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 Imaging ${\Bigl [}Thanks\ \ to\ \ DeGua{\Bigr ]}$

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

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

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

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

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

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

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

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

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

5 Coexisting Relationships BUT ONLY in TriRectangular Tetrahedrons

K = [ a 2 + b 2 + c 2 2 ] = [ x 2 + y 2 + z 2 ] = [ x 2 + c 2 ] = [ y 2 + b 2 ] = [ z 2 + a 2 ] {\displaystyle \ \ \ \ 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 Imaging [ T h a n k s t o D e G u a ] {\displaystyle {\Bigl [}Thanks\ \ to\ \ DeGua{\Bigr ]}}

G i v e n [ a , b , c ] G i v e n [ x , y , z ] {\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}a,b,c{\Bigr ]}\ \ \ \ \ \ \ \ \ \ \ \ \ Given{\Bigl [}x,y,z{\Bigr ]}}

U s e [ x 2 + c 2 ] x = K − c 2 c = K − x 2 {\displaystyle \ \ \ \ Use\ {\Bigl [}x^{2}+c^{2}{\Bigr ]}\ \ \ \ \ \ \ \ x={\sqrt {_{\ }K-c^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ c={\sqrt {_{\ }K-x^{2}\ \ }}}

U s e [ y 2 + b 2 ] y = K − b 2 b = K − y 2 {\displaystyle \ \ \ \ Use\ {\Bigl [}y^{2}+b^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ y={\sqrt {_{\ }K-b^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ b={\sqrt {_{\ }K-y^{2}\ \ }}}

U s e [ z 2 + a 2 ] z = K − a 2 a = K − z 2 {\displaystyle \ \ \ \ Use\ {\Bigl [}z^{2}+a^{2}{\Bigr ]}\ \ \ \ \ \ \ \ \ z={\sqrt {_{\ }K-a^{2}\ \ }}\ \ \ \ \ \ \ \ \ \ \ a={\sqrt {_{\ }K-z^{2}\ \ }}}

Volume of Tetrahedron & Enclosing Box

V T E T = x y z 6 V B O X = x y z {\displaystyle \ \ \ \ \ V_{TET}={\frac {\ xyz\ }{6}}\ \ \ \ \ \ \ \ V_{BOX}=xyz}

Height of Tetrahedron

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}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ }}}

2 Formulas For The Area of The Base

A A B C = x y z 2 H T E T A A B C = x 2 y 2 + z 2 [ x 2 + y 2 ] 2 {\displaystyle \ \ \ \ 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 A B C = x y z 2 H T E T A X Y = x y 2 A X Z = x z 2 A Y Z = y z 2 {\displaystyle \ \ \ \ 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 T E T = A A B C + A X Y + A X Z + A Y Z {\displaystyle \ \ \ \ A_{TET}=A_{ABC}+A_{XY}+A_{XZ}+A_{YZ}}

A T E T = x y + z [ x + y ] + x 2 y 2 + z 2 [ x 2 + y 2 ] 2 {\displaystyle \ \ \ \ 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 {\displaystyle \ \ \ \ a=14.4\ \ \ \ \ b=10\ \ \ \ \ c=12}

x = 9.037698822156002781875821128798 {\displaystyle \ \ \ \ x=9.037698822156002781875821128798}

y = 11.210709165793214885544254958172 {\displaystyle \ \ \ \ y=11.210709165793214885544254958172}

z = 4.2801869118065393196504607419769 {\displaystyle \ \ \ \ z=4.2801869118065393196504607419769}

The Other Way - Harder to Use But Still Works

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}\ }}}

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

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

Perfect Mirrored Imaging ${\Bigl [}Thanks\ \ to\ \ DeGua{\Bigr ]}$

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

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

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

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

$\ \ \ \ \ V_{TET}={\frac {\ xyz\ }{6}}\ \ \ \ \ \ \ \ V_{BOX}=xyz$

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

$\ \ \ \ A_{ABC}={\frac {xyz}{\ 2H_{TET}}}\ \ \ \ \ \ \ A_{ABC}={\frac {\ \ {\sqrt {x^{2}y^{2}+z^{2}{\Bigl [}x^{2}+y^{2}{\Bigr ]}\ }}\ \ \ }{2}}$

$\ \ \ \ A_{ABC}={\frac {xyz}{\ 2H_{TET}}}\ \ \ \ A_{XY}={\frac {\ xy\ }{2}}\ \ \ \ A_{XZ}={\frac {\ xz\ }{2}}\ \ \ \ A_{YZ}={\frac {\ yz\ }{2}}$

$\ \ \ \ A_{TET}=A_{ABC}+A_{XY}+A_{XZ}+A_{YZ}$

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

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

$\ \ \ \ x=9.037698822156002781875821128798$

$\ \ \ \ y=11.210709165793214885544254958172$

$\ \ \ \ z=4.2801869118065393196504607419769$

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

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