Then BN = B'n = t,

MB=MB'=2t。

MN=√5t。

draw

If the height of the right triangle MBN is h,

There is √ 5t * h = 2t * t.

So h=2t/√5,

BB'=2h=4t/√5，

AB' =BB '？ -AB？

= 16t？ /5-36.

Ann again? =(AB-BN)？ =(6-t)？ ,

Ann? =B'N？ AB？ =t？ -( 16t？ /5-36)

get

(6-t)？ =t？ -( 16t？ /5-36)

arrange

16t？ /5- 12t=0

solve

T=0 (give up), or t= 15/4.

therefore

The time when point B' happens to be on AD is 15/4 seconds.

During the whole movement, the area of rectangular ABCD is a constant value: 6 * 8 = 48;

The maximum area of the triangle MNB is:

MB'*NB'/2=(2t)*t/2=t？ =( 15/4)? =225/ 16.

In addition:

In the process of folding, the degrees of the two acute angles of △MNB remain unchanged, and the degrees and trigonometric function values of △ B' Na remain unchanged.

Find t= 15/4 seconds,

It is easy to find the length of each line segment.

Calculation method:

Starting from t= 15/4 seconds,

Get BN=B'N=t= 15/4，

MB=MB'=2t= 15/2。

MN=√5t= 15√5/4。

BB'=2h=4t/√5= 15√5/5=3√5。

AN=6-t=6- 15/4=9/4。

Through AB? =BB '？ -AB？

=(3√5)? -36

=9,

Get ab' = 3.