Puzzle on the triangle. In triangle top four figures that have to be repositioned to form the "triangle" without a unit square.
How to explain this?

Thank's.
 A: The three points along what looks like a hypotenuse of a big right triangle are actually not collinear. So since the "big triangle" is not really one right triangle the usual area formula doesn't apply.
Note: in the top "triangle" the three points are $(0,0),(8,3),(13,5)$, so that IF they were on a line the slopes 5/13 and 3/8 would have to match. But they don't.
ADDED: In the top figure, a careful diagram shows the "hypotenuse" is made of two segments which actually bend outwards a little, and the bottom figure's "hypotenuse" is also two segments that bend inwards a little. This accounts for the difference in apparant areas.
A: Note how the Fibonacci sequence provides a nice way of generating more such triangles, since for any four consecutive elements $S_n, S_{n+1}, S_{n+2},S_{n+3}$ we have:
$S_n \cdot S_{n+3} = S_{n+1} \cdot S_{n+2} \pm 1$ (sign alternates for even and odd values of n). 
Then make one of the two triangles on the "not-quite" hypotenuse have dimensions (width x height) $S_{n+2} \times S_n$ and the other $S_{n+3} \times S_{n+1}$. Then in one way of arranging, the rectangular vacancy is $S_{n+2} \times S_{n+1}$ and in the other it is $S_{n+3} \times S_n$, so there is exactly one square missing from the yellow/green rectangle in one of the two arrangements.
For example, to get a "one size smaller" arrangement, make the blue triangle $5 \times 2$ and the red triangle $3 \times 1$, to leave a $5 \times 1$ rectangle in the top half of the original diagram and a $3 \times 2$ rectangle in the bottom half of the diagram. Then use the middle row from the yellow and green pieces in the top half of the diagram, i.e. $2 \times 1$ and $3 \times 1$. 
