Volume of a TetrahedronDate: 02/27/2003 at 15:48:16 From: Haze Subject: Tetrahedron How do I find the volume of a tetrahedron? Date: 03/02/2003 at 18:43:35 From: Doctor Ian Subject: Re: Tetrahedron Hi Haze, The most straightforward way to find it is to apply the Pythagorean theorem several times. Here is an equilateral triangle, with vertices A, B, and C; center D; and midpoint of base E. It's just like the one at the base of your tetrahedron. A . . . . . . . . . . . . . . . D . . . . . . . . . C . . . .E. . . . B For brevity, we'll use a = AB = BC = CA h = AE d = DE r = CD The Pythagorean theorem tells us that h^2 + (a/2)^2 = a^2 h^2 = a^2 - (a/2)^2 h^2 = 3a^2/4 h = a sqrt(3)/2 So the area of the whole triangle is area = (1/2) * base * height = (1/2) * a * a sqrt(3)/2 = a^2 sqrt(3)/4 The area of triangle DBC is 1/3 the area of the whole triangle, so (1/3) a^2 sqrt(3)/4 = (1/2) * base * height = (1/2) a d (2/a)(1/3) a^2 sqrt(3)/4 = d a sqrt(3)/6 = d The Pythagorean theorem tells us that r^2 = (a/2)^2 + d^2 = (a/2)^2 + [a sqrt(3)/6]^2 = a^2/4 + 3a^2/36 = a^2/4 + a^2/12 = (3a^2 + a^2)/12 = 4a^2/12 = a^2/3 r = a/sqrt(3) So, if we make a tetrahedron from four of these triangles, and drop an altitude from the apex of the tetrahedron to the center of the base, we get a right triangle hypotenuse a, base r, and height H. The Pythagorean theorem tells us that r^2 + H^2 = a^2 H^2 = a^2 - r^2 = a^2 - a^2/3 = 2a^2/3 H = a sqrt(2/3) Since the volume of any pyramid is 1/3 the area of the base times the height, Volume of a Pyramid http://mathforum.org/library/drmath/view/55041.html the volume is volume = (1/3) * area of base * height = (1/3) * [a^2 sqrt(3)/4] * [a sqrt(2/3)] = a^3 (1/3) [sqrt(3)/4] [sqrt(2/3)] = a^3 sqrt(2)/12 Happily, this agrees with the formula in our Dr. Math Geometric Formulas FAQ: Pyramid & Frustum Formulas http://mathforum.org/dr.math/faq/formulas/faq.pyramid.html I hope this helps! - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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