Q: 6     A circular park of radius \small 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.
    

Answers (1)

Given: In the figure, A,S,D are positions Ankur, Syed and David respectively.

So, AS = SD = AD 

Radius of circular park = 20 m

  so, AO=SO=DO=20 m

Construction : AP\perp SD

Proof :

       

Let AS = SD = AD = 2x cm

In \triangleASD,

  AS = AD   and     AP\perp SD

So,  SP = PD = x cm 

In \triangleOPD, by pythagoras,

       OP^2=OD^2-PD^2

\Rightarrow OP^2=20^2-x^2=400-x^2

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

In \triangleAPD, by pythagoras,

   AP^2=AD^2-PD^2

\Rightarrow (AO+OP)^2+x^2=(2x)^2

\Rightarrow (20+\sqrt{400-x^2})^2+x^2=4x^2

\Rightarrow 400+400-x^2+40\sqrt{400-x^2}+x^2=4x^2

\Rightarrow 800+40\sqrt{400-x^2}=4x^2

\Rightarrow 200+10\sqrt{400-x^2}=x^2

\Rightarrow 10\sqrt{400-x^2}=x^2-200

Squaring both sides,

\Rightarrow 100(400-x^2)=(x^2-200)^2

\Rightarrow 40000-100x^2=x^4-40000-400x^2

\Rightarrow x^4-300x^2=0

\Rightarrow x^2(x^2-300)=0

\Rightarrow x^2=300

\Rightarrow x=10\sqrt{3}

Hence, length of string of each phone= 2x=20\sqrt{3}m

 

 

 

 

 

 

 

 

 

 

 

Preparation Products

Knockout NEET Sept 2020

An exhaustive E-learning program for the complete preparation of NEET..

₹ 15999/- ₹ 6999/-
Buy Now
Rank Booster NEET 2020

This course will help student to be better prepared and study in the right direction for NEET..

₹ 9999/- ₹ 4999/-
Buy Now
Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
Buy Now
Test Series NEET Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 11999/-
Buy Now
Exams
Articles
Questions