9. In $\triangle A B C$, it is known that $\tan A=\frac{5}{12}$, and the foot of the perpendicular from $A$ to $B C$ divides the segment $B C$ into two parts, with lengths 3 and 17, respectively. Try to find the perimeter of $\triangle A B C$.
$$
9.20+\sqrt{1212+48 \sqrt{627}}+\sqrt{1492+48 \sqrt{627}} .
$$

As shown in Figure 4, let $O$ be the circumcenter of $\triangle ABC$. Draw $AK \perp BC, OH \perp BC$, with the feet of the perpendiculars being $K$ and $H$ respectively. Then draw $OM \perp AK$, with $M$ being the foot of the perpendicular, and connect $OA, OB$, and $OC$.
$$
\begin{array}{l}
\because \tan A=\frac{5}{12}, \\
\therefore \cos A=\frac{12}{13}, \sin A=\frac{5}{13} .
\end{array}
$$

It is easy to know that $MO=KH=HB-KB=7$,
$$
OA=OB=OC=\frac{BC}{2 \sin A}=26 \text{. }
$$

By the Pythagorean theorem, we get $AM=\sqrt{627}$.
$$
\begin{array}{l}
\text{Also, } \because MK=OH=OB \cos A=24, \\
\therefore AK=AM+MK=24+\sqrt{627} .
\end{array}
$$

By the Pythagorean theorem, we get
$$
AB=\sqrt{1212+48 \sqrt{627}} \text{. }
$$

Similarly, we get
$$
AC=\sqrt{1492+48 \sqrt{627}} \text{. }
$$

Therefore, the perimeter of $\triangle ABC$ is
$$
20+\sqrt{1212+48 \sqrt{627}}+\sqrt{1492+48 \sqrt{627}} \text{. }
$$