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 the point P on the boundary of the park such that the distance of P from A is 35 m more than the distance of P from B. Find the distance of point P from A and B, respectively.

Answer 1

Given, diameter AB = 65 m

Let PB = x and PA = x + 35

∠APB = 90°  ....(∵ Angle in a semicircle is a right angle.)

In ΔAOB,

By Pythagorean theorem,

AB² = AP² + BP²

65² + (x + 35)² + x²

4225 = x² + 1225 + 70x + x²

4225 = 2x² + 70x + 1225

2x² + 70x + 1225 − 4225 = 0

2x² + 70x − 3000 = 0

x² + 35x − 1500 = 0   ...(Dividing by 2.)

x² + 60x − 25x − 1500 = 0

x(x + 60) − 25(x + 60) = 0

(x + 60) (x − 25) = 0

x = −60; x = 25

Distance can’t be negative.

∴ AP = x + 35

= 25 + 35

= 60 m

BP = x = 25 m