2012 AMC 10B Problem 19

Sketch the rectangle with $AB = 6$ horizontal and $AD = 30$ vertical. Then $E$ is on line $AB$ with $AE = 8$, and $F$ is where $ED$ crosses side $BC$.

Since $BC \parallel AD$, triangles $EBF$ and $EAD$ are similar, with ratio $\frac{EB}{EA} = \frac{2}{8} = \frac14$. Hence
$$BF = \frac14 \cdot AD = \frac{30}{4} = \frac{15}{2}.$$
Also $G$ is the midpoint of $AD$, so $GD = 15$.

Now look at $BFDG$: the sides $BF$ (on $BC$) and $GD$ (on $AD$) are parallel, so it is a trapezoid, and the distance between those parallel sides is $AB = 6$. Its area is
$$\frac{BF + GD}{2} \cdot 6 = \frac{\frac{15}{2} + 15}{2} \cdot 6 = \frac{45}{4}\cdot 6 = \frac{135}{2}.$$

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

Two of the quadrilateral's vertices sit on each of the rectangle's long sides, so the figure is automatically a trapezoid with height equal to the rectangle's width; only the two parallel side lengths are needed. One of them is a midpoint fact, the other comes from the similar triangles created by extending a side. Identify the shape before reaching for area formulas.

Coordinates: $A(0,0)$, $B(6,0)$, $D(0,30)$, $E(8,0)$, $G(0,15)$. Line $ED$ is $y = 30 - \frac{15}{4}x$, so at $x = 6$ it gives $F(6, 7.5)$. Shoelace on $B(6,0)$, $F(6,7.5)$, $D(0,30)$, $G(0,15)$: $\frac12|45 + 180 + 0 - 90| = \frac{135}{2}$.

• Using the wrong similarity ratio for BF (EB/EA = 2/8, not 2/6), or computing rectangle minus pieces and dropping one piece.
• Mistaking $F$ for the intersection of $ED$ with $BG$ or with $AC$, which changes the figure entirely.