Area triangle (possible) The area of a right triangle is an integer greater than $ 85 $. If the hypotenuse measures $ 20 $, what is the area of this triangle?
It’s just a 3-4-5 right triangle, so the area is $\boxed{96}$.
This is true? If so, how to prove it?
But he said the area is integer, not the sides
Can you use Weitzenbock inequality in this problem? 
 A: 
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 $100$. 

A: 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 $a=b=10 \sqrt 2$.
A: If the hypotenuse is $20$, the ratio is then $4$ times that of a $3-4-5$ triangle or $12:16:20$. Now the area is $\frac{1}{2} \times 12 \times 16 = 96$ which is indeed greater than $85$.
