9. Given $\triangle A B C$ with sides $A B=\sqrt{29} \mathrm{~cm}, B C=\sqrt{13} \mathrm{~cm}, C A=\sqrt{34} \mathrm{~cm}$, find the area of $\triangle A B C$.
$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} a b \sin C=\frac{1}{2} a b \sqrt{1-\cos ^{2} C}=\frac{1}{2} a b \sqrt{1-\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}} \\
=\frac{1}{2} \sqrt{a^{2} b^{2}-\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{4}}=\frac{1}{4} \sqrt{4 a^{2} b^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2}} \\
=\frac{1}{4} \sqrt{4 \cdot 13 \cdot 34-(13+34-29)^{2}}=\frac{1}{2} \sqrt{361}=\frac{19}{2} .
\end{array}
$$

The translation is as follows:

$$
\begin{array}{l}
S_{\triangle A B C}=\frac{1}{2} a b \sin C=\frac{1}{2} a b \sqrt{1-\cos ^{2} C}=\frac{1}{2} a b \sqrt{1-\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)^{2}} \\
=\frac{1}{2} \sqrt{a^{2} b^{2}-\frac{\left(a^{2}+b^{2}-c^{2}\right)^{2}}{4}}=\frac{1}{4} \sqrt{4 a^{2} b^{2}-\left(a^{2}+b^{2}-c^{2}\right)^{2}} \\
=\frac{1}{4} \sqrt{4 \cdot 13 \cdot 34-(13+34-29)^{2}}=\frac{1}{2} \sqrt{361}=\frac{19}{2} .
\end{array}
$$