ABCD is a square. E,F,G,H is the midpoint of the sides of square. J and K is the midpoint of the sides HG and FG . L is a point on sides EF such that LF= 1/3 EF. Find the ratio of area of triangle JKL to the area of square ABCD.

• ratio :- 1:8
• 7 years agoHelpfull: Yes(8) No(2)
• ratio = 5: 48
  
• 7 years agoHelpfull: Yes(3) No(4)
• root 13:24 root 2
  
  
  
• 7 years agoHelpfull: Yes(3) No(2)
• let AB= a
  area of ABCD=a^2
  FG=HG=EF=root(DG^2+HD^2)= root((a/2)^2+(a/2)^2)= root(a^2/4+a^2/4)= a/root2
  JG=GK= a/2root2
  JK= root((a/2root2)^2+(a/2root2)^2)= a/2
  LF=EF/3
  KL= root(a^2/8+(a/6root2)^2)= a (root5)/6
  area of JKL=1/2*JK*KL= 1/2*a/2*a root5/6= a^2 root5/24
  ratio=root5 : 24
• 7 years agoHelpfull: Yes(2) No(1)
• Let the side of square be 'a'
  Length of KJ=a/root(2)
  Area of triangle=(1/2)(a/root2)(a/3(root2))=a^2/12
  Area of Square=a^2
  Ratio=1:12
  
• 7 years agoHelpfull: Yes(1) No(8)
• Ratio= 5:48
• 7 years agoHelpfull: Yes(1) No(1)
• ans: 1/6
  
  
• 7 years agoHelpfull: Yes(0) No(2)
• 1:32
  
  
  
  
  
  
  
• 7 years agoHelpfull: Yes(0) No(3)
• please explain solution please
• 7 years agoHelpfull: Yes(0) No(0)
• Here LF = 1/3 * EF
  NOW,EF=ROOT(EB SQUARE + BF SQUARE)
  = a/root(2)
  So,
  LK=ROOT(LF SQUARE + KF SQUARE)
  = (a*root26)/12
  
  Area of triangle= 1/2 * base(lk) * height(jk)
  (a square* root 13)/24
  So ratio = root 13 : 24
• 7 years agoHelpfull: Yes(0) No(6)
• ratio = 5:48
• 7 years agoHelpfull: Yes(0) No(0)