a) \(S_{ABCD}=8\cdot3\cdot2=72\left(cm^2\right)\)
b) Ta có: \(S_{AOB}=S_{COD}=\frac{3\cdot8}{2}=12\left(cm^2\right)\)và \(S_{AOD}=S_{COB}\)
Mà \(S_{ABCD}=S_{AOB}+S_{COD}+S_{AOD}+S_{BOC}\)
\(\Leftrightarrow2\cdot12+2\cdot2BC=72\Leftrightarrow4BC=48\Leftrightarrow BC=12\left(cm\right)\)