![](https://ai-shiksha.com/wp-content/uploads/2024/02/maharashtra-board-sol-class-10-maths-p2-chapter-2-8.png)
(1) QS = ½ QR ……(i) [S is the midpoint of QR]
SR = ½ QR ……(ii)
QS = SR [From (i) and (ii)]
PT ⊥QR [Given]
PSR is an obtuse angle. [From figure]
PR2 = SR2+PS2+2SR×ST …..(iii) [Application of Pythagoras theorem]
Substitute SR = ½ QR in (iii)
PR2 =[(½)QR]2+PS2+2(1/2)QR×ST
PR2 =[(½)QR]2+PS2+QR×ST
PR2 = PS2 + QR×ST +(QR/ 2 )2
Hence proved.
(ii) PT⊥QS [Given]
![](https://ai-shiksha.com/wp-content/uploads/2024/02/maharashtra-board-sol-class-10-maths-p2-chapter-2-9.png)
PSQ is an acute angle [From figure]
PQ2 = QS2+PS2-2QS×ST [Application of Pythagoras theorem]
PR2 = [(1/2)QR]2+PS2-2(1/2 )QR×ST
PR2 = (QR/2)2+PS2-QR×ST
PR2 = PS2-QR×ST+( QR /2)2
Hence proved.