Question

please answer Q21 i will mark brainliest​

Answer 1

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