Since the area of ABCD is 40 cm2 more than that of square EFGH,

Sum of areas of 4 triangles AGH, BFG, CEF, DHE is given by

AGH + BFG + CEF + DHE = 40

But all four triangles have the same idea due to the rotational symmetry.

So, AGH = BFG = CEF = DHE = 10 cm2

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Since the area of EFGH is three times the area of square JKLM, by a very similar idea,

GHM + FGL + EFK + HGM is twice of JKLM

So, each triangle GHM is half the area of JKLM.

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So, area of triangle AGH is 10 cm2.
But the area of triangle GHM is clearly the same as the area of triangle AGH, simply because they are part of a rectangle AGMH being cut along the diagonal GH.

So, area of GHM is 10 cm2.

The eight triangles have a total area of 80 cm2.

Since the area of GHM is 10 cm2, the area of JKLM must be 20 cm2.

So, the total area of big square ABCD is 80 + 20 = 100 cm2.

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Length AB
= square root of 100 cm2
= 10 cm

so the perimeter of ABCD is 40 cm.

An assumption that GHM and the other 3 mentioned are triangles has to be made in this question, since it is not given that GLM, LKF, KJE and HMJ are straight lines.