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.