# Tiling a square with right-angled triangles

Tile a square with twenty congruent right-angled triangles. For each triangle, one leg is of length 1 and the other leg is of length 2.

Note first that

the area of the square is 20 times the area of the triangle, namely 20, so the side length is $$2\sqrt5$$.

That's

twice the hypotenuse of the triangle

which immediately suggests the overall shape of the thing. With apologies for the horrific ASCII art:

.   .   .   .   .   .   .
..''|:
.   .   ..''.___. : .   .
..''|       |  :
..''.___.___.___.___:   .
:  |       |       |:
. : .___.___.___.___. : .
:|       |       |  :
.   :___.___.___.___.___:   .   .
:  |       |   ..''
.   . : .___.___..''.   .   .   .
:|   ..''
.   .   :.''.   .   .   .   .   .

where

each of those 2x1 rectangles is of course divided into two triangles by one of its diagonals.

Here, have some slightly less horrific not-ASCII not-art:

• I like this answer because it explains the math involved plus it gives two views of a solution in two different ways: ASCII art vs Graphic image AND Tilted vs Non-tilted square. I also appreciate the guide dots and the guidelines. Feb 1 at 3:42
• That ascii art is fantastic! Feb 1 at 12:58
• I upvoted this just for the ASCII art :) Feb 1 at 13:03

May I offer a more aesthetic tiling?

• More aesthetic than ASCII at you mean? :)` Feb 1 at 5:27
• @MOehm Certainly - Gareth added the image about seven minutes after I posted. Even so, I find the mirror symmetry more pleasing to the eye. Feb 1 at 5:48
• Which Transformer is this? Feb 1 at 10:27
• reminds me a bit of a rot13(fgne qrfgeblre) shape... (saw that as the first after unspoilering the solution) Feb 1 at 18:43

Using the solution of this classic puzzle, "If the dotted lines connect the midpoint of each edge with a vertex, find the fraction of the area of the yellow square to the large red one":

namely,

1/5 i.e. the same as each of the 4 red triangles.

Each red triangle and the yellow square can be divided into 4 of the triangles given in the question, resulting in this neat solution: