2 identical circles intersect so that their centre and point at which they intersect form a square of 1 cm.the area of portion that is common two circles is
π2-1
4
√2-1
√5

• the answer is pi/2 -1
  the area of the sector is pi*r^2*90/360 (as it is a square, so the angle is 90)
  =pi*r^2/4
  =pi/4 (as the area of the square is 1cm^2, so the sides are 1 cm each. the sides of the square are the radii of the circle)
  now, the area of the triangle formed by two radii and a line which joins the point of intersection of the two circles = 0.5*1*1=0.5
  So, half of the portion of our desired area= (pi/4 - 0.5)
  therefore, the area of our desired portion is 2* (pi/4 - 0.5)
  = pi/2 - 1
• 9 years agoHelpfull: Yes(9) No(1)
• area of portion that common to the circle = 2 (area of sector - area of triangle)........ (1)
  
  where area of triangle = 1/2(area of square) = 1/2
  area of sector = pie*1^2/4
  put in equation 1 nd get result = pie/2 -1
• 9 years agoHelpfull: Yes(5) No(1)
• area of the sector if central angle is known
  area= pi*(r^2)*(c/360)
  where c is the central angle or angle of the sector.
  now it is a square so c=90 deg
  so area of one sector=pi*(1^2)*(90/360) [as r=1 length of a side of square]
  then if we take the area of the two sector then the area of the intersected portion will be taken twice.
  so are area of the intersected portion = (2*area of sector)-area of square
  =(2*pi*(1/4))-1
  =pi/2 - 1
• 9 years agoHelpfull: Yes(1) No(0)
• can you please explain it ?
• 9 years agoHelpfull: Yes(0) No(0)
• area of sector????
  can you plz explain?
• 9 years agoHelpfull: Yes(0) No(0)
• answer 2pi-1..........................area secot= pi mul 1pow2/4 =pi
  are triangle=1/2
  common area=2(pi - 1/2) = 2pi - 1
• 9 years agoHelpfull: Yes(0) No(2)