Question

Seg PM is a median of ΔPQR. If PQ=40,PR=42 and PM=29, find QR.

Answer 1

B appolonius theorem,
PQ^2+PR^2 = 2(PM)^2+2(MR)^2
1600+1764 = 1682+2(MR)^2
3364 = 1682 + 2(MR)^2
2(MR)^2 = 3364-1682
2(MR)^2 = 1682
MR^2 = 1682÷2
MR^2 = 841
MR = √841 = 29
QR = 29×2 = 58

Answer 2

Answer:

Seg PM is a median of ΔPQR. If PQ=40,PR=42 and PM=29,

then QR = 58

Step-by-step explanation:

Let draw an perpendicular from P at QR  at point D

Then

PR² = PD² + RD²

PR² = PD² + (RM + MD)²

PQ² = PD² + QD²

=> PQ² = PD² + (QM - MD)²

Adding both

PR² + PQ² = PD² + (RM + MD)² + PD² + (QM - MD)²

=> PR² + PQ² = PD² + RM² + MD² +2RM.MD + PD² + QM² + MD² - 2QM.MD

as RM = QM  (PM is median)

=> PR² + PQ² = PD² + RM² + MD²  + PD² + RM² + MD²

=> PR² + PQ² = 2(PD² + MD²)  + 2RM²

PD² + MD² = PM²

=> PR² + PQ² = 2PM² + 2RM²

PQ = 40

PR = 42

PM = 29

=> 42² + 40² = 2 * 29² + 2RM²

=> 1764 + 1600 = 1682 + 2RM²

=> 2RM² = 1682

=> RM² = 841

=> RM = 29

QR = RM + QM

=> QR = 29 + 29 = 58

QR = 58