Question

In △PQR, ∠PQR= 90º, seg QN⊥seg PR, PN= 9, NR=16. find QN


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Answer 1

Answer:

PN×NR=QN

2

Given: PN=QN,QN=NR

In ΔPNQ and ΔQNR

∠PNQ=∠QNR=90

o

PN=QN,QN=NR

ΔPNQ∼ΔQNR by S.A.S

∴∠PQN=∠QRN

In ΔQNR,

∠QNR+∠QRN+∠NQR=180

o

by angle sum property

⇒90

o

+∠PQN+∠NQR=180

o

⇒∠PQN+∠NQR=180

o

−90

o

=90

o

⇒∠PQR=90

o

Hence proved.