Sue Liu, S5, Madras College sent in two solutions to this problem which both use the same method. In this one she divides the quadrilateral into two triangeles by joining $\mathrm{PQ}$ and in the other solution she divides the quadrilateral into triangles by joining $\mathrm{AO}$. As $\Delta \mathrm{BOC}$ and $\Delta \mathrm{OQC}$ have the same height their areas are in the same ratio as their bases, that is

${\displaystyle \frac{\mathrm{BO}}{\mathrm{OQ}}=\frac{10}{5}=2.}$

Similarly, $\Delta \mathrm{BOP}$ and $\Delta \mathrm{OQP}$ also have the same height, so their areas are in the ratio

${\displaystyle \frac{\Delta \mathrm{BOP}}{\Delta \mathrm{OQP}}=\frac{\mathrm{BO}}{\mathrm{OQ}}=2.}$

Hence

${\displaystyle \Delta \mathrm{OQP}=8/2=4.}$

Let the area of $\Delta \mathrm{APQ}=x$, then

${\displaystyle \frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{\Delta \mathrm{ABQ}}{\Delta \mathrm{QBC}}=\frac{x+12}{15}}$

but

${\displaystyle \frac{\mathrm{AQ}}{\mathrm{QC}}=\frac{\Delta \mathrm{APQ}}{\Delta \mathrm{QPC}}=\frac{x}{9}.}$

So

${\displaystyle \frac{x+12}{15}=\frac{x}{9}}$

${\displaystyle 9x+108=15x}$

Hence $x=18$ and the area of the quadrilateral $\mathrm{APOQ}$ is 22 square units. The same argument could be applied to the side $\mathrm{AB}$.