14. As shown in Figure 8, in the right trapezoid $A B C D$, $\angle A$ is a right angle, $A B / / C D, A B=7$, $C D=5, A D=2$. A moving line $l$ intersects $A B$ at point $P$, and intersects $C D$ at point $Q$, and divides the trapezoid $A B C D$ into two parts of equal area. Then the maximum value of the distance from point $A$ to the moving line $l$ is
14. $\sqrt{10}$.

According to the problem, we have $A P+D Q=6$.
Let the midpoints of $A D$ and $P Q$ be $M$ and $N$ respectively, then $M N$ $=3$. Therefore, $N$ is a fixed point. The distance from point $A$ to the moving line $l$ is maximized when and only when $P Q \perp A N$, and the maximum value is
$$
\sqrt{A M^{2}+M N^{2}}=\sqrt{10} .
$$