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

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Question

(C) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(C) According to Apollonius's Theorem,for any triangle $\Delta ABC$ with median $\overline{AD}$ to side $\overline{BC}$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $AB = 8$,$AC = 15$,and $AD = 8.5$.Substituting the values:$8^2 + 15^2 = 2(8.5^2 + BD^2)$$64 + 225 = 2(72.25 + BD^2)$$289 = 2(72.25 + BD^2)$$144.5 = 72.25 + BD^2$$BD^2 = 144.5 - 72.25 = 72.25$$BD = \sqrt{72.25} = 8.5$Since $\overline{AD}$ is a median,$D$ is the midpoint of $\overline{BC}$,so $BC = 2 	imes BD$.$BC = 2 	imes 8.5 = 17$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 8$,$AC = 15$ and $AD = 8.5$,find $BC$.

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