The $12$-sided polygon $ABCDEFGHIJKL$ is shown. Each pair of consecutive sides is perpendicular. Every side has length $4$, except $BC$ and $FG$, which each have length $5$, and $DE$, which has length $2$. Segments $AG$ and $CH$ meet at $M$. What is the area of quadrilateral $ABCM$?

1. Use coordinates (in units of $1$): $A=(0,0)$, $G=(4,-12)$, $C=(4,-5)$, and $H=(0,-12)$. The lines $AG$ and $CH$ intersect at $M=(48/19,-144/19)$, so $AM:MG=12:7$.
2. Since $GC/GB=7/12$ and $GM/GA=7/19$, $[GCM]/[GAB]=(7/12)(7/19)=49/228$. Also $[GAB]=\frac12(4)(12)=24$.
3. Thus $[ABCM]=[GAB]-[GCM]=24\left(1-\frac{49}{228}\right)=\boxed{\frac{358}{19}}$.