在矩形ABCD中,AB＝a,BC＝2b,M为BC的中点,DE⊥AM于点E,且a,b满足a＋b＝7,ab＝12,求DE的长?

答:DE=4.8

解：由已知条件
a＋b＝7......(1)
ab＝12......(2)
解上方程组,得
a=4,b=3或a=3,b=4
(1)a=4,b=3
AB=CD=a=4,BC=AD=2b=2*3=6
CM=BM=BC/2=2b/2=2*3/2=3
S△ABM=S△CMD=AB*BM/2=4*3/2=6
AM=√(AB^2+BM^2)=√(4^2+3^2)=5
S△AMD=S矩形ABCD-S△ABM-S△CMD=4*6-2*6=12
S△AMD=AM*DE/2=5*DE/2=12
DE=4.8

(2)a=3,b=4
AB=CD=a=3,BC=AD=2b=2*4=8
CM=BM=BC/2=2b/2=2*4/2=4
S△ABM=S△CMD=AB*BM/2=3*4/2=6
AM=√(AB^2+BM^2)=√(3^2+4^2)=5

S△AMD=S矩形ABCD-S△ABM-S△CMD=3*8-2*6=12

S△AMD=5*DE/2=12
DE=4.8