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MATH HIGH SCHOOL

CONGRUENCE AND RIGID MOTION

EXERCISES

Copyright © 2015 Pearson Education, Inc. 2

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High School: Congruence and Rigid Motion

CONTENTS EXERCISES

EXERCISES

LESSON 1: RIGID MOTION �� 5

LESSON 2: REFLECTIONS �� 7

LESSON 3: ROTATIONS �� 11

LESSON 4: TRANSLATIONS �� 17

LESSON 5: BEING PRECISE �� 21

LESSON 6: COMPOSITE TRANSFORMATIONS �� 25

LESSON 7: PUTTING IT TOGETHER 1 �� 31

LESSON 11: DEFINING CONGRUENCE �� 33

LESSON 12: TRIANGLE CONGRUENCE CRITERIA �� 37

LESSON 13: USING CONGRUENCE CRITERIA �� 41

LESSON 14: SYMMETRIES OF POLYGONS �� 45

LESSON 15: PUTTING IT TOGETHER 2—DAY 1 �� 49




CONTENTS EXERCISES

ANSWERS

LESSON 2: REFLECTIONS �� 57

LESSON 3: ROTATIONS �� 60

LESSON 4: TRANSLATIONS �� 63

LESSON 5: BEING PRECISE �� 66

LESSON 6: COMPOSITE TRANSFORMATIONS �� 68

LESSON 7: PUTTING IT TOGETHER 1 �� 71

LESSON 11: DEFINING CONGRUENCE �� 73

LESSON 12: TRIANGLE CONGRUENCE CRITERIA �� 76

LESSON 13: USING CONGRUENCE CRITERIA �� 78

LESSON 14: SYMMETRIES OF POLYGONS �� 80





EXERCISESLESSON 1: RIGID MOTION

1. Write what you already know about geometric transformations—reflections, rotations, and translations.

Share your summary with a classmate.

Did you write about the same concepts?

2. Write what you would like to learn about geometric transformations.

Share your ideas with a classmate.

3. Write a goal stating what you plan to accomplish in this unit.

Write your goals for all units in the same place so you can review your past goals as you write new goals for this unit.

4. Based on your previous work in math, write three things that you will do during this unit to increase your success. Write your strategies for all units in the same place so you can review your past strategies as you write new strategies for this unit.

For example, consider ways you will participate in classroom discussions, your study habits, how you will organize your time, what you will do when you have a question, and so on.



EXERCISES

EXERCISES

1. Draw the reflection of this image across the dashed line.



LESSON 2: REFLECTIONS



EXERCISESLESSON 2: REFLECTIONS

4. This figure ABCD is reflected across the line y = –1. Draw a figure in the grid to represent the reflected figure. Label it A'B'C'D'.

–4

–6

–8

–6 2 4 6 8 10–2–4–10 –8

–10

2

–2

4

6

8

10

x

y

C

DB

A

–444

–666

88

5. What kind of triangle would you get if you reflected the slanted line segment across the vertical line and then connected the endpoints of the slanted lines?

A Scalene triangle

B Obtuse triangle

C Isosceles triangle

D Equilateral triangle




6. Which of these graphs shows the reflection of figure C across the line of reflection AB?

A

A

B

C

B

A

B

C

C

A

B

C

D

A

B

C

7. a. Draw the line of reflection for this reflection.

–1

–2

–3

–4

–2–3–4 4–5–6 5 6 72 31

3

2

4

1

5

6

7

8

9

x

y

444 555 666222 333111

333

444555

666

777

b. Write the equation of the line of reflection.




8. Draw the reflection of figure ABC across the line of reflection y = 0.

2

4

6

8

8642–2–4–6–8

–2

–4

–6

–8

x

y

444

B

C

A

Challenge Problem

9. Look at the function shown. Reflect all negative values (all values for which y < 0) across the x-axis.

8 9 10–1–2–3–4–5–6–7–8–9–10

–2–3–4–5–6–7–8–9–10 6 74 52 31

32

4

1

56789

10

x

y



EXERCISES

EXERCISES

1. Which of these graphs shows a figure rotated 120º about the origin starting with figure RSTUV? Remember that a positive rotation is in the counterclockwise direction.

A

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–8

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–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

R S

TU

V

R'

S'T'

U'V'

B

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

R S

TU

V

R'

S'

T'

U'

V'

C

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

R S

TU

VR'S'

T' U'

V'

D

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

R S

TU

V

R'

S'T'

U'V'

LESSON 3: ROTATIONS



EXERCISES

2. Rotate this figure 90º about the origin.

–2

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–6

–2–4–6 642

2

4

6

x

y

A

B

C

3. Figure ABCD is rotated 180º about the origin. Draw a figure in the grid to represent the rotated figure. Label it A'B'C'D'.

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

DC

AB

LESSON 3: ROTATIONS



EXERCISESLESSON 3: ROTATIONS

4. ∆ABC is rotated 270º about the origin. Draw a figure in the grid to represent the translated figure. Label it A'B'C'.

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

C

AB22

5. Rotate figure ABC 270º about the origin.

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–2–4–6 642

2

4

6

x

y

A

B

C



EXERCISESLESSON 3: ROTATIONS

6. Look at the triangles in the coordinate plane.

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2 4 6 8–8 –6 –4 –2

2

4

6

8

x

y

I

J

KI'

J'

K'

Complete the sentence by selecting the correct choice.

The transformation shown is a rotation of _____ about the origin.

A 90° B 180° C 270° D 360°

7. Select all statements that are true about the rotation of ST about the fixed point P.

P

S' S

T'

T

A The length of S'T' could be different from the length of ST.

B To find the angle of rotation you could measure the angle between PS and PS'.

C The angle of rotation is positive.

D The rotation does not change the size of the line.

E When you rotate ST, it maps each point of ST directly onto S'T', so the mapping is a function.



EXERCISES

8. Rotate figure ABCD 270º about the point (2, 1).

–2

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–6

–8

2 4 6 8–8 –6 –4 –2

2

4

6

8

x

y

22

44

A

B C

D

9. Rosa was asked to describe this transformation.

–1–2–3–4–5–6

–2–3–4–5–6–7 6 74 52 31

32

4

1

56

x

y

A

CB

A'

C'

B'

Here is Rosa’s response.

It looks like ∆ABC is rotated 90o about point (0, 0).

Is Rosa’s conclusion correct? Justify why you think she is correct or incorrect.

LESSON 3: ROTATIONS



EXERCISES

Challenge Problem

10. a. Draw a quadrilateral on a coordinate plane and label the axis.

b. Draw its mapping after it is rotated 90º clockwise about the point (3, –2).

c. Then reflected across the x-axis.

LESSON 3: ROTATIONS



EXERCISESLESSON 4: TRANSLATIONS

EXERCISES

1. What single transformation does the graph show?

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

A

A'

C' B

B' C

A Translation

B Rotation

C Reflection

D This is not a transformation.

2. Figure ABCD is translated x – 4, y – 3. Draw a figure in the grid to represent the translated figure. Label it A'B'C'D'.

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

DC

BA




3. A certain translation T involves every x-value travelling a distance of h and every y-value travelling three times the distance that the x-values travel.

Which of these functions correctly represents the translation T?

A T(x, y) = (x + h, 3y)

B T(x, y) = (3x, y + h)

C T(x, y) = (x + 3h, y + h)

D T(x, y)= (x + h, y + 3h)

4. Describe a translation that will shift figure ABCD entirely into quadrant IV.

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–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A

B

C

D




5. Translate figure ABC using the following translation: T(x, y) = (x – 4, y + 3). Draw a figure in the grid to represent the translated figure. Label it A'B'C'.

–2

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–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A

B

C

6. Select all statements that are true about the translation of ∠ABC to ∠A'B'C'.

B

C

A

B'

C'

A'

A The measure of ∠ABC could be different from the measure of ∠A'B'C'.

B The length of the sides of ∠ABC could be different from the length of the sides of ∠A'B'C'.

C ∠ABC is congruent to ∠A'B'C'.

D AB is parallel to A'B'.

E Translating ∠ABC maps each point of ∠ABC directly onto ∠A'B'C', so the mapping is a function.



EXERCISES

7. Describe the translation that is shown.

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–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A

A'

D'C'

B'

D C

B

8. Translate ∆ABC 2 units in the positive x-direction and 3 units in the negative y-direction.

–2

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–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

22222A

B

C

Challenge Problem

9. Create your own figure with at least 4 points. Then translate it as follows: T(x, y) = (x – 8, y + 4).

LESSON 4: TRANSLATIONS



EXERCISESLESSON 5: BEING PRECISE

EXERCISES

1.

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–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–4444–4

444

666

–2222–222–22–2 22222

222

What type of transformation is shown in this graph?

A Reflection

B Rotation

C Translation

D Vertical stretch

2. Reflections, rotations, and translations are called rigid motions because they have certain characteristics. Which of these characteristics define a rigid motion transformation? There may be more than one characteristic.

A Rigid motions expand figures.

B Rigid motions preserve all angle measures and segment lengths.

C Rigid motions do not alter the shape of the figure.

D Rigid motions distort the original image.

E Rigid motions produce two congruent figures.





–2

–4

–6

–8

–10

–2–4 6 8 10 12 14 16 18 20 22 2442

2

4

6

8

10

x

y

22222 88888 101010 12212212 14144 161616 181818 202202206666644444

A Reflection B Rotation C Translation D Horizontal stretch


–2

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–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A Rotation

B Dilation

C Horizontal stretch

D Vertical stretch




5. Figure A' is the image after a translation (a rigid motion) of figure A 6 units to the right.

Figure A'' is the image after a dilation (a non-rigid motion) of figure A centered at the origin with a scale factor of 2.

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–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–44–4

–22–244 444222

444

66

666 84444–22–2

222

222

A''

A'A

What observations can you make about figure A' and figure A'' compared to figure A?


–2

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–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–22–2 222

222

444

–44–4

–66–6

–888

–22–2

A''

A

A Reflection B Translation C Vertical stretch D Horizontal stretch




7. Write each type of transformation in the correct column to identify whether it is a rigid motion or a non-rigid motion.

Dilation

Horizontal stretch

Rotation

Reflection

Translation

Vertical stretch

Rigid Motion Transformation Non-Rigid Motion Transformation

Challenge Problem

8. For the figure shown, perform a transformation such that each x-value is transformed to 3x + 2 and each y-value is transformed to 4y – 3.

Then describe what type of motion this transformation describes.

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–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

12

x

y

22222–22–2

A



EXERCISESLESSON 6: COMPOSITE TRANSFORMATIONS

EXERCISES

1.

–1–2–3–4–5–6

–2–3–4–5–6–7 6 74 52 31

32

4

1

56

x

y

Reflecting the figure across the x-axis and rotating the figure 180º about the point (2, 0) will result in an image that is ________ the original figure.

A identical to B a rotation from

C a reflection from D a translation from

2. Suppose this figure is rotated 90º about the origin and then reflected across the y-axis. How far will the image of point A be from its original location, (5, 5)?

–1–2–3–4–5–6–7

–2–3–4–5–6–7 6 74 52 31

32

4

1

567

x

y

A (5, 5)

A 0 units B 5 units C 125 units D 10 units




3. Perform the following transformations on figure ABCD, and then graph the final transformed figure.

T(3, 4)

Translate the quadrilateral to the right 3 units and up 4 units.

R(0, 0), 180°

Then rotate the translated figure 180º about the origin to get the final transformed figure.

–2

–4

–6

–8

–10

–2–4–6–8–10–12 6 8 10 1242

2

4

6

8

10

x

y

666

A

B

C

D

Have a classmate check your work.

4. Perform a series of two transformations that moves figure ABC completely into quadrant II.

–2

–4

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

E

F

G

Have a classmate check your work.




5. Figure ABC has been translated 6 units to the right (A'B'C') and then rotated (A"B"C"). Find another sequence of transformations that will result in the same final image.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–4444–4

A

C

B

666

A'

A"

C"

B"

C'

B'

6. Determine the sequence of two transformations to transform figure JKL to the image J'K'L' shown.

–2

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–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

88

JL

K

J'

L'

K'




7. Figure ABC has been rotated 45º, then 30º, and then 95º around the origin. What one compound rotation will result in the same final image?

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–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

666

A

C

B

A"'

C"'

B"'

8. Figure DEF has been translated 4 units to the left, then 3 units down, then 5 units to the right, and then 6 units up.

Which transformation will result in the same final image?

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

D

F2266

44

444

22E""

D""

F""

E

A Reflect the figure over the line y = –x.

B Translate the figure up 3 units and to the right 1 unit.

C Reflect the figure over the line y = 0.

D Translate the figure up 1 unit and over 3 units.



EXERCISES

Challenge Problem

9. Perform the following sequence of transformations (in the order listed) on figure GHJK, and graph the final image.

T(7, 0)

R(0, 0), –90°

ry = 0

–2

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–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–44–466–66–6

GH

JK

Does the order in which you do these transformations matter for determining the final image? Explain.

LESSON 6: COMPOSITE TRANSFORMATIONS



EXERCISES

1. Look at your work on the Self Check and the card match. What would you do differently if you were starting the Self Check task or the card match task now?

2. Make a chart of all the rigid transformations you have learned about and everything you know about them. You might want to make a table with the type of transformation, the definition, the properties and an example.

3. Write a definition of a rigid motion.

4. Complete any exercises from this unit you have not finished.

LESSON 7: PUTTING IT TOGETHER 1



EXERCISESLESSON 11: DEFINING CONGRUENCE

EXERCISES

1. There are two definitions for triangle congruence. Match each picture to the definition it shows.

P

U

S

S'

T

T'

U'

Two triangles are congruent if their

corresponding pairs of sides and corresponding

pairs of angles are congruent.

C

F

E

D

B

A

ABC DEF

Two triangles are congruent if there exists a sequence of rigid motions that maps one figure to coincide with the other.

2. Select the transformation that, when applied to ∆ABC, shows that ∆ABC and ∆DEF are congruent.

2

4

6

8

10

–8

–6

–2

–4

108642–2–4–6–8–10 x

y

C

B

A

F

E

D

A Rotation of 90° about the origin B Reflection across the x-axis

C Translation D Vertical stretch with scale factor 7




3. Select the transformation that, when applied to ∆ABC, shows that ∆ABC and ∆DEF are congruent.

–4

–2

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

DA

B

FC

E

A Rotation of 90° about the origin

B Reflection across the line x = –1

C Translation

D Rotation of 270° about the origin

4. Determine if these two triangles are congruent by describing a sequence of rigid motions that maps ∆ABC onto ∆DEF.

–2–2 6 8 1042

2

4

6

8

x

y

E

A

B

C

FD




5. Determine if these two triangles are congruent. Describe your method and justify your response.

–2

–4

–6

–6 2 4 6–2–4

2

4

6

x

y

E

A

B

C

FD

6. Determine if ∆JKL and ∆MNO are congruent. Describe your method and justify your response.

–2

–4

–6

–8

–2–4–6–8 6 842

2

4

x

y

J

L

N

O

M

K

7. Show that ∆ABC and ∆DBC are congruent. Describe your method and justify your response.

6

6

10

10

65°65°

85°

85°

30°

30°

D

C

A

B



EXERCISES

Challenge Problem

8. Show that ∆DEF and ∆GHI are congruent by showing that corresponding pairs of sides and angles are congruent.

–2

–4

–6

–6 2 4 6–2–4

2

4

6

x

y

E

I

H

G

F

D

LESSON 11: DEFINING CONGRUENCE



EXERCISES

EXERCISES

1. What congruence criterion determines that the two triangles are congruent?

30˚

A SAS B SSS C SSA D AAA

2. What congruence criterion determines that the two triangles are congruent?

56°

60°

64°

A SAS B SSS C SSA D AAA

3. Determine whether ∆ABC is congruent to ∆ADC. Explain which congruence criterion you used.

D

C

A

B

5 cm

5 cm

90°90°

LESSON 12: TRIANGLE CONGRUENCE CRITERIA



EXERCISESLESSON 12: TRIANGLE CONGRUENCE CRITERIA

4. Consider this figure in which C is the center of the circle.

DE

C

A B5 cm

8 cm

Determine whether ∆ABC is congruent to ∆EDC. Explain which congruence criterion you used.

5. Determine whether ∆ABC is congruent to ∆DCB. Explain which congruence criterion you used.

D

C

A

B

22°

22°

40°

40°

6. Point C is the center of the large circle. BE is a diameter of the circle.

AB

D

C

E

Determine whether the two triangles shown are congruent. Explain which congruence criterion you used.



EXERCISESLESSON 12: TRIANGLE CONGRUENCE CRITERIA

7. Determine whether the triangles are congruent. Explain which congruence criterion you used.

4 cm

4 cm

3 cm

3 cm

5 cm

5 cm

8. Which missing piece (angle or side measurement) do you need to be sure that these two triangles are congruent?

D

FC

EB

A

5 cm5 cm

4 cm4 cm

Challenge Problem

9. Prove that trapezoids ABCD and EFGH are congruent—either by showing a rigid motion that links the two figures or by showing that all corresponding sides and angles are congruent.

–2–2 6 8 1042

2

4

6

8

x

y

EA

H

B

C

F

G

D



EXERCISESLESSON 13: USING CONGRUENCE CRITERIA

EXERCISES

1. Determine whether each of these criteria determines that two triangles are congruent and write it in the correct column.

SSS AAA SAS SSA ASA

Determines Triangles Congruent Does Not Determine Triangles Congruent

2. What congruence criteria determines that the two triangles are congruent?

30°

30°

A SAS B SSS C SSA D AAS

3. Find b.

60°

45°

a

d

c

15 cme

b

A 45º B 60º C 75º D 105º




4. Find a.

60°

45°

a

d

c

15 cme

b

A 45º B 60º C 75º D 105º

5. AD is parallel to BC, and AB is parallel to DC.

A

B

D

C


6. Determine whether the two triangles shown are congruent. Explain which congruence criterion you used.

A

B

D C




7. Figure ABCDE is a regular pentagon, with center point P.

A

C

B

D

E

P


8. m∠ABC = 60°

A

D

BC60°




EXERCISES

Challenge Problem

9. Figure ABCDEFGH is a regular cube. ∆ABD is on the base of the cube, and ∆DBH ggoes through the interior of the cube.

A

D

E

H

F

G

B

C


LESSON 13: USING CONGRUENCE CRITERIA



EXERCISES

EXERCISES

1. The four smaller circles are the same size.

Determine the number of lines of symmetry in this figure.

A 1 line of symmetry

B 2 lines of symmetry

C 3 lines of symmetry

D 4 lines of symmetry

2. The four smaller circles are the same size.

Determine the order of rotational symmetry in this figure.

A Order 1 rotational symmetry

B Order 2 rotational symmetry

C Order 3 rotational symmetry

D Order 4 rotational symmetry

LESSON 14: SYMMETRIES OF POLYGONS



EXERCISESLESSON 14: SYMMETRIES OF POLYGONS

3. The four smaller squares are the same size.






4. Determine the number of lines of symmetry in this figure.







EXERCISESLESSON 14: SYMMETRIES OF POLYGONS

5. Each triangle is half of a square.






6. The three concentric hexagons shown are regular.

a. Determine the number of lines of symmetry in this figure.

b. Determine the order of rotational symmetry in the figure.



EXERCISES

7. A circle is inscribed in an equilateral triangle.

a. Determine the number of lines of symmetry in this figure.

b. Determine the order of rotational symmetry in the figure.

Challenge Problem

8. This is a regular tetrahedron; a real-life example is a four-sided number die. Think carefully about what rotational and reflective symmetry might mean in 3-D space.

Determine how many axes of rotational symmetry this 3-D figure has.




EXERCISES

EXERCISES

1. Which congruence criterion determines that the two triangles are congruent?

B

C

E

F

DA

40 mm40 mm

86 mm 86 mm

A SAS B SSS C SSA D SAA

2. Which congruence criterion determines that the two triangles are congruent?

B

C E

F

D

A

98 mm

98 mm

A SAS B ASA C SSA D SAA

LESSON 15: PUTTING IT TOGETHER 2—DAY 1



EXERCISESLESSON 15: PUTTING IT TOGETHER 2—DAY 1

3. ∆EFD is reflected over line w to form ∆HGD.

100°

100°

12 cm

12 cm

30°

30°

E

F

D

G

H

line of reflection

w

What do you know about the triangles? There may be more than one true statement.

A The triangles are congruent because it is a reflection.

B The triangles are congruent by ASA.

C The triangles are congruent because all the angles are equal.

D The triangles are congruent because ∆HGD is a rigid motion transformation of ∆EFD.

E The triangles are congruent because they look congruent.

4. Determine whether ∆JKL and ∆MNO are congruent by showing whether the corresponding sides and angles are congruent. If they are not congruent, show how you know.

–2–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

J

L

N

O

MK



EXERCISESLESSON 15: PUTTING IT TOGETHER 2—DAY 1

5. Point M is the midpoint of line segment AB.

C

BA M


6. Two triangles both have side lengths of 10 cm and 8 cm and an angle with a measure of 120º.

Explain why the two triangles are not necessarily congruent.

7. Fill in the reasoning to prove ∆ABD is congruent to ∆CDB.

A

B

C

D

∠ABD ≅ ∠CDB and ∠BDA ≅ ∠DBC a. ______________________________

BD ≅ BD b. ______________________________

∆ABD ≅ ∆CDB c. ______________________________

Challenge Problem

8. Using construction, explore how many different triangles labeled ABC can be made with each set of given measurements.

a. AB = 10.0 cm, ∠A = 30°, and BC = 12.0 cm

b. AB = 10.0 cm, ∠A = 30°, and BC = 6.0 cm

c. AB = 10.0 cm, ∠A = 30°, and BC = 5.0 cm

d. AB = 10.0 cm, ∠A = 30°, and BC = 4.0 cm



EXERCISES

EXERCISES

1. Read your Self Check and think about your work in this unit.

Write three things you have learned during the unit.

Share your answer with a classmate. Did your classmate understand what you wrote?

2. Write two statements that describe how you know that two figures are congruent. Your statements should start with “If…” and end with “…then the two figures are congruent.”

3. Write two statements that describe what is true if figures are congruent. Your statements should start with “If two figures are congruent, then…”.

4. Review the notes you took during the lessons about congruence and rigid motion. Add any additional ideas you have about congruence and rigid motion to your notes.

5. Complete any exercises from this unit that you have not finished.


MATH HIGH SCHOOL

CONGRUENCE AND RIGID MOTION

ANSWERS FOR EXERCISES



ANSWERS

ANSWERS

G.CO.5 1.

G.CO.5 2.

G.CO.5 3.

LESSON 2: REFLECTIONS



ANSWERSLESSON 2: REFLECTIONS

G.CO.2 4.

–4

–6

–8

–6 2 4 6 8 10–2–4–10 –8

–10

2

–2

4

6

8

10

x

y

C

DB

A

–44444

66–666

88

222

444

666

C'

D'

B'

A'

G.CO.5 5. C Isosceles triangle

G.CO.5 6. B

A

B

C

G.CO.5 7. a.

–1

–2

–3

–4

–2–3–4 4–5–6 5 6 72 31

3

2

4

1

5

6

7

8

9

x

y

444 555 666222 333111

333

444555

666

777

b. The line of reflection is y = x + 2.



ANSWERSLESSON 2: REFLECTIONS

G.CO.5 8. The reflected image is shown as figure A'B'C'.

2

4

6

8

8642–2–4–6–8

–2

–4

–6

–8

x

y

4444

B

C

C'

B'

A

A'

Challenge Problem

G.CO.5 9. The resulting graph, with all negative portions of the graph reflected up, will look like this.

8 9 10–1–2–3–4–5–6–7–8–9–10

–2–3–4–5–6–7–8–9–10 6 74 52 31

32

4

1

56789

10

x

y



ANSWERS

ANSWERS

G.CO.5 1. A

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

R S

TU

V

R'

S'T'

U'V'

G.CO.5 2.

–2

–4

–6

–2–4–6 642

2

4

6

x

y

44

6

A

A'

C'B'

B

C

G.CO.2 3.

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

D

D'

C

C'

A

A'

B

B'

LESSON 3: ROTATIONS



ANSWERSLESSON 3: ROTATIONS

G.CO.2 4.

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

CC'

AA'

B

B'22

G.CO.5 5.

–2

–4

–6

–2–4–6 642

2

4

6

x

y

A

B

C

A'

C'B'

444

G.CO.3 6. C 270°

G.CO.2 G.CO.4

7. B To find the angle of rotation you could measure the angle between PS and PS'.

C The angle of rotation is positive.

D The rotation does not change the size of the line.

E When you rotate ST, it maps each point of ST directly onto S'T', so the mapping is a function.



ANSWERSLESSON 3: ROTATIONS

G.CO.5 8.

–2

–4

–6

–8

2 4 6 8–8 –6 –4 –2

2

4

6

8

x

y

22

44

A

B C

D

B'A'

D'C'

G.CO.3 9. No, Rosa is not correct. If the triangle was rotated 90° about point (0, 0), ∆A'B'C' would be located in the second quadrant. Triangle ABC was actually rotated 90° about point (2, 0).

Challenge Problem

G.CO.5 10. Here is one possible mapping. In this example, figure ABCD is rotated 90º clockwise about the point (3, –2) to get figure A'B'C'D. This figure is then reflected over the x-axis to get figure A"B"C"D".

–2

–4

–6

–8

–10

–2–4–6–8–10–12 6 8 10 1242

2

4

6

8

10

x

y

AA'

A"

C"

B"D"

B'

B

C

C'

D



ANSWERS

ANSWERS

G.CO.2 1. B Rotation

G.CO.5 2.

–4

–6

–6 2 4 6–2–4

2

–2

4

6

x

y

DC

BD'

C'

A'B'

22

A

G.CO.3 3. D T(x, y) = (x + h, y + 3h)

G.CO.2 4. To shift the figure entirely into the fourth quadrant, you need to shift the figure so that point A is below the x-axis and point B is to the right of the y-axis. Any translation T(x, y) = (x + h, y + k) with h > 6 and k < –4 brings the entire figure into quadrant IV. T(x, y) = (x + 7, y – 5) is one example.

G.CO.5 5. Figure A'B'C' is the translated image.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A

B

C

A'

B'

C'444222

LESSON 4: TRANSLATIONS



ANSWERSLESSON 4: TRANSLATIONS

G.CO.2 G.CO.4

6. C ∠ABC is congruent to ∠A'B'C'.

D AB is parallel to A'B'.

E Translating ∠ABC maps each point of ∠ABC directly onto ∠A'B'C', so the mapping is a function.

G.CO.3 G.CO.2

7. The transformation is T(x, y) = (x + 5, y + 6). This translation can be determined by comparing one point from the original and its image point, such as A(–6, 2) and A'(–1, 8). The other points can be used to verify that the translation is correct.

G.CO.5 8. Here is the original figure ABC and the translated image A'B'C'.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

222A

B

C

444222A'

B'

C'



ANSWERSLESSON 4: TRANSLATIONS

Challenge Problem

G.CO.5 9. Here is one possible translation.

The figure I created, ABCD, has vertices at these coordinates: (–4, 3), (–3, 0), (4, 2), and (–1, 6).

After the translation T(x, y) = (x – 8, y + 4), figure A'B'C'D' has these new coordinates: (–12, 7), (–11, 4), (–4, 6), (–9, 10). Both figures are shown on this graph.

–2–2–4–6–8–10–12–14 2 4

2

4

6

8

10

12

x

y

A

A'

D'

C'

B'

D

CB



ANSWERS

ANSWERS

G.CO.2 G.CO.5

1. D Vertical stretch

G.CO.2 G.CO.4

2. B Rigid motions preserve all angle measures and segment lengths.

C Rigid motions do not alter the shape of the figure.

E Rigid motions produce two congruent figures.

G.CO.2 G.CO.5

3. D Horizontal stretch

G.CO.2 G.CO.5

4. B Dilation

G.CO.2 G.CO.4

5. The translation preserved all side lengths, while the dilation made them longer.

The translation preserved the figure’s area, while the dilation made it greater.

The two sets of parallel lines were preserved in both cases.

G.CO.2 G.CO.5

6. B Translation

G.CO.2 G.CO.4

7.Rigid Motion Transformation Non-Rigid Motion Transformation

Rotation

Reflection

Translation

Dilation

Horizontal stretch

Vertical stretch

LESSON 5: BEING PRECISE



ANSWERSLESSON 5: BEING PRECISE

Challenge Problem

G.CO.2 8. The starting points are (–4, 1), (1, 1), (–1, –2), and (–3, –1). It may be easier to use algebra to determine the new coordinates, and then plot the four new points.

If each x-value is transformed to 3x + 2 and each y-value is transformed to 4y – 3, the new points will be (–10, 1), (5, 1), (–1, –11), and (–7, –7).

Figure A' shows the transformed image, and figure A shows the original.

The resulting motion is non-rigid. It has changed the edge lengths and area a great deal. This transformation is a mix of stretches and translations. It stretches the horizontal coordinates by a factor of 3, and stretches the y-coordinates by a factor of 4. It also shifts the (resulting) x-coordinates 2 units to the right and the resulting y-coordinates 3 units down.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

12

x

y

–22–2

–44–4

–66–6

–888

–1010

–44–4–66–6–88–8 444222222

–22–2

A'

A



ANSWERS

ANSWERS

G.CO.2 1. C a reflection from

G.CO.2 2. A 0 units

G.CO.5 3. Figure A'B'C'D' shows the image after the translation, and figure A''B''C''D'' shows the image after the translation and rotation.

–2

–4

–6

–8

–10

–2–4–6–8–10–12 6 8 10 1242

2

4

6

8

10

x

y

66666

A

A''A'

B''

B'

C''

C'

D''

D'

B

C

D

G.CO.5 4. Here is one possible series of transformations.

I used a translation and then a rotation to get the figure into quadrant II. First, the translation moves the figure 5 units to the left (figure A'B'C'), and then the rotation rotates it –90º (clockwise) into quadrant II (figure A''B''C'').

–2

–4

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

E

E''

F''

G''

F

G

E'

F'

G'

LESSON 6: COMPOSITE TRANSFORMATIONS



ANSWERSLESSON 6: COMPOSITE TRANSFORMATIONS

G.CO.5 5. There are many possible solutions. As long as the final image has the same coordinates as A"B"C" in the original graph, the solution is valid. Here is one possible series of transformations.

I also used a translation and then a rotation, but both different from the original transformations. First, I translated the figure up 8 units. Then I rotated the figure 180º about the point (–3, 4), which is the center of the second image. These transformations result in the exact same image as A"B"C".

G.CO.5 6. There are many possible sequences of transformations, including a sequence in two steps. Here is one possible sequence of transformations.

First, rotate the figure 180º about the origin: R(0, 0), 180°.

Second, translate the figure 3 units up and 3 units right: T(x, y) = (x + 3, y + 3).

G.CO.5 7. A single rotation of 170º will result in the same final image. To determine this, simply add up all of the degree measures of each individual rotation.

G.CO.2 8. B Translate the figure up 3 units and to the right 1 unit.



ANSWERSLESSON 6: COMPOSITE TRANSFORMATIONS

Challenge Problem

G.CO.5 9. Here is the final image after all three transformations.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–44–466–66–6

222

GH

J

H'

K' G'

J'

K

The order matters quite a bit. Here is an example of the figure that would result if you changed the given order of the above transformations.

1. Rotate the figure –90º about the origin to figure G'H'J'K'.

2. Reflect that figure across the x-axis to figure G''H''J''K''.

3. Translate that figure 7 units to the right to figure G'''H'''J'''K'''.

–2

–6

–8

–10

–2–4–6–8–10 6 8 10 1242

2

4

6

8

10

x

y

–44–466–66–6

444

666

–6666–6

G

G'' G'''

G'

H'

H'''H''J''

K''

J'''

K'''

J'

K'

H

JK



ANSWERSLESSON 7: PUTTING IT TOGETHER 1

ANSWERS

G.CO.2 2.

Rigid Transformation Definition Properties Examples

Reflection A reflection, rm, is a

transformation that maps every point Q in a plane to a point Q' across a fixed line of reflection m, such that line m is the perpendicular bisector of segment QQ'.

Properties of reflections:

• Reflections do not change segment lengths.

• Reflections do not change angle measures.

• Parallel lines remain parallel.

• Circles remain as circles.

• Points on the line of reflection are not moved by the reflection.

• Reflections reverse the orientation of the vertices of a polygon.

U

S

S'T'

U'

T

m

Rotation A rotation, RP,f,

is a transformation that moves every point Q in a plane through a fixed angle f about a given center P, producing a new point Q'. Point Q' is on the circle with center P and radius PQ.

Positive rotations are counterclockwise—the same way that the quadrants are numbered.

Properties of rotations:

• Rotations do not change segment lengths.

• Rotations do not change angle measures.

• Parallel lines remain parallel.

• Circles remain as circles.

• The center point is the only point not moved by a rotation.

• Rotations do not change the orientation of the vertices of a polygon.

P

U

S

S'

T

T'

U'



ANSWERSLESSON 7: PUTTING IT TOGETHER 1

G.CO.2 3. A rigid motion is a transformation of a figure that:

• Maps lines onto lines

• Maps a segment onto a segment of the same length

• Maps an angle to an angle of the same measure

• Maps parallel lines onto parallel lines

A rigid motion is a transformation that does not change the size or shape of a figure. The original figure and the transformed figure are congruent.

Rigid Transformation Definition Properties Examples

Translation A translation, Th, k

, is a transformation that moves a point P = (x, y) in the coordinate plane to the point P' = (x + h, y + k).

Properties of translations:

• Translations do not change segment lengths.

• Translations do not change angle measures.

• Parallel lines stay parallel.

• Circles stay circles.

• There are no points that are not moved by a translation.

• Translations do not change the orientation of vertices of a polygon.

A'

B'

C'

A

B

C

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y



ANSWERS

ANSWERS

G.CO.6 G.CO.7 G.CO.8

1.

P

U

S

S'

T

T'

U'

Two triangles are congruent if there exists a sequence of rigid motions that maps one figure to coincide with the other.

C

F

E

D

B

A

ABC DEF

Two triangles are congruent if their

corresponding pairs of sides and corresponding

pairs of angles are congruent.

G.CO.6 2. A Rotation of 90° about the origin

G.CO.6 3. B Reflection across the line x = –1

G.CO.6 4. The two triangles are congruent. You can translate ∆ABC up 2 units and to the right 3 units to map it directly onto ∆DEF.

G.CO.6 5. The two triangles are not congruent. The side lengths of ∆ABC are 26, 17, and 5, whereas ∆DEF has side DE with length 2 2. There is no corresponding side of ∆ABC that has the same length as DE, so the two triangles cannot be congruent.

LESSON 11: DEFINING CONGRUENCE



ANSWERSLESSON 11: DEFINING CONGRUENCE

G.CO.6 6. ∆JKL and ∆MNO are congruent. There are many different sequences of rigid motions that prove their congruence. Here is one possible method.

Rotate ∆JKL 180º about the origin and then translate the image 4 units down. The dotted triangle is ∆JKL after the first transformation (the rotation).

–2

–4

–6

–8

–2–4–6–8 6 842

2

4

x

y

J

L

O

M

KN

G.CO.7 7. You can determine that each pair of sides and each pair of angles are congruent.

AB ≅ BD and AC ≅ DC. BC is shared by both triangles.

∠A and ∠D have the same measure, and since they are between corresponding sides of equal length, they are corresponding angles. Likewise, ∠ABC and ∠DBC have the same measure and are corresponding angles because they are between two pairs of corresponding sides. And the remaining corresponding pair of angles, ∠DCB and ∠ACB, also have equal measures.

Therefore, ∆ABC and ∆DBC are congruent because the corresponding angles and sides are congruent.



ANSWERSLESSON 11: DEFINING CONGRUENCE

Challenge Problem

G.CO.6 G.CO.7 G.CO.8

8. First, determine the coordinates of each point. This will enable you to find the distances between each vertex.

D = (–5, 2), E = (–3, 1), F = (–1, 5) and G = (5, –4), H = (3, –5), I = (1, –1)

You can find the distance between each point to determine if the corresponding pairs of sides are of equal length.

DE GH

EF HI

DF GI

≅ = ≈

≅ = ≈≅ =

5 2 24

2 5 4 47

5

.

.

units

units

uniits

You can also determine the angle measures by comparing the slopes of the sides. For example, you can show that both triangles are right triangles by comparing the slope of side DE to side EF and the slope of side GH to side HI.

The slope of side DE is −1

2, and the slope of side EF is 2. Since these slopes are

negative reciprocals of each other, the angle they make must measure 90°.

Similarly, the slope of side GH is 1

2, and the slope of side HI is –2. These slopes are

also negative reciprocals of each other, so the angle they make must measure 90°.

You know the m∠DFE = m∠GIH because that is given. Because the sum of the angles of a triangle is 180°, and you know two pairs of corresponding angles are equal, the third pair of corresponding angles must also be equal.



ANSWERS

ANSWERS

G.CO.6 1. A SAS

G.CO.6 2. A SAS

G.CO.7 3. The triangles are congruent by SAS. Sides BC and DC are both equal to 5 cm. The 90° angles are congruent, and both triangles share the side CA, which must be congruent to itself. Thus, ∆ABC and ∆ADC are congruent.

G.CO.7 4. The triangles are not congruent. Although both have two sides that are radii of the circle, the given segment lengths of AB = 5 cm and DE = 8 cm make it impossible for the triangles to be congruent.

G.CO.7 5. The triangles are congruent by ASA. The given 40° and 22° angles match on both triangles: ∠ABC ≅ ∠DCB and ∠ACB ≅ ∠DBC. Also, the two triangles share the side BC, which must be congruent to itself. Therefore, ∆ABC and ∆DCB are congruent.

G.CO.7 6. The triangles are congruent by SAS. You know all four segments from the center to the circle are congruent, since they are all radii of the circle. This means that CA ≅ CB ≅ CE ≅ CD.

Since BE is a diameter, BC and CE form a straight line, making ∠BCA supplementary to ∠ACE and ∠ECD supplementary to ∠DCB. Since ∠ACE ≅ ∠DCB, that means ∠BCA ≅ ∠ECD.

So, CB ≅ CD, ∠BCA ≅ ∠ECD, and CA ≅ CE. Therefore, the triangles are congruent by SAS.

G.CO.7 7. The triangles are congruent by SSS. All three corresponding side lengths are congruent. Therefore, the triangles must be congruent.

G.CO.7 8. Knowing that AC ≅ DF shows congruence by SSS, and knowing that ∠ABC ≅ ∠DEF shows congruence by SAS.

LESSON 12: TRIANGLE CONGRUENCE CRITERIA



ANSWERSLESSON 12: TRIANGLE CONGRUENCE CRITERIA

Challenge Problem

G.CO.6 9. You could show the congruence using either method. The simpler one might be to show that trapezoid ABCD can be translated 5 units to the right to map directly onto trapezoid EFGH. The coordinates are as follows:

A = (2, 4), B = (4, 3), C = (4, 0), D = (2, –1)

By adding 5 to every x-coordinate, you get the exact coordinates of trapezoid EFGH:

E = (7, 4), F = (9, 3), G = (9, 0), H = (7, –1)

You could also use the coordinate plane to help determine all of the side lengths and angles in order to prove the congruence.

AD EH

BC FG

AB EF

CD GH

= == =

= =

= =

5 units

3 units

units5

5 uunits

Then using the fact that AD, BC, EH, and FG are vertical and that the other pairs of corresponding sides have the same slope, you can say that corresponding pairs of sides meet at congruent angles. Since each pair of corresponding sides or corresponding angles is congruent, the two polygons are congruent.

∠ = ∠ = °∠ = ∠ = °∠ = ∠ =

DAB HEF

CDA GHE

ABC EFG

63.43

63.43

116.577

116.57

°∠ = ∠ = °BCD FGH



ANSWERS

ANSWERS

G.CO.7 1.Determines Triangles Congruent Does Not Determine

Triangles Congruent

SSS

SAS

ASA

AAA

SSA

G.CO.6 2. D AAS

G.CO.7 3. B 60º

G.CO.7 4. C 75º

G.CO.7 5. The two triangles are congruent by ASA. Line AC is a transversal that cuts through both pairs of parallel lines in the figure. The transversal creates opposite interior angles that are congruent:

∠BAC ≅ ∠ACD and ∠DAC ≅ ∠BCA.

Also, the two triangles share AC as a side, which is in between the two known congruent angle pairs. AC must be congruent to itself, so the triangles are congruent.

G.CO.7 6. You cannot prove the triangles congruent with the given information. Therefore the triangles are not considered congruent.

You can get two congruent corresponding parts with the given ∠DAC ≅ ∠CAB and the shared side length AC. However, this information is not enough to conclude that the triangles are congruent.

G.CO.7 7. The triangles are congruent by SSS. You know all segments from the center P to a vertex of the pentagon are congruent, and you know that all edges of the pentagon are congruent as well. This gives you PC ≅ PB ≅ PD ≅ PE and also BC ≅ DE, which shows congruence by SSS.

You can also use the central angles of the pentagon to show congruence. Since it is a regular pentagon, you know all of the central angles are congruent, so ∠CPB ≅ ∠DPE. Therefore, you have congruence by SAS.

G.CO.7 8. You cannot prove the triangles congruent with the given information. Therefore the triangles are not considered congruent. They share the side length BC. The measure of ∠ABC is given, but there is no other information that can be obtained. The 60° angle is not duplicated anywhere since no lines are specified as being parallel.

LESSON 13: USING CONGRUENCE CRITERIA



ANSWERSLESSON 13: USING CONGRUENCE CRITERIA

Challenge Problem

G.CO.7 9. The triangles are not congruent. Suppose that the side length of the cube is 1 unit. ∆ABD can then be drawn as follows.

A

D

B

45º

90º 45º

1

12

∆DBH can be drawn in 2-D space. The base of ∆DBH shares the longer side of ∆ABD.

DB is in both triangles, but it is not a corresponding side. Using the length DB, and applying the Pythagorean Theorem again, ∆DBH can be drawn as follows.

H

D B90º

1

2

3

The triangles are not congruent because the corresponding sides are not congruent.

Here is another solution to this problem.

The triangles are not congruent. Both ∆ABD and ∆DBH have right angles, because the sides of a cube are perpendicular to each other, and since a triangle can only have one right angle, the two triangles can only be congruent if the right angles ∠BDH and ∠BAD correspond to each other. We know that AD ≅ AB ≅ DH because all edges of a cube are congruent, so DH could correspond to either leg of ∆ABD, and we just need to show that DB also has the same length. But DB is the hypotenuse of the right triangle ∆ABD, so DB > AB and DB > AD. DB is not congruent to either of the sides it could possibly correspond to, so the triangles are not congruent.



ANSWERS

ANSWERS

G.CO.5 1. D 4 lines of symmetry

G.CO.5 2. D Order 4 rotational symmetry

G.CO.5 3. D 4 lines of symmetry

G.CO.5 4. B 2 lines of symmetry

G.CO.5 5. A 1 line of symmetry

G.CO.5 6. a. There are 6 lines of symmetry.

b. The figure has a rotational symmetry of order 6.

G.CO.5 7. a. There are 3 lines of symmetry.

b. The figure has a rotational symmetry of order 3.

Challenge Problem

G.CO.5 8. A line drawn through one vertex, and intersecting the center of the opposite face, is an axis of rotational symmetry of order 3. There are four such axes of rotational symmetry.

Another type of axis of rotational symmetry can be found by drawing a line connecting the midpoint of one edge to the midpoint of the opposite edge in the tetrahedon. This axis of rotational symmetry has order 2, and there are three such axes of symmetry. Therefore there are seven axes of rotational symmetry..




ANSWERS

ANSWERS

G.CO.7 1. A SAS

G.CO.7 2. B ASA

G.CO.8 3. A The triangles are congruent because it is a reflection.

B The triangles are congruent by ASA.

D The triangles are congruent because ∆HGD is a rigid motion transformation of ∆EFD.

G.CO.8 4. ∆JKL is not congruent to ∆MNO. There are many ways to show this. As long as you show that one of the corresponding pairs of sides or angles do not match, you prove that the triangles are not congruent.

The length of side JK is 4 units, and the length of side MN = ≈2 5 4 47. units, which you can find by using the Pythagorean Theorem between points M and N.

Let length of side MN = c.

22 + 42 = c2

4 + 16 = c2

20 = c2

20 4 47≈ .

Since these corresponding sides do not have the same length, the triangles are not congruent with vertices in the order stated. You can confirm that they are not congruent in any orientation by finding the lengths of the other two sides of ∆JKL, 5 and 41, neither of which equals 2 5, so MN has no possible corresponding side in ∆JKL. Thus, the triangles are not congruent in any orientation.

G.CO.7 5. The triangles are congruent by SAS. Since point M is the midpoint of line segment AB, you know AM ≅ BM. Since ∠BMC and ∠AMC are supplementary and ∠BMC is given as a right angle, ∠BMC ≅ ∠AMC. They have one side in common, MC. This information is sufficient to say the triangles are congruent by SAS.

G.CO.7 6. Corresponding sides and corresponding angles must be congruent. One of these triangles could have the 120º angle included between the two given sides, while the other triangle does not.

G.CO.8 7. a. When parallel lines are cut by a transversal, alternate interior angles are congruent.

b. BD is congruent to itself.

c. ASA




ANSWERSLESSON 15: PUTTING IT TOGETHER 2—DAY 1

Challenge Problem

G.CO.12 G.CO.7

8. a. One triangle with obtuse angle B

b. Two triangles

In one case, ∠C is an acute angle; in the other case, ∠C is an obtuse angle.

c. One triangle with right angle C

d. No triangles

The 4.0 cm length of BC is too short. You cannot form AC to create a triangle.



ANSWERSLESSON 16: PUTTING IT TOGETHER 2—DAY 2

ANSWERS

G.CO.8 G.CO.6

2. If all corresponding sides are equal (same size) and all corresponding angles are equal (same shape), then the two figures are congruent.

If a figure can be mapped to another figure by a rigid motion transformation or series of rigid motions, then the two figures are congruent.

G.CO.8 G.CO.6

3. If two figures are congruent, then all corresponding sides are equal (same size) and all corresponding angles are equal (same shape).

If two figures are congruent, then one figure can be mapped to another figure by a rigid motion transformation or series of rigid motions.

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