4. Let positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=49 \\
y^{2}+y z+z^{2}=36 \\
z^{2}+z x+x^{2}=25
\end{array}\right.
$$

Then $x+y+z=$ $\qquad$
4. $\sqrt{55+36 \sqrt{2}}$.
(1) + (2) + (3) gives
$$
2(x+y+z)^{2}-3(x y+y z+z x)=110 \text {. }
$$

Also, (1) - (2) gives
$$
(x-z)(x+y+z)=13 \text {, }
$$
(2) - (3) gives
$$
(y-x)(x+y+z)=11 \text {, }
$$
(1) - (3) gives
$$
(y-z)(x+y+z)=24 \text {. }
$$

From $\left.\frac{1}{2}(\text { (5) })^{2}+(6)^{2}+(7)^{2}\right)$ we get
$$
\begin{array}{l}
(x+y+z)^{2}\left[(x+y+z)^{2}-3(x y+y z+z x)\right] \\
=433 .
\end{array}
$$

From equation (4) we have
$$
(x+y+z)^{4}-110(x+y+z)^{2}+433=0 \text {. }
$$

Therefore, $(x+y+z)^{2}=55 \pm 36 \sqrt{2}$.
Let $t=x+y+z$.
From equations (5) and (7) we get
$$
x=z+\frac{13}{t}, y=z+\frac{24}{t} \text {. }
$$

Thus, $t=3 z+\frac{37}{t}$.
Since $z>0$, we know $t>\frac{37}{t}$, i.e., $t^{2}>37$.
Hence, $t=\sqrt{55+36 \sqrt{2}}$.