Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre.

Solution:

Draw $O M \perp A B$ and $O N \perp C D$. Join $O B$ and $O D$.



$\mathrm{BM}=\frac{\mathrm{AB}}{2}=\frac{5}{2}$ (Perpendicular from the centre bisects the chord)

$\mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{11}{2}$

Let $O N$ be $x$. Therefore, $O M$ will be $6-x$.

In $\triangle \mathrm{MOB}$

$\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$

$(6-x)^{2}+\left(\frac{5}{2}\right)^{2}=\mathrm{OB}^{2}$

$36+x^{2}-12 x+\frac{25}{4}=\mathrm{OB}^{2}$...(1)

In $\triangle N O D$,

$\mathrm{ON}^{2}+\mathrm{ND}^{2}=\mathrm{OD}^{2}$

$x^{2}+\left(\frac{11}{2}\right)^{2}=\mathrm{OD}^{2}$

$x^{2}+\frac{121}{4}=\mathrm{OD}^{2}$..(2)

We have OB = OD (Radii of the same circle)

Therefore, from equation (1) and (2),

$36+x^{2}-12 x+\frac{25}{4}=x^{2}+\frac{121}{4}$

$12 x=36+\frac{25}{4}-\frac{121}{4}$

$=\frac{144+25-121}{4}=\frac{48}{4}=12$

$x=1$

From equation (2),

$(1)^{2}+\left(\frac{121}{4}\right)=\mathrm{OD}^{2}$

$\mathrm{OD}^{2}=1+\frac{121}{4}=\frac{125}{4}$

$\mathrm{OD}=\frac{5}{2} \sqrt{5}$

Therefore, the radius of the circle is $\frac{5}{2} \sqrt{5} \mathrm{~cm}$.