28. As shown in Figure 11, $\triangle A B C$ is translated along side $A B$ to the position of $\triangle A^{\prime} B^{\prime} C^{\prime}$. The overlapping part (the shaded area in the figure) is half the area of $\triangle A B C$. If $A B=\sqrt{2}$, then the distance the triangle has moved, $A A^{\prime}$, is ( ).
(A) $\sqrt{2}-1$
(B) $\frac{\sqrt{2}}{2}$
(C) 1
(D) $\frac{1}{2}$
28. A.

Let $A^{\prime} C^{\prime}$ intersect $B C$ at point $O$. Then $\triangle A B C \backsim \triangle A^{\prime} B O$. Also, $\frac{S_{\triangle A^{\prime} B O}}{S_{\triangle A B C}}=\frac{1}{2}$, so $\frac{A^{\prime} B}{A B}=\frac{1}{\sqrt{2}} \Rightarrow A^{\prime} B=1$.

Thus, $A A^{\prime}=\sqrt{2}-1$.