In Delta XYZ,overline{XM} is a median. If XY | vedclass

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(D) According to Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$ to side $\overline{YZ}$:$XY^2 + XZ^2 = 2(XM^2 + YM^2)$Given $XY = 2

Vedclass · Accepted Answer

$29$

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(D) According to Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$ to side $\overline{YZ}$:$XY^2 + XZ^2 = 2(XM^2 + YM^2)$Given $XY = 20$,$XZ = 21$,and $XM = 14.5$.Substituting the values:$20^2 + 21^2 = 2(14.5^2 + YM^2)$$400 + 441 = 2(210.25 + YM^2)$$841 = 2(210.25 + YM^2)$$420.5 = 210.25 + YM^2$$YM^2 = 420.5 - 210.25 = 210.25$$YM = \sqrt{210.25} = 14.5$Since $\overline{XM}$ is a median,$M$ is the midpoint of $\overline{YZ}$,so $YZ = 2 	imes YM$.$YZ = 2 	imes 14.5 = 29$.

In $\Delta XYZ$,$\overline{XM}$ is a median. If $XY = 20$,$XZ = 21$ and $XM = 14.5$,find $YZ$.

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