Oxford University MAT 2016 Solved Papers

Call the length of one of the sides of the hexagon $\mathit{p}$, then the side of the square is equal to $\mathit{p}+\left(1-\mathit{p}\right)=1$. Then as the hexagon side forms a triangle in each corner of the square, using Pythagoras, ${\mathit{p}}^2={\left(1-\mathit{p}\right)}^2+{\left(1-\mathit{p}\right)}^2$. Solving this quadratic results in $\mathit{p}=2\pm \sqrt{2}$, but as the length must be less than 1 the answer is (b).