In the Given Figure, 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) - Geometry Mathematics 2

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Sum

In the given figure, point T is in the interior of rectangle PQRS, Prove that, TS+ TQ= TP+ TR(As shown in the figure, draw seg AB || side SR and A-T-B)

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Solution

`square`PQRS is a rectangle  ...(Given)

∴ PQ || SR   ...(1)   ...[Opposite sides of a rectangle]

AB || SR   ...(2)   ...[Construction]

∴ AB || SR || PQ     ...[From 1 & 2]

∴ PQ ⊥ PS and PQ ⊥ QR    ...(3)  ...(Sides of a rectangle)

A line Perpendicular to one of 2 parallel lines is also Perpendicular to the other line 

∴ AB ⊥ PS and AB ⊥ QR    ...(4)

∴ `square`ABRS is rectangle

∴ AS = BR   ...(5)  ...[Opposite sides of a rectangle]

Similarly we can prove that AP = BQ  ...(6) 

Now, AB ⊥ PS and AB ⊥ QR    ...[From 4]

∴ ∠SAT = ∠RBT = ∠PAT = ∠QBT = 90°  ...[A-T-B]

In ΔPAT, PT2 = PA2 + AT2  ... (7)

In ∆ATS, TS2 = AT2 + AS ...(8)

In ∆QBT, QT2 = QB2 + BT2  ...(9)

In ∆BRT, TR2 = BT2 + BR2 ...(4)

Adding Equations 7 & 8

∴ TS2 + TQ2

= AT2 + AS2 + BT2 + BQ2

=  AT2 + BR2 + BT2 + AP2    ...[From 5 & 6]

= AT2 + AP2 + BT2 + BR2

= TP2 + TR2   ...[From 9 & 10]

Hence proved.

  Is there an error in this question or solution?
Chapter 2: Pythagoras Theorem - Practice Set 2.2 [Page 43]

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