1. **Problem statement:** Calculate the percentage of the area of the circle that is shaded, given the areas of triangles and sectors in the circle with diameters AC and BD intersecting at O with an angle of 130°. 2. **Formula used:** - Area of triangle: $\frac{1}{2}ab\sin C$ - Area of sector: $\frac{\theta}{360^\circ} \times \pi r^2$ - Percentage area: $\frac{\text{shaded area}}{\text{total area}} \times 100$ 3. **Step-by-step solution:** (a) Area of $\triangle AOB$: $$\frac{1}{2} \times 6 \times 6 \times \sin 130^\circ = 13.79\, \text{cm}^2$$ (b) Angle $\angle AOD = 180^\circ - 130^\circ = 50^\circ$ Area of sector $AOD$: $$\frac{50}{360} \times \pi \times 6^2 = 15.71\, \text{cm}^2$$ (c) Area of shaded segment $AD$: $$15.71 - \frac{1}{2} \times 6 \times 6 \times \sin 50^\circ = 15.71 - 13.79 = 1.92\, \text{cm}^2$$ Total shaded area: $$2 \times 13.79 + 2 \times 1.92 = 31.42\, \text{cm}^2$$ Area of circle: $$\pi \times 6^2 = 113.1\, \text{cm}^2$$ Percentage shaded: $$\frac{31.42}{113.1} \times 100 = 27.8\%$$ **Final answer:** The shaded area is 27.8% of the circle's area.