1. Problem statement: We have a circle with center O and radius OS = 7 cm. R is the midpoint of OS, so OR = RS = $\frac{7}{2} = 3.5$ cm. We have two sectors ORU (40°) and OST (50°). We need to find: (a) the perimeter of the shaded region (likely the difference between the two sectors or the sector formed by these arcs). (b) the area of the shaded region. 2. Understanding the problem and shaded region: Since the problem mentions sectors ORU and OST with central angles 40° and 50°, and R is midpoint of OS, the sectors are parts of the circle with different radii: ORU with radius OR = 3.5 cm, and OST with radius OS = 7 cm. 3. Calculating perimeter (a): The shaded region is between two arcs – smaller arc of radius OR and bigger arc of radius OS, both spanning certain angles. Assuming shaded region is sector OST minus sector ORU (or the lune between arcs): - Arc length of bigger sector OST with radius 7 cm and angle 50°: $$L_{OST} = \frac{50}{360} \times 2 \pi \times 7 = \frac{50}{360} \times 2 \times \frac{22}{7} \times 7 = \frac{50}{360} \times 44 = \frac{2200}{360} = 6.11 \text{ cm}$$ - Arc length of smaller sector ORU with radius 3.5 cm and angle 40°: $$L_{ORU} = \frac{40}{360} \times 2 \pi \times 3.5 = \frac{40}{360} \times 2 \times \frac{22}{7} \times 3.5 = \frac{40}{360} \times 22 = \frac{880}{360} = 2.44 \text{ cm}$$ - The perimeter of the shaded region consists of: 1) The larger arc of sector OST = 6.11 cm 2) The smaller arc of sector ORU = 2.44 cm 3) The two straight-line edges, between points U and S on the bigger radius (OS=7) and points U and R on the smaller radius (OR=3.5), which form the radial sides of sectors. But since R is midpoint of OS, the distance RS = RO = 3.5 cm, the straight edges for the shaded region are OU and US. The problem implies the shaded area is bounded by these two arcs and the two radii connecting their endpoints. The total perimeter = bigger arc + smaller arc + RS + RU Since RS = RU = radius difference = OS - OR = 7-3.5 = 3.5 cm, But actually RU = OR = 3.5 cm, RS = OS - OR = 7 - 3.5 = 3.5 cm. Hence, perimeter = $6.11 + 2.44 + 3.5 + 3.5 = 15.55$ cm 4. Calculating area (b): Area of sector OST (radius 7 cm, angle 50°): $$A_{OST} = \frac{50}{360} \times \pi \times 7^2 = \frac{50}{360} \times \frac{22}{7} \times 49 = \frac{50}{360} \times 154 = 21.39 \text{ cm}^2$$ Area of sector ORU (radius 3.5 cm, angle 40°): $$A_{ORU} = \frac{40}{360} \times \pi \times 3.5^2 = \frac{40}{360} \times \frac{22}{7} \times 12.25 = \frac{40}{360} \times 38.5 = 4.28 \text{ cm}^2$$ Shaded area = $A_{OST} - A_{ORU} = 21.39 - 4.28 = 17.11$ cm² 5. Final answers correct to 2 decimal places: (a) Perimeter = 15.55 cm (b) Area = 17.11 cm²