Answer :


(i)


Let radius OB = OD = 5 cm


Chord AB = 8 cm


Chord CD = 6 cm


BP = (1/2) AB


BP = (1/2) 8 = 4 cm


DQ = (1/2) CD


DQ = (1/2) 6 = 3 cm


In triangle OPB,


OP2 = OB2 - BP2


OP2 = 52 - 42


OP2= 25 - 16


OP2= 9


OP = 3 cm


In triangle OQD,


OQ2 = OD2 - DQ2


OQ2 = 52 - 32


OQ2= 25 - 9


OQ2= 16


OQ = 4 cm


Now,


PQ = OQ – OP = 4 – 3 = 1


Hence, distance between chords = 1 cm.


(ii)


Let radius OA = OC = 5 cm


Chord AB = 8 cm


Chord CD = 6 cm


AP = (1/2) AB


AP = (1/2) 8 = 4 cm


CQ = (1/2) CD


CQ = (1/2) 6 = 3 cm


In triangle OAP,


OP2 = OA2 - AP2


OP2 = 52 - 42


OP2= 25 - 16


OP2= 9


OP = 3 cm


In triangle OQD,


OQ2 = OC2 - CQ2


OQ2 = 52 - 32


OQ2= 25 - 9


OQ2= 16


OQ = 4 cm


Now,


PQ = OP + OQ = 3 + 4 = 7


Hence, distance between chords = 7 cm.


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