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

Construct a right triangle $\triangle A B C$
As shown in Figure 2, where
$$
A B=\frac{13}{2}, B C=\frac{5}{2} \text {, }
$$
$C A=\frac{12}{2}$. Let $P$ be
a point inside $\triangle A B C$ such that
$$
\begin{array}{l}
P B=x, P C=y, P A=z, \\
\angle B P C=\angle C P A=\angle A P B=120^{\circ} .
\end{array}
$$

Since $S_{\triangle B P C}+S_{\triangle C P A}+S_{\triangle A P B}=S_{\triangle A B C}$, then
$$
\frac{1}{2}(x y+y z+z x) \sin 120^{\circ}=\frac{1}{2} \times \frac{5}{2} \times \frac{12}{2} \text {. }
$$

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