∴AE= 12DE,
∫ square ABCD area is 20, ∴AD=25,
In Rt△ADE, let AE=x, then DE=2x,
According to Pythagorean Theorem, AD2=AE2+DE2, that is, 20=x2+4x2.
Solution: x=2,
∴AE=EF=2,
The EFGH area of a square is 4,
∵ square MNQP is the midpoint quadrilateral of square EFGH,
The area of square MNQP is 2.
So the answer is: 2