∴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