Question

in the given figure,a right triangle PQR,right angled at Q. X and Y are the points on PQ and PR respectively such that PX:PQ=1:2 and QY:YR=2:1. Prove that 9 (PY^2+XR^2)=13PR^2​

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Given that X divides PQ in 2:1

= QX = 2/3*PQ. ....(1)

Similarly, Y divides QR in 2:1

= RY = 2/3*QR ......(2)

In ∆PYQ,

PY^2 = PQ^2 + QY^22

PY^2 = PQ^2 + (2/3*QR)^2

9PY^2 = 9PQ^2 + 4QR^2 .......(3)

In ∆XQR,

XR^2 = XQ^2 + QR^2

XR^2 = (2/3*PQ)^2 + QR^2

XR^2 = 4/9*PQ^2 + QR^2

9XR^2 = 4PQ^2 + 9QR^2. ........(4)

On adding (3)+(4), we get

9(PY^2 + XR^2) = 13(PQ^2 + QR^2)

9(PY^2 + XR^2) = 13PR^2

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