Answer :

In the given figure:

Radius AB = 17cm

Chord CD = 17cm

Now we draw a perpendicular bisector of CD from A which is AB

Hence CE = DE = 8cm

We have to find the distance of chord from centre i.e. AE

In the triangle ACE

AC^{2} = AE^{2} + EC^{2}

Therefore AE^{2} = 17^{2}– 8^{2}

AE^{2} = 225

AE = 15cm

Hence the distance of chord from the centre is 15cm

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