Question

In a ∆PQR, N is a point on PR, such that QN ⊥PR. If PN.NR = QN^2, then prove that ∠PQR = 90°.

Class:- X. [ NCERT Exemplar ]


Triangles chapter

Answer 1

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Answer 2

Given :

QN ⊥ PR

PN × NR = QN²

∠QNP = 90°

In Δ PQN and Δ QNR

QN² = PN × NR

⇒ QN / PN = NR / QN

∠ PNQ = ∠ RNQ [ 90° each ]

Δ PNQ ≈ Δ RNQ [ S.A.S criteria ]

Hence ∠NRQ = ∠PQN

           ∠NQR = ∠ NPQ

Adding both we get :

∠NRQ + ∠NQR = ∠PQN + ∠NPQ

But ∠PQN + ∠NPQ = ∠PQR

So ∠NRQ + ∠NQR = ∠PQR .....( 1 )

Also in Δ NQR , ∠NQR + ∠NRQ + ∠RNQ = 180°

[ ∠ Sum Property ]

∠NQR + ∠NRQ + 90° = 180°

⇒ ∠NQR + ∠NRQ = 180° - 90°

⇒ ∠NQR + ∠NRQ = 90°

Hence in ( 1 ) we see :

∠PQR = 90° [ Proved ]