Question

The radius of a circle is 6cm and the length of one of its chord is 6cm. Find the distance of the chord from the centre.​

Answer 1

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.

Answer 2

Answer:

Hence, The distance of the Chord from the centre is :- $3\sqrt{3} \: cm.$

figure in 2nd attachment...