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(B) Let the length of the big rectangle be $L$ and the width be $W_1 = 2 \ m$.Area of the big rectangle $= L 	imes W_1 = L 	imes 2 = 2L \ m^2$.Let t

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$2 \ m$

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(B) Let the length of the big rectangle be $L$ and the width be $W_1 = 2 \ m$.Area of the big rectangle $= L 	imes W_1 = L 	imes 2 = 2L \ m^2$.Let the length of the small rectangle be $L$ (as given,lengths are equal) and the width be $W_2$.Area of the small rectangle $= L 	imes W_2$.According to the problem,the area of the big rectangle is equal to the area of the small rectangle.Therefore,$2L = L 	imes W_2$.Dividing both sides by $L$ (assuming $L 
eq 0$),we get $W_2 = 2 \ m$.Thus,the width of the small rectangle is $2 \ m$.

The area of a big rectangle is equal to the area of a small rectangle. If the length of the big rectangle is equal to the length of the small rectangle and the width of the big rectangle is $2 \ m$,what is the width of the small rectangle?

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