在直线o顺次取ABCD四点,是AB:BC:CD=2:3:4,若AB中点M与CD中点N的距离是15cm,则AB=

问题描述:

在直线o顺次取ABCD四点,是AB:BC:CD=2:3:4,若AB中点M与CD中点N的距离是15cm,则AB=

MN=MB+BC+CN=AB/2+BC+CD/2=(2/9AD)/2+3/9AD+(4/9AD)/2=2/3AD=2/3*9/2AB=3AB
AB=MN/3=15/3=5cm

设AB=2x,BC=3x,CD=4x
则有
x+3x+2x=15
x=2.5
则AB=5