Q. 14.1( 49 Votes )

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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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