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(B) Let the side of the square be $a$. The length of the diagonal of a square is $d = a\sqrt{2}$.Given $d = 120\, m$,so $a\sqrt{2} = 120$,which implie

Vedclass · Accepted Answer

$2052$

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(B) Let the side of the square be $a$. The length of the diagonal of a square is $d = a\sqrt{2}$.Given $d = 120\, m$,so $a\sqrt{2} = 120$,which implies $a = \frac{120}{\sqrt{2}} = 60\sqrt{2}\, m$.The diagonals of a square bisect each other at $90^\circ$. Thus,the distance from the center $O$ to each vertex is $r = \frac{120}{2} = 60\, m$.Each flower bed is a minor segment of a circle with radius $r = 60\, m$ and central angle $	heta = 90^\circ$.The area of one minor segment is given by the formula: $	ext{Area} = \frac{	heta}{360^\circ} 	imes \pi r^2 - \frac{1}{2} r^2 \sin	heta$.Substituting the values: $	ext{Area} = \frac{90}{360} 	imes 3.14 	imes (60)^2 - \frac{1}{2} 	imes (60)^2 	imes \sin(90^\circ)$.$	ext{Area} = \frac{1}{4} 	imes 3.14 	imes 3600 - \frac{1}{2} 	imes 3600 	imes 1$.$	ext{Area} = 3.14 	imes 900 - 1800 = 2826 - 1800 = 1026\, m^2$.Since there are two such flower beds,the total area is $2 	imes 1026 = 2052\, m^2$.

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