smc

$20^2+21^2=841=29^2$. Therefore the triangle is a right triangle. But then its hypotenuse is a diameter of the circumcircle, and thus $C$ is exactly one half of the circle. Moreover, the area of the triangle is $\frac{20\cdot 21}{2}=210$. Therefore the area of the other half of the circumcircle can be expressed as $A+B+210$. Thus the answer is $\boxed{A+B+210=C}$. To complete the solution, note that $(\mathrm{A})$ is clearly false. As $A+B<C$, we have $A^2+B^2<(A+B{)}^2<C^2$ and thus $(\mathrm{C})$ is false. Similarly $20A+21B<21(A+B)<21C<29C$, thus $(\mathrm{D})$ is false. And finally, since $0<A<C$, $\frac{1}{C^2}<\frac{1}{A^2}<\frac{1}{A^2}+\frac{1}{B^2}$, thus $(\mathrm{E})$ is false as well.