Mensa Tetrahedron Solution

 The 4 vertices of a particular tetrahedron are also half the vertices of a unit cube.

The base vertices are one unit away from the fourth (apex) vertex as shown in the diagram above.   Given this information find the height (H) and volume (V) of the tetrahedron.

Hint: The volume of a tetrahedron is one third the area of the base times the height H.

Solution:

The edges connecting the base vertices are face diagonals of the cube. Therefore, the base of the tetrahedron is an equilateral triangle  of edge length  Ö2. 

 

The height h of the equilateral triangle can be found from the Pythagorean theorem:

( Ö2 / 2) ²    +   h ²    =   Ö(2) ²   Þ  (½)   +   h ²    =   2

h ²    =   3/2  Þ   h     =   Ö6/ (2)

The distance from a base vertex to the center of the base is: Ö6/ (2)  -   x

Then:

( Ö2 / 2) ²    +   x ²    =  (Ö6/ (2)  -   x )   x ²

And:

(½)   +   x ²    =   3/2   -   Ö6 x )  -  x ²

^{Ö6 x )  =   1}

^Þ 

  ^{x  =    1  /} Ö6  =   ( Ö6/ 6  )

The distance from a base vertex to the center of the base is now:

Ö6/ (2)  -   ( Ö6/ 6  )  =   2 ( Ö6/ 6  ) =  ( Ö6/ 3  )

The height H of the tetrahedron can then be found from  Pythagorean theorem, i.e.:

( Ö6/ 3  ) ²  +    H  ²  =  1 ²

^Þ 

 

2/3 +    H  ²  =   1    

H  ²  =   1    -  2/3   = 1/3

H  =   Ö3/ 3  )

The volume  V of the tetrahedron can then be found from  

 V  = 1/3 Area of the base x height H

Then:

V = 1/3    [½] (Ö2) (Ö6/ 2)   (Ö3/ 3  )

V =  1/6  cubic units