In the regular octagon ABCDEFGH, we can get: HE⊥BG at point M, AD⊥BG at point N,

Each inner angle of a regular octagon is (8-2) × 180/8 = 135.

∴∠HGM=45，

∴MH=MG，

Let MH=MG=x,

Then HG=AH=AB=GF=√2 x.

∴BG×GF=2(√2+ 1)x^2=20

The area of quadrilateral ABGH =1/2 (ah+BG) × hm = (√ 2+1) x2 =10.

The area of a regular octagon is 10× 2+20 = 40 (cm2).

So the answer is: 40.