Question

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.

Answer 1

It is given that AS = SD = DA 

Therefore, ∆ASD is an equilateral triangle.

OA (radius) = 20 m

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ASD. We also know that medians intersect each other in the ratio 2: 1. 

As AB is the median of equilateral triangle ASD, we can write

⇒ \(\frac{OA}{OB}=\frac{2}{1}\)

⇒ \(\frac{20\,m}{OB}=\frac{2}{1}\)

⇒ OB=\((\frac{20}{2})\)m=10 m

∠AB=OA+OB=(20+10)m=30m

In ∆ABD,

AD²=AB²+BD²

AD²=(30)²+ \((\frac{AD}{2})^2\)

AD²=900+ \(\frac{1}{4}\) AD²

\(\frac{3}{4}\) AD²=900

AD²=1200

AD=20 \(\sqrt3\)

Therefore, the length of the string of each phone will be \(20\sqrt3\) m.