tính sai rồi phải là 100%-32%-33%=35%

$ABssCD\Rightarrow \frac{AB}{CD}=\frac{OB}{OD}=\frac{OA}{OC}=\frac{2}{3}$

a)$S_{AOD}=\frac{1}{2}OA.OD.sinAOB$

$S_{BOC}=\frac{1}{2}OB.OC.sinBOC$

$\Rightarrow \frac{S_{AOD}}{S_{BOC}}=\frac{OA.OD}{OB.OC}$ vì $\widehat{AOD}=\widehat{BOC}\Rightarrow sinAOD=sinBOC$

$\iff \frac{S_{AOD}}{S_{BOC}}=\frac{2}{3}.\frac{3}{2}=1$

b) vì $ABssCD\Rightarrow \frac{OH}{OK}=\frac{2}{3}\Rightarrow \frac{OH}{HK}=\frac{2}{5}$

$S_{AOB}=\frac{1}{2}.OH.AB{S}_{ABCD}=\frac{1}{2}\left(AB+CD\right).HK=\frac{1}{2}\left(AB+\frac{3}{2}AB\right).HK=\frac{1}{2}.\frac{5}{2}AB.HK$

$\Rightarrow \frac{S_{AOB}}{S_{ABCD}}=\frac{\frac{1}{2}OH.AB}{\frac{1}{2}HK.\frac{5}{2}AB}=\frac{2}{5}.\frac{1}{\frac{5}{2}}=\frac{4}{25}$

$\Rightarrow S_{ABCD}=\frac{4}{\frac{4}{25}}=25$