We borrowed the book Fortune Wookiee from the library. It is part of the Origami Yoda series and is written by Tom Angleberger.

The seven year old can read to himself so I haven’t read the book and I can’t vouch for what is inside it, but I was delighted when he came to me with a math problem. At the back of the book is instructions on how to make two origami figures:  Han Foldo and Fortune Wookiee. After making both he didn’t like their relative sizes and wanted to figure out what sizes of paper would provide better matching characters.

The Fortune Wookiee is a basic “cootie catcher” or “fortune teller.” You’ve probably made one before. Fold all the corners in till they touch the middle and you’ve created a second square. Then flip it over and repeat.

If you count the length of one side of your square as one unit in length, then the diagonal of your square will be the square root of two. But because you’ve folded your square over a couple of times the diagonal of each of those front panels of the wookie is going to be 1/4 the square root of two and the diagonal of the wookie itself would be 1/2 the square root of two. When my original sheet was 8.5 centimeters, then the wookie had a diagonal of 6.

Working backwards, you can choose what size you want the wookie to be, and divide that number by 0.707 (which is for the purpose of folding paper an okay approximation of 1/2 of the square root of two). That tells you what length you need the edges of your sheet of paper to be!

If you enjoyed this post you might enjoy my post on Origami fish and the Pythagorean theorem.