6. 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.

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)

i.e., \({x}^2 = {PC}^2 + (x/2)^2 \)

i.e., \({PC}^2 = {x}^2 - (x/2)^2\)

i.e., \({PC}^2 = {x}^2 - {x}^2/4 = 3{x}^2/4\) ...(Since, QC = 1/2 QR = x/2)

Therefore, PC = \(\sqrt{3}\)x/2

Now, MC = PC - PM = \(\sqrt{3}\)x/2 - 20 ...(Since, PM = radius)

Now, in \(\triangle{QCM}\),

\({QM}^2 = {QC}^2 + {MC}^2 \) ...(By Pythagoras theorem)

i.e., \({20}^2 = {x/2}^2 + (\sqrt{3}x/2 - 20)^2 \) ...(Since, QM = radius)

i.e., \(400 = {x}^2/4 +( \sqrt{3}x/2)^2 - 20\sqrt{3}x + 400\)

i.e., \(0 = {x}^2 - 20\sqrt{3}x\)

i.e., \({x}^2 = 20\sqrt{3}x\)

Therefore, \(x = 20\sqrt{3}\)

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