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

Given: Radius r = 14 cm, θ = 60°, π = 22/7

Step 1: Area of minor sector

$$\text{Area of sector} = \frac{\theta}{360} \times \pi r^2 = \frac{60}{360} \times \frac{22}{7} \times 14 \times 14 = \frac{1}{6} \times \frac{22}{7} \times 196 = \frac{308}{3} \approx 102.67 \text{ cm}^2$$

Step 2: Area of triangle OAB

Draw OM ⊥ AB. Since θ = 60°, ∠AOM = 30°.

$$OM = 14\cos 30° = 14 \times \frac{\sqrt{3}}{2} = 7\sqrt{3} \text{ cm}$$

$$AM = 14\sin 30° = 14 \times \frac{1}{2} = 7 \text{ cm} \Rightarrow AB = 14 \text{ cm}$$

$$\text{Area of } \triangle OAB = \frac{1}{2} \times 14 \times 7\sqrt{3} = 49\sqrt{3} \approx 49 \times 1.732 = 84.87 \text{ cm}^2$$

Step 3: Area of minor segment

$$= 102.67 - 84.87 = 17.80 \text{ cm}^2$$

Step 4: Area of major segment

$$= \pi r^2 - \text{minor segment} = \frac{22}{7} \times 196 - 17.80 = 616 - 17.80 = 598.20 \text{ cm}^2$$

Source: Areas Related to Circles, Section 11.1

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• Key formula: Area of segment = Area of sector − Area of triangle OAB.
• For θ = 60°, triangle OAB is equilateral (OA = OB = r, and the half-angle gives sin 30° = ½, making AB = r = 14 cm), which is a useful shortcut to verify.
• Use √3 ≈ 1.732 unless the question provides a specific value.
• Major segment = Total circle area − minor segment area. Show all steps clearly; marks are awarded for each stage of working.