In the parallelogram ABCD, BD⊥AD, AB=DC= 10, AD=BC=6, BD=8,

∠DAE=∠ABD，

Rt⊿DAE= Rt⊿ABD，

AE/AD=BD/AB，AE=AD？ BD/AB=6×8/ 10=4.8，

DE/AD=AD/AB，DE=AD？ /AB=6？ / 10=3.6,

AD = A′D′= 6，AE = A′E′= 4.8，DE = D′E′= 3.6，

Rt⊿D'A'E' moves in the DC direction at a speed of 5 units per second, stops moving when the point e' falls on the edge of BC, and A'D' passes through BD at m,

( 1)∵a′d′∥ad,∴∠dmd′=∠adb=90，

And ∠MDD'=∠ABD,

∴rt⊿d′md∽rt⊿abd，

DM/DD′= BD/AB = 8/ 10 = 4/5

DM =(4/5)DD′，

∫DD′= 5t，

∴DM=4t。

(2) As shown in figure 1, find the value of t when e falls on BD.

∵rt⊿dmd∽rt⊿a′me′∽rt⊿aed，

DD:DM :D’M = A’E’:A’M:E’M = AD:AE:DE

=6:4.8:3.6=5:4:3

d′M = 0.6 DD′，A′M = 0.8A′E′= 0.8×4.8 = 3.84，

D′M = A′D′-A′M = 6-3.84 = 2. 16，

0.6 DD′=(5/3)D′M = 5/3×2. 16 = 3.6

Or ∠ME'D'=∠MA'E' (the complementary angles of equal angles ∠A'D'E' are equal),

∠MDD′=∠ME′D′，DD′= D′E′= 3.6，

∴dd′=3.6=5 t

∴t=0.72。

(3)

(1) in the case of figure 1, S=SRt⊿ME'D',

MD′= 0 = 0.6 DD′= 3t，ME′= 4t

Srt⊿me'd'= doctor of medicine? ME'/2=6t (t = 0.72 at this time

② In the case of Figure 2, S=S quadrilateral MNE'D'

=srt⊿a′e′d′-srt⊿a′nm，

srt⊿a′e′d′=4.8×3.6/2=8.64，

MD = 3t，A′D′= 6，A′M = 6-3t，

MN =(4/3)A′M =(4/3)(6-3t)= 8-4t

∴srt⊿a′nm=a′m？ MN/2=(6-3t)(8-4t)/2=24-24t+6t？

S=8.64-(24-24t+6t？ )=- 15.36+24t-6t？ ,

S=24-24t+6t？ ;

③ In the case of Figure 3, S= SRt⊿MND'=6t,

S=6 tons.

(4) in the case of figure 1, rt⊿DMD'≌SRT⊿e'MD', at which time t = 0.72.

In the case of fig. 3, Rt⊿DMD'≌SRt⊿NMD',

The value range of ∴t is: (0,0.72).