Four. (15 points) The base of the pyramid $S-ABCD$ is a rectangle $ABCD$ with center $O$, where $AB=4$, $AD=12$, $SA=3$, $SB=5$, $SO=7$. A section of the pyramid is made through vertex $S$, the center $O$ of the base, and a point $N$ on edge $BC$. For what value of $BN$ is the area of the resulting section $\triangle SMN$ minimized? What is the minimum area of this section $\triangle SMN$?
$$
\begin{array}{l}
S . A^{2}+12^{2}=S B^{2}, S A^{2}+A 0^{2}=S C^{2}, \\
\angle S A B=90^{\circ}, \angle S A O-90^{\circ} .
\end{array}
$$
$\therefore S A \perp$ plane $A B C D . C B \perp A B, D A \perp A B \Rightarrow C B \perp$ plane $S A B, D A \perp$ plane $S A B, \angle N B S=90^{\circ}, \angle S A M=90^{\circ}$.

Let $B N=x$. By the central symmetry of the base rectangle about $O$,
$$
\begin{array}{l}
A M=C N=B C-B N=12-x, \\
S N=\sqrt{S B^{2}+B N^{2}}=\sqrt{25+x^{2}}, \\
S M=\sqrt{S A^{2}+A M^{2}}=\sqrt{x^{2}-24 x+153}, \\
\cos \angle D B C=\frac{B C}{B D}=\frac{12}{4 \sqrt{10}}=\frac{3 \sqrt{10}}{10}, \\
O N=\sqrt{O B^{2}+B N^{2}-2 O B \cdot B N \cos \angle D B C} \\
=\sqrt{40+x^{2}-2 \cdot 2 \sqrt{10} \cdot x \cdot \frac{3 \sqrt{10}}{10}} \\
=\sqrt{40+x^{2}-12 x}, \\
\therefore M N=2 O N=2 \sqrt{x^{2}-12 x+40}. \\
\therefore \cos \angle M S N=\frac{S M^{2}+S N^{2}-M N^{2}}{2 M S \cdot S N} \\
=\frac{-x^{2}+12 x+9}{\sqrt{x^{2}-24 x+153} \cdot \sqrt{25+x^{2}}}, \\
\sin \angle M S N=\sqrt{1-\cos ^{2} \angle M S N} \\
=\sqrt{1-\frac{\left(-x^{2}+12 x+9\right)^{2}}{\left(x^{2}-24 x+153\right)\left(25+x^{2}\right)}}. \\
\therefore S_{\triangle S M N}=\frac{1}{2} \cdot S M \cdot S N \cdot \sin \angle M S N \\
=\frac{1}{2} \sqrt{x^{2}-24 x+153} \cdot \sqrt{25+x^{2}} \\
- \sqrt{\frac{\left(x^{2}-24 x+153\right)\left(25+x^{2}\right)-\left(-x^{2}+12 x+9\right)^{2}}{\left(x^{2}-24 x+153\right)\left(25+x^{2}\right)}} \\
=\frac{\sqrt{\left(x^{2}-24 x+153\right)\left(25+x^{2}\right)-\left(-x^{2}+12 x+9\right)^{2}}}{2} \\
=\frac{1}{2} \sqrt{52 x^{2}-816 x+3744} \\
=\sqrt{13 x^{2}-204 x+936} \\
=\sqrt{13\left(x-\frac{102}{13}\right)^{2}+\frac{1764}{13}}. \\
\end{array}
$$
$\therefore$ When $x=\frac{102}{13}$, i.e., $B N=\frac{102}{13}$, the area of the section $\triangle S M N$ reaches its minimum value, $\sqrt{\frac{1764}{13}}=\frac{42 \sqrt{13}}{13}$.