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:** The area of intersection of two circles with radii $r_1$, $r_2$ and center distance $d$ is given by: $$\text{Area} = r_1^2 \cos^{-1}\left(\frac{d^2 + r_1^2 - r_2^2}{2 d r_1}\right) + r_2^2 \cos^{-1}\left(\frac{d^2 + r_2^2 - r_1^2}{2 d r_2}\right) - \frac{1}{2} \sqrt{(-d + r_1 + r_2)(d + r_1 - r_2)(d - r_1 + r_2)(d + r_1 + r_2)}$$ 3. **Step-by-step solution:** 1. Assign values: $r_1 = 6$, $r_2 = 8$, $d = 10$. 2. Calculate the first angle: $$\theta_1 = \cos^{-1}\left(\frac{10^2 + 6^2 - 8^2}{2 \times 10 \times 6}\right) = \cos^{-1}\left(\frac{100 + 36 - 64}{120}\right) = \cos^{-1}\left(\frac{72}{120}\right) = \cos^{-1}(0.6)$$ 3. Calculate the second angle: $$\theta_2 = \cos^{-1}\left(\frac{10^2 + 8^2 - 6^2}{2 \times 10 \times 8}\right) = \cos^{-1}\left(\frac{100 + 64 - 36}{160}\right) = \cos^{-1}\left(\frac{128}{160}\right) = \cos^{-1}(0.8)$$ 4. Evaluate the square root term: $$\sqrt{(-10 + 6 + 8)(10 + 6 - 8)(10 - 6 + 8)(10 + 6 + 8)} = \sqrt{(4)(8)(12)(24)}$$ Calculate inside: $$4 \times 8 = 32, \quad 12 \times 24 = 288$$ So: $$\sqrt{32 \times 288} = \sqrt{9216} = 96$$ 5. Substitute all values into the formula: $$\text{Area} = 6^2 \times \cos^{-1}(0.6) + 8^2 \times \cos^{-1}(0.8) - \frac{1}{2} \times 96$$ 6. Calculate angles in radians: $$\cos^{-1}(0.6) \approx 0.9273, \quad \cos^{-1}(0.8) \approx 0.6435$$ 7. Calculate each term: $$36 \times 0.9273 = 33.3828, \quad 64 \times 0.6435 = 41.184$$ 8. Calculate final area: $$33.3828 + 41.184 - 48 = 26.5668$$ 4. **Answer:** The area common to both circles is approximately $26.57$ square centimeters.