Let a=BC, b=CA, c=AB be the side lengths of a | vedclass

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(B) Let $AD$ be the median to side $BC$. By Apollonius Theorem,we have:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $a = BC = 8$,so $BD = DC = 4$. Also $m = AD

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$8$

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(B) Let $AD$ be the median to side $BC$. By Apollonius Theorem,we have:$AB^2 + AC^2 = 2(AD^2 + BD^2)$Given $a = BC = 8$,so $BD = DC = 4$. Also $m = AD = 6$.Let $AC = b$ and $AB = c$. Given $b - c = 2$,so $c = b - 2$.Substituting these values into the theorem:$(b - 2)^2 + b^2 = 2(6^2 + 4^2)$$b^2 - 4b + 4 + b^2 = 2(36 + 16)$$2b^2 - 4b + 4 = 2(52)$$2b^2 - 4b + 4 = 104$$2b^2 - 4b - 100 = 0$$b^2 - 2b - 50 = 0$Using the quadratic formula $b = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-50)}}{2(1)}$:$b = \frac{2 \pm \sqrt{4 + 200}}{2} = \frac{2 \pm \sqrt{204}}{2} = 1 \pm \sqrt{51}$Since $b$ must be positive,$b = 1 + \sqrt{51} \approx 1 + 7.14 = 8.14$.The nearest integer to $b$ is $8$.

Let $a=BC, b=CA, c=AB$ be the side lengths of a $\triangle ABC$ and $m$ be the length of the median through $A$. If $a=8, b-c=2, m=6$,then the nearest integer to $b$ is

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