CHAPTER 6 Lesson 6.1
Developing Concepts Activity 6.1 (p. 321) Drawing Conclusions
Think & Discuss (p. 319) 1. As mB increases, A and D move apart. Therefore AD
1. squall, snow, altostratus cloud, hail, showers, lightning 2. squall, hail, showers
increases. 2. The platform of the scissors’ lift will be raised higher. Skill Review (p. 320) 1. If two parallel lines are cut by a transversal, then the
pairs of consecutive interior angles are supplementary (Theorem 3.5 Consecutive Interior Angles). 2. If two parallel lines are cut by a transversal, then the
pairs of alternate interior angles are congruent (Theorem 3.4 Alternate Interior Angles).
3. Process, input/output, manual operation, decision, and
extract are polygons because they are made up of line segments and each line segment intersects exactly two other line segments, one at each endpoint. 4. The fewest number of sides a polygon can have is three,
because each segment must intersect exactly two other line segments. 6.1 Guided Practice (p. 325) 1. vertices
3. AAS Congruence Postulate
2. octagon, 15-gon
3. Yes, because the length of the string would be equal to Q
Y
the sum of each of the line segments in the polygon for both convex and concave polygons. 4. yes
P
R
X
Z
4. SSS Congruence Postulate Q
5. No, because one side is not a line segment.
6. No, because two sides intersect only one other side. 7. equilateral
8. none of these
9. regular
10. mA 360 mB mC mD
Y
360 125 60 70 P
R
X
Z
105 11. mA 360 mB mC mD
5. AB 3 22 4 82
360 105 113 75
52 122
67
25 144 169
6.1 Practice and Applications (pp. 325–328)
13
12. yes
m
20. heptagon; concave
12 5
16. yes
17. no
19. heptagon; concave 21. octagon
23. MP, MQ, MR, MS, MT
26. equiangular
Geometry Chapter 6 Worked-out Solutions Key
24. regular
25. equilateral
27. quadrilateral; regular
28. pentagon; none of these
29. triangle; regular
30. octagon; regular 31. Sample answer:
108
15. no
22. Sample answers: MNPQRSTL, NPQRSTLM,
2 3 8 4 , 2 2
14. no
18. pentagon; convex
8 4 2 3
1 , 2 2
13. no
32. Sample answer:
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 33. Sample answer:
34. Sample answer:
45. 82 25x 2 20x 1 25x 1 360
82 25x 2 20x 1 25x 1 360 70x 180 360 70x 280 x4
35. Yes; to be concave, a polygon must have an angle greater
than 180.
46. x 90 90 99 360 2
36. mABC is 90 because a regular quadrilateral has four
x 2 279 360
equal angles whose sum is 360.
x 2 81
x x x x 360
x9
4x 360
47. Tri–means three; sample answers: triangle is a polygon
x 90 37. mA mB mC mD 360
mA 95 100 90 360 mA 285 360 mA 75 38. mA mB mC mD 360
mA 55 110 124 360 mA 289 360 mA 71 39. mA mB mC mD 360
mA 85 63 87 360 mA 235 360 mA 125 40. The sum of the measures of the interior angles remains
with three angles, tricycle is a child’s vehicle with three wheels, tripod is an object with three legs, and triathlon is an athletic competition with three events. 48. hexagon; convex; regular 49. octagon; concave; equilateral 50. pentagon; concave; none of these 51. 17–gon; concave; none of these 52. a. 18–gon; no it is concave. b. Step 2: decagon; convex
Step 3: heptagon; concave Step 4: quadrilateral; convex c. These are line segments which are not part of the
segments in blue forming the square in Step 1. 53. 3x 3x 3y 3y 360
6x 6y 360
constant.
6x y 360
41. x 100 87 106 360
x y 60
x 293 360
x 60 y
x 67 42. 3x 150 60 90 360
4x 5 4x 5 3y 20 3y 20 360
3x 300 360
4x 5 4x 5 3y 20 3y 20 360 8x 6y 30 360
3x 60
8x 6y 390
x 20 43. 2x 2x 84 100 360
Substituting 60 y for x gives
4x 184 360
860 y 6y 390
4x 176
480 8y 6y 390
2y 90
x 44
y 45
44. 4x 10 108 3x 67 360
4x 10 108 3x 67 360 7x 185 360 7x 175 x 25
135° 45° 65° 115°
135° 45° 65° 115°
x 60 45 15 6.1 Mixed Review (p. 328) 54. 120
55. 63
56. x 2x 180
3x 180 x 60 Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solutions Key
109
Chapter 6 continued 57. 20x 2 9x 4 180
slope of LM
20x 2 9x 4 180
6 1 1 3 4
Find the slope of MN and draw a line through point L which is parallel to MN.
29x 6 180 29x 174
slope of MN
x6 58. x x 180
59. 25x 11x 180
2x 180
36x 180
x 90
x5
8 6 14 2 3 5
The points where the lines intersect are the vertices of the triangle: 1, 13, 5, 1, and 9, 15. y
60. Plot the midpoints of the triangle on a graph. Connect the
midpoints to form the midsegments. LN will be parallel to the side containing point M. LM will be parallel to the side containing point N. MN will be parallel to the side containing point L. Find the slope of LN and draw a line through point M which is parallel to LN. slope of LN
M
L
87 1 8 3 5
2
2
x
N
Find the slope of LM and a draw a line through point N which is parallel to LM. slope of LM
17 3 5 3
62. Plot the midpoints of the triangle on a graph. Connect the
Find the slope of MN and draw a line through point L which is parallel to MN. slope of MN
81 7 8 5 3
The points where the lines intersect are the vertices of the triangle: 0, 0, 10, 2, and 6, 14.
midpoints to form the midsegments. LN will be parallel to the side containing point M. LM will be parallel to the side containing point N. MN will be parallel to the side containing point L. Find the slope LN and draw a line through point M which is parallel to LN. slope of LN
Find the slope of LM and draw a line through point N which is parallel to LM.
y
slope LM
N
24 2 1 2 3
Find the slope of MN and draw a line through point L which is parallel to MN.
L
slope of MN M
2
74 3 02 2
2
x
72 5 0 1
The points where the lines intersect are the vertices of the triangle: 3, 9, 1, 1, and 3, 5. y
61. Plot the midpoints of the triangle on a graph. Connect the
midpoints to form the midsegments. LN will be parallel to the side containing point M. LM will be parallel to the side containing point N. MN will be parallel to the side containing point L. Find the slope of LN and draw a line through point M which is parallel to LN. 8 1 7 slope of LN 2 4 2
N
L M 1
2
x
Find the slope of LM and draw a line through point N which is parallel to LM.
110
Geometry Chapter 6 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 63. Plot the midpoints of the triangle on a graph. Connect the
midpoints to form the midsegments. LN will be parallel to the side containing point M. LM will be parallel to the side containing point N. MN will be parallel to the side containing point L. Find the slope of LN and draw a line through point M which is parallel to LN. 5 3 slope of LN 2 3 1
32 42 9 16 25 CE 2 42 1 12 62 02 36
73 4 6 1 7
CE 6
Find the slope of MN and draw a line through point L which is parallel to MN. slope of MN
BD 0 32 1 32
BD 5
Find the slope of LM and draw a line through point N which is parallel to LM. slope of LM
BE 29
5 7 4 36
Technology Activity 6.2 (p. 329) 1. Yes, ABFC is always a parallelogram because the slopes
of the opposite sides are always equal.
The points where the lines intersect are the vertices of the triangle: 2, 15, 4, 9, and 10, 1. y
2. Opposite sides are congruent. 3. The opposite sides are always congruent. 4. Opposite sides are congruent and have the same slope. 5. Opposite angles are congruent. If one angle increases by
x, the opposite angle also increases by x. Consecutive angles are supplementary. If one angle increases by x, its consecutive angles decreased by x.
M L
6. Opposite angles of a parallelogram are congruent. 2
2
x
Lesson 6.2 N
6.2 Guided Practice (p. 333) 1. A parallelogram is a quadrilateral with both paris of 64.
opposite sides parallel.
y
A E
B
1 1
2. No; only one pair of opposite sides is parallel. C
3. Yes; both pairs of opposite sides are parallel.
x
4. LM; opposite sides of a parallelogram are congruent.
D
5. KN; diagonals of a parallelogram bisect each other. 6. MJK; opposite angles of a parallelogram are congruent.
AC 1 32 4 02
7. JML; opposite angles of a parallelogram are congruent.
22 42
8. LN; diagonals of a parallelogram bisect each other.
4 16
9. LM; opposite sides of a parallelogram are congruent.
20
10. KNJ; vertical angles are congruent.
AC 25
11. KMJ; alternate interior angles are congruent.
AD 0 02 1 32
12. 13; opposite sides of a parallelogram are congruent.
02 42
13. 7; diagonals of a parallelogram bisect each other.
16
14. 8; opposite sides of a parallelogram are congruent.
AD 4
15. 8.2, diagonals of a parallelogram bisect each other.
BE 2 3 1 3 2
52 22
2
16. 80; consecutive angles of a parallelogram are
supplementary.
25 4 Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solutions Key
111
Chapter 6 continued 35 3t 15 180
17. 80; consecutive angles of a parallelogram are
supplementary.
3s 3t 15 180
18. 100; opposite angles of a parallelogram are congruent.
3s 325 15 180
19. 29; alternate interior angles are congruent.
60 35 180
340 4r
3s 120
120 4r
s 40
30 r
6.2 Practice and Applications (pp. 334–337) 20. 10; diagonals of a parallelogram bisect each other. 21. 11; opposite sides of a parallelogram are congruent.
38. a. AD BC or AB DC b. AB DC or AD BC
22. 12; opposite sides of a parallelogram are congruent.
c. DB DB
23. 60; consecutive angles of a parallelogram are
d. SSS
supplementary.
e. corresponding
24. 60; consecutive angles of a parallelogram are
f. AC
supplementary 25. 120; opposite angles of a parallelogram are congruent.
39. 1. UKLM is a . 2. Opposite of a are .
26. x 14, y 10 27. a 180 101
28. r 6
a 79
3. 360
p5
29.
s 3.5
4. Substitution property of equality
q36
b 101
5. mJ; mK
q9
30. 2m 70
k7
n 110
m8
32. 2x 4 8
33. 2u 2 5u 10
2x 4
12 3u
x2
4u
3y 9
6
y3
v 3
18 v 4w w 3
6 2z
3w 3
3z
w1
35. b 10 b 10 180
b 10 b 10 180 2b 180 b 90 c b 10
d b 10
c 90 10
d 90 10 100
80
6. Division property of equality
31. k 4 11
m 35
34. 2z 1 4z 5
3s 4r
36. 2f 5 5f 17
12 3f
gf2 g42
4f
7. Definition of supplementary angles 40. c, 0 43.
41. a c, b
a 2 c, b2
a 2 c, b2
42.
44. They share a common midpoint.
45. Opposite sides of a parallelogram are parallel. If two
parallel lines are cut by a transversal, then the alternate interior angles are congruent. Therefore 3 5. Consecutive angles of a parallelogram are supplementary, so m5 m6 180. By the substitution property of equality, m3 m6 180. Therefore 3 is supplementary to 6. 46. 4 and 3 are supplementary by Theorem 6.4, so
m4 m3 180. By the Alternate Interior Angles Theorem, 3 5, so m3 m5. By substitution m4 m5 180, thus 4 is supplementary to 5.
47. 4
48. 5; 8
49. Corresponding Postulate If 2 lines are cut by a
transversal, then corresponding are .
50. Yes. 1 5, 3 7, 2 6, and 4 8. If
the balusters are all the same height and distance apart, then the sides of the two s would also be . 51. mB 180 120 60 53. It increases.
52. It increases.
54. It increases.
6
37. 3t 15 2t 10
3t 15 2t 10 t 25
112
Geometry Chapter 6 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 55. Statements
Reasons
1. ABCD is a .
1. Given.
2. AB CD
2. If a quadrilateral is a , then its
64. three; 4, 0, 2, 4, 8, 8 (See graphs below.)
Graphs for Exercises 62 and 64. y
opposite sides are .
3. CEFD is a .
3. Given
4. CD FE
4. If a quadrilateral is a , then its
1
D (4, 0)
Reasons
are s.
y
2. 1 2
2. If a quadrilateral is a , then
3. 1 3
3. Transitive Property of .
2 3
1. Given.
2. WZ XY
2. Definition of a
WZ XY 3. ZWM XYM
3. If two lines are cut by a
MZW XYM
transitive, then the alternate interior are .
4. WMZ MXY
4. ASA Congruence Postulate
58. Statements 1. ABCD, EBGH,
C A
2. 1 2
1
65. AB 6 22 9 12
42 82 16 64 80 45
its opposite are .
x
6.2 Mixed Review (p. 337)
1. Given. 2. If a quadrilateral is a , then
3 4 1 4
1
Reasons
and HJKD are .
66. AB 2 42 1 22
62 32 36 9
3. 2 3
3. Transitive Property of .
Opposite are .
C 75 60 105 G 135
135 F
x
B
Reasons
1. WXYZ is a .
75 45
1
D (8, 8)
its opposite sides are .
57. Statements
A
x
A
1. Given.
1. PQRS and TUV’s
B 105
C
(2, 4)
A
1
59.
D
1
5. Transitive Property of .
56. Statements
B C
opposite sides are .
5. AB FE
y
B
120
120 D 60
Adjacent are supp. The sum of the interior of a quad. is 360.
45 E
45 35 67. AB 1 82 3 42
72 12 49 1 50 52
60. B
2s 30 3s 50 180
68. m
5s 80 180 5s 100 s 20 61. B 62. Sample answer: 4, 0 (See graph.) 63. Calculate the slope of each segment and the length of
each segment.
Copyright © McDougal Littell Inc. All rights reserved.
y2 y1 x2 x1 91 62
2 70. m
y2 y1 x2 x1
3 4 1 8
1 7
69. m
y2 y1 x2 x1 1 2 2 4
3 2
Geometry Chapter 6 Worked-out Solutions Key
113
Chapter 6 continued 71. Yes; if two coplanar lines are to the same line, then
12. Yes; if one of a quad. is supp. to both of its
72. B 180 65 35 80. If one of a is
13. No; one pair of opposite sides which are to each other
they are to each other.
larger than another , then the side opposite the larger is longer than the side opposite the smaller . So AB is the shortest side and AC is the longest side.
73. D 180 90 55 35. If one of a is
larger than another , then the side opposite the larger is longer than the side opposite the smaller . So EF is the shortest side and DF is the longest side.
74. H 180 45 60 75. If one of a is
larger than another , then the side opposite the larger is longer than the side opposite the smaller . So GH is the shortest side and GJ is the longest side.
Lesson 6.3 Activity 6.3 (p. 338)
Yes, it is always a parallelogram. 6.3 Guided Practice (p. 342)
consecutive , then the quad. is a .
and to a diagonal isn’t sufficient to prove that the quad. is a . 14. Yes; if one pair of opposite sides of a quad. are both
and then the quad. is a .
15. Sample answer: By AB CD and AD CB. If both
pairs of opposite sides of ABCD are , ABCD is a .
s are , 16. Sample answer: By parts of
ABX CDX and XAB XCD. If alternate interior formed by two lines and a transversal are , the lines are . Both pairs of opposite sides are , so ABCD is a .
17. x 70 18. 2x x 180
3x 180 x 60 19. x 10 x 10 180
2x 180
1. No; a parallelogram is a quadrilateral. 2. Yes; if both pairs of opposite in a quad. are , then
the quad. is a .
x 90 20. Sample answer:
3. Yes; if an of a quad. is supplementary to both of its
consecutive s, then the quad. is a .
4. Yes; if both pairs of opposite sides of a quad. are , then
the quad. is a .
5. Prove BCA DAC. Then BC AD and AB CD
s are . Since both pairs of oppobecause parts of site sides are , then ABCD is a .
6. Since ACB CAD and BAC DAC, and these
four are alt. interior , then AB DC and BC AD. Since both pairs of opposite sides of ABCD are , ABCD is a .
7. CDA and its exterior form a linear pair and are
supplementary. By substitution, BAD BCD and ADC CBA. Therefore, the two pairs of opposite are and ABCD is a .
8. Sample answer: Determine the slopes of each side.
Calculate the length of each side. Show that the diagonals bisect each other. 6.3 Practice and Applications (pp. 342–345) 9. Yes; if both pairs of opposite sides of a quad. are , then
the quad. is a .
10. Yes; if the diagonals of a quad. bisect each other, then the
quad. is a . 11. No; by the vertical thm., this would be true for any
quad.
114
Geometry Chapter 6 Worked-out Solutions Key
21. AB 3 12 5 62
42 12 17 BC 5 32 3 52 22 82 68 217 CD 1 52 2 32 42 12 17 AD 1 12 2 62 22 82 68 217 AB CD and BC AD 22. Midpoint AC
Midpoint BD
12 5, 6 23 2, 23 3 2 1, 5 22 2, 23
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 23. Slope of AB
56 1 3 1 4
26. Sample answer:
3 5 4 Slope of BC 53 1 2 3 Slope of CD 15 4 2 6 4 Slope of AD 1 1 56 1 24. Slope of AB 3 1 4 3 5 Slope of BC 4 53 1 2 3 Slope of CD 15 4
42 12
45 1 82 6
Slope of QR
4 4 8 98
Slope of RS
1 3 4 39 6
Slope of PS
3 5 8 32
Since both pairs of opposite sides have the same slope, they are and PQRS is a . 27. AB CD and BC DA. Since the opposite pairs of
sides are , ABCD is a and AB CD.
28. You could measure the alternate interior formed by the
slanted I and the guidelines. If the are congruent, then the two sides of the I would be . If the two sides of the I are both and , then the I is a .
2 6 Slope of DA 4 1 1 AB 3 12 5 62
Slope of PQ
29. If the diagonals of a quad. bisect each other, then the
quad. is a .
17 BC 5 32 3 52
30. Sample answer: Draw point P and Q on a line M. Then
↔ draw PQ. Draw an arc with the compass at point Q so → → ↔ that it crosses QP and QR. Copy PQR on QP. Make sure the two s are corresponding. Label the new angle ↔ TPS. Draw PS. Set the compass to the length of QR and draw QR onto PS.
22 82 68 217 CD 1 52 2 32 42 12
31. Make AD BC and AB DC. This will form a in
which case BC will remain to AD while the binoculars are being raised or lowered.
17 DA 1 12 2 62
32. Sample answer: Assume that RSTU is a , and given
that R T and S U, then mR mS mT mU 360. By the substitution property of equality and distributive property, 2ms 2mT 360. By the division property of equality, mS mT 180, so they are supplementary. Because opposite s are and consecutive s are supplementary, RSTU is a .
22 82 68 217 1 Slope of AB Slope of CD 4 and AB CD 17 or Slope of BC Slope of DA 4 and BC DA 217.
33.
25. Sample answer:
Midpoint of JL
2 26, 32 2 2, 21
Midpoint of KM
1 2 3, 3 24
2,
1 2
Statements
Reasons
1. P is supplementary
1. Given.
2. QR PS and QP RS
2. Consecutive Interior Thm.
to Q and S.
Conv. 3. PQRS is a .
3. Definition of a .
34. 0, a; the diagonals bisect each other at the origin. The
length of OM is a so the length of OP must also be a, however, it is below the x-axis.
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solutions Key
115
Chapter 6 continued 35. b, c; the diagonals of a bisect each other, so
44. The slope of the line through point 1, 2 is 4 . 1
0, 0 is the midpoint of QN. Let Q x, y. By the xb yc Midpoint Formula, 0, 0 , , so x b 2 2 and y c.
36. Slope of MQ
Slope of NP
y mx b
ca c a b 0 b
z 14 1 b b 94 y 14 x 94 45. x 180 52 68
180 120
c a c a b0 b
Slope of MN
ca ca b0 b
Slope of QP
a c c a 0 b b
ca Slope of MQ slope of NP and slope of b ca MN slope of QP . b
60 46. x 2x 14 50 180
3x 36 180 3x 144 x 48 47. x 85 2x 50
x 35 x 35
37. a. mAEF mEAF mAFE 180
63 90 mAFE 180 mAFF 27 b. mFGD mGDF mDFG 180
Quiz 1 (p. 346) 1. convex, equilateral, equiangular, regular 2. 2x 2x 110 110 360
mFGD 90 27 180
4x 220 360
mFGD 63
4x 140
c. mGHC 27; mEHB 27
x 35
d. mE 54, mF 126, mG 54,
mH 126; EFGH is a parallelogram because opposite sides are and consecutive are supplementary.
38. PTRU and PQRS share a common diagonal PR. PTRU
and QTSU share common diagonal TU. Since PR bisects SQ and SQ bisects TU, PR must also bisects TU which would make PTRU a parallelogram. 6.3 Mixed Review (p. 345) 39. If x 2 2 2, then x 0. If x 0, then x 2 2 2. 40. If 4x 7 x 37, then x 10. If x 10, then
4x 7 x 37.
41. If a quadrilateral is a parallelogram then each pair of
opposite sides is parallel. If each pair of opposite sides is parallel, then the quadrilateral is a parallelogram. 42. A point is on the perpendicular bisector of a segment if
and only if it is equidistant from the endpoints of the segment. 43. A point is on the bisector of an angle if and only if it is
equidistant from the two sides of the angle.
The sum of the interior angles of a quadrilateral is 360. 3. Statements
Reasons
1. ABCG and CDEF
1. Given.
are s. 2. A BCG,
2. If a quad. is a , then the
3. BCG DCF
3. Vertical Angles Theorem
4. A E
4. Transitive Property of Congruence.
DCF E
opposite are .
4. Sample answer: Use slopes to show that both pairs of
opposite sides are parallel, use the Distance Formula to show that both pairs of opposite sides are congruent, use slope and the Distance Formula to show that one pair of opposite sides are both parallel and congruent, use the Midpoint Formula to show that the diagonals bisect each other. Math and History (p. 346) 1. A 2 b1 b2h 1
12 2800 1800500 1,150,000 ft2
116
Geometry Chapter 6 Worked-out Solutions Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 6.4 Guided Practice (p. 351)
27.
B
C
28. B
C
A
D
1. rhombus 2. If a parallelogram is a rhombus, then each diagonal
A
bisects a pair of opposite angles. If a diagonal bisects a pair of opposite angles, then a parallelogram is a rhombus. Diagonal PR would bisect P and R, and SQ would bisect S and Q. 3. always 7. C, D
4. sometimes 8. B, D
5. sometimes
9. B, D
D
Always; if a quad. is a , then the opp. are .
6. always
10. A, B, C, D 29.
11. 2x 90
B
Sometimes; ABCD would have to be a square because A and B are and supplementary so each would measure 90.
30.
C
B
C
x 45 A
6.4 Practice and Applications (pp. 351–355) 12. B
C
13. B
C
A
D
A
D
Always; a rectangle has four right angles and right angles are congruent. C
15. B
C
A
D
A
D
A
Always; each diagonal of a rhombus bisects a pair of opposite . 31. B
C
A
D
Sometimes; if rectangle ABCD is also a rhombus (a square), then AB BC.
14. B
D
Always; a rhombus is a parallelogram with four congruent sides. 32.
16. rectangle, square
Sometimes; if rectangle ABCD is also a rhombus (a square), then the diagonals of ABCD are .
17. square
Sometimes; if a rhombus is also a rectangle (square).
FG HI, FI GH, FG HI, GH FI, G I, F H, FH and GI bisect each other.
I
P
S
24. B
C
A
D
25. rectangle
38. 8x 13 7x 11
x 24
x1 H
R
x5
x 50
18. rhombus, square
21. rhombus, square
Q
36. 2x 10
3x 150
2x 2
23.
50 3x 30 180
20. parallelogram, rectangle, rhombus, square
F
D
34. x 180 130
35. x 40 2x 10 180
37. 3x x 2
G
C
Always; a rhombus is a parallelogram with four congruent sides.
x 18
19. parallelogram, rectangle, rhombus, square
22.
B
A
33. 5x 90
Always; ABCD could also be a square.
D
PQ RS, PS QR, PQ QR RS SP, P R, S Q, PR and QS bisect each other, QS RP, QS bisects PSR and PQR. PR bisects SPQ and SRQ. All four sides are congruent, all four angles are right angles; its diagonals are congruent, bisect each other, and bisect opposite angles.
39. 22
40. 90
41. 45
42. P HJ KJ KH 2 2 22 4 22 43.
24 5x 1 13 x 12 4x 3x WY XZ 13 3 10
44. You would need to know that A, B, C, and D are
right angles. 45. Assume temporarily that MN PQ, 1 2, and that
MQ PN. By the definition of a , MNPQ is a . This contradicts the given information that 1 2. It follows that MQ is not parallel to NP.
26. square
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
117
Chapter 6 continued 46. Statements
Reasons
1. RSTU is a ,
1. Given
2. RSTU is a rhombus.
2. A is a rhombus if its
SU RT.
diagonals are .
3. mSTR mUTR
3. The diagonals of a
4. STR UTR
4. Def. of
rhombus bisect opp. .
47. If a is a rectangle, then its diagonals are congruent.
If the diagonals of a are congruent, then the is a rectangle. JL KM
52. Statements
Reasons
1. FGHJ is a , FH
1. Given
2. HFG HFJ,
2. Def. of bisector
3. FH FH, GJ GJ
3. Reflexive Prop. of
bisects JFG and GHJ, JG bisects FJH and HGF.
FHG FHJ; FGJ HGJ, FJG HJG
Congruence 4. FHJ FHG,
4. ASA Cong. Post.
sides. (A rhombus by definition has four congruent sides.)
5. JH GH, FG GH,
5. Corresponding parts of
If a quadrilateral has four congruent sides, then it is a rhombus. (By definition a with four congruent sides is a rhombus.)
6. JH FG GH FJ
48. If a quadrilateral is a rhombus, then it has four congruent
49. If a quadrilateral is a rectangle, then it has four right
angles. (def. of a rectangle) If a quadrilateral has four right angles, then it is a rectangle. (Both pairs of opp. are , so the quad. is a . Since all 4 are congruent and the sum of the measures of the interior angles of a quadrilateral is 360, the measure of each angle is 90, and the quadrilateral is a rectangle.) 50. If a quadrilateral is a square, then it is a rhombus and a
rectangle. (A square has four sides which makes it a rhombus and four rt. which makes it a rectangle.) If a quadrilateral is a rhombus and a rectangle, then it is a square. (By definition, a rhombus has four sides and a rectangle has four rt. . The only quad. which has four sides and four rt. is a square.) 51. Sample answer:
Statements
Reasons
1. PQRT is a rhombus.
1. Given
2. PQ PT QR RT
2. A quad. is a rhombus if
3. PR PR, QT QT
and only if it has 4 sides. 3. Reflexive Prop. of
Congruence 4. PRQ PRT;
4. SSS Cong. Postulate
PTQ RTQ
5. TPR QPR,
TRP QRP; PTQ RTQ, PQT RQT
6. PR bisects TPQ and
QRT, QT bisects PTR and RQP.
118
FGJ HGJ
JH FJ, FG FJ
congruent triangles are congruent. 6. Transitive Property of
Congruence 7. FGHJ is a rhombus.
7. Def. of a rhombus
53. Sample answer: Draw a line f and GH where f is not
to GH. f should intersect GH at H. Construct HJ on f so that HJ GH. With centers G and J, construct two arcs with radius GH which intersect at K. Draw GK and JK. Since all four sides of GHJK are , GHJK is a rhombus. It is not a square or rectangle because GH and HJ are not . 54. Sample answer: Draw a segment AB. Copy a 90 onto
AB with vertex A. Mark a segment on the side that is not AB and that is not to AB. Label the endpoint D. Copy another right angle onto AB that has vertex B. Copy AD onto this new side. Label the endpoint C, connect C and D. 55. Rectangle;
PR 3 2 2 1 3 2 52 42 41 QS 3 2 2 3 12 52 42 41 If the diagonals of a are , then the is a rectangle.
y
S (2, 1)
s 5. Corresp. parts of
are .
P (3, 1) 1
R (2, 3)
1
x
Q (3, 3)
6. Def. of bisector
Geometry Chapter 6 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 56. Rhombus; slope of PR
slope of QS
22 0 0 5 3 8
9 5 14 und. 11 0
PR QS
slope of PS
21 3 1 52 3
slope of RS
21 3 1 1 2 3 y
So, PQ QR, PS RS.
If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus.
y
R (3, 2)
Q (2, 5)
A quadrilateral is a square if it has 4 congruent sides and 4 right angles.
Q (1, 9)
P (5, 2)
R (1, 2)
P (5, 2)
2
1
2
1
x
S (1, 5)
x
S (2, 1)
59. b, a; KM ON, so KM b and KO MN, so
MN a.
60. OM b 02 a 02
57. Rectangle;
b2 a2
PR 1 22 4 32
KN b 02 a 02
32 72
b2 a2
58
OM KN
QS 3 42 2 1 2 7 2
61. Sample answer: ABCD is a since cross braces AD and
32
58 If the diagonals of a parallelogram are , then the is a rectangle.
y
P (1, 4)
S (4, 1)
1 1
x
Q (3, 2) R (2, 3)
BC bisect each other. AD BC so ABCD is a rectangle. Since a rectangle has four right angles, mCAB mDBA 90. Then, mBAC mBAE and mABD mABF. By substitution mBAE mABF 90. So table top AB is perpendicular to legs AE and BF by the def. of perpendicular.
62. Sample answer: Since ABCD is a rectangle, AB CD, AB
and CD are to AC and BD, respectively. They would also be to AE and BF. Therefore, EF AE and BF and EF AB.
63. Rhombus; AE CE AF CF; AECF remains a
58. Square;
PQ 5 2 2 5 2
2
32 32 32 QR 2 1 2 5 22 32 32
rhombus. 64. When A or C is dragged, mFAC mEAC and
mAEF mCEF.
65. Each diagonal of a rhombus bisects a pair of opp. .
(Theorem 6.12) 66. D
32
3x 12
RS 1 22 2 1 2 32 32 32 SP 2 52 1 22 b32 32 32 So, PQ QR RS SP. slope of PQ
52 1 25
slope of QR
52 1 2 1
Copyright © McDougal Littell Inc. All rights reserved.
7x 3 4x 9 x4
67. B
9x 9 90 9x 81 x9
2xy 90 29y 90 y5
68. In a parallelogram, opposite are and consecutive
are supp. Therefore, all four must measure 90 in which case, the would be a rectangle.
69. The diagonals of a bisect each other. So, OA OC
and OD OB. AO OC AC and DO OB DB, then AO OC DO OB by substitution. AO AO DO DO and OC OC OB OB, again by substitution, giving AO DO and OC OB. By the Transitive Property of Congruence, OA OB OC OD. Geometry Chapter 6 Worked-out Solution Key
119
Chapter 6 continued 70. OB 0 b2 0 02 b2 b
2
1
82. PD 3 AD
81. PB 3 FB
Since OB OA, OA b and let A a, y.
6 23FB
1339
b a 02 y 02
9 FB
13
b a y 2
2
83. Assume temporarily that ABCD is a quad. with four acute
, that is, mA < 90, mB < 90, mC < 90, and mD < 90. Then mA mB mC mD < 360. This contradicts the Interior Angles of a Quadrilateral Theorem. Therefore, no quad. has 4 acute .
b a y 2
2
2
b2 a2 y2 b2 a2 y
Aa, b2 a2 71. Since OB OD, OD b, and OD lies on the x-axis,
D b, 0. Since OA OC, OC b, and C is in quadrant III, the coordinates must be negative. So, C a, b2 a2 .
72. Slope of AB:
Slope of BC:
b2 a2
ab
b2 a2 0
ab
b2 a2
ab
1. CD and AB
congruent sides, but opp. sides are not . In a rhombus, both pairs of opp. sides are , so a rhombus can’t be a kite.
0 b2 a2 b2 a2 b a ba
b2 a2
ba
a b baa b 2
2
b ab a a ba b
a b 1 ab
3. isosceles trapezoid
Slope of DA:
6. Show that AC BD. 8.
b2 a2 0
a b
2
b2 a2
ab
ab baa b 2
2
b ab a a ba b
mL mK 136
17. mJ 180 78 102; mL 180 132 48
19. MN 2 7 9 2 16 8
3 2
AD
77. yes
1
6.6
78. yes
2 3 CE 2 3 CE
9.9 CE
PD AD AP
EP PC CE
PD 32 1
EP 6.6 9.9
12
120
EP 3.3
Geometry Chapter 6 Worked-out Solution Key
1
20. MN 2 16 14 2 30 15
22. 80. PC
15. base angles
16. mM mJ 44, mK 180 44 136;
21. MN 76. no
12. consecutive sides
14. opposite angles
1
6.4 Mixed Review (p. 355)
1
11. legs
13. diagonals
S
mK mL 98
ABCD is a parallelogram with four right angles, or it is a rectangle.
75. no
R
18. mJ mM 82, mL 180 82 98;
CD DA
2 3 AD 2 3 AD
Q
P
10. bases
ab
1
79. AP
3 1210 5
5. trapezoid
1 1 7. 2 11 7 2 18 9 1 1 19 9. 2 12 7 2 19 2
6.5 Practice and Applications (pp. 359–362)
b2 a2
74. no
1 2 7
b2 a2 0 b2 a2 a b a b
ab ba
73. yes
4. kite
Figure for Exs. 10–15:
Slope of CD:
2
6.5 Guided Practice (p. 359)
2. A kite is a quadrilateral that has two pairs of consecutive
AB BC
Lesson 6.5
7
1
1 2 15
1 2 x
9
9
14 x 9 5x 24.
1 2 24
12
23.
7 12x 4 14 x 4 10 x
8 1211 x 16 11 x 5x
25. Yes; X is equidistant from the vertices of the dodecagon,
so XA XB and XAB XBA by the Base Angles Theorem. Since trapezoid ABPQ has a pair of congruent base angles, ABPQ is isosceles. 360 26. mAXB 12 30 Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 27. In ABX, mAXB 30, and A B, so
mA mB 75, mP 180 75 105; mQ mP 105
28. AD AB 22 32
29. EH GH 52 72
13
74
3.61
8.60
BC CD 32 42
EF FG 52 42
25
41
5
6.40
30. JK JM 122 82
208 14.42 KL LM 82 52 89 9.43 31. mE mG
2mG 360 120 50 2mG 190 mG 95 32. mG 360 110 90 90
70 33. mF mH 100
mG 360 100 100 70 90 34. A parallelogram is a quad. in which both pairs of opp.
sides are congruent and parallel. In an isosceles trapezoid, the bases are parallel but not congruent, and the legs of the isosceles trapezoid are congruent but not parallel. 35. Any two numbers whose sum is 10. 36. AB 3 42 3 52
72 22 53 BC 3 62 3 132 32 162 265 CD 6 62 13 22 122 112 265 DA 6 42 2 52 22 72 53 AB DA, so AB and DA are consecutive sides. DC BC, so DC and BC are consecutive sides. So, the points are vertices of a kite.
37. Slope of AB
40 4 2 0 2 2
Slope of BC
44 0 0 50 5
Slope of CD
04 4 85 3
Slope of AD
00 0 0 8 2 10
AB 2 02 0 42 22 42 20 25 CD 5 82 4 02 32 42 25 5 ABCD is a trapezoid; slope of BC slope of AD 0, so BC AD; slope of AB 2 and slope of CD 43, so AB is not to CD. ABCD is not isosceles; AB 25 and CD 5. 38. Slope of EF
29 7 41 3
Slope of EG
22 0 0 54 1
Slope of GH
92 7 85 3
Slope of EH
99 0 0 81 7
EF 4 12 2 92 32 72 58 GH 8 52 9 22 32 72 58 ABCD is a trapezoid; slope of FG slope of EH 0, so FG EH; slope of EF 73 and slope of GH 73, so EF is not to GH. ABCD is isosceles; EF GH 58. 39. BE 2 CD AF 1
1210 22 1232 16 in. 40. It is given that ABCD is an isosceles trapezoid with
AB to DC and AD BC. If AE is drawn so that ABCE is a parallelogram, since opposite sides of a parallelogram are congruent, BC AE and by the Transitive Property of Congruence, AE AD. By the Base Angles Theorem D AED. Since ABCD is an isosceles trapezoid, the base angles are congruent, so D C. By the Consecutive Interior Angles Theorem, —CONTINUED—
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
121
Chapter 6 continued 50. Yes. Both pairs of base are , but opposite angles are
40. —CONTINUED—
mB mC 180 and mDAB mD 180. By the Substitution property of equality, mB mC mDAB mD; mB mC mDAB mC. Finally mB mDAB by the Subtraction property of equality. So, B DAB by definition of congruence. 41. TQRS is an isosceles trapezoid so QTS RST
because base of an isosceles trapezoid are . QT RS and by the Reflexive Prop. of Cong., TS TS, QTS RST by the SAS Congruence Postulate. Then TR SQ because corresp. parts of triangles are . 1
42. BG 2 CD and BG CD by the Midsegment Thm. and
EG 12AF and EG AF by the Midsegment Thm. BE BG GE by substitution BE 12CD 12AF and BE is parallel to both CD and AF. 43. If AC BC then ACBD is a kite; AC CD and
BC BD, so the quadrilateral has two pairs of congruent sides, but opposite sides are not congruent. (If AC BC then ABCD is a rhombus.)
44. If dragged to the left, the stay . If dragged in other
directions, the change. If it is a kite, then only one pair of opp. will be congruent at a time.
45. If a quad. is a kite, then exactly one pair of opp. are
. (Theorem 6.19)
46. Statements
Reasons
1. AB CB, AD CD
1. Given
2. BD BD
2. Reflexive Prop. of
3. BCD BAD
3. SSS Congruence Post.
4. CBX ABX
s 4. Corresp. parts of
5. BX BX
5. Reflexive Prop. of
Congruence
are .
Congruence 6. CBX ABX
6. SAS Congruence Post.
7. CXB AXB
s 7. Corresp. parts of
are .
8. CXB and AXB are 8. Given
not congruent. 51. 15 2 3x 2 2x 2 1
30 5x 6x 53. Draw a perpendicular segment from Q to PS and label
the point of intersection M. Draw a perpendicular segment from R to PS and label the point of intersection N. QMP RNS because perpendicular lines form right angles and all right angles are congruent. If two lines are perpendicular to the same line, then they are parallel to each other, so QM RN. Therefore, QRNM is a parallelogram because both pairs of opposite sides are parallel. If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other, so QM QR and RN QR. We know that MQR NRQ because perpendicular lines form right angles and all right angles are congruent. QRNM is a rectangle because it is a parallelogram with four right angles. By definition of a rectangle, QM RN. We know that QMS and RNP are right triangles, and it is given that PR SQ, so QMS RNP by the HL Congruence Theorem. QSP RPS because corresponding parts of congruent triangles are congruent. PS PS by the Reflexive Property of Congruence. QPS RSP by SAS Congruence Postulate. Therefore, QP RS because corresponding parts of congruent triangles are congruent. 6.5 Mixed Review (p. 363) 54. If a triangle is scalene, then it has no congruent sides. 55. If a quadrilateral is a kite, then it has perpendicular diag-
onals. 56. If a polygon is a pentagon, then it has five sides. 57. 5.6
58. 10
59. 7
9. If two lines intersect to
form a linear pair of congruent angles, then the lines are . 47. Draw BD. (Through any 2 points, there is exactly 1 line.)
Since BC BA and CD AD, BCD BAD by the s SSS Congruence Postulate. A C because corresp. parts of are . Assume temporarily that B D, then both pairs of opp. of ABCD would be making ABCD a parallelogram. This contradicts the def. of a kite. Therefore B D. 48. Yes; The legs are and it has one pair of sides. 49. Yes; The diagonals are and it has one pair of sides.
122
Geometry Chapter 6 Worked-out Solution Key
60. 11.2
61. 80
62. 100
63. Yes; Sample answer:
Slope of AB
88 0 0 52 7
Slope of BC
80 8 52 3
Slope of CD
00 0 0 25 7
Slope of AD
80 8 2 5 3
a linear pair. 9. AC BD
52. C
AB 2 52 8 82 49 7 BC 5 22 8 02 32 82 73 CD 2 52 02 49 7 AD 2 52 8 02 32 82 73 Both pairs of opposite sides are parallel and congruent, so ABCD is a parallelogram.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 3. AB 3 02 6 92
64. Yes; Sample answer:
Slope of PQ
1 3 2 94 5
32 32
6 1 5 5 Slope of QR 89 1 8 6 2 2 Slope of RS 38 5 5 Slope of PS
8 3 5 5 34 1
PQ 4 9 3 1 2
2
52 22 29 QR 9 82 1 62
12
26
52
RS 8 32 6 82 52 22 29 PS 4 32 3 82
12
52
26
Both pairs of opposite sides are parallel and congruent, so PQRS is a parallelogram. Quiz 2 (p. 363) 1. A is a rhombus if and only if each diagonal bisects a
pair of opp. , BJ is a diagonal of rhombus EBFJ so EJB FJB and FBJ EBJ. By ASA Cong. Post. BEJ DFJ. By corresponding parts of congruent triangles are congruent, BFJ BEJ, KFJ JEH by the Congruent Supplements Thm. Since HE JE JF KF, JFK HEJ by the SAS Congruence Postulate and since corresponding parts of congruent triangles are congruent, HJ JK.
2. PQ 2 42 5 52 36 6
RS 2 (42 7 72 36 6 PR 2 22 5 72 144 12 QS 4 42 7 52 144 12 QR 4 22 5 72 62 122 180 65 PS 2 42 5 72 62 122 180 65 Opposite sides are congruent and diagonals are congruent, so PQSR is a rectangle.
18 32 BC 0 32 9 62 32 32 18 32 CD 3 02 6 102 32 162 265 AD 3 02 6 102 32 162 265 Two pairs of consecutive sides are congruent, so ABCD is a kite. 4. JK 5 42 6 22
12 82 65 KL 4 42 2 12 82 12 65 LM 4 32 1 72 12 82 65 MJ 5 32 6 72 82 12 65 MK 4 32 2 72 72 92 130 JL 5 42 6 12 92 72 130 All sides are congruent and diagonals are congruent, so JKLM is a square. 5. Slope of PS
3 9 12 1 5 7 12
Slope of SR
93 6 6 76 1
Slope of RQ
3 2 5 1 61 5
—CONTINUED—
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
123
Chapter 6 continued 5. —CONTINUED—
Slope of QP
14. trapezoid
2 3 1 1 5 6
17. square
21. , rectangle, rhombus, square 22. , rectangle, square, rhombus, isosceles trapezoid 23. parallelogram, rectangle, rhombus, square 24. square 25. Show that the quad. has two pairs of consecutive congru-
ent sides, but opposite sides are not congruent (def. of a kite). 26. Show that the quad. has four rt. with four sides (def.
of a square). 27. Show that the quad. has four rt. ; show that the quad. is
a and that the diagonals are .
28. Show that only one pair of opp. sides is (def. of a trape-
6.6 Guided Practice (p. 367)
zoid).
1. Draw segment DB. By the Midsegment Theorem for
29. Show that exactly 2 sides are parallel and that the nonpar-
Triangles, EF DB and HG DB, so EF HG. Property
5. 6.
20. trapezoid
Exactly one pair of opposite sides are parallel, so PQRS is a trapezoid.
Lesson 6.6
4.
18. kite
19. , rectangle, square, rhombus, kite
D C. Draw AE BC, so ABCE is a parallelogram. AED C by the Corr. Postulate. AED D by the Trans. Prop. of Cong. AD AE by the Converse of Base Thm. AE BC because opp. sides of a parallelogram are congruent. By the Trans. Prop. of Congruence, AD BC.
3.
16. trapezoid
PS RQ
6. It is given that ABCD is a trapezoid with AB DC and
2.
15. isosceles trapezoid
Both pairs of opp. sides are .
Rectangle Rhombus Square Kite X
X
X
X
allel sides are congruent (def. of isosceles trap.); show that the quad. is a trapezoid and that the pair of base are ; show that the quad. is a trapezoid and that its diagonals are . 30. AB CD; the nonparallel sides must be ;
Exactly 1 pair of opp. sides are .
X
mC 110 and mD 70; each pair of base must be .
31. BE DE; the diagonals of a parallelogram bisect
Diagonals are .
X
X
X
each other. 32. Sample answer: ABD CBD; the diagonals of a
Diagonals are . Diagonals bisect each other.
Trapezoid
X X
X
rhombus bisect opposite .
X X
33. Sample answer: BD AC; diagonals of a rectangle
X
are .
34. BC DC; use SAS to prove that ABC ADC,
which implies that AB AD. A quad. that has two pairs of consecutive congruent sides is a kite.
7. parallelogram, rectangle, rhombus, square
35. Sample answer: AB AD; us transitive property to
6.6 Practice and Applications (pp.367–370) Property 8.
9.
10. 11.
12.
13.
Both pairs of opp. sides are .
Rectangle Rhombus Square Kite Trapezoid X
X
X
36. PQ 0 02 0 22
X
4 2 QR 0 52 2 52
Exactly 1 pair of opp. sides are . All sides are . Both pairs of opp. are .
52 32 34
X
Exactly 1 pair of opp. are . All are .
124
show that all four sides are congruent. A rectangle that has four congruent sides is a square.
X
X
X
X
X
RS 5 22 5 02 32 52 34 X
X
X
Geometry Chapter 6 Worked-out Solution Key
PS 0 22 0 02 4 2 Kite; PQ PS and QR RS, two pairs of consecutive sides are congruent.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 37. Slope of SR
0 88 0 2 4 2
7 81 7 Slope of RQ 4 5 1 11 0 0 Slope of QP 51 4 Slope of PS
81 7 7 21 1
SR 4 2 8 8 2
2
4
17 QR 5 42 2 82 12 62 37 RS 3 12 2 12 42 12 SP 1 02 1 72
RQ 5 4 1 8 2
42 12
17
2 12
39. PQ 4 02 8 72
2
7
2
50
12 62 37 Slope of PQ
87 1 40 4
Slope of QR
82 6 6 4 5 1
Slope of RS
21 1 51 4
50
Slope of SP
71 6 6 0 1 1
52
Parallelogram; Sample answer: PQ RS and QR SP.
52 QP 1 52 1 12 16 4 PS 1 22 1 82 12 72
Isosceles trapezoid; SR QP, and PS and RQ are congruent but not parallel. 38. PQ 2 72 1 12
25 5 QR 7 72 1 72 36 6 RS 7 22 7 52 52 22 29 SP 2 22 5 12 16
97 2 1 51 4 2
Slope of QR
3 9 6 2 85 3
Slope of RS
31 2 1 84 4 2
Slope of SP
71 6 2 1 4 3
PR 8 12 3 72 72 42 65 QS 5 42 9 12 12 82 65
4 Slope of PQ
40. Slope of PQ
11 0 0 72 5
71 6 undefined Slope of QR 77 0 Slope of RS
75 2 72 5
Slope of SP
51 4 undefined 22 0
Trapezoid; QR and SP are the only pair of parallel sides.
Copyright © McDougal Littell Inc. All rights reserved.
Rectangle; Sample answer: PQ QR and QR SP, and the diagonals PR and QS are congruent. 41. PQ 5 92 1 62
42 52 41 QR 9 52 6 112 42 52 41 —CONTINUED—
Geometry Chapter 6 Worked-out Solution Key
125
Chapter 6 continued 4. PF FQ QG GR
41. —CONTINUED—
RS 5 12 11 62 42 52 41 SP 5 12 1 62 42 52
and square
5. EPF, FQG, GRH,
5. Definition of
HSE are isosceles triangles.
isosceles triangle
6. EPF FQG
6. SAS Congruence
GRH HSE
41
Postulate
7. EF FG GH HE
s 7. Corresp. parts of
8. EFGH is a rhombus.
8. Definition of rhombus
9. FEP EFP
9. Base Angles Theorem;
are .
Rhombus; Sample answer: PQ QR RS SP. 42. trapezoid
43. isosceles trapezoid
44. The segment AB and the perpendicular bisector are the
diagonals of a rhombus which are to each other. 45. ; if the diagonals of a quad. bisect each other, the quad.
is a parallelogram. Since the diagonals are not perpendicular, the is not a rhombus and since the diagonals are not congruent, the is not a rectangle. 46. Rhombus; because the diagonals are perpendicular. Since
AC BD, the quad. is not a square. 47. Kite; AC BD and AC bisects BD, you can use congrus to prove that AB AD and that CB CD. BD ent does not bisect AC, so ABCD is not a . Opp. sides are not , so ABCD is a kite.
48. EFLM is a parallelogram. EF HG and HG JK and
JK LM. Therefore EF LM.
49. Draw a line through C to DE. Draw a line through E
parallel to CD. Label the intersection F. CDEF is a by def. of a . CDE and DEF are rt. because consec. of a are supplementary. DCF and CFE are rt. because opp. of a are and CDEF is a rectangle. The diagonals of a parallelogram bisect each other, so DM 12DF and CM 12CE. The diagonals of a rectangle are , so DF CE, 12DF 12CE, and DM CM. By the definition of congruence, DM CM.
50. ABCD is a quad. with diagonals BD CA. The diagonals
intersect at N with BN ND CN NA. ABN NBC BCN NCD CDN NDA DAN NAB by the base thm. Let mABN x then 8x 360 because the sum of the int. of a quad. is 360. Therefore x 45. By the Angle Add. Postulate, mABN mBNC 2x or 90 mBCD mBCN mNCD or 2x 90. By the same reasoning mBAD 90 and mCDA 90.
51. Statements 1. PQRS is a square;
Reasons 1. Given
E, F, G, and H are midpoints of the sides of the square.
GFQ FGQ RGH RHG SHE SEH
2. Definition of square
3. P Q R S
3. Right Angle
P, Q, R, and S are right angles.
corresponding parts of s are .
10. mFEP mEFP
10. Triangle Sum Theorem
11. mFEP mHES
11. Angle Addition
mGFQ mFGQ mRGH mRHG mSHE mSEH 45
mHEF 180, mEFP mGFQ mGFE 180, mQGF mRGH mHGF 180, mRHG mSHE mGHE 180
12. mHEF mEFG
mHGF mGHE 90
13. EFGH is a square.
Postulate
12. Subtraction and
substitution properties of equality 13. A rhombus with four
right angles is a square. 52. Sample answer: Statements
Reasons
1. JK LM, E, F, G,
1. Given
and H are the midpoints of JL, KL, KM, and JM. 2. JK LM
2. Def. of
3. EH, EF, FG, and GH
3. Def. of midsegment
are midsegments. 1
1
4. EH 2 LM, EF 2 JK,
FG
1 2 LM,
GH
4. Midsegment Thm.
1 2 JK
5. EH EF FG GH
2. PQ QR RS SP;
4. Definition of midpoint
RH HS SE EP
5. Substitution prop. of
eq. and trans. prop. of eq. 6. EH EF FG GH
6. Def. of
7. EFGH is a rhombus.
7. Def. of rhombus
Congruence Theorem
126
Geometry Chapter 6 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 53. a. Isosceles; since JKLM is a
L
M
regular pentagon, LK KJ, and each angle is 108. In JKL, P KLJ LJK by the Base K N Theorem. 2mKJL mK 180, so 2mKJL 72 J and mKJL 36. mKJL mLJN 108 and mKLJ mJLM 108. By the transitive property of equality, mKJL mLJN mKLJ mJLM. We know that mKJL mKLJ, so by the subtraction property of equality, mLJN mJLM and LJN JLM by the definition of congruence. b. mNJL mJLM 72, mLMN mMNJ
108. If two lines are cut by a transversal so that consecutive interior are supplementary, then the lines are parallel so JL NM.
2. The area of each is
1 2
the area of the . The bases and heights are the same for both the and the .
3. The area of each trapezoid is
1 2
the area of the parallelogram. The base of the parallelogram is equal to the sum of the bases of the trapezoids. The heights are the same.
6.7 Guided Practice (p. 376) 1. The midsegment of a trapezoid is the segment that
connects the midpoints of the legs. 2. 6
3. A
8. A
1 2 bh
9. A
s2
4. E
5. C
1 2 74
52
6. B
7. D
14 sq units
25 sq units
10. A bh 94 36 sq units 11. A 2 d1d2 2 108 40 sq units 1
1
c. parallelogram
12. A 2 d1d2 2 1212 72 sq units
d. Yes; because JPMN is a parallelogram, opposite sides
13. A 2 hb1 b2 2 68 4 36 sq units
1
1
1
must be congruent. We know that JN MN (JKLMN is a regular pentagon), so JN PM and JP MN. By the Transitive Property of Congruence, JN PM JP MN, so JPMN is a rhombus. 54. ABCD is an isosceles trapezoid.
A
Plan for proof: prove that AND BNC. Then prove AB OC. Prove AC BD.
B N D
6.7 Practice and Applications (pp. 376–379) 14. A 2 bh 2 57 17.5 sq units 1
15. A s2 72 49 sq units 16. A bh 59 45 sq units 17. A bh 158 120 sq units
C
18. A bh 2221 462 sq units 1
20. A
55. A s2 42 16 sq units
56. A s2 72 49 sq units
22. A 2 b1 b2h 1
58. A lw 96 54 sq units 1
60. A 2 bh
1 2 1 2
1
1 2 b1 b2h 1 2 6 108
1
21. A 2 d1d2
123819 361 units2
64 units2
57. A lw 53 15 sq units 1
1
19. A 2 bh 2 54 10 sq units
6.6 Mixed Review (p. 370)
59. A 2 bh
1
12 5 30 sq units 12 8 48 sq units
61. 32x 15 133 80 44x 1 360
x 1.75
16 15
372
240 units2
units2
24. A 2 b1 b2 1
1
25. A 2 d1d2
128 1614
76x 227 360 76x 133
23. A bh
1224 724
121410 70 units2
168 units2 26. bh A
27.
1 2 8x
7x 63
62. mA 32x 15
x 9 cm
321.75 15 1
64. midsegment 2 AD BC 2 4 8 6 1
65. midsegment 2 AD BC 2 2 8 5 1
48
x 12 ft
63. midsegment 2 AB DC 2 4 10 7 1
A
4x 48
71 1
1 2 d1d2
1
Lesson 6.7
28.
1 dd A 2 1 2 1 2x16 48 2
29.
1 A bh 2 2A b h
16x 48 x 3 in.
Developing Concepts Activity 6.7 (p. 371) 1. The areas, bases and heights are the same.
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
127
Chapter 6 continued 30.
1 A d1d2 2
1 A b1 b2h 2
31.
2A d1 d2
1 2
46. A 5 32 8 ft2
2A b1 b2 h
8 ft2
1152 in.2 576 daisies 2 in.2
2A b2 b1 h 1
32. A bh
33. A 2 d1d2
34
12 units
1 2 24
1 2
47. A 38 12 ft2
4 units2
2
12 ft2
34. A s2 2
5 units
1
48. A 2 2 b1 b2h 22 bh
2
1
36. A bh 2 bh
1232
4832 124812
3 ft2
1824 in.2
37. A 2 2 b1 b2h
38. A bh 2 b1 b2h 1
16 3012
2016 129 205
552 in.2
392.5 in.2
39. No; although the bases are the same, the angles of the
two parallelograms may not be congruent. 40. Yes. By definition, a rectangle has four right angles. So
the angles of one rectangle will always be congruent to the of any other rectangle. Since A bh, the height for all rectangles with area 24 ft2 and base 6 ft must have a height of 4 ft. Thus, making the sides of one rectangle congruent to the sides of another rectangle. 41. 102 62 b2
100 36 b2 64 b2
561,216 41,076 602,292 in.2 or about 4182.6 ft2 49.
602,292 in.2 6023 shakes 100 in.2
50. s 32 62
45 35 Area of blue s2 35 2 45 units2
Area of yellow 412bh
236 36 units2 51. Area of blue 22 bh 1
128
A 12bh 1286 24 units2
96 units2 Area of yellow bh Area of blue 1216 96
8b 42. 132 122 b2
169 144 b2 25 b2
A 12d1d2 122010 100 units2
5b 43. h2 202 162
A bh
h2 400 256
1612
h2 144
192 units2
h 12 44. The area would be doubled if the length of one of the
diagonals was doubled. The area of the kite would be four times as much if both diagonals were doubled. 45. A 25 10 ft2
10 ft2
144 in.2 1440 in.2 1 ft2
1440 in.2 480 carnations 3 in.2 128
Geometry Chapter 6 Worked-out Solution Key
1
1908 1644158 252163
1
35. A 2 d1d2
1
144 in.2 1728 in.2 1 ft2
1728 in.2 432 chrysanthemums 4 in.2
S 1 02 3 12 12 22 5
A 5
144 in.2 1152 in.2 1 ft2
96 units2 52. ATotal
1 2
112 112
121 units2 Area of yellow 1242 42 1272 72 16 49 65 units2 Area of blue bh 42 72 56 units2 53. Square, square; Sample answer: In quad. EBFJ, E, J,
and F are right angles by the Linear Pair Postulate and B is a right angle by the Interior Angles of a Quadrilateral Theorem. Then EBFJ is a rectangle by the Rectangle Corollary EJ FJ because they are corresp. parts of . Then, by the definition of a and the Transitive Property of Congruence, EBFJ is a rhombus and, therefore, a square. Similarly, HJGD is a square.
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 54. Square; each side has length b h and each is a rt. . 55. Each side b h and the area of ABCD would be
Quiz 3 (p. 380) 1. ON 0 02 0 12
b h2 b2 2bh h2.
1 1
56. The area of EBFJ h2. The area of HJGD b2.
OP 0 12 0 02
57. b2 2bh h2 b2 h2 2A
1 1
2bh 2A
PM 132 0 32
bh A
22 32
58. Show that the area of KMNQ b1 b2h. Show that
the Area of KLPQ Area of LMNP Area of KMNQ. Then, use the distributive and division properties of equality to show that the Area of LPQK 12b1 b2h. b2. Since EBCF and GHDF are , Area of ABCD Area of AEFD Area of EBCF Area of AEFD Area GHDF Area of AEGH 12hb2 b2.
59. Show that the area of AEGH
1 2 hb2
60. A 2 13 84 42 in.2 1
13 MN 0 32 1 32 32 22 13 Kite; ON OP and PM MN, but opposite sides are not congruent. 21 1 undefined 22 0
2. Slope of QR
D 61. A 53 15 cm2
Slope of RS
10 1 24 2
Slope of ST
04 4 undefined 44 0
Slope of TQ
42 2 1 42 2
B 62. Area of PQRS Area of PQR Area of PRS. Area
of PQR 12PRQT. The area of PRS 12PRTS. So, the area of PQRS 12PRQT TS, but PR is a diagonal, d1, and QT TS is another diagonal, d2. Therefore, A 12d1d2.
6.7 Mixed Review (p. 380)
RS 2 42 1 02 22 12 5
63. obtuse; about 140
64. right; 90
66. Sample answer:
67. Sample answer:
y
65. acute; about 15
TQ 4 22 4 22 22 22 8 22
(0, 5) y (4, 4)
Trapezoid; QR ST, but TQ and RS are not parallel. 3. WX 2 42 0 12
1
(0, 0)
68. 4
1
(3, 0) x
1
2 3 4
2x
6 4 2x 2 2x
69. 14
x 1
2 3 x
14
21 x 14 7x
(5, 0)
22 12 5 XY 4 22 1 32 22 22 8 22 YZ 2 02 3 22
1x 70. x 6 3 2x 6 2
x 6 43x 4 2
(5, 0)
1 3x
6x
22 12 5 ZW 0 22 2 02 22 22 8 22 Slope of WX
10 1 42 2
Slope of XY
1 3 2 1 42 2
—CONTINUED— Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
129
Chapter 6 continued 3. —CONTINUED—
Slope of YZ
4.
13. rhombus, square
32 1 20 2
14. parallelogram, rectangle, rhombus, square 15. rhombus, square
2 20 1 Slope of ZW 0 2 2
16. midsegment 2 6 16
Parallelogram; Sample answer: WX YZ, XY ZW, WX YZ, and XY ZW.
17. mABC mDAB 112
bh A
1 2 bh
5.
1 2 x10
12x 60 x 5 in. A
6.
60
1
A 60
x 12 in.
1 2 d1d2 1 2 15x
7. A
1 2 8.3
115.4
52.11 cm2
11 units mADC 180 mDAB 180 112 68 mBCD mADC 68 18. It is given that AD BC and ADC BCD.
DC DC by the Reflexive Property of Congruence. ADC BCD, by the SAS Congruence Postulate. s are . ACD BDC by corresponding parts of
19. PQ 5 02 6 32
8 in. x
52 32 Chapter 6 Review (p. 382) 1. Sample answer:
2. Sample answer:
34 RS 2 32 11 82 52 32 34 PR 2 02 11 32 22 82
3. 67 115 63 x 360
x 360 67 115 63 115 4. 5x 3x 90 90 360
8x 180 360 8x 180 x 22.5 5. 6x 9x 75 90 360
15x 165 360 15x 195 x 13 6. FH DH 9.5
DF DH FH 9.5 9.5 19 7. mEFG mGDE 65
mDEF 180 mEFG 180 65 115 8. P 2l 2w
217 QR 5 22 6 112 32 52 34 PS 0 32 3 82 32 52 34 QS 5 32 6 82 82 22 68 217 Square; Sample answer: PQ RS QR PS and PR QS (the diagonals), PQRS is a rhombus and a rectangle. A quad. that is both a rhombus and a rectangle is a square. 20. PQ 6 02 8 02
212 210
36 64
44 units
100
9. No; you are not given information about opposite sides. 10. Yes; opposite pairs of are . 11. Yes; you can prove PQT SRT and opposite sides
are .
12. Yes; two angles which are consecutive to the same angle
are also supplementary, so opposite angles are congruent.
130
68
Geometry Chapter 6 Worked-out Solution Key
10 QR 8 62 5 82 22 32 13 —CONTINUED—
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued QR 3 12 6 62
20. —CONTINUED—
RS 8 4 5 6 2
42
42
2
112
4
137
RS 1 12 6 22
SP 4 02 6 02
42
16 36
4
52 213
SP 1 52 2 02
80 8 4 60 6 3
Slope of PQ
62 22 40
6 5 11 11 Slope of RS 48 4 4 Slope of PS
3 6 0 6 40 4 2
Slope of QR
58 3 86 2
210 Kite; two pairs of consecutive sides are congruent.
25. A
21. PQ 4 2 5 1 2
22
1
12812 7
Trapezoid; PS QR, but PQ and RS are not parallel.
24. A 2 b1 b2h
1
23. A 2 bh
2
4
123 63
2934 in.2 1 2 d1d2 1 2 46
13.5 ft2
12 units2
2
20 25 QR 4 02 5 32
42
2
2
Chapter 6 Test (p. 385) 1. Sample answer:
20 25 RS 0 22 3 12 22 42 20 25
2. x 100 70 75 360
x 245 360
SP 2 22 1 12 42 22 20 25 Slope of PR Slope of QS
3 1 2 1 02 2 1 5 6 1 2 4 6
PR 2 02 1 32 2 2 2
2
8 22 QS 4 22 5 12 62 62 72 62 Rhombus; PQ QR RS SP and the diagonals are but not . 22. PQ 5 32 0 62
22 62
x 115 3.
1 2y
3x 5x 6 2x 6
4
y8
x3 4. x 110
y 180 110 70 5. 2y 7
x 6 10
y 72 9. yes
6. yes
7. yes
8. no
x4
10. sometimes
11. never
12. always
13. Trapezoid; it appears to have exactly one pair of parallel
sides. 14. Rhombus; there are four congruent sides with diagonals
that bisect each other, but are not congruent. 15. Rectangle; opposite sides are parallel and there are four
right angles. 16. Square; there are 4 congruent sides and diagonals are
congruent.
40 210
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
131
Chapter 6 continued 17. WX a 02 0 b2
PQ 8 62 3 72
a2 b2
22 102
XY 0 a2 b 02
104 226 QM 8 32 3 42
a b 2
2
YZ 0 a2 b 02
52 12
a2 b2
26
ZW 0 a2 b 02
Opposite sides are congruent.
a b 2
2
C 7. RS 5 32 7 92
So, WX XY YZ ZW. To show the diagonals bisect each other, find the midpoints of the diagonals. They should be equal. To show the diagonals are perpendicular, find the slope of each diagonal and multiply them to get 1. 18.
22 22 8 22 ST 3 12 9 72 22 22
19. x 2 6 15 1
B
10.5 in. A
C
8 22 TU 1 32 7 112 22 182
D
AC BD
20. A 2
1 2 b1
328 282 b2h 2
1 2 bh
UR 3 52 11 72
32 2217 2015
22 182
918 300
328 282 Two pairs of consecutive sides are congruent.
1218 ft2
B Chapter 6 Standardized Test (pp. 386–387) 1. D 2. 7x 1 8x 27 108 5x 18 360
20x 100 360 20x 260 x 13
8. Column A is a trapezoid with base lengths of 8 and
12 and height 9. The area is 90 sq units. Column B is a parallelogram with base 11 and height 9 so its area is 99 sq units. B 9. 21
21 42
C 3. 8p 17 7p 8
4. D
p9
45
1 2 x 4 1 3 2 2x 3 3 2x 3 3 2x
12x 1
30 x
8p 17 6q 17 180
D
79 8 6q 17 180
10. The base 13 units and height 10 units.
So, A 21013 65 sq units. 1
6q 72 180
D
6q 108
11. Yes; FB DH, and BD FH so FBDH is a parallelogram
q 18
and the diagonals of a parallelogram bisect each other. Therefore, BE HE.
C 5. MN 3 12 4 62
6. A
22 102 104 226 NP 1 62 6 72 52 12
12. mDHF mFBD 180 mABF mCBD
180 40 40 100 mBDH 180 mDHF 180 100 80
26 132
Geometry Chapter 6 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 13. Yes; show that BEF HED, and that EF ED.
4. Sample answers:
14. It would be a rhombus because it would have four con-
a. QPR and SPT
b. SQT and QRS
gruent sides and the diagonals are to each other.
c. RQS and TSQ
d. RQS and TSQ
15. YZ 2 QR UT 1
5. QPR QPS and QPR TPS; 1 2,
QP QP, and mQPS 90 mQPR, so QPR QPS by the ASA Congruence Postulate. Therefore, s are , PR PS. since corresponding parts of 2 T and QRP TSP by the Alternate Interior Angles Theorem, so QPR TPS by the AAS Congruence Theorem.
9 123x 10 2x 3 18 5x 7 25 5x 5x UT PQ QR RS
16.
PT 2UW
17.
6y 1 2y 5 2y
3a 2 214 a
6y 1 4y 5
3a 2 28 2a
2y 4
5a 30
y2
a6
UY TZ
18.
6. PR PS, by the Segment Addition Postulate,
RS PS PR. By the Substitution property of equality, RS PS PS 2PS. → → 7. P is equidistant from QS and QR by the Angle Bisector Theorem. 8. concave pentagon
19. convex pentagon
9. 5x 18 14x 3 21x 1
3b 4 4b 5
19x 15 21x 1
ab 20. Statements
2x 14
Reasons
1. PSTU is a rectangle,
14 2x
1. Given
PQ SR.
7x
2. UPQ and TSR are
2. Rectangle Corollary
3. UPQ TSR
3. All rt. are .
4. PU ST
4. If a quad is a , then
5. PQU SRT
5. SAS Congruence Post.
57 18 53
rt. .
147 3 95
opp. sides are .
21. ABCD is an isosceles trapezoid. To prove that ABCD is a
trapezoid, you could show that only one pair of sides is parallel. To prove that ABCD is an isosceles trapezoid, show AD BC but that AB DC. 22. A 2 b1 b2h 1
1 2 8
1.
0.04;
So,
89 99
18 99
0.18;
10. no
11. no
12. yes; SAS Congruence Postulate
13. yes; HL Congruence Theorem
↔
14. slope of AB
61 5 3 11 8
y mx b 4 15 8 b
2
17 8 b
Cumulative Practice, Chs. 1–6 (pp. 388–389) 4 99
obtuse
4 583 b
412
72 units
180 217 1 32
35 99
0.35 2.
89
A
3. mRQP mSQP
2mRQP 180 mQST 2mRQP 180 90 2mRQP 90 mRQP 45 mPTS mURQ mRQP mQPR
B
15. AB 11 32 1 62
82 52
D
would be 0.89.
y 58x 17 8
C
BC 3 32 6 42 102 10 CA 11 32 1 42 82 52 89 Since AB CA, ABC is an isosceles .
45 90 135 Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
133
Chapter 6 continued 16. B C by the Base Angles Theorem.
↔
17. slope of AC
23. mx > mz; the angle opposite the longer side is larger
than the angle opposite the shorter side.
4 1 5 3 11 8
24. Use the converse of the Hinge Theorem in triangles ACD
↔ 11 3 1 4 , midpoint AC 2 2
7,
3 2
and WYZ. Since AC > WY, mD > mZ.
25. rhombus 26. mZ 25; mW mY 180 25 155
27. slope of PQ
y mx b
34 1 80 8
3 8 7 b 2 5
slope of QR
83 5 98
3 56 b 2 5
slope of RS
98 1 19 8
slope of SP
94 5 10
97 b 10
Sample answer: PQRS is a parallelogram because both pairs of opposite sides are parallel.
8 97 y x 5 10 18. The circumcenter, or the intersection of the three
28. Trapezoid; mH 113. Since E and F are supple-
19. The centroid is two thirds of the distance from each ver-
29. a. rhombus, kite, square
bisectors of the ABC.
tex to the midpoint of the opposite side. Use the median from vertex A to BC. The midpoint of BC is 3, 1. Call this D. Find the distance from vertex A to midpoint D. The distance from midpoint D to vertex A is 8 units. The coordinates of the centroid are 238 units left from vertex A along the median AD. Centroid
17 3,
1.
y
mentary, EH FG, but the other two sides are not . b. rectangle, isosceles trapezoid, square
30. Length of midsegment 2 b1 b2 2 12 5 1
1
17 2 units Area
1 2 b1 b2h 1 2 12 56
51 sq units 31. Sample answer: Show that mC mF 65 by the
B (3, 6)
Linear Pair Post. mABC 90 mDEF, so ABC DEF by the AAS Congruence Theorem.
A (11, 1) 1
( , 1) 17 3
D (3, 1) 1
x
C (3, 4) 1
20. midsegment 2 BC
123 32 6 42 12102 1210 5 21. If two angles are supplementary, then they form a linear
pair; false; Sample answer: two consecutive angles of a parallelogram are supplementary, but they do not form a linear pair.
32. Rectangle; since ABC DEF, AB DE. AB and DE
are both perpendicular to BE, so AB DE by the Property of Perpendicular Lines. One pair of opposite sides of quad. ADEB are both congruent and parallel, so ADEB is a parallelogram. ABE and DEB are both right angles, so all four angles are right angles and quad. ADEB must be a rectangle.
33. 180 42 138
138 2 69 34. The middle shelf is one half the length of the distance
between the supports on the floor. So, the distance between supports on the floor is 60 in. The length of the top shelf 1230 15 in.
22. 4 < xz < 20
134
Geometry Chapter 6 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.
Chapter 6 continued 35. A 2 b1 b2h 1
25.
1230 151912
t 4 27 9
26.
9t 108
438.75 in.2
5r 24 r
t 12 Algebra Review (pp. 390–391) 20 4 1. 25 5 5.
824 360
27 2 25 3. 30 3 27
7 2. 4
103 45
6.
220 180
11 9
7.
8 2
41
20 16
8.
54
10. 2x 7 20
11. 8x 6 24
2x 14 20
8x 48 24
2x 6
8x 24
x3
x 3
9.
12. 6x 2 24
13. 10y 8 40
6x 12 24
10y 80 40
6x 36
10y 40
x6
15. 72 x 5x
48 16d 4
14 7x 5x
16d 52
14 12x
d 13 4
7 6
17. 95 3x 9
4x 24 28
45 27x 9
4x 4 9x 32
w 29.
x2 19.
5 92x 32 92x 72 x 79
2 3 m 4 8 2 8 3m 3 8 2 16 3m 3 2 3m
31.
33.
3a 6
21. 3x 1 3 4x 2
22.
3x 3 3 4x 8
6 6 5m 25 150 30m
d4
5m
19 9 x 5
34.
3w 6 3 28 4
9x 95
12w 24 84
95 9
12w 60
6 2z 10 45 15
37.
3 21 8 2y 1
w5 36.
3a 54 11 22 66a 594 a9
38.
1 5 18 4x 1
6y 6 168
4x 4 90
6y 174
4x 86
y 29
a 2
32.
4 z
9
15a 10 12a 16
m 30
360 90z
11 3
6 m 19 95 19m 570
90 90z 450
11 3
20. 53a 2 26a 8
30.
15 5 27 9
6d 24
x
35.
27 3 5 z
z
3 2
3 3 8 2d
24 5
27z 15
42 17
y 3 50 100
y
11 3
m 27 2
28.
100y 150
27x 54
x 1
w 7 6 17 17w 42
11
x
16. 4x 6 28
1 2 10
16 16
y 4
14. 163 d 4
18.
27.
22 11 4. 8 4
5 4 6 r
x
86 43 4 2
x 1 20 5 5x 20
3x 4x 8
x4
8x 23.
2 4 q 18
24.
7 14 100 y
4q 36
7y 1400
q9
y 200
Copyright © McDougal Littell Inc. All rights reserved.
Geometry Chapter 6 Worked-out Solution Key
135
Chapter 6 continued 39.
3 9 m 4 14
40.
42 9m 36
41.
3 1 p6 p 3p p 6
6 9m
2p 6
2 m 3
p 3
r 2 3r 1 3 3r 6r 2 3r 2
42.
w 9 4 w w2 36 w ±6
2 r 3
136
Geometry Chapter 6 Worked-out Solution Key
Copyright © McDougal Littell Inc. All rights reserved.