Surface Area Relationship of Pythagorean Tetrahedra | Steve Wait – November 15, 2019

Delineated is an oblique irregular (orthocentric) tetrahedron ABCD made of four right triangle faces where two vertices are formed by two right angle faces. In this example, triangle ABC lies in the x,y plane and vertice D is above. The height of the tetrahedron, perpendicular to and above its base, is given as h.

The sum of the squares of the area BCD and ACD less the same of ABD equals that of ABC. This relationship could be considered analogous to De Gua’s Theorem, but it is not the same. Having said this, perhaps there is something new here?

This by:

$.25 {a}^2 {b}^2=\left(.25 {a}^2\left({h}^2+ {b}^2\right)\right)+\left(.25 {b}^2 {h}^2\right)-\left(.25 {c}^2 {h}^2\right)$

Where:

$a=the\ cathetus\ opposite\ vertice\ A$

$b=the\ cathetus\ opposite\ vertice\ B$

$c=the\ hypotenuse\ opposite\ vertice\ C$

$h=tetrahedron\ height\ (along\ line\ AD)$