In Delta XYZ,mangle Y = 90^{circ} and M is th | vedclass

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(B) In $\Delta XYZ$,since $m\angle Y = 90^{\circ}$,the triangle is a right-angled triangle.Using the Pythagorean theorem,$XZ^2 = XY^2 + YZ^2$.Substitu

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$2.5$

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(B) In $\Delta XYZ$,since $m\angle Y = 90^{\circ}$,the triangle is a right-angled triangle.Using the Pythagorean theorem,$XZ^2 = XY^2 + YZ^2$.Substituting the given values,$XZ^2 = 3^2 + 4^2 = 9 + 16 = 25$.Therefore,$XZ = \sqrt{25} = 5$.In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Since $M$ is the midpoint of $\overline{XZ}$,$\overline{YM}$ is the median to the hypotenuse $\overline{XZ}$.Thus,$YM = \frac{1}{2} 	imes XZ = \frac{1}{2} 	imes 5 = 2.5$.

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $M$ is the midpoint of $\overline{XZ}$. If $XY = 3$ and $YZ = 4$,then $YM = \ldots$

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