In two previous blog posts I talked about a puzzle made up of five two by two squares, where each square was cut in two along a line from a vertex to the midpoint of a side.

The challenge, which I gave in the first post, was to put the ten pieces together to form one large square. In the second post I gave a few hints to help anyone who might not have succeeded. Jesse, who is our videographer, was behind the camera for the first two videos. He worked at the puzzle without success, so we decided to do a video together in which he would talk briefly about his experience and then together we would solve the puzzle. While we focused on some of the mathematics involved in the puzzle, and it is clearly helpful to solving the puzzle, there is nothing to prevent one from solving the puzzle by simple trial and error. In fact, when I have done this in workshops with teachers, trial and error is generally the first way they solve it, and only afterwards do we do the mathematics.

In the second blog post I posed a second question about these pieces. How many similar right triangles can be made using these ten pieces? Remember that triangles are similar if corresponding sides are proportional. The answer is that six such triangles are possible. The ratios of the legs of these six triangles are 2/1, 4/2, 2√5/√5, 6/3, 8/4, and 4√5/2√5, respectively.

Using all ten of the puzzle pieces, we know we can build a square and we also know we can build a right triangle. I wonder if it’s possible to build a rectangle or parallelogram or trapezoid using all of the pieces.

It turns out that there are many other questions we might ask of these puzzle pieces, but we’ll save that for another time and place.

Awesome! What a fun activity to teach students (and teachers)!

Please solve