A square ABCD is inscribed into the circle. A triangle is formed inside the square with one side CD and other two sides meet at the mid point of AB. Find the ratio of Area of circle to the area of Triangle.
(A) pi (B) pi/2 (C) 2 pi. (D) not remember

• Pi is the correct answer
  Let X be the side of square
  Diagonal will be X*2^1/2
  Radius=Diagonal/2
  Area of circle=pi*X^2/2
  Area of the triangle = 1/2*X*X
  So the ratio is pi
• 9 years agoHelpfull: Yes(37) No(7)
• A of circle=pi*r^2
  a of triangle=1/2*b*h
  here b=s,h=s
  s^2=2r^2
  thus ans will be pi
• 9 years agoHelpfull: Yes(3) No(1)
• area of circle/ area of triangle = pi r^2/(1/2*b*h)=pi* (cd/2)^2/(1/2*cd*cd)
  
  PI/2 ANS
• 9 years agoHelpfull: Yes(3) No(7)
• let r be radius of circle
  ar(circle)=pi*r*r
  side of square
  a
  
  a*a+a*a=(2*r)^2
  a=r*sqrt(2)
  area(triangle)=1/2*b*h=(1/2)*r*sqrt2*r*sqrt2=r*r
  area(circle)/area(triangle)=pi
• 9 years agoHelpfull: Yes(2) No(0)
• ans = c) 2 pi
• 9 years agoHelpfull: Yes(1) No(3)
• area of triangle=1/2*side*side=side^2/2
  area of circle=pi*(diagonal/2)^2
  diagonal=sqrt(2)side
  area of circle=pi*(side)^2/2
  ratio=pi(ans)
• 7 years agoHelpfull: Yes(0) No(0)
• ans will be pi
• 6 years agoHelpfull: Yes(0) No(0)