Bài giải:

Từ giả thiết ta có:

PB = PC; DC = 3IC; QD = QC = $\frac{3}{2}$IC; QC = 3 QI

Ta có: $\frac{OI}{OB}=\frac{{{S}_{POI}}}{{{S}_{POB}}}=\frac{{{S}_{QOI}}}{{{S}_{QOB}}}=\frac{{{S}_{POI}}+{{S}_{QOI}}}{{{S}_{POB}}+{{S}_{QOB}}}=\frac{{{S}_{IPQ}}}{{{S}_{BPQ}}}=\frac{\frac{1}{3}{{S}_{PQC}}}{{{S}_{PQC}}}=\frac{1}{3}$

Do ${{S}_{OPI}}=3c{{m}^{2}}$ nên suy ra: ${{s}_{OPB}}=9c{{m}^{2}}$ và ${{S}_{IPB}}={{S}_{OPB}}+{{S}_{OPI}}=12c{{m}^{2}}$

${{S}_{IBC}}=2{{\text{S}}_{IPB}}=24c{{m}^{2}};{{S}_{BCQ}}=\frac{3}{2}{{S}_{BIC}}=36c{{m}^{2}}$ 

\[{{S}_{QCP}}=\frac{1}{2}{{S}_{QCB}}=18c{{m}^{2}}\] và ${{S}_{ABC\text{D}}}=4\times {{\text{S}}_{BCQ}}=4\times 36=144c{{m}^{2}}$

Vậy: ${{S}_{APQ}}={{S}_{APCQ}}-{{S}_{CPQ}}=144:2-18=54c{{m}^{2}}$