2. In a non-isosceles triangle, one interior angle is equal to the difference of the other two interior angles, and one interior angle is twice another interior angle. If the shortest side of the triangle is 1, then the area of the triangle is ( ).
(A) 1
(B) $\frac{\sqrt{3}}{2}$
(C) $\frac{\sqrt{5}}{2}$
(D) 2
2. B.

Let the interior angles of the triangle be $\angle A, \angle B, \angle C$, and $\angle B=\angle C-\angle A$. Then
$$
\begin{array}{l}
\angle A+(\angle C-\angle A)+\angle C=180^{\circ} \\
\Rightarrow \angle C=90^{\circ} .
\end{array}
$$

If the largest angle $\angle C$ is twice another interior angle, it is easy to see that the triangle is an isosceles right triangle, which does not match the given conditions.
Assume $\angle A=2 \angle B$. Then
$$
\angle A=60^{\circ}, \angle B=30^{\circ} \text {. }
$$

Thus, $A C=1, B C=\sqrt{3}$.
Therefore, the area of the triangle is
$$
\frac{1}{2} \times 1 \times \sqrt{3}=\frac{\sqrt{3}}{2} \text {. }
$$