Dividing a square
Divide a 5 in by 5 in square into five parts of equal area with four straight segments that all go through one of the square’s vertices. Show that the five areas are equal!
Here is the solution:
We give three different decompositions of the square, and show in each case that the solution above is correct.
(a) One proof showing 4 triangles and one quadrilateral (“tie shape”):
Each of the four triangles has base 2 inches and height 5 inches, and thus area 5 square inches. The whole square has area 25 square inches, so the “tie shape” has area 25 - 20 = 5 square inches. We could also bisect the “tie” into two triangles, each with base 1 inch and height 5 inches, and thus area 2.5 square inches, as shown below!
(b) Another proof showing a decomposition into six triangles:
The four “outside” triangles each have area 5 square inches as above, and the two that make up the “tie” shape together have area 5 square inches. The smaller part of the tie, the small isosceles right triangle in the corner, has legs 1 inch and thus area 1/2 square inch. And the slender isosceles triangle above it has base √2 and height
√ (52 + 52) - √ (2)/2
So its area is 1/2 * √2 * (√50 - √ (2)/2 ) = √ (100)/2 - 1/2 = 4.5 square inches.
So once again we see that the solution is correct.
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(c) A third decomposition, this time showing six triangles and a square:
The “tie shape” now consists of a square with area 1 square inch, and two triangles. Each triangle has base 1 inch and height 4 inches, for an area of 2 square inches. So once again we have shown that the area of the tie shape is 5 square inches.
As mentioned above, there is also a simple proof by subtraction, after you know that four of the triangles have area five square inches: The whole square has area 25 square inches, so the middle shape has area 25-20=5 square inches also.