An irregular hexagon with all sides of equal length is placed inside a square of side length \(1\), as shown below (not to scale). What is the length of one of the hexagon sides?

(a) \(\sqrt{2} - 1,\quad\) (b) \(2 - \sqrt{2},\quad\) (c) \(1,\quad\) (d) \(\dfrac{\sqrt{2}}{2},\quad\) (e) \(2 + \sqrt{2}\).

Let’s call the length of the hexagon sides \(x\). We can now add some lengths to the diagram.

the irregular hexagon inside the unit square with lengths added

The triangles in the corners are isosceles and right-angled so their sides are in the ratio \(1:1:\sqrt{2}\). So we have \[\frac{x}{1-x} = \sqrt{2} \quad\implies\quad x=\frac{\sqrt{2}}{1+\sqrt{2}} = 2-\sqrt{2}\] and the answer is (b).