2009 AMC 10B Problem 18

$ME$ is perpendicular to the diagonal, so triangle $AME$ has a right angle at $M$. Look for a bigger right triangle with the same shape.

1. The diagonal $AC$ of an $8$ by $6$ rectangle has length $\sqrt{8^2 + 6^2} = \sqrt{100} = 10$. So $AM = 5$.
2. Triangle $ABC$ has a right angle at $B$. Triangle $AME$ has a right angle at $M$. Both have the same angle at $A$ (the angle between side $AB$ and the diagonal). Two triangles with two equal angles have the same shape (they are similar).
3. In triangle $ABC$, the side next to angle $A$ is $AB = 8$ and the side across from angle $A$ is $BC = 6$. So the side across is $\frac68 = \frac34$ of the side next to it.
4. In triangle $AME$, the side next to angle $A$ is $AM = 5$ and the side across from angle $A$ is $ME$. Same shape, so
$$ME = \tfrac34 \times 5 = \tfrac{15}{4}.$$
5. $AM$ and $ME$ are the two legs of the right triangle $AME$, so
$$\text{area} = \tfrac12 \times 5 \times \tfrac{15}{4} = \tfrac{75}{8}.$$

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

A perpendicular dropped inside a right triangle makes a smaller right triangle that shares an angle with the big one, so the two have the same shape. Then every side of the small one is found from the $6$-$8$-$10$ triangle by one scale factor. Always match sides by where they sit relative to the shared angle.

Use coordinates: $A = (0, 0)$, $B = (8, 0)$, $C = (8, 6)$, so $M = (4, 3)$. The diagonal has slope $\frac68 = \frac34$, so a line perpendicular to it has slope $-\frac43$. Starting at $M$ and going down $3$ to reach the $x$-axis means going right $3 \times \frac34 = \frac94$. So $E = \left(4 + \frac94, 0\right) = \left(\frac{25}{4}, 0\right)$. Triangle $AME$ has base $AE = \frac{25}{4}$ on the $x$-axis and height $3$ (the height of $M$), so its area is $\frac12 \times \frac{25}{4} \times 3 = \frac{75}{8}$.