∵E和点F分别是长方形ABCD的边AD和CD的中点
∴AE=DE=1/2AD=1/2BC
DF=CF=1/2CD=1/2AB
∴S△ABE=1/2AB×AE=1/2AB×1/2AD=1/4AB×AD=1/4S矩形ABCD
S△BCF=1/2BC×CF=1/2AD×1/2AB=1/4AB×AD=1/4S矩形ABCD
S△DEF=1/2DE×DF=1/2×1/2AD×1/2AB=1/8AB×AD=1/8S矩形ABCD
∴S△BEF=S矩形ABCD-(S△ABE+S△BCF+S△DEF)
=S矩形ABCD-(1/4S矩形ABCD+1/4S矩形ABCD+1/8S矩形ABCD)
=3/8S矩形ABCD
∴S矩形ABCD=15×8/3=40dm