Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

- 2 : 5
- 4 : 9
- 3 : 8
- 1 : 3

Let the area of square ABCD be 100.

Side of ABCD = 10

Area of EFGH = 62.5

Side of EFGH = √62.5

Triangles AEH, BFE, CGF and DHG are congruent by ASA.

Let AE = BF = CG = DH = x

EB = FC = DG = AH = 10 - x

AE^{2} + AH^{2} = EH^{2}

x^{2} + (10 - x)^{2} = (√62.5)^{2}

Solving,

x = 2.5 or 7.5

Since, CG is longer than EB,

CG = 7.5 and EB = 2.5

Therefore, EB : CG = 1 : 3

**The correct option is D.**

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