Let AB = BC = AC = a m
Hence ABC is an equilateral triangle.
Draw AD ⊥ BC.
Now AD is the median of ΔABC and passes through the center O.
Also, O is the centroid of ΔABC and OA is the radius of the triangle, OA = 40 m.
OA = 2/3 AD
[The centroid of an equilateral triangle divides each median in 2: 1]
Let the sides of the triangle be BD = a / 2 m.
In ΔABD,
AB² = BD² + AD² (from Pythagoras theorem)
⇒ AD² = AB² - BD²
⇒ AD² = a² - (a / 2) ²
⇒ AD² = a² - a² / 4
⇒ AD² = (4a² - a²) / 4
⇒ AD² = 3a² / 4
⇒ AD = √3a / 2
OA = 2/3 AD
40 m = 2/3 × √3a / 2
40 = 2√3a / 6
40 = √3a / 3
40 × 3 = √3a
120 = √3a
a = 120 / √3
a = (120 × √3) / (√3 × √3)
[By rationalising the denominator]
a = 120√3 / 3
a = 40√3 m
Hence, the length of the string of each phone is 40√3 m.
