A circular park of radius 20 m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other.
Find the Length of the string of each phone.

Let Ankur, Syed and David standing on the point P, Q and R.

Let PQ = QR = PR = X



\(\therefore \) DPQR is an equilateral traingle.

Drawn altitudes PC, QD and RN from vertices to the sides of a traingle and intersect these altitudes at the centre of a circle M.

As PQR is an equilateral, therefore these altitudes bisects their sides.

In \(\triangle{PQC}\),
\({PQ}^2 = {PC}^2 + {QC}^2 \)
(By Pythagoras theorem)
\(\Rightarrow \) \({x}^2 = {PC}^2 + ( \frac{x}{2} )^2 \)
\(\Rightarrow \)\({PC}^2 = {x}^2 - (\frac{x}{2} )^2\)
\(\Rightarrow \) \({PC}^2 = {x}^2 - \frac{x^2}{4} = 3\frac{x^2}{4} \)
(Since, QC = \( \frac{1}{2} \) QR = \(\frac{x}{2} \) )
\(\therefore \) PC = \(\frac{\sqrt{3}x}{2} \)
Now, MC = PC - PM = \(\frac{\sqrt{3}x}{2} \) - 20
(Since, PM = radius)

Now, in \(\triangle{QCM}\),
\({QM}^2 = {QC}^2 + {MC}^2 \)
(By Pythagoras theorem)
\(\Rightarrow \) \( 20^2 = (\frac{x}{2} )^2 + (\frac{\sqrt{3}x}{2} - 20)^2 \)
(Since, QM = radius)
\(\Rightarrow \) \(400 = \frac{x^2}{4} +( \frac{\sqrt{3}x}{2})^2 - 20\sqrt{3}x + 400\)
\(\Rightarrow \) \(0 = {x}^2 - 20\sqrt{3}x\)
\(\Rightarrow \) \({x}^2 = 20\sqrt{3}x\)
\(\therefore \) \(x = 20\sqrt{3}\)

Hence, PQ = QR = PR = \(20\sqrt{3}\)m.