6. A six-pointed star is drawn by constructing six isosceles triangles with a vertex angle of 30 degrees on the sides of a regular hexagon. Knowing that the circumference passing through the points of the star has a radius of 1, calculate the area of the star itself.
(A) $\sqrt{3}$
(B) $\frac{3(\sqrt{3}-1)}{2}$
(E) none of the above.
(C) $\frac{3(\sqrt{3}+1)}{2}$
(D) $\frac{3(\sqrt{3}-2)}{2}$
6. The answer is (B). The six isosceles triangles have two angles of $75^{\circ}$. The angle between two sides of two isosceles triangles sharing a vertex is therefore $360^{\circ}-\left(75^{\circ}+75^{\circ}+120^{\circ}\right)=90^{\circ}$. Referring to the figure on the side, it is observed that the area $S$ required can be obtained by subtracting the area of 6 isosceles right triangles with side $\frac{1}{\sqrt{2}}$ from the area of the regular hexagon (which has a side equal to the radius of the circumscribed circle). Therefore, we have $S=\frac{6 \cdot \frac{\sqrt{3}}{2}}{2}-6 \cdot \frac{\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}}}{2}$, which simplifies to $S=\frac{6 \sqrt{3}}{4}-\frac{6}{4}=\frac{3(\sqrt{3}-1)}{2}$

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