Question

• A circular park of radius 20m 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 their hands to talk each other. Find the length of the string of each phone.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that,

• A circular park of radius 20m 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 their hands to talk each other.

Let assume that

• Ankur, Syed and David at A, B and C respectively.

Let we draw angle bisector of BAC intersecting BC at D.

As AB and AC are equal, so angle bisector passes through centre of the circle O and perpendicular bisector of BC.

So, BD = DC and AD is perpendicular to BC.

Let assume that AB = BC = CA = 2x

So, BD = DC = x.

Now, In right triangle ABD

Using Pythagoras Theorem, we have

$\rm \: {AB}^{2} = {BD}^{2} + {AD}^{2}$

$\rm \: {(2x)}^{2} = {x}^{2} + {AD}^{2}$

$\rm \: 4 {x}^{2} = {x}^{2} + {AD}^{2}$

$\rm \: 4 {x}^{2} - {x}^{2} = {AD}^{2}$

$\rm \: 3{x}^{2} = {AD}^{2}$

$\red{\rm\implies \:AD = \sqrt{3} \: x }\:$

Now, Consider

$\rm \: AD = OA + OD$

$\rm \: \sqrt{3}x = 20 + OD$

$\red{\rm\implies \:OD = \sqrt{3}x - 20}$

Now, In right triangle OBD

$\rm \: {OB}^{2} = {OD}^{2} + {BD}^{2}$

$\rm \: {20}^{2} = {(\sqrt{3}x - 20)}^{2} + {x}^{2}$

$\rm \: 400 = {3x}^{2} + 400 - 40 \sqrt{3}x + {x}^{2}$

$\rm \: 0 = {3x}^{2} - 40 \sqrt{3}x + {x}^{2}$

$\rm \: 4 {x}^{2} - 40 \sqrt{3}x = 0$

$\rm \: 4x(x - 10 \sqrt{3}) = 0$

$\rm\implies \:\rm \: x - 10 \sqrt{3} = 0 \: \: \: \{as \: x \: \ne \: 0 \}$

$\rm\implies \:x = 10 \sqrt{3} \: m$

Hence,

$\rm\implies \:AB = 2x = 20 \sqrt{3} \: m$