2} \sqrt{(-d + r_1 + r_2)(d + r_1 - r_2)(d - r_1 + r_2)(d + r_1 + r_2)}$$ This formula comes from summing the areas of two circular segments minus the overlapping triangle area. 3. **Substitute values:** $r_1 = 6$, $r_2 = 8$, $d = 10$ Calculate each term: $$\cos^{-1}\left(\frac{10^2 + 6^2 - 8^2}{2 \times 10 \times 6}\right) = \cos^{-1}\left(\frac{100 + 36 - 64}{120}\right) = \cos^{-1}\left(\frac{72}{120}\right) = \cos^{-1}(0.6)$$ $$\cos^{-1}(0.6) \approx 0.9273 \text{ radians}$$ Similarly, $$\cos^{-1}\left(\frac{10^2 + 8^2 - 6^2}{2 \times 10 \times 8}\right) = \cos^{-1}\left(\frac{100 + 64 - 36}{160}\right) = \cos^{-1}\left(\frac{128}{160}\right) = \cos^{-1}(0.8)$$ $$\cos^{-1}(0.8) \approx 0.6435 \text{ radians}$$ Calculate the square root term: $$\sqrt{(-10 + 6 + 8)(10 + 6 - 8)(10 - 6 + 8)(10 + 6 + 8)} = \sqrt{(4)(8)(12)(24)}$$ $$= \sqrt{9216} = 96$$ 4. **Calculate area:** $$6^2 \times 0.9273 + 8^2 \times 0.6435 - \frac{1}{2} \times 96 = 36 \times 0.9273 + 64 \times 0.6435 - 48$$ $$= 33.3828 + 41.184 - 48 = 26.5668 \text{ cm}^2$$ **Final answer:** The area common to both circles is approximately $26.57$ cm$^2$.