Question 1193785: Given: isosceles trapezoid MNPQ with QP = 18 and m∠M = 120°
the bisectors of ∠s MQP and NPQ meet at point T on MN
Find: the perimeter of MNPQ
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
Here is a diagram with some lines added...
QT bisects angle MQP and PT bisects angle NPQ; angles MQP and NPQ are each 60 degrees, so angles PQT and TPQ are each 30 degrees. So triangle PTQ is isosceles, with base angles 30 degrees.
TU is the altitude of triangle PTQ; it divides triangle PTQ into congruent 30-60-90 right triangles PTU and QTU. Since QP is 18, PU and QU are each 9.
PU with length 9 is the long leg of a 30-60-90 right triangle, so TU is 9/sqrt(3) = 3*sqrt(3), and PT is twice TU or 6*sqrt(3).
Angle MNP is 120 degrees, so angle NPQ is 60 degrees; PT bisects angle NPQ, so angle NPT is 30 degrees; since angle MNP is 120 degrees, triangle TNP is also isosceles with base angles 30 degrees.
Altitude NV of triangle TNP divides triangle TNP into two congruent 30-60-90 right triangles, each with long leg equal to half of TP, which is 3*sqrt(3). So in triangle TVN NV is 3 and TN is 6.
Then triangles QMT and PNT are both isosceles with legs of length 6, and the perimeter of trapezoid MNPQ is 6+6+9+9+6+6 = 42.
ANSWER: 42
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