. In the figure 2.22, M is themidpoint of QR. < PRo-90",Prove thot, PO

1.	. In the figure 2.22, M is themidpoint of QR. < PRo-90",Prove thot, PO 4PM-PR
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__________

. In the figure 2.22, M is themidpoint of QR. < PRo-90",Prove thot, PO 4PM-PR

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__________