Do two right triangles with the same length hypotenuse have the same area? I watched computer monitors and I asked myself, do two monitors with the same display diagonal have the same display area?

I managed to find out that the answer is yes, if two right triangles with the same length hypotenuse have the same area. The answer is trivial if the two triangles are identical, and according to Thales' theorem I know that there are cases when two right triangles have the same length hypotenuse, but they are not identical (the legs length and the angles are different).

So my final question:


  Do two, not identical right triangles with the same length hypotenuse have the same area?


I guess that the answer is yes, but I can't give a proof that I'm 100% sure.
No. Consider the following picture: 

$\hspace{90pt}$

All the triangles have the same hypothenuse, but one of them has area of $r^2$ while others might have it arbitrarily small.

I hope this helps $\ddot\smile$
A basic algegra approach lets us not only demonstrate that there can be multiple different values of area for a given hypotenuse, it lets us work out the range of possible values for that area and calculate the corresponding lengths of the other two sides.

Define $A$ as area of the triangle, $c$ as the length of the hypotonuse and $a$ and $b$ as the length of the other two sides. $A &gt; 0$, $a &gt; 0$, $b &gt; 0$, $c &gt; 0$

$$a^2 + b^2 = c^2$$ 
$$A = \frac{ab}{2}$$
$$2A = ab$$
$$b = \frac{2A}{a}$$

Substituting

$$a^2 + 4\frac{A^2}{a^2} = c^2$$

$$a^4 + 4A^2 = a^2c^2$$

$$a^4 - c^2a^2 + 4A^2 = 0$$

Solving the quadratic in $a^2$.

$$a^2 = \frac{c^2\pm\sqrt{c^4-4*4A^2} }{2}$$

$$a = \sqrt{\frac{c^2\pm\sqrt{c^4-16A^2} }{2}}$$


If $c^4 &lt; 16A^2$ (i.e. if $c^2 &lt; 4A$) there are no soloutions.
If $c^4 = 16A^2$ (i.e. if $c^2 = 4A$) there is one soloution.
If $c^4 &gt; 16A^2$ (i.e. if $c^2 &gt; 4A$) there are two soloutions.