Question

In triangle XYZ XM is a median if XY= 20 XZ= 21 and XM= 14.5 find YZ.​

Answer 1

Given:

In triangle XYZ, XM is a median if XY= 20, XZ = 21 and XM = 14.5

To find:

YZ

Solution:

We know that → a median from a vertex of a triangle bisects its third side.

∴ YM = ZM = $\frac{1}{2} YZ$

Apollonius theorem → $\boxed{\bold{AB^2 + AC^2 = 2(AD^2 + BD^2)}} [considering\:AD \:is \:the\:median\:of\:\triangle ABC]$

Now, using the above Apollonius theorem for Δ XYZ where XM is the median, we get

$XY^2 + XZ^2 = 2 [XM^2+(\frac{YZ}{2} )^2]$

on substituting the values of XY = 20, XZ = 21 and XM = 14.5

$\implies 20^2 + 21^2 = 2 [14.5^2+\frac{YZ^2}{4} ]$

$\implies 841 = 2 [210.25+\frac{YZ^2}{4} ]$

$\implies 841 = 420.5 +\frac{YZ^2}{2}$

$\implies \frac{YZ^2}{2} = 420.5$

$\implies YZ^2 = 841$

$\implies \bold{YZ = 29}$

Thus, the value of YZ is → 29.

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