In a Tri-Rectangular Tetrahedron it's possible to extract the 3 right angled elements | x,y,z | directly from the triangle | a,b,c | The "S" equation looks strikingly similar to Heron's semi-parameter equation that he developed 2000 years ago to calculate the area of any triangle.

These equations were derived from 5 relationships that hide inside Tri-Rectangular Tetrahedrons. I'm astonished that I have never seen any of them in the textbooks I studied when I was in college. Just in case somebody else needs them here they are.

1_st - Striking similar to Heron's Theorem

$S={\Biggl [}{\frac {\ a^{2}+b^{2}+c^{2}\ }{2}}{\Biggr ]}={\Bigl [}a^{2}+z^{2}{\Bigr ]}={\Bigl [}b^{2}+y^{2}{\Bigr ]}={\Bigl [}c^{2}+x^{2}{\Bigr ]}$

$x={\sqrt {S-c^{2}\ }}\ \ \ \ \ \ \ y={\sqrt {S-b^{2}\ }}\ \ \ \ \ \ \ z={\sqrt {S-a^{2}\ }}$

2_nd - Striking similar to Pythagorean's Theorem

$S={\Bigl [}x^{2}+y^{2}+z^{2}{\Bigr ]}={\Bigl [}a^{2}+z^{2}{\Bigr ]}={\Bigl [}b^{2}+y^{2}{\Bigr ]}={\Bigl [}c^{2}+x^{2}{\Bigr ]}\$

$a={\sqrt {S-z^{2}\ }}\ \ \ \ \ \ \ \ b={\sqrt {S-y^{2}\ }}\ \ \ \ \ \ \ \ c={\sqrt {S-x^{2}\ }}$

$A_{abc}={\tfrac {\ {\sqrt {\ x^{2}y^{2}+z^{2}(x^{2}+y^{2})\ }}\ }{2}}\$

This equation can be used to calculate the area of the Tri-Rectangular Tetrahedron's BASE. Heron's equation uses the three sides | a,b,c | but if you have already calculated | x,y,z | this equation is easier.

$h_{tet}={\frac {xyz}{{\sqrt {\ x^{2}y^{2}+z^{2}(x^{2}+y^{2})\ \ }}\ }}$

This equation can be used to calculate the internal height of a Tetrahedron is from the point of Origin at | x, y, z | to its base inside of Heron's Triangle | a, b, c |. This is a real gem originally presented using a,b,c designations. I had to changed their a,b,c designations to an x,y,z form. This gem is found at https://handwiki.org/wiki/Trirectangular_tetrahedron

Once you have calculated | x,y,z | from | a,b,c | it's easy to back-check the accuracy of the results by using Pythagorean's Theorem Where: a = 14.4 b = 10 c = 12 x = 9.03769882216 y = 11.2107091658 z = 4.28018691181

$a={\sqrt {x^{2}+y^{2}\ }}\ \ \ \ \ \ b={\sqrt {x^{2}+z^{2}\ }}\ \ \ \ \ \ c={\sqrt {y^{2}+z^{2}\ }}$

I struggled my way to the | x,y,z | elements because I'm trying to build Bass Reflex speaker boxes using only a set of diagonal ratios & the required Bass Box internal volume that's acoustically determined by DB Keele's equations. I succeeded. The idea of constructing Bass Boxes for a given internal volume using it's diagonals was a kick in the head European idea. Right now only the Golden Ratio Boxes won't resonate internally . ^[1]Golden ratio

^ "Golden ratio", Wikipedia, 2024-04-12, retrieved 2024-04-24

x = 9.0376988221560027818758211287983

y = 11.210709165793214885544254958172

z = 4.2801869118065393196504607419769

[1] "Golden ratio", Wikipedia, 2024-04-12, retrieved 2024-04-24

[1]

1st - Striking similar to Heron's Theorem

S = [ a 2 + b 2 + c 2 2 ] = [ a 2 + z 2 ] = [ b 2 + y 2 ] = [ c 2 + x 2 ] {\displaystyle S={\Biggl [}{\frac {\ a^{2}+b^{2}+c^{2}\ }{2}}{\Biggr ]}={\Bigl [}a^{2}+z^{2}{\Bigr ]}={\Bigl [}b^{2}+y^{2}{\Bigr ]}={\Bigl [}c^{2}+x^{2}{\Bigr ]}}

x = S − c 2 y = S − b 2 z = S − a 2 {\displaystyle x={\sqrt {S-c^{2}\ }}\ \ \ \ \ \ \ y={\sqrt {S-b^{2}\ }}\ \ \ \ \ \ \ z={\sqrt {S-a^{2}\ }}}

2nd - Striking similar to Pythagorean's Theorem

S = [ x 2 + y 2 + z 2 ] = [ a 2 + z 2 ] = [ b 2 + y 2 ] = [ c 2 + x 2 ] {\displaystyle S={\Bigl [}x^{2}+y^{2}+z^{2}{\Bigr ]}={\Bigl [}a^{2}+z^{2}{\Bigr ]}={\Bigl [}b^{2}+y^{2}{\Bigr ]}={\Bigl [}c^{2}+x^{2}{\Bigr ]}\ }

a = S − z 2 b = S − y 2 c = S − x 2 {\displaystyle a={\sqrt {S-z^{2}\ }}\ \ \ \ \ \ \ \ b={\sqrt {S-y^{2}\ }}\ \ \ \ \ \ \ \ c={\sqrt {S-x^{2}\ }}}

A a b c = x 2 y 2 + z 2 ( x 2 + y 2 ) 2 {\displaystyle A_{abc}={\tfrac {\ {\sqrt {\ x^{2}y^{2}+z^{2}(x^{2}+y^{2})\ }}\ }{2}}\ }

h t e t = x y z x 2 y 2 + z 2 ( x 2 + y 2 ) {\displaystyle h_{tet}={\frac {xyz}{{\sqrt {\ x^{2}y^{2}+z^{2}(x^{2}+y^{2})\ \ }}\ }}}

a = x 2 + y 2 b = x 2 + z 2 c = y 2 + z 2 {\displaystyle a={\sqrt {x^{2}+y^{2}\ }}\ \ \ \ \ \ b={\sqrt {x^{2}+z^{2}\ }}\ \ \ \ \ \ c={\sqrt {y^{2}+z^{2}\ }}}