6. To improve living conditions for urban residents, a renovation program is being implemented in some cities - replacing housing stock that cannot be preserved by demolishing it and constructing new buildings on the freed-up land. Typically, the dismantling of old buildings and clearing the site for new construction occurs on a limited area. Therefore, equipment (rotary excavators) is needed that can clear the largest area with the least number of

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-15.jpg?height=320&width=389&top_left_y=511&top_left_x=251)
movements.

In the diagram, the working area and the movement scheme of the excavator are shown. Starting from point $E$, the excavator first clears the nearest area to itself (mainly containing metal scrap). This area is a sector of a circle bounded by the large arc $F G$ of a circle centered at point $E$ with a radius of $2 \sqrt{2}$, and the radii segments $E F$ and $E G$. Then the excavator moves to point $O$ and clears the area farthest from itself (mainly containing brick scrap). This area is bounded externally by the large arc $C D$ of a circle centered at point $O$ with a radius of 6 meters, the segments $C G$ and $F D$ (points $F$ and $G$ are on the corresponding radii $O C$ and $O D$), and internally by the large arc $F G$ of the first circle centered at point $E$.

Determine the area under the brick scrap (excluding metal residues) if point $O$ is 2 meters from point $E$ in the direction of the center of the metal debris, and the angle $E O D$ is $45^{\circ}$. (20 points)
Solution. By the Law of Sines for triangle $O E F$, we have

$$
\begin{aligned}
& \frac{O E}{\sin \angle O F E}=\frac{E F}{\sin \angle E O F^{\prime}}, \frac{2}{\sin \angle O F E}=\frac{2 \sqrt{2}}{\sin 45^{\circ}}, \sin \angle O F E=\frac{1}{2^{\prime}} \\
& \angle O F E=30^{\circ}, \angle O E F=105^{\circ} \\
& \frac{O E}{\sin 30^{\circ}}=\frac{O F}{\sin 75^{\circ}}, O F=4 \sqrt{2} \cdot \frac{\sqrt{2}}{2}\left(\frac{1+\sqrt{3}}{2}\right)=\sqrt{2}(1+\sqrt{3})
\end{aligned}
$$

Area of sector $COD$: $S_{1}=\frac{\pi \cdot 6^{2}}{360} \cdot 270=27 \pi$.

![](https://cdn.mathpix.com/cropped/2024_05_06_b3b045c4417b92ef0023g-15.jpg?height=531&width=682&top_left_y=1359&top_left_x=1235)

Area of sector $F E G$: $S_{2}=\frac{\pi \cdot 8}{360} \cdot 210=\frac{14 \pi}{3}$.

$S=S_{1}-S_{2}+2 S_{O E F}=27 \pi-\frac{14 \pi}{3}+2 \cdot \frac{1}{2} \cdot 2 \cdot \sqrt{2}(1+\sqrt{3}) \cdot \frac{\sqrt{2}}{2}=\frac{67 \pi}{3}+2(1+\sqrt{3})$.

Answer: $\frac{67 \pi}{3}+2(1+\sqrt{3})$.