Q.2 identical circles intersect so hat their centers and the points at which they intersect, from a square of side 1cm. the area in sq. cm of the portion that is common to the two circle is
a. (pi/2)-1
b. 4
c. 1.414-1
d. 2.23606

• a.(pi/2)-1
  sol- circle r=1cm. area of common portion = 2[{(pi*r*r*90)/360}-{(1/2)*r*r* sin 90}]
  =pi/2-1
• 9 years agoHelpfull: Yes(6) No(3)
• According to the diagram said above
  the side of the square is 1, which means the radius of each circle is also 1.
  Area of squares = 1
  Area of two circles not included in the square = 3π / 2
  Area of oval overlap region of circles = π/2 - 1
  where as Area of entire diagram = 3π/2 + 1
  so π/2 - 1 is the answer.
• 9 years agoHelpfull: Yes(5) No(0)
• a. pi/2-1
  area formed by sector in circle =pi/4
  area formed by triangle in circle =1/2*1*1=1/2
  req area =2(pi/4-1/2)
  
• 9 years agoHelpfull: Yes(2) No(3)
• Plz explain
• 9 years agoHelpfull: Yes(0) No(0)
• we know area of sector of a circle is pi*r^2*theta/360
  a/c to given question u may draw it nd found that redii of circle is 1cm so area of sector of two circle is
  2*pi*r^2*theta/360
  after that subtract with whole square area to it ...area of square is 1 sq.cm
  so pi/2-1
• 9 years agoHelpfull: Yes(0) No(0)