II. (15 points) As shown in Figure 2, in $\triangle ABC$, it is known that $\angle ACB=90^{\circ}, AC=3, BC=4$. With point $B$ as the center, $\triangle ABC$ is rotated clockwise so that point $A$ lands on point $A_{1}$ on the extension of $CB$, and at this time, point $C$ lands on point $C_{1}$. Connect $AA_{1}$ and $CC_{1}$ intersecting at point $O$, $CC_{1}$ intersects $AB$ at point $D$, and $AA_{1}$ intersects $BC_{1}$ at point $E$. Find the area of quadrilateral $BDOE$.
It is known that, $\angle A B A_{1}=\angle C B C_{1}$.
Therefore, $\triangle A B A_{1} \backsim \triangle C B C_{1}$, and
$\angle O A B=\angle O C B$.
Thus, points $B, C, A, O$ are concyclic.
Connecting $B O$. Then $\angle A O B=90^{\circ}, O$ is the midpoint of $A B$, and points $B, O, C_{1}, A_{1}$ are also concyclic.
By the Pythagorean theorem,
$A A_{1}=\sqrt{A C^{2}+C A_{1}^{2}}=\sqrt{3^{2}+9^{2}}=3 \sqrt{10}$.
Therefore, $A O=\frac{3 \sqrt{10}}{2}$.
It is known that, $\triangle C D B \backsim \triangle A_{1} B E \sim \triangle A D O$.
Let $S_{\triangle C D B}=x$. Then
$S_{\triangle M_{1} E B}=\frac{25}{16} x, S_{\triangle \triangle D O}=\frac{45}{32} x$.
Let $S_{\text {quadrilateralRDOE }}=y$. Thus,
$S_{\triangle 1 C A_{1}}=\frac{1}{2} \times 3(4+5)=\frac{27}{2}$.
Since $C O$ bisects the area of $\triangle A C A_{1}$, we have
$S_{\triangle C O A_{1}}=\frac{1}{2} S_{\triangle \triangle A_{1}}=\frac{27}{4}$.

And $S_{\triangle \triangle B N_{1}}=\frac{1}{2} \times 5 \times 3=\frac{15}{2}$, then we get the system of equations
$$
\left\{\begin{array} { l } 
{ ( 1 + \frac { 2 5 } { 1 6 } ) x + y = \frac { 2 7 } { 4 } , } \\
{ ( \frac { 2 5 } { 1 6 } + \frac { 4 5 } { 3 2 } ) x + y = \frac { 1 5 } { 2 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
x=\frac{24}{13} \\
y=\frac{105}{52}
\end{array}\right.\right.
$$

Therefore, the area of quadrilateral $B D O E$ is $\frac{105}{52}$.