2003 AMC 10B Problem 20

Since $CD = AB = 5$, the segment $FG$ has length $5 - 1 - 2 = 2$. Because $\overline{FG} \parallel \overline{AB}$, triangles $EFG$ and $EAB$ are similar (equal corresponding angles), with ratio $\frac{FG}{AB} = \frac{2}{5}$.

Let $h$ be the distance from $E$ to line $AB$. The distance from $E$ to line $CD$ is then $h - 3$, and the similarity ratio applies to these heights:
$$\frac{h - 3}{h} = \frac{2}{5} \quad\Longrightarrow\quad 5h - 15 = 2h \quad\Longrightarrow\quad h = 5.$$

Therefore
$$[AEB] = \frac{1}{2} \cdot AB \cdot h = \frac{1}{2} \cdot 5 \cdot 5 = \frac{25}{2}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{25}{2}}$.

A line parallel to one side of a triangle cuts off a similar triangle, and similar triangles have all linear measurements (including altitudes) in the same ratio. The rectangle's height $3$ is the difference of the two altitudes, which pins down the scale. Whenever two lines cross a pair of parallels, look for the small-triangle-inside-big-triangle configuration.

Coordinates with $A = (0,0)$, $B = (5,0)$, $F = (1,3)$, $G = (3,3)$: line $AF$ is $y = 3x$ and line $BG$ is $y = -\frac{3}{2}(x - 5)$. Solving, $x = \frac{5}{3}$, $y = 5$, so the height is $5$ and the area is $\frac{25}{2}$.

• Using $DF = 1$ and $GC = 2$ as if $FG$ were $3$ (it is $5 - 1 - 2 = 2$), or adding the trapezoid's area to the wrong small triangle.
• Setting the height ratio as $\frac{3}{h} = \frac{2}{5}$ (treating the rectangle's height as the small triangle's altitude), which gives $h = 7.5$ and area $\frac{75}{4}$.