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

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Question

(A) In a right-angled triangle $\Delta XYZ$ where $\angle Y = 90^{\circ}$,$\overline{YM}$ is the altitude to the hypotenuse $\overline{XZ}$.According

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(A) In a right-angled triangle $\Delta XYZ$ where $\angle Y = 90^{\circ}$,$\overline{YM}$ is the altitude to the hypotenuse $\overline{XZ}$.According to the geometric mean theorem for right triangles,the altitude to the hypotenuse divides the triangle into two triangles similar to the original triangle and to each other.Specifically,$YM^2 = XM \cdot MZ$.Given $YM = 12$ and $XM = 8$,we substitute these values into the equation:$12^2 = 8 \cdot MZ$$144 = 8 \cdot MZ$$MZ = 144 / 8 = 18$.The length of the hypotenuse $XZ$ is the sum of the segments $XM$ and $MZ$:$XZ = XM + MZ = 8 + 18 = 26$.Therefore,the correct option is $A$.

In $\Delta XYZ$,$m\angle Y = 90^{\circ}$ and $\overline{YM}$ is an altitude to the hypotenuse $\overline{XZ}$. If $YM = 12$ and $XM = 8$,find $XZ$.

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