A circle of radius $\sqrt6+\sqrt2$ is centered at $O$. Square $OABC$ has side length $1$. Sides $AB$ and $CB$ are extended past $B$ to meet the circle at $D$ and $E$, respectively. What is the area of the region bounded by $BD$, $BE$, and the minor arc connecting $D$ and $E$?

1. Put $O=(0,0)$, $A=(1,0)$, $B=(1,1)$, $C=(0,1)$ and $R=\sqrt6+\sqrt2$, so $R^2=8+4\sqrt3$. Point $D$ is on the line $x=1$, so $D=(1,y)$ with $y^2=R^2-1=7+4\sqrt3=(2+\sqrt3)^2$; thus $D=(1,2+\sqrt3)$ and, by symmetry, $E=(2+\sqrt3,1)$.
2. $\tan\angle AOD=2+\sqrt3$, so $\angle AOD=75^\circ$ and $\angle AOE=15^\circ$. Hence $\angle DOE=75^\circ-15^\circ=60^\circ$.
3. The sector $ODE$ has area $\frac16\pi R^2=\frac{(8+4\sqrt3)\pi}6=\frac{(4+2\sqrt3)\pi}3$.
4. The sector is the shaded region together with the quadrilateral $ODBE$, which splits into $\triangle ODB$ and $\triangle OBE$. Here $BD=BE=2+\sqrt3-1=1+\sqrt3$, and the distance from $O$ to line $AB$ (and to line $CB$) is $1$. So $[ODB]=[OBE]=\frac12(1+\sqrt3)\cdot1$, and $[ODBE]=1+\sqrt3$.
5. The shaded area is $\dfrac{(4+2\sqrt3)\pi}3-1-\sqrt3\approx5.08$, written $\boxed{\dfrac{(4+2\sqrt3)\pi}3-1-\sqrt3}$.