Let the side length of the octagon be $s$. The side lengths of the isosceles right triangles that were removed are then $\frac{1-s}2$. By the Pythagorean Theorem, $s^2=2\cdot\left(\frac{1-s}2\right)^2=\frac{(1-s)^2}2$. $\implies2s^2=1-2s+s^2$ $\implies s^2+2s-1=0$ $\implies s=\frac{2\pm\sqrt{4+4}}2=1\pm\sqrt2$ Assume $s=1+\sqrt2$, so $s^2=3+2\sqrt2$. Then the area of the octagon is: $\left(2+2\sqrt2\right)\left(3+2\sqrt2\right)=14+10\sqrt2$, which is greater than $1$ (the area of the square) so $s$ cannot be $1+\sqrt2$. Thus, $s=1-\sqrt2$. $\implies s^2=3-2\sqrt2$ So the answer is: $\left(2+2\sqrt2\right)\left(3-2\sqrt2\right)=\boxed{-2+2\sqrt2}\approx0.828$.

Quote: By the Pythagorean Theorem, $s=2\cdot\left(\frac{1-s}2\right)^2=\frac{(1-s)^2}2$. It should be $s^2$ on the LHS Also it's pretty suspicious that you got the area of the octagon as a number greater than $1$...

Similarly to jasperE3, we let the side of the octagon be $s$. We then have $2 \cdot \frac{s}{\sqrt2}+s=1$, or $s+s\sqrt2=1 \implies s(1+\sqrt2)=1$. Therefore $s=\sqrt2-1$. Applying the area formula for regular octagons gives $2(\sqrt2+1)(\sqrt2-1)(\sqrt2-1)=2 \cdot 1 \cdot (\sqrt2-1)=\boxed{2\sqrt2-2}$.

Also, jasperE3's answer is less than $1$. It also matches mine after an edit, so I believe it's correct.