Answer :

Let radius OA = OC = 17 cm


Chord AB = 30 cm and CD = 16 cm



Draw OL and OM


Therefore,


AP = (1/2) AB


AP = (1/2) 30 = 15 cm


CQ = (1/2) CD


CQ = (1/2) 16 = 8 cm


In triangle OAP,


OP2 = OA2 - AP2


OP2 = 172 - 152


OP2= 289 - 225


OP2= 64


OP = 8 cm


In triangle OQD,


OQ2 = OC2 - CQ2


OQ2 = 172 - 82


OQ2= 289 - 64


OQ2= 225


OQ = 15 cm


Now,


PQ = OP + OQ = 8 + 15 = 23


Hence, distance between chords = 23 cm.


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