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| A Parallelogram Puzzle (Posted on 2007-05-01) |
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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.
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Submitted by K Sengupta
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Solution:
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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.
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