Question

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‎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

Answer:

Let Ankur be represented as A, Syed as S and David as D.

The boys are sitting at an equal distance.

Hence, △ASD is an equilateral triangle.

Let the radius of the circular park be r meters.

∴OS=r=20m.

Let the length of each side of △ASD be x meters.

Draw AB⊥SD

$\sf \: ∴ SB=BD= \frac{1}{2} SD = \frac{x}{2} m$

In △ABS,∠B=90°

By Pythagoras theorem,

$\sf \: AS {}^{2} =AB {}^{2} +BS {}^{2}$

$\sf \: AB {}^{2} =A S {}^{2} - BS {}^{2} - x {}^{2} - ( \frac{x}{2} ) {}^{2} = \frac{3x {}^{2} }{4}$

$\sf \: Ab = \sqrt{ \frac{3x}{2} }m$

Now, AB=AO+OB

OB=AB−AO

$\sf \: OB = ( \sqrt{ \frac{3x}{2} - 20)m }$

$In △OBS, \\ </p><p>OS {}^{2} =OB {}^{2} +SB {}^{2}$

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

$\boxed{ \sf \: 400 = \frac{3}{4} x {}^{2} + 400 - 2(20) \: ( \frac{ \sqrt{3x} }{2}) + \frac{x {}^{2} }{4} }$

$\sf \: 0=x {}^{2} −20 \sqrt{3x}$

$\sf \: ∴x=20 \sqrt{3} m$

$\sf \: The \: length \: of \: the \: string \: of \: each \: phone \: is 20 \sqrt{3m}$

Hence, Vertified

Answer 2

Solution:—

• 20√3 m

Step by Step Explanation:—

Given,

→ The positions of Ankur, Syed and David are represented as A, B and C respectively.

→ As they are sitting at equal distances, the triangle is equilateral

→ AD ⊥ BC is drawn.

→ AD is the median of ΔABC and it passes through the centre O.

→ O is the centroid of the ΔABC. OA is the radius of the triangle.

→ OA = 2/3 AD

Let the side of a triangle a metres then BD = a/2 m.

On applying Pythagoras theorem in ΔABD,

→ AB² = BD² + AD²

→ AD² = AB² - BD²

→ AD² = a² -( a/2)²

→ AD² = 3a²/4

→ AD = √3a/2

OA = 2/3 AD

→ 20 m = 2/3 × √3a/2

→ a = 20√3 m

Answer

The length of the string of the toy is 20√3 m.

$@vikramjeeth$