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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

  1. $2:5$
  2. $4:9$
  3. $3:8$
  4. $1:3$
in Quantitative Aptitude 11.6k points 221 1489 2361
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Given, Area(EFGH) = 62.5% Area(ABCD)


Area(EFGH) = 5/8Area(ABCD)


Let EB = 1 and CG = r


Similarly, the rest other dimensions are also of lengths 1 or r units
Applying pythagoras theorem on △DHG we get GH = √(1+r2)
Area(ABCD) = (1+r)^2 and Area(EFGH) = (√(1+r2))^2


=> (1+r2) = 5/8(1+r)2

1 + r2 = 58581+r2+2r)
8 + 8r2 = 5 + 5r2 + 10r
3r2 – 10r + 3 = 0
(3r-1)×(r-3) = 0
r = 1313 or r = 3
As CG > EB
r = 3 and ratio (EB : CG)= (1:r) = (1:3)

Option (D)

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