Question

PQR is a triangle right angled at P and M is a point on QR such that PM pendicular QR that PM2 =QM xMR.​

Answer 1

Step-by-step explanation:

In △PMR,

By Pythagoras theorem,

(PR)

2

=(PM)

2

+(RM)

2

.......(1)

In △PMQ,

By Pythagoras theorem,

(PQ)

2

=(PM)

2

+(MQ)

2

.......(2)

In △PQR,

By Pythagoras theorem,

(RQ)

2

=(RP)

2

+(PQ)

2

........(3)

∴ (RM+MQ)

2

=(RP)

2

+(PQ)

2

∴ (RM)

2

+(MQ)

2

+2RM.MQ=(RP)

2

+(PQ)

2

....(4)

Adding 1) and 2) we get,

(PR)

2

+(PQ)

2

=2(PM)

2

+(RM)

2

+(MQ)

2

...(5)

From 4) and 5) we get,

2RM.MQ=2(PM)

2

∴ (PM)

2

=RM.MQ

Answer 2

given,

PM PERPENDICULAR QR

To prove,

PM²=QM.MR

Prove :

Using Pythagoras theorem,

In ∆QPR,

QR²=PQ²+PR²

In ∆PMQ,

PQ²=QM²+PM²

In ∆PMR,

PR²=PM²+MR²

Put the value of equation (ii) and (iii) in equation (I)

QR² = QM²+PM²+PM²+MR²

QR² = 2PM²+QM²+MR²

(QM+MR)² = 2PM²+QM²+MR² (QR=QM+MR)

QM²+MR²+2QM.MR = 2PM²+QM²+MR²

2QM.MR = 2PM²

QM.MR = PM²

proved.