Two identical circles intersect so that their centres, and the points at which they intersect,
form a square of side 1 cm. The area in sq. cm of the portion that is common to the two
circles is:
a) (π/2) – 1
b) 4
c) √2 – 1
d) √5

• area of sector is (angle)/360*3.14*r*r
  so angle is 90 and r is 1
  so area of sector will be 3.14/4
  now we will have to subtract area of triangle from it which is 0.5*1*1
  answer will be (3.14/2)-1
• 8 years agoHelpfull: Yes(3) No(0)
• a) first draw the figure.....radius of circle is 1, first we have to find the area of arc which is
  360 degree makes (pie*1*1)area so 90 degree makes=pie/4. now subtract this area from square to get the bound portion by square but not bound by arc.
  now (area of square-(2*(bound region of square but unbound region by arc))give you the area of common region which is =1-(2*(1-pie/4)=option a
• 8 years agoHelpfull: Yes(1) No(0)
• option a)
  area of square is 1
  area of sector of each circle of angle 90 is 3.14/4 as radius is 1cm
  if we add areas of both sectors it will be 3.14/2 ie area of squre+ area of common portion of two circles
  hence area of common portion will be option a) pie/2 -1
• 8 years agoHelpfull: Yes(1) No(0)
• option (a) is the answer.
• 8 years agoHelpfull: Yes(0) No(2)