In Delta ABC,AB = 4,BC = 2sqrt{3},and AC = 2s | vedclass

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(C) First,identify the longest side of $\Delta ABC$. We have $AB = 4$,$BC = 2\sqrt{3} \approx 3.46$,and $AC = 2\sqrt{7} \approx 5.29$. Since $AC$ is t

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$\sqrt{7}$

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(C) First,identify the longest side of $\Delta ABC$. We have $AB = 4$,$BC = 2\sqrt{3} \approx 3.46$,and $AC = 2\sqrt{7} \approx 5.29$. Since $AC$ is the longest side,we need to find the length of the median $BD$ drawn to side $AC$.Using Apollonius Theorem for $\Delta ABC$ with median $BD$ on side $AC$:$AB^2 + BC^2 = 2(BD^2 + AD^2)$Since $D$ is the midpoint of $AC$,$AD = AC/2 = (2\sqrt{7})/2 = \sqrt{7}$.Substitute the values: $4^2 + (2\sqrt{3})^2 = 2(BD^2 + (\sqrt{7})^2)$$16 + 12 = 2(BD^2 + 7)$$28 = 2(BD^2 + 7)$$14 = BD^2 + 7$$BD^2 = 7$$BD = \sqrt{7}$.

In $\Delta ABC$,$AB = 4$,$BC = 2\sqrt{3}$,and $AC = 2\sqrt{7}$. Then,in $\Delta ABC$,the length of the median on the longest side is $\ldots$

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