Volume of a tetraedron in function of the areas

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SUMMARY

The volume V of a tetrahedron can be calculated using the areas of its surfaces, specifically the areas of its four triangular faces (X, Y, Z, W). However, according to the referenced PDF, three additional pseudo-faces (H, J, K) are necessary for accurate volume computation. This indicates that merely knowing the areas of the four faces is insufficient due to the ambiguity in the shape of the tetrahedron, which can lead to multiple configurations with the same face areas but different volumes.

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Bruno Tolentino
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Given any tetraedron, I want to calculate the volume V in function of the areas of the surfaces of the solid. I found this pdf that explain this:

http://daylateanddollarshort.com/mathdocs/Heron-like-Results-for-Tetrahedral-Volume.pdf

But, o pdf says that beyond of the 4 faces (X, Y, Z, W) is necessary more 3 pseudo-faces (H, J, K). But, is it correct? Is really necessary 7 areas for compute the volume? With just 4 is not possible?
 
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Have you tried working it out for the areas of 4 faces alone?

The trouble is that the area of a triangle does not determine it's shape - so you can construct many differently volumed tetrahedra out of four triangles knowing only their areas (but not their shapes).
 
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