Chapter 5.5, Problem 34E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# Note: Exercises preceded by an asterisk are of a more challenging nature. Given: Isosceles trapezoid M N P Q with Q P = 12 and m ∠ M = 120 ° ; the bisectors of ∠ s M Q P and Find: The perimeter of M N P Q

To determine

To find:

The perimeter of the isosceles trapezoid MNPQ with QP=12 and mM=120°; the bisecfors of SMQP and NPQ meet at point T on MN¯.

Explanation

Definition:

The perimeter of any closed figure is given by the sum of all the lengths of its boundary.

Calculation:

Given,

An isosceles trapezoid MNPQ with QP=12

mâˆ M=120Â°

The bisectors of âˆ MQP and âˆ NQP meet at point T on MNÂ¯

In the above given isosceles trapezoid, we have the following congruent measurement.

A pair of the non-parallel sides of any isosceles trapezoid is congruent.

MQÂ¯â‰…NPÂ¯

Each pair of base angles are congruent.

mâˆ QMNâ‰…mâˆ PNM

Thus,

mâˆ QMN=120Â°=mâˆ PNM

Also

mâˆ MQPâ‰…mâˆ NPQ

Since, the above congruent angles mâˆ MQP and mâˆ NPQ are bisected by the line segments QTÂ¯ and PTÂ¯, all the four angles formed by bisector are congruent.

mâˆ MQTâ‰…mâˆ TQPâ‰…mâˆ NPTâ‰…mâˆ TPQ

In a trapezoid sum of the adjacent angles which lie on the opposite bases is 180Â°

mâˆ MQP+mâˆ QMN=180Â°

mâˆ MQP+120Â°=180Â°

mâˆ MQP+120Â°+-120Â°=180Â°-120Â°

mâˆ MQP+120+-120=180Â°-120Â°

mâˆ MQP=60Â°

Thus, mâˆ MQP=mâˆ NPQ=60Â°

Also,

mâˆ MQT=mâˆ TQP=mâˆ NPT=mâˆ TPQ=12Â·mâˆ MQP (or) 12Â·mâˆ NPQ

mâˆ MQT=mâˆ TQP=mâˆ NPT=mâˆ TPQ=30Â°

By the property that the sum of all the three inferior angle of any triangle is 180Â°, we have

In âˆ†PQT

mâˆ P=30Â°

mâˆ Q=30Â° and

Thus,

mâˆ T=120Â°

Also, in âˆ†MTQ

mâˆ M=120Â°

mâˆ Q=30Â°, and

Thus,

mâˆ T=30Â°

Further, in âˆ†TNP

mâˆ P=30Â°,

mâˆ N=120Â°, and

Thus,

mâˆ T=30Â°

Now consider the âˆ†TPQ and draw TSÂ¯âŠ¥PQ.

Since, TPQ is an isosceles triangle with sides TQÂ¯ and TPÂ¯ congruent, the line segment TSÂ¯ bisects the angle âˆ T of the âˆ†TPQ and also the side PQÂ¯.

Thus, we have two right triangle âˆ†TSQ and âˆ†TSP of 30Â°-60Â°-90Â° type.

In âˆ†TSQ

TQ= Hypotenuse (Opposite to 90Â°)

QS= Linger leg (Opposite to 60Â°)

TS= shorter leg = 6 (Opposite to 30Â°)

By the application of 30-60-90 theorem

30Â°-60Â°-90Â° theorem

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