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A Parallelogram Puzzle (Posted on 2007-05-01) Difficulty: 2 of 5
Consider the parallelogram PQRS with PS parallel to QR and PQ parallel to SR. The bisector of the angle PQR intersects PS at T.

Determine PQ, given that TS = 5, QT = 6 and RT = 6.

  Submitted by K Sengupta    
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Solution: (Hide)
Let PQ = PT = y; QR = z

Now, Angle PQT = Angle TQR, since QT bisects Angle PQR
Angle TQR = Angle QRT, since Triangle QRT is isosceles as QT=TR = 6
Angle QRT = Angle RTS, as QR is parallel to PS
Angle QTP = Angle TQR, as QR is parallel to PS

So, QTP is also an isosceles triangle which is similar to Triangle QRT, and:
z = 5+y, since QR = PS

From similar triangles BAP and PBC, it follows that:
QR/QT = QT/PT
Or, z/6 = 6/y
Or, 36 = yz
Or, y^2 + 5y -36 = 0, as z= 5+y
Or, y = 4, ignoring the negative root which is inadmissible.

Hence the required length of the side PQ is 4.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Harder solutionDennis2007-05-01 21:39:09
SolutionSolutionhoodat2007-05-01 15:57:46
SolutionSolutionBractals2007-05-01 12:50:29
Harder solutionJer2007-05-01 12:13:28
Solutiona solutionDennis2007-05-01 10:14:00
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