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Answers
Question:
In the given figure, S and T trisect the side QR of a right triangle PQR. Prove that
8PT² = 3PR² + 5PS².
Answer:
8PT² = 3PR² + 5PS²
Step-by-step-explanation:
NOTE: Refer to the attachment for the diagram.
In figure, △PQR is a right-angled triangle.
We have given that, points S and T trisect the side QR.
∴ QS = ST = TR = 1 / 3 * QR
⇒ QS = ST = TR = QR / 3 - - ( 1 )
⇒ QR = 3 * ( QS = ST = TR )
⇒ QR = 3QS = 3ST = 3TR - - ( 2 )
Now, in △PQR, m∟PQR = 90°
∴ PR² = PQ² + QR² - - [ Pythagoras Theorem ]
⇒ PR² = PQ² + ( 3QS )² - - [ From ( 2 ) ]
⇒ PR² = PQ² + 9QS²
⇒ 3 * PR² = 3 * PQ² + 3 * 9QS² - - [ Multiplying each term by 3 ]
⇒ 3PR² = 3PQ² + 27QS² - - ( 3 )
Now, in △PQS, m∟PQS = 90°
∴ PS² = PQ² + QS² - - [ Pythagoras Theorem ]
⇒ PS² = PQ² + ( QR / 3 )² - - [ From ( 1 ) ]
⇒ PS² = PQ² + QR² / 9
⇒ 5 * PS² = 5 * PQ² + 5 * ( QR² / 9 ) - - [ Multiplying each term by 5 ]
⇒ 5PS² = 5PQ² + 5QR² / 9 - - ( 4 )
Now, in △PQT, m∟PQT = 90°
∴ PT² = PQ² + QT² - - [ Pythagoras Theorem ]
⇒ PT² = PQ² + ( QS + ST )² - - [ Q - S - T ]
⇒ PT² = PQ² + QS² + 2QS. ST + ST²
⇒ PT² = PQ² + ( QR / 3 )² + 2 * ( QR / 3 ) * ( QR / 3 ) + ( QR / 3 )² - - [ From ( 1 ) ]
⇒ PT² = PQ² + QR² / 9 + 2 * QR² / 9 + QR² / 9
⇒ PT² = PQ² + QR² / 9 + 2QR² / 9 + QR² / 9
⇒ PT² = PQ² + ( QR² + 2QR² + QR² ) / 9
⇒ PT² = PQ² + 4QR² / 9
⇒ 8 * PT² = 8 * PQ² + 8 * ( 4QR² / 9 ) - - [ Multiplying each term by 8 ]
⇒ 8PT² = 8PQ² + 32QR² / 9 - - ( 5 )
We have to prove that,
8PT² = 3PR² + 5PS²
⇒ ( 8PQ² + 32QR² / 9 ) = ( 3PQ² + 27QS² ) + ( 5PQ² + 5QR² / 9 ) - - [ From ( 3 ) & ( 4 ) & ( 5 ) ]
⇒ 8PQ² + 32QR² / 9 = 3PQ² + 27QS² + 5PQ² + 5QR² / 9
⇒ 8PQ² + 32QR² / 9 = 3PQ² + 5PQ² + 27QS² + 5QR² / 9
⇒ 8PQ² + 32QR² / 9 = 8PQ² + 27QS² + 5QR² / 9
⇒ 8PQ² + 32 * 3QS² / 9 = 8PQ² + 27QS² + 5 * 3QS² / 9 - - [ From ( 2 ) ]
⇒ 8PQ² + 32 * QS² / 3 = 8PQ² + 27QS² + 5 * QS² / 3
⇒ 8PQ² + 32QS² / 3 = 8PQ² + 27QS² + 5QS² / 3
⇒ 8PQ² + 32QS² = 8PQ² + 27QS² + 5QS² - - [ Cancelling 3 from both sides ]
⇒ 8PQ² + 32QS² = 8PQ² + 32QS²
∴ 8PT² = 3PR² + 5PS²