A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Let P, D, Q denote the positions of Ankur, David, and Syed, respectively.

∆PDQ is an equilateral triangle since all the 3 boys are equidistant from one another.

Let B denote the mid-point of DQ, and hence PB is the median and perpendicular bisector of DQ.

Hence ∆PBQ is a right-angled triangle with ∠PBQ = 90º.

O (centroid) divides the line PB in the ratio 2 : 1. So OP : OB = 2 : 1.

OP/OB = 2/1

Since OP = 20

thus, OB = 10m

PB = OP + OB = 20 + 10 = 30 m ….(1)

Let the side of the equilateral triangle ∆PDQ be 2x.

PD = DQ = QP = 2x …. (2)

Since B is the mid-point of DS, we get BQ = BD = x …. (3)

Applying Pythagoras theorem to ∆PBD, we get:

PD² = PB² + BD²

(2x)² = 30² + x²

4x² = 900 + x²

3x² = 900

x² = 300

x = 10√3

x = 17.32

PD = DQ = QP = 2x = 34.64 m

Length of the string = Distance between them = PD or DQ or QP = 34.64 m.