In parallelogram ABCD,AB^{2} + AD^{2} = 200 a | vedclass

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(C) In parallelogram $ABCD$,let $\overline{AC} \cap \overline{BD} = \{O\}$.Since the diagonals of a parallelogram bisect each other,$O$ is the midpoin

Vedclass · Accepted Answer

$16$

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(C) In parallelogram $ABCD$,let $\overline{AC} \cap \overline{BD} = \{O\}$.Since the diagonals of a parallelogram bisect each other,$O$ is the midpoint of $\overline{AC}$ and $\overline{BD}$.Therefore,$BO = \frac{1}{2} BD = \frac{1}{2}(12) = 6$.In $\Delta ABD$,$\overline{AO}$ is the median to side $\overline{BD}$.By Apollonius' theorem,$AB^{2} + AD^{2} = 2(AO^{2} + BO^{2})$.Substituting the given values: $200 = 2(AO^{2} + 6^{2})$.$100 = AO^{2} + 36$.$AO^{2} = 100 - 36 = 64$.$AO = \sqrt{64} = 8$.Since $O$ is the midpoint of $\overline{AC}$,$AC = 2 AO = 2 	imes 8 = 16$.

In parallelogram $ABCD$,$AB^{2} + AD^{2} = 200$ and $BD = 12$. Then $AC = \ldots$

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