CHAPTER 7mrjaffe.net/GeomRef/unit7solutions.pdf · Chapter 7 Worked-out Solutions Key Copyright ©...
Transcript of CHAPTER 7mrjaffe.net/GeomRef/unit7solutions.pdf · Chapter 7 Worked-out Solutions Key Copyright ©...
CHAPTER 7
Geometry 137Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Think & Discuss (p. 393)
1. The image in box A is flipped to get the image in box B.The image in box C is turned to get the image in box D.
2. Sample answer: If you look at the picture as a whole, theright half is the image of the left half flipped over thecenter vertical red line.
Skill Review (p.394)
1.
2.
and are not congruent.
3.
AB � BC
� �34
� �9 � 25
� �32 � ��5�2
BC � ��7 � 4�2 � �1 � 6�2
� 34
� �9 � 25
� �32 � 52
AB � ��4 � 1�2 � �6 � 1�2
BCAB
� 5
� �25
� �16 � 9
� �42 � 32
BC � ��7 � 3�2 � �4 � 1�2
� �13
� �9 � 4
� �32 � ��2�2
AB � ��3 � 0�2 � �1 � 3�2
AB � BC
� 5�2
� �25�2
� �50
� �49 � 1
� �72 � 12
BC � ��8 � 1�2 � �4 � 3�2
� 5�2
� �25 � �2
� �50
� �49 � 1
� �72 � ��1�2
AB � ��1 � ��6��2 � �3 � 4�2
4. 5. 6.
7. 8.
9.
Lesson 7.1
Developing Concepts Activity 7.1 (p. 395)
Exploring the Concept
1. a. corresponds to b. corresponds to
corresponds to corresponds to
corresponds to corresponds to
corresponds to
corresponds to
c. corresponds to d. corresponds to
corresponds to corresponds to
corresponds to corresponds to
corresponds to corresponds to
corresponds to
2. a. Turn to get
b. Flip figure ABCDE to get figure JKLMN.
c. Slide figure WXYZ to get figure MNPQ.
d. Turn figure NPQRS to get figure TUVWX.
3. The three types of motion that preserve the congruenceof a figure when it is moved in the plane are flip, slide,and turn.
4. Figure ABCD is flipped over line n to get figure EFGH.
5. Yes, EFGH is congruent to JKLM.m
n
AB
DC
H G
E F
L M
KJ
�LKJ.�FGH
TX.NS
WX.RSMQ.WZ
VW.QRPQ.YZ
UV.PQNP.XY
TU.NPMN.WX
JN.AE
MN.DE
LM.CDJL.GH
KL.BCKJ.FG
JK.ABKL.FH
QR � 7.0
�QR�2 � 48.84
�QR�2 � 100 � 148.84
�QR�2 � 102 � 12.22
�QR�2 � �RP�2 � �PQ�2
m�Z � 90�
m�Z � 90� � 180�
m�Z � 55� � 35� � 180�
YZ � QRm�Z � m�Y � m�X � 180�
m�Q � 55�m�X � 35�XZ � 10
MCRBG-0701-SK.qxd 5-25-2001 11:19 AM Page 137
Chapter 7 continued
6. ABCD is turned to get JKLM.
Yes, ABCD is congruent to JKLM.
7. No; there is no line over which ABCD can be flipped togive JKLM.
7.1 Guided Practice (p. 399)
1. An operation that maps a preimage onto an image iscalled a transformation.
2. The preimage and image of a transformation are some-times congruent.
3. A transformation that is an isometry always preserveslength.
4. An isometry never maps an acute triangle onto an obtusetriangle.
5. translation 6. reflection 7. rotation 8. 9.
10. Sample answer: 11.
and
7.1 Practice and Applications (pp. 399–402)
12. Figure ABCDE Figure JKLMN.
13. This transformation is a rotation about the origin. Thefigure ABCDE is turned about the origin.
14. Sample answer: 15. Sample answer:
and and
16.
17. Sample answer:
corresponds to
So,
18. true 19. false 20. true
21. reflection in the line flip over the line and
22. translation, slide 6 units to the right; and
23. Yes; the preimage and image appear to be congruent.
24. Yes; the preimage and image appear to be congruent.
25. No; the preimage and image are not congruent.
26. 27. �DEF → �LKJ�ABC → �PQR
M��3, 4�N��5, �2�,L��2, �2�,
D��6, �1�.C��3, �1�,B��3, 4�,x � 1,x � 1;
AB � JK.
� �5
� �4 � 1
� ���2�2 � 12
JK � ���3 � ��1��2 � �2 � 1�2
� �5
� �1 � 4
� �12 � 22
AB � ��2 � 1�2 � �3 � 1�2
JK.AB
�2, 4�.
�C�LJNAE
→
�VWX�QRS
�WXY
VWST
28. 29.
30. 31.
32.
So,
So,
So,
33.
—CONTINUED—
� 3�2
� �9 � �2
� �18
� �9 � 9
� ���3�2 � 32
AB � ���5 � ��2��2 � �1 � ��2��2
FH � RT.
� 3
� �9
� �9 � 0
� �32 � 02
RT � ��4 � 1�2 � ��1 � ��1��2
� 3
� �9
� �0 � 9
� �02 � 32
FH � ���1 � ��1��2 � �4 � 1�2
GH � ST.
� �13
� �4 � 9
� �22 � ��3�2
ST � ��4 � 2�2 � ��1 � 2�2
� �13
� �9 � 4
� �32 � 22
GH � ���1 � ��4��2 � �4 � 2�2
FG � RS.
� �10
� �1 � 9
� �12 � 32
RS � ��2 � 1�2 � �2 � ��1��2
� �10
� �9 � 1
� ���3�2 � 12
FG � ���4 � ��1��2 � �2 � 1�2
�RQP → �CBA�LJK → �DFE
�PRQ → �ACB�KJL → �EFD
138 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0701-SK.qxd 5-25-2001 11:19 AM Page 138
Geometry 139Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
33. —CONTINUED—
So,
So,
So,
34.
36. translation 37. translation 38. reflection
39. rotation
40. Yes, a point or a line segment can be its own preimagewhen it is rotated or when it is reflected. Points or linesegments on a line of reflection are their own preimages.A center of rotation is its own preimage.
41. From A to B, the stencil is reflected. From A to C, thestencil is reflected. From A to D, the stencil is eitherrotated or reflected twice.
d � 2
3d � 6
c � 7
b � 92
a � 48
2a� � 96�
AC � XZ.
� 4
� �16
� �0 � 16
� �02 � 42
XZ � ��0 � 0�2 � �2 � ��2��2
� 4
� �16
� �0 � 16
� �02 � 42
AC � ���2 � ��2��2 � �2 � ��2��2
BC � YZ.
� �10
� �9 � 1
� ���3�2 � 12
YZ � ��0 � 3�2 � �2 � 1�2
� �10
� �9 � 1
� �32 � 12
BC � ���2 � ��5��2 � �2 � 1�2
AB � XY.
� 3�2
� �9�2
� �18
� �9 � 9
� �32 � 32
XY � ��3 � 0�2 � �1 � ��2��2
42. The letters b, d, p, and q can be formed from each otherby reflection or rotation; the letters n and u can beformed from each other by rotation or repeated reflection.
43. Sample answer: The lower right corner is the horizontalreflection of the upper right corner. Then reflect the lowerright corner vertically to get the pattern for the lower leftcorner. From there, reflect horizontally to get the upperleft corner.
44. B 45. D
46. Statements Reasons
1. and 1. Given
are isometries.
2. 2. Definition of isometry
3. 3. Transitive property of equality
4. 4. Definition of and congruent segments
5. is an 5. Definition of isometryisometry.
7.1 Mixed Review (p. 402)
47.
48.
49.
50.
� 2�34
� �4 � �34
� �136
� �36 � 100
� ���6�2 � ��10�2
GH � ��0 � 6�2 � ��7 � 3�2
� �89
� �64 � 25
� �82 � 52
EF � ��0 � ��8��2 � �8 � 3�2
� 5�17
� �25 � �17
� �425
� �256 � 169
� �162 � ��13�2
CD � ��5 � ��11��2 � ��7 � 6�2
� 13
� �169
� �25 � 144
� �52 � 122
AB � ��3 � ��2�2 � �10 � ��2��2
�ABC → �XYZ
AC � XZBC � YZ,AB � XY,
AC � XZBC � YZ,AB � XY,
QR � YZ, and PR � XZAC � PR, PQ � XY,AB � PQ, BC � QR,
�PQR → �XYZ
�ABC → �PQR
35.
w � 35
2w� � 70�
x �133 � 41
3
3x � 13
3x � 1 � 14
y � 3
2y � 6
MCRBG-0701-SK.qxd 5-25-2001 11:19 AM Page 139
Chapter 7 continued
51. polygon 52. polygon
53. Not a polygon because one side is not a line segment.
54. Not a polygon because one side is not a line segment.
55. Not a polygon because two of the sides intersect only oneother side.
56. polygon
57. Sample answers:
(1) slope of
slope of
slope of
slope of
and
So, PQRS is a parallelogram because both pairs of oppo-site sides are parallel.
(2)
and
PQRS is a parallelogram because opposite sides are congruent.
QR � PSPQ � RS
� �65
� �1 � 64
� ���1�2 � 82
PS � ��0 � 1�2 � �4 � ��4��2
� �65
� �1 � 64
� ���1�2 � 82
QR � ��7 � 8�2 � �6 � ��2��2
� �53
� �49 � 4
� �72 � 22
RS � ��8 � 1�2 � ��2 � ��4��2
� �53
� �49 � 4
� ���7�2 � ��2�2
PQ � ��0 � 7�2 � �4 � 6�2
QR � PSPQ � RS
PS �4 � ��4�
0 � 1�
8�1
� �8
QR �6 � ��2�
7 � 8�
8�1
� �8
RS ��4 � ��2�
1 � 8�
�2�7
�27
PQ �6 � 47 � 0
�27
58. Sample answers:
(1) slope of
slope of
slope of
slope of
and
WXYZ is a parallelogram because opposite sides are parallel.
(2)
and
WXYZ is a parallelogram because opposite sides are congruent.
XY � WZWX � YZ
� 3�5
� �9 � �5
� �45
� �9 � 36
� �32 � 62
WZ � ��1 � ��2��2 � �5 � ��1��2
� 3�5
� �9 � �5
� �45
� �9 � 36
� �32 � 62
XY � ��9 � 6�2 � �5 � ��1��2
� 8
� �64
� �64 � 0
� �82 � 02
YZ � ��6 � ��2��2 � ��1 � ��1��2
� 8
� �64
� �64 � 0
� �82 � 02
WX � ��9 � 1�2 � �5 � 5�2
XY � WZWX � YZ
WZ �5 � ��1�1 � ��2� �
63
� 2
XY �5 � ��1�
9 � 6�
63
� 2
YZ ��1 � ��1�6 � ��2� �
08
� 0
WX �5 � 59 � 1
�08
� 0
140 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0701-SK.qxd 5-25-2001 11:19 AM Page 140
Geometry 141Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
Lesson 7.2
Developing Concepts Activity 7.2 (p. 403)
Exploring the Concept
Sample answer:
Investigate
1. Measurements will vary, but , and.
2. The measure of each angle is
3. The line m is the perpendicular bisector of each segment.
Exploring the Concept
Sample answer:
Make a Conjecture
4. Line m is the perpendicular bisector of each segment;Sample answer: and
and
5. The line of reflection is the perpendicular bisector of thesegment connecting a point and its image.
7.2 Guided Practice (p. 407)
1. A line of symmetry is a line in which a figure can bereflected onto itself.
2. When a point is reflected in the x-axis, the x-coordinatesof the point and its image are the same and the y-coordinates are opposites.
3. not a reflection 4. not a reflection 5. reflection
6. 7.
8. 9. 10.
11. 12. 3 lines of symmetry
13. 4 lines of symmetry 14. 5 lines of symmetry
DC → DG
�CBA → �GFED → DC → G
�DAB → �DEFAB → EF
BD � B�D.AC � A�Cm�ACD � m�BDC � 90�
C
BB'
A A'
Dm
90�.
YC � CY�ZB � BZ�XA � AX�,
Y Y'
Z Z'
X X'A
B
Cm
7.2 Practice and Applications (pp. 407–410)
15. 16.
17.
18. True; N is 2 units above the line so its image is 2 units below the line.
19. True; M is 3 units to the right of the line so itsimage is 3 units to the left of the line.
20. False; W is one unit below the line so its imageis one unit above the line. Its image should be
21. True; U is 4 units to the right of the line so itsimage is 4 units to the left of the line.
22. 23. 24. 25.
26. 27.
28. 29.
y
x2
2
R'(�7, �2) R(7, �2)
y
x1
1
Q(�3, �3) Q'(3, �3)
R���7, �2�Q��3, �3�
y
x2
2
T(3, 8)
T'(3, �8)
y
x1
1
S(0, 2)
S'(0, �2)
T��3, �8�S��0, �2�
EFFECDGH
x � 1,
W���6, �1�.
y � �2,
x � 3,
y � 2,
k
k
k
MCRBG-0701-SK.qxd 5-25-2001 11:19 AM Page 141
Chapter 7 continued
30. Sample answer:
The coordinates of the vertices of the image of arereversals of the coordinates of the vertices of the preim-age,
31. 32.
33. Draw and intersecting line m at points S and T.intersects line m at R. By the definition
of reflection, and and and and by theReflexive Property of Congruence. Angles
and RTQ are right angles (definition of perpendicular) and are congruent (all right angles arecongruent). It follows that and
by the SAS Congruence Postulate. Since corresponding parts of s are and So and Since
and by theSegment Addition Postulate, we get by substitution
or
34. Since P is on m, then by definition of a reflec-tion. By definition of reflection, and where R is the point where m and intersect. Since
and are right angles.because all right angles are congruent.
by the Reflexive Property of Congruence. Soby SAS Congruence Postulate.
Therefore or because correspond-ing parts of congruent triangles are congruent. Finally,
by definition of congruent segments.
35. By definition of a reflection, m is the perpendicular bisec-tor of Q is on m, and Then bythe definition of reflection. But so
36. Reflect H in line n to obtain its image Then draw aline This will intersect n in a point K. Then the distance traveled, HK � KJ, will be as small as possible.
↔H�J.
H�.
PQ � P�Q�.Q � Q�,PQ � P�QQ � Q�.PP�,
PQ � P�Q�
PQ � P�Q�PQ � PQ��PRQ � �PRQ�PR � PR�PRQ � �PRQ�
�PRQ��PRQm�QQ�,QQ�
QR � Q�Rm�QQ�P � P�
PQ � P�Q�.PQ � P�Q�,
PQ � PR � QRP�Q� � P�R � Q�RQ�R � QR.P�R � PRQ�R � QR.
P�R � PR�,���Q�TR � �QTR
�P�SR � �PSR
RSP, RTQ�P�SR,
SR � SRRT � RTRT�QQ�.Q�T � QTRS�PP�P�S � PS
�and P�Q��PQQQ�PP�
k
j
�y, x�.
�x, y�
y
x2
2A(9, 3)
C'(7, 5)
B(11, 9)
B'(9, 11)A'(3, 9)
C(5, 7)
37.
39.
41. The two molecules are reflections of each other.
42. Triangle 2 is a reflection of triangle 1; triangle 3 is atranslation of triangle 1.
43. Triangles 2 and 3 are reflections of triangle 1. Triangle 4is a rotation of triangle 1.
44.
46.
47. Drawings will vary. The distance between each vertex ofthe preimage and line m is equal to the distance betweenthe corresponding vertex of the image and line m.
48.
50. B 51. B
z � 3
2z � 6
2z � 1 � 5
y � 36
12 y � 18
12 y � 10 � 8
x �43
3x � 4
m�A � 60�
3�m�A� � 180�
n�m�A� � 180�
m�A � 45�
4�m�A� � 180�
n�m�A� � 180�
�3, 0�
y
x4
2
A'(�1, �4)
A(�1, 4) B(6, 3)
C(3, 0)
�6, 0�
y
x2
2
A'(1, �5)
A(1, 5)
B(7, 1)
C(6, 0)
142 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
45.
m�A � 90�
2�m�A� � 180�
n�m�A� � 180�
38.
40.
��1, 0�
y
x2
2
A'(�4, �6)
A(�4, 6)B(3.5, 9)
C(�1, 0)
�5, 0�
y
x2
2A'(2, 2)
A(2, �2)B(11, �4)
C(5, 0)
49.
w � 5
3w � 15
v �295
5v � 29
5v � 10 � 19
u � 6
2u � 12
2u � 1 � 13
MCRBG-0702-SK.qxd 5-25-2001 11:19 AM Page 142
Geometry 143Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
52. 53.
54. slope of
The slope of is So the slope of the line perpendic-
ular to is because
55.
56. (52)
(53)
(54) slope of
The slope of is so a line perpendicular tohas slope 1 because
(55)
7.2 Mixed Review (p. 410)
57. 58. 59.
60. 61.
62.
63.
65.
21 < c < 45
33 � 12 < c < 12 � 33
b � a < c < a � b
10 < c < 24
17 � 7 < c < 7 � 17
b � a < c < a � b
�R � �C
m�Q � m�B � 101�m�C � m�R � 35�
QR � BCPQ � AB � 12�A � �P
y � x
y �12 � x �
12
y �12 � 1�x �
12�
�1 � 1 � �1.RR��1RR�
� �1�5
�5�
3 � ��2��2 � 3
RR�
Q � �12
, 12�
Q � �3 � ��2�2
, �2 � 3
2 �
y
x
1
R(�2, 3)
R'(3, �2)
1
Q( , )12
12
y � �2x
y � 0 � �2�x � 0�
�2 �12
� �1.�2RR�
12
.RR�
�12
�24
�1 � ��1�2 � ��2�RR�
Q � �0, 0�
Q � �02
, 02�
Q � ��2 � 22
, �1 � 1
2 �y
x
1
R'(�2, �1)
R(2, 1)
Q(0, 0)
67.
68.
69.
70.
71.
Lesson 7.3
Technology Activity 7.3 (p. 411)
1.
2. a rotation about the point of intersection of the lines
3. Answers will vary.
4. The measure of the acute angle is half the measure of
5. twice the measure of the acuteangle formed by lines m and k
6. The measure of the angle of rotation is twice the measureof the acute angle formed by the two lines.
Extension
The conjecture is correct.
Activity 7.3 (p. 413)
Answers may vary.
Sample answer:
P
A
BCA'
B'
C'
m�BPB� � m�CPC� �
�APA�.
�ABC � �A�B�C�.
m�A � 106� m�C � 61�
m�A � 74 � 180 m�C � 119� � 180�
m�A � m�D � 180� m�C � m�B � 180�
� 180� � 115� � 65�
m �A � 180� � m�B
m�D � 90�
� 119�
� 180� � 61�
m�A � m�B � 180� � m�D
m�C � m�D � 61�
4.8 < c < 19
11.9 � 7.1 < c < 7.1 � 11.9
b � a < c < a � b
25.7 < c < 56.7
41.2 � 15.5 < c < 41.2 � 15.5
a � b < c < a � b
64.
66.
20 < c < 32
26 � 6 < c < 26 � 6
a � b < c < a � b
12 < c < 30
21 � 9 < c < 9 � 21
b � a < c < a � b
MCRBG-0702-SK.qxd 5-25-2001 11:19 AM Page 143
Chapter 7 continued
7.3 Guided Practice (p. 416)
1. A center of rotation is the fixed point about which a figure being rotated is turned.
2. counterclockwise
3. Yes, because a rotation is an isometry.
4. No, because the distance between any pointand its image after a rotation is not fixed.
5. The measure of the acute angle between k and m wouldbe half the measure of the angle of rotation. So it wouldbe
6. A clockwise rotation of about P maps R onto S.
7. A counterclockwise rotation of about P maps Ronto Q.
8. A clockwise rotation of about Q maps R onto W.
9. A counterclockwise rotation of about P maps Vonto R.
10. The figure has rotational symmetry about its center witha rotation of either clockwise or counterclockwise.
11. The figure has rotational symmetry about its center witha rotation of either clockwise or counterclockwise.
12. The figure does not have rotational symmetry.
7.3 Practice and Applications (pp. 416–419)
13. 14. 15. 16. 17.
18. 19.
20. By definition of a rotation, and By the definition of congruent segments and
By the Segment Addition Postulate,, so by the
subtraction property of equality. Finally bydefinition of congruent segments.
21. By definition of rotation Since P and R arethe same point and R and are the same point, then
22. 23.
P
Q
Q�
T
T�
R
R�
S
S�
A
C�
A�
B
C
B�
P
QR � Q�R�.R�
QP � Q�P.
RQ � R�Q�RQ � R�Q�PR� � R�Q�PR � RQ �
PQ � PQ�.PR � PR�
PQ � PQ�.PR � PR�
�CPA�FGL
�MABBMGELHCD
180�,
180�,
180�
120�
60�
60�
12 � �90�� � 45�.
AA� BB�
AB � A�B�
24.
25.
26.
27.
28. the x-coordinate of theimage is the y-coordinate of the preimage. The y-coordi-nate of the image is the opposite of the x-coordinate ofthe preimage.
29. and the x-coordinate of theimage is the opposite of the x-coordinate of the preimage.The y-coordinate of the image is the opposite of the y-coordinate of the preimage.
30. The measure of the angle of rotation from tois twice the measure of the acute angle of the
intersecting lines, which is or
31. The measure of the angle of rotation from tois twice the measure of the acute angle of the
intersecting lines, which is or
32. The measure of the angle of rotation about D is or
33. The measure of the acute angle between lines m and n isor
34.
36. The wheel hub can be mapped onto itself by a clockwiseor counterclockwise rotation of or about its center.
180�135�,45�, 90�,
a � 55
2a� � 110�
c � 14
c2
� 7
d � 8
d � 2 � 10
b � 4
3b � 12
e �74
4e � 7
4e � 2 � 5
81�.12 � 162�
72�.2 � 36�
30�.2 � 15��A�B�C�
�ABC
70�.2 � 35��A�B�C�
�ABC
Z���3, 4�;X��2, 3�,O��0, 0�,
C��2, �5�;B��4, �2�,A��1, 1�,
F��2, 5�.E��0, 2�,D��4, 1�,S���2, 0�.R���4, �2�,Q���3, �5�,P���1, �3�,
J��1, 2�, K��4, 1�, L��4, �3�, M��1, �3�
PW
W �
Z
Z �
X
X �
Y
Y �
144 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
35.
s � 11
u � 2
2u � 4
r � 5
2r � 10
t � 1
3t � 3
q � 30
2q� � 60�
MCRBG-0702-SK.qxd 5-25-2001 11:19 AM Page 144
Geometry 145Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
37. The wheel hub can be mapped onto itself by a clockwiseor counterclockwise rotation of or (which is and respectively)about its center.
38. The wheel hub can be mapped onto itself by a clockwiseor counterclockwise rotation of and (which is
and respectively) about its center.
39. Yes, the image can be mapped onto itself by a clockwiseor counterclockwise rotation of about its center.
40. Yes; the answer would change to a clockwise or counter-cockwise rotation of or about its center. This isbecause the white figures can be mapped onto the blackfigures.
41. The center of rotation is the point of intersection of thediagonals of the square.
42. Yes, it is possible for the piece to be hung upside downbecause the rotational symmetry has an angle of rotationof This would make the picture the same right sideup and upside down.
43. a. Graph for a–c.
b.
c.
d. A single transformation that maps ontowould be a counterclockwise rotation of
about the origin.
e. Any polygon can be rotated counterclockwiseabout the origin by doing two reflections of the poly-gon. First, reflect the polygon in one of the axis. Thenreflect the result of the first reflection in the line
or Then the measure of the acuteangle between the two lines is and the angle ofrotation is 90�.
45�y � x.y � �x
90�
90��R�S�T��RST
T���1, 5�S���3, 4�,R���1, 1�,
T���5, 1�S���4, 3�,R���1, 1�,
y
x1
1
S
TR''
T''
T'
S'S''
R' R
180�.
180�90�
180�
360 � 25,360 � 1
5
144�72�
360� � 37,360� � 2
7,360� � 17,
15427�1026
7�,5137�,
44. a. By definition of a reflection, and where A is the point of intersection of k and
and are right angles, andbecause all right angles are congru-
ent. By the Reflexive Property of Congruence,So, by the SAS
Congruence Postulate. By corresponding parts of con-gruent triangles are congruent, By defini-tion of a reflection, and where Bis the point of intersection of m and and are right angles, and since all right angles are congruent. by theReflexive Property of Congruence. So
by the SAS Congruence Postulate.Since corresponding parts of congruent triangles arecongruent, Since then
by the Transitive Property of Congruence.is a rotation of Q about point P.
b. and becausecorresponding parts of s are By the definition of congruent angles, and
By the Angle AdditionPostulate,
Then,by
substitution. By the Distributive property,Finally
by substitution,
7.3 Mixed Review (p. 419)
45. 46.
47.
48. 49.
50.
51. Sample answer: 52. Sample answer:
The circumcenter is outside the triangle when The circumcenter of a the triangle is obtuse. right triangle is always
on the triangle.
m�8 � m�6 � 98�
m�4 � 98� m�6 � 98�
m�4 � 82� � 180� m�6 � 82� � 180�
m�4 � m�1 � 180� m�6 � m�5 � 180�
m�3 � m�1 � 82�
m�7 � m�5 � 82�m�5 � m�1 � 82�
m�QPQ� � 2�m�APB�.m�QPQ� � 2�m�Q�PA � m�Q�PB�.
m�Q�PA � m�Q�PB � m�Q�PBm�QPQ� � m�Q�PA �m�Q�PB.
m�QPQ� � m�QPA � m�Q�PA � m�Q�PB �m�APB � m�Q�PA � m�Q�PB,
m�QPA � m�Q�PA.m�Q�PB � m�Q�PB
�.���Q�PB � �Q�PB�QPA � �Q�PA
Q�QP � Q�P
QP � Q�PPQ� � PQ�.
�Q�BP � �Q�BP
BP � BP�Q�BP � �Q�BP�Q�BP
�Q�BPQ�Q�.Q�B � Q�Bm�Q�Q�
PQ � PQ�.
�QAP � �Q�APAP � AP.
�QAP � �Q�AP�Q�AP�QAP
QQ�.QA � Q�Ak�QQ�
MCRBG-0702-SK.qxd 5-25-2001 11:19 AM Page 145
Chapter 7
53. Sample answer:
The circumcenter of an acute triangle is always inside thetriangle.
54. Only one pair of sides are given as parallel, which is notenough information to show that the figure is a parallelo-gram.
Quiz 1 (p. 420)
1. Figure Figure RSTQ.
2. The transformation is a reflection in line m. The figure isflipped over line m.
3. Yes, the reflection is an isometry because it preserveslength.
4. 5. 6.
7.
8. The rotations that map the knot onto itself are rotationsby multiples of clockwise or counter-clockwise about the center of the knot where the ropestarts to unravel.
Math & History (p. 420)
1. The design has 2 lines of symmetry.
2. The design has rotational symmetry. It can be mappedonto itself by a rotation of 180° clockwise or counter-clockwise about its center.
Lesson 7.4
7.4 Guided Practice (p. 425)
1. A vector is a quantity that has both direction and magnitude.
2. Sample answer: The direction is incorrect. starts atP and ends at Q. So the vector from P to Q is
3. 4.
5. 6.
7. If maps onto then maps
8. If maps onto then maps onto
9. If maps onto then maps onto
10. If maps onto then maps onto
11.–14. Sample figures are given.
�0, 0�.�8, 5���8, �3�,�0, 2�
�5, �2�.�8, 5���3, �5�,�0, 2�
�3, 7�.�8, 5���5, 4�,�0, 2�
�8, 3�.�8, 5��0, 0�,�0, 2�
�x, y� → �x � 5, y � 8�.�x, y� → �x � 7, y � 1�.
�x, y� → �x � 4, y � 3�.�x, y� → �x � 6, y � 2�.
�6, �2�.PQ
\
120� �360� 3�
P���8.2, �3�
N���4, 0�M��2, �4�L��2, �3�
ABCD →
11. 12.
13.
14.
7.4 Practice and Applications (pp.425–428)
15. a. b.
16. a. b.
17. 18. 19.
20. 21. k and m
22. Sample answer: and
23.
24. Yes, the distance from to m is the same as the distancefrom to m because is reflected in m onto and bythe definition of reflection, the distances are equal.
25. The image of is or
26. The image of is or
27. The preimage of is or
28. The preimage of is or
29. The image of is or
30. The preimage of is or
31. y
x�1
1
R�
S�
P�
Q�
R
S
P
Q
�1
��17.5, 1.5�.��5.5 � 12, �5.5 � 7���5.5, �5.5�
�12.5, �4.5�.�0.5 � 12, 2.5 � 7��0.5, 2.5�
��12, 1�.�0 � 12, �6 � 7��0, �6�
��14, 8�.��2 � 12, 1 � 7���2, 1�
�11, �9�.��1 � 12, �2 � 7���1, �2�
�17, �4�.�5 � 12, 3 � 7��5, 3�
B�B�B�B�
CC� � 1.4 � 1.4 � 2.8 inches
CC�AA�
�A�B�C�
�5, 0�MN\
;��4, �4�LK\
;�4, 2�HJ\
;
��5, 2��x, y� → �x � 5, y � 2�
��3, �4��x, y� → �x � 3, y � 4�
146 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0702-SK.qxd 5-25-2001 11:19 AM Page 146
Geometry 147Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
32.
33.
34.
35. true 36. false 37. true. 38. true
39.
40. y
x
1
1
C
A
A� B
C�
B�
y
x1
1
C
A
A�
B
C�
B�
y
x�1
1R�
S�
P�
Q�
R
S
P
Q
y
x�1
1R�
S�
P�
Q�
R
S
P
Q
�1
y
x�1
1
R�
S�
P�
Q�
R
S
P
Q
�1
41.
42.
43. We are given and Suppose has coordi-nates Then
or The
slope of or If and
as given, then and the slope
of is So, the coordinates of are
By the Distance Formula,
and
or Thus, by the substitution prop-erty of equality,
44. C 45. D 46. A 47. B
48. yes 49. no 50. yes
51. Samples might include photographs of floor tiles or fabric patterns.
52. C: D: E:F:
53. The two vectors are and
54. To arrive at D from C, the vector is
55. To go straight from town A to town D, the vector wouldbe
56. The correct answer is C because a translation preserveslength.
AD\
�18, 12�.
CD\
�8, 2�.
BC\
�4, 6�.AB\
�6, 4�
�x, y� → �x � 12, y � 6��x, y� → �x � 6, y � 6�.�x, y� → �x, y � 6�,�x, y� → �x � 12, y�,
PQ � P�Q�.��a � c�2 � �b � d�2.
P�Q� � ���a � r� � �c � r��2 � ��b � s� � �d � s��2
PQ � ��a � c�2 � �b � d�2
�c � r, d � s�.Q�sr.QQ�
QQ� � �r2 � s2PP� QQ�
PP� � QQ�sr.PP� �
b � s � ba � r � a
�r2 � s2.PP� � ��a � r � a�2 � �b � s � b�2
�a � r, b � s�.P�Q�c, d�.P�a, b�
C
AA�
BC�
B�
y
x
2
1
C
A
A�
B
C�
B�
y
x
1
1
MCRBG-0703-SK.qxd 5-25-2001 11:19 AM Page 147
Chapter 7 continued
57.
B
58.
A
59.
A
� 4�2
� �16 � �2
� �32
� �16 � 16
� ���4�2 � ��4�2
A�B� � ��4 � 8�2 � ��1 � 3�2
� 2�13
� �4 � �13
� �52
� �36 � 16
� �62 � ��4�2
A�B� � ��9 � 3�2 � ��1 � 3�2
� 5
� �25
� �25 � 0
� �52 � 02
A�A� � ��8 � 3�2 � �3 � 3�2
� �34
� �25 � 9
� �52 � ��3�2
BB� � ��4 � ��1��2 � ��1 � 2�2
� �34
� �25 � 9
� �52 � ��3�2
AA� � ��3 � ��2��2 � �3 � 6�2
� �17
� �1 � 16
� �12 � ��4�2
AB � ���1 � ��2��2 � �2 � 6�2 60.
7.4 Mixed Review (p. 428)
62.
64.
66.
68. If then 69. If then
70. If then 71. true 72. false 73. false
Lesson 7.5
Developing Concepts Activity 7.5 (p. 429)
Exploring the Concept
y
x1
1 A
A' A''
B
B'
B''
C
C'C''
QS � 6.RL � 6,
JL � 12.QR � 6,SR � 6.JK � 12,
�53
�106
m �10 � 0
0 � ��6�
� 0
�09
m �1 � 1
�1 � ��10�
�67
��6�7
m ��8 � ��2�
�7 � 0
23 � z
2 � 3z
1 � 1 � 3z
y � 0
8y � 0
8y � 3 � 3
8y � 1 � 4 � 3
w � 3
w � 1 � 4
1 � x
2 � 2x
3 � 2x � 1
�1 � 4 � 2x � 1
148 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
61.
2 � u
�6 � �3u
u � 6 � �2u
t � 2
2t � 4
2t � 1 � 5
2t � 2 � 3 � 5
1 � s
2 � s � 1
8 � 6 � s � 1
r � 1
r � 2 � 3
r � 1 � 3 � 3
63.
65.
67.
�34
�912
m �6 � ��3�9 � ��3�
� �6
��61
m �6 � 12
�1 � ��2�
� �5
�15�3
m �18 � 3�1 � 2
MCRBG-0703-SK.qxd 5-25-2001 11:19 AM Page 148
Geometry 149Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
Investigation
1.
2.
3. Yes, the order in which transformations are completedaffects the final image.
Investigate
4.
5.
the coordinates arethe same; switching the order of the transformations didnot affect the image in this example.
K��11, �8�;J��8, �7�,H��11, �3�,
y
x2
1
H
J
J' J''K' K''
KH' H''
K��11, �8�H��11, �3�, J��8, �7�,
y
x2
1
H
J
J''K''
KH''
J'K'
H'
C��13, 4�B��11, 5�,A��10, 2�,
y
x1
1 A
A'
A''B
B'
B''C
C'
C''
C��5, �4�B��7, �5�,A��8, �2�,
Extension
So,
So,
So,
The transformation that maps onto is anisometry, by definition.
7.5 Guided Practice (p. 433)
1. In a glide reflection, the direction of a the translationmust be parallel to the line of reflection.
2. The order in which two transformations are performedsometimes affects the resulting image.
3. In a glide reflection, the order in which the two transfor-mations are performed never matters.
4. A composition of isometries is always an isometry.
5. 6.
7. The line of reflection is the y-axis.
8. �x, y� → �x, y � 3�
A�B�A�B�
�A�B�C��ABC
BC � B�C�.
� �5
� �4 � 1
� �22 � ��1�2
B�C� � ��13 � 11�2 � �4 � 5�2
� �5
� �1 � 4
� ���1�2 � 22
BC � ��3 � 4�2 � �4 � 2�2
AC � A�C�.
� �13
� �9 � 4
� �32 � 22
A�C� � ��13 � 10�2 � �4 � 2�2
� �13
� �4 � 9
� �22 � 32
AC � ��3 � 1�2 � �4 � 1�2
AB � A�B�.
� �10
� �1 � 9
� �12 � 32
A�B� � ��11 � 10�2 � �5 � 2�2
� �10
� �9 � 1
� �32 � 12
AB � ��4 � 1�2 � �2 � 1�2
MCRBG-0703-SK.qxd 5-25-2001 11:19 AM Page 149
Chapter 7 continued
7.5 Practice and Applications (pp. 433–436)
9. A 10. C 11. B
12. 13.
14. 15.
16. 17.
18.
19.
20. The order does affect thefinal image.
y
x1
1
F��(�4, 4)
G ��(2, �1)
F ��(4, �4)
G��(�2, 1)
reflection,thenrotation
rotation,thenreflection
P �(�2, 7)
Q �(�7, 6)
R �(1, 2)
y
x
1
1�1
y
x
1
P �(2, 7)
Q �(2, 4)
R �(6, 5)
1
y
x
2
�22�2
P �(�4, �2)
Q �(�7, �6)
R �(�8, 1)
y
x
1
�1�1
P �(2, 1)
Q �(0, �2)
R �(3, �4)
A''(2, �6)
A'(�6, 2)
y
x2
2
A(�3, 5)
y � x
A'(�9, 4)A''(7, 4)
y
x2
2
A(�3, 5)
x � �1
A''(1, �10)
A'(1, 6)
y
x2
2
A(�3, 5)
y � �2A'(�3, 1) A''(3, 1)
y
x1
1
A(�3, 5)
21.
The order does affect the final image.
22. reflection in the line followed by a clockwiserotation of about the origin
23. reflection in the line followed by a reflection inthe line
24. clockwise rotation of about the origin, followed bythe translation
25. counterclockwise rotation of about the point (0, 1),followed by the translation
26. A glide reflection is an isometry because it is a composi-tion of a translation and a reflection, both of which areisometries. The composition of two isometries is an isometry.
27. A, B, and C are preserved in a glide reflection.
28. Answers will vary.
29.
So,
30.
So,
31. After each part was painted, the stencil was movedthrough a glide reflection. The translation moved it to the right and the reflection in a horizontal line through its center flipped the design.
32. 5 33. 1, 4, 5, 6 34. 2, 7
35. The pattern can be made by a horizontal translation,rotation, vertical line reflection, or horizontal glide reflection.
36. The pattern can be made by a vertical translation.
37. The pattern can be made by a translation or rotation.
38. Sample answer: The X tile needs to be rotated clock-wise, then reflected in a horizontal line. The Y tile needsto be reflected in a vertical line, then rotated counter-clockwise.
90�
90�
180�
180�
�x, y� → �x � 15, y�
�x � 8, y � 4� → �x � 15, y�
�x � 9, y � 4� → �x � 9 � 6, y � 4 � 4�
�x, y� → �x � 9, y � 4�
�x, y� → �x � 6, y � 1�.
�x � 7, y � 2� → �x � 6, y � 1�
�x � 7, y � 2� → �x � 7 � 1, y � 2 � 3�
�x, y� → �x � 7, y � 2�
�x, y� → �x � 2, y � 3�90�
�x, y� → �x, y � 3�90�
x � �2y � 2,
90�y � �
12,
y
x1
1
G�(4, 8) G�(8, 8)
F �(1, 7) F �(5, 7)
translation,thenreflection
reflection,thentranslation
150 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0703-SK.qxd 5-25-2001 11:19 AM Page 150
Geometry 151Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
39. a. Answers will vary.
b. Conjecture: The midpoint of the segment connectingthe point and its image is on the x-axis.
c. Let be the original point, since the translationmust be parallel to the line of reflection, the coordi-nates of the image are for some numbera. The coordinates of the midpoint are
Then the midpoint is on the x-axis
d. Yes; the midpoint is the point where the segments andthe line of reflection intersect.
40.
7.5 Mixed Review (p. 436)
41. 42. 43.
44.
45.
—CONTINUED—
� �17
� �1 � 16
� �12 � ��4�2
QR � ��6 � 5�2 � ��5 � ��1��2
� �17
� �16 � 1
� �42 � 12
PQ � ��5 � 1�2 � ��1 � ��2��2
�3 � g
�9 � 3g
�4 � 3g � 5
2 � d
10 � 5d
�1 � 5d � 11
�4 � 3 � 5d � 11
a � 2
2a � 4
�1 � f
1 � �f
���1� � �f
0 � c
1 � c � 1
�2 � 3 � c � 1
x � a � x2
, �y � y
2 � � 2x � a2
, 0�.
�x � a, �y�
�x, y�
45. —CONTINUED—
so PQRS is a rhombus.
Since the diagonals of PQRS are congruent, soPQRS is a rectangle. By the Square Corollary, PQRS is asquare.
46.
Since opposite sides are congruent, PQRS is a parallelogram.
—CONTINUED—
� 6
� �36
� �0 � 36
� �02 � ��6�2
PS � ��10 � 10�2 � �1 � 7�2
� 5
� �25
� �25 � 0
� ���5�2 � 02
RS � ��10 � 15�2 � �1 � 1�2
� 6
� �36
� �0 � 36
� �02 � ��6�2
QR � ��15 � 15�2 � �1 � 7�2
� 5
� �25
� �25 � 0
� �52 � 02
PQ � ��15 � 10�2 � �7 � 7�2
PR � QS,
� �34
� �9 � 25
� �32 � 52
QS � ��5 � 2�2 � ��1 � ��6��2
� �34
� �25 � 9
� ���5�2 � 32
PR � ��1 � 6�2 � ��2 � ��5��2
PQ � QR � RS � PS
� �17
� �1 � 16
� �12 � ��4�2
PS � ��2 � 1�2 � ��6 � ��2��2
� �17
� �16 � 1
� ���4�2 � ��1�2
RS � ��2 � 6�2 � ��6 � ��5��2
�2 � h
2 � h � 4
32 � e
64 � e
6 � 4e
b � 5
b � 3 � 2
b � 6 � 3 � 2
MCRBG-0703-SK.qxd 5-25-2001 11:19 AM Page 151
Chapter 7 continued
46. —CONTINUED—
Since the diagonals are congruent, so PQRS isa rectangle.
47.
All four sides are congruent, but the diagonals are notcongruent, so PQRS is a rhombus but not a rectangle or a square.
� 4
� �16
� �42 � 02
QS � ��10 � 6�2 � ��7 � 7��2
� 6
� �36
� �02 � 62
PR � ��8 � 8�2 � ��4 � ��10��2
� �13
� �4 � 9
� ���2�2 � ��3�2
PS � ��6 � 8�2 � ��7 � ��4��2
� �13
� �4 � 9
� ���2�2 � 32
RS � ��6 � 8�2 � ��7 � 10��2
� �13
� �4 � 9
� ���2�2 � ��3�2
QR � ��8 � 10�2 � ��10 � ��7�2
� �13
� �4 � 9
� �22 � ��3�2
PQ � ��10 � 8�2 � ��7 � ��4��2
PR � QS,
� �61
� �25 � 36
� �52 � 62
QS � ��15 � 10�2 � �7 � 1�2
� �61
� �25 � 36
� ���5�2 � 62
PR � ��10 � 15�2 � �7 � 1�2
48.
49.
50.
51.
52.
53.
C���5, 8.5�B���9, 3.5�,A���9, 9.5�,
�1
1
y
xB
B�
C
C �
A
A�
C���9, 2�B���13, �3�,A���13, 3�,
�2
2
y
xB
C A
B�
C �
A�
A���3, 7�, B���3, 1�, C��1, 6�
1
1
y
xB B�
A A�
C C �
C���6, 1�B���10, �4�,A���10, 2�,
�2
2
y
x
BC A
A�
B�
C �
C���2, 8�A���6, 9�, B���6, 3�,
�1
1
y
xB
B�
C
C �A
A�
�1, 7�.�8, 3�,
152 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
MCRBG-0704-SK.qxd 6-15-2001 3:22 PM Page 152
Geometry 153Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
54.
Lesson 7.6
7.6 Guided Practice (p. 440)
1. A frieze pattern is a pattern that extends to the left andright in such a way that the pattern can be mapped ontoitself by a horizontal translation.
2. The pattern is an example of TV because there is no rotational symmetry. But there is a vertical line symmetrybecause the triangle is equilateral.
3. translation, vertical line reflection
4. translation, rotation, horizontal line reflection, verticalline reflection, horizontal glide reflection
5. translation, rotation, vertical line reflection, horizontalglide reflection
6. translation, rotation, vertical line reflection, horizontalglide reflection
7. The five possible transformations that can be found in afreeze pattern are translation (T), rotation (R), hori-zontal glide reflection (G), vertical line reflection (V),and horizontal line reflection (H).
7.6 Practice and Applications (pp. 440–443)
8. C 9. D 10. A 11. B
12. translation, horizontal line reflection, horizontal glidereflection
13. translation, rotation 14. translation, rotation
15. translation, rotation, horizontal line reflection,vertical line reflection, horizontal glide reflection
16. Yes; there is a reflection in any vertical line that lies midway between two figures.
17. Yes; there is a reflection in a the x-axis.
18. The transformation that maps A onto F is a reflection inthe x-axis, followed by a horizontal translation describedby
19. The transformation that maps D onto B is a rotationabout
20. The frieze pattern is TRHVG.
21. The pattern on the collar can be classified as TRHVG.
�8, 0�.180�
�x, y� → �x � 14, y�.
180�
180�180�
180�
180�
C���3.5, 1.5�B���7.5, �3.5�,A���7.5, 2.5�,
�2
2
y
xB
C A
B�
A�
C �
22. The pattern on the collar can be classified as TG.
23. The pattern on the collar can be classified as T.
24. Answers will vary.
25. Answers will vary.
26.–31. Sample patterns are given.
26.
27.
28.
29.
30.
31.
32. T
33. TRHVG
34. TR
35. There are three bands of frieze patterns visible.
36. The patterns near the top and bottom of the jar are T. Thepattern in the middle of the jar is TR.
37. Answers will vary.
38. Sample answer:
The band around the middle is an example of THG.
39.
The circumference of the base is about 29.83 inches. Ifyou want 10 repetitions of the design, the design shouldbe about or 2.98 inches wide.29.83 � 10
� 29.83 in.� 3.14 � 9.5 C � d
d � 9.5 in.
MCRBG-0704-SK.qxd 6-15-2001 3:22 PM Page 153
Chapter 7 continued
40.
41.
42.
43. Sample answer: The design on the tiles limits what clas-sifications of patterns can be made. For instance, inExercise 40, the design on the tile would not allow thecreation of THG in a single row because there is not ahorizontal line of symmetry in the tile. The same wouldbe true for the tile in Exercise 41.
44. If a pattern can be mapped onto itself by a horizontalglide reflection and by a vertical line reflection, it can bemapped onto itself by a rotation about the pointwhere the lines of reflection intersect.
45. If a pattern can be mapped onto itself by a horizontal linereflection and by a vertical line reflection, it can bemapped onto itself by a 180° rotation about the pointwhere the lines intersect. It can also be mapped ontoitself by a horizontal glide reflection involving the givenhorizontal line reflection and any translation.
46. If a pattern can be mapped onto itself by a 180° rotationabout a point and a horizontal glide reflection, the centerof rotation must be on the line of reflection for the glidereflection. Then the pattern can be mapped onto itself byreflection in a vertical line through the center of rotation.
47. Sample answer:
48. Sample answer:
49. Sample answers:
a.
—CONTINUED—
y
x1
1
D
E
FG
H
y
x1
1A
C B
180�
49. —CONTINUED—
b.
c.
50. a. T b. TVG c. TRVG
51. Design A does not have a rotational symmetry.
Sample answer:
52. If it has rotational symmetry, then its classificationmust be at least TR. The symmetry means that thepattern can be mapped onto itself with a rotation.Therefore, it must at least be TR.
53.
7.6 Mixed Review (p. 444)
54. girls 55.
56. 57.
58. 59.
girlsboys
�1010
� 1girlsboys
�7
11
20 � 10 � 10 girls18 � 11 � 7 girls
girlsboys
�1619
girlsboys
�103
35 � 19 � 16 girls13 � 3 � 10 girls
girlsboys
�138
girlsboys
�1112
21 � 8 � 13 girls23 � 12 � 11
180�180�
180�
154 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Yes
Yes
Yes
No
No Yes No Yes No
No
Is there a rotation?
180�
Is there a line of reflection?
Is there a line ofreflection?
Is the reflectionin a horizontal
line?
Is the reflectionin a horizontal
line?
Is there a glidereflection?
Yes No
TRTRHVG THG TV TG TTRVG
MCRBG-0704-SK.qxd 6-15-2001 3:22 PM Page 154
Geometry 155Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
60.
62.
square units
64.
Quiz 2 (p. 444)
1.
2.
3.
4.
5. 6.
7. Yes, the frieze pattern is TR.
Chapter 7 Review (pp. 446–448)
1. Yes; it is an isometry because the figure and its imageappear to be congruent.
2. No; it is not an isometry because the figure and its imageare not congruent.
3. Yes; it is an isometry because the figure and its imageappear to be congruent.
y
x�1 1
1
�1
P (5, �3)
Q (4, 1)
R (2, �2)P (�3, �3)
Q (�1, 0)
R (2, �3)
y
x
1
1
C��6, 3�B��9, 5�,A��4, 4�,
C���1, �3�B��2, �1�,A���3, �2�,
C���2, 5�B��1, 7�,A���4, 6�,
C��2, 4�B��5, 6�,A��0, 5�,
� 442 square units
� 12 � 17 � 52
� 12 � 17 � �17 � 35�
A �12 h�b1 � b2�
� 360
� 12 � 30 � 24
� 12 �15 � 15��12 � 12�
A �12 d1d2
z � 2
13 z �23
7z �203 z �
23
7z � 2 �203 z �
43
7z � 2 �23 �10z � 2�
7z � 2 �23 �7z � 2 � 3z�
x � 5
13 x �53
3x �83 x �
53
3x � 5 �83 x �
103
3x � 5 �23 �4x � 5�
3x � 5 �23 �3x � 5 � x� 4. 5.
6. 7.
8. 9.
10. C 11. A 12. D 13. B
14. followed by clockwise rota-tion about the origin
15. reflection in the x-axis, followed by a counterclock-wise rotation about the origin
16. The rainbow boa’s snakeskin has a frieze pattern classifiedas TR.
17. The gray-banded kingsnake’s snakeskin has a frieze pat-tern classified as TRHVG.
Chapter 7 Test (p. 449)
1. The transformation that maps onto is areflection in the y-axis.
2. Yes, because a reflection preserves length.
3. The image of T is Z. 4. The preimage of Y is S.
5. Sample answer: 6. Sample answer:
7. The transformation that maps figure T onto figure is areflection in line m.
8. The transformation that maps figure T onto is a reflec-tion in line m followed by a reflection in line n. Or it is arotation about the point of intersection of lines m and n.
9. The measure of the angle of rotation is twice the measureof the acute angle formed by lines m and n. So, the measure of the angle of rotation is or 170�.2 � 85�
T
T�
RT � XZ
�XYZ�RST
90�
90��x, y� → �x � 4, y � 5�,
K
M�
P
L� N
L
N�
M
K�
P
F
H� G�
G
H
F�
A�
A B�
BP
k
k
k61.
63.
� 288 square units
� 12 � 24 � 24
� 12 �12 � 12��4 � 20�
A �12 d1d2
w � 8
�13 w � �
83
w �43w �
83
w �23 �2w � 4�
w �23 �w � w � 4�
y � 2
�43 y � �
83
2y �103 y �
83
2y �23 �5y � 4�
2y �23 �2y � 3y � 4�
MCRBG-0704-SK.qxd 6-15-2001 3:22 PM Page 155
Chapter 7 continued
10. The transformation that maps figure R onto figure is areflection in line k.
11. The transformation that maps figure R onto is a reflec-tion in line k, followed by a reflection in line m, or it is atranslation.
12. The distance between corresponding parts of figure R andfigure is twice the distance between lines k and m. Thedistance is or 10 units.
13. A glide reflection is a composition of a translation fol-lowed by a reflection in a line parallel to the translationvector.
14. Sample answer:
is the final image when is rotated clockwiseabout the origin, then reflected in the y-axis. is thefinal image when is reflected in the y-axis, then rotated clockwise about the origin.
15. Sample answer:
is reflected in the x-axis, then translatedThe same image results if the
transformations are performed in reverse order.
16. The flag of Switzerland has a vertical line of symmetry,a horizontal line of symmetry, two diagonal lines of symmetry and rotational symmetry. It can be mappedonto itself by a clockwise or counterclockwise rotation of about the center.
17. The flag of Jamaica has a vertical line of symmetry, ahorizontal line of symmetry, and rotational symmetry. Itcan be mapped onto itself by a clockwise or counter-clockwise rotation of about the center.180�
180�
�x, y� → �x � 5, y�.PQ
y
x
1
1
Q(4, 6)
Q�(9, �6)
P(1, 2)
P�(6, �2)
90�AB
AB90�ABA�B�
B
A
B�
A�B
A
y
x
1
1�1�1
2 � 5R
R
R� 18. The flag of the United Kingdom has a vertical line ofsymmetry, a horizontal line of symmetry, and rotationalsymmetry. It can be mapped onto itself by a clockwise or counterclockwise rotation of about its center.
19. translation, 180° rotation, horizontal line reflection,vertical line reflection, glide reflection
20. translation, 180° rotation, horizontal line reflection,vertical line reflection, glide reflection
21. translation, vertical line rotation
Chapter 7 Standardized Test (pp. 450–451)
1. B 2. D 3. E
4.
A
5. or
or
or
B
6. D
7. A counterclockwise rotation maps onto Areflection in the x-axis maps onto So the correct answer is D.
8. C
9. With the reflection in is mapped ontoand is mapped onto A
clockwise rotation about the point mapsonto and onto So
the correct answer is E.
10. The letters that have a vertical line of symmetry are A, H,I, M, O, T, W, X, and Y.
11. The letters that have a horizontal line of symmetry are C,E, H, K, O, and X.
12. The letters with a rotational symmetry are H, N, O, S, X,and Z.
T�2, 2�.T���3, 7�S��1, 5�S���6, 4���3, 2�90�
T���3, 7�.T��3, �5�S���6, 4�S��6, �2�y � 1,
AB.A�B�A�B�.AB90�
��3, �8�Y��5, 2� → �5 � 8, 2 � 10�
��1, �4�X��7, 6� → �7 � 8, 6 � 10�
��5, �2�W�3, 8� → �3 � 8, 8 � 10�
x � 41
2x � 82
�2x � 19�� � 101�
y � 6
8y � 48
�8y � 6� � 42�
180�
156 GeometryChapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
z � 6
3z � 18
MCRBG-0704-SK.qxd 6-15-2001 3:22 PM Page 156
Geometry 157Chapter 7 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 7 continued
13. a. b.
14. The coordinates of the vertices of figure WXYZ areor or
or andor
15. clockwise rotation about the origin
16. Sample answer:
Project Chapters 6–7 (pp. 452–453)
Investigation
1. No; the quadrilateral is not a regular polygon.
2. Some transformations that would map the pattern ontoitself are translations and rotations.
3. The sum of the measures of the angles at any vertex of aquadrilateral tessellation is Any quadrilateral willtessellate because the sum of the measures of its angles is
If all four angles are placed so they are adjacent,the full 360° rotation is covered.
4. Square and equilateral triangle tessellations are regulartessellations.
5. Tessellations will vary.
A translation and a rotation map the pattern onto itself.
6. Sample answer:
This is a semiregular tessellation.
7. These shapes cannot be used together to create a tessella-tion because there is no combination of multiples of and multiples of that add together to get
8. These shapes can be used together to create a nonregulartessellation.
Sample answer:
360�.90�108�
360�.
360�.
y � 212
90�
Z�7, 1�.Z�0 � 7, 3 � 2�Y�12, 1�,Y�5 � 7, 3 � 2�X�11, 4�,
X�4 � 7, 6 � 2�W�9, 4�,W�2 � 7, 6 � 2�
��2, 7��x, y� → �x � 2, y � 7�
MCRBG-0704-SK.qxd 6-15-2001 3:22 PM Page 157