|Pythagoras puzzle [message #6476] Sat, 12 September 2009 16:36 UTC
In the package I received yesterday there was also the Pythagoras puzzle.
It allows to demonstrate the Pythagoras theorem: in a right-angled triangles, the sum of the areas of the two squares on the legs (A and B) equals the area of the square on the hypotenuse (C).
You can draw squares on all the sides of a right-angled triangle, to visualize this.
As A²+B²=C², there exist at least one dissection of the square C² whose pieces can be rearranged to fill the square A² and B².
This puzzle is such a dissection (5 pieces), with a support consisting of 3 square-shaped boxes surrounding empty right-angled triangle.
You can put all the pieces in the two smaller squares:
or into the biggest one:
By the way, the biggest triangle of this dissection has the same shape as the initial righ-angled triangle.
The biggest dimension of the smaller piece is approximately 1cm. Except this one, all the other pieces are hollowed, and I had to clean-up some support material only on the smallest hollowed part.
(Size: 119.74KB, Downloaded 761 time(s))
(Size: 130.38KB, Downloaded 678 time(s))
(Size: 134.13KB, Downloaded 702 time(s))
[Updated on: Sun, 13 September 2009 13:39 UTC]