## Engage NY Eureka Math Geometry Module 1 Lesson 13 Answer Key

### Eureka Math Geometry Module 1 Lesson 13 Exploratory Challenge Answer Key

Exploratory Challenge

You need a pair of scissors and a ruler.

Cut out the 75° angle on the right, and use it as a guide to rotate the figure below 75° counterclockwise around the given center of rotation (Point P).

→ Place the vertex of the 75° angle at point P.

→ Line up one ray of the 75° angle with vertex A on the figure. Carefully measure the length from point P to vertex A.

→ Measure that same distance along the other ray of the reference angle, and mark the location of your new point, A’.

→ Repeat these steps for each vertex of the figure, labeling the new vertices as you find them.

→ Connect the six segments that form the sides of your rotated image.

Discussion

In Grade 8, we spent time developing an understanding of what happens in the application of a rotation by participating in hands-on lessons. Now, we can define rotation precisely.

The notion of the entire plane being subject to the transformation is new and should be reinforced. However, note that neither the figure nor the plane is actually moved in the transformation. The image represents the output after applying the transformation rule to the input points.

In the definition below, students may benefit from some discussion using a visual of a clock. Discuss the intuitive notion of directions on a clock before the more formal definition.

First, we need to talk about the direction of the rotation. If you stand up and spin in place, you can either spin to your left or spin to your right. This spinning to your left or right can be rephrased using what we know about analog clocks: spinning to your left is spinning in a counterclockwise direction, and spinning to your right is spinning in a clockwise direction. We need to have the same sort of notion for rotating figures in the plane. It turns out that there is a way to always choose a counterclockwise half-plane for any ray: The counterclockwise half-plane of \(\overrightarrow{C P}\) is the half-plane of \(\overrightarrow{C P}\) that lies to the left as you move along \(\overrightarrow{C P}\) in the direction from C to P. (The clockwise half-plane is then the half-plane that lies to the right as you move along \(\overrightarrow{C P}\) in the direction from C to P.) We use this idea to state the definition of rotation.

For 0°<θ°<180°, the rotation of θ degrees around the center C is the transformation R_{C,θ} of the plane defined as follows:

1. For the center point C, R_{C,θ} (C)=C, and

2. For any other point P, R_{C,θ}(P) is the point Q that lies in the counterclockwise half-plane of \(\overrightarrow{C P}\), such that CQ=CP and m∠PCQ=θ°.

A rotation of 0 degrees around the center C is the identity transformation (i.e., for all points A in the plane, it is the rotation defined by the equation R_{C,0}(A)=A).

A rotation of 180° around the center C is the composition of two rotations of 90° around the center C. It is also the transformation that maps every point P (other than C) to the other endpoint of the diameter of a circle with center C and radius CP.

→ A rotation leaves the center point C fixed. R_{C,θ}(C)=C states exactly that. The rotation function R with center point C that moves everything else in the plane θ° leaves only the center point itself unmoved.

→ Any other point P in the plane moves the exact same degree arc along the circle defined by the center of rotation and the angle θ°.

→ Then R_{C,θ}(P) is the point Q that lies in the counterclockwise half-plane of ray \(\overrightarrow{C P}\) such that CQ=CP and such that m∠PCQ=θ°. Visually, you can imagine rotating the point P in a counterclockwise arc around a circle with center C and radius CP to find the point Q.

→ All positive angle measures θ assume a counterclockwise motion; if citing a clockwise rotation, the answer should be labeled with CW.

A composition of two rotations applied to a point is the image obtained by applying the second rotation to the image of the first rotation of the point. In mathematical notation, the image of a point A after a composition of two rotations of 90° around the center C can be described by the point R_{C,90} (R_{C,90}(A)). The notation reads, “Apply R_{C,90} to the point R_{C,90}(A).” So, we lose nothing by defining R_{C,180}(A) to be that image. Then, R_{C,180}(A) = R_{C,90}(R_{C,90}(A)) for all points A in the plane.

In fact, we can generalize this idea to define a rotation by any positive degree: For θ°>180°, a rotation of θ° around the center C is any composition of three or more rotations, such that each rotation is less than or equal to a 90° rotation and whose angle measures sum to θ°. For example, a rotation of 240° is equal to the composition of three rotations by 80° about the same center, the composition of five rotations by 50°, 50°, 50°, 50°, and 40° about the same center, or the composition of 240 rotations by 1° about the same center.

Notice that we have been assuming that all rotations rotate in the counterclockwise direction. However, the inverse rotation (the rotation that undoes a given rotation) can be thought of as rotating in the clockwise direction. For example, rotate a point A by 30° around another point C to get the image R_{C,30}(A). We can undo that rotation by rotating by 30° in the clockwise direction around the same center C. Fortunately, we have an easy way to describe a rotation in the clockwise direction. If all positive degree rotations are in the counterclockwise direction, then we can define a negative degree rotation as a rotation in the clockwise direction (using the clockwise half-plane instead of the counterclockwise half-plane). Thus, R_{C,-30} is a 30° rotation in the clockwise direction around the center C. Since a composition of two rotations around the same center is just the sum of the degrees of each rotation, we see that

R_{C,-30} (R_{C,30}(A))=R_{C,0}(A)=A,

for all points A in the plane. Thus, we have defined how to perform a rotation for any number of degrees—positive or negative.

As this is our first foray into close work with rigid motions, we emphasize an important fact about rotations. Rotations are one kind of rigid motion or transformation of the plane (a function that assigns to each point P of the plane a unique point F(P)) that preserves lengths of segments and measures of angles. Recall that Grade 8 investigations involved manipulatives that modeled rigid motions (e.g., transparencies) because you could actually see that a figure was not altered, as far as length or angle was concerned. It is important to hold onto this idea while studying all of the rigid motions.

Constructing rotations precisely can be challenging. Fortunately, computer software is readily available to help you create transformations easily. Geometry software (such as Geogebra) allows you to create plane figures and rotate them a given number of degrees around a specified center of rotation. The figures in the exercises were rotated using Geogebra. Determine the angle and direction of rotation that carries each pre-image onto its (dashed-line) image. Assume both angles of rotation are positive. The center of rotation for Exercise 1 is point D and for Figure 2 is point E.

(Remind students that identifying either CCW or CW degree measures is acceptable; if performing a 30° rotation in the clockwise direction, students can label their answers as 30° CW or -30°.)

### Eureka Math Geometry Module 1 Lesson 13 Exercise Answer Key

Exercises 1–3

Exercise 1.

Answer:

285° or 75° CW or -75°

To determine the angle of rotation, you measure the angle formed by connecting corresponding vertices to the center point of rotation. In Exercise 1, measure ∠AD’A’. What happened to ∠D? Can you see that D is the center of rotation, therefore, mapping D’ onto itself? Before leaving Exercise 1, try drawing ∠BD’B’. Do you get the same angle measure? What about ∠CD’C’?

Try finding the angle and direction of rotation for Exercise 2 on your own.

Remind students that the solid-line figure is the pre-image, and the dotted-line figure is the image.

Exercise 2.

Answer:

60° or 300° CW or -300°

Did you draw ∠DED’ or ∠CEC’?

Now that you can find the angle of rotation, let’s move on to finding the center of rotation. Follow the directions below to locate the center of rotation, taking the figure at the top right to its image at the bottom left.

Exercise 3.

a. Draw a segment connecting points A and A’.

b. Using a compass and straightedge, find the perpendicular bisector of this segment.

c. Draw a segment connecting points B and B’.

d. Find the perpendicular bisector of this segment.

e. The point of intersection of the two perpendicular bisectors is the center of rotation. Label this point P.

Justify your construction by measuring ∠APA’ and ∠BPB’. Did you obtain the same measure?

Answer:

Yes

This method works because a chord of a circle is a segment joining two points on a circle. The endpoints of the chord are equidistant from the center of the circle. The perpendicular bisector of a chord (being the set of all points equidistant from the endpoints) includes the center of the circle. Since students may have had limited experience studying circles, they may have difficulty understanding why this works. Consider pointing out or sketching the circle that has the center of rotation as its center, for each of the examples, to supply some justification for why it works.

Exercises 4–5

Find the centers of rotation and angles of rotation for Exercises 4 and 5.

Exercise 4.

Answer:

Either 244° or 116° CW or -116°

Exercise 5.

Answer:

90° or 270° CW or -270°

### Eureka Math Geometry Module 1 Lesson 13 Problem Set Answer Key

Question 1.

Rotate triangle ABC 60° around point F using a compass and straightedge only.

Answer:

Question 2.

Rotate quadrilateral ABCD 120° around point E using a straightedge and protractor.

Answer:

Question 3.

On your paper, construct a 45° angle using a compass and straightedge. Rotate the angle 180° around its vertex, again using only a compass and straightedge. What figure have you formed, and what are its angles called?

Answer:

The figure formed is an X, and the angles are called vertical angles.

Question 4.

Draw a triangle with angles 90°, 60°, and 30° using only a compass and straightedge. Locate the midpoint of the longest side using your compass. Rotate the triangle 180° around the midpoint of the longest side. What figure have you formed?

Answer:

The figure formed is a rectangle.

Question 5.

On your paper, construct an equilateral triangle. Locate the midpoint of one side using your compass. Rotate the triangle 180° around this midpoint. What figure have you formed?

Answer:

The figure formed is a rhombus.

Question 6.

Use either your own initials (typed using WordArt in Microsoft Word) or the initials provided below. If you create your own WordArt initials, copy, paste, and rotate to create a design similar to the one below. Find the center of rotation and the angle of rotation for your rotation design.

Answer:

### Eureka Math Geometry Module 1 Lesson 13 Exit Ticket Answer Key

Find the center of rotation and the angle of rotation for the transformation below that carries A onto B.

Answer:

270° or 90° CW or -90°