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Volume of a tetrahedron regular

  1. Jun 14, 2015 #1
    See the image that I uploaded...


    I want to write the surface S (bounded by edges u, v and w) in terms of x, y and z, u, v and w and A, B and C. And I got it!

    [tex]S(A,B,C) = \sqrt{A^2+B^2+C^2}[/tex]
    [tex]S(x,y,z) = \sqrt{\frac{1}{4}( (yz)^2 + (zx)^2 + (xy)^2 )}[/tex]
    [tex]S(u,v,w) = \sqrt{(+u+v+w)(-u+v+w)(+u-v+w)(+u+v-w)}[/tex]

    And the Volume V:
    [tex]V(x,y,z) = \frac{1}{6} xyz [/tex]
    But, I don't know how to write V in terms of A, B, C neither u, v, w. Can you help me with this, please?
  2. jcsd
  3. Jun 14, 2015 #2


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    There are some complicated Heron-type formulas for computing the volume of a tetrahedron which are similar to those for computing the area of a triangle. However, these formulas are much more complex.

    The paper at the following link shows the derivation of these formulas ad gives references for further study:


    Tetrahedrons are discussed starting at p. 11, but the previous material provides a good refresher.
  4. Jun 14, 2015 #3
    I thank you for this answer. Actually, this no answer my question, but I'll intend to ask this in another thread. I'll intend to ask this and more one thing, that's the following:

    Given a tetrahedron irregular (any tetrahedron), how to write the volume V in terms of the areas A, B, C and S?

    OBS: my first question in this thread still no be answered.
  5. Jun 14, 2015 #4


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    If the edges x, y, and z are mutually perpendicular, you can write expressions for the areas A, B, and C using those lengths.
  6. Jun 14, 2015 #5
    I don't understand you explanation...

    I discovered how to write V in terms of u, v and w:

    [tex]V(u,v,w) = \sqrt{\frac{1}{288} (+u^2+v^2-w^2) (+u^2-v^2+w^2) (-u^2+v^2+w^2) }[/tex]
  7. Jun 15, 2015 #6
    I discovered too: [tex]V(A,B,C) = \sqrt{\frac{2}{9} A B C} [/tex]
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