In a circle with centre O, AB and CD are parallel chords on the oppos…

Perpendicular drawn from the center of circle to chord which bisects the chord.

For chord AB,

OP is he perpendicular drawn from the center and OB is the radius.

AP = PB = \(\frac{12}{2}\) = 6 cm.

Using pythagoras theorem

= \( { OB}^{2 } \) = \( { PB}^{2 } \) + \( { OP}^{2 } \)

= \( { OB}^{2 } \) = \( { 6}^{2 } \) + \( { OP}^{2 } \)

Similarly, for chord CD.

CQ = QD = \(\frac{18}{2}\) = 9 cm

In right angled triangle QDC,

= \( { OD}^{2 } \) = \( { QD}^{2 } \) + \( { OQ}^{2 } \)

= \( { OD}^{2 } \) = \( { 9}^{2 } \) + \( { OQ}^{2 } \)

Then,

= OP = OQ + QP

= OP = OQ = 3

But,

OD = OB   (Radii of circle)

= \( {6 }^{ 2} \) + \( {OP }^{ 2} \) = \( {9 }^{ 2} \) + \( {OQ }^{ 2} \)

= \( {(OQ + 3) }^{ 2} \) - \( {OQ }^{ 2} \) = 45

= 6OQ = 36

= OQ = 6

Then,

= \( {OD}^{ 2} \) = \( {9 }^{ 2} \) + \( {6 }^{ 2} \) = 117

= OD = 3\(\sqrt {13 }\)

Therefore, radius of circle is 3\(\sqrt {13 }\).