Question

$\huge \mathfrak{ \fcolorbox{black}{red}{ question}}$
A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on it's 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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Answer 1

Answer:

• Length of string of each phone is 20√3 m.

Explanation:

Given :-

• Radius of circular part is 20 m.
• Ankur, syed and David are sitting at equal distance on boundary of circle.

To find :-

• Length of string of each phone.

Solution :-

Let, Points where Ankur, Syed and David are sitting be A, S and D respectively.

Radius = AO = 20 m

Construction :- Draw a median AP on SD.

Three boys are sitting at equal distance. So,

• AS = SD = DA

By this, ∆ASD is an equilateral triangle.

We know,

Centroid divides the median in the ratio 2:1. [Centroid is a point where all three median of equilateral triangle intersect each other.]

So,

$\sf \longrightarrow \dfrac{OA}{OP} = \dfrac{2}{1}$

• OA is 20 m.

$\sf \longrightarrow \dfrac{20}{OP} = \dfrac{2}{1}$

• Cross multiply

$\sf \longrightarrow 2 OP = 20$

$\sf \longrightarrow OP = \dfrac{20}{2}$

$\sf \longrightarrow \bold{OP = 10}$

Measure of OP is 10 m.

• AP = AO + OP

$\sf \longrightarrow AP = 20 + 10$

$\sf \longrightarrow \bold{AP = 30}$

Measure of AP is 30 m.

Median will divide SD in two equal parts.

PD = $\sf \dfrac{SD}{2}$

• SD = AD

PD = $\sf \dfrac{AD}{2}$

In equilateral triangle medians divides the line in equal parts and median is always perpendicular on that line.

So,

∆APD and ∆APS are two right angle triangles.

In ∆APD :

By Pythagoras theorem :

$\sf \longrightarrow AD^{2} = AB^{2} + PD^{2}$

$\sf \longrightarrow AD^{2} = AB^{2} + (\dfrac{AD}{2})^{2}$

$\sf \longrightarrow AD^{2} = (30)^{2} + (\dfrac{AD}{2})^{2}$

$\sf \longrightarrow AD^{2} = 900 + \dfrac{1}{4}AD^{2}$

$\sf \longrightarrow AD^{2} - \dfrac{1}{4} AD^{2} = 900$

$\sf \longrightarrow \dfrac{4 \: AD^{2} - AD^{2}}{4} = 900$

$\sf \longrightarrow \dfrac{3}{4} AD^{2} = 900$

$\sf \longrightarrow 3 \: AD^{2} = 900 \times 4$

$\sf \longrightarrow 3 \: AD^{2} = 3600$

$\sf \longrightarrow AD^{2} = \dfrac{3600}{3}$

$\sf \longrightarrow AD^{2} = 1200$

$\sf \longrightarrow AD = \sqrt{1200}$

$\sf \longrightarrow AD = \sqrt{2 \times 2 \times 2 \times 2 \times 5 \times 5 \times 3}$

$\sf \longrightarrow AD = 2 \times 2 \times 5 \sqrt{3}$

$\sf \longrightarrow \bold{AD = 20 \sqrt{3}}$

AD will be the length of string of each phone.

Therefore,

Length of string of each phone is 20√3 m.

Answer 2

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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

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

In △ABS,∠B=90⁰

By Pythagoras theorem,

AS² = AB² + BS²

$∴ {AB}^{2} = {AS}^{2} - {BS}^{2} = {x}^{2} - ( \frac{x}{2} {)}^{2} = \frac{3 {x}^{2} }{4}$

$∴AB= \frac{ \sqrt{3x} }{2} m$

Now, AB = AO + OB

OB = AB − AO

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

In △OBS,

OS² =OB² + SB²

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