Question

In the figure, M is the midpoint of QR. Angle PRQ = 90°. Prove that PQ² = 4PM² - 3PR²

Answer 1

Given in right triangle PQR, QS =  SR
By Pythagoras theorem, we have
PR2 = PQ2 + QR2 → (1)
In right triangle PQS, we have
PS2 = PQ2 + QS2
       = PQ2 + (QR/2)2    [Since QS =  SR = 1/2 (QR)]
       = PQ2 + (QR2/4)
4PS2 =  4PQ2 + QR2
∴ QR2 = 4PS2 −  4PQ2 → (2)
Put (2) in (1), we get
PR2 = PQ2 + (4PS2 −  4PQ2)
∴ PR2 = 4PS2 −  3PQ2

Answer 2

Hey!!!! Here is your answer...

---> Given- PRQ=90°
M is mid point of QR so,
RM=MQ=1/2RQ

----> To find- PQ^2 =4PM^2-3PR^2

SOLUTION---> In ∆PRM ( By Pythagoras Theorem)

◆ PM^2= PR^2+ RM^2

◆PM^2=PR^2+ (1/2RQ)^2 ( As M is mid point)

◆PM^2=PR^2+RQ^2/4

◆RQ^2=4(PM^2-PR^2)

◆RQ^2=4PM^2-4PR^2------(1)

◆ In ∆PRQ ( By Pythagoras Theorem)

◆PQ^2=PR^2 +RQ^2

◆PQ^2=PR^2+(4PM^2-4PR^2). { from equation 1}

◆PQ^2= PR^2+4PM^2-4PR^2

◆PQ^2=4PM^2-3PR^2

______H.PROVED__________


◆hope it helps you☺️