Document 1805777

Chapter 9,
continued
5. The x-coordinate is now 24 and the y-coordinate is
y G0
now 3.
6. J9
F0
y
K9
J
1
(2, 0)
L9
F9
22
L
K
x
P
G9
40. The distance from the origin to A never changes. It will
always equal 2. For a 908 rotation, you would add 908
to the angle. You would do the same for 1808 and 2708
rotations.
(2, 1208), (2, 2108), (2, 3008)
1
21
x
J9(25, 9), K9(23, 9), L9(23, 7)
Lesson 9.5
Mixed Review for TAKS
Investigating Geometry Activity 9.5 (p. 607)
41. C;
Explore 1
The maximum value of the parabola occurs at (0, 3).
42. H;
Each number shown can be written as mm.
Step 3: nD0E0F 0 > nDEF; DD0E0F 0 is a translation
of nDEF.
Step 4: Yes; The relationship hold true.
1 5 11
Explore 2
4 5 22
Step 2: n A0B0C 0 > n ABC; n A0B0C 0 is a rotation
of n ABC.
27 5 3
3
256 5 44
Because 5 5 3125, the next number that would be in the
sequence is 3125.
5
Step 3: The measure of the acute angle is half the measure of
Ž APA0. Yes, the relationship hold true.
1. A translation can map a figure onto the same image as a
reflection in two parallel planes.
Problem Solving Workshop 9.4 (p. 606)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1.
reflection in two intersecting lines.
y
C9
2. A rotation can map a figure onto the same image as a
D9
9.5 Guided Practice (pp. 608–611)
B9
1. A(3, 2), B(6, 3), C(7, 1)
A9
1
21
x
A
B
A(3, 2) l A9(3, 2 2 4) 5 A9(3, 22)
B(6, 3) l B9(6, 3 2 4) 5 B9(6, 21)
C(7, 1) l C9(7, 1 2 4) 5 C9(7, 23)
D
C
2.
Translation: (x, y) l (x, y 2 4)
Reflection: y-axis, (x, y) l (2x, y)
A9(3, 22) l A0(23, 22)
y R
B9(6, 21) l B0(26, 21)
S
C9(7, 23) l C 0(27, 23)
T
T9
2. A0(29, 22) l (29 1 x, y) l A(3, 2)
1
R9
21
x
S9
3. 908 clockwise and 2708 counterclockwise are rotations
29 1 x 5 3
x 5 12
(x, y) l (x 1 12, y) followed by a reflection in the x-axis.
in opposite directions placing you at 908 below the
reference line.
4. Draw the preimage and trace the preimage on the tracing
paper. To reflect in the y-axis, flip the paper vertically and
line up the axes and trace. To reflect in the x-axis, flip the
paper horizontally and line up the axes and trace.
Geometry
Worked-Out Solution Key
283
Chapter 9,
continued
3. R(1, 23), S(2, 26)
4. C(2, 25), D(4, 0)
C(2, 25) l C9(21, 25)
(a, b) l (2b, a)
R(1, 23) l R9(3, 1)
S(2, 26) l S9(6, 2)
Reflection y 5 21
C 0(21, 3)
(a, b) l (2a, b)
D0(1, 22)
R9(3, 1) l R0(23, 1)
2
x
D
C
C
y D
D
(x, y) l (x, y 1 4)
y
C(2, 25) l C9(2, 21)
S9
1
D
D(4, 0) l D9(4, 4)
R9
22
x
Reflection: x 5 3
R
x
1
C
C
C 0(4, 21)
S
D0(2, 4)
}
Yes, the order of transformations matter, because R0S0 is
}
not the same as the R0S 0 in Example 2.
4. GG0 5 HH 0 because opposite sides of a parallelogram
are equal.
5. A translation would map the blue figure to the green figure.
6. PP0 5 2d 5 2(1.6) 5 3.2
The distance between P and P 0 is 3.2 cm. If P is not
}
on line k, then k is a perpendicular bisector of PP9.
}
So, PP9 is perpendicular to line k.
C
6. C(2, 25), D(4, 0)
(x, y) l (x 1 2, y 1 2)
C(2, 25) l C9(4, 23)
y
D
C
D
D(4, 0) l D9(6, 2)
1
(x, y) l (y, x)
D
x
1
C9(4, 23) l C 0(23, 4)
D9(6, 2) l D0(2, 6)
C
C
7. A single transformation that maps the blue figure onto
the green figure is a rotation of 2(808) or 1608 about
point P.
8. The angle formed by the intersecting lines of reflection
768
or 388.
is }
2
1 2
9.5 Exercises (pp. 611–615)
1. In a glide reflection, the direction of the translation must
be parallel to the line of reflection.
2. A glide reflection is an isometry because it is a
composition of two transformations that are isometries.
P
P(2, 4) l P9(2, 21)
Q(6, 0) l Q9(6, 25)
R(7, 2) l R9(7, 23)
Q
P9(2, 21) l P0(22, 21)
R9(7, 23) l R0(27, 23)
8. P(2, 4), Q(6, 0), R(7, 2)
Q(6, 0) l Q9(3, 2)
D(4, 0) l D9(4, 21)
R(7, 2) l R9(4, 4)
(x, y) l (2x, y)
(x, y) l (2y, x)
C9(2, 26) l C 0(22, 26)
P9(21, 6) l P0(26, 21)
D9(4, 21) l D0(24, 21)
Q9(3, 2) l Q 0(22, 3)
y
R9(4, 4) l R0(24, 4)
D
x
1
D
D
C
C
Geometry
Worked-Out Solution Key
8 x
R
Q
Q9(6, 25) l Q 0(26, 25)
C(2, 25) l C9(2, 26)
1
Q
P P
R
y
P
P(2, 4) l P9(21, 6)
(x, y) l (x, y 2 1)
R
2
(x, y) l (x 2 3, y 1 2)
3. C(2, 25), D(4, 0)
C
y
(x, y) l (x, y 2 5)
(x, y) l (2x, y)
Skill Practice
284
7. P(2, 4), Q(6, 0), R(7, 2)
R
R
Q P
P
2
R
Q
2
Q
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
R0
D
2
5. C(2, 25), D(4, 0)
S9(6, 2) l S 0(26, 2)
4
D
D(4, 0) l D9(1, 0)
Reflection: in y-axis
S0
y
C
(x, y) l (x 2 3, y)
Rotation: 908 about the origin
Chapter 9,
continued
12. F(21, 28), G(26, 23)
9. P(2, 4), Q(6, 0), R(7, 2)
(x, y) l (x 1 12, y 1 4)
Reflection: y 5 2
P(2, 4) l P9(14, 8)
F9(21, 12), G9(26, 7)
Q(6, 0) l Q9(18, 4)
Rotation: (a, b) l (2b, a)
R(7, 2) l R9(19, 6)
F 0(212, 21), G 0(27, 26)
(x, y) l (x 2 5, y 2 9)
y
1
P9(14, 8) l P0(9, 21)
22
F0
Q9(18, 4) l Q 0(13, 25)
x
R9(19, 6) l R0(14, 23)
y
P
G0
R
P
Q
R
2
Q
2
F(21, 28), G(26, 23)
Rotation 908: F9(8, 21), G9(3, 26)
x
P
Reflection y 5 2: F 0(8, 5), G0(3, 10)
R
Yes, the order does affect the final image.
Q
13. To go from A to A9, move 5 units to the right and
10. P(2, 4), Q(6, 0), R(7, 2)
y
(x, y) l (x, 2y)
P(2, 4) l P9(2, 24)
(x, y) l (x 1 5, y 11)
Q
Q(6, 0) l Q9(6, 0)
2
R(7, 2) l R9(7, 22)
P
P
2
(x, y) l (2y, x)
P
P9(2, 24) l P0(4, 2)
14. First, there is a reflection in the y-axis followed by a
reflection in the x-axis.
15. A translation maps n ABC onto n A0B0C 0.
}
}
18. CC 0 5 2d 5 2(2.6) 5 5.2
11. F(25, 2), G(22, 4)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
following by a 1808 rotation about the origin.
x
17. Sample answer: AA9 and AA0 are parallel to @###$
BB0.
R9(7, 22) l R0(2, 7)
(x, y) l (x 1 3, y 2 8)
F(25, 2) l F9(22, 26)
G(22, 4) l G9(1, 24)
(x, y) l (x, 2y)
CC 0 is 5.2 inches.
The length of @###$
19. The distance from B9 to M is the same as the distance
from B0 to M because of the definition of a reflection
in a line.
20. The angle of rotation is 2(55)8 5 1108.
F9(22, 26) l F 0(22, 6)
G9(1, 24) l G0(1, 4)
21. The angle of rotation is 2(15)8 5 308.
22. Because the y-axis, which is the line of reflection, is
not parallel to the direction of the translation, this is not
a glide reflection. Sample answer: (x, y) l (x, y 1 3)
followed by a reflection in the y-axis is a glide reflection.
y
G0
23. P(1, 4), Q(3, 22), R(7, 1)
1
21
R
Q
Q
R
@###$.
16. Lines k and m are perpendicular to AA9
Q9(6, 0) l Q 0(0, 6)
F0
1 unit up. To go from A9 to A0, you move from (1, 3)
to (21, 23) which is 1808 rotation.
R
x
(x, y) l (x, 2y)
F(25, 2) l F9(25, 22)
G(22, 4) l G9(22, 24)
(x, y) l (x 1 3, y 2 8)
P
1
4
R
7
0
1
1
5
P9 Q9 R9
0
1 3 7
5
5
9 3 6
F GF GF G
F GF G F
G
0
5
P9 Q9 R9
P0
1 3 7
21
5
9 3 6
9
21 0
0 1
F9(25, 22) l F 0(22, 210)
Q0
23
3
R0
27
6
y
P 0 P9
G9(22, 24) l G0(1, 212)
Yes, the order does affect the final image.
Q
3
22
P
R0
Q0 2
R9
Q9
22
R
x
Q
Geometry
Worked-Out Solution Key
285
Chapter 9,
continued
24. P(1, 4), Q(3, 22), R(7, 1)
P
1
4
Q
3
22
26. J(1, 23), K(2, 2), L(3, 0)
R
7
5
1
F GF G F
F
GF
F
G
1
0
0 21
P9 Q9 R9
1 3
7
29
1
24 2 21
24
P0
28
5
28
Q0
26
22
P9 Q9
1 3
24 2
29
24
29
24
R9
7
21
1808 Rotation about (22, 2):
G
J9(25, 7), K9(26, 2), L9(27, 4)
Reflection in y 5 2x
(x, y) l (2y, 2x)
G
J 0(27, 5), K0(22, 6), L0(24, 7)
R0
22
25
y
L
J
K
J
L
K
K
1
y
P
Q9
2
Q0
R
R9 x
22
L
x
1
J
Q
R0 P9
Problem Solving
P0
27.
18 in.
25. Sample answer:
A(2, 5), B(2, 1), C(8, 1)
(x, y) l (x 1 9, y) followed by a reflection over a
horizontal line that separates the left and right prints
of the bald eagle’s legs
Reflection in the x-axis:
(x, y) l (x, 2y)
A(2, 5) l A9(2, 25)
28.
15 in.
B(2, 1) l B9(2, 21)
(x, y) l (x 1 2, y 2 3)
A9(2, 25) l A0(4, 28)
B9(2, 21) l B0(4, 24)
C9(8, 21) l C 0(10, 24)
Rotation 2708 about origin:
(x, y) l ( y, 2x)
29. C; After gliding to the next key, a reflection through a
horizontal line will not place you on another key.
30. a horizontal translation and a reflection along the
horizontal line through the middle of the boat
31. reflection and translation
A0(4, 28) l A09(28, 24)
32. rotation and translation
B0(4, 24) l B09(24, 24)
33. translation and reflection
C 0(10, 24) l C 09(24, 210)
A
2
2
A
B
B
B
34. First, reflect the object across two parallel lines. Then,
reflect the object across a third line that is perpendicular
to the first two lines.
y
B
C
C x
C
A
A
C
Because the three transformations are isometries, the
preimage and the final image are congruent because
isometries preserve length and angle measure.
286
(x, y) l (x 1 7.5, y) followed by a reflection over a
horizontal line that separates the left and right prints of
an armadillo’s legs
Geometry
Worked-Out Solution Key
35. Use the Rotation Theorem followed by the Reflection
Theorem.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
C(8, 1) l C9(8, 21)
Chapter 9,
continued
36. Rotation and reflection, rotation and translation, rotation
and rotation, reflection and translation, reflection and
reflection, reflection and rotation, translation and
reflection, translation and rotation, translation and
translation
Sample Proof:
Given: A reflection in m maps Q to Q9 and R to R9.
A translation maps Q9 to Q0 and R9 to R0.
Prove: QR 5 Q0R0
m
Q
Q9
Q0
R
R
R
Statements
Reasons
Statements
1. Given
2. A reflection in line
k maps Q to Q9 and a
reflection in line m maps
Q9 to Q0.
} } }
3. K > QQ9, QA > Q9A
2. Given
3. Definition of a reflection
4. Ž QAP and Ž Q9AP are
right angles.
4. Definition of
perpendicular lines
5. Ž QAP > Ž Q9AP
5. Right Angle Congruence
Theorem
} }
6. AP > AP
1. Reflection Theorem
7. nQAP > nQ9AP
2. R9Q9 5 R0Q 0
2. Translation Theorem
3. RQ 5 R0Q 0
3. Transitive Property of
Equailty
} }
8. m > Q9Q0, Q9B > BQ0
Statements
1. A reflection in line *
} }
maps JK to J9K9, a
reflection in line m maps
} }
J9K9 to J 0K0, and * i m.
}
}
2. * > KK9, m > K9K0
}
}
3. KK9 > m, K9K0 > *
4. @##$
KK is perpendicular to *
and m.
Reasons
1. Given
3. Perpendicular Transversal
Theorem
4. Perpendicular Postulate
6. Reflexive Property of
Congruence
7. SAS Congruence
Postulate
8. Definition of a reflection
9. Ž Q9BP and Ž Q0BP
are right angles.
9. Definition of
perpendicular lines
10. Ž Q9BP > Ž Q0BP
10. Right Angle Congruence
Theorem
} }
11. BP > BP
12. nQ9BP > nQ0BP
2. Definition of a reflection
Reasons
1. Lines k and m intersect
at point P. Q is any point
not on k or m.
1. RQ 5 R9Q9
37. a.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
38. a.
} }
13. QP > Q9P,
} }
Q9P > Q0P
} }
14. QP > Q0P
15. Q0 is the image of Q.
11. Reflexive property of
congruency
12. SAS Congruence
Postulate
13. Corresponding parts of
congruent triangles are
congruent.
14. Transitive Property of
Congruence
15. Definition of a rotation
b. Let K*, K9* be the distances from K, K9 to line *.
K9m, K0m be the distances from K9, K0 to line m.
Statements
Reasons
5. KK9 5 K* 1 K9*
K9K0 5 K9m 1 K0m
5. Segment Addition
Postulate
6. K* 5 K9*, K9m 5 K0m
6. Definition of a reflection
7. KK9 5 2K9*,
K9K0 5 2K9m
7. Substitution Property of
Equality
8. KK0 5 KK9 1 K9K0
8. Segment Addition
Postulate
9. KK0 5 2K9* 1 2K9m
9. Substitution Property of
Equality
10. KK0 5 2(K9* 1 K9m)
5 2d
10. Distributive Property
and Substitution
Property of Equality
Geometry
Worked-Out Solution Key
287
Chapter 9,
continued
43. F;
b.
Statements
Reasons
16. mŽ QPQ9 5 mŽ QPA
1 mŽ APQ9
mŽ Q9PQ0 5 mŽ Q9PB
1 mŽ BPQ0
16. Angle Addition
Postulate
17. mŽ QPA 5 mŽ APQ9
mŽ Q9PB 5 mŽ BPQ0
17. Corresponding parts of
congruent triangles are
congruent.
P(2 and then a number greater than 4)
5 P(2) + P(a number greater than 4)
1
2
2
1
5}
5 }6 + }6 5 }
36
18
Quiz 9.3–9.5 (p. 615)
18. mŽ QPQ9 5 2mŽ APQ9 18. Substitution Property of
mŽ Q9PQ0 5 2mŽ Q9PB
Equality
19. mŽ QPQ0 5 mŽ QPQ9
1 mŽ Q9PQ0
19. Angle Addition
Postulate
20. mŽ QPQ0 5 2mŽ APQ9
1 2mŽ Q9PB
20. Substitution Property of
Equality
21. mŽ QPQ0 5 2
(mŽ APQ9 1 mŽ Q9PB)
21. Distributive Property
1. A(7, 1), B(3, 5), C(10, 7)
y
(x, y) l (2x, y)
C
C
A(7, 1) l A9(27, 1)
B
B(3, 5) l B9(23, 5)
2
A
They are done simultaneously.
40. a. A(2, 0, 0), B(2, 3, 0)
x
2
2. A(7, 1), B(3, 5), C(10, 7)
A9(215, 1), B9(211, 5), C9(218, 7)
y
C
B
B
2
A
A
x
2
39. a. Translation and a rotation
b. One transformation is not followed by the second.
A
C(10, 7) l C9(210, 7)
C
22. mŽ QPQ0 5 2(mŽ APB) 22. Angle Addition
Postulate
B
3. A(7, 1), B(3, 5), C(10, 7)
(x, y) l (2y, 2x)
A(7, 1) l A9(21, 27)
B(3, 5) l B9(25, 23)
z
C(10, 7) l C9(27, 210)
y
B
B(2, 3, 0)
2
A
x
Ÿx, y, z® l Ÿx 1 4, y 1 0, z 2 1®
A9Ÿ2, 0, 0® l A0Ÿ6, 0, 21®
B9Ÿ2, 0, 3® l B0Ÿ6, 0, 2®
41. The conjecture is not always true. Consider a reflection
of a point (a, b) in the x-axis followed by a reflection in
y 5 x.
Mixed Review for TAKS
42. D;
x
A
Rotate 908, A9(2, 0, 0), B9(2, 0, 3)
b.
A
2
B
C
4. P(2, 23)
5. P(2, 23)
(x, y) l (2x, 2y)
P(2, 23) l P9(22, 3)
(x, y) l (2y, x)
P(2, 23) l (3, 2)
6. P(2, 23)
(x, y) l (y, 2x)
P(2, 23) l (23, 22)
7. P(28, 8), Q(25, 0), R(21, 3)
(x, y) l (x 1 6, y)
2x 1 y 5 10
y
2x 1 3y 5 9
P(28, 8) l P9(22, 8)
Q(25, 0) l Q9(1, 0)
(3, 4)
(x, y) l (2x, y)
1
y
R(21, 3) l R9(5, 3)
1
x
The y-coordinate of the solution to the system of linear
equations is 4.
P
P
P
P9(22, 8) l P0(2, 8)
Q9(1, 0) l Q 0(21, 0)
R9(5, 3) l R0(25, 3)
R
R
Q
R
Q Q
4
288
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
A(2, 0, 0)
C
B
y
Geometry
Worked-Out Solution Key
4
x
Chapter 9,
continued
8. P(28, 8), Q(25, 0), R(21, 3)
P9(28, 212), Q9(25, 24), R9(21, 27)
Rotation 908: (x, y) l (2y, x)
P 0(12, 28), Q0(4, 25), R0(7, 21)
y
P
9.5 Extension (p. 617)
1. Yes; The equilateral triangle tesselates. The tesselation is
regular because it uses congruent regular triangles.
2. A circle does not tesselate because it leaves gaps and
overlaps.
3. A kite tesselates. The tesselation is not regular because
the kite is not a regular polygon.
R
2
4. Sample answer:
Q
Q
x
R
2
Q
R
P
5. a. The sum of the measures of the angles add up to 3608:
P
Then, the sum of the angle measures at any vertex
adds up to 3608.
9. P(28, 8), Q(25, 0), R(21, 3)
b. A quadrilateral will tesselate because the sum of the
(x, y) l (x 2 5, y)
measures of the interior angles adds up to 3608.
P(28, 8) l P9(213, 8)
6. Sample answer:
Q(25, 0) l Q9(210, 0)
7. Sample answer:
R(21, 3) l R9(26, 3)
(x, y) l (x 1 2, y 1 7)
P9(213, 8) l P0(211, 15)
Q9(210, 0) l Q 0(28, 7)
R9(26, 3) l R0(24, 10)
8. Sample answer:
y
P
R
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
P
P
Q
R
R
Q
2
Q
x
2
9. a.
F
A
10. P(28, 8), Q(25, 0), R(21, 3)
E
(x, y) l (2x, 2y)
P(28, 8) l P9(8, 28)
Q(25, 0) l Q9(5, 0)
b.
B
D
C
F
A
R(21, 3) l R9(1, 23)
(x, y) l (x 1 4, y 2 3)
E
B
P9(8, 28) l P0(12, 211)
Q9(5, 0) l Q 0(9, 23)
c.
R9(1, 23) l R0(5, 26)
D
C
F
A
y
P
E
B
R
2
Q
D
Q
2
Q
C
x
R
R
P
P
Geometry
Worked-Out Solution Key
289
Chapter 9,
continued
Lesson 9.6
d.
9.6 Guided Practice (pp. 619–621)
1.
10. Sample answer:
2.
The object appears to have
8 lines of symmetry.
The object appears to
have 5 lines of symmetry.
4. An example of a
3.
hexagon with no lines
of symmetry is
11. Sample answer:
12. Sample answer:
The object appears to have
1 line of symmetry.
5. Yes; The center is the intersection of the diagonals.
A rotation of 1808 about the center maps the rhombus
to itself.
6. Yes; The center is the intersection of the diagonals. A
7. No; The right triangle does not have rotational symmetry,
because a rotation of 1808 or less does not map the right
triangle to itself.
The line of symmetry is the altitude.
There is no rotational symmetry.
8.
a
a
h
14. Rotation and translation
15. Translation
16. Translation and reflection
17. Rotations
b
9.6 Exercises (pp. 621–624)
18. Translation and reflection
Skill Practice
1. The center of symmetry is the point about which the
figure is rotated if the figure has rotational symmetry.
2. The arrow has one line of symmetry and no rotational
symmetry.
No. A figure that has two
lines of symmetry also has
rotational symmetry.
290
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
rotation of 908 or 1808 about the center maps the octagon
to itself.
13. Sample answer: