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

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(B) According to Apollonius's Theorem,for any triangle $ABC$ with median $AD$ to side $BC$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $AB = 10$,$AC = 24$,and

Vedclass · Accepted Answer

$26$

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(B) According to Apollonius's Theorem,for any triangle $ABC$ with median $AD$ to side $BC$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $AB = 10$,$AC = 24$,and $AD = 13$.Substituting the values:$10^2 + 24^2 = 2(13^2 + BD^2)$$100 + 576 = 2(169 + BD^2)$$676 = 2(169 + BD^2)$$338 = 169 + BD^2$$BD^2 = 338 - 169 = 169$$BD = \sqrt{169} = 13$Since $AD$ is a median,$D$ is the midpoint of $BC$,so $BC = 2 	imes BD$.$BC = 2 	imes 13 = 26$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 10$,$AC = 24$,and $AD = 13$,find $BC$.

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