The following from an article:

I'm going to prove the formula for the volume of a truncated rectangular pyramid with a square base and square top. It's much easier to work with a square base instead of a rectangular one. Liu Hui (220 - 280 AD, China) discovered the volume of a truncated pyramid with a square base and square top. He called it a fangting. First, he broke the shape into 9 pieces: one rectangular prism (box), four pyramids and four triangular prisms (wedges). Then, he calculated the volume of each piece and came up with a formula for the overall shape.

Let's say the top is a square with side length a, the bottom is a square with side length b and the height is h. Now, let's look at the volume of each piece.

Rectangular Prism: V = l w h,
so our rectangular prism's volume will be (a)(a)(h) or a2h

Pyramid: V = (1/3)(base)(height),
so our pyramid's volume will be V = (1/3)((b - a)/2)2(h)

Triangular Prism: V = (1/2)(length)(width)(height),
so our triangular prism's volume will be V = (1/2)((b - a)/2)(a)(h)

Now, we need to add all the pieces together to get the whole truncated pyramid. Since we have one rectangular prism, four identical pyramids and four identical triangular prisms, our formula will be:

V = (a2h) + 4(1/3)((b - a)/2)2(h) + 4(1/2)((b - a)/2)(a)(h)
= a2h + (1/3)(b - a)2(h) + (b - a)(a)(h)
= a2h + (1/3)(b2 - 2ab + a2)(h) + abh - a2h
= a2h + (1/3)(b2h - 2abh + a2h) + abh - a2h (Now, factor out a (1/3))
= (1/3)(3a2h + b2h - 2abh + a2h + 3abh - 3a2h)
= (1/3)(a2h + abh + b2h) (Now, factor out an h)
= (1/3)(a2 + ab + b2)(h)

So, our formula for a truncated pyramid with a square base and top is
V = (1/3)(a2 + ab + b2)(h)

Cheers
Don Barone



"There is nothing as impenetrable as a closed mind"
and ..." if everything is a coincidence what is the point of studying or measuring or analyzing anything ?" db