smc

Name the trapezoid $ABCD$, where $AB$ is parallel to $CD$, $AB<CD$, and $AD<BC$. Draw a line through $B$ parallel to $AD$, crossing the side $CD$ at $E$. Then $BE=AD$, $EC=DC-DE=DC-AB$. One needs to guarantee that $BE+EC>BC$, so there are only three possible trapezoids: $$AB=3,BC=7,CD=11,DA=5,CE=8$$ $$AB=5,BC=7,CD=11,DA=3,CE=6$$ $$AB=7,BC=5,CD=11,DA=3,CE=4$$ In the first case, by Law of Cosines, $\cos (\angle BCD)=(8^2+7^2-5^2)/(2\cdot 7\cdot 8)=11/14$, so $\sin (\angle BCD)=\sqrt{1-121/196}=5\sqrt{3}/14$. Therefore the area of this trapezoid is $\frac{1}{2}(3+11)\cdot 7\cdot 5\sqrt{3}/14=\frac{35}{2}\sqrt{3}$. In the second case, $\cos (\angle BCD)=(6^2+7^2-3^2)/(2\cdot 6\cdot 7)=19/21$, so $\sin (\angle BCD)=\sqrt{1-361/441}=4\sqrt{5}/21$. Therefore the area of this trapezoid is $\frac{1}{2}(5+11)\cdot 7\cdot 4\sqrt{5}/21=\frac{32}{3}\sqrt{5}$. In the third case, $\angle BCD=90^{\circ }$, therefore the area of this trapezoid is $\frac{1}{2}(7+11)\cdot 3=27$. So $r_1+r_2+r_3+n_1+n_2=17.5+\mathrm{10.666...}+27+3+5$, which rounds down to $\boxed{63}$.