Ta có:

\[{S_{\Delta ADC}} = {S_{ABCD}} - {S_{\Delta ABC}} = 25c{m^2}\]

Dễ dàng chứng minh được \[\Delta DAC\~\Delta DMN\]

Suy ra

\[\frac{{{S_{\Delta ADC}}}}{{{S_{\Delta DMN}}}} = {\left( {\frac} \right)^2} = {k^2}\]

Kẻ \[AH \bot MN\]

Đặt \[{S_{\Delta DMN}} = S,\,\,{S_{\Delta ADC}} = {S_1},\,{S_{ACNM}} = {S_2}\] thì ta có:

\[{S_1} = {k^2}S \Rightarrow S = \frac{}{{{k^2}}} = \frac{{{k^2}}}\]

\[{S_{\Delta ABC}} = \frac{1}{2}AH.AC\]

\[{S_2} = {S_{\Delta AMB}} + {S_{\Delta BCN}} + {S_{\Delta ABC}} = \frac{1}{2}AH.MB + \frac{1}{2}AH.NB + \frac{1}{2}AH.AC\]

\[ = \frac{1}{2}AH(MN + AC) = \frac{1}{2}AH.\left( {\frac{k} + AC} \right) = \frac{k}{S_{\Delta ABC}}\]

\[ \Rightarrow {S_2} = \frac{{11(k + 1)}}{k}\]

Mặt khác \[S = {S_1} + {S_2} \Rightarrow \frac{{{k^2}}} = 25 + \frac{{11(k + 1)}}{k} \Leftrightarrow 25{k^2} + 11k(k + 1) - 25 = 0 \Leftrightarrow k = \frac\]

Vậy \[S = 51,84c{m^2}\]