5. Given $\triangle A B C$ with three sides $A B=\sqrt{34}, B C$ $=5 \sqrt{10}, C A=2 \sqrt{26}$. Then the area of $\triangle A B C$ is $\qquad$
5.10.

As shown in Figure 4, with $BC$ as the hypotenuse, construct a right triangle $\triangle BCD$ on one side of $\triangle ABC$ such that $\angle BDC = 90^\circ$, $BD = 5$, and $CD = 15$.

Then construct a rectangle $DEA'F$ such that $DE = 2$ and $DF = 5$. Thus, $BE = 3$ and $CF = 10$. At this point,
$$
\begin{array}{l}
A'C = \sqrt{104} = AC, \\
A'B = \sqrt{34} = AB.
\end{array}
$$

Therefore, point $A$ coincides with point $A'$. 
Hence,
$$
\begin{array}{l}
S_{\triangle ABC} = S_{\triangle BCD} - S_{\triangle ABE} - \\
\quad S_{\triangle ACF} - S_{\text{rect } EAF} \\
= \frac{1}{2} \times 5 \times 15 - \frac{1}{2} \times 3 \times 5 - \frac{1}{2} \times 2 \times 10 - 2 \times 5 \\
= 10.
\end{array}
$$