Ex.10.4 Q6 Circles Solution - NCERT Maths Class 9

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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.

 Video Solution
Ex 10.4 | Question 6

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.


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.



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 \(SA = 34.64\,\rm m.\)

 Video Solution
Ex 10.4 | Question 6
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