(1)过D作DE⊥AB于E,交MN于F,由勾股定理,得AD=5.
由△DMF∽△DAE,得DM/DA=MF/AE,
即,x/5=(y-3)/3
∴y=(3/5)x+3 (0＜x＜5)
(2)(2)由平行线的性质可得DF=4x/5,根据梯形的面积公式可得
S=(6/25)x^2+(12/5)x (0
(1)过D作DE⊥AB于E,交MN于F,由勾股定理,得AD=5.
由△DMF∽△DAE,得DM/DA=MF/AE,
即,x/5=(y-3)/3
∴y=(3/5)x+3 (0＜x＜5)
(2)(2)由平行线的性质可得DF=4x/5,根据梯形的面积公式可得
S=(6/25)x^2+(12/5)x (0