jmc

Recall that the ratio of the hypotenuse to the leg of an isosceles right triangle is $\sqrt{2}$. Looking at the removed triangles, we see that the three segments that make up the side of square are $s/\sqrt{2}$, $s$, and $s/\sqrt{2}$, where $s$ is the side length of the octagon. Setting the sum of these three side lengths equal to 10 inches, we find $$\begin{array}{rl}\frac{s}{\sqrt{2}}+s+\frac{s}{\sqrt{2}}& =10⟹\\ s\left(\frac{1}{\sqrt{2}}+1+\frac{1}{\sqrt{2}}\right)& =10⟹\\ s(\sqrt{2}+1)& =10⟹\\ s& =\frac{10}{\sqrt{2}+1}=10(\sqrt{2}-1),\end{array}$$ where we have rationalized the denominator twice: $$\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2},$$ and $$\frac{10}{\sqrt{2}+1}=\frac{10}{\sqrt{2}+1}\cdot \frac{\sqrt{2}-1}{\sqrt{2}-1}=10(\sqrt{2}-1).$$ To the nearest hundredth, $s=\boxed{4.14}$ inches.