In parallelogram ABCD,AB^{2} + BC^{2} = 260 a | vedclass

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(B) In a parallelogram $ABCD$,the sum of the squares of the diagonals is equal to the sum of the squares of its four sides.This is known as the Parall

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$14$

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(B) In a parallelogram $ABCD$,the sum of the squares of the diagonals is equal to the sum of the squares of its four sides.This is known as the Parallelogram Law: $AC^{2} + BD^{2} = 2(AB^{2} + BC^{2})$.Given $AB^{2} + BC^{2} = 260$ and $AC = 18$.Substituting the values into the formula: $18^{2} + BD^{2} = 2(260)$.$324 + BD^{2} = 520$.$BD^{2} = 520 - 324$.$BD^{2} = 196$.$BD = \sqrt{196} = 14$.

In parallelogram $ABCD$,$AB^{2} + BC^{2} = 260$ and $AC = 18$. Then $BD = \ldots$

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