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 to each other. Find the length of the string of each phone.

Let Ankur be represented as A, Syed as S and David as D. The boys are sitting at an equal distance. Hence, △ASD is an equilateral triangle. Let the radius of the circular park be r meters. ∴OS=r=20m. Let the length of each side of △ASD be x meters. Draw AB ⊥ SD ∴SB=BD=12SD=x2m In △ABS, ∠B=90o By Pythagoras theorem, AS2=AB2+BS2 ∴AB2=AS2−BS2=x2−(x2)2=3x24 ∴AB=√3x2m Now, AB=AO+OB OB=AB−AO OB=(√3x2−20)m In △OBS, OS2=OB2+SB2 202=(√3x2−20)2+(x2)2 400=34x2+400−(20)(√3x2)+x24 400=34x2+400−20√3x+x24 0=x2−20√3x 40=x(20−√3x) 40=x(20−√3x) ∴x=20√3m The length of the string of each phone is 20√3m.