Show that the ratio of the volumes of two similar pyramids

is equal to the ratio of the cubes of corresponding edges- not using modern

calculus,but by comparing magnitudes

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- September 12th 2008, 11:31 AMrmpatel5ratio of volumes
Show that the ratio of the volumes of two similar pyramids

is equal to the ratio of the cubes of corresponding edges- not using modern

calculus,but by comparing magnitudes - September 12th 2008, 03:42 PMticbol
Pyramid.

V = (1/3)(area of base)(perpendicular height of the apex from the base)

area of base is in square units.

If all its sides are "a", then area = k*a^2, where k is a constant.

If its sides are in "a" and "b", the area = j*a*b, where j is a constant. But it can be solved that b = m*a, where m is a constant, so area = n*a^2, where n is a constant.

If the base has many edges, it can be shown that the area will be p*a^2, where p is a constant and "a" is one of the edges of the base

The h, or perpendicular height of the apex.

It can also be solved that h is proportional to the "a" in the base. So, h = r*a, where r is a constant.

Therefore, the volume of the pyramid, in terms of "a" only is

V = (1/3)(p*a^2)(r*a)

V = (pr/3)(a^3) ------remember that p and r are constants, so the unit of the volume is still cubic.

Now we get two similar pyramids, where "a" is an edge of the base of one, and "b" is an edge of the base of the other, and that "a" and "b" are corresponding edges of the two similar pyramids.

V1 = (pr/3)(a^3)

V2 = (pr/3)(b^3)

V1 / V2

= [(pr/3)(a^3)] / [(pr/3)(b^3)]

= a^3 / b^3

Shown.

Even on any corresponding slant edges of the similar pyramids it can be shown just the same.