# Ex.10.4 Q6 Circles Solution - NCERT Maths Class 9

## Question

A circular park of radius \(20\,\rm m\) is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of string of each phone.

## Text Solution

**What is known?**

Three boys are on boundary of a circular park. Distance between them is equal. Radius of circular park is given.

**What is unknown?**

Length of string of each phone.

**Reasoning:**

Centre and Centroid are the same for an equilateral triangle. And it divides the median in the ratio \(\begin{align}2:1.\end{align}\) The median is also the perpendicular bisector for the opposite side.

**Steps:**

Let \(\begin{align}{A, D, S}\end{align}\) denote the positions of Ankur, David and Syed respectively.

\(\begin{align}\Delta{ADS}\end{align}\) is an equilateral triangle since all the **\(3\)** boys are on equidistant from one another.

Let \({B}\) denote the mid-point of \({DS}\) and hence \({AB}\) is the median and perpendicular bisector of \({DS.}\) Hence \( \Delta {.ABS}\) is a right angled triangle with \(\begin {align} \angle {ABS}=90^{\circ} \end {align}\)

**\(O\)** (centroid) divides the line \({AB}\) in the ratio \(\text{2:1.}\) So \({OA : OB = 2 : 1.}\)

\(\begin{align}\frac{{OA}}{{OB}} &= \frac{2}{1}\,\end{align}\)

\(\begin{align} \text{Since } OA &= 20 \; \rm {then}\\OB &=10\rm{m}\end{align}\)

\(\begin {align} {AB}&={OA}+{OB}\\ &=20+10 \\&=30 \, \rm{m}\qquad \ldots .(1) \end {align}\)

Let the side of equilateral triangle \(\Delta {ADS}\) be \(2x.\)

\(\begin{align}{AD}&={DS} \\ &={SA}\\ &=2 x \ldots .(2) \end {align}\)

Since \({B}\) is the mid-point of \({DS,}\) we get

\(\begin {align}{BS}={BD}=x \dots(3)\end {align}\)

Applying Pythagoras theorem to \(\Delta {ABS,}\) we get:

\[\begin {align}{AD}^{2}&={AB}^{2}+{BD}^{2} \\ {(2 x)^{2}}&={30^{2}+x^{2}} \\ 4 x^{2}&=900+x^{2}\\ 3 x^{2}&=900 \\ x^{2}&=\frac{900}{3}=300 \\ x&=\sqrt{300} \\ &=17.32 \, \rm {m}\end{align}\]

\[\begin {align}{AD}={DS}={SA}=2 x=34.64 \, \mathrm{m} \end {align}\]

Length of the string \(=\) Distance between them

\(= AD\) or \(DS\) or \(SA = 34.64\,\rm m.\)