In the adjoining figure, M is the midpoint of QR. ∠PQR = 90°
.
Prove that, PQ²
= 4 PM²
– 3 PR²
Attachments:
Answers
Answered by
0
In ΔPRM
PM
2
=PR
2
+RM
2
(Pythagoras theorem) ⟶(1)
Also,
In ΔPQR
PQ
2
=PR
2
+RQ
2
=PR
2
+(RM+MQ)
2
=PR
2
+(RM+RM)
2
(∵RM=MQ)
=PR
2
+4RM
2
⟶(2)
From (1),
RM
2
=PM
2
−PR
2
⟶(3)
Putting (3) in (2) we get
PQ
2
=PR
2
+4(PM
2
−PR
2
)
PQ
2
=PR
2
+4PM
2
−4PR
2
PQ
2
=4PM
2
−3PR
2
Hence proved.
Similar questions