S and T trisect the side QR of a right triangle PQR, prove that : 8PT^

1.	S and T trisect the side QR of a right triangle PQR, prove that : 8PT^2 = 3PR^2 + 5PS^2
Answer» Given that a right angle\xa0{tex}\\triangle PQR{/tex}\xa0in which\xa0{tex}\\angle P Q R = 90 ^ { \\circ }{/tex}And S and T are the points of trisection of QR. Thus, QS = ST = TRTo Prove: 8PT2 = 3PR2 + 5PS2.Proof: Let QS = ST = TR = x unitsThen, QS = x units, QT = 2x units and QR = 3x units.\xa0From right angle triangles PQS, PQT and PQR where\xa0{tex}\\angle Q= 90^°{/tex} in all, by using Pythagoras Theorem, we havePS2 = PQ2 + QS2 ... (i)\xa0PT2 = PQ2 + QT2... (ii)\xa0and PR2 = PQ2 + QR2\xa0\u200b\u200b\u200b\u200b\u200b\u200b... (iii)Multiplying (iii) by 3, (i) by 5 and (ii) by 8Adding thus obtained (iii) and (i) and Subtracting (ii) from it, we get{tex}\\therefore{/tex}\xa03PR2 + 5PS2 - 8PT2=3(PQ2 + QR2) + 5(PQ2 + QS2) - 8(PQ2 + QT2)= 3QR2 + 5QS2 - 8QT2{tex}= 3 \\times ( 3 x ) ^ { 2 } + 5 ( x ) ^ { 2 } - 8 \\times ( 2 x ) ^ { 2 }{/tex}\xa0[{tex}\\because{/tex}\xa0QR = 3x, QS = x and QT = 2x](27x2\xa0+ 5x2\xa0- 32x2\xa0) = 0Thus, 3PR2 + 5PS2 - 8PT2 = 0Hence, 8PT2 = 3PR2 + 5PS2Hence Proved\xa0\xa0

S and T trisect the side QR of a right triangle PQR, prove that : 8PT^2 = 3PR^2 + 5PS^2