In Delta ABC,overline{AD} is a median. If AB^ | vedclass

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(D) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$,the following relation holds:$

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$10$

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(D) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$,the following relation holds:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given that $AB^2 + AC^2 = 122$ and $AD = 6$,we substitute these values into the formula:$122 = 2(6^2 + BD^2)$$122 = 2(36 + BD^2)$Divide both sides by $2$:$61 = 36 + BD^2$$BD^2 = 61 - 36$$BD^2 = 25$$BD = \sqrt{25} = 5$Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 	imes BD$.$BC = 2 	imes 5 = 10$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB^2 + AC^2 = 122$ and $AD = 6$,find $BC$.

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