Derived From De Gua's Theorem - Works like Magic !!!

5 Coexisting Relationships BUT ONLY in TriRectangular Tetrahedrons

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

Perfect Mirrored Imaged Equations

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}a,b,c{\Bigr ]}\ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}x,y,z{\Bigr ]}$

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

$\ \ \ \ Use\ {\Bigl [}x^{2}+c^{2}{\Bigr ]}\ \ \ \ \ \ \ \ x={\sqrt {_{\ }K-c^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ c={\sqrt {_{\ }K-x^{2}\ \ }}$

$\ \ \ \ Use\ {\Bigl [}y^{2}+b^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ y={\sqrt {_{\ }K-b^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ b={\sqrt {_{\ }K-y^{2}\ \ }}$

$\ \ \ \ Use\ {\Bigl [}z^{2}+a^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ z={\sqrt {_{\ }K-a^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ a={\sqrt {_{\ }K-z^{2}\ \ }}$

Volume of Tetrahedron & Enclosing Box

$\ \ \ \ \ V_{tet}={\frac {\ xyz\ }{6}}\ \ \ \ \ \ \ \ V_{box}=xyz$

Height of Tetrahedron

$\ \ \ \ \ h_{tet}\ ={\frac {xyz}{\ \ {\sqrt {\ x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }}$

2 Formulas For The Area of The Base

$\ \ \ \ A_{abc}={\frac {xyz}{\ 2h_{tet}}}\ \ \ \ \ \ \ A_{abc}={\frac {\ \ {\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ \ }{2}}$

Individual Areas of All Four Surfaces

$\ \ \ \ A_{abc}={\frac {xyz}{\ 2h_{tet}}}\ \ \ \ A_{xy}={\frac {\ xy\ }{2}}\ \ \ \ A_{xz}={\frac {\ xz\ }{2}}\ \ \ \ A_{yz}={\frac {\ yz\ }{2}}$

Total Surface Area of a Tetrahedron

$\ \ \ \ A_{tet}=A_{abc}+A_{xy}+A_{xz}+A_{yz}$

$\ \ \ \ A_{tet}={\frac {\ xy+z{\Bigl [}x+y{\Bigr ]}+{\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }{2}}$

Example Parameters

$\ \ \ \ a=14.4\ \ \ \ \ b=10\ \ \ \ \ c=12$

$\ \ \ \ x=9.037698822156002781875821128798$

$\ \ \ \ y=11.210709165793214885544254958172$

$\ \ \ \ z=4.2801869118065393196504607419769$

The Other Way - Harder to Use But Still Works

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

$Given\ {\Bigl [}a,b,c{\Bigr ]}\ \ \ \ \ \ \ \ x={\sqrt {{\frac {\ a^{2}+b^{2}-c^{2}\ }{2}}\ }}\ \ \ \ \ \ \ \ y={\sqrt {{\frac {\ a^{2}+c^{2}-b^{2}\ }{2}}\ }}\ \ \ \ \ \ \ \ z={\sqrt {{\frac {\ b^{2}+c^{2}-a^{2}\ }{2}}\ }}$

User:Tet-Math2/sandbox

Contents

Derived From De Gua's Theorem - Works like Magic !!!

5 Coexisting Relationships BUT ONLY in TriRectangular Tetrahedrons

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

Perfect Mirrored Imaged Equations

$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}a,b,c{\Bigr ]}\ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}x,y,z{\Bigr ]}$

$\ \ \ \ Use\ {\Bigl [}x^{2}+c^{2}{\Bigr ]}\ \ \ \ \ \ \ \ x={\sqrt {_{\ }K-c^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ c={\sqrt {_{\ }K-x^{2}\ \ }}$

$\ \ \ \ Use\ {\Bigl [}y^{2}+b^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ y={\sqrt {_{\ }K-b^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ b={\sqrt {_{\ }K-y^{2}\ \ }}$

$\ \ \ \ Use\ {\Bigl [}z^{2}+a^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ z={\sqrt {_{\ }K-a^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ a={\sqrt {_{\ }K-z^{2}\ \ }}$

Volume of Tetrahedron & Enclosing Box

$\ \ \ \ \ V_{tet}={\frac {\ xyz\ }{6}}\ \ \ \ \ \ \ \ V_{box}=xyz$

Height of Tetrahedron

$\ \ \ \ \ h_{tet}\ ={\frac {xyz}{\ \ {\sqrt {\ x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }}$

2 Formulas For The Area of The Base

$\ \ \ \ A_{abc}={\frac {xyz}{\ 2h_{tet}}}\ \ \ \ \ \ \ A_{abc}={\frac {\ \ {\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ \ }{2}}$

Individual Areas of All Four Surfaces

$\ \ \ \ A_{abc}={\frac {xyz}{\ 2h_{tet}}}\ \ \ \ A_{xy}={\frac {\ xy\ }{2}}\ \ \ \ A_{xz}={\frac {\ xz\ }{2}}\ \ \ \ A_{yz}={\frac {\ yz\ }{2}}$

Total Surface Area of a Tetrahedron

$\ \ \ \ A_{tet}=A_{abc}+A_{xy}+A_{xz}+A_{yz}$

$\ \ \ \ A_{tet}={\frac {\ xy+z{\Bigl [}x+y{\Bigr ]}+{\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }{2}}$

Example Parameters

$\ \ \ \ a=14.4\ \ \ \ \ b=10\ \ \ \ \ c=12$

$\ \ \ \ x=9.037698822156002781875821128798$

$\ \ \ \ y=11.210709165793214885544254958172$

$\ \ \ \ z=4.2801869118065393196504607419769$

The Other Way - Harder to Use But Still Works

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

$Given\ {\Bigl [}a,b,c{\Bigr ]}\ \ \ \ \ \ \ \ x={\sqrt {{\frac {\ a^{2}+b^{2}-c^{2}\ }{2}}\ }}\ \ \ \ \ \ \ \ y={\sqrt {{\frac {\ a^{2}+c^{2}-b^{2}\ }{2}}\ }}\ \ \ \ \ \ \ \ z={\sqrt {{\frac {\ b^{2}+c^{2}-a^{2}\ }{2}}\ }}$