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It is known since ancient times how the volume of triangle can be expressed in terms of its edge lengths (Heron's formula). Is there a generalization for simplices that specifies the volume of an $n$-dimensional simplex in terms of the areas of its facets?

(I'm not looking for the Cayley-Menger determinant which gives the $n$-volume in terms of all edge lengths.)

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    $\begingroup$ The big issue is to define what you call the "area" of $n$-dimensional facets... Besides, have a look at De Gua's theorem and its generalization (en.wikipedia.org/wiki/De_Gua%27s_theorem) $\endgroup$
    – Jean Marie
    Mar 10, 2021 at 18:08
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    $\begingroup$ See the answer to this question $\endgroup$
    – Jean Marie
    Mar 10, 2021 at 18:10
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    $\begingroup$ Even in three dimensions, there's a problem: a tetrahedron admits six degrees of freedom, but faces account for only four. My personal "hedronometric" research introduces the notion of three pseudo-faces to make up the deficit. (There's a relation between face and pseudo-face areas that reduces the degrees of freedom to six.) With these, volume can be uniquely determined. See this answer of mine to "Does knowing the surface area of all faces uniquely determine a tetrahedron?" $\endgroup$
    – Blue
    Mar 10, 2021 at 18:14
  • $\begingroup$ @JeanMarie What's meant by "surface area" is clear: The sum of the areas of all facets. The question you link is mostly concerned with the fact that a common formula only works if the spatial dimension matches that of the simplex. $\endgroup$ Mar 10, 2021 at 19:01
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    $\begingroup$ @NicoSchlömer: Read the "extreme example" in my answer. (See also the introduction to my "Heron-like Results for Tetrahedral Volume" (PDF link via daylateanddollarshort.com Tetrahedra with four equal face areas range in volume from $0$ to a maximum for regular tetrahedra. Four areas simply aren't enough to determine a tetrahedron's volume. $\endgroup$
    – Blue
    Mar 10, 2021 at 19:25

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Promoting comments to an answer, as requested:


Even in three dimensions, there's a problem: a tetrahedron admits six degrees of freedom, but faces account for only four. My personal "hedronometric" research introduces the notion of three pseudo-faces to make up the deficit. (There's a relation between face and pseudo-face areas that reduces the degrees of freedom to six.) With these, volume can be uniquely determined.

See my answer to the question "Does knowing the surface area of all faces uniquely determine a tetrahedron?".

I'll echo here that my note "Heron-like Results for Tetrahedral Volume" (PDF link via daylateanddollarshort.com) considers this topic at length. You don't have to take my own word for it, though; as mentioned in the note (as Theorem 7), Gerber proved in 1975 that a tetrahedron whose opposite edges are orthogonal maximizes volume for a given set of face-areas. We wouldn't need that result if face-areas determined volume uniquely.

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