In Delta XYZ,mangle Y = 90^{circ} and overlin | vedclass

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Question

(A) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle

Vedclass · Accepted Answer

$\sqrt{6}$

Vedclass · Answer

(A) In a right-angled triangle,the altitude drawn to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.According to the geometric mean theorem,the altitude $\overline{YM}$ to the hypotenuse $\overline{XZ}$ satisfies the relation: $YM^2 = XM \cdot ZM$.Given $XM = \sqrt{12}$ and $ZM = \sqrt{3}$.Substituting these values into the formula:$YM^2 = \sqrt{12} \cdot \sqrt{3}$$YM^2 = \sqrt{12 \cdot 3}$$YM^2 = \sqrt{36}$$YM^2 = 6$Therefore,$YM = \sqrt{6}$.

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $XM = \sqrt{12}$ and $ZM = \sqrt{3}$,find $YM$.

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