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(A) Let the side of the original square be $x 	ext{ cm}$.The diagonal of the original square is $d_1 = \sqrt{2}x 	ext{ cm}$.The area of the original

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Become $4$-fold

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(A) Let the side of the original square be $x 	ext{ cm}$.The diagonal of the original square is $d_1 = \sqrt{2}x 	ext{ cm}$.The area of the original square is $A_1 = \frac{d_1^2}{2} = \frac{(\sqrt{2}x)^2}{2} = x^2 	ext{ cm}^2$.According to the question,the diagonal of the new square is doubled,so $d_2 = 2 	imes d_1 = 2\sqrt{2}x 	ext{ cm}$.The area of the new square is $A_2 = \frac{d_2^2}{2} = \frac{(2\sqrt{2}x)^2}{2} = \frac{8x^2}{2} = 4x^2 	ext{ cm}^2$.Comparing the areas,$A_2 = 4 	imes A_1$.Therefore,the area of the new square becomes $4$-fold.

If the diagonal of a square is doubled to make another square,the area of the new square will

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