In a circle with centre $O, A B$ and $C D$ are two parallel chords on the same side of the diameter. If $A B=12 \mathrm{~cm}, C D=18$ $\mathrm{cm}$ and distance between the chords $\mathrm{AB}$ and $\mathrm{CD}$ is $3 \mathrm{~cm}$, then find the radius of the circle (in $\mathrm{cm}$ ).

3$\sqrt{13}$

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

For chord AB,

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

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

In right angled triangle POB,

Using Pythagoras theorem,

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

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

Similarly, for chord CD

CO = OD = \(\frac{18}{2}\) = 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)

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

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

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