# CAT 2018 Set-1 | Question: 80

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

1. $2:5$
2. $4:9$
3. $3:8$
4. $1:3$

Given that$, \text{Area (EFGH)} = 62.5\% \;\text{Area (ABCD)}$

$\Rightarrow \text{Area (EFGH)} = \frac{5}{8}\;\text{Area(ABCD)}$

Let $\text{EB} = 1$ and $\text{CG} = r.$

Similarly, the rest other dimensions are also of lengths $1$ or $r$ units

Applying pythagoras theorem on $\triangle \text{DHG}$ we get $\text{GH} = \sqrt{(1+r^{2})}$

$\text{Area(ABCD)} = (1+r)^2$ and $\text{Area(EFGH)} = \left(\sqrt{(1+r^{2})}\right)^2 = (1 + r^{2})$

$\Rightarrow (1+r^{2}) = \frac{5}{8}(1+r)^{2}$

$\Rightarrow (1 + r^{2}) = \frac{5}{8}(1+r^{2}+2r)$

$\Rightarrow 8 + 8r^{2} = 5 + 5r^{2} + 10r$

$\Rightarrow 3r^{2} – 10r + 3 = 0$

$\Rightarrow 3r^{2} – 9r – r + 3 = 0$

$\Rightarrow 3r(r – 3) – 1(r – 3) = 0$

$\Rightarrow (r-3)(3r-1) = 0$

$\Rightarrow r = 3,\frac{1}{3}$

As $\text{CG} > \text{EB}$

So, $r = 3$

$\therefore$ The ratio $\text{EB} : \text{CG} = 1 : r = 1 : 3$

Option (D)

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