Let O is center and r is radius of circle r = 10 cm chord AB = 12 cm and chord CD = 16 cm.
Draw OP ⊥ AB which cuts chord CD at Q
Since AB || CD
Thus, OQ || CD
AP = BP
= \(\frac { 1 }{ 2 }\)AB
= \(\frac { 1 }{ 2 }\) × 12 = 6 cm
and CQ = QD = \(\frac { 1 }{ 2 }\) × CD = \(\frac { 1 }{ 2 }\) × 16 = 8 cm.
In right angled triangle OPA By Pythagoras theorem
OA2 = AP2 + OP2
(10)2 = (6)2 + (OP)2
100 = 36 + (OP)2
OP2 = 100 – 36 = 64
OP = \(\sqrt { 64 }\) = 8 cm
Similarly on right angled triangle OCQ
By Pythagoras theorem
OC 2 = CQ2 + OQ2
(10)2 = (8)2 + OQ2
100 = 64 + OQ2
OQ2 = 100 – 64 = 36
OQ = \(\sqrt { 36 }\) = 6 cm
(a) Hence, distance between AB and CD
PQ = OP – OQ = 8 – 6 cm = 2 cm
(b) Hence, distance between two chords AB and CD
PQ = OP + OQ = 8 + 6 cm PQ = 14 cm
Thus, Distance between two chords is 14 cm