Rectangle $ABCD$ has $AB=8$ and $BC=6$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$? $\text{(A) } \frac{65}{8} \qquad$ $ \text{(B) } \frac{25}{3} \qquad$ $ \text{(C) } 9 \qquad$ $ \text{(D) } \frac{75}{8} \qquad$ $ \text{(E) } \frac{85}{8}$
Solution
$AC=10$, so $AM=5$.
Triangles $AME$ and $ABC$ have right angles at $M$ and $B$ and share $\angle A$, so $\triangle AME\sim \triangle ABC$ with $AM:AB=5:8$.
Thus $[AME]:[ABC]=25:64$. Since $[ABC]=24$, $[AME]=\frac{75}8$.
Rectangle $ABCD$ has $AB=12$ and $BC=5$. Point $M$ is the midpoint of diagonal $\overline{AC}$, and $E$ is on $AB$ with $\overline{ME}\perp\overline{AC}$. What is the area of $\triangle AME$?
