Question

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

Answer 1

Given:- Circular park with radius $20m$.

Let Ankur Syed and David be donated by points $A,S\&D$ respectively. Given all there are sitting at equal distance i.e $AS=SD=AD$

To find: Length of $AS=SD=AD$

Explanation: LEt $AS=SD=AD=2x$ [Ref. image 1]

In $ASD$, all sides are equal,

$\therefore ASD$ is equilateral triangle

We draw $OP\perp SD$ (perpendicular drawn from centre of circle to a chord of the chord)

So, $SP=DP=\frac{1}{2}SD$ [Ref. image 2]

$\Rightarrow SP=DP=\frac{2x}{2}=x$

Join $OS$ & $AO$

In $\Delta OPS$

By Pythagoras theorem

$O{S}^2=O{P}^2+P{S}^2$

$(20{)}^2=O{P}^2+x^2$

$\left(400\right)=OP+x^2$

$400-x^2=O{P}^2$

$OP=\sqrt{400-x^2}$

In $△APS$

By Pythagoras theorem

$A{S}^2=A{P}^2+P{S}^2$

$(2x{)}^2=A{P}^2+x^2$

$4x^2 = AP^2+x^2

$4x^2-x^2=A{P}^2$

$3x^2=A{P}^2$

$AP=\sqrt{3x}$

Now

$AP=AO+OP$ (Medium of equilateral triangle passes through its circum center)

$\sqrt{3}x=20+\sqrt{400-x^2}$ [Ref. image 3]

$\sqrt{3}x-20=\sqrt{400-x^2}$

$\sqrt{400-x^2}=\sqrt{3}x-20$

Squaring on both sides

$(\sqrt{400-x^2{)}^2}=(\sqrt{3}x-20{)}^2$

$400-x^2=(3x^2+(20)\times 20-2\times (\sqrt{3}x)\times 20$

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

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

$\Rightarrow 40\sqrt{3}x=4x^2$

$\Rightarrow 4x^2=40\sqrt{3}x$

$\Rightarrow \frac{x^2}{x}=\frac{40}{4}\sqrt{3}$

$\Rightarrow x=10\sqrt{3}m$