<s>[INST] Find the area of ​the CEN triangular region For reference: If ABCD is a square, AB = DE.
Calculate the area of ​​the CEN triangular region.(Answer:$6m^2$)

My progress:
$S_{\triangle MCE}=X=\frac{MC⋅DC}{2}=\frac{MC⋅ℓ}{2}(I)\\
I+S_{\triangle MBA}=\frac{l^2}{2}\\
\therefore X+5 = \frac{l^2}{2}\implies 9 + S_{\triangle CEN} = \frac{l^2}{2}\\
\frac{4}{MN}=\frac{S_{\triangleÇNE}}{EN}=\frac{CN.DE}{2EN}\\
\frac{5}{BM}=\frac{4}{MC}\\
S_{ABCE} = \frac{3AB^2}{2}$
...?
 [/INST] You can use a system of $4$ equations :
$\begin{cases} \frac{l \cdot BM}{2} = 5 \\ \frac{(l -BM)\cdot CN}{2} = 4  \\  \frac{(l -BM)\cdot l}{2} = A_{CNE}+4 \\ \frac{l\cdot CN}{2} = A_{CNE} \end{cases} $
and solve for $A_{CNE}$.
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