$In △ PQR, seg PM is the median . If PM = 9 and {PQ}^{2} + {PR}^{2} = 290, f ind QR .$

Open in App

Solution

$In △ PQR, seg PM is the median . Therefore, by Appollonius' T heorem, {PQ}^{2} + {PR}^{2} = 2 {PM}^{2} + 2 {MR}^{2} 290 = 2 \times 9^{2} + 2 {MR}^{2} \Rightarrow 290 = 162 + 2 {MR}^{2} \Rightarrow 162 = 2 {MR}^{2} \Rightarrow {MR}^{2} = 64 \Rightarrow MR = 8 cm [Sides cannot be negative] ∵ QR = 2 MR ∴ QR = 2 \times 8 = 16 cm$

Suggest Corrections

Similar questions

△

P M = 9

P Q^{2} + P R^{2}

= 290

Q R

Δ P Q R, P M

P Q^{2} + P R^{2}

△ P Q R, Q M

P R

P R^{2} - P Q^{2} = Q R^{2}

Q M^{2} = P M \times M R

∠ Q P R = 90^{\circ}, s e g P M ⊥ s e g Q R

P M = 10, Q M = 8,

Δ

P Q R

Join BYJU'S Learning Program