Question

theorem 6.7 ch.triangles proof

Answer 1

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∆PQR IS A RIGHT ANGLED TRIANGLE

TO PROVE :- PR^2 = PQ^2 + QR^2

CONSTRUCTION:- QM | PR IS DRAWN.

PROOF :- FROM ∆PQR AND ∆PQM

<PQR = <PMQ [90° EACH]

<QPR = <QPM [COMMON ANGLE]

THEREFORE, ∆PQR ~ ∆PQM [A.A]

$\frac{pq}{pm} = \frac{qr}{qm} \: = \frac{pr}{pq} ........(1)$



AGAIN, FROM ∆PQR AND ∆QMR

<PQR = <QMR [90° EACH]

<PRQ= <QRM [COMMON ANGLE]

THEREFORE, ∆PQR ~ ∆QMR [A.A]

$\frac{pq}{qm} = \frac{qr}{mr} \: = \frac{pr}{qr} ........(2)$
FROM 1

$\frac{pq}{pm} = \frac{pr}{pq} \\ = > {pq}^{2} = pm \times pr \: - - - - (3)$

FROM 2

$\frac{qr}{mr} \: = \frac{pr}{qr} \\ = > {qr}^{2} = mr \times pr \: - - - (4)$
3 +4

PQ^2 + QR^2 = PM×PR + MR×PR
=PR(PM+MR)
=PR ×PR
=PR^2










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Answer 2

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