a regular hexagon inscribed in a square

The diagram shows a regular hexagon, with sides of length \(1\), inside a square. Two vertices of the hexagon lie on a diagonal of the square and the other four lie on the edges.

What is the area of the square?

(A) \(\quad 2+\sqrt{3}\qquad\) (B) \(\quad 4\qquad\) (C) \(\quad 3+\sqrt{2}\qquad\) (D) \(\quad 1+\dfrac{3\sqrt{3}}{2}\qquad\) (E) \(\quad\dfrac{7}{2}\)

Let’s draw in the diagonal referred to and label the diagram as follows.

a labelled regular hexagon inscribed in a square

We need to find the edge length of the square, which we have labelled as \(x+y\).

Firstly, notice that since the hexagon is regular \(EF\) is parallel to the diagonal \(AC\) and is therefore at \(45^\circ\) to the sides of the square. Triangle \(BEF\) is isosceles right-angled and so \[y=\frac{1}{\sqrt{2}}.\]

In triangle \(EAD\) the angle at \(A\) is \(45^\circ\) and the angle at \(D\) is \(120^\circ\) (because \(\angle EDC=60^\circ\)). So we can use the sine rule to find \(x\). \[\begin{align*} \frac{1}{\sin 45^\circ} &= \frac{x}{\sin 120^\circ} \\ \implies\quad x &= \frac{\sqrt{3}}{\sqrt{2}} \end{align*}\]

So now the area of the square is \[(x+y)^2 = \left(\frac{\sqrt{3}}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)^2 = 2+\sqrt{3}\] and the answer is (A).