Volume of a Pyramid
Date: 5/20/96 at 17:56:33 From: Anonymous Subject: Geometry Hi. I have a problem that I'm trying to solve. The problem reads: All six edges of a triangular pyramid are 4 inches long. Find the volume of the pyramid. I can't figure out a way to find the altitude. I tried the Pythagorean theorem by creating a rectangle. I used four as the hypotenuse and half of the bottom edge which would be 2. That is what I am not sure about. I came up with the answer of (4/3) times the square root of 13. Is that right? Thanks, Tim
Date: 7/17/98 at 17:20:53 From: Doctor Ken Subject: Re: Geometry [Note: one of our volunteers answered this question, but his answer contained a mathematical error, and was incorrect. A while later, a person reading our archives wrote to us to point out the error, and gave us a correct solution. This solution follows.] The volume of a tetrahedron in which all six edge lengths are represented by x is [(x^3)/12]*sqrt(2). If the edge lengths are 6, then the volume is 18*sqrt(2). In the given problem the edge lengths are 4. Therefore the volume should be [16/3]*sqrt(2). I will begin by proving the statement in the first sentence of this paragraph. Call the length of an edge of the regular tetrahedron x. Volume = (1/3)*[Area of base]*[Height] (for any pyramid) First we'll find the area of the base. Base has altitude = (x/2)*sqrt(3) (it's a 30-60-90 triangle) Area of base = (1/2)*[edge]*[altitude] (formula for area of a triangle) Area of base = (1/2)*x*(x/2)*sqrt(3) (substitution) Area of base = [(x^2)/4]*sqrt(3) (simplification) Notice that the Area of the base has a factor of sqrt(3). The center of base divides the altitude into two segments, whose lengths are in a 2 to 1 ratio. Therefore, the distance from the center of base to any vertex of the base is 2/3 of one of the base's altitudes, or (2/3)*(x/2)*sqrt(3) = x/sqrt(3). Thus: (Height)^2 = x^2 - [x/sqrt(3)]^2 (by Pythagorean Theorem) (Height)^2 = x^2 - [x^2]/3 (apply the square to the last term) (Height)^2 = [x^2]*[1 - (1/3)] (factor out x^2) (Height)^2 = [2/3]*[x^2] (simplify) Height = [sqrt(2)/sqrt(3)]*x (take square root of both sides) Notice that the Height has a denominator of sqrt(3). The two instances of sqrt(3) should disappear in the computation. Volume = (1/3)*[Area of base]*[Height] Volume = (1/3)*[(x^2/4)*sqrt(3)]*[(sqrt(2)/sqrt(3))*x] Volume = [(x^3)/12]*sqrt(2) Thus, the volume of the pyramid in your problem is [16/3]*sqrt(2). Robert H. Becker email@example.com
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