In the figure, AB = BC = CD = DE = EF = FG = GA. Then Angle DAE is approximately:

1) 15
2) 20
3) 30
4) 25

• DAE = x
  => BCA = x (isoceles triangle)
  
  => CBD = x+x = 2x (external angle of triangle ABC)
  => CDB = 2x (isoceles triangle)
  
  
  DCE= DAE + CDA (external angle of triangle ADC)
  = x+2x = 3x
  
  => AED = 3x (isoceles triangle)
  
  similarly,
  DFE = DAE + FEA (external angle of triangle ADC)
  =DAE + FGE
  =DAE +(DAE+GFA)
  = 3*DAE = 3x
  
  => EDF = EDA = 3x (isoceles triangle)
  
  
  in triangle ADE, x+3x+3x = 180
  => x = 180/7=25.7 ≈25
• 10 years agoHelpfull: Yes(0) No(0)