解：∵$∠B = 90°$，$AB = 3$，$BC = 4$
∴$AC^2=AB^2+BC^2=25$，$AC=5$
$ $在$\triangle ACD$中，$AC^2=25$，$CD = 12$，$AD = 13$
∵$AC^2+CD^2=25+12^2= 169$，
$AD^2=13^2=169$
∴$AC^2+CD^2=AD^2$
∴$\triangle ACD$是直角三角形，且$∠ACD = 90°$
∴$S_{四边形ABCD}= S_{\triangle ABC}+S_{\triangle ACD}$
$ =\frac 12×AB×BC+\frac 12×AC×CD$
$ =\frac 12×3×4+\frac 12×5×12=36$
