Explain clearly.......What is side of the largest possible regular octagon that can be cut out of a square of side 1 cm?

• sorry,that was the wrong solution.The correct one is
  length of side of octagon=sqrt(2)*[1/(2+sqrt(2))] [draw the geometrical figure]
  =sqrt(2)*0.293
  =0.41 cm
• 9 years agoHelpfull: Yes(8) No(0)
• let side of small triangle be x
  then
  x+sqrt(2)x+sqrt(2)x=1
  on solving
  x=0.3
  
• 9 years agoHelpfull: Yes(4) No(1)
• length of side of octagon =sqrt(2)[(1/3)*side of Square]
  =sqrt(2)*(1/3)
  =1.414*(1/3)
  =0.47 cm
• 9 years agoHelpfull: Yes(3) No(2)
• let side of octagon be 'y'
  one side of octagon occupies y length of square on respective side
  part left on side of a square be (x+x) {i.e.,on both sides of its side equal part remains)
  Therefore, 2x+y=1 cm
  also there forms a right angled triangle with square two adjacent sides and a side of octagon
  Hence by Pythagoras theorem,y=sqrt(x^2+x^2)=1.414*x
  so, x=0.707*y
  sub this in 2x+y=1,we get
  2(0.707y)+y=1
  Implies,y=0.414
• 9 years agoHelpfull: Yes(3) No(0)
• .44 cm
  let side=x;
  x+x/(sqrt(2)+x/(sqrt(2)=1
  x=.44;
• 9 years agoHelpfull: Yes(2) No(3)
• it can be best understand by drawing octagon in the square
  let the side of octagon=x
  renaining side of sq on each side of octagon=y,y
  now y+x+y=1cm
  x+2y=1, x=1-2y
  now y^2+y^2=x^2
  x^2=2y^2
  (1-2y)^2=2y^2
  y=1+/-0.707
  taking value of y=1-0.707=0.293
  so x=1-2y=1-2*0.293=0.414(ans)
• 9 years agoHelpfull: Yes(2) No(0)
• sqrt((1/3)^2 + (1/3)^2)=0.47
• 9 years agoHelpfull: Yes(1) No(1)
• Draw a square and divide it into 9 equal blocks by drawing 2 horizontal and 2 vertical lines. Draw an octagon inside that square by joining points.
  Now length of one side is simply 1/3=.33
• 9 years agoHelpfull: Yes(1) No(0)
• @rishabh can you say why x,x/sq2 please
• 9 years agoHelpfull: Yes(0) No(0)
• draw the diagram
  one side of squre in term of octagon =x+x+(1-2x)
  angle made by octagon to square=45
  by formula (x^2)*2=y
  in triangle ,
  cos45=x/y
  x=0.707
  y=.414cm
• 9 years agoHelpfull: Yes(0) No(0)