(1)
∵BF=1/5AC,∴AC=1.5÷1/5=7.5
又∵点D、E分别为AC、AB的中点,
∴EB=AB/2,BF=BC/2
∴EF=EB+BF=AB/2+BC/2=(AB+BC)/2=AC/2=3.75
(2)
设AC=4a,BC=3a
AC+BC=AB
即 4a+3a=14
得,AC=8,BC=6
∴AD=AC/2=4,AE=AB/2=7
∴DE=AE-AD=7-4=3
(手打的酸死了,一定要采纳哈)
(1)
∵BF=1/5AC,∴AC=1.5÷1/5=7.5
又∵点D、E分别为AC、AB的中点,
∴EB=AB/2,BF=BC/2
∴EF=EB+BF=AB/2+BC/2=(AB+BC)/2=AC/2=3.75
(2)
设AC=4a,BC=3a
AC+BC=AB
即 4a+3a=14
得,AC=8,BC=6
∴AD=AC/2=4,AE=AB/2=7
∴DE=AE-AD=7-4=3
(手打的酸死了,一定要采纳哈)