Starting from a pyramid with a heigth that equals only half of the height of the cube, we even don't have to deform the pyramids to find a relationship between the volume of a cube and a pyralid.

Let s be the length of the side of a cube. In the applet we can see that the volume of a pyramid with a square base and a height that equals half of the the heigth of the cube can be calculated as [b]V = 1/6 . s³[/b] [br]In the opposite way we can say that you an divide a cube into 6 equal pyramids.[br]as a formula for the volume of a pyramid we find:[br][table][tr][td]6 . I[/td][td]= s³[/td][/tr][tr][td]I[/td][td]= 1/6 . s³[/td][/tr][tr][td]I[/td][td]= 1/3 . s² . 1/2 . s[/td][/tr][tr][td][b]I[/b][/td][td][b]= 1/3 area base . heigth[/b][/td][/tr][/table]