CAT 2018 Question Paper Shift 1 Solved Paper

Question : 75 of 100

Marks: +1, -0

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 Options

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

Validate

Solution:

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}

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

x^{2}+(10-x)^{2}=(√{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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