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

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Question

(C) In a right-angled triangle $\Delta PQR$ where $m \angle Q = 90^{\circ}$,$\overline{QD}$ is the altitude drawn to the hypotenuse $\overline{PR}$.Ac

Vedclass · Accepted Answer

$\angle DRQ$

Vedclass · Answer

(C) In a right-angled triangle $\Delta PQR$ where $m \angle Q = 90^{\circ}$,$\overline{QD}$ is the altitude drawn to the hypotenuse $\overline{PR}$.According to the property of similar triangles in a right triangle,the altitude to the hypotenuse divides the triangle into two triangles that are similar to the original triangle and to each other.Specifically,$\Delta PQD \sim \Delta QRD$.In $\Delta PQD$,the angles are $\angle P$,$\angle PQD$,and $\angle PDQ = 90^{\circ}$.In $\Delta QRD$,the angles are $\angle R$,$\angle RQD$,and $\angle RDQ = 90^{\circ}$.Since $\angle P + \angle R = 90^{\circ}$ and $\angle P + \angle PQD = 90^{\circ}$,it follows that $\angle PQD = \angle R$.Looking at the options,the angle $\angle DRQ$ (which is the same as $\angle R$) is the correct corresponding angle to $\angle PQD$ in the context of the geometric properties of the altitude of a right triangle.

In $\Delta PQR$,$m \angle Q = 90^{\circ}$ and $\overline{QD}$ is an altitude to the hypotenuse $\overline{PR}$. Then,$\angle PQD \cong \ldots \ldots \ldots$

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