## 11. "Wipers"

On the windshield of a car, there are two "wipers" of length $L$ each, rotating around two points that are also $L$ apart. Each "wiper" "sweeps" one semicircle. What area do both wipers sweep?
11. The area swept by the "windshield wipers" is equal to the sum of the areas of two semicircles with radius $L$ minus the area of the "curvilinear triangle" bounded by the straight base $O O^{\prime}$ of length $L$ and two arcs $O A$

![](https://cdn.mathpix.com/cropped/2024_05_21_fe999c0fe2ad81fc5164g-155.jpg?height=346&width=943&top_left_y=918&top_left_x=568)

and $O^{\prime} A$ with radius $L$ and centers at $O$ and $O^{\prime}$. If the area of half of this curvilinear triangle is $S$, then the area of the sector $O A O^{\prime}$ is $(1 / 6) \pi L^{2}$, and the area of the triangle $O A H$ (where $H$ is the midpoint of $O O^{\prime}$) is

$$
\frac{1}{2} \cdot \frac{\sqrt{3}}{2} L \cdot \frac{L}{2}=\frac{\sqrt{3}}{8} L^{2}
$$

Therefore,

$$
S=(\pi / 6-\sqrt{3} / 8) L^{2}
$$

Thus, the desired area is

$$
(2 \pi / 3+\sqrt{3} / 4) L^{2} \approx 2.527 L^{2}
$$