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

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(D) In $\Delta XYZ$,$\overline{XM}$ is a median to the side $\overline{YZ}$.By definition of a median,$M$ is the midpoint of $\overline{YZ}$,so $YM =

Vedclass · Accepted Answer

$60$

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(D) In $\Delta XYZ$,$\overline{XM}$ is a median to the side $\overline{YZ}$.By definition of a median,$M$ is the midpoint of $\overline{YZ}$,so $YM = MZ = 30.5$.Thus,the length of the side $YZ = YM + MZ = 30.5 + 30.5 = 61$.We use Apollonius's Theorem for $\Delta XYZ$ with median $\overline{XM}$:$XY^2 + XZ^2 = 2(XM^2 + YM^2)$.Substitute the given values: $11^2 + XZ^2 = 2(30.5^2 + 30.5^2)$.$121 + XZ^2 = 2(930.25 + 930.25)$.$121 + XZ^2 = 2(1860.5)$.$121 + XZ^2 = 3721$.$XZ^2 = 3721 - 121$.$XZ^2 = 3600$.$XZ = \sqrt{3600} = 60$.

Part $I$	Part $II$
$1.$ In $\Delta ABC, AB=3, BC=4, AC=5$	$a.$ In-radius $= 1$
$2.$ In $\Delta PQR, PQ=5, QR=12, PR=13$	$b.$ In-radius $= 2$
$3.$ In $\Delta XYZ, XY=8, YZ=15, XZ=17$	$c.$ In-radius $= 3$
$4.$ In $\Delta MNP, MN=20, NP=21, MP=29$	$d.$ In-radius $= 6$

In $\Delta XYZ$,$\overline{XM}$ is a median. If $XY = 11$ and $XM = YM = 30.5$,find $XZ$.

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