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 Take Unit Length, Get PR (Posted on 2007-06-17)
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?

 Submitted by K Sengupta Rating: 3.0000 (1 votes) Solution: (Hide) Let the lengths of the three sides QR, PR and PQ be respectively denoted by a, b and c. Since PS is perpendicular to QR, it follows that the triangles PQR and PSR are similar, giving: PR/ QR = RS/ PR Or, RS = b^2/a = 1, giving: a = b^2……..(i) Also from Triangle QPT, w observe that: QT^2 = PQ^2 + PT^2 = PQ^2 + (PR – TR)^2 Or, 1 = c^2 + (b-1)^2 But, we know that: c^2 = a^2 – b^2 Therefore: 1 = c^2 + (b-1)^2 = a^2 – b^2 +(b-1)^2 = a^2 – 2b + 1 = b^4 – 2b + 1 This yields, b^4 – 2b = 0, so that: b = 3√(2) Consequently, the required length of PR is 3√(2)

 Subject Author Date Solution (other way) Praneeth Yalavarthi 2007-07-10 07:13:14 Solution Bractals 2007-06-17 13:26:44

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