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?

• 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...
• 10 years agoHelpfull: Yes(25) No(0)
• the answer is 5:48 how many of u are getting this? my answer is coming 5:48 i wanna verify
• 10 years agoHelpfull: Yes(6) No(7)
• 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
• 10 years agoHelpfull: Yes(4) No(3)
• 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?
• 10 years agoHelpfull: Yes(3) No(0)
• 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
• 10 years agoHelpfull: Yes(2) No(0)
• answer is 5:48
• 10 years agoHelpfull: Yes(2) No(0)
• ques is wrong.
• 10 years agoHelpfull: Yes(2) No(0)
• sqrt(26):48 = 5:48 (approx) i got this ans
• 10 years agoHelpfull: Yes(2) No(0)
• 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
• 10 years agoHelpfull: Yes(1) No(2)
• 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...
• 10 years agoHelpfull: Yes(0) No(1)
• @prakash can u please clarify me the question?i thnk the question is wrongly typed.
• 10 years agoHelpfull: Yes(0) No(0)
• i think the proper soln is 1:16.......anyone can state what the right ans is??
  
• 10 years agoHelpfull: Yes(0) No(0)