Question:
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
Correct Answer: 4
Let the area of ABCD be 100. Side of ABCD = 10 Area of EFGH is 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 -xx
$\mathrm{AE}^{2}+\mathrm{AH}^{2}=\mathrm{EH}^{2}$
$\mathrm{x}^{2}+(10-\mathrm{x})^{2}=(\sqrt{62.5})^{2}$
Solving, x = 2.5 or 7.5
Since it’s given that CG is longer than EB, CG = 7.5 and EB = 2.5.
Therefore, EB : CG = 1 : 3
Let the area of ABCD be 100. Side of ABCD = 10 Area of EFGH is 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 -xx
$\mathrm{AE}^{2}+\mathrm{AH}^{2}=\mathrm{EH}^{2}$
$\mathrm{x}^{2}+(10-\mathrm{x})^{2}=(\sqrt{62.5})^{2}$
Solving, x = 2.5 or 7.5
Since it’s given that CG is longer than EB, CG = 7.5 and EB = 2.5.
Therefore, EB : CG = 1 : 3
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