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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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,$\angle PQD \cong \ldots \ldots \ldots$

Similar Questions

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BC = 20$,find $BM$.

$\square ABCD$ is a trapezium in which $\overline{ AD } \| \overline{ BC }, \overline{ AC } \cap \overline{ BD }=\{ P \} .$ If $PD =9, PA =5$ and $PB =7.2,$ then $AC =\ldots \ldots \ldots \ldots$

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM : AB = 5 : 7$ and $AN = 4$,find $NC$.

In $\Delta ABC,$ $\overline{AD}$ is a median. If $AD=7,$ $AC=7$ and $BD=4,$ find the perimeter of $\Delta ABC.$

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude. If $AC = 13$ and $CM = 9$,then $BM = \ldots \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module