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| Take Unit Length, Get PR (Posted on 2007-06-17) |
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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?
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Submitted by K Sengupta
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Rating: 3.0000 (1 votes)
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Solution:
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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)
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