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

• area of circle= 2/pie
  pi *r^2=2/pie
  r^2=2/(pie)^2
  r=root(2)/pie
  so diameter= 2* root(2)/pie
  if a quadrilateral is circumscribed by a circle then the product of its diagonal is equal to the product of opposite sides of the quadrilateral.That is if ABCD is a quadrilateral then AC*BD=AB*CD=AD*BC
  as it is given that rectangle is circumscribed in the circle so AC*BD=AB^2=AD^2
  the two diagonal of the rectangle is the diameter of the circle so
  (2*root(2)/pi)*(2*root(2)/pi) =AB^2=AD^2
  AB=AD=(2*root(2)/pi)
  now perimeter=2*((2*root(2)/pi) +(2*root(2)/pi) )
  =(8*root(2))/pi
• 9 years agoHelpfull: Yes(12) No(8)
• Data insufficient because infinite types of rectangle we can draw in a circle and all having different parameter. SO dear "AKANKSHA" u r nt correct ....
• 9 years agoHelpfull: Yes(5) No(0)
• Ans: 8/pi
  rectangle in the circle should be square then
  diameter of the circle equal to diagonal of rectangle =2*root(2)/pi
  Side of rectangle (L)=2/pi
  so therefore parameter of rectangle =4*L
  =4*2/pi
  =8/pi
• 9 years agoHelpfull: Yes(3) No(2)
• AC>AB & BD>CD . so how can AC*BD =AB*CD ??
• 9 years agoHelpfull: Yes(2) No(3)
• ans is 8/pie
  bcz inside a circle any quadrilateral is a square get the diameter, it is diagonal of the square apply pythagoras theorem u'll get the ans...
• 9 years agoHelpfull: Yes(2) No(3)
• inside circle it will be a square.
  so area=pie(r)^2=2/pie.
  r^2=2/(pie)^2
  r=root(2)/pie
  so diameter of circle will be diagonal of square=2*root(2)/pie
  now by pythagoras theorem
  2*root(2)/pie=root(l^2+l^2) supposing side of square be l.
  so l=2/pie
  now parameter=4*2/pie
  8/pie
  for better understanding draw diagram
• 9 years agoHelpfull: Yes(2) No(4)
• You may be wrong :(
  
• 9 years agoHelpfull: Yes(1) No(0)
• All the solutions are wrong. They just changed the question . How u can say a rectangle will be a square inside a circle. Are u guys crazy ?????
• 2 years agoHelpfull: Yes(0) No(0)