Question

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

Answer 1

Answer:

Step-by-step explanation:

Let 

QS = a

Thus

QT = 2a

and QR = 3a

In triangle PQR

PR^2 = PQ^2 + QR^2

PR^2 = PQ^2 + 3a^2

PR^2 = PQ^2 + 9a^2         Eqn 1

In triangle PQT

PT^2 = PQ^2 + 4a^2          Eqn 2

In Triangle PQS

PS^2 = PQ^2 + a^2            Eqn 3

Thus 

3 PR^2 + 5 PS^2 = 3 PQ^2 + 27a^2 + 5 PQ^2 + 5a^2

                           = 8 [ PQ^2 + 4a^2 ]

                            = 8 PT^2

Hence proved

Answer 2

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