Two Triangles in a Dodecagon

Let $S$ be the area of the regular decagon.

The difference between the two areas is equal to the difference between the area of $△ABC$ and the area of $△ABD$:

$S(BCM) - S(AMD) = S(ABC) - S(ABD)$

Since the hexagon is regular, we have that $△ABD$ and $△ABC$ are right-angle, and also $∠ABD = 15°$ and $∠ABC = ∠CAB = 45°$.

If $AB = 2R$, then $AD = 2\sqrt{2+\sqrt{3}}R, BD = 2\sqrt{2-\sqrt{3}}R, AC = BC = 2\sqrt{2}R$. Therefore, $S(ABC) = 4R^2$ and $S(ABD) = 2R^2$. If $S(ABC) - S(ABD) = 2$, then $2R \times R = 2$, and $R = 1$. Finally, $S = 12S(AOD) = 6S(ABD) = 12$.