12. A. As shown in Figure 5, in rectangle $A B C D$, $A B=a$, $B C=b$, and $a<b$. A line through the center $O$ of rectangle $A B C D$ intersects segments $B C$ and $D A$ at points $E$ and $F$, respectively. The quadrilateral $E C D F$ is folded along $E F$ to the plane of quadrilateral $B E F A$, such that point $C$ coincides with point $A$, resulting in quadrilateral $E F G A$.
(1) Prove that the area of pentagon $A B E F G$ is $\frac{a\left(3 b^{2}-a^{2}\right)}{4 b}$;
(2) If $a=1$ and $b$ is a positive integer, find the minimum value of the area of pentagon $A B E F G$.
12. A. (1) It is easy to know that quadrilateral $A E C F$ is a rhombus. Since $\angle A O F=\angle A D C=90^{\circ}$, we have $\triangle A O F \backsim \triangle A D C \Rightarrow \frac{A F}{A C}=\frac{A O}{A D}$. Given $A O=\frac{A C}{2}=\frac{\sqrt{a^{2}+b^{2}}}{2}, A D=B C=b$, we know $A F=\frac{A C \cdot A O}{A D}=\frac{a^{2}+b^{2}}{2 b}$,
$$
F D=A D-A F=b-\frac{a^{2}+b^{2}}{2 b}=\frac{b^{2}-a^{2}}{2 b} \text {. }
$$

By symmetry, $\triangle A F G \cong \triangle C F D$.
Thus, $S_{\triangle A F G}=S_{\triangle C F D}=\frac{1}{2} C D \cdot F D$
$$
=\frac{a}{2} \cdot \frac{b^{2}-a^{2}}{2 b}=\frac{a\left(b^{2}-a^{2}\right)}{4 b} \text {. }
$$

Also, $S_{\text {quadrilateral } A B E F}=\frac{1}{2} S_{\text {quadrilateral } A B C D}=\frac{a b}{2}$, so
$S_{\text {pentagon } A B E F G}=S_{\text {quadrilateral } A B E F}+S_{\triangle A F G}$
$$
=\frac{a b}{2}+\frac{a\left(b^{2}-a^{2}\right)}{4 b}=\frac{a\left(3 b^{2}-a^{2}\right)}{4 b} \text {. }
$$
(2) Given $a=1$, and $b$ is a positive integer greater than 1, we know $b \geqslant 2$, and $S_{\text {pentagon } A B E F G}=\frac{3 b^{2}-1}{4 b}=\frac{3 b}{4}-\frac{1}{4 b}$.
When $b=2$, $S_{\text {pentagon } A B E F G}$ reaches its minimum value of $\frac{3 \times 2}{4}-\frac{1}{4 \times 2}=\frac{11}{8}$.