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.

x = 5 would not make PSRQ a rectangle since (5,5,6) is not a pythagorean triple.

x = 5 must be rejected because 5/6 is not equal to 6/10 (contradicting triangle QPT similar to triangle QTR).

Posted by Dennis on 2007-05-01 21:39:09