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 = 28Perimeter=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, alphaα and betaβ, are congruent (alpha=betaα=β). 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^@-alphaγ=180α and delta=180^@-betaδ=180β; since alpha =betaα=β then gamma=deltaγ=δ.
In the present case, gamma=delta=120^@γ=δ=120 and alpha=beta=60^@α=β=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 PMˆQP would be equal to 120^@120 and a line bisecting this angle would be 60^@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+x=60 => x=30^@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=BPBQ=BP
Since PQ=BQ+BP=12PQ=BQ+BP=12 => BQ=BP=6BQ=BP=6
Since AQBT is a rectangle, AT=BQ=6AT=BQ=6

In the triangle AQT
tan A hat Q T=tan (30^@+x)=tan (60^@)="AT"/"AQ"=6/"AQ"tanAˆQT=tan(30+x)=tan(60)=ATAQ=6AQ => "AQ"= 6/sqrt(3)=2*sqrt(3)AQ=63=23

In the triangle AMQ
cos 30^@= "AQ"/"MQ"cos30=AQMQ => 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