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(A) The area of the square-shaped pond is given by $side^2 = 8 \, m 	imes 8 \, m = 64 \, m^2$.Given that the area of the pond is $\frac{1}{8}$ of the

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$32$

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(A) The area of the square-shaped pond is given by $side^2 = 8 \, m 	imes 8 \, m = 64 \, m^2$.Given that the area of the pond is $\frac{1}{8}$ of the area of the field,the area of the field is $8 	imes 64 \, m^2 = 512 \, m^2$.Let the length of the rectangular field be $L$ and its width be $W$.According to the problem,$L = 2W$,which implies $W = \frac{L}{2}$.The area of the rectangular field is $L 	imes W = L 	imes \frac{L}{2} = \frac{L^2}{2}$.Equating this to the calculated area: $\frac{L^2}{2} = 512$.$L^2 = 512 	imes 2 = 1024$.$L = \sqrt{1024} = 32 \, m$.

The length of a rectangular field is double its width. Inside the field,there is a square-shaped pond $8 \, m$ long. If the area of the pond is $\frac{1}{8}$ of the area of the field,what is the length of the field? (in $m$)

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