In given figure, is a right triangle PQR, right angled at Q. X and Y are the points on PQ and QR such that PX : XQ=

1 : 2 and QY : YR= 2 : 1. Prove that 9(PY 2+ XR 2 )= 13PR 2 .

In triangle QXR,

   XR^2 = (2x)^2 + (3x)^2

             = 13x^2

In triangle PQY,

   PY^2 = 13x^2

IN triangle PQR ,

PR^2= 18 x^2

now substitute the value of both in the equation

     9( 13 x^2 + 13x^2 ) = 13 * 18x^2

234 x^2 = 234 x^2

since LHS =RHS 

it is proved....

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In triangle QXR,

XR^2 = (2x)^2 + (3x)^2

= 13x^2

In triangle PQY,

PY^2 = 13x^2

IN triangle PQR ,

PR^2= 18 x^2

now substitute the value of both in the equation

9( 13 x^2 + 13x^2 ) = 13 * 18x^2

234 x^2 = 234 x^2

since LHS =RHS

  • -27
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