7. A circle with a radius of 1 has six points, these six points divide the circle into six equal parts. Take three of these points as vertices to form a triangle. If the triangle is neither equilateral nor isosceles, then the area of this triangle is ( ).
(A) $\frac{\sqrt{3}}{3}$
(B) $\frac{\sqrt{3}}{2}$
(C) 1
(D) $\sqrt{2}$
(E) 2
7. B.

Consider three cases:
(1) If two vertices of the triangle are adjacent, let's assume they are $A_{1}$ and $A_{2}$ (as shown in Figure 2), then the third vertex can only be $A_{4}$ or $A_{5}$. Therefore, the triangle is a right triangle.
(2) If two vertices of the triangle are separated, let's assume they are $A_{1}$ and $A_{3}$ (as shown in Figure 3), then the third vertex can only be $A_{4}$ or $A_{6}$. Therefore, the triangle is a right triangle.
(3) If two vertices of the triangle are opposite, let's assume they are $A_{1}$ and $A_{4}$ (as shown in Figure 4), then the third vertex can be $A_{2}$, $A_{3}$, $A_{5}$, or $A_{6}$. Therefore, the triangle is a right triangle.
In summary, $S=\frac{1}{2} \times 1 \times \sqrt{3}=\frac{\sqrt{3}}{2}$.