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Equal Tetrahedron

  1. Aug 6, 2006 #1
    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?
     

    Attached Files:

  2. jcsd
  3. Aug 7, 2006 #2

    Integral

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    Show me that the base of the triangle is [itex] \frac h 2 [/itex]

    Please take the time to read this thread. Your equations are not very clear.
     
  4. Aug 7, 2006 #3

    danago

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    Im finding it kinda hard to understand your working, but shouldnt the second line be:
    L2 = H2 + (H2/4)
    ?
     
  5. Aug 7, 2006 #4

    HallsofIvy

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    The center of an equilateral triangle is on the altitude, 1/3 of the way from the vertex to the base. That is not "H/2". If an equilateral triangle has sides of length L, what is the length of the altitude? What is 1/3 of that?
     
  6. Aug 9, 2006 #5
    In an equilateral triangle the Height is always side times sq(3)/2, that's basic Geometry knowledge, but do you mean the height of the whole figure, or the height of one of the faces? That...is not the same.
     
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