jmc

We draw the longest altitude of this triangle, which breaks the triangle into two right triangles. The right triangles are congruent because the original triangle is isosceles; each right triangle has one leg length $8/2=4$ and hypotenuse length 9. The other leg, which is also the height of the original triangle, has length $\sqrt{9^2-4^2}=\sqrt{65}$. Thus, the original triangle has base length 8, height $\sqrt{65}$, and area $$\frac{1}{2}(8)(\sqrt{65})=\boxed{4\sqrt{65}}.$$