Question

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

Answer 1

$\huge\sf\green{✯Answer✯}$

Given :-

Radius of circular park = 20m

Find :-

Length of string of each phone

Solution :-

★Let Ankur, Syed and David be A ,S and D

★These Boys are sitting at equal distance

So, ∆ASD is an equilateral triangle

★Let the radius be r metres

Hence, OS = r = 20m

★Let the length of each side of ∆ASD be X metres

⇛Draw AB ⊥ SD

Hence, SB = BD

⇛1/2 × SB

⇛X/2m

In ∆ABS ∠B=90°

By pythagoras theorem,

⇛AS² = AB² + BS²

⇛AB² = AS² - BS²

⇛AB² = X² - (X/2)²

⇛AB² = 3x²/4

⇛AB = √3x/2m

★AB = AO + OB

⇛OB = AB - AO

⇛OB = (√3x/2 - 20)m

★In ∆OBS

⇛OS² = OB² + SB²

⇛20² = (√3x/2 - 20)² + (X/2)²

⇛400 = 3/4x² + 400 - 2(20)(√3x/2) + x²/4

⇛0 = x² - 20√3x

⇛x = 20√3m

Hence,The length of the string of each phone is 20√3m

Answer 2

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$\large\bold\green{Given :-}$Radius of circular park = 20m

$\large\bold\red{Find :-}$Length of string of each phone

$\large\bold\blue{Solution :-}$

■Let Ankur, Syed and David be A ,S and D. As these Boys are sitting at equal distance

So, ∆ASD is an equilateral triangle

■Let the radius be r metres

Hence, OS = r = 20m

■Let the length of each side of ∆ASD be 'x' metres

Draw angle bisector of angle A.

As AS = AD

So, AB passes through centre of circle and perpendicular bisector of SD.

⇛AB ⊥ SD and SB = BD = 1/2 × SB = x/2 m

■Now, In right triangle, ∆ABS

Using pythagoras theorem,

⇛AS² = AB² + BS²

⇛AB² = AS² - BS²

⇛AB² = x² - (x/2)²

⇛AB² = 3x²/4

⇛AB = √3x/2 m

■AB = AO + OB

⇛OB = AB - AO

⇛OB = (√3x/2 - 20) m

■ In right triangle ∆OBS

⇛OS² = OB² + SB²

⇛20² = (√3x/2 - 20)² + (x/2)²

⇛400 = 3/4x² + 400 - 2(20)(√3x/2) + x²/4

⇛0 = x² - 20√3x

⇛0 = x(x - 20√3)

⇛x = 20√3m

Hence, the length of the string of each phone is 20√3 m.

$\huge \fcolorbox{black}{cyn}{♛Hope it helps U♛}$