Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.2 | Q 5 | Page 43
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)
In ∆ATS
In ∆QBT
In ∆BRT
Now,
\[{TS}^2 + {TQ}^2 = \left( {AS}^2 + {AT}^2 \right) + \left( {QB}^2 + {BT}^2 \right) \left( \text{From} \left( 2 \right) \text{and} \left( 3 \right) \right)\]
\[ \Rightarrow {TS}^2 + {TQ}^2 = \left( {BR}^2 + {AT}^2 \right) + \left( {PA}^2 + {BT}^2 \right) \left( \because AS = BR \text{ and } PA = QB \right)\]
\[ \Rightarrow {TS}^2 + {TQ}^2 = \left( {BR}^2 + {BT}^2 \right) + \left( {PA}^2 + {AT}^2 \right)\]
\[ \Rightarrow {TS}^2 + {TQ}^2 = \left( {TR}^2 \right) + \left( {PT}^2 \right) \left( \text{From} \left( 1 \right) \text{and} \left( 4 \right) \right)\]
\[ \Rightarrow {TS}^2 + {TQ}^2 = {TR}^2 + {PT}^2\]
Hence, TS2 + TQ2 = TP2 + TR2.
Explanation:-
Given:
- PT² = PA² + AT² …(1)
- TS² = AT² + AS² …(2)
- QT² = QB² + BT² …(3)
- TR² = BT² + BR² …(4)
We need to prove that TS² + TQ² = TP² + TR²
From (2) and (3), we have: TS² + TQ² = (AT² + AS²) + (QB² + BT²)
Since AS = BR and QB = PA, we can rewrite the above equation as: TS² + TQ² = (AT² + BR²) + (PA² + BT²)
From (1) and (4), we have: PT² = PA² + AT² TR² = BT² + BR²
Adding these two equations, we get: PT² + TR² = PA² + AT² + BT² + BR²
Substituting the values of PT² and TR² from above, we get: PA² + AT² + BT² + BR² = PT² + TR² = TP²
Therefore, we can rewrite the equation as: TS² + TQ² = TP² + BT² + BR² + PA²
Since PT² = PA² + AT² and TR² = BT² + BR², we can rewrite the above equation as: TS² + TQ² = TP² + PT² + TR²
Therefore, TS² + TQ² = TP² + TR².
Hence, the given statement is proved.
Chapter 2 – Pythagoras Theorem- Text Book Solution
Practice Set 2.2 | Q 5 | Page 43