Right triangle inscribed in a square. Find the square area? I hope it's valid to ask for "a more neat solution" of a problem on this network, despite the fact that I don't have a strict definition of the word "neat".
Here is the square and the right triangle inscribed in it.

I did the following:
$$AC = Ah + hB$$
$$4\sin\theta = 4\cos\theta + 3\sin\theta$$
So
$$\tan\theta = 4$$
But
$$\sin\theta = \frac{\tan\theta}{\sqrt{1+\tan^2\theta}}$$
Therefore
$$AC = 4\sin\theta = \frac{16}{\sqrt {17}}$$
$$\text{Area} = \left(\frac{16}{\sqrt {17}}\right)^2$$

Denote: $AB=a, BF=x, BH=y$. Then:
$$\begin{cases}a^2+(a-x)^2=25 \ (1)\\
(a-y)^2+a^2=16 \ (2)\\
x^2+y^2=9 \ (3)\end{cases}$$
$(1)-(2)$:
$$(a-x)^2-(a-y)^2=x^2+y^2 \Rightarrow 2a(y-x)=2y^2 \Rightarrow a=\frac{y^2}{y-x} \ \ \ \ \ \ \ \ \ \ \ (4)$$
$(4)\to (2)$:
$$\begin{align}\left(\frac{y^2}{y-x}-y\right)^2+\left(\frac{y^2}{y-x}\right)^2&=16 \Rightarrow \\
\frac{y^2(x^2+y^2)}{(y-x)^2}&=16 \stackrel{(3)}\Rightarrow \\
9y^2&=16(9-2xy) \stackrel{(3)}\Rightarrow \\
144-9y^2&=32y\sqrt{9-y^2} \Rightarrow \\
1104y^4-11808y^2+20736&=0 \Rightarrow \\
y_1&=\frac{12}{\sqrt{17}}; y_2=\frac{12}{\sqrt{65}} \Rightarrow \\
x_1&=\frac3{\sqrt{17}}; x_2=\frac{21}{\sqrt{65}}.\end{align}$$
Hence, from $(4)$:
$$a=\frac{y^2}{y-x}=\frac{\frac{144}{17}}{\frac9{\sqrt{17}}}=\frac{16}{\sqrt{17}}\\
y-x<0 \Rightarrow a\in \emptyset$$ 
Reference: WA answer.
