CAT 2018

Slot 1

QA

Question & Solution

It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1. Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.

Let us assume that 'x+1' is the length of side of square ABCD. 

Area of square ABCD = $(x+1)^2$ sq. units.

Therefore, area of square EFGH = $\dfrac{62.5}{100}*(x+1)^2$ = $\dfrac{5(x+1)^2}{8}$ ... (1)

In right angle triangle EBF,

$EF^2 = EB^2 + BF^2$

$\Rightarrow$ $EF = \sqrt{1^2+x^2}$

Therefore, the area of square EFGH = $EF^2$ = $x^2+1$ ... (2)

By equating (1) and (2),

$x^2+1 = \dfrac{5(x+1)^2}{8}$

$\Rightarrow$ $8x^2+8 = 5x^2+10x+5$

$\Rightarrow$ $3x^2-10x+3 = 0$

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

$\Rightarrow$ $x = 3$ or $1/3$

The ratio of length of EB to that of CG = 1 : x 

EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer.