Question

In triangle PQR, PR^2- PQ^2= QR^2 and M is a point on side PR such that QM is perpendicular to PR. Prove that QM^2 =PM×MR.

Answer 1

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

Answer 2

Step-by-step explanation:

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

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