Question

PQR is a triangle in which QM perpendicular to PR and PR square - PQ square = QR square. Prove that QM square = PM*MR

Answer 1

Given that
                          PQR is a triangle and QM is perpendicular on PR

Also,
                      PR^2 - PQ^2 = QR^2

Now in traingle QMR
                                      QR^2 = QM^2+MR^2

Thus fro  above two equations for QR
 We get
                     PR^2 - PQ^2 = QM^2 + MR^2
                         QM^2 = PR^2 - PQ^2 -MR^2
                          QM^2 = (PM+MR)^2 - PQ^2 - MR^2
                            QM^2 = PM^2 + MR^2 + 2PM*MR - PQ^2 - MR^2
                             QM^2 = PM^2 + 2 PM^MR - PQ^2
                             QM^2 = PQ^2 - QM^2 + 2PM*MR - PQ^2
                            thus,         2QM^2 = 2 PM * MR
       Hence prooved

Answer 2

Answer:

Step-by-step explanation:

In a triangle PQR, PR2-PQ2 ​=QR2 and M is a point on side PR such that QM is perpendicular to PR. Prove that QM2=PM*MR

Answer.....

Given:

In ∆ PQR, PR²-PQ²= QR² & QM ⊥ PR

To Prove: QM² = PM × MR

Proof:

Since, PR² - PQ²= QR²

PR² = PQ² + QR²

So, ∆ PQR is a right angled triangle at Q.

In ∆ QMR & ∆PMQ

∠QMR = ∠PMQ [ Each 90°]

∠MQR = ∠QPM [each equal to (90°- ∠R)]

∆ QMR ~ ∆PMQ [ by AA similarity criterion]

By property of area of similar triangles,

ar(∆ QMR ) / ar(∆PMQ)= QM²/PM²

1/2× MR × QM / ½ × PM ×QM = QM²/PM²

[ Area of triangle= ½ base × height]

MR / PM = QM²/PM²

QM² × PM = PM² × MR

QM² =( PM² × MR)/ PM

QM² = PM × MR

HOPE THIS WILL HELP YOU...