**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

AC^{2} = AO^{2} + DC^{2}

AC^{2} = (60)^{2} + **\((\frac{AC}{2})^2\) **

AC^{2} = 3600 + **\(\frac{AC\,\times\,AC}{4}\) **

\(\frac{3}{4}\)** **AC^{2} = 3600

AC^{2} = 4800

AC = 40\(\sqrt 3\) m

**So,**

Length of string of each phone will be 40\(\sqrt 3\) m.