Two circles both of radii 1cm intersect each other such that the circumference of each one passes through the center of the circle of the other. What is the area(in sq cm) of the intersecting region?

a. π/3 - √3/4

b. 2π/3 - √3/2

c. 4π/3 - √3/2

d. 4π/3 + √3/2

• 
  Let A and B be the centres of the circles respectively. Let C and D be the points of intersection. Let line AB and line CD meet at E.
  Funda here is: “The line joining centres of the circles perpendicularly bisects the line joining points of intersection” => CE = ½CD and CD ⊥ AB
  By symmetry, E is midpoint of AB => AE = ½(AB) = ½(radius of circle with centre A) = ½(1) = ½
  AC = radius = 1
  Applying Pythagoras to the right triangle ACE, CE2 = AC2-AE2 = 12 – (½)2 = ¾
  
  => CE = √3/2 => CD = 2*CE = √3
  
  
  cos(∠CAE) = AE/AC = ½ => ∠CAE = 600 => ∠CAD = 2*600 = 1200
  Area covered under arc CBD = 1200/3600 (π*12) = π/3
  Similarly, area covered under arc CAD = π/3
  Area of intersection of the two circles
  = (Area covered under arc CBD) + ( Area covered under arc CAD) - (Area of rhombus ACBD)
  = (π/3)+( π/3) - (½ * product of diagonals) = 2π/3 – ½ * (AB)(CD) = 2π/3 – ½ * (1)(√3)
  = 2π/3 – √3/2
  
• 10 years agoHelpfull: Yes(1) No(1)
• i think, option B is right
• 10 years agoHelpfull: Yes(0) No(0)