Question

I have to confirm tht the ancient Egyptian formula for the volume of a truncated pyramid is correct. The formula I am given is V=h/3(a^2+ab+b^2). The base of the pyramid is an a x a square, and the top is a b x b square. I am just not sure how to figure this one out. Can you help me?

Answer 1

Height of vertex to base, H= ha/(a-b)
Volume of complete pyramid
=(a^2H/3)
Volume of truncated part
=(b^2(H-h))/3
Difference
=a^2*H/3-b^2*(H-h)/3
=(h/3)(a^2+ab+b^2)
exactly what the Egyptians did.

Answer 2

Ok...I am c(H-h)onfused...Why is the height of the vertex to the base ha/(a-b)...and isn't the formula the volume of a complete pyramid 1/3b^2h? Why b^2(H-h)?

Answer 3

The given dimension h is the truncated pyramid, with the top missing.
We can calculate the height H of the untruncated pyramid by proportion, namely
a=side length of base
b=side length of base of truncated part
The difference (a-b) is proportional to the truncated height as a is proportional to the full height H, thus:
h:(a-b) = H:a
which gives
H=ah/(a-b)