For the side opposite of the right angle, is there a better way to calculate it than Heron's?

3d analytic geometry leads to different paths

Let the lengths of the edges with one endpoint at the vertex with the right angles be $a,b,c$. Then, the areas of the three adjacent faces is $\frac{ab}{2},\frac{bc}{2},\frac{ca}{2}$. By the Pythagorean Theorem, the side lengths of the fourth face are $\sqrt{a^2+b^2},\sqrt{b^2+c^2},\sqrt{c^2+a^2}$. Let these be $x,y,z$. Then, by Heron's formula the area of this triangle is $$\sqrt{\frac{-x^4-y^4-z^4+2(x^2y^2+y^2z^2+z^2x^2)}{16}}=\frac{\sqrt{-(a^2+b^2)^2-(b^2+c^2)^2-(c^2+a^2)^2+2\left((a^2+b^2)(b^2+c^2)+(b^2+c^2)(c^2+a^2)+(c^2+a^2)(a^2+b^2)\right)}}{4}$$$$=\frac{\sqrt{-2a^4-2b^4-2c^4-2a^2b^2-2b^2c^2-2c^2a^2+2\left(a^4+b^4+c^4+3a^2b^2+3b^2c^2+3c^2a^2\right)}}{4}$$$$\frac{\sqrt{4a^2b^2+4b^2c^2+4c^2a^2}}{4}=\frac{\sqrt{(ab)^2+(bc)^2+(ca)^2}}{2}.$$From here it is evident that the problem is true because $$\left(\frac{\sqrt{(ab)^2+(bc)^2+(ca)^2}}{2}\right)^2=\left(\frac{ab}{2}\right)^2+\left(\frac{bc}{2}\right)^2+\left(\frac{ca}{2}\right)^2.$$

Alternatively, let the coordinates of the vertices of the tetrahedron be $(0,0,0)$, $(x,0,0)$, $(0,y,0)$ and $(0,0,z)$. It's easy to compute the three non-opposite areas as $\tfrac{1}{2}xy$, $\tfrac{1}{2}yz$, and $\tfrac{1}{2}xz$. For the opposite face, we will use the fact that $$||\vec{v}\times\vec{w}||=||\vec v||\cdot||\vec w||\cdot\sin(\theta_{\text{bet}}).$$Note that the area of the triangle defined by these vectors is half the cross product. Starting at $(x,0,0)$, we can get to $(0,y,0)$ and $(0,0,z)$ by adding vectors $(-x,y,0)$ and $(-x,0,z)$. Taking the cross product, we get $$\left[\begin{matrix}\hat i & \hat j & \hat k \\ -x & y & 0 \\ -x & 0 & z \end{matrix}\right]=\langle yz,xz,xy\rangle,$$which has magnitude $\sqrt{(yz)^2+(xz)^2+(xy)^2}$. Taking half of this quantity and squaring gives $$\left(\frac{\sqrt{(yz)^2+(xz)^2+(xy)^2}}{2}\right)^2=\left(\frac{xy}{2}\right)^2+\left(\frac{yz}{2}\right)^2+\left(\frac{xz}{2}\right)^2,$$as desired. $\blacksquare$