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

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(A) According to Apollonius' Theorem for $\Delta ABC$ with median $\overline{AD}$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Substitute the given values: $18^2 + 1

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) According to Apollonius' Theorem for $\Delta ABC$ with median $\overline{AD}$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Substitute the given values: $18^2 + 14^2 = 2(14^2 + BD^2)$$324 + 196 = 2(196 + BD^2)$$520 = 2(196 + BD^2)$$260 = 196 + BD^2$$BD^2 = 260 - 196 = 64$$BD = 8$Since $\overline{AD}$ is a median,$BD = DC = 8$.Therefore,$BC = BD + DC = 8 + 8 = 16$.The perimeter of $\Delta ABC = AB + AC + BC = 18 + 14 + 16 = 48$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 18$,$AC = 14$,and $AD = 14$,find the perimeter of $\Delta ABC$.

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