The length of the hour hand of a clock is 10 cm. Find the area of the minor sector swept by the hour hand of the clock between 5 a.m. to 8 a.m. Also, find the area of the major sector.
Generated by claude-sonnet-4-6 · 2026-06-15 10:36 · grounding rag
Model Answer
From 5 a.m. to 8 a.m. = 3 hours. The hour hand completes 360° in 12 hours, so in 3 hours it sweeps:
$$\theta = \frac{3}{12} \times 360° = 90°$$
Given: radius $r = 10$ cm
Area of minor sector:
$$= \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times \frac{22}{7} \times 10 \times 10$$
$$= \frac{1}{4} \times \frac{2200}{7} = \frac{550}{7} \approx 78.57 \text{ cm}^2$$
Area of major sector:
$$= \pi r^2 - \text{Area of minor sector} = \frac{22}{7} \times 100 - \frac{550}{7}$$
$$= \frac{2200 - 550}{7} = \frac{1650}{7} \approx 235.71 \text{ cm}^2$$
Source: Chapter 11, Section 11.1
---
Explanation
- Key step: Convert time to angle. Hour hand = 360° in 12 h → 30° per hour → 3 hours = 90°.
- Use the formula: Area of sector = $\frac{\theta}{360} \times \pi r^2$.
- Major sector = Total circle area − Minor sector area.
- Use $\pi = \frac{22}{7}$ as instructed unless told otherwise.
- Examiners award marks for: correct angle (1 mark), minor sector area (1 mark), major sector area (1 mark).