5. As shown in Figure 1, the vertices of $\triangle O A B$ are $O(0,0)$, $A(2,1)$, and $B(10,1)$. The line $C D \perp x$-axis and bisects the area of $\triangle O A B$. If the coordinates of point $D$ are $(x, 0)$, then the value of $x$ is . $\qquad$
$5.10-2 \sqrt{10}$.
Let $CD$ intersect $AB$ at point $E$, and intersect $OB$ at point $F$.
Since the equation of line $OB$ is $y=\frac{1}{10} x$, the coordinates of point $F$ are $\left(x, \frac{1}{10} x\right)$ (here $0<x<10$). Then
$$
\begin{array}{l}
EF=1-\frac{1}{10} x=\frac{10-x}{10}, \\
EB=10-x, AB=10-2=8 . \\
\text { Therefore, } S_{\triangle B B P}=\frac{1}{2} \times \frac{10-x}{10}(10-x)=\frac{(10-x)^{2}}{20} . \\
\text { Also, } S_{\triangle A A B}=\frac{1}{2} AB \times 1=4, \text { so } \frac{(10-x)^{2}}{20}=2 .
\end{array}
$$