Two chords AB and

Draw OM perpendicular AB and ON perpendicular CD

Join OB and OD

BM = (The perpendicular drawn from the center bisects the chord)

Let ON be x

Therefore, OM will be 6 –x

In ΔMOD,

OM² + MB² = OB²

(6 – x)² + ² = OB²

36 + x² -12x + OB² (i)

In ΔNOD,

ON² + ND² = OD²

(x)² + ² = OD²

(x)² + = OD² (ii)

We have OB = OD (radii of the same circle)

Therefore,

From (i) and (ii),

36 + x² -12x + (x)² +

36 -12x +

x = 1

From (ii),

(1)² + = OD²

OD = cm