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=90o?
Applying the Pathagorean theorem to right triangles:
PSQ: |PS|^2 + |SQ|^2 = |PQ|^2
|PS|^2 + (|QR| - 1)^2 = |PQ|^2 (1)
PSR: |PS|^2 + |SR|^2 = |PR|^2
|PS|^2 + 1 = |PR|^2 (2)
QPR: |QP|^2 + |PR|^2 = |QR|^2
|PQ|^2 + |PR|^2 = |QR|^2 (3)
QPT: |QP|^2 + |PT|^2 = |QT|^2
|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