A chord of a circle of radius 10 cm subtends a right angle at the centre of the circle. Find the area of the corresponding minor segment. [Use $\pi = 3.14$]
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
Given: radius $r = 10$ cm, angle $\theta = 90°$, $\pi = 3.14$
Area of minor sector:
$$= \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 314 = 78.5 \text{ cm}^2$$
Area of triangle OAB (right-angled at O, with OA = OB = 10 cm):
$$= \frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2$$
Area of minor segment:
$$= \text{Area of sector} - \text{Area of } \triangle OAB = 78.5 - 50 = \textbf{28.5 cm}^2$$
Source: Areas of Circles and Sectors, Chapter 11, Exercise 11.1 Q.4
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Explanation
- The chord subtends 90° at the centre, so the triangle OAB is right-angled at O with both legs equal to the radius. This makes the triangle area calculation straightforward: $\frac{1}{2} \times r \times r$.
- Always use the formula: Area of segment = Area of sector − Area of triangle.
- Show each step clearly for full 3-mark credit (1 mark each for sector area, triangle area, and final segment area).