Let the length and height of S3 rectangle be x and y respectively.

BE=HM=3，BF=MN=4

So:

AB=HM+BE-y=6-y

BC=BF+MN-x=8-x

So:

AE=AB-BE=6-y-3=3-y

AH=AD-HD=8-x-4=4-x

AE and AH are integers, then y= 1 or y = 2；; X= 1 or x=2 or x=3.

It can be proved that AE=CN=3-y and cf = ah = 4-X.

So: S 1=S2.

Because: 4S3=S 1+S2=2S 1.

So: S 1=S2=2S3.

So: S 1=(3-y)(4-x)=2S3=2xy.

So: 12-3x-4y+xy=2xy.

So: xy= 12-3x-4y.

1) When y = 1: x= 12-3x-4, 4x=8 and x=2.

2) When y = 2: 2)y=2 12-3x-8, 5x=4 and x=4/5 are not integers, this is contradictory.

To sum up, x=2, y= 1.

So: blank area = ab * BC-S3 = (6-y) (8-x)-xy = 5 * 6-2 = 28.

So: the blank area is 28.