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 Subject: "Demonstration for the volume of the pyramid?" Previous Topic | Next Topic
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ricardo
Member since Aug-7-04
Apr-03-06, 07:28 AM (EST)    "Demonstration for the volume of the pyramid?"

 For some time I ask myself why the volume of the pyramid is 1/3*B*h. That the volume is directly proportional to the Area "B" and the height "h" seems reasonable enough but the 1/3 puzzled me. I know some different demonstrations for it but I wasn't completely satisfied with them. Most probably this "new" demonstration is not new but since I liked it and can't find it on the web I think it may be of interest to someone. (If you already seen this demonstration somewhere else please tell me so) First let's pick a rectangular prism with square base ABCD and truncate it with a plane not parallel to the base. Let's say that at each vertice ABCD the solid has heights h_a, h_b, h_c and h_d. We can show using some geometry that (h_a + h_c)/2 = (h_b + h_d)/2 = "height at the center of the square"then "average height" = (h_a + h_b + h_c + h_d)/4 = (h_a + h_c)/2 = (h_b + h_d)/2. The volume of the truncated solid is most reasonably B*(average height).(For more rigor this can be demonstrated dividing the solid in prisms or by some cutting and gluing to make it a rectangular prism)And what happens if the base is an equilateral triangle ABC? The same idea should apply so its volume is B*(h_a + h_b + h_c)/3. If the we have h_a = h_b = 0 and h_c = h we have a pyramid so it's volume should beB*(0 + 0 + h)/3 = B*h/3Unfortunately this demonstration is not rigorous.Another idea for a more rigorous demonstration isThere are three different pyramids we can make namely when: h_a = h_b = 0 and h_c = h h_b = h_c = 0 and h_a = h h_c = h_a = 0 and h_b = hThese pyramids are all equal so they have the same volume each pyramid is determined by a different plane. We can think the volume as the integral of the heights of the pyramid over ABC (under the linear function of R^2 in R) if we add the functions determined by each plane (and so the volumes) we arrive at a linear function where h_a = h_b = h_c = h That is a prism so the volume of the pyramid must be 1/3 of that of the prism.Obs. I made the base of the pyramid an equilateral triangle to ease things up. We can use Cavalieri�s Principle to get to the other cases.

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