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A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

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Given that,

 AB = BC = CA 


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


AO = OA + OD

= 40 + 20 

= 60 m


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


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

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