Đáp án:
\[{S_{AOD}} = 40\sqrt 5 \left( {c{m^2}} \right)\]
Giải thích các bước giải:
Qua O kẻ \(OH \bot AB\,\,\left( {H \in AB} \right);\,\,\,OK \bot CD\left( {K \in CD} \right)\)
Do AB//CD nên O,H,K thẳng hàng.
Ta có:
\(\begin{array}{l}
\left. \begin{array}{l}
AB//CD \Rightarrow \frac{{AB}}{{CD}} = \frac{{OA}}{{OC}} = \frac{{OB}}{{OD}}\\
AH//KC \Rightarrow \frac{{OA}}{{OC}} = \frac{{OH}}{{OK}}
\end{array} \right\} \Rightarrow \frac{{OH}}{{OK}} = \frac{{AB}}{{CD}}\\
\frac{{{S_{OAB}}}}{{{S_{OCD}}}} = \frac{{\frac{1}{2}OH.AB}}{{\frac{1}{2}OK.CD}} = \frac{{OH.AB}}{{OK.CD}} = {\left( {\frac{{OH}}{{OK}}} \right)^2}\\
\Rightarrow {\left( {\frac{{OH}}{{OK}}} \right)^2} = \frac{{64}}{{125}} \Rightarrow \frac{{OH}}{{OK}} = \frac{{8\sqrt 5 }}{{25}}\\
\frac{{{S_{AOD}}}}{{{S_{DOC}}}} = \frac{{AO}}{{OC}} = \frac{{OH}}{{OK}} = \frac{{8\sqrt 5 }}{{25}} \Rightarrow {S_{AOD}} = 40\sqrt 5 \left( {c{m^2}} \right)
\end{array}\)
