In Delta MNP,overline{MX} is a median. If MN^ | vedclass

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(A) According to Apollonius's Theorem for $\Delta MNP$ where $\overline{MX}$ is the median to side $\overline{NP}$:$MN^2 + MP^2 = 2(MX^2 + NX^2)$Given

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$8$

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(A) According to Apollonius's Theorem for $\Delta MNP$ where $\overline{MX}$ is the median to side $\overline{NP}$:$MN^2 + MP^2 = 2(MX^2 + NX^2)$Given $MN^2 + MP^2 = 50$ and $MX = 3$,substitute these values into the formula:$50 = 2(3^2 + NX^2)$$50 = 2(9 + NX^2)$$25 = 9 + NX^2$$NX^2 = 25 - 9 = 16$$NX = \sqrt{16} = 4$Since $\overline{MX}$ is a median,$X$ is the midpoint of $\overline{NP}$,so $NP = 2 	imes NX$.$NP = 2 	imes 4 = 8$.

In $\Delta MNP$,$\overline{MX}$ is a median. If $MN^2 + MP^2 = 50$ and $MX = 3$,then $NP = \ldots$

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