1. **State the problem:** We have a circle with center O and radius $r$ cm. Points A and B lie on the circumference, and the central angle $\angle AOB$ is 140°. The area of the shaded segment is 65 cm$^2$. We need to find the radius $r$. 2. **Relevant formulas:** - Area of sector $AOB$ with central angle $\theta$ (in degrees) is given by: $$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2$$ - Area of triangle $AOB$ (formed by two radii and chord AB) when $\theta$ is known: $$\text{Area of triangle} = \frac{1}{2} r^2 \sin(\theta)$$ - Area of segment (shaded area) is: $$\text{Area of segment} = \text{Area of sector} - \text{Area of triangle}$$ 3. **Apply the formulas:** Given: $$\theta = 140^\circ$$ $$\text{Area of segment} = 65$$ Write the equation: $$65 = \frac{140}{360} \pi r^2 - \frac{1}{2} r^2 \sin(140^\circ)$$ 4. **Simplify the equation:** Calculate constants: $$\frac{140}{360} = \frac{7}{18}$$ $$\sin(140^\circ) = \sin(180^\circ - 40^\circ) = \sin(40^\circ) \approx 0.6428$$ Substitute: $$65 = \frac{7}{18} \pi r^2 - \frac{1}{2} r^2 \times 0.6428$$ 5. **Factor out $r^2$:** $$65 = r^2 \left( \frac{7}{18} \pi - \frac{1}{2} \times 0.6428 \right)$$ Calculate inside the parentheses: $$\frac{7}{18} \pi \approx 1.2217$$ $$\frac{1}{2} \times 0.6428 = 0.3214$$ So: $$65 = r^2 (1.2217 - 0.3214) = r^2 (0.9003)$$ 6. **Solve for $r^2$:** $$r^2 = \frac{65}{0.9003} \approx 72.21$$ 7. **Find $r$:** $$r = \sqrt{72.21} \approx 8.5$$ **Final answer:** $$\boxed{r \approx 8.5 \text{ cm}}$$