# How do I cut a square in half?

I have a square that's $10\mathrm{m} \times 10\mathrm{m}$. I want to cut it in half so that I have a square with half the area. But if I cut it from top to bottom or left to right, I don't get a square, I get a rectangle!

I know the area of the small square is supposed to be $50\mathrm{m}^{2}$, so I can use my calculator to find out how long a side should be: it's $7.07106781\mathrm{m}$. But my teacher said I should be able to do this without a calculator. How am I supposed to get that number by hand?

• Actually, the exact side length is the square root of 50, which is irrational. Jul 20, 2010 at 21:40
• cut trought both diagonals! Feb 10, 2017 at 16:16

Does this give you any ideas?

• Wow, looking at the picture, I see that the square is evenly split into the black and white areas. I wasn't convinced at first, but then I printed out the picture and saw that the four black triangles can be 'folded' over to cover the small white square perfectly. Jul 20, 2010 at 23:23
• yup, each of the quadrants of the outer square is split in half. Jul 20, 2010 at 23:24
• ...of course, if you're working in wood and the direction of the grain is important, this method would mess that up. Jul 20, 2010 at 23:25
• This is exactly the same figure that Plato used in his dialogue Meno (search for the first instance of "Boy", and read from there) to answer the very same question as the OP asked here. Oct 17, 2012 at 11:10

Take a pair of compasses and draw an arc between two opposite corners, centred at another corner; then draw a diagonal that bisects the arc. If you now draw two lines from the point of intersection, parallel to the sides of the square, the biggest of the resulting squares will have half the area of the original square.

Here's an illustration:

• Really? Why does that work? Jul 20, 2010 at 21:47
• It works because the diagonal of the inner square is equal to the side of the outer square. The ratio of the diagonal to the side of any square is root 2, so the ratio of the diagonals of the two squares is root 2, so the ratio of the areas is 2. Jul 20, 2010 at 22:11
• farm5.static.flickr.com/4082/4813660948_8e8caa2ec9.jpg should be a picture to match this answer. Because C is on the circle with center at A and passing through the vertices adjacent to A, the length of AC is equal to the length of a side of the square. &ang;CAD has measure 45 degrees, so AD and DC both have length 1/sqrt(2) times the length of a side of the original square, so the shaded square has half the area of the original square. @Will, feel free to edit my image into your answer. Jul 21, 2010 at 0:03
• While the original post mentions only needing one square of half the size, I'm left wondering if there's a nice dissection of the remaining 'L' into some number of constructible pieces that can be reassembled into a second square... Nov 26, 2011 at 23:09
• @Steven, indeed there is. Cut along the diagonal through C and its perpendiculars on both sides, giving two trapezoids and two isosceles right triangles. Join the two trapezoids like a miter joint, and fill in the square hole using the triangles. (I'd post a picture, but I don't have any drawing software installed on this machine...)
– user856
Nov 27, 2011 at 1:07

One more approach:

Consider a square of side length a. Label it A,B,C and D, counterclockwise starting from the left lower corner (A).

Draw the diagonals intersecting at M, at right angles, bisecting each other.

Pythagoras: $a^2 +a^2 = 2a^2$.

Length of diagonal = $√2 a$.

Length AM = length BM = $(1/2) √2 a$.

As mentioned before the diagonals bisect each other, and intersect perpendicularly.

These are 2 sides of the new square with half the area of the original square.

Area of reduced square : Length AM × length BM = $(1/2) a^2$.

To complete your reduced square draw a parallel to AM through B, and a parallel to BM through A.