Question

segment pm is a median of triangle pqr if PQ is equals to 40 PR is equal to 42 and pm is equal to 29 find QR​

Answer 1

Answer:

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

then QR = 58

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

Answer 2

Answer:

In PQR, PM is medium

By apollonius theorem

${pq}^{2} + {pr}^{2} = 2 {pm}^{2} \ + 2 {qm}^{2} ..... \\ {40}^{2} + {42}^{2} = 2 \times {29}^{2} + 2 \times {qm}^{2} \\ 1600 + 1764 = 2 \times 841 + 2 \times {qm}^{2} \\ 3364 = 1682 + 2 {qm}^{2} \\ 3364 - 1685 = 2 {qm}^{2} \\ 1682 = 2 \times {qm}^{2} \\ \frac{1682}{2} = {qm}^{2} \\ 841 = {qm}^{2} \\ taking \: square \: root \: \\ qm = 29 \\ \\ qm = 29 \: cm \\ \\ qr = 2 \times qm \\ = 2 \times 29 \\ = 58 \\ \\ qr = 58cm \:$

QR=58 cm.