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Answers
Answer:
\begin{gathered}{\bf{Given\::}} \\ \end{gathered}
Given:
Radius of a circular park is 20 m.
Three boys Ankur, Syed and David are sitting at equal distances on its boundary.
\begin{gathered} \\ {\bf{To\: Find\::}} \\ \end{gathered}
ToFind:
Length of the string of each phone.
\begin{gathered} \\ {\bf{Solution\::}} \\ \end{gathered}
Solution:
Let,
As shown in figure,
Point A denotes Ankur.
Point B denotes Syed.
Point C denotes David.
It is given that,
They are sitting at equal distances.
\begin{gathered}:\implies\:\bf{AB\:=\:BC\:=\:AC\:=\:x\:m} \\ \end{gathered}
:⟹AB=BC=AC=xm
\bf\red{\therefore\:\triangle{ABC}\:is\:an\: equilateral\:\triangle.}∴△ABCisanequilateral△.
Point O denotes the centre of circular park.
➣ We draw OD, which is perpendicular to BC.
\begin{gathered}\bf{OD\perp{BC}} \\ \end{gathered}
OD⊥BC
Thus,
\begin{gathered}:\implies\:\bf{BD\:=\:CD\:=\:\dfrac{x}{2}\:} \\ \end{gathered}
:⟹BD=CD=
2
x
\red\checkmark✓ Join OB & OA.
Now,
In ∆OBD,
According to Pythagoras theorem,
➠ \tt{(OB)^2\:=\:(BD)^2\:+\:(OD)2\:}(OB)
2
=(BD)
2
+(OD)2
Where,
OB is the radius of the circle = 20 m
➠ \tt{(20)^2\:=\:\left(\dfrac{x}{2}\right)^2\:+\:(OD)2\:}(20)
2
=(
2
x
)
2
+(OD)2
➠ \tt{400\:=\:\dfrac{x^2}{4}\:+\:(OD)2\:}400=
4
x
2
+(OD)2
➠ \tt{(OD)^2\:=\:400\:-\:\dfrac{x^2}{4}\:}(OD)
2
=400−
4
x
2
➠ \tt{OD\:=\:\sqrt{400\:-\:\dfrac{x^2}{4}}\:}OD=
400−
4
x
2
Again,
In ∆ABD,
➠ \tt{(x)^2\:=\:(AD)^2\:+\:\left(\dfrac{x}{2}\right)^2\:}(x)
2
=(AD)
2
+(
2
x
)
2
➠ \tt{x^2\:=\:(AD)^2\:+\:\dfrac{x^2}{4}\:}x
2
=(AD)
2
+
4
x
2
➠ \tt{(AD)^2\:=\:x^2\:-\:\dfrac{x^2}{4}\:}(AD)
2
=x
2
−
4
x
2
➠ \tt{AD\:=\:\sqrt{\dfrac{3x^2}{4}}\:}AD=
4
3x
2
➠ \tt{AD\:=\:\dfrac{\sqrt{3}x}{2}\:}AD=
2
3
x
Now,
As we know that,
\begin{gathered}:\implies\:\bf{AD\:=\:OA\:+\:OD\:} \\ \end{gathered}
:⟹AD=OA+OD
Where,
OA is the radius of the circle = 20 m
\begin{gathered}:\implies\:\tt{\dfrac{\sqrt{3}x}{2}\:=\:20\:+\:\sqrt{400\:-\:\dfrac{x^2}{4}}\:} \\ \end{gathered}
:⟹
2
3
x
=20+
400−
4
x
2
\begin{gathered}:\implies\:\tt{\dfrac{\sqrt{3}x}{2}\:-\:20\:=\:\sqrt{400\:-\:\dfrac{x^2}{4}}\:} \\ \end{gathered}
:⟹
2
3
x
−20=
400−
4
x
2
Squaring both sides, we get
\begin{gathered}:\implies\:\tt{\left(\dfrac{\sqrt{3}x}{2}\:-\:20\right)^2\:=\:\left(\sqrt{400\:-\:\dfrac{x^2}{4}}\right)^2\:} \\ \end{gathered}
:⟹(
2
3
x
−20)
2
=(
400−
4
x
2
)
2
\begin{gathered}:\implies\:\tt{\left(\dfrac{\sqrt{3}x}{2}\right)^2\:+\:(20)^2\:-\:2\times{\dfrac{\sqrt{3}x}{2}}\times{20}\:=\:400\:-\:\dfrac{x^2}{4}\:} \\ \end{gathered}
:⟹(
2
3
x
)
2
+(20)
2
−2×
2
3
x
×20=400−
4
x
2
\begin{gathered}:\implies\:\tt{\dfrac{3x^2}{4}\:+\:\cancel{400}\:-\:20\sqrt{3}x\:=\:\cancel{400}\:-\:\dfrac{x^2}{4}\:} \\ \end{gathered}
:⟹
4
3x
2
+
400
−20
3
x=
400
−
4
x
2
\begin{gathered}:\implies\:\tt{\dfrac{3x^2}{4}\:-\:20\sqrt{3}x\:=\:-\:\dfrac{x^2}{4}\:} \\ \end{gathered}
:⟹
4
3x
2
−20
3
x=−
4
x
2
\begin{gathered}:\implies\:\tt{\dfrac{3x}{4}\:-\:20\sqrt{3}\:=\:-\:\dfrac{x}{4}\:} \\ \end{gathered}
:⟹
4
3x
−20
3
=−
4
x
\begin{gathered}:\implies\:\tt{\dfrac{3x}{4}\:+\:\dfrac{x}{4}\:=\:20\sqrt{3}\:} \\ \end{gathered}
:⟹
4
3x
+
4
x
=20
3
\begin{gathered}:\implies\:\tt{\dfrac{3x\:+\:x}{4}\:=\:20\sqrt{3}\:} \\ \end{gathered}
:⟹
4
3x+x
=20
3
\begin{gathered}:\implies\:\tt{\dfrac{4x}{4}\:=\:20\sqrt{3}\:} \\ \end{gathered}
:⟹
4
4x
=20
3
\begin{gathered}:\implies\:\bf\pink{x\:=\:20\sqrt{3}\:m\:=\:34.64\:m\:} \\ \end{gathered}
:⟹x=20
3
m=34.64m
\bf{\therefore}∴ Length of the string of each phone is 34.64 m.
⠀⠀⠀⠀⠀
Explanation: