Question

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.

Hint:

If a triangle is in a semicircle, the angle opposite diameter is $90^\circ$—apply Pythagoras theorem directly.

Answer 1

Let the distance of point P from B be $x$ meters.
Then, distance from A = $x + 35$ meters
Since $\triangle APB$ is inscribed in a semicircle (angle in semicircle is right), use Pythagoras theorem: \[ AB^2 = AP^2 + PB^2\\ 65^2 = (x + 35)^2 + x^2\\ 4225 = x^2 + 70x + 1225 + x^2 = 2x^2 + 70x + 1225 \] Rearranging: \[ 2x^2 + 70x + 1225 - 4225 = 0\\ 2x^2 + 70x - 3000 = 0\\ x^2 + 35x - 1500 = 0 \] Solve using quadratic formula: \[ x = \frac{-35 \pm \sqrt{35^2 + 4 \cdot 1500}}{2} = \frac{-35 \pm \sqrt{1225 + 6000}}{2}\\ x = \frac{-35 \pm \sqrt{7225}}{2} = \frac{-35 \pm 85}{2} \] Taking positive root: \[ x = \frac{50}{2} = 25 \Rightarrow \text{PB} = 25\,\text{m}, \text{PA} = 60\,\text{m} \] Answer: Distance from A = 60 m, from B = 25 m