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 PQ and in the other solution she divides the quadrilateral into triangles by joining AO. As ∆BOC and ∆OQC have the same height their areas are in the same ratio as their bases, that is

BO OQ =  10 5 = 2.

Similarly, ∆BOP and ∆OQP also have the same height, so their areas are in the ratio

∆BOP ∆OQP =  BO OQ = 2.

Hence

∆OQP = 8/2 = 4.

Let the area of ∆APQ = x, then

AQ QC =  ∆ABQ ∆QBC =  x + 12 15

but

AQ QC =  ∆APQ ∆QPC =  x 9 .

So

x + 12 15 =  x 9

9x + 108 = 15x

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