For reference: On sides $AB$ and $BC$ of a triangle $ABC$, they are built outwardly the $ABFL$ and $BCQR$ squares; the segment that joins the centers of the squares measures $8m$. Calculate the area of the $AFRC$ quadrilateral.(Answer:$64m^2$)
My progress:
$S_{FBA}=\frac{S_{FBMA}}{2}=\frac{l^2}{2}\\ S_{CBR} = \frac{S_{BRQC}}{2}=\frac{L^2}{2}\\ S_{AFRC} = \frac{L^2+l^2}{2}+S_{FBR}+S_{ABC}$
I can't see the use of the DE segment
If $\angle ABC = \theta, \angle CBF = \angle ABR = \angle DBE = 90^\circ + \theta$
We also have $AB = BF, BC = BR$ and $BD:BE = BF:BC = AB:BR$
So $\triangle ABR \cong \triangle FBC$ and $\triangle ABR \sim \triangle DBE$
That leads to $AR \perp CF$ and $AR = CF = 8 \sqrt2$
Therefore $S_{AFRC} = \frac{1}{2} \cdot 8 \sqrt2 \cdot 8 \sqrt2 = 64$