I1.2 Let $x, y, z, w$ be four consecutive vertices of a regular $A$-gon. If the length of the line segment $x y$ is 2 and the area of the quadrilateral $x y z w$ is $a+\sqrt{b}$, find the value of $B=2^{a} \cdot 3^{b}$.
Let $O$ be the centre of the regular dodecagon.
Let $O x=r=O y=O z=O w$
$\angle x O y=\angle y O z=\angle z O w=\frac{360^{\circ}}{12}=30^{\circ}$ ( $\angle \mathrm{s}$ at a point $)$
In $\Delta x O y, r^{2}+r^{2}-2 r^{2} \cos 30^{\circ}=2^{2}$ (cosine rule)
$(2-\sqrt{3}) r^{2}=4 \Rightarrow r^{2}=\frac{4}{2-\sqrt{3}}=4(2+\sqrt{3})$
Area of $x y z w=$ area of $O x y z w-$ area of $\Delta O x w$
$=3 \times \frac{1}{2} \cdot r^{2} \sin 30^{\circ}-\frac{1}{2} r^{2} \sin 90^{\circ}=\left(\frac{3}{4}-\frac{1}{2}\right) \cdot 4(2+\sqrt{3})=2+\sqrt{3}$
$a=2, b=3, B=2^{2} \cdot 3^{3}=4 \times 27=108$