The perimeter of a bigger square is 12,m more | vedclass

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(C) Let the side of the smaller square be $x\,m$ and the side of the bigger square be $y\,m$.The perimeter of the smaller square is $4x$ and the perim

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$8$

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(C) Let the side of the smaller square be $x\,m$ and the side of the bigger square be $y\,m$.The perimeter of the smaller square is $4x$ and the perimeter of the bigger square is $4y$.According to the problem,$4y = 4x + 12$,which simplifies to $y = x + 3$.The area of the smaller square is $x^2$ and the area of the bigger square is $y^2$.According to the problem,$3x^2 = y^2 + 11$.Substitute $y = x + 3$ into the second equation:$3x^2 = (x + 3)^2 + 11$$3x^2 = x^2 + 6x + 9 + 11$$2x^2 - 6x - 20 = 0$$x^2 - 3x - 10 = 0$$(x - 5)(x + 2) = 0$Since the side length cannot be negative,$x = 5$.Therefore,the side of the bigger square is $y = x + 3 = 5 + 3 = 8\,m$.

The perimeter of a bigger square is $12\,m$ more than the perimeter of a smaller square. Three times the area of the smaller square is $11\,m^2$ more than the area of the bigger square. Find the length of the side of the bigger square.

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