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.