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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)
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"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 be

B*(0 + 0 + h)/3 = B*h/3

Unfortunately this demonstration is not rigorous.

Another idea for a more rigorous demonstration is

There 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 = h

These 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 Cavalieriís Principle to get to the other cases.


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