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Is there any formula for the volume of a tetrahedron in terms of the areas of its four faces ?

Need any assumptions be necessarily made ( like each face altitude, circum-radius) etc?

EDIT

Like to know... if it is possible to find area of a triangle in terms of its sides then in higher dimensions why in terms of enveloping areas it is not possible to find enclosed volume.

I want to know the mathematical logic about possibility of finding it or lack of it.

Thanks.

Narasimham
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  • Since the areas of the faces can be determined by the edge lengths, this amounts to asking for a formula for the volume based on edge lengths. The Wikipedia article on tetrahedra discusses properties analogous to those of triangles. – Théophile Feb 06 '21 at 06:01
  • You have three very different questions here. What have you tried? The second seems simple. For the third, what computation do you mean? If it is just the volume of the tetrahedron, is the $\frac 16$ an arbitrary constant? I don't think so. -1 – Ross Millikan Feb 06 '21 at 06:01
  • @Théophile: So, wanted to avoid basic lengths in order to attempt a direct relation if conditionally possible. – Narasimham Feb 06 '21 at 06:08
  • @RossMillikan: Thanks. edited somewhat.. keeping it simple for a start..Vague thought about Gauss divergence thm, or its relevance if any.. – Narasimham Feb 06 '21 at 06:36
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    I have asked similar question. maybe you can check the question and answer . https://math.stackexchange.com/questions/160258/is-the-volume-of-a-tetrahedron-determined-by-the-surface-areas-of-the-faces – Mathlover Feb 06 '21 at 07:05
  • There are 4 faces but 6 edges so I would be surprised if the areas of the faces fully determined the volume – hacatu Feb 06 '21 at 07:15
  • Me too... a plane triangle where sides are given its area can be found... so what undetermined constant may be involved when face areas are given while finding volume of the tetrahedron. – Narasimham Feb 06 '21 at 08:04
  • According to @Jim Belk 's answer , we cannot determine the volume of tetrahedron via using surface areas. Please see his counterexamples in his answer. https://math.stackexchange.com/questions/160258/is-the-volume-of-a-tetrahedron-determined-by-the-surface-areas-of-the-faces – Mathlover Feb 06 '21 at 08:16
  • If $r$ is the radius of the inscribed sphere, then $$V=\frac{1}{3}r\left(A_1+A_2+A_3+A_4\right)$$ – Raffaele Feb 06 '21 at 16:32
  • Interesting. but how does $r$ depend on these four areas? – Narasimham Feb 06 '21 at 16:58

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