2012 AMC 10B Problem 19

1. Draw the rectangle with $AB = 6$ along the bottom and $AD = 30$ going up the left side. So $B$ is the bottom-right corner, $C$ is the top-right corner, and $D$ is the top-left corner.
2. Extend the bottom side $2$ past $B$ to reach $E$. Then $AE = 6 + 2 = 8$.
3. Draw the line from $E$ up to $D$. It crosses the right side $BC$ at $F$.
4. Find $BF$ with similar triangles. Triangle $EBF$ (small) and triangle $EAD$ (big) both have a right angle at the bottom and share the angle at $E$, so they have the same shape.
5. The small one's bottom side is $EB = 2$ and the big one's is $EA = 8$, so the small triangle is $\frac{2}{8} = \frac14$ the size. Its height $BF$ is $\frac14$ of $AD$:
$$BF = \frac14 \times 30 = 7.5.$$
6. $G$ is the midpoint of $AD$, so $GD = 15$.
7. Now look at $BFDG$. Side $BF$ lies on the right edge and side $GD$ lies on the left edge, so these two sides are parallel. The shape is a trapezoid, and the distance between the parallel sides is the rectangle's width, $6$.
8. Trapezoid area = average of the parallel sides times the distance between them:
$$\frac{7.5 + 15}{2} \times 6 = \frac{22.5}{2} \times 6 = 67.5 = \frac{135}{2}.$$

The answer is $\boxed{\textbf{(C)}\ \frac{135}{2}}$.

Before choosing an area formula, name the shape. Two of the corners sit on each long side of the rectangle, so the shape is a trapezoid whose height is simply the rectangle's width. Then only the two parallel sides are needed: one is a midpoint fact, and the other comes from the similar triangles made by extending a side.

Coordinates. Put $A = (0,0)$, $B = (6,0)$, $D = (0,30)$, $E = (8,0)$ and $G = (0,15)$. The line from $E$ to $D$ drops $30$ over a run of $8$, so moving $2$ to the left from $E$ to $x = 6$ raises it by $30 \times \frac{2}{8} = 7.5$. So $F = (6, 7.5)$, and the trapezoid area is again $\frac{7.5 + 15}{2} \times 6 = \frac{135}{2}$.