For reference:
In the figure below, the quadrilaterals $AMNB, AEFC$ and $APQR$ are squares; $S_1=9\ \mathrm{m^2}$, $S_2(BRE)=6\ \mathrm{m^2}$, calculate the area of the triangular region $ABC$. (Answer: $15\ \mathrm{m^2}$)
Here is a trigonometric solution
Let $a=AC$.
Considering $△ABC$:
$AB=a\sin(α)⟹\tan(α)=\dfrac{PB}{AB}$
$⟹PB=a\sin(α\tan(α)$
$S_1=a^2\sin^2(α)-\dfrac{a^2}{2}\sin^2(α)\tan(α)$
$S_{ARE}=S_2+S_{ABE}+S_{ABR}$
$S_{ARE}=\dfrac{a⋅a⋅\tan α}{2}$
$S_{ABR}=\dfrac{(a\tan(α))⋅(a\sin(α)\cos(α))}{2}=\dfrac{a^2\sin^2(α)}{2}$
$S_{EAB}=\dfrac{a⋅a⋅\sin2(α)}{2}=S_{ABR}$
$\therefore S_2=\dfrac{a^2}{2}(\tan(α)−2\sin^2(α)))$
$\therefore \tan(α)=1−\sqrt{\dfrac{2}{5}}$
From the equations above,
$\implies 12=a^2(1−\sqrt{\dfrac{2}{5}}−\dfrac{1}{26}(22−5\sqrt{10})))$
$\implies a^2=\dfrac{12⋅130}{20−\sqrt{10}}=4⋅(20+\sqrt{10})$
$S_{ABC}=\dfrac{a^2}{2}\sin(α)\cos(α)=\dfrac{a^2}{2}\dfrac{\sin^2(α)}{\tan(α)}=\dfrac{a^2}{2}\dfrac{(20−\sqrt{10})}{52}$
$=\dfrac{2}{52}(20^2−10)=15$
But there must be a better solution simple...??? The solution would be to combine the two areas...
Drop perp $~RH$ from $R$ to $AB$. Also as $\angle MAP = \angle BAC$ and $AB = AM$, $M, N, E~$ are collinear.
Notice $~\triangle ARH \cong \triangle PAB~$ (by $A$-$S$-$A$) and hence $RH = AB$. So,
$[ARB] = [AMN] \tag1$
Also, $~\triangle APB \sim \triangle EPN \implies \frac{PB}{AB} = \frac{PN}{EN}$,
$PB \cdot EN = AB \cdot PN \implies [EPB] = [MPN] \tag2$
Adding $(1)$ and $(2)$,
$[ARB] + [EPB] = S1$
$[ARB] + [EPB] + [ERP] = S1 + S2$
But also,
$[ARB] + [EPB] + [ERP] = [ARE] - [ABP]$
$ = [ACP] - [ABP] = [ABC]$