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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.
Answer :

Let Ankur, Syed and David standing on the point P, Q and R.
Let PQ = QR = PR = X
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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.