Question

A triangle PQR is right angeled at Q and points S and T trisect side QR.prove that 8PT^2= 3PR^2+5PS^2.
if any body know how to solve it then please help me out with this

Answer 1

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Let QR = 3a
Given QR is trisected at S and T, hence QS = ST = TR = a
In right DPQR, PR2 = PQ2 + QR2
⇒ PR2 = PQ2 + (3a)2
⇒ PR2 = PQ2 + 9a2   à (1)
In right DPQT, PT2 = PQ2 + QT2
⇒ PT2 = PQ2 + (2a)2
⇒ PT2 = PQ2 + 4a2   à (2)
In right DPQS, PS2 = PQ2 + QS2
⇒ PS2 = PQ2 + a2   à (3)
Consider, 3PR2 + 5PS2 = 3[PQ2 + 9a2] + 5[PQ2 + a2]     
                                            = 3PQ2 + 27a2 + 5PQ2 + 5a2        
                                            = 8PQ2 + 32a2
                                            = 8[PQ2 + 4a2]
                                            = 8[PQ2 + (2a)2]
                                            = 8[PQ2 + QT2]
                                            = 8PT2   [Since PT2 = PQ2 + QT2]
∴ 3PR2 + 5PS2 = 8PT2





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Answer 2

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