jmc

Name the vertices of the hexagon so that hexagon $ABCDEF$ has $AB=1$, $BC=7$, $CD=2$, and $DE=4$. The hexagon is equiangular, so each interior angle measures $180(6-2)/6=120$ degrees. Extend sides $AB$, $CD$, and $EF$ and call their intersection points $G$, $H$, and $J$ as shown. The exterior angles of the hexagon each measure $180-120=60$ degrees, so triangles $JDE$, $CBH$, $FGA$, and $JHG$ are all equilateral. It follows that $JD=DE=4$ units and $CH=CB=7$ units. Therefore the side length $JH$ of triangle $JGH$ is $4+2+7=13$ units. Turning to side $HG$, we find that $AF=AG=13-(7+1)=5$ units. Finally, we solve $JG=JE+EF+FG$ for $EF$ to get $EF=13-(4+5)=4$ units. The sum of the missing sides is $5+4=\boxed{9}$ units.