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  • In a D PQR, PR2–PQ2 = QR2 and M is a point on side PR such that QM perpendicular to PR. Prove that : QM2 = PM × MR.

In a D PQR, PR2–PQ2 = QR2 and M is a point on side PR such that QM perpendicular to PR.

Prove that : QM2 = PM × MR.

 

Answers (1)

Given : PR^{2} - PQ^{2} = QR^{2} \; and\; QM \perp PR

To prove :  QM^{2} = PM \times MR

Proof : PR^{2} - PQ^{2} = QR^{2}      (given)

PR^{2} = PQ^{2} + QR^{2}

Because \DeltaPQR holds Pythagoras theorem, therefore, \DeltaPQR is right-angled triangle right angle at Q.

In\; \Delta QMR \; and \; \Delta PMQ

\angle M = \angle M       {each angle is 90°}

\angle MQR = \angle QPM       [each equal to 90° – angle R]

As we know that if the two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.

\therefore \Delta QMR \sim \Delta PMQ

Now using the property of area of similar triangles

\frac{ar(\Delta QMR)}{ar(\Delta PMQ)}=\frac{(QM)^{2}}{(PM)^{2}}

\frac{\frac{1}{2}\times RM \times QM}{\frac{1}{2}\times PM \times QM}=\frac{(QM)^{2}}{(PM)^{2}}                 \left \{ \text {Q area of triangle}=\frac{1}{2}\times base \times height \right \}

\Rightarrow QM^{2} = PM \times RM

 

Posted by

infoexpert23

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