congruence and rigid Motion - · PDF filecongruence and rigid Motion exerciSeS. Copyright © 2014 Pearson Education, Inc. 2 ... LEssON 11: dEfiNiNg cONgruENcE

Math high School

congruence and rigid Motion

exerciSeS

Copyright © 2014 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 �� 6

LEssON 3: rOtatiONs �� 10

LEssON 4: traNsLatiONs �� 14

LEssON 5: BEiNg PrEcisE �� 17

LEssON 6: cOMPOsitE traNsfOrMatiONs �� 21

LEssON 7: PuttiNg it tOgEthEr �� 25

LEssON 11: dEfiNiNg cONgruENcE �� 26

LEssON 12: triaNgLE cONgruENcE critEria �� 29

LEssON 13: usiNg cONgruENcE critEria �� 32

LEssON 14: sYMMEtriEs Of POLYgONs �� 35

aNswErs

LEssON 2: rEfLEctiONs �� 40

LEssON 3: rOtatiONs �� 43

LEssON 4: traNsLatiONs �� 46

LEssON 5: BEiNg PrEcisE �� 49

LEssON 6: cOMPOsitE traNsfOrMatiONs �� 55

LEssON 11: dEfiNiNg cONgruENcE �� 58



CONTENTS ExErCiSES

aNswErs

LEssON 12: triaNgLE cONgruENcE critEria �� 61

LEssON 13: usiNg cONgruENcE critEria �� 63

LEssON 14: sYMMEtriEs Of POLYgONs �� 65

Note: Some of these problems are designed to be delivered electronically.



ExErcisEsLEssON 1: rigid MOtiON

Review your end of unit assessment from the previous unit.

Write your wonderings about geometric transformations.

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

Based on your previous work, write three things you will do differently during this unit to increase your success.



ExErcisEs

ExErcisEs

1. Draw the reflections of the following images across the dotted line.

LEssON 2: rEfLEctiONs



ExErcisEsLEssON 2: rEfLEctiONs

2. Come up with several words that are symmetric (e.g. MOM). Describe the line of symmetry for the words you pick.

3. What kind of triangle would you get if you reflected the green line across the black line and then connected their endpoints?




4. Which of the following graphs shows the proper reflection of the figure across the line segment AB?

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y

A

B

A

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y

A

B

B

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y

A

B

C

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y

A

B

D

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y

A

B




5. Draw in the line of symmetry for the following reflection:

–1–2–3–4

–2–3–4 4–5–6 5 6 72 31

32

4

1

56789

x

y

444 555 666222 333111

333444555666777

6. Draw the shape given by these coordinates, (5, 2) (4, –1) (1, 6), and then reflect that shape across the line x = 0.

challenge Problem

7. If you reflect all negative values (all values for which y < 0) of the function drawn below across the x-axis, what will the result look like?

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. Rotate the following figure 90º around the origin.

–1–2–3–4–5–6

–2–3–4–5–6–7 6 74 52 31

32

4

1

56

x

y

2. Reflecting the object below across the x-axis and rotating the object 180º about the point (2, 0) will result in images that are:

–1–2–3–4–5–6

–2–3–4–5–6–7 6 74 52 31

32

4

1

56

x

y

A Identical

B Rotations

C Reflections

D Translations

LEssON 3: rOtatiONs



ExErcisEs

3. This object is rotated clockwise 90º and then reflected across the y-axis. How far will 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)

LEssON 3: rOtatiONs



ExErcisEsLEssON 3: rOtatiONs

4. Which of the following graphs shows the correct image of the initial graph rotated 120º counter clockwise about the origin?

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

666 8

A

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

66666 8

B

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

66666 8

––

––

C

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

66666 8–44–66–66–688–88–8

D

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

66666 8

444



ExErcisEsLEssON 3: rOtatiONs

5. Construct the figure with the given coordinates, (3, 4) (6, 2) (5, 4) and then rotate the figure 270º around the origin.

challenge Problem

6. Draw a quadrilateral in the grid below and label the vertices. Draw its mapping after it is rotated 90º clockwise around the point (3, –2) and then reflected across the x-axis.

–2

–4

–6

–8

–10

–2–4–6–8–10–12 6 8 10 1242

2

4

6

8

10

x

y



ExErcisEsLEssON 4: traNsLatiONs

ExErcisEs

1. A translation involves every x-value traveling a distance of h and every y-value traveling three times the distance that the x-values travel. Which translation represents this correctly?

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)

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

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A

B

C

D




3. Translate the figure EFG using the following translation: T(x, y) = (x – 4, y + 3).

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

E

F

G

4. Describe the translation that is shown:

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

A

A'

D'C'

B'

D C

B




5. Translate the following figure 2 units in the positive x-direction, and 3 units in the negative y-direction.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

22

challenge Problem

6. Create your own figure with at least 4 points, and then translate it with the following translation T(x, y) = (x – 8, y + 4).



ExErcisEsLEssON 5: BEiNg PrEcisE

ExErcisEs

1. Perform a vertical stretch on the following figure, using the stretch that takes the y-coordinate of each point and doubles it. (Note that the x-coordinate of each point is unchanged.)

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–22–2–22–2 222

222

2. Perform a horizontal stretch on the following figure. Use the stretch that takes the x-coordinate of each point and triples it. (Note that the y-coordinate of each point is unchanged.)

–2

–4

–6

–8

–10

–2–4–6–8–10–12–14 6 8 10 12 1442

2

4

6

8

10

x

y

6666644422222




3. Perform a dilation on the following figure, with the origin as the center point, and a scale factor of 4.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

4. Create a figure with at least 4 points, and demonstrate a rigid motion transformation and a non-rigid transformation on the same figure. Explain the different effects you notice.




5. Perform a horizontal stretch and a vertical translation on the following figure. Show an image for each transformation separately, and then one image with both transformations done.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–22–2–22–2 222

222

444




challenge Problem

6. Perform the following transformation T(x, y) = (3x + 2, 4y – 3) on the given figure, then describe what types of motion this transformation describes.

–2

–4

–6

–8

–10

–12

–2–4–6–8–10 6 8 10 12–12 42

2

4

6

8

10

12

x

y

22222–22–2



ExErcisEsLEssON 6: cOMPOsitE traNsfOrMatiONs

ExErcisEs

1. Perform the following transformations on the figure ABCD, then graph the final image.

T(3, 4)

R(0, 0), 45°

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 10 1242

2

4

6

8

10

x

y

666

A

B

C

D

2. Perform a series of two transformations that moves the figure EFG completely into quadrant II.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

666E

F

G




3. The 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'

4. Determine the sequence of transformations with the fewest steps required to move the figure JKL to the image J'K'L' shown.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

8

JL

K

J'

L'

K'




5. The 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?

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

666

A

C

B

A"'

C"'

B"'

6. Figure DEF has been translated 4 units to the left, then 3 units down, then 5 units to the right, then 6 units up. What one compound translation 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




challenge Problem

7. Perform the following transformations on the figure GHIJ, and graph the final image.

T(7, 0)

R(0, 0), –90°

r y = 0

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

–44–466–66–6

GH

IJ

Explain whether or not the order in which you do these transformations matters for determining the final image.



ExErcisEs

Read through your Self Check and think about your work in this lesson.

Write down what you have learned during the lesson.

What would you do differently if you were starting Self Check task now?

Which method would you prefer to use if you were doing the task again? Why?

Compare the new approaches you learned about with your original method.

Record your ideas— keep track of problem-solving strategies.

Complete any exercises from this unit you have not finished.

LEssON 7: PuttiNg it tOgEthEr



ExErcisEsLEssON 11: dEfiNiNg cONgruENcE

ExErcisEs

1. Determine if these two triangles are congruent by showing a sequence of rigid motions that maps ABC to DEF.

–2–2 6 8 1042

2

4

6

8

x

y

E

A

B

C

FD

2. 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




3. Determine if triangle JKL and triangle 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

4. Show that the triangle ABC and triangle DBC are congruent. Describe your method and justify your response.

6

6

10

10

65°65°

85°

85°

30°30°

D

C

A

B




5. Show that figures DEF and GHI are congruent by showing each pair of sides and each pair of angles are congruent.

–2

–4

–6

–6 2 4 6–2–4

2

4

6

x

y

EI

H

G

F

D

6. Determine whether or not triangles JKL and MNO are congruent by showing if each pair of sides and each pair of angles is 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

challenge Problem

7. Create a triangle that lies completely in quadrant III, and then create a congruent triangle that is rotated from the original and lies completely within quadrant I.



ExErcisEs

ExErcisEs

For problems 1–5, determine whether or not each pair of triangles is congruent. Explain which congruence criteria you used for each case.

1.

D

C

A

B

5 cm

5 cm

90°90°

2.

DE

C

A B5 cm

8 cm

3.

D

C

A

B

22°

22°

40°

40°

LEssON 12: triaNgLE cONgruENcE critEria



ExErcisEsLEssON 12: triaNgLE cONgruENcE critEria

4.

C

A

D

E

FB

120°

120°

12 cm

15 cm

15 cm

12 cm

5. 4 cm

4 cm

3 cm

3 cm

5 cm

5 cm

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

D

FC

EB

A

5 cm5 cm

4 cm4 cm



ExErcisEsLEssON 12: triaNgLE cONgruENcE critEria

challenge Problem

7. Prove that trapezoids ABCD and EFGH are congruent, either by showing a rigid motion that links the two figures, or by showing that every matching pair of sides and angles is 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

For problems 1–6, determine whether or not the two triangles shown are congruent. Explain the congruence criteria that you used for each figure.

1. AD is parallel to BC, and AB is parallel to DC:

A

B

D

C

2. C is the center of the green circle. BE is a diameter of the circle.

AB

D

C

E




3. A

B

D C

4. ABCDE is a regular pentagon, with center point P.

A

C

B

D

E

P

5. M is the midpoint of line segment AB.

A

C

BM




6. ∠ABC = 60°

A

D

BC60°

challenge Problem

7. ACBDEFGH is a regular cube. The green triangle ABD is on the base of the cube, and the orange triangle DBH goes through the interior of the cube. Use triangle congruence criteria to show whether or not the two triangles are congruent.

A

D

E

H

F

G

B

C



ExErcisEs

ExErcisEs

For each figure, determine the number of lines of symmetry and the order of rotational symmetry.

1.

2.

3.

LEssON 14: sYMMEtriEs Of POLYgONs



ExErcisEsLEssON 14: sYMMEtriEs Of POLYgONs

4.

5.

6.



ExErcisEsLEssON 14: sYMMEtriEs Of POLYgONs

challenge Problem

7. Determine how many planes of symmetry and what order or rotational symmetry this 3-D object has. It is a regular tetrahedron; a real life example is a 4-sided number die. Think carefully about what rotational and reflective symmetry might mean in 3-D space.

Math high School

congruence and rigid Motion

exerciSeSanSWerS

For exerciSeS



ANSWERS

aNswErs

1. The reflected images are shown:

2. BOB BOX CEDE have a horizontal line of symmetry.

TOT MUM YAY have a vertical line of symmetry.

LEssON 2: rEfLEctiONs



ANSWERSLEssON 2: rEfLEctiONs

3. The resulting triangle will be Isosceles.

4. B

–1–1

–2

–3

–4

–5

–2–3–4 4 52 31

3

2

4

1

5

x

y

A

B

5. The line of symmetry is y = x + 2.

–1–2–3–4

–2–3–4 4–5–6 5 6 72 31

32

4

1

56789

x

y

444 555 666222 333111

333444555666777



ANSWERSLEssON 2: rEfLEctiONs

6. The original figure (yellow) is an obtuse triangle, the reflected image is in purple.

2

4

6

8

8 96 74 52 31–1–2–3–4–5–6–7

1

3

5

7

x

y

444433 44

challenge Problem

7. 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

1. Here is the resulting figure:

–1–2–3–4–5–6

–2–3–4–5–6–7 6 74 52 31

32

4

1

56

x

y

5556

2. C Reflections The reflections are across the line x = 2.

–2

–4

–6

–2–4–6 642

2

4

6

x

x = 2

y

reflection rotation

3. You can see that the object passes through the points (0, 0), (5, 0), (5, 5), and (0, 5), which if connected would create a perfect square inside the circle. Rotating the object clockwise 90º would place point A at (5, 0). Reflecting it across the y-axis would place A at (–5, 0). The distance between (–5, 0) and (5, 5) is:

5 5 5 0 1252 2− −( )( ) + −( ) =

LEssON 3: rOtatiONs



ANSWERSLEssON 3: rOtatiONs

4. A

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

66666 8

5. The following graph shows the original figure in green, and the rotated image figure in orange.

–2

–4

–6

–2–4–6 642

2

4

6

x

y



ANSWERSLEssON 3: rOtatiONs

challenge Problem

6. Answer will vary depending with the figure chosen. In this example, the figure ABCD is rotated 90º around the point (3, –2) to get A'B'C'D. This figure is then reflected over the x-axis to get 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

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

2. 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) = T(x + h, y + k) with x > 6 and y < –4 brings the entire figure into quadrant IV. T(x, y) = (x + 7, y – 5) is one example.

3. E'F'G' is the translated image, as shown in the following graph:

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

E

F

G

E'

F'

G'

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

LEssON 4: traNsLatiONs



ANSWERSLEssON 4: traNsLatiONs

5. Here is the original and translated figure. The translation is the red figure.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

22

444222



ANSWERSLEssON 4: traNsLatiONs

challenge Problem

6. The figure I created ABCD has vertices at the following coordinates: (–4, 3), (–3, 0), (4, 2), and (–1, 6). After the translation T(x, y) = (x – 8, y + 4), the figure A'B'C'D' has the new coordinates (–12, 7), (–11, 4), (–4, 6), (–9, 10). Both figures are shown on the following 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

C

B



ANSWERS

aNswErs

1. The results of the vertical stretch are shown. The new points are (–4, 8), (3, 6), (–2, –6), and (3, –4). The grey figure is the resulting image.

–2

–4

–6

–8

–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

2.

–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

LEssON 5: BEiNg PrEcisE



ANSWERSLEssON 5: BEiNg PrEcisE

3.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

4. The shape I started with was a parallelogram. It has the coordinates (–2, 2), (3, 5), (–2, –3), and (3, 0). The yellow figure is an image after a translation (a rigid motion) of 6 units to the right. The green figure is an image after a dilation (a non-rigid motion) centered at the origin with a scale factor of 2.




–2

–4

–6

–8

–10

–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

You can make many observations about these two transformations. The translation preserved all side lengths, while the dilation made them longer. The translation preserved the shape’s area, while the dilation made it larger. The two sets of parallel lines were preserved in both cases.

5. From the original figure, first I did the horizontal stretch, with a scale factor of 3. The orange figure is the image after the stretch.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

66666444–2222–2

22–22–2 22222

22222

444

–44–44–466–66–6

44




Then I performed only the vertical translation, I used a shift of 5 units downward.

–2

–4

–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

And then I made the image after both transformations, one after the other.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 842

2

4

6

8

10

10 x

y

22

22222

444

–4444–4

–6666–6

–88888

– 444–2222–2

22–22–2 22222

Here is a graph that includes all three images for comparison.




–2

–4

–6

–8

–10

–2–4–6–8–10 6 842

2

4

6

8

10

10 x

y

666664422

222

444

44–44–466–66–6

44

4442222222–22–2–

–4444–4

–6666–6

–88888

–2222–2

Again, we can make many observations about the different types of transformations. In this case, we can also see that the set of parallel lines remained parallel even after the horizontal stretch (although they both have new slopes, they both have the same new slope) It is also interesting to note that it doesn’t matter which order you do the two transformations in, you will end up with the same final image.




challenge Problem

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

T(x, y) = (3x + 2, 4y – 3)

So the new points will be (–10, 1), (5, 1,) (–1, –11) ,and (–7, –7).

The grey figure shows the transformed image, and the red figure shows the original:

–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 44422222

–22–2

The resulting motion is definitely non-rigid. It has changes 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 to the right, and the resulting y-coordinates 3 downward.



ANSWERS

aNswErs

1. The orange figure shows the image after just the translation, then the purple figure shows the image after the translation and rotation.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 10 1242

2

4

6

8

10

x

y

666

A

A'B'

C'D'

B

C

D

LEssON 6: cOMPOsitE traNsfOrMatiONs



ANSWERSLEssON 6: cOMPOsitE traNsfOrMatiONs

2. 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 (the red image) and then the rotation rotates it –90º (clockwise) into quadrant II (the green image).

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

666E

E"

F"

G"

F

G

3. There are many possible solutions. As long as the final image has the same coordinates at A"B"C" in the original graph, the solution is valid. 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 rotated the figure 180º around the point (–3, 4), which is the center of the second image. This results in the exact same A"B"C" image.

4. There are many possible sequences of transformations, but the shortest sequence is two steps. Here is one possible sequence of transformations.

First, rotate the figure 180º around the origin. R(0, 0), 180°. Second, translate the figure 3 units up and 3 units right. T(x, y) = (x + 3, y + 3).

5. One compound 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.

6. One compound translation up 3 units, and right 1 unit, will take DEF to D""E""F"". This is determined by the net change done by all of the individual translations.



ANSWERSLEssON 6: cOMPOsitE traNsfOrMatiONs

challenge Problem

7. 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

I

H"'

J"' G"'

I"'

J

Interestingly, the order matters quite a bit, here is an example where changing the order of the transformations results in a totally different final image.

1. Rotate –90º around the origin to the red figure.

2. Reflect across the x-axis to the yellow figure.

3. Translate 7 units to the right to the purple figure.

–2

–4

–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

4444

–6666–6

GH

IJ



ANSWERS

aNswErs

1. The two triangles are indeed congruent. You can translate ABC up 2 units and right 3 units to map directly onto DEF.

2. 2. These two triangles are not congruent. You can show this by attempting to map ABC onto DEF, but there is no way to do it with only rigid motions. You can get close by reflecting ABC over the x-axis, but here is the resulting image, which shows that you cannot map the triangles exactly onto each other.

–2

–4

–6

–6 2 4 6–2–4

2

4

6

x

y

E

A

B

B'

C

FD

3. JKL and MNO are congruent. There are a number of different sequences of rigid motions that prove this. One such method is to rotate 180º around the origin and then translate 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

N

O

M

K

LEssON 11: dEfiNiNg cONgruENcE



ANSWERSLEssON 11: dEfiNiNg cONgruENcE

4. ∆ABC and ∆DBC are indeed congruent. You can determine that each pair of sides and each pair of angles is congruent, therefore the two triangles are congruent. AB = BD and AC = DC. BC is shared by both triangles. Therefore, ∆ABC and ∆DBC are congruent.

5. First determine the coordinates of each point. This will lead to finding the distances between each vertex and angles between each side. D = (–5, 2), E = (–3, 1), F = (–1, 5) and G = (5, –4), H = (3, –5), I = (1, –1) You can determine the distance between each point to determine if the matching pairs of sides are equal length. DE GH

EF HI

DF GI

≅ =≅ =≅ =

2 24

4 47

5

.

.

units

units

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

The slope of DE is − 12

, and the slope of EF is 2. Since these are negative reciprocals of

each other, the angle they make must be 90°.

Similarly, the slope of GH is 12

, and the slope of HI is –2, these are also negative reciprocals of each other, so the angle they make must be 90°.

All that is left is to show one more pair of angles (the third pair would solve itself once we know the second pair) you can measure it using a protractor or an electronic graphing tool. You will find that ∠ ≅ ∠ = ° ∠ ≅ ∠ = °FED IGH DFE GIH63 43 26 57. .and

6. ∆JKL is not congruent to ∆MNO. There are many ways to show this. As long as you show one of the corresponding pairs of sides or angles does not match, this proves the triangles are not congruent.

The length of JK is 4 units, the length of MN is 4.47 units, found by using the Pythagorean theorem between points M and N. Since these matching sides do not have the same length, the triangles are not congruent.



ANSWERSLEssON 11: dEfiNiNg cONgruENcE

challenge Problem

7. Here is one pair of congruent triangles that meets the criteria. These were found by graphing a right triangle in the third quadrant, and then rotating it 180º into the first quadrant. It is certain that they are congruent, since the rigid motion of the rotation links the two.

–2

–4

–6

–8

–10

–2–4–6–8–10 6 8 1042

2

4

6

8

10

x

y

C

B

F

ED

A



ANSWERS

aNswErs

1. The triangles are congruent by SAS. BC and CD are given as both equal to 5 cm. The 90° angles are congruent, and both triangles share the side CA, which must be congruent to itself. This shows that ∆ABC and ∆ADC are congruent.

2. The triangles are not congruent. Although they share sides that are all 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.

3. The triangles are congruent by ASA. The given 40° and 22° angles match on both triangles, ∠ABC matches∠DCB and ∠ACB matches with ∠DBC. Also the two triangles share the side BC, which must be congruent to itself. Therefore ∆ABC and ∆DCB are congruent.

4. The triangles are not necessarily congruent. Although the 120° angles match, and two of the side lengths are given to match, the position of the matching pieces does not fit any of the congruence criteria. This set up is an example of SSA, where the congruent sides are adjacent, but the congruent angle is not in between them. You could conceivably make different triangles using the given information.

5. The triangles are congruent by SSS. All three sides are given as congruent lengths, therefore the triangles must be congruent.

6. Knowing AC DF≅ shows congruence by SSS, and knowing ∠ABC = ∠DEF shows congruence by SAS.

LEssON 12: triaNgLE cONgruENcE critEria



ANSWERSLEssON 12: triaNgLE cONgruENcE critEria

challenge Problem

7. You could show the congruence using either method, the simpler one might be to show that ABCD can be translated 5 units to the right to map directly onto EFGH. The coordinates are as follows: A = (2, 4), B = (4, 3), C = (4, 0), D = (2, –1)

By adding 4 to every x-coordinate, you get the exact coordinates of 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 by comparing slopes, using circles or a graphing program, you can measure each angle, since you know all of the coordinates.

∠ ≅ ∠ = °∠ ≅ ∠ = °∠ ≅ ∠ =

DAB HEF

CDA GHE

ABC EFG

63.43

63.43

116.577

116.57

°∠ ≅ ∠ = °BCD FGH



ANSWERS

aNswErs

1. The two triangles are congruent, by ASA. AC is a transversal that cuts through both pairs of parallel lines in the figure. This creates opposite interior angles at ∠ ≅ ∠CAB ACD and at ∠ ≅ ∠DAC BCA . Also the two triangles share the segment AC, which is in between the two known angle congruence pairs. AC must be congruent to itself, so the triangles are congruent.

2. 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, you know it cuts a line exactly through the center of the circle. So ∠BCE must be 180º, the given angles make supplementary angles with ∠BCA and ∠ECD, so by a transitive relationship ∠ ≅ ∠BCA ECD .

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

3. The triangles are not congruent. You can get two congruent pieces, with the given angles ∠ ≅ ∠DAC CAB , and the shared side length AC. But this is not quite enough information to conclude that the triangles are congruent.

4. The triangles are congruent by SAS, or by SSS. You know each segment from the center P to a vertex of the pentagon will be congruent, and you know each edge of the pentagon is congruent as well. This gives you PC PB PD PE≅ ≅ ≅ and also BC DE≅ which shows congruence by SSS. You can also use the interior angles of the pentagon to show congruence. Since it is a regular pentagon, you know all of the interior angles are congruent, so ∠ ≅ ∠CPB DPE . This shows congruence using SAS.

5. The triangles are congruent by SAS. Since M is the midpoint of AB, you know AM BM≅ . Since ∠BMC and ∠AMC are complementary and ∠BMC is given as a right angle, ∠ ≅ ∠BMC AMC . This is sufficient to say the triangles are congruent by SAS.

6. The triangles are not congruent. They share the side length BC, and the ∠ABC is given, but there is no other information to add to this. The 60° angle is not duplicated anywhere since no lines are specified as parallel.

LEssON 13: usiNg cONgruENcE critEria



ANSWERSLEssON 13: usiNg cONgruENcE critEria

challenge Problem

7. The triangles are not congruent. Say that the side length of the cube is 1. The base ∆ABD can then be drawn as:

A

D

B

45º

90º 45º1

12

The orange triangle can be drawn in two-dimensional space. The base of triangle DBH shares the longer side of triangle ABD. Segment DB is in both triangles, but they are not corresponding sides. Using the Pythagorean theorem, you can fill in the hypotenuse of the triangle DBH.

H

D B90º

1

2

3

The triangles are not congruent.



ANSWERS

aNswErs

1. 4 lines of symmetry Order 4 rotational symmetry


3. 2 line of symmetry Order 2 rotational symmetry

4. 1 line of symmetry Order 1 rotational symmetry



challenge Problem

7. One way to think about planes of symmetry is to imagine a 2-D mirror slicing through a 3-D figure. If the reflection in the mirror completes the figure, then that is a plane of symmetry. A regular tetrahedron has three planes of symmetry, each bisecting a side and also going through the vertex opposite that side. Students should discuss and debate their methods for analyzing the 3-D shape, here is one example analysis:

There are 7 axes of symmetry. 4 of them connect each vertex to the center of the opposite face. The figure could rotate around each of these axes three times to map to itself.

3 of them connect the midpoints of opposite edges. The figure is symmetric across the planes that cut through the figure this way.

LEssON 14: sYMMEtriEs Of POLYgONs

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