Area triangle (possible)

The area of the right triangle with hypotenuse $|AB|=20$ can be any real number in a range $(0,100]$, so if the only limiting condition is that it must be an integer greater than $85$, the answer is that it could be any integer $n$ from $86$ to $100$.

We need

$$a^2+b^2=20^2\implies a^2=400-b^2$$

which by inspection, taking $a$ and $b$ integers, leads to the unique solution $a=12$ and $b=16$ such that $\frac12 ab\ge 85$ and the area in that case is equal to $96$.

For $a$ and $b$ reals we have

$$\frac12ab=\frac12a\sqrt{400-a^2}=86 \implies a\sqrt{400-a^2}=172 $$$$\implies a=\sqrt{186}\pm\sqrt{14},\quad b=\sqrt{186}\mp\sqrt{14}$$

which is the unique solution for the area equal to $86$.

In a similar way we can obtain integer solutions up to $100$ which is the largest area we can obtain for $a=b=10 \sqrt 2$.