Solution :
In $\Delta COD$ and $\Delta AOB$,
$\angle DOC = \angle AOB$ [vertically opposite ]
$\angle DOC = \angle AOB$ [Alternate angles ]
⇒$\Delta COD \approx \Delta AOB$ [AA similarity]
Let $OB = xx\; cm$
$ \large\frac{AB}{CD} = \frac{OB}{OD}$
$\large\frac{9}{6} = \frac{x}{12-x}$
=> $108 -9x =6x$
=> $15 x =108$
=> $x = 7.2 \;cm$