Given : In figure, points A, S and D are the positions of Ankur, Syed and David respectively.
Therefore AS = SD = AD.
Radius of circular park = 20 m, therefore AO = SO = DO = 20
Construction: Draw AP ⊥SD
Proof: Let AS = SD = AD = 2x cm
In ΔASD,
AS = AD and AP⊥SD [∵ By construction]
Therefore, SP = PD = x cm [∵ SD = 2x cm]
In ΔOPD, using Pythagoras theorem
OP² = OD² – PD²
⇒ OP² = 20² – x² = 400 – x²
⇒ OP = √(400 – x²)
Now, in ΔAPD, using Pythagoras theorem
AP² + PD² = AD²
⇒ (AO + OP)² + x² = (2x)² ⇒ (20 + √400 – x²)² + x² = 4x²
⇒ 400 + 400 – x² + 2 × 20 × √(400 – x²) + x² = 4x²
⇒ 800 + 40√(400 – x²) = 4x² ⇒ 200 + 10√(400 – x²) = x²
⇒ 10√(400 – x²) = x² – 200
Squaring both sides
100 (400 – x²) = (x² – 200)²
⇒ 40000 – 100x² = x⁴ + 40000 – 400x²
⇒ x⁴ – 300 x² = 0 ⇒ x²(x² – 300) = 0
⇒ x² = 300 ⇒ x = 10√3
Hence, the length of the string of each phone = 2x = 2 × 10√3 = 20√3 m.