Question

如图,矩形ABCD中,AB＝2,AD＝3,点M为CD上一动点(与点C不重合),将矩形沿某一直线对折,使点B与点M重合,折

Answer 1

1）设CM=x,则DM=2-x,AE=y,则DE=3-y

在直角三角形ABE中,由勾股定理,得,

BE^2=AB^2+AE^2=4+y^2,

在直角三角形DEM中,由勾股定理,得,

EM^2=DE^2+DM^2=(3-y)^2+(2-x)^2=9-6y+y^2+4-4x+x^2=13-4x-6y+x^2+y^2

因为对折

所以BE=ME

即4+y^2=13-4x-6y+x^2+y^2

整理,得y=x^2/6-2x/3+3/2(0