Given a square $A B C D$ with side length $a$ and a segment $K L$ of length $5 a$ such that $A \equiv K$ and side $A B$ lies on segment $K L$. The square $A B C D$ moves along segment $K L$ by "rolling" (rotating around the lower right corner of the square by $90^{\circ}$) until side $A B$ again coincides with a part of segment $K L (B \equiv L)$.

a) Sketch how the point $S$, which is the midpoint of side $A B$, will move.

b) Determine the length of the curve traced by point $S$ during its motion.
a) At the beginning of the motion, we marked the position of the square as $A_{0} B_{0} C_{0} D_{0}$. Then we gradually marked the vertices with primes, with its final position being marked as $A B C D$. The path of point $S$ is marked in bold on Fig. 1.

![](https://cdn.mathpix.com/cropped/2024_04_17_a2239cbea6160263a4b2g-2.jpg?height=426&width=1328&top_left_y=815&top_left_x=387)

b) Length of the curve: To determine the length of the curve, it is necessary to realize that it consists of four quarter-circles. There are only two different radii of the quarter-circles: the radius $B_{0} S_{0}=\frac{1}{2} a$ and the radius $\left|S_{1} C^{\prime}\right|=\frac{1}{2} \sqrt{5} a$. Therefore, the length of the curve is

$$
l=\frac{1}{4} 2 \pi \frac{a}{2}+\frac{1}{4} 2 \pi \sqrt{5} \frac{a}{2}+\frac{1}{4} 2 \pi \sqrt{5} \frac{a}{2}+\frac{1}{4} 2 \pi \frac{a}{2}=\frac{\pi a}{2}(1+\sqrt{5}) .
$$

[Drawn path ... 2 points

radii ... 2 points

$\frac{1}{2} \pi a(1+\sqrt{5}) \ldots 2$ points]