Eureka Math Geometry Module 1 Lesson 33 Answer Key

Engage NY Eureka Math Geometry Module 1 Lesson 33 Answer Key

Eureka Math Geometry Module 1 Lesson 33 Review Exercise Answer Key

Review Exercises:

We have covered a great deal of material in Module 1. Our study has included definitions, geometric assumptions, geometric facts, constructions, unknown angle problems and proofs, transformations, and proofs that establish properties we previously took for granted.

In the first list below, we compile all of the geometric assumptions we took for granted as part of our reasoning and proof-writing process. Though these assumptions were only highlights in lessons, these assumptions form the basis from which all other facts can be derived (e.g., the other facts presented in the table). College-level geometry courses often do an in-depth study of the assumptions.

The latter tables review the facts associated with problems covered in Module 1. Abbreviations for the facts are within brackets.

Geometric Assumptions (Mathematicians call these axioms.):

1. (Line) Given any two distinct points, there is exactly one line that contains them.

2. (Plane Separation) Given a line contained in the plane, the points of the plane that do not lie on the line form two sets, called half-planes, such that
a. Each of the sets is convex.
b. If P is a point in one of the sets and Q is a point in the other, then \(\overline{P Q}\) intersects the line.

3. (Distance) To every pair of points A and B there corresponds a real number dist (A, B) ≥ 0, called the distance from A to B, so that
a. dist(A,B) = dist(B,A)
b. dist(A, B) ≥ 0, and dist(A, B) = 0 ⇔ A and B coincide
.
4. (Ruler) Every line has a coordinate system.

5. (Plane) Every plane contains at least three noncollinear points.

6. (Basic Rigid Motions) Basic rigid motions (e.g., rotations, reflections, and translations) have the following properties:
a. Any basic rigid motion preserves lines, rays, and segments. That is, for any basic rigid motion of the plane, the image of a line is a line, the image of a ray is a ray, and the image of a segment is a segment.
b. Any basic rigid motion preserves lengths of segments and angle measures of angles.

7. (180° Protractor) To every ∠AOB, there corresponds a real number m∠AOB, called the degree or measure of the
angle, with the following properties:
a. 0°< m∠AOB < 180°
b. Let \(\overline{O B}\) be a ray on the edge of the half-plane H. For every r such that 0° <r° < 180°, there is exactly one ray \(\overline{O A}\) with A in H such that m∠AOB = r°.
c. If C is a point in the interior of ∠AOB, then m∠AOC + m∠COB = m∠AOB.
d. If two angles ∠BAC and ∠CAD form a linear pair, then they are supplementary (e.g., m∠BAC + m∠CAD = 180°).

8. (Parallel Postulate) Through a given external point, there is at most one line parallel to a given line.

Eureka Math Geometry Module 1 Lesson 33 Review Exercise Answer Key 1

Answer:

Eureka Math Geometry Module 1 Lesson 33 Review Exercise Answer Key 3

Eureka Math Geometry Module 1 Lesson 33 Review Exercise Answer Key 4

The facts and properties in the immediately preceding table relate to angles and triangles. In the table below, we review facts and properties related to parallel lines and transversals.

Eureka Math Geometry Module 1 Lesson 33 Review Exercise Answer Key 2

Answer:

Eureka Math Geometry Module 1 Lesson 33 Review Exercise Answer Key 5

Eureka Math Geometry Module 1 Lesson 33 Problem Set Answer Key

Use any of the assumptions, facts, and/or properties presented in the tables above to find x and y in each figure below. Justify your solutions.

Question 1.
x = ___
y = ___

Eureka Math Geometry Module 1 Lesson 33 Problem Set Answer Key 6

Answer:
x= 52°, y= 56°
m∠AEB is 72° Linear pairs form supplementary angles.
m∠FEB is 56° Linear pairs form supplementary angles.
x = 52° if two parallel lines are cut by a transversal, then the corresponding angles are congruent.
y = 56° Angles in a triangle add up to 180°.

Question 2.
You need to draw an auxiliary line to solve this problem.
x =
y =

Eureka Math Geometry Module 1 Lesson 33 Problem Set Answer Key 7

Answer:
x = 45° y = 45°
∠ABC and ∠DCB are alternate interior angles because \(\overline{A B}\) || \(\overline{C D}\) ii ; x = 45°
Angles x and y are also alternate interior angles because \(\overline{B C}\) || \(\overline{E G}\) y = 45°

Question 3.
x =
y =

Eureka Math Geometry Module 1 Lesson 33 Problem Set Answer Key 8

Answer:
x= 73°, y = 39°
∠HIK and ∠JKI are supplementary because they are same side interior angles and \(\overline{J K}\) || \(\overline{H I}\) therefore, the value of x is 73°. ∠MKL and ∠JKI are vertical angles. So, using the fact that the sum of angles in a triangle is 180°, we find that the value of y is 39°.

Question 4.
Given the labeled diagram at the right, prove that ∠VWX ≅ ∠XYZ.

Eureka Math Geometry Module 1 Lesson 33 Problem Set Answer Key 9

Answer:
∠VWX ≅∠YXW When two parallel lines are cut by a transversal, then alternate interior angles are congruent.
∠XYZ ≅∠YXW When two parallel lines are cut by a transversal, then alternate interior angles are congruent.
∠VWX ≅ ∠XYZ Substitution property of equality

Eureka Math Geometry Module 1 Lesson 33 Exit Ticket Answer Key

Question 1.
Which assumption(s) must be used to prove that vertical angles are congruent?
Answer:
The “protractor postulate” must be used. 1f two angles, BAC and CAD, form a linear pair, then they are supplementary (e.g., m∠BAC + m∠CAD = 180°).

Question 2.
If two lines are cut by a transversal such that corresponding angles are NOT congruent, what must be true? Justify your response.
Answer:
The lines are not parallel. Corresponding angles are congruent if and only If the lines are parallel. The “if and only if” part of this statement requires that, if the angles are NOT congruent, then the lines are NOT parallel.

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