# Cutting a 10-by-2 rectangle

How does one dissect a $10\times2$ rectangle into four pieces that can be reassembled to form a square?

• can the rectangle only be cut into unit sizes? does the square need to be completely filled in? an area of 20 square units can't be evenly divided into a square with whole-number sides. need more info Dec 9 '15 at 15:54
• @dperry: This is a standard dissection problem, without any strange requirements. This site contains many other puzzles of this type, as for instance puzzling.stackexchange.com/questions/22615/the-challenge-square Dec 9 '15 at 16:13
• @dperry: I'm assuming the square needs to be completely filled in, otherwise you could just do four 2 x 2.5 blocks and make a 4.5 x 4.5 square with a hole in the middle.
– orp
Dec 9 '15 at 16:13
• @Hugh >!yourText
– user14478
Dec 9 '15 at 16:18
• @Hugh: take a look at meta.stackexchange.com/questions/72877/… for a nice little guide Dec 9 '15 at 16:26

Index your rectangle from (0,0) to (10,2). Then cut from

(3,0) to (4,2); (4,2) to (8,0); (8,0) to (9,2).

These four pieces can be used to make the square.

Note that this dissection works without any flipping or even any rotation of the pieces! To show that it's a square is also relatively simple. Easy geometry shows that the angles are right angles, and that the sides are $2\sqrt{5}$.

• Any reason why you do not shift all the cut to the left by one unit so that the outer parts are symmetric? =) Dec 10 '15 at 5:21
• That's a good observation, which probably reveals something about my brain. Or at least the way I came to the solution! Dec 10 '15 at 5:43
• If you make the long cut first and slide the pieces appropriately, you can make this in two cuts instead of three! :D Dec 10 '15 at 17:40
• That's true. Also, the initial long cut can be anywhere between a vertical line at 1 and a vertical line at 9. All it does is shift where the line in the central parallelogram of the square goes. Dec 10 '15 at 17:43
• @corsiKa: But it's still four pieces, and the piece count is what the puzzle specifies. Dec 10 '15 at 18:24

Well, I can do it with 4 cuts, 5 pieces

Cut two 4x2 sections (2 cuts). Cut them diagonally (4 total cuts). The diagonals have length $\sqrt{20}$. Assemble them so the diagonals are the sides of the new square, length 4 against length 2. That leaves a 2x2 hole in the middle, for your remaining 2x2 piece.

• So inspired by this answer I tried it myself Dec 9 '15 at 16:40

Big square: Actually 2 - Inside (10x10) and outside (10.5x10.5).

2 cuts (or 3 without stacking) Cut piece in half parallel with the long side, giving 2 10x1 pieces. Stack these pieces together, and cut in the same manner, giving 4 10x0.5 pieces. Arrange these pieces so that the short ends are against then end of the next piece, forming a large square

Picture:

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• I like this answer! I guess it's a square perimeter, but you can't argue that this doesn't answer this question. Dec 9 '15 at 21:46
• You can make it two cuts without stacking by simply cutting along the horizontal middle then the vertical middle to make four 5x1s. Dec 9 '15 at 22:09

Here's a four-piece solution, taken from here (the link only has pictures, but I supplemented with cut locations):

Cut from (0, 2) to (4, 0), from (4, 0) to (4 + $\sqrt{3}$, 2), and from (10 - $\sqrt{3}$, 0) to (10, 2).

Image of the cut-up rectangle:

Then just slide the four pieces into a (rotated) square:

Despite the irrational locations of the cuts in the original rectangle, the vertices of all pieces in the joined square are at integer locations.