A round table has radius $4$. Eight rectangular place mats, each with width $1$ and length $x$, are placed on the table so that each mat has two corners on the edge of the table, these two corners being endpoints of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?

1. With 8 congruent mats, each pair of adjacent inner corners $D,E$ subtends $\angle DOE=45^\circ$ at the center $O$. Let $M$ be the midpoint of $DE$; since $\triangle ODE$ is isosceles with $DE=x$ (the mat's own length), $OM=\dfrac x2\cot22.5^\circ=\dfrac x2(1+\sqrt2)$, using the exact value $\cot22.5^\circ=1+\sqrt2$.
2. The outer corner $C$ (directly outward from $M$ by the mat's width $1$) lies on the table's edge, so by the Pythagorean theorem (theorem 18): $\left(\dfrac x2\right)^2+\left(\dfrac x2(1+\sqrt2)+1\right)^2=R^2=16$.
3. Expanding: $(2+\sqrt2)x^2+2(1+\sqrt2)x-30=0$.
4. Solving, $x=\dfrac{\sqrt{63+32\sqrt2}-(1+\sqrt2)}{2+\sqrt2}$.
5. The answer is $\boxed{x=\dfrac{\sqrt{63+32\sqrt2}-(1+\sqrt2)}{2+\sqrt2}}\approx2.340$.