2004 AMC 12B Problem 14

Since $5^2 + 12^2 = 13^2,$ $\triangle ABC$ is right-angled at $C$ with area $\tfrac12 (5)(12) = 30.$ The small right triangles $\triangle AMJ$ and $\triangle NBK$ are each similar to $\triangle ABC,$ with hypotenuses $AM = 5 - 4 = 1$ and $BN = 12 - 4 = 8.$ Their areas are $\left(\dfrac{1}{13}\right)^2 (30)$ and $\left(\dfrac{8}{13}\right)^2 (30).$ The pentagon is what remains: $$\begin{gathered} \left(1 - \dfrac{1}{169} - \dfrac{64}{169}\right)(30) \\ {}= \dfrac{104}{169}(30) = \dfrac{240}{13}. \end{gathered}$$ Thus, the correct answer is D.