Japan 2007

Since $AB=3$, $BC=4$ and $\angle ABC=90^{\circ }$, we get $AC=5$. Let $M$ be the midpoint of $AD$. Because $AM=DM$, $MC=MC$, $AC=5=DC$ and $\angle ABC=\angle AMC=90^{\circ }$, $AB=AM$, $AC=AC$, the triangles $ABC$, $AMC$ and $DMC$ are all congruent.

Therefore, the area of $ABCD$ is $\frac{3\times 4}{2}\times 3=18$.