In rectangle $ABCD, AB=5$ and $BC=3$. Points $F$ and $G$ are on $\overline{CD}$ so that $DF=1$ and $GC=2$. Lines $AF$ and $BG$ intersect at $E$. Find the area of $\triangle AEB$. $\text {(A) } 10 \qquad$ $ \text {(B) } \frac{21}{2} \qquad$ $ \text {(C) } 12 \qquad$ $ \text {(D) } \frac{25}{2} \qquad$ $ \text {(E) } 15$
Solution
Place $D=(0,0), C=(5,0), B=(5,3), A=(0,3)$, so $F=(1,0), G=(3,0)$.
Line $AF: y=3-3x$. Line $BG: y=1.5x-4.5$. Setting equal gives $x=\frac53$, so $y=-2$.
Triangle $AEB$ has base $AB=5$ (at height $y=3$) and height $|3-(-2)|=5$, so $[AEB]=\frac12\cdot5\cdot5=\frac{25}2$.
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 $\triangle AEB$.
$\text {(A) } 10 \qquad$ $ \text {(B) } \frac{21}{2} \qquad$ $ \text {(C) } 12 \qquad$ $ \text {(D) } \frac{25}{2} \qquad$ $ \text {(E) } 15$ 