Question

In figure 2.30, point T is in the interior of rectangle PQRS, Prove that, TS²+TQ²=TP²+TR² (As shown in the figure, draw seg AB || side SR and A-T-B)

Answer 1

Through ' T ' draw AB // SR so that

A lies on PS and B lies on QR .

Now AB // SR

Therefore ,

AB perpendicular to PS and

AB perpendicular to QR

[ Since , <S = <R = 90° ]

Therefore ,

SABR and PABQ are both rectangles.


Now from ∆TAS ,

TS² = SA² + TA² ----- ( 1 )

Similarly from ∆TBQ , we have

TQ² = TB² + BQ² ------( 2 )

From ∆TBR , we have

TR² = TB² + RB² ------( 3 )

And from ∆TPA , TP² = PA² + TA²

Adding ( 1 ) and ( 2 )

TS² + TQ²

= SA² + TA² + TB² + BQ²

= RB² + TA² + TB² + PA²

[ Since , SA = RB and QB = PA ]

= (RB² + TB² )+ (TA² + PA²)

= TR² + TP² [ from ( 3 ) & ( 4 ) ]

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