6. As shown in Figure 1, point $C$ is a trisection point of the semicircular arc $\overparen{A B}$ with a radius of 1. Two semicircles are constructed outward on the chords $A C$ and $B C$, and points $D$ and $E$ are also trisection points of these two semicircular arcs. Four more semicircles are then constructed outward on the chords $A D$, $D C$, $C E$, and $B E$. The area of the shaded region (4 crescent shapes) in the figure is ( ).
(A) $\frac{\sqrt{3}}{4}$
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{3 \sqrt{3}}{4}$
(D) $\sqrt{3}$
6. B.

It is easy to know that points $D$, $C$, and $E$ are collinear,
\[
\begin{array}{l}
A C=\frac{1}{2} A B=1, B C=A B \cos 30^{\circ}=\sqrt{3}, \\
B E=B C \cos 30^{\circ}=\frac{3}{2}, C E=D C=\frac{\sqrt{3}}{2}, A D=\frac{1}{2},
\end{array}
\]

and quadrilateral $A B E D$ is a right trapezoid, with the outer 4 semicircles having no overlap. Therefore,
\[
\begin{array}{l}
\left.+\frac{1}{4} B E^{2}\right)-S_{\text {¿ }}-\frac{\pi}{2}\left(\frac{1}{4} A C^{2}+\frac{1}{4} B C^{2}\right) \\
=S_{\triangle A D C}+S_{\triangle B C E} \\
=\frac{1}{2} \times \frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times \frac{3}{2} \times \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \text {. } \\
\end{array}
\]