Newsgroups: sci.math From: firstname.lastname@example.org (Joseph O'Rourke) Subject: Re: pyramid volume Message-ID: <1993Jan21.email@example.com> Organization: Smith College, Northampton, MA, US References: <1993Jan21.firstname.lastname@example.org> Date: Thu, 21 Jan 1993 17:34:23 GMT
In article <1993Jan21.email@example.com> firstname.lastname@example.org (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.