a circle whose area is 2/pie.a rectangle inside the circle.then find the parameter of that rectangle.

• A=2/pi
  pi*d^2=8/pi
  =>d^2=8/pi^2
  as d^2=a^2+a^2
  =>a=2/pi
  =>P=4a=8/pi
• 11 years agoHelpfull: Yes(12) No(27)
• Area=2/pie
  so ,
  pie*r^2=2/pie (r=radius of the circle)
  =>r^2=2/pie^2
  =>r=>sqrt(2)/pie
  so , diameter of the circle=2*sqrt(2)/pie
  for the rectangles whose diagonal is diameter of the circle, following equation is applicable
  l^2+b^2=(2*sqrt(2)/pie)^2
  =>l^2+b^2=8/pie^2
  Apart from this many rectangles are possible. for them parameters will be different!
• 11 years agoHelpfull: Yes(5) No(6)
• Many rectangle are possible in circle of area 2/pie and hence radius (sqrt2)/Pie.
• 11 years agoHelpfull: Yes(4) No(2)
• area of circle = pie*r^2
  2/pie = pie*r^2;
  therefore r = underroot2/pie;
  since rectangle is inside the circle, therefore rectangle diagonal made a righ angle triangle suppose abc where ab and bc are lenght and base of a triangle and also equal to radius.
  hence ab and bc = underrot2/pie.
  hence ac^2 = underrot2/pie^2 + underrot2/pie^2 ;
  ac = 2/pie.
  where ac is the lenght of a rectangle.
  the parameter of rectangle is 4 times of ac = 4*2/pie = 8/pie
  
  
  
• 11 years agoHelpfull: Yes(2) No(6)
• @ sambhit Ghosh can u explain ur answer?
  
• 11 years agoHelpfull: Yes(1) No(0)
• ans should be 6/pi
• 11 years agoHelpfull: Yes(0) No(3)