There is a circular park of diameter 65 m as shown in the following figure, where AB is a diameter. An entry gate is to be constructed at a point P on the boundary of the park such that distance of P from A is 35 m more than the distance of P from B. Find distance of point P from A and B respectively.
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
Given: Diameter AB = 65 m, PA = PB + 35
Step 1: Let PB = x m. Then PA = (x + 35) m.
Step 2: Since AB is a diameter of the circle, angle APB = 90° (angle in a semicircle).
Step 3: Applying Pythagoras theorem in △APB:
$$PA^2 + PB^2 = AB^2$$
$$(x + 35)^2 + x^2 = 65^2$$
Step 4: Expanding:
$$x^2 + 70x + 1225 + x^2 = 4225$$
$$2x^2 + 70x + 1225 - 4225 = 0$$
$$2x^2 + 70x - 3000 = 0$$
$$x^2 + 35x - 1500 = 0$$
Step 5: Factorising:
$$x^2 + 60x - 25x - 1500 = 0$$
$$x(x + 60) - 25(x + 60) = 0$$
$$(x - 25)(x + 60) = 0$$
So x = 25 or x = −60 (rejected, as distance cannot be negative).
Therefore, PB = 25 m and PA = 60 m.
Source: Chapter 6, Triangles (Pythagoras Theorem / angle in semicircle)
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Explanation
- The key fact used is: angle in a semicircle = 90° (∠APB = 90° when AB is a diameter). This allows applying the Pythagoras theorem directly.
- Set one unknown (PB = x), express the other in terms of it, substitute into $PA^2 + PB^2 = AB^2$, and solve the quadratic.
- Always reject the negative root since distance is non-negative.
- Examiners award marks for: correct setup (1), forming the equation (1), solving the quadratic (2), stating both answers (1) — totalling 5 marks.