In the adjoining figure, seg $P S$ is the median of $A P Q R$ and $P T ⊥ Q R$ . Prove that,
i. ${P R}^{2} = {P S}^{2} + Q R \times S T + {(\frac{Q R}{2})}^{2}$
ii. $P^{2} = {P S}^{2} - Q R \times S T + {(\frac{Q R}{2})}^{2}$

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Solution

i. $Q S = S R = \frac{1}{2} Q R$ (i) [S is the midpoint of side $Q R]$
$∴$ In $△ P S R, ∠ P S R$ is an obtuse angle [Given] and $P T ⊥ S R [$ Given, $Q - S - R]$ $∴ {P R}^{2} = {S R}^{2} + {P S}^{2} + 2 S R \times S T$ (ii) [Application of Pythagoras theorem] $∴ {P R}^{2} = {(\frac{1}{2} Q R)}^{2} + {P S}^{2} + 2 (\frac{1}{2} Q R) \times S T$ [From (i) and (ii)] $∴ {P R}^{2} = {(\frac{Q R}{2})}^{2} + {P S}^{2} + Q R \times S T$ $∴ {P R}^{2} = {P S}^{2} + Q R \times S T + {(\frac{Q R}{2})}^{2}$
ii. In. $△ P Q S, ∠ P S Q$ is an acute angle and [Given]

$P T ⊥ Q S [$ Given, $Q - S - R]$ $∴ {P Q}^{2} = {Q S}^{2} + {P S}^{2} - 2 Q S \times S T$ (iii) [Application of Pythagoras theorem] $∴ {P R}^{2} = {(\frac{1}{2} Q R)}^{2} + {P S}^{2} - 2 (\frac{1}{2} Q R) \times S T [$ From (i) and (iii)] $∴ {P R}^{2} = {(\frac{Q R}{2})}^{2} + {P S}^{2} - Q R \times S T$ $∴ {P R}^{2} = {P S}^{2} - Q R \times S T + {(\frac{Q R}{2})}^{2}$

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