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

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(A) In $\Delta ABC$,$\overline{AD}$ is a median to side $\overline{BC}$.Therefore,$D$ is the midpoint of $\overline{BC}$.$BD = \frac{1}{2} BC = \frac{

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) In $\Delta ABC$,$\overline{AD}$ is a median to side $\overline{BC}$.Therefore,$D$ is the midpoint of $\overline{BC}$.$BD = \frac{1}{2} BC = \frac{1}{2}(16) = 8$.By Apollonius' theorem for $\Delta ABC$ with median $\overline{AD}$:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Substituting the given values:$10^2 + AC^2 = 2((\sqrt{58})^2 + 8^2)$$100 + AC^2 = 2(58 + 64)$$100 + AC^2 = 2(122)$$100 + AC^2 = 244$$AC^2 = 244 - 100$$AC^2 = 144$$AC = \sqrt{144} = 12$.

In $\Delta ABC$,$\overline{AD}$ is a median. If $AB = 10$,$BC = 16$ and $AD = \sqrt{58}$,find $AC$.

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