Từ đề bài, ta biết:
$\text{PB = PC ; DC = 3 IC ; QD = QC = $\dfrac{3}{2}$ IC ; QC = 3 QI}$
Ta có:
$\text{$\dfrac{OI}{OB}$=$\dfrac{S_{POI}}{S_{POB}}$=$\dfrac{S_{QOI}}{S_{QOB}}$}$
$\Longleftrightarrow$ $\dfrac{S_{POI}+S_{QOI} }{S_{POB}+S_{QOB}}$=$\dfrac{S_{IPQ}}{S_{BPQ} }$=$\dfrac{\dfrac{1}{3}S_{PQC}}{S_{PQC}}$=$\dfrac{1}{3}$
Vì:
$S_OPI$$\text{=3cm²}$ $\rightarrow$ $\text{$S_OPB$=9cm² ; $S_IPB$=$S_OPB$+$S_OPI$=12cm²}$
$\rightarrow$ $\text{$S_IBC$=$2S_IPB$=24cm² ; $S_BCQ$=$\dfrac{3}{2}$$S_BIC$=36cm²}$
$\Longrightarrow$ $\text{$S_QCP$=$\dfrac{1}{2}$$S_QCB$=18cm²}$
$\Longrightarrow$ $\text{$S_ABCD$=4×$S_BCQ$=4×36=144cm²}$
Vậy $\text{$S_APQ$=$S_APCQ$-$S_CPQ$=144:2-18=54cm²}$
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