10.110. A square with side a is cut at the corners so that a regular octagon is formed. Determine the area of this octagon.
## Solution.

Let $A E=x$ (Fig. 10.106). Then $A B=2 A E+E F$ or $2 x+x \sqrt{2}=a$, from which $x=\frac{a}{2+\sqrt{2}}=\frac{a(2-\sqrt{2})}{2}$. Therefore, the desired area is

$$
S=S_{A B C D}-4 S_{\triangle A E N}=a^{2}-\frac{4 x^{2}}{2}=a^{2}-\frac{4 a^{2}(4-4 \sqrt{2}+2)}{8}=2 a^{2}(\sqrt{2}-1)
$$

Answer: $\quad 2 a^{2}(\sqrt{2}-1)$