$In the fig ure, ∠ PQR = 90 ° . T is the mid - point of the side QR . Prove that {PR}^{2} = 4 {PT}^{2} - 3 {PQ}^{2} .$

Open in App

Solution

$In △ PQR, S eg PT is the median . ∴ {PQ}^{2} + {PR}^{2} = 2 {PT}^{2} + 2 {QT}^{2} [By Appollonius' T heorem] \Rightarrow {PR}^{2} = 2 {PT}^{2} + 2 {QT}^{2} - {PQ}^{2} . . . (i) In △ PQT, ∠ PQT = 90 ° (Given) ∴ {PT}^{2} = {PQ}^{2} + {QT}^{2} (Pythagora s' T heorem) \Rightarrow {QT}^{2} = {PT}^{2} - {PQ}^{2} Substituing the value in (i), we get: {PR}^{2} = 2 {PT}^{2} + 2 [{PT}^{2} - {PQ}^{2}] - {PQ}^{2} \Rightarrow {PR}^{2} = 2 {PT}^{2} + 2 {PT}^{2} - 2 {PQ}^{2} - {PQ}^{2} ∴ {PR}^{2} = 4 {PT}^{2} - 3 {PQ}^{2}$

Suggest Corrections

Similar questions

∠ P Q R = 90^{o}

P R^{2} = 4 P T^{2} - 3 P Q^{2}

3 P Q^{2} = S N^{2} + N R^{2} + Q X^{2} + X R^{2} + P Y^{2} + Y S^{2}

P R^{2} = 4 P S^{2} - 3 P Q^{2}

△ P Q R, P R^{2} - P Q^{2} = Q R^{2}

M

P R

Q M ⊥ P R

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

D

Q R

P Q R

P D R

Q P R

Q R . D R = {P R}^{2} .

Join BYJU'S Learning Program