What is the ratio of volumes of a regular tetrahedron and the smallest cube that can encompass it?

Let's start with getting the volume of a tetrahedron. A tetrahedron is a pyramid on a triangle base, so the volume of the object would be V=1/3*Ab*H (where Ab=area of base, and H=height). The height is obtained by H=1/3*√6*a (where a=the length of a side of the tetrahedron). The area of the base is obtained by Ab=1/4*√3*a². Substituting into and compounding volume equation we obtain V=1/12*√2*a³.

Now we will find the volume of a cube. That's pretty easy as it is the side of the cube, we'll call it s, cubed. Therefore V=s³.

Now, we must realize that the largest tetrahedron that will fit inside a cube will do so when the sides of the tetrahedron, a, run along the diagonal of the faces of the cube. Therefore the relationship between s & a is ==> a=√2*s.

Finally, we put the ratio together and substitute with the a=√2*s relationship. From this we get:

Ratio=(1/12*√2*(√2*s)³) / (s³)

and reduce that down to Ratio=1/3. Sorry about the poor description, but hopefully this might help someone.

Posted by DuCk on 2003-04-28 05:02:00