# A circular park of radius 20m 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 each other. Find the length of string of each phone

**Solution:**

Centre and Centroid are the same for an equilateral triangle, and it divides the median in the ratio 2 : 1.

Let A, D, S denote the positions of Ankur, David, and Syed, respectively.

∆ADS is an equilateral triangle since all the 3 boys are equidistant from one another.

Let B denote the mid-point of DS, and hence AB is the median and perpendicular bisector of DS.

Hence ∆ABS is a right-angled triangle with ∠ABS = 90º.

O (centroid) divides the line AB in the ratio 2 : 1. So OA : OB = 2 : 1.

OA/OB = 2/1

Since OA = 20 (given)

thus, OB = 10m

AB = OA + OB = 20 + 10 = 30 m ….(1)

Let the side of the equilateral triangle ∆ADS be 2x.

AD = DS = SA = 2x …. (2)

Since B is the mid-point of DS, we get BS = BD = x …. (3)

Applying Pythagoras theorem to ∆ABD, we get:

AD^{2} = AB^{2} + BD^{2}

(2x)^{2} = 30^{2} + x^{2}

4x^{2} = 900 + x^{2}

3x^{2} = 900

x^{2} = 300

x = 10√3

x = 17.32

AD = DS = SA = 2x = 34.64 m

Length of the string = Distance between them = AD or DS or SA = 34.64 m.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 10

**Video Solution:**

## A circular park of radius 20m 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

### Maths NCERT Solutions Class 9 - Chapter 10 Exercise 10.4 Question 6:

**Summary:**

If a circular park of radius of 20m 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, then the length of the string of each phone is 34.64m.

**☛ Related Questions:**

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