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On a square cardboard of length 10 cm, two identical circles of radius 2 cm are drawn such that the two centres and the two points of intersection of the circle form a square. Find the total area of the cardboard covered by the two circles?
Let A and C be the centers of the two circles and B and D be the points of intersection. ABCD is a square with side 2 cm. Required area of the two circles = Sum of the areas of the two circles - Common area of the two circles
Common area = 2*(Area of sector ABD - Area of triangle ABD)
Area of sector ABD = pi * $$2^2$$ / 4
Area of triangle ABD = 1/2 * 2 * 2
Substituting these values, we get area of the two circles = 6*pi + 4
As two circles intersecting points are making square with their centres,the angle made by the each sector is 360°-90°=270°
area covered by each sector is
270°×π(2)^2/360°=3π.
Area covered by the square is 2^2=4.
Hence total area is 3π+3π+4=22.85(appro.).
