Segment Area Percentage Cf3720

1. **Problem Statement:** We need to find the percentage of the area of the segment ACB cut off by the straight line AC relative to the remaining area of the sector AOC. 2. **Understanding the problem:** - The sector AOC is a part of a circle with center O and radius OA = OC. - The segment ACB is the area bounded by the chord AC and the arc ACB. - The area cut off is the segment area, and the remaining area is the sector area minus the segment area. 3. **Formulas:** - Area of sector AOC: $$\text{Area}_{sector} = \frac{\theta}{360} \pi r^2$$ where $\theta$ is the central angle AOC in degrees and $r$ is the radius. - Area of segment ACB: $$\text{Area}_{segment} = \text{Area}_{sector} - \text{Area}_{triangle}$$ - Area of triangle AOC (isosceles triangle): $$\text{Area}_{triangle} = \frac{1}{2} r^2 \sin(\theta)$$ 4. **Calculate the segment area:** $$\text{Area}_{segment} = \frac{\theta}{360} \pi r^2 - \frac{1}{2} r^2 \sin(\theta)$$ 5. **Calculate the remaining area:** $$\text{Area}_{remaining} = \text{Area}_{sector} - \text{Area}_{segment} = \frac{1}{2} r^2 \sin(\theta)$$ 6. **Find the percentage:** $$\text{Percentage} = \frac{\text{Area}_{segment}}{\text{Area}_{remaining}} \times 100 = \frac{\frac{\theta}{360} \pi r^2 - \frac{1}{2} r^2 \sin(\theta)}{\frac{1}{2} r^2 \sin(\theta)} \times 100$$ 7. **Simplify the expression:** $$= \left(\frac{\frac{\theta}{360} \pi r^2}{\frac{1}{2} r^2 \sin(\theta)} - 1\right) \times 100 = \left(\frac{2 \theta \pi}{360 \sin(\theta)} - 1\right) \times 100$$ 8. **Final formula for percentage:** $$\boxed{\text{Percentage} = \left(\frac{2 \theta \pi}{360 \sin(\theta)} - 1\right) \times 100}$$ This formula gives the percentage of the segment area cut off relative to the remaining area of the sector AOC for any central angle $\theta$.