A $3$-$4$-$5$ right $\triangle ABC$ ($AC=5$) fits in a square such that $A$ is also vertex of the square. Find the side of square. 
A right angle triangle ABC of sides $3$, $4$, and $5$ ($AC=5$) is fit in a square such that $A$ is also vertex of square . Find the side of square.

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My attempt:
I tried to use Pythagoras by assuming side $x$. But I think I am missing something. There should be an easy way to finish it.
\color{red}{ Is this square unique ?}


Establish the ratio below from the two similar right triangles with 3 and 4 as hypotenuses,
$$\frac ax=\frac{a-x}{\frac34 x}$$
Use $x=\sqrt{16-a^2}$ to get,
$$\sqrt{16-a^2}=\frac14a$$
which yields
$$a=\frac{16}{\sqrt{17}}$$
