home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: sci.math
- Path: sparky!uunet!news.smith.edu!orourke
- From: orourke@sophia.smith.edu (Joseph O'Rourke)
- Subject: Re: pyramid volume
- Message-ID: <1993Jan21.173423.11339@sophia.smith.edu>
- Organization: Smith College, Northampton, MA, US
- References: <1993Jan21.140402.25519@mr.med.ge.com>
- Date: Thu, 21 Jan 1993 17:34:23 GMT
- Lines: 29
-
- In article <1993Jan21.140402.25519@mr.med.ge.com> carl@crazyman.med.ge.com (Carl Crawford) writes:
- >
- >how do show that the volume of a pyramid is
- >
- > 1/3 * area of base * altitude
- >
- >without using calculus?
-
- Nice question! Although I don't have an answer, I would like to
- make an observation: three identical tetrahedra pack half a cube.
- Take the unit cube
- 0 <= x <= 1
- 0 <= y <= 1
- 0 <= z <= 1
- and intersect it with the halfspace x+y <= 1. The result is a
- unit-height triangular prism, the convex hull of
- 000, 100, 010,
- 001, 101, 011,
- where "000" means "(0,0,0)" etc. Suppose you believe the volume of
- this is V = A * h = A, where A is the base area and h=1 the height.
- Now partition the prism into three right tetrahedra, the hulls of
- T1: 000, 100, 010, 001;
- T2: 001, 101, 011, 100;
- T3: 000, 001, 011, 100.
- T1 has base A on the plane z=0, and height 1;
- T2 has base A on the plane z=1, and height 1;
- T3 has base A on the plane x=0, and height 1.
- Since these three are congruent, each has volume V/3.
- I looked for this construction in several books without luck.
-