Latest Elitmus Aptitude Question SOLUTION: ABCD is a square.E,F,G,H are midpoints of CD,BD,AB,CA respectivly and J and K are the midpoints of GH and GF. point L is such that LF=EF/3.then what is the rario of area of triangl

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(#M40019208) ELITMUS Numerical Ability question Keep an EYE

ABCD is a square.E,F,G,H are midpoints of CD,BD,AB,CA respectivly and J and K are the midpoints of GH and GF. point L is such that LF=EF/3.then what is the rario of area of triangle JKL to area of square ABCD?

Asked In Elitmus Prakash Gupta (11 years ago)

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sushmita nigam  (11 years ago)

since BD and CA are the diagonals so F and H must be the same points because diagonals of a square are equal and bisect each other at midpoint?? So i think the question is wrongly typed...

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sushmita nigam  (11 years ago)

the answer is 5:48 how many of u are getting this? my answer is coming 5:48 i wanna verify

Is this solution Helpfull? Yes (6) | No (7)  

shekhar pankaj  (11 years ago)

draw your figure ler side of sqr =2a
A G B
|---------|
H| |F
|_________|
c E d
now connect ghef it is also n square of side (2)^1/2where j and k are mid points
draw a perp to ef that bisects ef let say it x
so lf=ef/3 and lx= ef/6

are of tri jkl = .5*are of square ghef -2*tria klf - tria jlx
a^2 - 2 * a^2/6 - a^2/3

Hint tria klf =tria gjk =.5 * lf*kf (both known)
tria jlx =.5*jl(=gf)*lx (both known)

So total =1/3*a^2

Ratio =1/3)/4
= 1/12

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Prakash Gupta  (11 years ago)

CORRECTION...!!!!
ABCD is a square.E,F,G,H are midpoints of AD,CD,BC,AB respectivly and J and K are the midpoints of GH and GF. point L is such that LF=EF/3.then what is the rario of area of triangle JKL to area of square ABCD?

Is this solution Helpfull? Yes (3) | No   

Prakash Gupta  (11 years ago)

do construt the figure according to the question.
now further consturction is needed,
1. assume two midpoints M,N of EF and EH. A new square is made JKME.
2. construct a perpendicular KP, P is at JL.
angle KJN =90'
MF= EF/2
ML= MF - LF = EF(1/2-1/3)= EF/6
angle KJM is divided in the ratio of LF : ML = 2:1
so angle KJL = 45*2/3=30'
so altitude KP =JK sin30= JK/2

now assume side ABCD as anything like I assume it 16,
AB=BC=CD=DA=16
SO, GH=GL=8sqrt2
GJ=GK= 8(sqrt2)/2 =4sqrt2
JK= 4*sqrt2*sqrt2= 8
KP= JK/2 =8/2 = 4
area of triangle JKL =1/2*JK*KP = 1/2*8*4 =16
area of square ABCD =16*16= 256
ratio JKL/ABCD = 16/256 =1:16

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siddhant  (11 years ago)

answer is 5:48

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satya prakash  (11 years ago)

ques is wrong.

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Vijay Singh Adhikari  (10 years ago)

sqrt(26):48 = 5:48 (approx) i got this ans

Is this solution Helpfull? Yes (2) | No   

Prakash Gupta  (11 years ago)

do construt the figure according to the question.
now further consturction is needed,
1. assume two midpoints M,N of EF and EH. A new square is made JKME.
2. construct a perpendicular KP, P is at JL.
angle KJN =90'
MF= EF/2
ML= MF - LF = EF(1/2-1/3)= EF/6
angle KJM is divided in the ratio of LF : ML = 2:1
so angle JKL = 45*2/3=30'
so altitude KP =JK sin30= JK/2

now assume side ABCD as anything like I assume it 16,
AB=BC=CD=DA=16
SO, GH=GL=8sqrt2
GJ=GK= 8(sqrt2)/2 =4sqrt2
JK= 4*sqrt2*sqrt2= 8
KP= JK/2 =8/2 = 4
area of triangle JKL =1/2*JK*KP = 1/2*8*4 =16
area of square ABCD =16*16= 256
ratio JKL/ABCD = 16/256 =1:16

Is this solution Helpfull? Yes (1) | No (2)  

suryakant kumar  (11 years ago)

5/48..
area of jkl=area of square-(area of two triangle )-area of trapazoidal

find all triangle area and by fig. you can get easly...

finally divide by ar. of bigger square...

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SAMBIT KUMAR GHOSH  (11 years ago)

@prakash can u please clarify me the question?i thnk the question is wrongly typed.

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SAMBIT KUMAR GHOSH  (11 years ago)

i think the proper soln is 1:16.......anyone can state what the right ans is??

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