2003 AMC 10B Problem 20

1. The area of triangle $AEB$ is $\frac{1}{2} \times 5 \times (\text{height})$, where the height is the distance from $E$ down to $AB$. So we need that height; call it $h$.
2. Side $CD$ is also $5$. Taking away $DF = 1$ and $GC = 2$ leaves $FG = 5 - 1 - 2 = 2$.
3. $FG$ lies on the top side of the rectangle, parallel to $AB$. So triangle $EFG$ (the small triangle above the rectangle) has the same angles as triangle $EAB$. It is a shrunken copy, and the shrink factor is $\frac{FG}{AB} = \frac{2}{5}$.
4. In a shrunken copy, every length shrinks by the same factor, including the height. The big triangle has height $h$. The small triangle's height is the part above the rectangle, $h - 3$. So
$$\frac{h - 3}{h} = \frac{2}{5}.$$
5. Cross-multiply: $5(h - 3) = 2h$, so $5h - 15 = 2h$, $3h = 15$, and $h = 5$.
6. Area:
$$\frac{1}{2} \times 5 \times 5 = \frac{25}{2}.$$

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

A line parallel to one side of a triangle cuts off a smaller triangle of the same shape, and in two triangles of the same shape all lengths, heights included, are in the same ratio. The rectangle's height is exactly the difference of the two heights, which pins down the size. Look for this "small triangle on top of a big one" picture whenever two lines cross a pair of parallel lines.

Use coordinates: $A = (0,0)$, $B = (5,0)$, $F = (1,3)$, $G = (3,3)$. Line $AF$ goes up $3$ for every $1$ across, so it is $y = 3x$. Line $BG$ goes from $(5,0)$ to $(3,3)$, up $3$ for every $2$ to the left, so $y = \frac{3}{2}(5 - x)$. Setting them equal: $3x = \frac{3}{2}(5 - x)$, so $2x = 5 - x$ and $x = \frac{5}{3}$. Then $y = 5$, so the height is $5$ and the area is $\frac{25}{2}$.