Pythagorean Theorem in Three Dimensions
Date: 05/18/2001 at 00:15:28 From: Erin Avery Subject: Tetrahedron with trirectangular vertex Pythagorean proof Suppose a tetrahedron has a trirectangular vertex S. Let A, B and C be the areas of the 3 faces that meet at S, and D be the area of the face opposite S. Prove that D^2 = A^2 + B^2 + C^2 (This is a three-dimensional version of the Pythagorean theorem.)
Date: 05/18/2001 at 08:55:09 From: Doctor Rick Subject: Re: Tetrahedron with trirectangular vertex Pythagorean proof Hi, Erin. I haven't heard the term "trirectangular" before, but I can guess that it means a vertex such that each of the three faces meeting at the vertex has a right angle at the vertex. Thus we can place this tetrahedron with S at the origin of a Cartesian coordinate system, and the three edges meeting at S lie along the x-, y- and z-axes. One way to prove the theorem follows these lines. I will take face A to be in the y-z plane; B in the x-z plane; and C in the x-y plane. The lengths of the edges along the three coordinate axes will be called x, y and z. Express the areas A, B, and C in terms of x, y and z. Now you can express A^2 + B^2 + C^2 in terms of x, y and z. Now you can find the lengths of the other three edges of the tetrahedron in terms of x, y and z, using the Pythagorean theorem. Finally, use Heron's formula for the area of a triangle in terms of its sides, to get the area D; and show that this expression is equal to the expression for A^2 + B^2 + C^2. You can find Heron's formula in our Dr. Math FAQ (under Formulas), if you don't know it. You might have trouble simplifying the area D. I suggest that you look for a way to write Heron's formula in terms of the SQUARES of the three sides; it can be done, using the factorization of a difference of squares in reverse. Then you won't have to think at all about square roots. This is a nice theorem, isn't it? When you get such a simple result, you wonder if it's necessary to go through this much algebra along the way. Maybe there's a more elegant solution. Let us know if you learn one. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/
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