2008 AMC 10A Problem 20

Triangles $AKB$ and $CKD$ are similar with ratio $\dfrac{9}{12} = \dfrac{3}{4}.$ Since $\triangle AKD$ and $\triangle KCD$ have bases $AK$ and $KC$ on the same line and share the same altitude from $D,$ $\dfrac{[KCD]}{[AKD]} = \dfrac{KC}{AK} = \dfrac{4}{3},$ so $[KCD] = 32.$ Similarly $[AKB] = 18.$ Also $[BKC] = [AKD] = 24.$ The total is $24 + 32 + 18 + 24 = 98.$ Thus, the correct answer is D.