Answer :
Let AB be a chord of a circle with center O. OCAB, then
AB = 16 cm, and OA = 10 cm.
OCAB
Therefore,
OC bisects AB at C
AC = (1/2) AB
⇒ AC = (1/2) 16
⇒ AC = 8 cm
In triangle OAC,
OA2 = OC2 + AC2
⇒ 102 = OC2 + 82
⇒ 100 = OC2 + 64
⇒ OC2= 36
⇒ OC= 6
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