Peaceful Day on Annulus Court

It seems plausible that we would walk across the street in a straight line tangent to the inner circle and then safely walk around the loop on the inner side of the road, doing the same thing in reverse at the end. The length of this path will be $2\sqrt{44^2-40^2}+2*40\left(\pi-\arccos(40/44)\right)\approx \boxed{253.6}$ (replace $40$ by $r_i$ and $44$ by $r_o$ to get the general formula.)

It is not hard to see that this is indeed the shortest path. If we write the path in polar coordinates, $r(\theta)$ , then the length of the long arc between the purple lines is $\int (r(\theta)^2 + r'(\theta)^2)d\theta,$ which is minimised when $r(\theta)$ is held constant, at $40$ . Furthermore, the points on the purple lines at $r=40$ are closest to Brilli's initial position.