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

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(D) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$.By the geom

Vedclass · Accepted Answer

$2$ or $8$

Vedclass · Answer

(D) In a right-angled triangle $\Delta PQR$ with $\angle Q = 90^{\circ}$,$\overline{QM}$ is the altitude to the hypotenuse $\overline{PR}$.By the geometric mean theorem (or properties of similar triangles),we have $QM^2 = PM \cdot MR$.Let $PM = x$. Since $PR = 10$,then $MR = 10 - x$.Given $QM = 4$,we substitute these values into the equation:$4^2 = x(10 - x)$$16 = 10x - x^2$$x^2 - 10x + 16 = 0$Factoring the quadratic equation:$(x - 2)(x - 8) = 0$Thus,$x = 2$ or $x = 8$.Therefore,$PM$ can be $2$ or $8$.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QM}$ is an altitude to the hypotenuse $\overline{PR}$. If $QM = 4$ and $PR = 10$,find $PM$.

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