Qua O kẻ \(OH\perp AB\left(H\in AB\right)\) \(;OK\perp CD\left(K\in CD\right)\)
Do \(AB//CD\) nên \(O,H,K\) thẳng hàng.
Ta có:
\(AB//CD\rightarrow\frac{AB}{CD}=\frac{OA}{OC}=\frac{OB}{OD}\)
\(AH//KC\rightarrow\frac{OA}{OC}=\frac{OH}{OK}\)
\(\rightarrow\frac{OH}{OK}=\frac{AB}{CD}\)
\(\frac{S_{OAB}}{S_{OCD}}=\frac{\frac{1}{2}OH.AB}{\frac{1}{2}OK.CD}=\frac{OH.AB}{OK.CD}=\left(\frac{OH}{OK}\right)^2\)
\(\rightarrow\left(\frac{OH}{OK}\right)^2=\frac{64}{125}=\frac{OH}{OK}=\frac{8\sqrt{5}}{25}\)
\(\frac{S_{AOD}}{S_{DOC}}=\frac{AO}{OC}=\frac{OH}{OK}=\frac{8\sqrt{5}}{25}\rightarrow S_{AOD}=50\sqrt{5}\left(cm^2\right)\)
