Finding the area of a region in an isosceles triangle that contains squares of area $256$ and $49$ There was a quiz posted on F*cebook by someone. Here's the problem.

And here's my attempt:

First, as you can see there, I drew a line from one of the corner of the pink square to the one of the corner of the green square. The degree has been symbolized as alpha ($\alpha$).
$$\begin{align}
\alpha &= \tan^{-1}\left(\frac{16}{23}\right)\\
\end{align}$$
That means the degree between the line (almost diagonally to the green square) and the side line of the green square is $90-\alpha$. Let's say the side green square is $x$
We know $\cot(x)=\tan(90^{0} - x)$, so
$$\begin{align}
\tan(90^{0} - \alpha) = \cot(\alpha) &= \frac{16}{x}\\
x &= 16 \tan(\alpha)\\
&= 16 \tan\left(\tan^{-1}\left(\frac{16}{23}\right)\right)\\
&= \frac{16^2}{23} = \frac{256}{23}
\end{align}$$
Now, I assume (I can't tell reason, it's just my guesswork) that $x$ I've found earlier has the same length as the blue side one. I also assume those two blue are right triangles.
Finally, the last calculation is:
$$\begin{align}
& \left(\frac{1}{2}\cdot \frac{256}{23}\cdot 16\right) + \left(\frac{1}{2}\cdot 7 \left( \frac{256}{23} + 9\right)\right)\\
 &= 159.5
\end{align}$$
The poster gave me cry emoji and didn't say anything, I conclude my answer is incorrect. Where's the mistake?
 A: Let $\alpha$ is the base angle. Then the area of one (bigger) blue triangle is $\frac{16^2}{2} \tan (2\pi-2\alpha)$, the area of the second blue triangle is $\frac{7^2}{2} \tan \alpha$. On the other hand, $\tan \alpha=\frac{16\tan (2\pi-2 \alpha)+9}{7}$ or (using double-angle formula and considering that $\tan (2\pi-2\alpha)=-\tan 2\alpha$) $$7\tan \alpha \cdot (1-\tan^2 \alpha)=-32 \tan \alpha-9 \tan^2 \alpha +9 \implies \tan \alpha =3, \tan 2\alpha=-\frac{3}{4}$$ Thus, the blue area is $8 \cdot 16 \cdot \frac{3}{4}+\frac{7^2 \cdot 3}{2}=169.5$
Your assumption is incorrect as the blue side is $12 \ne \frac{256}{23}$
A: Let $\Delta ABC$ be our triangle, $AB=BC$, $PQRL$ be a square,
where $P\in AB$, $Q\in BC$, $R\in QC$, $L\in AR$, $PQ=16.$
Also, let $MRNK$ be a square, $M\in LR$, $N\in RC$ and $K\in AC$.
Let $AL=x$.
Thus, since $\Delta ALP\sim\Delta PQB,$ we obtain $$\frac{BQ}{16}=\frac{16}{x}$$ or $$BQ=\frac{256}{x}.$$
Also, since $\Delta KNC\sim\Delta AMK,$ we obtain:
$$\frac{NC}{7}=\frac{7}{x+9}$$ or
$$NC=\frac{49}{x+7}.$$
Thus, $$BC=\frac{256}{x}+23+\frac{49}{x+9}$$ and from $\Delta ARB$ we obtain:
$$(x+16)^2+\left(16+\frac{256}{x}\right)^2=\left(\frac{256}{x}+23+\frac{49}{x+9}\right)^2$$ or
$$(x+16)^2+\left(16+\frac{256}{x}\right)^2=\left(\frac{256}{x}+16+\frac{7(x+16)}{x+9}\right)^2$$ or
$$1+\frac{256}{x^2}=\left(\frac{16}{x}+\frac{7}{x+9}\right)^2$$ or
$$(x-12)(x^2+30x+168)=0,$$ which gives $x=12.$
Can you end it now?
I got $169.5.$
