3 Tutor System
Starting just at 265/hour

# 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

Now, in $$\triangle{QCM}$$,
$${QM}^2 = {QC}^2 + {MC}^2$$
(By Pythagoras theorem)
$$\Rightarrow$$ $$20^2 = (\frac{x}{2} )^2 + (\frac{\sqrt{3}x}{2} - 20)^2$$
$$\Rightarrow$$ $$400 = \frac{x^2}{4} +( \frac{\sqrt{3}x}{2})^2 - 20\sqrt{3}x + 400$$
$$\Rightarrow$$ $$0 = {x}^2 - 20\sqrt{3}x$$
$$\Rightarrow$$ $${x}^2 = 20\sqrt{3}x$$
$$\therefore$$ $$x = 20\sqrt{3}$$
Hence, PQ = QR = PR = $$20\sqrt{3}$$m.