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- Path: sparky!uunet!ogicse!das-news.harvard.edu!husc-news.harvard.edu!birkhoff!kubo
- From: kubo@birkhoff.harvard.edu (Tal Kubo)
- Newsgroups: sci.math
- Subject: Re: Area of five sided polygon needed
- Message-ID: <1992Oct13.040617.16321@husc3.harvard.edu>
- Date: 13 Oct 92 08:06:16 GMT
- Article-I.D.: husc3.1992Oct13.040617.16321
- References: <1992Oct12.200125.826@altair.selu.edu>
- Organization: Dept. of Math, Harvard Univ.
- Lines: 31
- Nntp-Posting-Host: birkhoff.harvard.edu
-
- In article <1992Oct12.200125.826@altair.selu.edu> fcs$1224@altair.selu.edu writes:
- >Given a five sided polygon, is there a formula, in terms
- >of its sides, for the area?
-
- No, the sides don't uniquely determine the area, but...
-
- 1. There is a formula (due to Moebius?) for the area, S, in terms of the
- areas of the triangles A,B,C,D,E cut off by the diagonals:
-
- S^2 - (A+B+C+D+E)S + (AB+BC+CD+DE+EA) = 0
-
- This is almost certainly an exercise in Hobson's "Plane and Advanced
- Trigonometry".
-
- 2. There is an even more arcane formula for the product of areas of two
- polygons, in terms of the squared intervertex distances. When the two
- polygons coincide, this gives the area in terms of these distances.
- I don't remember the exact formula, but I remember where I saw it: a
- centenarian French book, "Theorie des Equipollences" by Laisant.
- If people are sufficiently interested, I could look up the precise
- formula.
-
- 3. I've heard of Brahmagupta-type polynomial equations for the areas of
- *cyclic* N-gons, in terms of their sides. For N=3 (Heron) and N=4
- (Brahmagupta), the equations are well-known and of degree 4 in the
- sides. For N=5,6,7 the degrees were said to be 7,7, and 38.
- Unfortunately I didn't see the formulas. Can someone reading this
- provide more information?
-
-
- -Tal kubo@math.harvard.edu
-