imc

In the diagram, $\overline{DE}\perp \overline{AB},\overline{FC}\perp \overline{AB}$. Denote $\overline{AE}=x$ and $\overline{DE}=h$. In right triangle $AED$, we have from the Pythagorean theorem: $x^2+h^2=100$. Note that since $EF=DC$, we have $BF=33-DC-x=12-x$. Using the Pythagorean theorem in right triangle $BFC$, we have $(12-x{)}^2+h^2=196$. We isolate the $h^2$ term in both equations, getting $h^2=100-x^2$ and $h^2=196-(12-x{)}^2$. Setting these equal, we have $100-x^2=196-144+24x-x^2⟹24x=48⟹x=2$. Now, we can determine that $h^2=100-4⟹h=4\sqrt{6}$. The two diagonals are $\overline{AC}$ and $\overline{BD}$. Using the Pythagorean theorem again on $△AFC$ and $△BED$, we can find these lengths to be $\sqrt{96+529}=25$ and $\sqrt{96+961}=\sqrt{1057}$. Since $\sqrt{96+529}<\sqrt{96+961}$, $25$ is the shorter length, so the answer is $\boxed{25}$. Or, alternatively, one can notice that the two triangles have the same height but $△AFC$ has a shorter base than $△BED$.