A circular park of radius 20,m is situated in | vedclass

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Question

(A) Let the circle with centre $O$ represent the park and the points $A, S,$ and $D$ represent the positions of Ankur,Syed,and David respectively. Sin

Vedclass · Accepted Answer

$20 \sqrt{3} \, m$

Vedclass · Answer

(A) Let the circle with centre $O$ represent the park and the points $A, S,$ and $D$ represent the positions of Ankur,Syed,and David respectively. Since Ankur,Syed,and David are sitting at equal distances from each other,$\Delta ASD$ is an equilateral triangle.Draw the perpendicular bisector of $SD$ from its midpoint $M$. It will pass through the centre $O$ and vertex $A$.Let $SM = x \, m$. Then $SD = 2SM = 2x \, m$.In $\Delta OMS$,$\angle M = 90^{\circ}$. Using the Pythagorean theorem:$OM^2 = OS^2 - SM^2 = (20)^2 - x^2 = 400 - x^2$$OM = \sqrt{400 - x^2}$Since $O$ is the centroid of the equilateral triangle $ASD$,the distance from the vertex to the centroid is twice the distance from the centroid to the midpoint of the opposite side.$AM = AO + OM = 20 + \sqrt{400 - x^2}$In $\Delta ASD$,the height $AM = \frac{\sqrt{3}}{2} 	imes 	ext{side} = \frac{\sqrt{3}}{2} 	imes (2x) = x\sqrt{3}$.Equating the two expressions for $AM$:$x\sqrt{3} = 20 + \sqrt{400 - x^2}$$x\sqrt{3} - 20 = \sqrt{400 - x^2}$Squaring both sides:$(x\sqrt{3} - 20)^2 = 400 - x^2$$3x^2 - 40\sqrt{3}x + 400 = 400 - x^2$$4x^2 - 40\sqrt{3}x = 0$$4x(x - 10\sqrt{3}) = 0$Since $x 
eq 0$,$x = 10\sqrt{3}$.The length of the string is the side of the triangle $SD = 2x = 2(10\sqrt{3}) = 20\sqrt{3} \, m$.

$A$ 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 with each other. Find the length of the string of each phone.

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