In rectangle ABCD,AB^{2} + BC^{2} + CD^{2} + | vedclass

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(A) In rectangle $ABCD$,we know that opposite sides are equal,so $CD = AB$ and $DA = BC$.Also,$\angle B = 90^{\circ}$.Given the equation: $AB^{2} + BC

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(A) In rectangle $ABCD$,we know that opposite sides are equal,so $CD = AB$ and $DA = BC$.Also,$\angle B = 90^{\circ}$.Given the equation: $AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$.Substituting the equal sides: $AB^{2} + BC^{2} + AB^{2} + BC^{2} = 338$.$2(AB^{2} + BC^{2}) = 338$.$AB^{2} + BC^{2} = 169$.In $	riangle ABC$,by Pythagoras theorem: $AC^{2} = AB^{2} + BC^{2}$.Therefore,$AC^{2} = 169$.$AC = \sqrt{169} = 13$.

In rectangle $ABCD$,$AB^{2} + BC^{2} + CD^{2} + DA^{2} = 338$,then $AC = \ldots$

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