2009 AMC 10B Problem 18

By the Pythagorean Theorem, $AC=\sqrt{8^2+6^2}=10,$ so $AM=5.$ Right triangles $AME$ and $ABC$ share angle $A,$ so they are similar with $$\dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68,$$ giving $ME=\dfrac{15}{4}.$ Then $$\begin{gathered} [\triangle AME]=\dfrac12\cdot AM\cdot ME \\ = \dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8}. \end{gathered}$$ Thus, the correct answer is D.