A horse is tied to a peg at one corner of a square shaped grass field of side 15 m by means of a 5 m long rope. Find the area of that part of the field in which the horse can graze. Also, find the increase in grazing area if length of rope is increased to 10 m. (Use $\pi = 3 \cdot 14$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:37 · grounding rag
Model Answer
Given: Side of square field = 15 m, rope length (radius) = 5 m, angle at corner of square = 90°
Part (i): Grazing area with rope = 5 m
The horse can graze over a sector of radius 5 m and angle 90° (since the field corner is 90°).
$$\text{Area} = \frac{\theta}{360} \times \pi r^2 = \frac{90}{360} \times 3.14 \times 5 \times 5$$
$$= \frac{1}{4} \times 3.14 \times 25 = \mathbf{19.625 \text{ m}^2}$$
Part (ii): Increase in grazing area when rope = 10 m
With rope = 10 m (still less than 15 m, so horse stays within the field corner sector):
$$\text{Area} = \frac{90}{360} \times 3.14 \times 10 \times 10 = \frac{1}{4} \times 3.14 \times 100 = 78.5 \text{ m}^2$$
$$\text{Increase in grazing area} = 78.5 - 19.625 = \mathbf{58.875 \text{ m}^2}$$
Source: Chapter 11, Exercise 11.1, Q.8
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Explanation
- The horse is tied at a corner of the square, so it can only sweep a quarter circle (90° sector) inside the field.
- The rope length acts as the radius of the sector.
- Examiners expect you to clearly identify the angle as 90°, apply the sector area formula correctly, and subtract the two areas to find the increase.
- Show all substitution steps clearly — each step carries marks. The final answer for increase (58.875 m²) is key.