Review question

# What is the area of a square containing a regular hexagon? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6408

## Solution

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.

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).