Unizor - Creative Minds through Art of Mathematics - Math4Teens
Truncated Pyramids
Imagine a pyramid. Pick any point on any of its side edges and draw a plane through this point parallel to a base plane. Cut-off a piece of a pyramid between this new plane and an apex. Whatever is left is called a truncated pyramid.
Let's do a concrete construction of a truncated pyramid.
Assume that the original pyramid was SABCD with apex S and quadrilateral ABCD that belongs to a base plane β.
Pick some point A' on edge SA between apex S and vertex A on the base plane β.
Now draw a plane γ through point A' parallel to base β.
Plane γ intersects edges SB, SC and SD at points B', C' and D' correspondingly, so we have quadrilateral A'B'C'D' within plane γ that bounds our truncated pyramid from above.
We can say now that object ABCDA'B'C'D' is a truncated pyramid.
Since β ∥ γ,
the corresponding edges lying on parallel planes are parallel:
AB ∥ A'B',
BC ∥ B'C',
CD ∥ C'D',
AD ∥ A'D'.
Each side face of a truncated pyramid is, therefore, a trapezois.
Also, from the above property follows similarity of corresponding triangles:
ΔSAB ~ ΔSA'B',
ΔSBC ~ ΔSB'C',
ΔSCD ~ ΔSC'D',
ΔSAD ~ ΔSA'D'.
The center of scaling for all these similarities is apex S and, since pairs of these triangles share sides, they all share the same scaling factor.
That same scaling factor is between the bases of two pyramids - triangles ΔABC and ΔA'B'C'.
Therefore, two pyramids, small pyramid SA'B'C'D' that we cut-off and big original pyramid SABCD are similar with the same scaling center and factor.
Drop a perpendicular from apex S onto base β.
It falls into point H, so SH⊥β.
This same perpendicular is also perpendicular to plane γ since planes β and γ are parallel.
Let the intersection of this perpendicular with γ be point H', so H'∈AH and SH'⊥γ.
It's easy to prove that the same scaling with a center at apex S and the same factor as above for pyramids transforms point H' into H (good exercise for self-study).
We see that similarity between pyramids includes not only all edges but altitudes SH' and SH as well.
Now we can proceed with calculating the volume of a truncated pyramid as a difference between volumes of two pyramids - the original one SABCD and the one we cut-off from it, SA'B'C'D'.
Assume that the factor of scaling we discussed above is f.
Then all linear elements of two pyramids are in this ratio: LenSA/LenSA'=f (edge factor)
(the same for all other edges)
Let the lengths of altitudes SH and SH' be h and h' correspondingly.
Then h/h'=f (altitude factor).
Let the areas of bases of two pyramids, big SABCD and small SA'B'C'D' be s and s' correspondingly.
As was explained in the topic 3-D Similarity, the areas of similar objects are proportional to a square of a scaling factor:
Then s/s'=f².
The volumes of small and big pyramids are:
VolumeSA'B'C'D' = s'·h'/3
VolumeSABCD = s·h/3
Therefore, the volume v of a truncated pyramid ABCDA'B'C'D' equals to
v = s·h/3 − s'·h'/3
This is a good formula, but not good enough. It would be much cleaner to express the volume of an object in terms of elements of this object. In the formula above s and s' are such elements - areas of top and bottom bases of a truncated pyramid, but h and h' are not elements of a truncated pyramid since they involve a distance from apex S, which is outside of our truncated pyramid) to its bases.
Our goal now is to replace dependency on these parameters with a dependency on the height of a truncated pyramid itself - the distance between its bases a.
Let's summarize what we know and what has to be done in terms of elements of a truncated pyramid.
We know that
h/h' = f
h − h' = a
s/s' = f²
We have to express the volume v=s·h/3−s'·h'/3 in terms of s, s' and a.
Consider the top three equations above as a system of three equations with three unknown h, h' and f.
Solving it and substituting the solutions for h and h' into a formula for volume will produce the desired result.
v = (a/3)·(s+√s·s'+s')