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  • 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 his hands to talk each other. Find the length of the string of each phone.

Q: 6 A circular park of radius $20$ m is situated in a colony. Three boys Ankur, Syed and David are sitting at an equal distance 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.

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Given: In the figure, A, S, and D are positions Ankur, Syed and David respectively.
So, $\mathrm{AS}=\mathrm{SD}=\mathrm{AD}$
The radius of the circular park $=20 \mathrm{~m}$
so, $\mathrm{AO}=\mathrm{SO}=\mathrm{DO}=20 \mathrm{~m}$
Construction: $\mathrm{AP} \perp \mathrm{SD}$
Proof :

Let $\mathrm{AS}=\mathrm{SD}=\mathrm{AD}=2 \mathrm{x} \mathrm{cm}$
In $\triangle \mathrm{ASD}$,
$\mathrm{AS}=\mathrm{AD}$ and $\mathrm{AP} \perp \mathrm{SD}$
So, $\mathrm{SP}=\mathrm{PD}=\mathrm{x} \mathrm{cm}$
In $\triangle \mathrm{OPD}$, by Pythagoras,
$O P^2=O D^2-P D^2$
$\Rightarrow O P^2=20^2-x^2=400-x^2$
$\Rightarrow O P=\sqrt{400-x^2}$

In $\triangle \mathrm{APD}$, by Pythagoras,

$A P^2=A D^2-P D^2$
$\Rightarrow (A O+O P)^2+x^2=(2 x)^2$
$\Rightarrow \left(20+\sqrt{400-x^2}\right)^2+x^2=4 x^2$
$\Rightarrow 400+400-x^2+40 \sqrt{400-x^2}+x^2=4 x^2$
$\Rightarrow 800+40 \sqrt{400-x^2}=4 x^2$
$\Rightarrow 200+10 \sqrt{400-x^2}=x^2$
$\Rightarrow 10 \sqrt{400-x^2}=x^2-200$

Squaring both sides,
$\Rightarrow 100\left(400-x^2\right)=\left(x^2-200\right)^2$
$\Rightarrow 40000-100 x^2=x^4-40000-400 x^2$
$\Rightarrow x^4-300 x^2=0$
$\Rightarrow x^2\left(x^2-300\right)=0$
$\Rightarrow x^2=300$
$\Rightarrow x=10 \sqrt{3}$
Hence, the length of the string of each phone $=2 x=20 \sqrt{3} \mathrm{~m}$

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seema garhwal

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