Chapter 4, continued

Chapter 4,
continued
Mixed Review for TAKS
7. No; The moving part of the window is not a translation.
52. B; Cost of the 4 CDs: 4(14) 5 $56
8. Yes; The moving part of the window is a translation.
9. (x, y) l (x 1 2, y 2 3)
Discount: 0.15(56) 5 $8.40
Cost after the discount: 56 2 8.40 5 $47.60
A(23, 1) l (21, 22)
Sales tax: 0.075(47.60) 5 $3.57
B(2, 3) l (4, 0)
Total cost: 47.60 1 3.57 5 $51.17
C(3, 0) l (5, 23)
Sue does not have enough to buy the CDs. She needs
$51.17 2 $50 5 $1.17 more.
D(21, 21) l (1, 24)
y
B
A
1
C
23
x
D
53. G;
d
10. (x, y) l (x 2 1, y 2 5)
6
r 5 }2 5 }2 5 3 in.
y
B
A(23, 1) l (24, 24)
V 5 Bh 5 (:r 2)h 5 : (32)(7.5) ø 212.06
The volume of the can is about 212.06 cubic inches.
B(2, 3) l (1, 22)
A
1
C
3
C(3, 0) l (2, 25)
x
D
D(21, 21) l (22, 26)
Lesson 4.8
Investigating Geometry Activity 4.8 (p. 271)
1. In a slide, the x-coordinates are changed by the amount
the triangle was shifted up or down. The y-coordinates
are changed by the amount the triangle was shifted left
or right.
In a flip, only one coordinate of the triangle’s vertices
changes. The x-coordinate changes sign if the triangle is
flipped over the y-axis or the y-coordinate changes sign if
the triangle is flipped over the x-axis.
11. (x, y) l (x 1 4, y 1 1)
y
B
A(23, 1) l (1, 2)
B(2, 3) l (6, 4)
A
1
C
C(3, 0) l (7, 1)
x
3
D(21, 21) l (3, 0)
D
2. Yes; yes; When sliding or flipping a triangle, the size and
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
shape do not change, only the position changes. So the
original triangle is congruent to the new triangle.
B(2, 3) l (0, 6)
1. The transformation shown is a reflection.
x-coordinate and subtracting 1 from each y-coordinate.
A
D(21, 21) l (23, 2)
1
23
C
x
D
(x, y) l (x 1 1, y 2 1)
}
3. The y-coordinates are multiplied by 21, so RS was
reflected in the x-axis. (x, y) l (x, 2y)
4. nPQR is a 1808 rotation of nSTR.
}
5. PQ 5 ST 5 2, so PQ > ST. PR 5 SR 5 3, so
}
B
C(3, 0) l (1, 3)
2. The new coordinates are found by adding 1 to each
}
y
A(23, 1) l (25, 4)
4.8 Guided Practice (pp. 272–275)
}
12. (x, y) l (x 2 2, y 1 3)
PR > SR. nPQR and nSTR are right triangles, so
nPQR > nSTR by HL. Therefore the transformation is
a congruence transformation.
13. (x, y) l (x 2 4, y 2 2)
14. (x, y) l (x 1 6, y 1 3)
15. (x, y) l (x 1 2, y 2 1)
16. (x, y) l (x 2 7, y 1 9)
17. (x, y) l (x, 2y)
y
(1, 1) l (1, 21)
(4, 1) l (4, 21)
(4, 3) l (4, 23)
1
21
x
4.8 Exercises (pp. 276–279)
Skill Practice
1. The new coordinates are formed by subtracting 1 from
each x-coordinate and adding 4 to each y-coordinate.
2. The term congruence transformation is used because
when an object is translated, reflected, or rotated, the new
image is congruent to the original figure.
18. (x, y) l (x, 2y)
y
(1, 2) l (1, 22)
(3, 1) l (3, 21)
(4, 3) l (4, 23)
1
21
x
(4, 1) l (4, 21)
3. The transformation is a translation.
4. The transformation is a rotation.
5. The transformation is a reflection.
6. Yes; The moving part of the window is a translation.
Geometry
Worked-Out Solution Key
111
Chapter 4,
continued
19. (x, y) l (x, 2y)
29. (x, y) l (x 1 2, y 2 3)
y
(2, 3) l (2, 23)
(x, y) l (4, 0)
(4, 1) l (4, 21)
(5, 2) l (5, 22)
1
21
x
x1254
y2350
x52
y53
The point on the original figure is (2, 3).
30. (x, y) l (2x, y)
}
20. CD is a 908 clockwise
y
}
rotation of AB.
B
(x, y) l (23, 5)
2x 5 23
1
y55
x53
A
21
The point on the original figure is (3, 5).
x
C
31. (x, y) l (x 2 7, y 2 4)
(x, y) l (6, 29)
D
x2756
y
of AB because
mŽAOC > mŽBOD.
1
y 2 4 5 29
x 5 13
C
y 5 25
The point on the original figure is (13, 25).
D
21
x
32. Length of sides of upper triangle:
}}
A
}
}
22. CD is not a rotation
y
of AB because
the points are rotated in
different directions.
D
C
Length of sides of lower triangle:
}}
A
x
Both triangles have congruent side lengths, so the
triangles are congruent by SSS.
}
}
33. UV is a 908 clockwise rotation of ST about E.
}
}
34. AV is a 908 counterclockwise rotation of BX about E.
B
23. CD is not a rotation
y
of AB because
mŽAOC > mŽBOD.
A
1
21
}
(0 2 (21))2 1 (1 2 0)2 5 Ï2
Ï}}
}
(3 2 0)2 1 (21 2 1)2 5 Ï13
Ï}}}
}
Ï(3 2 (21))2 1 (21 2 0)2 5 Ï17
1
21
}
}
}
(2 2 1)2 1 (3 2 2)2 5 Ï2
Ï}}
}
(5 2 2)2 1 (1 2 3)2 5 Ï13
Ï}}
}
Ï(5 2 1)2 1 (1 2 2)2 5 Ï17
B
C
B
35. nDST is a 1808 rotation of nBWX about E.
x
36. nXYC is a 1808 rotation of nTUA about E.
37. (x, y) l (x 2 2, y 1 1)
D
A(2, 3) l (0, 4) 5 C(m 2 3, 4)
24. To find the angle of rotation, corresponding angles of
B(4, 2a) l (2, 2a 1 1) 5 D(n 2 9, 5)
the triangles should be used to find the rotation angle.
The red triangle is rotated 908 clockwise.
m2350
m53
25. Yes; Any point or line segment
can be rotated 3608 and be
its own image.
n2952
2a 1 1 5 5
n 5 11
P
2a 5 4
a52
y
3
(x, y) l (x, 2y)
C(0, 4) l (0, 24) 5 E(0, g 2 6)
23
x
3608
26. If (0, 3) translates to (0, 0), then (2, 5) translates to (2, 2).
27. If (0, 3) translates to (1, 2), then (2, 5) translates to (3, 4).
28. If (0, 3) translates to (23, 22), then (2, 5) translates to
(21, 0).
D(2, 5) l (2, 25) 5 F(8h, 25)
g 2 6 5 24
8h 5 2
g52
1
h 5 }4
1
So the variables are m 5 3, n 5 11, a 5 2, g 5 2, h 5 }4 .
Problem Solving
38. a. The designer can reflect the kite layout in the
horizontal line.
b. The width of the top half of the kite is 2 feet, so the
maximum width of the entire kite is 4 feet.
112
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}
21. CD is not a rotation
}
Chapter 4,
continued
39. Starting at A, you will move the stencil 908 clockwise
to get the design at B. To go from A to C, you move the
stencil 908 counterclockwise.
Mixed Review for TAKS
45. D; The volume depends on the area of the base and the
height.
40. a. Sample answer:
MOM
b. Sample answer:
TOT
OH
IO
D
E
B
41. a. The Black Knight moves up 2 spaces and to the left
1 space, so the translation is (x, y) l (x 2 1, y 1 2).
b. The White Knight moves down 1 space and to the
right 2 spaces, so the translation is (x, y) l
(x 1 2, y 2 1).
c. No; the White Knight is directly below the Black
Knight, so the Black Knight would miss the White
Knight because it would have to move at least one
square horizontally according to the rules.
223
21
}
42. slope FE 5 } 5 } 5 21
1
21 2 (22)
1. 6x 1 12 5 24
2. (3x 1 48)8 5 608
6x 5 12
3x 5 12
x52
x54
3. 4x 1 30 5 50
4x 5 20
x55
4. (x, y) l (x 1 4, y 2 1)
y
E
E(0, 2) l (4, 1)
F(2, 1) l (6, 0)
G(1, 0) l (5, 21)
F
1
21
x
G
The transformation is a translation.
5. (x, y) l (2x, y)
y
022
22
}
slope DE 5 }
5}
51
22
23 2 (21)
E(0, 2) l (0, 2)
E
So FE > DE and ŽFED is a right angle.
G(1, 0) l (21, 0)
}
122
21
21 2 1
x
The transformation is a reflection in the y-axis.
22
} }
So CB > AB and ŽCBA is a right angle.
E
E(0, 2) l (0, 22)
y
F
1
F(2, 1) l (2, 21)
G
21
x
G(1, 0) l (1, 0)
Length of hypotenuses:
}}}
}
FD 5 Ï(22 2 (23)) 1 (3 2 0) 5 Ï10
2
}}
}
CA 5 Ï(3 2 2)2 1 (2 2 (21))2 5 Ï10
} }
So FD > CA.
Length of corresponding legs:
}}}
}
FE 5 Ï(21 2 (22))2 1 (2 2 3)2 5 Ï2
}}
3
G
21
6. (x, y) l (x, 2y)
slope AB 5 }
5}
51
224
22
2
F
F(2, 1) l (22, 1)
}
slope CB 5 }
5}
5 21
423
1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Quiz 4.7–4.8 (p. 279)
}
CB 5 Ï(4 2 3)2 1 (1 2 2)2 5 Ï2
} }
So FE > CB.
By the HL Congruence Theorem, n ABC > nDEF, so
nDEF is a congruence transformation of nABC.
43. B; Figure B represents the reflection of the folded paper
over the folded line, or the unfolded paper.
44. Undo translation and rotation:
(x, y) l (x, y 2 3) l rotate 908 clockwise
A(24, 4) l (24, 1) l (1, 4)
B(21, 6) l (21, 3) l (3, 1)
C(21, 4) l (21, 1) l (1, 1)
The vertices of the original triangle are (1, 4), (3, 1), and
(1, 1). The final image would be different if the original
triangle was translated up 3 units and then rotated
counterclockwise 908. The final triangle would have
vertices (27, 1), (24, 3), and (24, 1).
The transformation is a reflection in the x-axis.
7. (x, y) l (x 2 3, y 1 2)
y
E(0, 2) l (23, 4)
E
F(2, 1) l (21, 3)
F
1
G(1, 0) l (22, 2)
21
G
x
The transformation is a translation.
8. No, Figure B is not a rotation of Figure A about the
origin because not all the angles formed by connecting
corresponding vertices are the same.
Mixed Review for TEKS (p. 280)
} }
} }
1. B; It is given that AJ > EJ and BJ > FJ and
ŽAJB > ŽEJF by the Vertical Angles Congruence
Theorem. So, nABJ > nEFJ by the SAS Congruence
Postulate.
2. H; The length of the side opposite the 348 angle would
not be enough additional information for Judith to
construct the triangle. Clint told her that the length of
one side is 8 centimeters and one of the angles formed
with this side is 348. In order for Judith to construct the
triangle, she would need the measure of one of the other
angles or the length of the other side that forms the
348 angle.
3. B; Figure A to Figure C describes a rotation. Figure C is
a 908 counterclockwise rotation of Figure A.
Geometry
Worked-Out Solution Key
113
Chapter 4,
continued
4. J; If a triangle has sides that are 5 centimeters and
3 centimeters long, and the side that is 5 centimeters long
forms a 288 angle with the third side, there is not enough
information to assert that this triangle is unique. Also
needed is the length of the third side or the measure of
one of the other angles.
5. The triangle that is marked with the 21 unit side length is
equiangular, so it is equilateral. The triangle marked with
the 2x 2 3 unit side length is isosceles by the Converse
of Base Angles Theorem. The two triangles share a
common side, which is congruent to the 2x 2 3 unit side.
2x 2 3 5 21
}
} }
}
}
}
15. True; XY > RS, YZ > ST, and XZ > RT, so
nXYZ > nRST by SSS.
}
}
16. Not true; AC 5 5 and DB 5 4, so AC À DB.
Therefore n ABC À nDCB.
17. True; Ž QSR > Ž TSU by the Vertical Angles Theorem.
} }
} }
Because QS > TS and RS > US, nQRS > nTUS
by SAS.
18. Not true; The triangle vertices are in the incorrect order.
}
} }
DE > HG and EF > GF, so nDEF > nHGF by HL.
} }
19. DE > GH, Ž D > Ž G, Ž F > Ž J
} }
20. DF > GJ, Ž F > Ž J, Ž D > Ž G
}
21. Show nACD and nBED are congruent by AAS, which
2x 5 24
}
}
makes AD congruent to BD. nABD is then an isosceles
triangle, which makes Ž1 and Ž2 congruent.
x 5 12
22. Show nFKH > nFGH by HL. So Ž1 > Ž2 because
Chapter 4 Review (pp. 282–285)
corresponding parts of congruent triangles are congruent.
1. A triangle with three congruent angles is
23. Show nQSV > nQTV by SSS. So Ž QSV > Ž QTV
called equiangular.
2. In an isosceles triangle, base angles are opposite the
congruent sides while the congruent sides form the
vertex angle.
3. An isosceles triangle has at least two congruent sides
while a scalene triangle has no congruent sides.
because corresponding parts of congruent triangles
are congruent. Using vertical angles and the Transitive
Property, you get Ž1 > Ž2.
24. ŽL > Ž N, so x 5 65.
25. n WXY is equilateral;
1 }32 x 1 30 28 5 608
4. Sample answer:
3
2
} x 5 30
2
1
}
x 5 20
}
26. TU > VU;
3
Ž R and Ž N
}
} }
}
Corresponding sides: PQ and LM, PR and LN,
}
}
QR and MN
(2x 2 25)8 5 x8 1 208
6.
8x8 5 2x8 1 908
6x 5 90
4x 5 36
x 5 15
x59
(9(9) 1 9)8 5 908
9. mŽB 5 1808 2 508 2 708 5 608
}
27. (x, y) l (x 2 1, y 1 5)
}
10. AB > UT, so AB 5 15 m.
11. Ž T > ŽB, so mŽ T 5 608.
Q(2, 21) l (1, 4)
13. (2x 1 4)8 5 1808 2 1208 2 208
2x 5 36
x 5 18
14. 5x8 5 1808 2 358 2 908
1
Q
21
x
R
S
28. (x, y) l (x, 2y)
y
Q(2, 21) l (2, 1)
R(5, 22) l (5, 2)
S(2, 23) l (2, 3)
1
21
Q
x
R
12. Ž V > ŽA, so mŽ V 5 508.
2x 1 4 5 40
y
(9x 1 9)8 5 5x8 1 458
8.
8(15)8 5 1208
}
x51
S(2, 23) l (1, 2)
(2(45) 2 25)8 5 658
7.
8x 5 8
R(5, 22) l (4, 3)
x 5 45
S
29. (x, y) l (2x, 2y)
y
Q(2, 21) l (22, 1)
R(5, 22) l (25, 2)
S(2, 23) l (22, 3)
1
21
Q
5x 5 55
x 5 11
114
Geometry
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
7x 1 5 5 13 2 x
5. Corresponding angles: Ž P and Ž L, Ž Q and Ž M,
x
R
S