Use half the chord length and the radius to identify the side lengths of right triangle $OMA$. Its side lengths fit a special right-triangle ratio.

The desired region is not the full sector. Find the sector area for the minor central angle, then subtract the area of triangle $AOB$.

First use the chord and radius to determine manually that the minor central angle is $120^\circ$. Desmos is most useful here for checking the final arithmetic rather than replacing the geometry.

In a Desmos expression line, enter

This evaluates the sector area minus the triangle area.

Compare the exact expression or its decimal value with the answer choices. The matching choice uses the $120^\circ$ sector and subtracts the triangle area once.

Let $M$ be the midpoint of chord $\overline{AB}$. Since $OM$ is perpendicular to the chord, it bisects the chord, so

In right triangle $OMA$, $OA=10$ and $AM=5\sqrt{3}$. Thus,

The side lengths of triangle $OMA$ are $5$, $5\sqrt{3}$, and $10$, so it is a $30^\circ$-$60^\circ$-$90^\circ$ triangle. Therefore, $\angle AOM=60^\circ$. Since $OA=OB$, segment $OM$ also bisects $\angle AOB$, so

The sector for the minor arc has central angle $120^\circ$, so its area is

Triangle $AOB$ has base $AB=10\sqrt{3}$ and height $OM=5$. Therefore,

The region bounded by the chord and the minor arc is the sector minus triangle $AOB$:

So the correct choice is $\frac{100\pi}{3}-25\sqrt{3}$.