Question 876658
Let x = one side of the inner square. Since the difference between the perimeters of the two squares is 12 meters then the side of the outer square is 12/4=3 m longer than the inner side or x+3.
(x+3)^2-x^2=39
x^2+6x+9-x^2=39
6x+9=39
6x=30
x=5 m.
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Ed