a chord of length 6 cm is drawn in a circle of radius 5 cm. calculate its distance from the center of the circle
Answers
Answer:
4 CM
Step-by-step explanation:
Refer the attached figure
Length of Chord AB = 6 cm
Perpendicular Line form the center of the circle to the chord = OC
OB = OA = Radius of circle = 5 cm
Theorem : A perpendicular dropped from the center of the circle to a chord bisects it. It means that both the halves of the chords are equal in length.
So, OB bisects AB
So, AC = CB =\frac{AB}{2} =\frac{6}{2} =3AC=CB=2AB=26=3
In ΔOCB
Hypotenuse^2=Perpendicular^2+Base^2Hypotenuse2=Perpendicular2+Base2
OB^2=OC^2+CB^2OB2=OC2+CB2
5^2=OC^2+3^252=OC2+32
25=OC^2+925=OC2+9
25-9=OC^225−9=OC2
16=OC^216=OC2
\sqrt{16}=OC16=OC
4=OC4=OC
Hence distance of chord from the center of the circle is 4 cm.
Hope this helps you
Answer:
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