Mathematics — Standard (041) — CBSE Class 10 board question

Given: Diameter = 20 cm → Radius, r = 10 cm
Circle divided into 5 equal sectors → Angle of each sector, θ = 360°/5 = 72°

Area of one sector:

$$\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{72}{360} \times \frac{22}{7} \times 10 \times 10$$

$$= \frac{1}{5} \times \frac{2200}{7} = \frac{2200}{35} = \frac{440}{7} \approx 62.86 \text{ cm}^2$$

Perimeter of one sector = length of arc + 2 radii

$$\text{Arc length} = \frac{72}{360} \times 2\pi r = \frac{1}{5} \times 2 \times \frac{22}{7} \times 10 = \frac{440}{35} = \frac{88}{7} \approx 12.57 \text{ cm}$$

$$\text{Perimeter} = \frac{88}{7} + 2 \times 10 = \frac{88}{7} + 20 = \frac{88 + 140}{7} = \frac{228}{7} \approx 32.57 \text{ cm}$$

Source: Chapter 11, Section 11.1 – Areas of Sector and Segment of a Circle

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• Examiners expect you to first find θ by dividing 360° by 5, then apply both formulas from the summary.
• Perimeter of a sector = arc length + two radii (a common mistake is forgetting to add the two straight sides).
• Use π = 22/7 as instructed in the exercise unless told otherwise.
• Show each step clearly; marks are awarded for correct formula, substitution, and final answer.