Question

(ii) Given : In A PQR, ZQ = 90° QM I PR Prove that: PR² = PQ+ QR?​

Answer 1

[FIGURE IS IN THE ATTACHMENT]

Given:

In A PQR, PR²-PQ²= QR² & QM 1 PR To Prove: QM² = PM × MR

Proof:

Since, PR2 - PQ²= QR²

PR² = PQ² + QR² So, A PQR is a right angled triangle at Q.

In A QMR & APMQ

ZQMR = 2PMQ [Each 90°]

ZMQR = ZQPM [each equal to (90°- ZR)] A QMR ~ APMQ [ by AA similarity criterion] By property of area of similar triangles, ar(A QMR) / ar(APMQ)= QM²/PM²

1/2x MR X QM / 12 x PM XQM = QM²/PM² [Area of triangle= ½ base x height]

MR / PM = QM²/PM²

QM² x PM = PM² x MR

QM² (PM² x MR)/ PM X QM² = PM x MR

HOPE THIS WILL HELP YOU...

Answer 2

Answer:

ple mark me brain list ple

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