23. Two squares are drawn inside a regular hexagon with side-length 2 , as shown. What is the area of the overlap of the two squares?

A 2
B $2-\sqrt{3}$
C $4-\sqrt{3}$
D $4-2 \sqrt{3}$
E $8-4 \sqrt{3}$
23. E The sum of the interior angles of a hexagon is $(6-2) \times 180^{\circ}=720^{\circ}$.
Hence each interior angle of a regular hexagon is $720^{\circ} \div 6=120^{\circ}$.
In the figure, $\sin 60^{\circ}=\frac{P Q / 2}{2}$. So $P Q=4 \sin 60^{\circ}=4 \times \frac{\sqrt{3}}{2}=2 \sqrt{3}$.
Therefore the height of the area of overlap is $2+2-2 \sqrt{3}=4-2 \sqrt{3}$.
The length of the area of overlap is 2 .
Hence the area of overlap of the two squares is $2(4-2 \sqrt{3})=8-4 \sqrt{3}$.