In a circle with centre O, AB and CD are parallel chords on the opposite sides of a diameter. If AB = 12 cm, CD = 18 cm and the distance between the chords AB and CD is 15 cm, then find the radius of the circle (in cm).

$3 \sqrt{13}$

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\(\sqrt {13 }\).