triangle QPR is a right angled triangle, right angled at Q and the points S and T trisect the side QR.prove that 8PT2 =3PR2 +5PS2.
Hi!
Here is the answer to your question.
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Since, S and T trisect the side QR.
∴ QS = ST = TR
Let QS = ST = TR = k (say)
∴ QT = 2k and QR = 3k
In right ∆PQS,
PS2 = PQ2 + QS2 (Pythagoras Theorem)
⇒ PS2 = PQ2 + k2
In right ∆PQT,
PT2 = PQ2 + QT2
⇒ PT2 = PQ2 + (2k)2
⇒ PT2 = PQ2 + 4k2
In right ∆PQR,
PR2 = PQ2 + QR2
⇒ PR2 = PQ2 + (3k)2
⇒ PR2 = PQ2 + 9k2
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Cheers!