In Figure, two circular flower beds have been shown on two sides of a square lawn ABCD of side 56m. If the centre of each circular flower bed is the point of intersection of the diagonals of the square lawn, find the sum of the areas of the lawns and the flower beds.

Explanation

We know that the diagonals of a square bisect each other perpendicularly
`:. Angle DOC = 90^(@)`
Also, diagonal BD = side `sqrt2 = 56 sqrt2 m`
`:. OD = (1)/(2) xx BD = 28 sqrt2 m`
`:.` Area of `Delta DOC = (1)/(2) xx 28 sqrt2 xx 28 sqrt2 = (28)^(2) m^(2)`
Area of sector ODCO `= pi (OD)^(2) xx (theta)/(360^(@))`
`= pi (28sqrt2)^(2) xx (90^(@))/(360^(@)) = (11 xx 28 xx 28)/(7) m^(2)`
`= 1232 m^(2)`
`:.` Area of 1 flower bed = Area of sector - Area of `Delta`
`:.` Area of 2 flower beds = `2 xx 448 = 896 m^(2)`
Area of square lawn `= (56)^(2) = 3136 m^(2)`
`:.` Required area `= 896 + 3136 = 4032 m^(2)`