Example 9 If three positive real numbers $x, y, z$ satisfy
$$
\begin{array}{l}
x^{2}+x y+y^{2}=\frac{25}{4}, \\
y^{2}+y z+z^{2}=36, \\
z^{2}+z x+x^{2}=\frac{169}{4},
\end{array}
$$

find the value of $x y+y z+z x$.
[7]
It is known that the three equations can be transformed into
$$
\begin{array}{l}
x^{2}+y^{2}-2 x y \cos 120^{\circ}=\left(\frac{5}{2}\right)^{2}, \\
y^{2}+z^{2}-2 y z \cos 120^{\circ}=\left(\frac{12}{2}\right)^{2}, \\
z^{2}+x^{2}-2 z x \cos 120^{\circ}=\left(\frac{13}{2}\right)^{2} .
\end{array}
$$

As shown in Figure 9, construct $\triangle P B C, \triangle P C A, \triangle P A B$, such that $P B=x, P C=y, P A=z$, and
$$
\begin{array}{l}
\angle B P C=\angle C P A \\
=\angle A P B=120^{\circ} .
\end{array}
$$

Then $A B=\frac{13}{2}$,
$$
\begin{array}{l}
B C=\frac{5}{2}, \\
C A=\frac{12}{2} .
\end{array}
$$

Thus, $\triangle P B C, \triangle P C A, \triangle P A B$ can be combined to form a right triangle $\triangle A B C$.
$$
\begin{array}{l}
\text { By } S_{\triangle P B C}+S_{\triangle P C A}+S_{\triangle P A B}=S_{\triangle B B C}, \text { we get } \\
\frac{1}{2}(x y+y z+z x) \sin 120^{\circ}=\frac{1}{2} \times \frac{5}{2} \times \frac{12}{2} .
\end{array}
$$

Therefore, $x y+y z+z x=10 \sqrt{3}$.