A circular park of radius 40 m is situated in a colony. Three boys Ankur,

Solution:

Given that AB = BC = CA

So, ABC is an equilateral triangle

OA (radius) = 40 m

Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC.

We also know that median intersect each other at the ratio 2: 1.

As AD is the median of equilateral triangle ABC, we can write:

OA/OD = 2/1

⇒ 4OM/OD = 2/1

⇒ OD = 20m

Therefore, AD = OA + OD = (40 + 20) m

= 60 m

In ΔADC

By using Pythagoras theorem

$A C^{2}=A D^{2}+D C^{2}$

$A C^{2}=60^{2}+\left(A C^{2}\right)^{2}$

$A C^{2}=3600+A C^{2} / 4$

$\Rightarrow 3 / 4 \mathrm{AC}^{2}=3600$

$\Rightarrow \mathrm{AC}^{2}=4800$

$\Rightarrow A C=40 \sqrt{3} \mathrm{~m}$

So, length of string of each phone will be 40√3 m.