9. Three points A,B and C are located on a circle which are equidistant from one another. If the radius of the circle is 20m the calculate the length of AB.
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Answers
let Ankur be represented as A, Syed as S and David as D.
The boys are sitting at an equal distance.
Hence, △ASD is an equilateral triangle.
Let the radius of the circular park be r meters.
∴OS=r=20m.
Let the length of each side of △ASD be x meters.
Draw AB⊥SD
∴SB=BD=
2
1
SD=
2
x
m
In △ABS,∠B=90
o
By Pythagoras theorem,
AS
2
=AB
2
+BS
2
∴AB
2
=AS
2
−BS
2
=x
2
−(
2
x
)
2
=
4
3x
2
∴AB=
2
3
x
m
Now, AB=AO+OB
OB=AB−AO
OB=(
2
3
x
−20) m
In △OBS,
OS
2
=OB
2
+SB
2
20
2
=(
2
3
x
−20)
2
+(
2
x
)
2
400=
4
3
x
2
+400−2(20)(
2
3
x
)+
4
x
2
0=x
2
−20
3
x
∴x=20
3
m
The length of the string of each phone is 20
3√m.
PLS MARK ME AS BRAINLIEST
Answer:
Let A,B and C be three non-collinear points
Join AB and BC
Draw perpendicular bisector of AB and BC
Let them meet at point O
Then O lie on the right bisector of AB
So, OA=OB
And O lies on the right bisector of BC
So, OB=OC
Hence, the point O is equidistant from A,B and C.
Now since the right bisector of AB and BC are non-parallel lines, therefore they have only one point in common.
So, O is the only point equidistant from A,B and C.
Hence, the required locus is the centre of the circle through the given non-collinear points.