Sửa đề: \(S_{OAB}=12\left(\operatorname{cm}^2\right);S_{OCD}=27\left(\operatorname{cm}^2\right)\)
Vì AB//CD
nên \(\frac{OA}{OC}=\frac{OB}{OD}\)
TA có: \(\frac{S_{BOA}}{S_{BOC}}=\frac{OA}{OC}\)
\(\frac{S_{BOC}}{S_{BOD}}=\frac{OB}{OD}\)
Do đó: \(\frac{S_{BOA}}{S_{BOC}}\cdot\frac{S_{BOC}}{S_{COD}}=\frac{OA}{OC}\cdot\frac{OB}{OD}\)
=>\(\left(\frac{OA}{OC}\right)^2=\frac{S_{OAB}}{S_{COD}}=\frac{12}{27}=\frac49=\left(\frac23\right)^2\)
=>\(\frac{OA}{OC}=\frac23\)
=>\(\frac{OB}{OD}=\frac23\)
Vì \(\frac{OA}{OC}=\frac23\)
nên \(\frac{S_{BOA}}{S_{BOC}}=\frac23\)
=>\(S_{BOC}=12\times\frac32=18\left(\operatorname{cm}^2\right)\)
Vì \(\frac{OB}{OD}=\frac23\)
nên \(\frac{S_{AOB}}{S_{AOD}}=\frac{OB}{OD}=\frac23\)
=>\(\frac{12}{S_{AOD}}=\frac23=\frac{12}{18}\)
=>\(S_{AOD}=18\left(\operatorname{cm}^2\right)\)
\(S_{ABCD}=S_{OAB}+S_{OAD}+S_{OBC}+S_{OCD}\)
\(=12+18+18+27=39+36=75\left(cm^2\right)\)
