Area triangle (possible)

Question

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?

Answer 1

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 $100$.

Answer 2

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 $a=b=10 \sqrt 2$.

Answer 3

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 $85$.