7.97 A flat corridor in the shape of a right angle has a width of 1 meter and extends infinitely in both directions. Given a rigid, unbendable metal wire (not necessarily straight), which can be dragged through the entire corridor, what is the maximum distance between the two endpoints of the wire?
[Solution] Take a quarter of the circumference of a circle with radius $r=2+\sqrt{2}$. The height of the segment formed by this $90^{\circ}$ arc is exactly 1. It is evident that when this arc reaches the corner, it can smoothly pass through the corner by rotating around the center $O$ as shown in the right figure. The distance between the two endpoints of this arc is $d=2+2 \sqrt{2}$, so the maximum value sought is no less than $2+2 \sqrt{2}$.

Assume the distance between the two endpoints of the metal wire $l$ is greater than $2+2 \sqrt{2}$. Draw a perpendicular line $M^{\prime} C^{\prime}$ from the midpoint $M^{\prime}$ of the line segment $A^{\prime} B^{\prime}$ connecting the two endpoints of $l$, intersecting $l$ at $C^{\prime}$. If $l$ can be dragged through the entire corridor, then $C^{\prime} M^{\prime} \leqslant 1$. Consider the case where $C^{\prime}$ lies on the angle bisector when $l$ passes through the corner. Clearly, when $A^{\prime}$ and $B^{\prime}$ are on the outer edges of the corridor, point $C^{\prime}$ is closest to point $P$. At this time, $P M^{\prime}=\frac{1}{2} A^{\prime} B^{\prime}>1+\sqrt{2}$, thus $P C^{\prime}>\sqrt{2}$. This indicates that $l$ cannot pass through the bend.
In summary, the maximum value sought is $2+2 \sqrt{2}$.