Anurag purchased a farmhouse which is in the form of a semicircle of diameter 70 m. He divides it into three parts by taking a point P on the semicircle in such a way that $\angle PAB = 30°$ as shown in the following figure, where O is the centre of semicircle. In part I, he planted saplings of Mango tree, in part II, he grew tomatoes and in part III, he grew oranges.
Based on given information, answer the following questions.
Generated by claude-sonnet-4-6 · 2026-06-15 10:38 · grounding stimulus
Model Answer
Given: Diameter AB = 70 m, so radius = 35 m. ∠PAB = 30°.
(i) Measure of ∠POA:
Since AB is diameter and P is on the semicircle, by the inscribed angle theorem:
∠POA = 2 × ∠PAB = 2 × 30° = 60°
(ii) Length of wire to fence entire land:
Perimeter = Diameter + Semicircular arc
$$= 70 + \pi r = 70 + \frac{22}{7} \times 35 = 70 + 110 = \textbf{180 m}$$
(iii) Area of region I (Mango saplings — sector POB):
∠POB = 180° − ∠POA = 180° − 60° = 120°
$$\text{Area of sector POB} = \frac{\theta}{360°} \times \pi r^2 = \frac{120}{360} \times \frac{22}{7} \times 35 \times 35$$
$$= \frac{1}{3} \times \frac{22}{7} \times 1225 = \frac{1}{3} \times 3850 = \textbf{1283.33 m}^2 \approx \frac{3850}{3} \text{ m}^2$$
Source: Areas Related to Circles, Case Study Application
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Explanation
- Part (i): The central angle is twice the inscribed angle subtending the same arc (Inscribed Angle Theorem), so ∠POA = 2 × 30° = 60°.
- Part (ii): Fencing = straight diameter + curved arc (half circumference). Don't forget the diameter!
- Part (iii): Mango region is sector POB. ∠POB = 180° − 60° = 120°. Use sector area formula $\frac{\theta}{360} \pi r^2$. Examiners award marks for correct angle identification and formula application.