Question

$\tt\red{Please \: help \: me: - }$
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Answer 1

Given :

In ∆ PQR ,

PQR = 90° and QS perpendicular on PR , Where S lies on PR .

To Prove :

PR² = PQ² + QR²

Proof :

In ∆ PQS and ∆ PQR

∠PSQ = ∠PQR = 90°

∠QPS = RPQ (Common)

∴ ∆ PQS ≅ PRQ (By AA similarity)

$\large\sf{{\red{ \frac{PQ}{PR} \: = \: \frac{SR}{PQ} }}} \: \: \: \: {\rm{{\purple{(Cross \: \: Multiplying)}}}}$

PQ² = SR × PR ----- (equation 1)

_______________________

In ∆ QSR and ∆ PQR

∠QSR = ∠PQR = 90°

∠QRS = ∠QRP (Common)

∴ ∠QRS ≅ ∠PQR (By AA similarity)

$\large\sf{{\orange{ \frac{QR}{PR} \: = \: \frac{SR}{QR} }}} \: \: \: \: {\rm{{\blue{(Cross \: \: Multiplying)}}}}$

QR² = PR × SR ----- (equation 2)

_______________________

Adding the relations obtained in equation (1) and equation (2).

We get ,

PQ² + QR² = SR × PR × PR × SR

PR (PS + SR)

PR × PR

PR²

Hence , PR² = PQ² + QR²

Answer 2

Answer:

Given :

In ∆ PQR ,

PQR = 90° and QS perpendicular on PR , Where S lies on PR .

To Prove :

PR² = PQ² + QR²

Proof :

In ∆ PQS and ∆ PQR

∠PSQ = ∠PQR = 90°

∠QPS = RPQ (Common)

∴ ∆ PQS ≅ PRQ (By AA similarity)

\large\sf{{\red{ \frac{PQ}{PR} \: = \: \frac{SR}{PQ} }}} \: \: \: \: {\rm{{\purple{(Cross \: \: Multiplying)}}}}

PR

PQ

=

PQ

SR

(CrossMultiplying)

PQ² = SR × PR ----- (equation 1)

_______________________

In ∆ QSR and ∆ PQR

∠QSR = ∠PQR = 90°

∠QRS = ∠QRP (Common)

∴ ∠QRS ≅ ∠PQR (By AA similarity)

\large\sf{{\orange{ \frac{QR}{PR} \: = \: \frac{SR}{QR} }}} \: \: \: \: {\rm{{\blue{(Cross \: \: Multiplying)}}}}

PR

QR

=

QR

SR

(CrossMultiplying)

QR² = PR × SR ----- (equation 2)

_______________________

Adding the relations obtained in equation (1) and equation (2).

We get ,

PQ² + QR² = SR × PR × PR × SR

PR (PS + SR)

PR × PR

PR²

Hence , PR² = PQ² + QR²