A circle with center $O$ has area $49\pi$. Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA=5\sqrt3$, and point $O$ is outside $\triangle ABC$. What is the side length of $\triangle ABC$?

1. Assume $AB=x$; then $BD=\frac x2, AD=\frac{\sqrt3}2x$ (theorem 22), where $D$ is the midpoint of $BC$.
2. $O$ and $A$ both lie on the perpendicular bisector of $BC$, and $O$ is outside the triangle, so either $A$ is between $O$ and $D$ ($OD=5\sqrt3+\frac{\sqrt3}2x$) or $D$ is between $A$ and $O$ ($OD=5\sqrt3-\frac{\sqrt3}2x>0$).
3. In the first case $\left(\frac x2\right)^2+\left(5\sqrt3+\frac{\sqrt3}2x\right)^2=49$ gives $x^2+15x+26=0$, with no positive root.
4. In the second case $\left(\frac x2\right)^2+\left(5\sqrt3-\frac{\sqrt3}2x\right)^2=49$ gives $x^2-15x+26=0$, so $x=2$ or $x=13$. For $x=13$, $AD=\frac{13\sqrt3}2>OA$, so $O$ would be inside the triangle. Hence $x=\boxed2$ (then $OD=4\sqrt3$ and $OB^2=1+48=49$).