Let $D$ be the point on the hypotenuse $BC$, and let $DE$ and $DF$ be the perpendiculars on $AC$ and $AB$, respectively. Let $x = |CE|$, $y = |BF|$ and use the standard notation for the side lengths, so that $a^2 = b^2 + c^2$ by Pythagoras.

The triangles $ABC$, $FBD$ and $EDC$ are similar, hence
$$\frac{x}{1} = \frac{c}{b} = \frac{1}{y}$$
and so $xy = 1$ and
$$x + y = \frac{c}{b} + \frac{b}{c} = \frac{b^2 + c^2}{bc} = \frac{a^2}{bc} = \frac{16}{bc}$$
Because $AFDE$ is a square, we have $|AF| = |AE| = 1$ and so $c = 1 + y$ and $b = 1 + x$. Using $xy = 1$ we obtain $(1 + x)(1 + y) = 1 + x + y + xy = 2 + (x + y)$. Therefore, the area $|ABC|$ of triangle $ABC$ is equal to
$$|ABC| = \frac{bc}{2} = 1 + \frac{x + y}{2} = 1 + \frac{4}{|ABC|}$$
where we have used that $\frac{x + y}{2} = \frac{b}{bc}$. This gives a quadratic equation for the area of $ABC$:
$$|ABC|^2 - |ABC| - 4 = 0,$$
which has $|ABC| = \frac{1 + \sqrt{17}}{2}$ as its only positive solution.