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(C) The side of the square $ABCD$ is $a = 56 \, m$.Area of the square lawn $ABCD = a^2 = 56 	imes 56 = 3136 \, m^2$.The diagonals of the square inter

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$4032$

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(C) The side of the square $ABCD$ is $a = 56 \, m$.Area of the square lawn $ABCD = a^2 = 56 	imes 56 = 3136 \, m^2$.The diagonals of the square intersect at $O$. The distance from $O$ to the sides is half the side length,i.e.,$28 \, m$. The radius of the circular sectors forming the flower beds is $r = OA = OB = OC = OD$. In $	riangle OAB$,$OA^2 + OB^2 = AB^2$,so $2r^2 = 56^2$,which gives $r^2 = \frac{56 	imes 56}{2} = 1568$.The area of one circular sector (flower bed) with central angle $90^\circ$ is $\frac{90}{360} 	imes \pi r^2 = \frac{1}{4} 	imes \frac{22}{7} 	imes 1568 = 1232 \, m^2$.There are two such flower beds,so their total area is $2 	imes 1232 = 2464 \, m^2$.However,the flower beds are segments of the circle. The area of one segment is (Area of sector $OAB$ - Area of $	riangle OAB$).Area of $	riangle OAB = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 56 	imes 28 = 784 \, m^2$.Area of one flower bed segment = $1232 - 784 = 448 \, m^2$.Total area = Area of square + $2$ $	imes$ Area of segment = $3136 + 2 	imes 448 = 3136 + 896 = 4032 \, m^2$.

In the figure,two circular flower beds are shown on two sides of a square lawn $ABCD$ of side $56 \, m$. If the centre of each circular flower bed is the point of intersection $O$ of the diagonals of the square lawn,find the sum of the areas of the lawn and the flower beds (in $m^2$).

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