Đáp án:
\[{S_{AMON}} = \frac{{39}}{5}\left( {{m^2}} \right)\]
Giải thích các bước giải:
Đặt \({S_{ANO}} = x;\,\,\,\,{S_{AOM}} = y\,\,\,\,\left( {x,y > 0} \right)\)
Ta có:
\(\begin{array}{l}
\frac{{{S_{ANO}}}}{{{S_{AOC}}}} = \frac{{NO}}{{OC}} = \frac{{{S_{BON}}}}{{{S_{BOC}}}} = \frac{3}{4} \Rightarrow \frac{x}{{2 + y}} = \frac{3}{4} \Leftrightarrow x = \frac{3}{4}\left( {2 + y} \right)\,\,\,\,\,\,\,\left( 1 \right)\\
\frac{{{S_{ABO}}}}{{{S_{AOM}}}} = \frac{{BO}}{{OM}} = \frac{{{S_{BOC}}}}{{{S_{OMC}}}} = 2 \Rightarrow \frac{{x + 3}}{y} = 2 \Leftrightarrow x + 3 = 2y\,\,\,\,\,\,\,\,\,\left( 2 \right)
\end{array}\)
Thay (1) vào (2) ta được:
\[\begin{array}{l}
\frac{3}{4}\left( {2 + y} \right) + 3 = 2y \Leftrightarrow \frac{3}{2} + \frac{3}{4}y + 3 = 2y \Leftrightarrow y = \frac{{18}}{5} \Rightarrow x = \frac{{21}}{5}\\
{S_{AMON}} = x + y = \frac{{39}}{5}\left( {{m^2}} \right)
\end{array}\]
