Given that,
AB = BC = CA
So,
ABC is an equilateral triangle
OA = 40 cm (Radius)
Medians of equilateral triangle pass through the circumference (O) of the equilateral triangle ABC
We also know that,
Median intersects each other at 2 : 1 as AD is the median of equilateral triangle ABC,
We can write :
\(\frac{OA}{OD}\) = \(\frac{2}{7}\)
\(\frac{40}{OD}\) = \(\frac{2}{7}\)
OD = 20 m
Therefore,
AO = OA + OD
= 40 + 20
= 60 m
In Δ ADC,
By using Pythagoras theorem
AC2 = AO2 + DC2
AC2 = (60)2 + \((\frac{AC}{2})^2\)
AC2 = 3600 + \(\frac{AC\,\times\,AC}{4}\)
\(\frac{3}{4}\) AC2 = 3600
AC2 = 4800
AC = 40\(\sqrt 3\) m
So,
Length of string of each phone will be 40\(\sqrt 3\) m.