1. **Problem statement:** Find the area, arc length, and perimeter of the given sectors. 2. **Formulas used:** - Area of sector: $$A = \frac{\theta}{360} \times \pi r^2$$ where $\theta$ is the central angle in degrees and $r$ is the radius. - Arc length: $$L = \frac{\theta}{360} \times 2 \pi r$$ - Perimeter of sector: $$P = 2r + L$$ (two radii plus arc length) 3. **Important notes:** - Use the given value of $\pi$ for each problem. - Radius and angle must be consistent with units. --- ### (i) $\pi = 3.14$, $\theta = 84^\circ$, $r = 6$ cm Area: $$A = \frac{84}{360} \times 3.14 \times 6^2 = \frac{84}{360} \times 3.14 \times 36$$ $$= 0.2333 \times 3.14 \times 36 = 26.36 \text{ cm}^2$$ Arc length: $$L = \frac{84}{360} \times 2 \times 3.14 \times 6 = 0.2333 \times 37.68 = 8.79 \text{ cm}$$ Perimeter: $$P = 2 \times 6 + 8.79 = 12 + 8.79 = 20.79 \text{ cm}$$ --- ### (ii) $\pi = \frac{22}{7}$, $\theta = 162^\circ$, $r = 12$ mm Area: $$A = \frac{162}{360} \times \frac{22}{7} \times 12^2 = \frac{162}{360} \times \frac{22}{7} \times 144$$ $$= 0.45 \times \frac{22}{7} \times 144$$ $$= 0.45 \times 22 \times \frac{144}{7}$$ $$= 0.45 \times 22 \times 20.5714 = 203.57 \text{ mm}^2$$ Arc length: $$L = \frac{162}{360} \times 2 \times \frac{22}{7} \times 12 = 0.45 \times 2 \times \frac{22}{7} \times 12$$ $$= 0.45 \times 2 \times 22 \times \frac{12}{7}$$ $$= 0.45 \times 44 \times 1.7143 = 34.00 \text{ mm}$$ Perimeter: $$P = 2 \times 12 + 34.00 = 24 + 34.00 = 58.00 \text{ mm}$$ --- ### (iii) $\pi = 3.14$, major sector with missing wedge $58^\circ$, so central angle $\theta = 360 - 58 = 302^\circ$, $r = 30$ m Area: $$A = \frac{302}{360} \times 3.14 \times 30^2 = 0.8389 \times 3.14 \times 900$$ $$= 0.8389 \times 2826 = 2369.5 \text{ m}^2$$ Arc length: $$L = \frac{302}{360} \times 2 \times 3.14 \times 30 = 0.8389 \times 188.4 = 158.0 \text{ m}$$ Perimeter: $$P = 2 \times 30 + 158.0 = 60 + 158.0 = 218.0 \text{ m}$$