Height of Tetrahedral PyramidDate: 09/12/2001 at 23:07:17 From: Bob Subject: Height of tetrahedral pyramid Dr. Math: I'm looking for a simple formula (and derivation) of the height of a tetrahedral pyramid with an equilateral triangle as a base. The "sides" making up the pyramid are isosceles triangles with 45-degree angles - essentially the pyramid is a portion of a standard cube. I realize that the volume of the pyramid is defined by V=bh/3, but is there a simple formula that relates the height, h, to the sides of the base? Thanks. Date: 09/13/2001 at 11:25:41 From: Doctor Rick Subject: Re: Height of tetrahedral pyramid Hi, Bob. Yes, there is a simple formula, and I'll show you how to derive it yourself - in two different ways. By mentioning the volume, you've given me an idea for an interesting way to find the height of this pyramid. Since it is the corner of a cube, we can regard one of the isosceles triangles as the base, and the other edge of the cube is perpendicular to the base, so it is the altitude. Thus the volume of the pyramid is V = (1/3)bh = (1/3)(s^2/2)s = s^3/6 What is s? It is the leg of an isosceles right triangle with hypotenuse equal to the side of the equilateral triangle, which I'll call a. Use the Pythagorean theorem to find s. Plug this expression for s into the equation for V to get an expression for V in terms of a. Now, turning the pyramid back so the equilateral triangle is the base, you can find a second equation for V in terms of a and h. This will require finding the area of an equilateral triangle, which in turn involves finding the length of the altitude of the equilateral triangle. If you don't know this, you can find it using the fact that the altitude of an equilateral triangle is also its median, so it bisects the side it is perpendicular to. This allows you to use the Pythagorean theorem. The volume must be the same however we calculate it, so you can set the two expressions for V equal to each other. Solve for h to get a formula for h in terms of a. You can also calculate the altitude h as follows. Find the side length s, as above. Find the median (or altitude) of the equilateral triangle, as above. Due to symmetry, the altitude h meets the base at the centroid of the triangle, which divides each median in 1/3:2/3 proportion. Therefore we have a right triangle consisting of a side of length s, the altitude, and 2/3 of a median of the base. You can use the Pythagorean theorem to find h, and you'll get the same answer as by the volume method. I'm not sure whether you really needed the formula for the altitude given the base side, or you just needed the volume of the pyramid, and you thought that this required finding the altitude. If you only needed the volume, then I've given you more than you needed, but I hope it's interesting anyway. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 09/13/2001 at 12:22:44 From: Doctor Jubal Subject: Re: Height of tetrahedral pyramid Hi Bob, The height of a pyramid depends on its shape, so unlike the volume formula V=bh/3 that works for all pyramids, there's no succinct formula that will give the height of any pyramid. However, you can derive formulas specific to pyramids of specific shapes, including your pyramid. To derive a formula for the height of your pyramid, all you need to do is apply the Pythagorean theorem again and again. Consider one of the triangles we could expose if we made a vertical slice cutting the pyramid in half: A |\ | \ | \ | \ s h | \ | \ | \ |_ \ |_|______\ C c B A is the apex of the pyramid, B is one of the corners of the base, and C is the center of the equilateral triangle that is the base. That makes s the length of one of the non-base edges of the pyramid's sides, h the pyramid's height, and c the distance from the center of the base to one of its vertices. If we can find s and c, we can use the Pythagorean theorem to find h. Let's start with s. The sides of the pyramid are isoceles right triangles. I'll use a for the length of an edge along the base of the pyramid. a is the hypoteneuse of one of the side faces, and s is the length of the other two sides, so by the Pythagorean theorem: s^2 + s^2 = a^2 s^2 = (a^2)/2 s = a * sqrt(1/2) Now for c. The base of the pyramid is an equilateral triangle: B ^ /|\ / | \ / |c \ a / | \ / |C \ / __/| \ /__/c |i \ //______|_______\ D H a/2 Points B and C are the same as before, H is the midpoint of the side opposite B, D is another corner of the triangle. c is the length of segment BC, as before. Since the triangle is equilateral, c is also the length of CD. i is the length of CH. Applying the Pythagorean theorem to triangle BDH: (a/2)^2 + (c + i)^2 = a^2 (a^2)/4 + (c + i)^2 = a^2 (c + i)^2 = (3/4)a^2 (c + i) = a * sqrt(3/4) c^2 + 2ic + i^2 = (3/4)a^2 We can get rid of i^2 by applying the Pythagorean theorem to triangle CDH: i^2 + (a/2)^2 = c^2 i^2 = c^2 - (a^2)/4 Substituting this back into the previous expression 2*c^2 + 2ic - (a^2)/4 = (3/4)a^2 2*c^2 + 2ic = a^2 2c * (c + i) = a^2 2c * [a * sqrt(3/4)] = a^2 c = a/(2sqrt(3/4)) = a/sqrt(3) So now we have c and s, and one final application the Pythagorean theorem gives us h. h^2 = s^2 - c^2 h^2 = (a^2)/2 - (a^2)/3 h^2 = (a^2)/6 h = a/sqrt(6) Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/ |
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