131 For a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length $a$, $E$ is the midpoint of $C D$, and $F$ is the midpoint of $A A_{1}$. The area of the section through $E, F, B_{1}$ is
A. $\frac{55 \sqrt{5}}{192} a^{2}$
B. $\frac{\sqrt{29}}{4} a^{2}$
C. $\frac{11 \sqrt{29}}{48} a^{2}$
D. $\frac{7 \sqrt{29}}{8} a^{2}$
131 C. As shown in Figure 3, on the plane $A B B_{1} A_{1}$, $B_{1} F$ intersects $B A$ at $P$. On the plane $A B C D$, $P E$ intersects $A D$ and $B C$ at $G$ and $Q$, respectively. On the plane $B_{1} B C C_{1}$, $B_{1} Q$ intersects $C C_{1}$ at $H$. The section is a pentagon $B_{1} F G E H$. It is easy to know that
$$
\begin{array}{c}
A P=a, D G=\frac{1}{3} a, C Q=D G=\frac{1}{3} a, C H=\frac{1}{4} a . \\
S_{B B_{1} P}=a^{2}, S_{B B_{1} Q}=\frac{1}{2}\left(1+\frac{1}{3}\right) a^{2}=\frac{2}{3} a^{2}, \\
S_{B P Q}=\frac{1}{2} \times 2 \times\left(1+\frac{1}{3}\right) a^{2}=\frac{4}{3} a^{2} . \\
S_{B_{1} P Q}=\sqrt{S_{B B_{1} P^{2}}+S_{B B_{1} Q^{2}}+S_{B P Q}{ }^{2}}=\sqrt{1+\left(\frac{2}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2}} a^{2}=\frac{\sqrt{29}}{3} a^{2} . \\
S_{\text {to }}=S_{B_{1} P Q}-S_{P F G}-S_{B Q H}=\left(1-\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{4}\right)^{2}\right) S_{B_{1} P Q}=\frac{11}{16} \times \frac{\sqrt{29}}{3} a^{2} \\
=\frac{11 \sqrt{29}}{48} a^{2} .
\end{array}
$$