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S AND T trisect the side QR of a right triangle PQR. Prove that 8PT^{2}=3PR^{2}+5PS^{2}^{}

S AND T trisect the side QR of a right triangle PQR. Prove that 8PT^{2}=3PR^{2}+5PS^{2}^{}

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

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,

PR^{2 }= PQ^{2} + QR^{2}

PR^{2 }= PQ^{2} +(3x)^{2}

PR^{2 }= PQ^{2} + 9x^{2}

In Right angled triangle PQS,

PS^{2} = PQ^{2} +QS^{2 }

PS^{2} = PQ^{2} +x^{2}

In Right angled triangle PQT,

PT^{2} = PQ^{2 }+ OT^{2}

PT^{2} = PQ^{2 }+ (2x)^{2}

PT^{2} = PQ^{2 }+ 4x^{2}

To prove that, 8PT^{2} =3 PR^{2} + 5PS^{2}

LHS = 8PT^{2} = 8(PQ^{2 }+ 4x^{2}) = 8PQ^{2 }+ 32x^{2}

RHS = 3 PR^{2} + 5PS^{2} = 3(PQ^{2} + 9x^{2}) – 5 (PQ^{2} + x^{2})= 3PQ^{2} + 27x^{2} + 5PQ^{2} + 5x^{2} = 8PQ^{2 }+ 32x^{2}

LHS = RHS

Hence, proved

^{ }

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

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