Answer:
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.
A circular park of radius 20 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
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
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:
AD2 = AB2 + BD2
(2x)2 = 302 + x2
4x2 = 900 + x2
3x2 = 900
x2 = 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.