Volume of Pyramid. Most of the informal proofs about volume of pyramid I have seen involves cutting cube into 3 like this:

Then it skips to statement that volume of any pyramid is $\frac{bh}{3}$ where $b$ is base and $h$ is height.
How does cutting a cube into 3 pieces proves that it is true for any pyramid(let's say star pyramid, rectangular pyramid,...) is $\frac{bh}{3}$?
 A: Consider what happens when you elongate any 3d shape along one axis by a factor of $n$: the volume is multiplied by $n$. Now consider what happens when you move the tip of the pyramid so that $h$ remains constant: all cross sections of the pyramid remain the same, so the area is the same. It follows that if one pyramid's area is $bh/3$, then all similar pyramids are $bh/3$.
There are full proofs out there, but I hope this gives a better intuitive understanding of why this works.
A: Suppose you know that any pyramid with a square base has volume $\frac13 bh$.
Next, suppose the base (with area $b$) is a shape made up of $k$ squares (with areas $b_1, \dots, b_k$). We can chop up the pyramid into $k$ pyramids with square bases, which have volumes $\frac13b_1h, \dots, \frac13b_kh$. The total volume will be $\frac13(b_1 + \dots + b_k)h$, or $\frac13bh$.
Next, suppose the base is any other shape. (The pyramid could be a cone, a pentagonal pyramid, whatever.) We can approximate the base arbitrarily well with shapes made up of many tiny squares (that's how your computer screen works). Since all those approximations have volume $\frac13bh$, so does the real pyramid.
A: Just adding some visual aids for  "when you elongate any $3$d shape along one axis by a factor of $u$"(This is not an answer).
Suppose you scale the unit pyramid by $u=7$ in $z$ axis. It will look like this.

Let's change $z$ axis to $u$ axis where every unit length on vertical axis is $u$ times the unit of $z$.

So in this axis our pyramid will have volume $\frac{1*1*1}{3}$ and changing back to original space we get $\frac{u*1*1}{3}$.
