In rectangle $ABCD$, $AB=6$ and $BC=4$. Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=1$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\triangle AEB$.

1. With $AB=a$, $BC=b$, $DF=p$, $GC=q$, we have $FG=a-p-q$. The lines $AF$ and $BG$ meet at $E$ beyond $CD$. Since $FG\parallel AB$, $\triangle EFG\sim\triangle EAB$; if $H$ is the distance from $E$ to $AB$, then $\frac{H-b}H=\frac{a-p-q}a$, so $H=\frac{ab}{p+q}$ and $[AEB]=\frac12aH=\dfrac{a^2b}{2(p+q)}$.
2. Here $a=6,b=4,p=q=1$: $[AEB]=\dfrac{36\cdot4}{2\cdot2}=\boxed{36}$.