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(D) $1$. The side of the square $ABCD$ is $a = 60\, m$.$2$. The diagonals of the square intersect at $O$. The distance from $O$ to any side is half th

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$1026$

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(D) $1$. The side of the square $ABCD$ is $a = 60\, m$.$2$. The diagonals of the square intersect at $O$. The distance from $O$ to any side is half the side length,i.e.,$h = 60/2 = 30\, m$.$3$. The radius $r$ of the circle forming the segment is the distance from $O$ to a vertex (e.g.,$OA$). In a square of side $60\, m$,the diagonal is $60\sqrt{2}\, m$. Thus,$r = OA = (60\sqrt{2})/2 = 30\sqrt{2}\, m$.$4$. The central angle $	heta$ subtended by the side $AD$ at the center $O$ is $90^\circ$ (since diagonals of a square are perpendicular).$5$. The area of one circular segment is given by: $	ext{Area} = \frac{	heta}{360^\circ} 	imes \pi r^2 - 	ext{Area of } 	riangle OAD$.$6$. $	ext{Area of } 	riangle OAD = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 60 	imes 30 = 900\, m^2$.$7$. $	ext{Area of segment} = (\frac{90}{360} 	imes 3.14 	imes (30\sqrt{2})^2) - 900 = (0.25 	imes 3.14 	imes 1800) - 900 = 1413 - 900 = 513\, m^2$.$8$. Since there are two such flower beds,the total area is $2 	imes 513 = 1026\, m^2$.

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