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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?

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# Homework Help: Equal Tetrahedron

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