24. A square has its vertices on the edges of a regular hexagon. Two of the edges of the square are parallel to two edges of the hexagon, as shown in the diagram. The sides of the hexagon have length 1 .
What is the length of the sides of the square?
A $\frac{5}{4}$
B $3-\sqrt{3}$
C $\frac{4}{3}$
D $\sqrt{2}$
E $\frac{3}{2}$

SolUTION
B

We label the vertices of the hexagon $P, Q, R, S$, $T$ and $U$, and the vertices of the square $V, W, X$ and $Y$, as shown in the diagram.

Also, we let $K$ and $L$ be the feet of the perpendiculars from $P$ and $U$, respectively, to $V Y$, and we let $M$ be the point where $V W$ meets $Q T$.
We let the length of the sides of the square be $s$.
The internal angles of the regular hexagon are $120^{\circ}$. The figure is symmetrical about the line $Q T$. Therefore $\angle V Q M=60^{\circ}, \angle V M Q=90^{\circ}$ and $V M=\frac{S}{2}$

From the right-angled triangle $V M Q$, we have $\frac{V M}{V Q}=\sin 60^{\circ}=\frac{\sqrt{3}}{2}$. Therefore $V Q=\frac{2}{\sqrt{3}} V M=$ $\frac{2}{\sqrt{3}} \times \frac{s}{2}=\frac{s}{\sqrt{3}}$
$K L U P$ is a rectangle. Therefore $K L=P U=1$. Therefore, $V K+L Y=V Y-K L=s-1$. By symmetry, $V K=L Y$. Hence $V K=\frac{s-1}{2}$.
$\angle P K V=90^{\circ}$ and $\angle P V K=60^{\circ}$. Therefore in the triangle $P V K$ we have $\frac{V K}{P V}=\cos 60^{\circ}=\frac{1}{2}$. Hence $P V=2 V K=s-1$.
Since $P Q=1$, we have $P V+V Q=1$, and hence $s-1+\frac{s}{\sqrt{3}}=1$. Hence $s\left(1+\frac{1}{\sqrt{3}}\right)=2$. That is, $s\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)=2$. Therefore $s=\frac{2 \sqrt{3}}{\sqrt{3}+1}=\frac{2 \sqrt{3}(\sqrt{3}-1))}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{6-2 \sqrt{3}}{2}=3-\sqrt{3}$.