Math, asked by atharva31217, 11 months ago

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​

Answers

Answered by aqibjawedkhan9p5p8sa
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


aqibjawedkhan9p5p8sa: Mark it as brainliest
Answered by Anonymous
3

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.

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