jmc

Let $x=BC$ be the height of the trapezoid, and let $y=CD$. Since $AC\perp BD$, it follows that $△BAC\sim △CBD$, so $\frac{x}{\sqrt{11}}=\frac{y}{x}⟹x^2=y\sqrt{11}$. Let $E$ be the foot of the altitude from $A$ to $\overline{CD}$. Then $AE=x$, and $ADE$ is a right triangle. By the Pythagorean Theorem, $$x^2+{\left(y-\sqrt{11}\right)}^2=1001⟹x^4-11x^2-11^2\cdot 9\cdot 10=0$$ The positive solution to this quadratic equation is $x^2=\boxed{110}$.