# In a circle of ra

Given that,

AB || CD (Chords on opposite side of centre)

OL and OM perpendicular bisector of CD and AB respectively

LM = 23 cm

AB = 16 cm

In OLB,

OB2 = OL2 + LB2

OL2 = 225

OL = 15 cm

OM = LM - OL

= 8 cm

In OMD,

OD2 = OM2 + MD2

MD2 = 225

MD = 15 cm

Now,

CD = 2 MD = 30 cm

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