The perpendicular from vertex P of the triangle PQR meets QR at the point S. A point T is located on PR such that QT=TR=RS=1.

What is the length of PR, given that Angle QPR=90^o?

Applying the Pathagorean theorem to right triangles:

  PSQ: |PS|^2 + |SQ|^2 = |PQ|^2

                  or

       |PS|^2 + (|QR| - 1)^2 = |PQ|^2        (1)

  PSR: |PS|^2 + |SR|^2 = |PR|^2

                  or

       |PS|^2 + 1 = |PR|^2                   (2)

  QPR: |QP|^2 + |PR|^2 = |QR|^2

                  or

       |PQ|^2 + |PR|^2 = |QR|^2              (3)

  QPT: |QP|^2 + |PT|^2 = |QT|^2

                  or

       |PQ|^2 + (|PR| - 1)^2 = 1             (4)

Solving the equations (1)-(4) for |PR| we get

  |PR| = cube-root(2) 

Posted by Bractals on 2007-06-17 13:26:44