In Delta PQR,m angle Q = 90^{circ} and overli | vedclass

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(A) In a right-angled triangle $\Delta PQR$,where $\angle Q = 90^{\circ}$ and $\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$,the ge

Vedclass · Accepted Answer

$4 \sqrt{6}$

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(A) In a right-angled triangle $\Delta PQR$,where $\angle Q = 90^{\circ}$ and $\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$,the geometric mean theorem states that the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.According to the property of the altitude to the hypotenuse in a right triangle:$QM^{2} = PM \cdot RM$Given $PM = 8$ and $RM = 12$:$QM^{2} = 8 	imes 12$$QM^{2} = 96$Taking the square root of both sides:$QM = \sqrt{96}$$QM = \sqrt{16 	imes 6}$$QM = 4 \sqrt{6}$

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude. If $PM = 8$ and $RM = 12$,then $QM = \dots$

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