Two identical circles intersect so thta 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

• Ans:pi/2 -1
  Area of the common portion can be calculated as = 2(Area of the sector of a circle having the angle of 90o - area of the right angled triangle of side 1 cm). Here the right angled triangle is obtained by joining the two intersecting points which is in turn the diagonal of the square.
  Now the area of common section is = 2( (90/360)* pi * 1*1 - (1/2)*1*1)
  =pi/2 -1
• 9 years agoHelpfull: Yes(18) No(1)
• area of the square = 1 sq cm
  area of the sector of a circle when angle is provided = pi r^2 * angle /360
  for a square the angle at all vertices is 90 deg
  so pi*(1)^2* 90 / 360
  = pi/4 ( this is for one circle)
  for 2 circles = pi/2
  
  total area of the 2 sectors of the circle = pi/2 ( this includes overlapped area)
  so overlapped area = area of 2 sectors of circle - area of square
  = pi/2 - 1
• 9 years agoHelpfull: Yes(11) No(2)
• π/2 -1
  common area is reduced from area of quarter sector
• 9 years agoHelpfull: Yes(1) No(1)
• 0.57 sq cm
• 9 years agoHelpfull: Yes(0) No(5)
• please tell me how to draw the figure i cant visualize the figure
• 9 years agoHelpfull: Yes(0) No(0)