Here's the problem: A regular tetrahedron is a three-dimensional object that has four faces, each of which is an equilateral triangle. Each of the edges of such an object has a length L. The height H of a regular tetrahedron is the perpendicular distance from one corner to the center of the opposite triangular face. Show that the ratio between H and L is H/L = sqrt (2/3). Here's what I've done so far: Take a look at the attachment. L2 = H2 + (H/2)2 L2 = H2 + (H2/2) L2 = ( (2H2)/2 ) + (H2/2) L2 = 3H2/2 2L2 = 3H2 (√2)L = (√3)H ((√2) / (√3)) / L = H ((√2) / (√3)) = H / L √(2/3) = H / L I was excited that I thought I had found the answer that I completly squared the 'b' in this euqation, thus throwing off the rest of the equation. I know I'm close, but I seem to be missing something. Will someone please help?