Here is the shape. The first thing we should do is label what we already know. We knew the side lengths of the paper, so let’s start with those. One long side is still intact and one short side is folded over, but still whole. Now we can start figuring out other measurements. The big triangle came from a right angle split evenly in half, so it is a 45° right triangle. We know one side of the triangle is length 1, so the other must be as well. We can also use Pythagorean Theorem to find the hypotenuse. hypotenuse = √( (1)² + (1)² ) = √(2). The hypotenuse is √(2), the same as the long side. We can also find the short side of the rectangle that is formed. It is √(2) - 1. We will label it as that. The rectangle is has the same length on the other short side, which is a side on the smaller triangle. This triangle is also a 45° right triangle, so the other side of the triangle is the same length. We can use Pythagorean Theorem to find this hypotenuse as well. the hypotenuse = √( (√(2) - 1)² + (² - 1)² ) = 2 - √(2). This triangle has a hypotenuse of 2 - √(2). The last side we need to find can be found by subtracting √(2) - 1 from 1. The side has a length of 2 - √(2), the same length as the shorter hypotenuse. Now that we have solved all of the sides, we can simply add them up. Some canceling out gives us 4 as our answer. What a nice and simple answer! We have solved the puzzle, but there is a different approach worth looking into. Instead of solving it as two triangles and a rectangle, lets look at the whole shape. We know that the bottom right corner is 90°. We can easily figure that the bottom left angle is 45° because of how it was folded. We also know that folding the two triangles over made two 45° angles with the imaginary edge of the paper. That makes the top angle is 90° The last angle must be 135° because it is a parallelogram. These two equivalent angles define this quadrilateral as a kite. Kites always have two pairs of equivalent sides. If we do some quick math to solve the sides of the kite, We can easily find the perimeter. When we add up the sides, it (again) comes out to the simple answer of 4.

Note: I did not invent the a4 puzzle! I don’t know who did, but good job to whoever did, because it is a good puzzle.

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