home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!snorkelwacker.mit.edu!linus!linus.mitre.org!gauss!bs
- From: bs@gauss.mitre.org (Robert D. Silverman)
- Subject: Re: Area of five sided polygon needed
- Message-ID: <1992Oct14.124037.9927@linus.mitre.org>
- Sender: news@linus.mitre.org (News Service)
- Nntp-Posting-Host: gauss.mitre.org
- Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
- References: <1992Oct12.200125.826@altair.selu.edu> <1992Oct13.030526.22048@galois.mit.edu>
- Date: Wed, 14 Oct 1992 12:40:37 GMT
- Lines: 20
-
- In article <1992Oct13.030526.22048@galois.mit.edu> jbaez@riesz.mit.edu (John C. Baez) writes:
- :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?
- :
- :Clearly not, because the lengths of the sides don't determine the
- :area. Pentagons are "floppy." So are quadrilaterals, for that matter.
- :
-
- This is clearly correct. However, there is such a formula if the pentagon
- is incribable in a circle.
-
- Hint: It is an extension of Heron's formula.
-
-
- --
- Bob Silverman
- These are my opinions and not MITRE's.
- Mitre Corporation, Bedford, MA 01730
- "You can lead a horse's ass to knowledge, but you can't make him think"
-