Question

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?
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Answer 1

GIVEN :–

• 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.

TO FIND :–

• Find the length of the string of each phone = ?

SOLUTION :–

• Let AP = SD = AD = x cm. , So SP = PD = x/2.

• In △SPO –

⇒ (SO)² = (OP)² + (SP)²

⇒ (20)² = (OP)² + (x/2)²

⇒ 400 = (OP)² + x²/4

⇒ (OP)² = 400 - x²/4

⇒ OP = √(400 - x²/4) ______eq.(1)

• Now in △APS –

⇒ (AS)² = (SP)² + (AP)²

⇒ (x)² = (x/2)² + (AP)²

⇒ (AP)² = x² - x²/4

⇒ (AP)² = 3x²/4

⇒ AP = √3x/2 ______eq.(2)

• We know that –

⇒ AP = AO + OP

• Using eq.(1) & eq.(2) –

⇒ √3x/2 = 20 + √(400 - x²/4)

⇒ √3x/2 - 20 = √(400 - x²/4)

• Square on both side –

⇒ (√3x/2 - 20)² = 400 - x²/4

⇒ 3x²/4 + 400 - 20√3x = 400 - x²/4

⇒ x² - 20√3x = 0

⇒ x = 20√3 cm

• Hence , The length of the string of each phone is 20√3 cm.

Answer 2

${\bf{Given\::}}$

• Radius of a circular park is 20 m.

• Three boys Ankur, Syed and David are sitting at equal distances on its boundary.

$\\ {\bf{To\: Find\::}}$

• Length of the string of each phone.

$\\ {\bf{Solution\::}}$

Let,

As shown in figure,

• Point A denotes Ankur.

• Point B denotes Syed.

• Point C denotes David.

It is given that,

• They are sitting at equal distances.

$:\implies\:\bf{AB\:=\:BC\:=\:AC\:=\:x\:m}$

$\bf\red{\therefore\:\triangle{ABC}\:is\:an\: equilateral\:\triangle.}$

• Point O denotes the centre of circular park.

➣ We draw OD, which is perpendicular to BC.

• $\bf{OD\perp{BC}}$

Thus,

$:\implies\:\bf{BD\:=\:CD\:=\:\dfrac{x}{2}\:}$

$\red\checkmark$ Join OB & OA.

Now,

In ∆OBD,

According to Pythagoras theorem,

➠ $\tt{(OB)^2\:=\:(BD)^2\:+\:(OD)2\:}$

Where,

• OB is the radius of the circle = 20 m

➠ $\tt{(20)^2\:=\:\left(\dfrac{x}{2}\right)^2\:+\:(OD)2\:}$

➠ $\tt{400\:=\:\dfrac{x^2}{4}\:+\:(OD)2\:}$

➠ $\tt{(OD)^2\:=\:400\:-\:\dfrac{x^2}{4}\:}$

➠ $\tt{OD\:=\:\sqrt{400\:-\:\dfrac{x^2}{4}}\:}$

Again,

In ∆ABD,

➠ $\tt{(x)^2\:=\:(AD)^2\:+\:\left(\dfrac{x}{2}\right)^2\:}$

➠ $\tt{x^2\:=\:(AD)^2\:+\:\dfrac{x^2}{4}\:}$

➠ $\tt{(AD)^2\:=\:x^2\:-\:\dfrac{x^2}{4}\:}$

➠ $\tt{AD\:=\:\sqrt{\dfrac{3x^2}{4}}\:}$

➠ $\tt{AD\:=\:\dfrac{\sqrt{3}x}{2}\:}$

Now,

As we know that,

$:\implies\:\bf{AD\:=\:OA\:+\:OD\:}$

Where,

• OA is the radius of the circle = 20 m

$:\implies\:\tt{\dfrac{\sqrt{3}x}{2}\:=\:20\:+\:\sqrt{400\:-\:\dfrac{x^2}{4}}\:}$

$:\implies\:\tt{\dfrac{\sqrt{3}x}{2}\:-\:20\:=\:\sqrt{400\:-\:\dfrac{x^2}{4}}\:}$

Squaring both sides, we get

$:\implies\:\tt{\left(\dfrac{\sqrt{3}x}{2}\:-\:20\right)^2\:=\:\left(\sqrt{400\:-\:\dfrac{x^2}{4}}\right)^2\:}$

$:\implies\:\tt{\left(\dfrac{\sqrt{3}x}{2}\right)^2\:+\:(20)^2\:-\:2\times{\dfrac{\sqrt{3}x}{2}}\times{20}\:=\:400\:-\:\dfrac{x^2}{4}\:}$

$:\implies\:\tt{\dfrac{3x^2}{4}\:+\:\cancel{400}\:-\:20\sqrt{3}x\:=\:\cancel{400}\:-\:\dfrac{x^2}{4}\:}$

$:\implies\:\tt{\dfrac{3x^2}{4}\:-\:20\sqrt{3}x\:=\:-\:\dfrac{x^2}{4}\:}$

$:\implies\:\tt{\dfrac{3x}{4}\:-\:20\sqrt{3}\:=\:-\:\dfrac{x}{4}\:}$

$:\implies\:\tt{\dfrac{3x}{4}\:+\:\dfrac{x}{4}\:=\:20\sqrt{3}\:}$

$:\implies\:\tt{\dfrac{3x\:+\:x}{4}\:=\:20\sqrt{3}\:}$

$:\implies\:\tt{\dfrac{4x}{4}\:=\:20\sqrt{3}\:}$

$:\implies\:\bf\pink{x\:=\:20\sqrt{3}\:m\:=\:34.64\:m\:}$

$\bf{\therefore}$ Length of the string of each phone is 34.64 m.

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