ABCD is a square having mid points E F G H respectively I J are the mid-points of HG GF

1/6root2

by taking the diagram,,,using the side as 'a' of a square.. u can solve it easily
10 years agoHelpfull: Yes(4) No(0)
48:5 will be correct answer,,,..suppose ab=bc=cd=da=6sqrt2,then area of square=72unit^2
then ef=fg=gh=he=6 unit
take a mid point on ef as o.....
join as per question ijk...
now,join i and taken point o
you will be able to identify three small right angle triangle
area of ijk=area of iofg-sum of area of(iok,ifj,ijg)=(6*3)-(3+3+4.5)=18-10.5=7.5.......(all values and notation are in sequence here)
ratio of area of abcd to area of ijk=72/7.5=48:5.....enjoyyyyyyyyyy
10 years agoHelpfull: Yes(2) No(2)
Are E,F,G and H mid points of each side?
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i think answer is=9/32
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@PIJUSH
yes they are the mid points of AB, BC, CD & DA
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@NISHANT...plz explain yr
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Answer is 7/24......s_naveen46@yahoo.com on Facebook.... For
10 years agoHelpfull: Yes(0) No(0)
@nishant pls explain how height of triangle=sa/12
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ans is 48:5
7 years agoHelpfull: Yes(0) No(0)

ABCD is a square having mid points E,F,G & H respectively.
I , J are the mid-points of HG & GF.
K is a point on EF such that KF= 1/3(EF).
what is the ratio of area of ABCD to the area of triangle IJK?