4. As shown in Figure 5, in $\triangle A B C$, it is known that $\tan \angle B A C=\frac{3}{4}$, $A D \perp B C$, with the foot of the perpendicular being $D$, and $B D$ $=9, C D=3$. Then
$$
S_{\triangle A B C}=
$$
$\qquad$
$4.48+6 \sqrt{91}$.
As shown in Figure 7, let O be
the circumcenter of $\triangle ABC$, and
draw $OE \perp BC$ at point $D$, $OF$
$\perp AD$ at point $F$, and connect $OA, OB, OC$.
Given $\tan \angle BAC = \frac{3}{4}$
$\Rightarrow \cos \angle BAC = \frac{4}{5}, \sin \angle BAC = \frac{3}{5}$.
It is easy to see that $BC = 12, BE = 6$. Thus, $OF = ED = 3$.
By the Law of Sines,
$OA = OB = OC = \frac{BC}{2 \sin \angle BAC} = 10$.
Therefore, $AF = \sqrt{10^2 - 3^2} = \sqrt{91}$.
Also, $DF = OE = OB \cos \angle BAC = 8$, so $AD = AF + DF = 8 + \sqrt{91}$.
Hence, $S_{\triangle ABC} = \frac{1}{2} \times 12(8 + \sqrt{91})$
$$
= 48 + 6 \sqrt{91} \text{. }
$$