12. A. As shown in Figure 5, in rectangle \(ABCD\), \(AB = a\), \(BC = b\), and \(a < b\). A line through the center \(O\) of rectangle \(ABCD\) intersects segments \(BC\) and \(DA\) at points \(E\) and \(F\), respectively. The quadrilateral \(ECDF\) is folded along \(EF\) to the plane of quadrilateral \(BEFA\), such that point \(C\) coincides with point \(A\), resulting in quadrilateral \(EFGA\).
Figure 5
(1) Prove that the area of pentagon \(ABEFG\) is \(\frac{a(3b^2 - a^2)}{4b}\);
(2) If \(a = 1\) and \(b\) is a positive integer, find the minimum value of the area of pentagon \(ABEFG\).
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, $\triangle A F G \cong \triangle C F D$.
$$
\begin{array}{l}
\text { Then } S_{\triangle A F G}=S_{\triangle C F D}=\frac{1}{2} C D \cdot F D \\
=\frac{a}{2} \cdot \frac{b^{2}-a^{2}}{2 b}=\frac{a\left(b^{2}-a^{2}\right)}{4 b} . \\
\text { Also, } S_{\text {quadrilateral } A B E F}=\frac{1}{2} S_{\text {quadrilateral } A B C D}=\frac{a b}{2}, \text { so } \\
S_{\text {pentagon } A B E F}=S_{\text {quadrilateral } A B E F}+S_{\triangle A F G} \\
=\frac{a b}{2}+\frac{a\left(b^{2}-a^{2}\right)}{4 b}=\frac{a\left(3 b^{2}-a^{2}\right)}{4 b} .
\end{array}
$$
(2) Given $a=1$, and $b$ is a positive integer greater than 1, we know $b \geqslant 2$, and
$$
S_{\text {pentagon } A B E F G}=\frac{3 b^{2}-1}{4 b}=\frac{3 b}{4}-\frac{1}{4 b} \text {. }
$$

When $b=2$, $S_{\text {pentagon } A B E F G}$ reaches its minimum value of
$$
\frac{3 \times 2}{4}-\frac{1}{4 \times 2}=\frac{11}{8} \text {. }
$$