square ABCD is a rectangle. If AB^{2} + BC^{2 | vedclass

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(D) In a rectangle $\square ABCD$,opposite sides are equal,so $AB = CD$ and $BC = DA$.Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$.Su

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$8$

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(D) In a rectangle $\square ABCD$,opposite sides are equal,so $AB = CD$ and $BC = DA$.Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$.Substituting the equal sides,we get: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 128$.This simplifies to: $2(AB^{2} + BC^{2}) = 128$.Dividing by $2$,we get: $AB^{2} + BC^{2} = 64$.In the right-angled triangle $	riangle ABC$,by the Pythagorean theorem: $AC^{2} = AB^{2} + BC^{2}$.Therefore,$AC^{2} = 64$.Taking the square root of both sides,$AC = \sqrt{64} = 8$.

$\square ABCD$ is a rectangle. If $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 128$,find the length of diagonal $\overline{AC}$.

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