解:连接AC,因为AB=4,BC=3,CD=13,DA=12,∠B=90°,
所以AC2=AB2+BC2 ,
=42+32,
=16+9,
=25,所以AC=5,
又因CD2-DA2,
=132-122,
=169-144,
=25,
=AC2,
所以△DAC为直角三角形,
因此S四边形ABCD的面积=S△ABC+S△DAC,
=AB×BC+AD×AC,
=×4×3+ ×12×5,
=6+30,
=36.
答:四边形ABCD的面积等于36.
故答案为:36.
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