Question

A circular park of radius 30m is situated in a colony. Three boys Amit, Naren and Mukesh 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.​

Answer 1

Given:

• The radius of the circular park is 30 m

In the attached image, radius = AS = SD = AD = 30m

Three boys Amit, Naren and Mukesh are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other.

So, let Amit be 'A', Naren is 'S' and Mukesh is 'S'.

And the string starting from A to S, S to D and D to A.

To find:

• The string of each phone.

As three boys are equidistance from each other so,

Chord AS = chord BD = chord DA

⇒ ΔASD is an equilateral triangle.

→ AS = SD = DA

Also,

Construction,

• AP ⊥ SD

• SR ⊥ AD

• DQ ⊥ AS

Let,  AS = SD = AD = 2x cm

In ΔASD,

AS = AD and AP ⊥ SD

Then,

PD = SP = x m

And,

In the triangle OPD,

We get,

OP² = OD² - PD²  

• [By Pythagoras theorem]

→ OP² = 30² - x²

→ OP² = 900 - x²

→ OP = $\sqrt{900-x^{2}}$

Now,

In triangle APD,

We get,

AP² = AD² - PD²

→ (AO+OP)² + x² = (2x)²

→ (30+ $\sqrt{900-x^{2}}$ )² + x² = 4x²

• (a+b)² = a² + b² + 2ab

→ (30)² + ($\sqrt{900-x^{2}}$)² + 2(30)($\sqrt{900-x^{2}}$) + x² = 4x²

→ 900 + 900 - x² + 60($\sqrt{900-x^{2}}$) + x² = 4x²

→ 1800 + 60($\sqrt{900-x^{2}}$) = 4x²

→ [1800 + 60($\sqrt{900-x^{2}}$) ] ÷ 4 = 4x² ÷ 4

→ 450 + 15($\sqrt{900-x^{2}}$) = x²

→ 15($\sqrt{900-x^{2}}$) = x² - 450

• By transporting (450) to RHS

Now,

• By squaring both the sides, we get,

→ [ 15($\sqrt{900-x^{2}}$) ]² = [ x² - 450 ]²

→ 225 (900 - x²) = x⁴ + 202500 - 2(x²)(450)

→ 202500 - 225x² = x⁴ + 202500 - 900x²

→ 202500 - 202500 - 225x² = x⁴ - 900x²

• Transporting 202500 to LHS

→ 900x² - 225x² = x⁴

• By transporting 225x² to LHS

→ 675x² = x⁴

→ x⁴ - 675x² = 0

• By trasporting 675x² to LHS

→ x²(x²-675) = 0

• Taking x² as common

→ x²- 675 = 0

→ x² = 675

• [By transporting 675 to RHS]

→ x = $\sqrt{675}$

• Taking the square roots of both the sides

→ x = $\sqrt{\underline{5*5}*\underline{3*3}*3}$

→ x = 15$\sqrt{3}$

Now,

Length pf each string = 2x = 2 (15$\sqrt{3}$) = 30$\bf \sqrt{3}$ m

Answer 2

☆ Given Question :-

A circular park of radius 30m is situated in a colony. Three boys Amit, Naren and Mukesh 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.

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Step by step Calculation

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☆ Given :-

• Radius of circular park, r = 30m

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Find :-

• Length of string of each phone.

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Solution :-

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★Let Amit, Naren and Mukesh 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 = 30m

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

⇛Draw AB ⊥ SD

Hence, SB = BD

$\sf \:  ⇛SB = \dfrac{1}{2} × SD$

$\sf \:  ⇛SB = \dfrac{x}{2} \: m$

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In ∆ABS, ∠B=90°

☆ By pythagoras theorem,

⇛AS² = AB² + BS²

⇛AB² = AS² - BS²

⇛AB² = x² - (x/2)²

⇛AB² = 3x²/4

⇛AB = √3x/2m

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★AB = AO + OB

⇛OB = AB - AO

⇛OB = (√3x/2 - 30)m

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★In ∆OBS

⇛OS² = OB² + SB²

⇛30² = (√3x/2 - 30)² + (X/2)²

⇛900 = 3/4x² + 900 - 2(30)(√3x/2) + x²/4

⇛0 = x² - 30√3x

⇛x = 30√3m

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Hence,The length of the string of each phone is 30√3m