An isosceles trapezoid MNPQ with QP=12 and measure of angle M = 120 degrees has bisectors of angles MQP and NPQ that meet at point T on line MN. What is the perimeter of MNPQ?

1 Answer
Jan 3, 2016

#Perimeter = 28#

Explanation:

First a discussion about the internal angles of the isosceles trapezoid.
Consider Figure 1

Figure generated in my computer using MS Excel

In a isosceles trapezoid, if its equal sides are extended, we get an isosceles triangle whose base is the bigger base of the trapezoid. In a isosceles triangle, the angles with the base, #alpha# and #beta#, are congruent (#alpha=beta#). Since the segments P2P3 and P1P4 are parallel (from the definition of trapezoid), by the Alternate Interior Angles Theorem we can conclude that #gamma=180^@-alpha# and #delta=180^@-beta#; since #alpha =beta# then #gamma=delta#.
In the present case, #gamma=delta=120^@# and #alpha=beta=60^@#.

Discarding impossible case
Consider Figure 2

Figure generated in my computer using MS Excel

Since the problem doesn't present a picture, there would be two possibilities of configuration. Case 1 in which the segment MN forms the trapezoid's smaller base and Case 2 in which the segment QM forms the smaller base.
But the second case is impossible since then the angle #M hat Q P# would be equal to #120^@# and a line bisecting this angle would be #60^@# to the segment QM: in that case this line would be parallel to the segment MN, and wouldn't intercept it, what doesn't satisfy the conditions of the problem. So Case 2 is discarded.

Resolving the problem for Case 1
Consider Figure 3

Figure generated in my computer using MS Excel

From the discussion of the internal angles of the trapezoid and from the conditions of the problem, we know that
#x+x=60^@# => #x=30^@#

Drawing a line from the vertex Q and perpendicular to the segment MN, it intercepts the line in which the segment MN stands in point A

Drawing a line from the point T and perpendicular to the segment MN, it intercepts the line in which the segment PQ stands in point B. Using the rule angle-side-angle, we can see the triangles MQT and NPT are congruents, then, using the rule side-angle-side, we can see that the triangles BQT and BPT are congruent what means that
#BQ=BP#
Since #PQ=BQ+BP=12# => #BQ=BP=6#
Since AQBT is a rectangle, #AT=BQ=6#

In the triangle AQT
#tan A hat Q T=tan (30^@+x)=tan (60^@)="AT"/"AQ"=6/"AQ"# => #"AQ"= 6/sqrt(3)=2*sqrt(3)#

In the triangle AMQ
#cos 30^@= "AQ"/"MQ"# => #MQ=("AQ"/cos 30^@)=(2*cancel(sqrt(3)))/(cancel(sqrt(3))/2)=2*2# => #MQ=4# (obs.: #NP=MQ=4#)

#tan 30^@= "AM"/"AQ"# => #AM="AQ*tan 30^@=2*sqrt(3)*sqrt(3)/3=2*cancel(3)/cancel(3)# => #AM=2#

#MT=AT-AM=6-2=4#
#MN=MT+NT# and since #MT=NT# because triangles MQT and NPT are congruent:
#MN=2*MT=2*4# => #MN=8#

#Perimeter = MN+NP+PQ+MQ=8+4+12+4=28#