jmc

Let $M$ be the midpoint of $BC$, so $BM=BC/2$. Since triangle $ABC$ is isosceles with $AB=AC$, $M$ is also the foot of the altitude from $A$ to $BC$. Hence, $O$ lies on $AM$.



Also, by Pythagoras on right triangle $ABM$, $AM=4$. Then the area of triangle $ABC$ is $$K=\frac{1}{2}\cdot BC\cdot AM=\frac{1}{2}\cdot 6\cdot 4=12.$$Next, the circumradius of triangle $ABC$ is $$R=\frac{AB\cdot AC\cdot BC}{4K}=\frac{5\cdot 5\cdot 6}{4\cdot 12}=\frac{25}{8}.$$Then by Pythagoras on right triangle $BMO$, $$\begin{array}{rl}MO& =\sqrt{B{O}^2-B{M}^2}\\ & =\sqrt{R^2-B{M}^2}\\ & =\sqrt{{\left(\frac{25}{8}\right)}^2-3^2}\\ & =\sqrt{\frac{49}{64}}\\ & =\frac{7}{8}.\end{array}$$Finally, the area of triangle $OBC$ is then $$\frac{1}{2}\cdot BC\cdot OM=\frac{1}{2}\cdot 6\cdot \frac{7}{8}=\boxed{\frac{21}{8}}.$$