A3. In a regular octagon, a square is inscribed with vertices at the vertices of the regular octagon (see figure). The area of the square is $2 \mathrm{~cm}^{2}$. How many square centimeters is the area of the regular octagon?
(A) $1+\sqrt{2}$
(B) $\frac{5}{2}$
(C) $2 \sqrt{2}$
(D) 3
(E) $3 \sqrt{2}-1$

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-05.jpg?height=297&width=311&top_left_y=1208&top_left_x=1606)
A3. A square with an area of $2 \mathrm{~cm}^{2}$ has a side length of $\sqrt{2} \mathrm{~cm}$ and a diagonal of $2 \mathrm{~cm}$. If we connect the vertices of the octagon to its center, we get 8 isosceles triangles with legs of length $r=1 \mathrm{~cm}$ and an internal angle between the legs of $\alpha=45^{\circ}$. The area of one such triangle is $\frac{1}{2} r^{2} \sin \alpha=\frac{1}{2} \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4} \mathrm{~cm}^{2}$, and the area of the octagon is $8 \cdot \frac{\sqrt{2}}{4}=2 \sqrt{2} \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-16.jpg?height=300&width=305&top_left_y=1363&top_left_x=887)