Question

Q5 Acircular park of radius 30 m is situated in a colony. Three boys Ankur, syed, and Mandeep are sitting at
equal distance on its boundary each having a toy telephone in his hand it talk each other. Find the length
of the string of each phone.​

Answer 1

Answer :

• Length = 17.4 m

S O L U T I O N :

Given,

• Radius of a circular park (r) = 30 m
• Three boys Ankur, syed, and Mandeep are sitting at equal distance.

To Find,

• The length of the string of each phone.

Explanation,

Let, ankur, syed, and Mandeep be donated by Point A & Point S & Point M.

Given, Three boys Ankur, syed, and Mandeep are sitting at equal distance.

⇒ AS = SM = AM -------(1)

Let, AS = SM = AM = 2x

In ΔASM,

⇒ AS = SM = AM -----[ (From equation (1) ]

.°. ΔASM is a equilateral triangle.

Now,

Draw OR ⊥ SM,

We know that,

Perpendicular drawn from centre of a circle to a chord bisects the chord.

⇒ SR = RM = ¹/2 × SM

⇒ SR = RM = ¹/2 × 2x

⇒ SR = RM = x

Join OS and AO,

In ∆ORS , m∠ORS = 90°

Applying Pythagoras theorem,

OS² = OR² + RS²

[ Put the values]

⇒ (30)² = OR² + x²

⇒ 900 = OR² + x²

⇒ OR² = 900 - x²

⇒ OR = √900 - x²

In ∆ARS, m∠ARS = 90°

Applying Pythagoras theorem,

AS² = AR² + RS²

[ Put the values]

⇒ (2x)² = AR² + x²

⇒ 4x² = AR² + x²

⇒ AR² = 4x² - x²

⇒ AR² = 3x²

⇒ AR = √3x

Now,

AR = AO + OR (A - O - R)

[ Put the values ]

⇒ √3x = 30 + √900 - x²

⇒ √3x - 30 = √900 - x²

[ Squaring both sides, we get ]

⇒ (√3x - 30)² = 900 - x²

⇒ 3x² - 60√3x + 900 = 900 - x²

⇒ 3x² - 60√3x = -x²

⇒ - 60√3x = -x² - 3x²

⇒ 60√3 = 4x²

⇒ 15√3 = x²

⇒ 225 × 3x = x⁴

⇒ 675x = x⁴

⇒ x = 8.7 m

So,

AS = SM = AM = 2x

⇒ AS = SM = AM = 2 × 8.7

⇒ AS = SM = AM = 17.4 m

Therefore,

The length of the string of each phone is 17.4 m.