2009 AMC 10B Problem 18

The diagonal $AC$ of the $8 \times 6$ rectangle has length $10$, so $AM = 5$.

Triangles $AME$ and $ABC$ are both right triangles (at $M$ and at $B$) and share the angle at $A$, so they are similar. In triangle $ABC$ the leg adjacent to angle $A$ is $AB = 8$ and the opposite leg is $BC = 6$; the ratio opposite-to-adjacent is $\frac{6}{8} = \frac34$.

In triangle $AME$ the leg adjacent to angle $A$ is $AM = 5$, so the opposite leg is
$$ME = \frac34 \cdot 5 = \frac{15}{4}.$$

The area of the right triangle $AME$ is
$$\frac12 \cdot AM \cdot ME = \frac12 \cdot 5 \cdot \frac{15}{4} = \frac{75}{8}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{75}{8}}$.

A perpendicular dropped inside a right triangle always creates a smaller triangle similar to the original, since they share an acute angle. Once the similarity is spotted, every length is a fixed multiple of the corresponding length in the $6$-$8$-$10$ triangle, and the area follows from the two legs.

Coordinates: $A = (0,0)$, $C = (8, 6)$, $M = (4, 3)$. The line through $M$ perpendicular to $AC$ has slope $-\frac43$: $y - 3 = -\frac43(x - 4)$. Setting $y = 0$ gives $x = \frac{25}{4}$, so $E = \left(\frac{25}{4}, 0\right)$ and the area is $\frac12 \cdot \frac{25}{4} \cdot 3 = \frac{75}{8}$.

• Treating E as the midpoint of AB or assuming AE = AM, instead of using the similarity ratio to find ME.
• Pairing the sides of the similar triangles incorrectly (using $\frac{4}{3}$ instead of $\frac34$), which gives $ME = \frac{20}{3}$ and an area not among the choices.