S AND T trisect the side QR of a right triangle PQR. Prove that 8PT2=3PR2+5PS2

Asked by apporv1999 | 7th Jul, 2014, 09:32: PM

Expert Answer:

Since, S and T trisect the side QR

QS = TS = RT

Let’s assume QS = TS = RT = x

QR = 3x and QT = 2x

In Right angled triangle PQR,

PR2 = PQ2 + QR2

PR2 = PQ2 +(3x)2

PR2 = PQ2 + 9x2

In Right angled triangle PQS,

PS2 = PQ2 +QS2

PS2 = PQ2 +x2

In Right angled triangle PQT,

PT2 = PQ2 + OT2

PT2 = PQ2 + (2x)2

PT2 = PQ2 + 4x2

To prove that, 8PT2 =3 PR2 + 5PS2

LHS = 8PT2 = 8(PQ2 + 4x2) = 8PQ2 + 32x2

RHS = 3 PR2 + 5PS2 = 3(PQ2 + 9x2) – 5 (PQ2 + x2)= 3PQ2 + 27x2 + 5PQ2 + 5x2 = 8PQ2 + 32x2

LHS = RHS

Hence, proved

Answered by Anuja Salunke | 8th Jul, 2014, 11:14: AM