Download CHAPTER 6

CHAPTER 6
7. ma A ma B ma C ma D 360
Chapter Opener
Chapter Readiness Quiz (p. 302)
1. C
ma A 105 113 75 360
ma A 293 360
2. G; ma 4 180 125 55
ma A 67
3. C; ma J ma K ma L 180
55 90 ma L 180
145 ma L 180
ma L 35
6.1 Practice and Applications (pp. 306–308)
8. The figure is a polygon with five sides, so it is a pentagon.
9. The figure is not a polygon because it is not formed by
segments.
Lesson 6.1
10. The figure is a polygon with ten sides, so it is a 10-gon
6.1 Checkpoint (pp. 304–305)
(or decagon).
1. The figure is not a polygon because two of the sides
intersect only one other side.
2. The figure is a polygon with five sides, so it is a pentagon.
3. The figure is a polygon with four sides, so it is a
11. The fewest number of sides a polygon can have is three
because each side must intersect exactly two other sides.
A polygon with three sides is a triangle.
12. Sample answer:
13. Sample answer:
quadrilateral.
4. The figure is not a polygon because it has a side that is
not a segment.
5. ma A ma B ma C ma D 360
ma A 110 60 100 360
14. Sample answer:
ma A 270 360
ma A 90
6. ma A ma B ma C ma D 360
ma A 80 55 160 360
ma A 295 360
ma A 65
7. ma A ma B ma C ma D 360
ma A 72 108 108 360
ma A 288 360
ma A 72
ma A 55 110 124 360
ma A 289 360
ma A 71
16. ma A ma B ma C ma D 360
ma A 95 63 87 360
ma A 245 360
ma A 115
17. ma A ma B ma C ma D 360
6.1 Guided Practice (p. 306)
2. DA
**** and DB
****
1. octagon; 15-gon
15. ma A ma B ma C ma D 360
3. The figure is a polygon because it is formed by five
straight lines.
ma A 95 100 90 360
ma A 285 360
ma A 75
4. The figure is not a polygon because it has a side that is
not a segment.
5. The figure is not a polygon because two of the sides
intersect only one other side.
6. ma A ma B ma C ma D 360
ma A 125 60 70 360
ma A 255 360
ma A 105
Copyright © McDougal Littell Inc.
All rights reserved.
18. 65 80 110 (x 30) 360
x 285 360
x 75
19. 90 60 150 3x 360
300 3x 360
3x 60
x 20
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
87
Chapter 6 continued
20. 100 84 2x 2x 360
184 4x 360
4x 176
x 44
21. The polygon has eight sides, so it is an octagon.
22. Sample answers: PNMLTSRQ, SRQPNMLT
23. MP
**** , MQ
**&, MR
**&, MS
**** , MT
**&
24. The polygon has six sides, so it is a hexagon.
25. The polygon has eight sides, so it is an octagon.
2. The lengths of the opposite sides are equal.
3. The measures of the opposite angles are equal.
4. a. The opposite sides of a parallelogram are congruent.
b. The opposite angles of a parallelogram are congruent.
6.2 Checkpoint (pp. 310–311)
1. AB DC 9; AD BC 8
2. By Theorem 6.3, ma C ma A 60;
by Theorem 6.4, ma D ma A 180
ma D 60 180
26. The polygon has five sides, so it is a pentagon.
27. The polygon has 17 sides, so it is a 17-gon.
28. The sum of the interior angles of the quadrilateral is
360. The sum remains the same when you drag the
vertices of the quadrilateral.
ma D 120;
by Theorem 6.3, ma B ma D 120.
3. By Theorem 6.3, ma B ma D 105;
by Theorem 6.4, ma A ma D 180
ma A 105 180
6.1 Standardized Test Practice (p. 308)
29. a. Before it is folded, the envelope has 18 sides, so it is
an 18-gon.
b. In step 2, the envelope has ten sides, so it is a 10-gon
(or decagon). In step 3, the envelope has seven sides,
so it is a heptagon. In step 4, the envelope has four
sides, so it is a quadrilateral.
c. The sum of the measures of the four angles must
equal 360. So if the four angles are congruent, then
360
the measure of each angle must be 90.
4
ma A 75;
by Theorem 6.3, ma C ma A 75.
6.2 Guided Practice (p. 313)
1. A parallelogram is a quadrilateral with both pairs of
opposite sides parallel.
2. The figure is not a parallelogram because only one pair of
sides is parallel.
3. The figure is a parallelogram.
4. JK
**** ⬵ ML
**** ; by Theorem 6.2, opposite sides of a parallelo-
gram are congruent.
6.1 Mixed Review (p. 308)
30. neither
31. parallel
32. perpendicular
33. neither
34. ma 1 108 by the Alternate Interior Angles Theorem.
35. ma 2 72 by the Alternate Exterior Angles Theorem.
36. ma 3 97 180 by the Same-Side Interior Angles
Theorem. So, ma 3 83.
6.1 Algebra Skills (p. 308)
37. 4(x 3) 4(x) 4(3) 4x 12
38. (x 1)6 (x)6 (1)6 6x 6
39. 2(x 7) 2(x) 2(7) 2x 14
40. 5(2x 3) 5(2x) 5(3) 10x 15
41. 3(5x 2) 3(5x) 3(2) 15x 6
42. (4x 4)x (4x)x (4)x 4x2 4x
5. a MLK ⬵ a KJM; by Theorem 6.3, opposite angles of a
parallelogram are congruent.
6. a JKL ⬵ a LMJ; by Theorem 6.3, opposite angles of a
parallelogram are congruent.
7. JN
**** ⬵ NL
**** ; by Theorem 6.5, the diagonals of a parallelo-
gram bisect each other.
8. a MNL ⬵ a KNJ; vertical angles are congruent.
9. **&
NM ⬵ KN
**** ; by Theorem 6.5, the diagonals of a parallelo-
gram bisect each other.
10. By Theorem 6.3, ma C ma A 65.
11. By Theorem 6.5, HK KF 9.
12. By Theorem 6.4, ma X ma Y 180
110 ma Y 180
ma Y 70.
6.2 Practice and Applications (pp. 313–315)
Lesson 6.2
6.2 Activity (p. 309)
1. This type of quadrilateral is called a parallelogram
because the opposite sides are parallel.
88
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
13. B; diagonals of a parallelogram bisect each other.
14. C; opposite sides of a parallelogram are congruent.
15. D; diagonals of a parallelogram bisect each other.
16. A; opposite sides of a parallelogram are congruent.
Copyright © McDougal Littell Inc.
All rights reserved.
Chapter 6 continued
17. E; vertical angles are congruent.
37. When you decrease ma A, the overall height of the
18. H; opposite angles of a parallelogram are congruent.
38. a 4 ⬵ a 1 (Opposite angles in a parallelogram are
19. G; Alternate Interior Angles Theorem
20. F; opposite angles of a parallelogram are congruent.
21.
E
scissors lift increases.
F
congruent.)
39. a 5 and a 8 (Consecutive angles in a parallelogram are
supplementary.)
40. Corresponding Angles Postulate
H
G
41. No; the lengths of the vertical sides of the red parallelo-
EF
HG. From the sketch above, you can see
*& is parallel to **&
that EF
HG.
*& intersects GF
**& and EF
*& is parallel to **&
22. EF HG 8; FG EH 7
gram are shorter than the corresponding lengths in the
blue parallelogram.
6.2 Standardized Test Practice (p. 315)
23. EF HG 25; FG EH 17
42. D
24. EF HG 31; FG EH 31
43. H; (2x 3) 65 180
25. By Theorem 6.3, ma L ma J 51;
2x 62 180
by Theorem 6.4, ma J ma K 180
2x 118
51 ma K 180
x 59
ma K 129;
by Theorem 6.3, ma M ma K 129.
26. By Theorem 6.3, ma L ma J 120;
by Theorem 6.4, ma J ma K 180
120 ma K 180
ma K 60;
by Theorem 6.3, ma M ma K 60.
27. By Theorem 6.3, ma K ma M 90;
by Theorem 6.4, ma M ma L 180
90 ma L 180
ma L 90;
by Theorem 6.3, ma J ma L 90.
29. DE BE 13
1
1
30. DE BD (20) 10
2
2
31. x 12;
32. x 98 180
3y 9
x 82;
y 98
y3
33. 2x 4
x 2;
y16
y7
34. ma B ma A 180
ma B 120 180
ma B 60
35. When you decrease ma A, ma B increases because the
sum of the two measures must be 180.
36. When you decrease ma A, AD increases.
Copyright © McDougal Littell Inc.
All rights reserved.
44. Lines p and q are parallel by the Corresponding Angles
Converse.
45. Line p is not parallel to line q because the same-side
interior angles are not supplementary (110 80 190).
46. 50 65 115, so p 储 q by the Alternate Exterior
Angles Converse.
47. 2x 62
48. x 6 13
x 31
x7
49. 3x 15
x5
6.2 Algebra Skills (p. 315)
y2 y1
x2 x1
50. slope 28. DE BE 6
3y 1 10
6.2 Mixed Review (p. 315)
53
61
2
5
y2 y1
51. slope x2 x1
4 (8)
74
12
4
3
y2 y1
x2 x1
53. slope 1 2
5 (4)
3
9
1
3
y2 y1
x2 x1
52. slope 01
1 2
1
3
1
3
y2 y1
54. slope x2 x1
4 (2)
12 6
6
6
1
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
89
Chapter 6 continued
y2 y1
x2 x1
7. Yes; both pairs of opposite sides are parallel (definition of
55. slope parallelogram).
6 (3)
5 0
3
5
3
5
6.3 Practice and Applications (pp. 320–323)
8. Yes; both pairs of opposite sides are congruent.
9. Yes; both pairs of opposite sides are congruent.
10. No; opposite sides are not congruent.
11. Yes; both pairs of opposite angles are congruent.
Lesson 6.3
12. Yes; both pairs of opposite angles are congruent.
6.3 Geo-Activity (p. 316)
13. No; one pair of opposite angles is not congruent (or
consecutive angles are not supplementary).
Step 3. Yes; the quadrilateral is always a parallelogram.
consecutive angles, which measure 120.
6.3 Checkpoint (pp. 317–319)
1. The quadrilateral is a parallelogram because both pairs of
opposite sides are congruent.
2. The quadrilateral is not a parallelogram because the
opposite angles are not congruent.
3. No, WXYZ is not a parallelogram. From the figure below,
you can see that two pairs of sides are congruent, but
they are not opposite sides (WX XY and YZ ZW).
W
quadrilateral are not supplementary.
16. Yes; one angle is supplementary to both of its consecu-
tive angles.
17. Yes; the diagonals of the quadrilateral bisect each other.
18. Yes; the diagonals of the quadrilateral bisect each other.
19. No; the diagonals of the quadrilateral do not bisect each
20. Quadrilateral ABCD has two pairs of opposite sides that
20
15
X
15. No; 72 118 190, so the consecutive angles of the
other.
Z
20
14. Yes; the 60 angle is supplementary to both of its
15
Y
4. The quadrilateral is not a parallelogram because none of
the consecutive angles are supplementary (or opposite
angles are not congruent).
5. LMNK is a parallelogram because a K is supplementary
to both of its consecutive angles (a L and a N).
6. PQRS is a parallelogram because the diagonals bisect
are congruent, and so by Theorem 6.6, ABCD is a parallelogram. By definition of a parallelogram, opposite sides
are always parallel, so ****
AB and CD
**** will always be parallel when the derailleur moves.
21. In order to be a parallelogram, both pairs of opposite sides
must be parallel and congruent. A counterexample of the
students argument is shown in the figure below.
Quadrilateral ABCD has one pair of sides congruent and
the other pair of sides parallel, but it is not a parallelogram
because BC
**** is not congruent to AD
**** .
each other.
B
7. TUVW is not a parallelogram because its diagonals do not
bisect each other.
C
20
6.3 Guided Practice (p. 320)
1. EH
**** 储 FG
**** ; EF
**** 储 HG
****
20
20
A
2. a H ⬵ a F; a E ⬵ a G
3. a E supplementary to a H and a F;
a H supplementary to a E and a G;
a G supplementary to a H and a F;
a F supplementary to a G and a E.
4. Every parallelogram has four sides, so every parallelo-
gram is a quadrilateral. Some quadrilaterals have no pairs
of parallel sides or one pair of parallel sides, so not every
quadrilateral is a parallelogram.
D
30
22. No, this is not enough information. Since the sum of the
four angles of a quadrilateral must be 180, you can
conclude that the measure of the fourth angle is 105.
There are two sets of congruent angles, but you can’t be
sure that the opposite angles are congruent. For example,
quadrilateral ABCD shown below has the given angle
measures, but it is not a parallelogram.
B
105
C
105
5. No; you can tell that one pair of sides must be parallel,
but not both.
A
75
75
D
6. No; you do not have any information about the lengths of
two of the sides.
90
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
Chapter 6 continued
23. The diagonals of the resulting quadrilateral were drawn
to bisect each other. Therefore, the resulting quadrilateral
is a parallelogram.
d. ma BEA ma BEH ma HEF ma AFE 180
63 ma HEF 63 180
ma HEF 126 180
24. Yes. Consider the figure below.
C
B
ma HEF 54;
ma CGD ma CGH ma HGF ma FGD 180
63 ma HGF 63 180
D
A
We are given that BC
**** 储 DA
**** and BC
**** ⬵ DA
**** . By the
Alternate Interior Angles Theorem, a DAC ⬵ a BCA. By
the Reflexive Property of Congruence, ****
AC ⬵ ****
AC . Then,
by the SAS Congruence Postulate, T BAC ⬵ T DCA.
Because corresponding parts of congruent triangles are
congruent, ****
AB ⬵ CD
**** . Therefore, by Theorem 6.6, if
both pairs of opposite sides of a quadrilateral are
congruent, then the quadrilateral is a parallelogram.
3
41
1
12
41
3
Slope of HJ
**** : 3
1
56
44
0
Slope of GH
**** : 0
15
4
0
11
Slope of FJ
**** : 0
4
26
ma HGF 126 180
ma HGF 54;
ma AFD ma AFE ma EFG ma GFD 180
27 ma EFG 27 180
ma EFG 54 180
ma EFG 126;
ma BHC ma BHE ma EHG ma GHC 180
27 ma EHG 27 180
25. Slope of GF
**** : 3
The slopes of GF
**** and HJ
**** are the same so GF
**** 储 HJ
**** .
The slopes of GH
**** and FJ
**** are the same so GH
**** 储 FJ
**** .
Both pairs of opposite sides are parallel, so FGHJ is
a parallelogram.
52
3
20
2
52
3
Slope of RS
**** : 3
54
1
26. Slope of QP
**** : QP
**** and RS
**** are opposite sides, but their slopes are not
equal and therefore are not parallel. So, PQRS is not a
parallelogram.
ma EHG 54 180
ma EHG 126
EFGH is a parallelogram because both pairs of opposite
angles are congruent.
6.3 Mixed Review (p. 323)
28. ma A ma B ma C ma D 360
ma A 101 85 99 360
ma A 285 360
ma A 75
29. ma A ma B ma C ma D 360
ma A 70 90 114 360
ma A 274 360
ma A 86
30. ma A ma B ma C ma D 360
ma A 78 82 110 360
ma A 270 360
6.3 Standardized Test Practice (p. 322)
ma A 90
27. a. The adjacent sides of the pool table are perpendicular,
so ma EAF 90 and T EAF is a right triangle.
Since the acute angles of a right triangle are complementary, ma EAF ma AFE 90
31. By Theorem 6.3, ma K ma M 71
32. By Theorem 6.4, ma M ma J 180
71 ma J 180
63 ma AFE 90
ma AFE 27.
b. The ball bounces off each wall at the same angle at
which it hits the wall, so ma AFE ma GFD 27.
T DFG is a right triangle, and the acute angles of a
right triangle are complementary, so
ma FGD 90 ma DFG 90 27 63.
c. ma HCG 90
ma HGC ma FGD 63
So ma GHC 90 63 27;
a EHB ⬵ a GHC, so ma EHB 27.
Copyright © McDougal Littell Inc.
All rights reserved.
ma J 109
33. By Theorem 6.2, ML JK 14.
34. By Theorem 6.2, KL JM 11.
6.3 Algebra Skills (p. 323)
35. 2(5) 7 10 7 17
36. 4(2) 3 8 3 5
37. 13 3(3) 13 9 22
38. 1 (10) 11
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
91
Chapter 6 continued
39. 5 2(6) 5 (12) 17
4. By Theorem 6.10, the diagonals of a rhombus are perpen-
dicular, so x 90.
40. 8(4) 5 32 5 37
41. 12(1) (1)2 12 1 11
3
3
4
4
43. 5(2)3 4(2) 5(8) 8 40 8 32
42. (4)2 2 (16) 2 12 2 10
44. 15(3) (3)2 45 9 36
Quiz 1 (p. 323)
gruent, so x 12.
6. A square is also a rhombus, so by Theorem 6.10, the
diagonals of JKLM are perpendicular. Since the acute
angles of a right triangle are complementary, x x 90.
2x 90, so x 45.
6.4 Guided Practice (pp. 328)
1. The figure is a polygon with five sides, so it is a
pentagon.
2. The figure is a polygon with six sides, so it is a hexagon.
3. The figure is not a polygon because none of the sides are
segments.
4. x 16; y 15
5. x 122 180
5. By Theorem 6.11, the diagonals of a rectangle are con-
6. x 12;
x 58;
y9
1. A rhombus is a parallelogram with four sides.
2. D
3. B, C, D
4. A, D
5. A, B, C, D
6. By Theorem 6.11, PR QS 12.
A rectangle is a parallelogram and by Theorem 6.5,
the diagonals of a parallelogram bisect each other, so
PT TR.
PT TR PR
PT PT PR
y 122;
2PT 12
z 58
PT 6
7. Yes; the 115 angle is supplementary to both of its
consecutive angles.
8. No; one pair of opposite sides is parallel, and one pair of
consecutive angles are supplementary, but you don’t
know anything about the other two angles.
9. Yes; the quadrilateral has two pairs of opposite sides that
are parallel, so by definition, it is a parallelogram.
6.3 Technology Activity (p. 324)
6.4 Practice and Applications (pp. 328–330)
7. AB BC AD 4
8. ma E ma F ma G 90
9. ma W 90; YZ XY 3
10. This quadrilateral has four right angles, so it is a rectangle.
11. This quadrilateral has four congruent sides, so it is a
rhombus.
1. Slopes will vary; the slopes of the opposite sides are
equal.
12. This quadrilateral has four right angles and four congruent
sides, so it is a square.
2. The slopes tell you that the opposite sides of the quadri-
lateral are parallel.
3. Because ABCD has two pairs of opposite sides that are
13. No, you can not conclude that the frame is a rectangle. In
the figure below, ****
AC ⬵ DB
**** , but ABCD is not a parallelogram and therefore not a rectangle.
parallel, by definition, it is a parallelogram.
A
4. If both pairs of opposite angles of a quadrilateral are con-
1
In order to conclude that the
frame is a rectangle, you must
also know that both pairs of
opposite sides are congruent.
B
gruent, then it is a parallelogram.
5. When EI IG and FI IH, EFGH is a parallelogram; if
the diagonals of a quadrilateral bisect each other, then the
quadrilateral is a parallelogram.
D
5
C
14. rectangle, square
Lesson 6.4
15. parallelogram, rectangle, rhombus, square
6.4 Checkpoint (pp. 325–328)
16. parallelogram, rectangle, rhombus, square
1. By definition, a rhombus is a parallelogram with four
congruent sides, so QR RS SP PQ 6.
2. The quadrilateral has four congruent sides. So, by the
Rhombus Corollary, the quadrilateral is a rhombus.
3. The quadrilateral has four congruent sides and four right
angles. So, by the Square Corollary, the quadrilateral is a
square.
92
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
17. rhombus, square
18. LM NM
3x x 2
2x 2
19. ma B 90
5x 90
x 18
x1
Copyright © McDougal Littell Inc.
All rights reserved.
Chapter 6 continued
20. FH EG
Lesson 6.5
6x 3x 9
6.5 Technology Activity (p. 331)
3x 9
1. Answers will vary.
x3
21. The consecutive angles of a parallelogram are supple-
mentary, so the two angles consecutive to a J must also
be right angles. Opposite angles of a parallelogram
are congruent, so the fourth angle must also be a right
angle. By definition, a parallelogram with four right
angles is a rectangle.
22. By Theorem 6.10, the diagonals of a rhombus are
perpendicular, so a RPQ is a right angle and T RPQ
is a right triangle. By the Pythagorean Theorem,
(AB DC)
2
2. Answers will vary, but EF.
(AB DC)
2
4. The slopes may vary, but the slopes of ^AB
&(, DC
^&(, and EF
****
are always equal for trapezoid ABCD. Therefore, they
are parallel.
3. EF and are always equal for trapezoid ABCD.
⏐AB DC⏐
5. EF 2
6.5 Checkpoint (p. 333)
x2 42 52
1. a D and a C are a pair of base angles of an isosceles
x2 16 25
trapezoid, so ma C ma D 100; by the Same-Side
Interior Angles Theorem, ma D 180 100 80;
a A and a B are a pair of base angles of an isosceles
trapezoid, so ma B ma A 80.
x2 9
x 3.
23. By Theorem 6.5, LJ GL 兹2
苶.
2. a C and a D are a pair of base angles of an isosceles
GJ GL LJ 兹2苶 兹2苶 2兹2苶
By Theorem 6.11, HK GJ 2兹2苶.
24. By Theorem 6.10, ma KLJ 90.
25. ma HJG 45; by Theorem 6.10, a HLJ 90. By
Theorems 6.5 and 6.11, HL
**** ⬵ LJ
**** . By the Base Angles
Theorem a LJH ⬵ a LHJ. By the Corollary to the
Triangle Sum Theorem, ma LJH ma LHJ 90. Since
those two angles are congruent, each angle measures 45.
26. By definition of a square, HJ JK GH 2. From
Exercise 23, HK 2兹2苶. So the perimeter of T HJK is
2 2 2兹2苶 4 2兹2苶 ⬇ 6.8.
6.4 Standardized Test Practice (p. 330)
27. D
trapezoid, so ma D ma C 70; by the Same-Side
Interior Angles Theorem, ma B 180 70 110;
a B and a A are a pair of base angles of an isosceles
trapezoid, so ma A ma B 110.
3. a A and a B are a pair of base angles of an isosceles
trapezoid, so ma B ma A 75; by the Same-Side
Interior Angles Theorem, ma D 180 75 105;
a D and a C are a pair of base angles of an isosceles
trapezoid, so ma C ma D 105.
1
1
4. MN (8 14)
5. MN (10 6)
2
2
1
1
(22)
(16)
2
2
11
8
1
2
1
(42)
2
21
6. MN (24 18)
28. G; 7x 3 25
7x 28
x4
6.4 Mixed Review (p. 330)
29. ma 1 90 by the Corresponding Angles Postulate.
30. ma 2 120 by the Alternate Exterior Angles Theorem.
31. ma 3 35 180 by the Same-Side Interior Angles
Theorem. So, ma 3 145.
9 cm
9
5 cm
5
18 m
3
34. 12 m
2
Copyright © McDougal Littell Inc.
All rights reserved.
1. ****
AB and DC
****
2. AD
**** and BC
****
3. isosceles trapezoid
4. neither
5. trapezoid
1
2
1
2
6. (7 11) (18) 9
6.4 Algebra Skills (p. 330)
32. 6.5 Guided Practice (p. 334)
14 in.
7 in.
2
1
33. 1
2
1
2
1
2
7. (7 3) (10) 5
1
2
8. (19 15) (34) 17
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
93
Chapter 6 continued
6.5 Practice and Applications (pp. 334–336)
9. D
10. C
11. B
12. E
13. A
14. a J and a M are a pair of base angles of an isosceles
trapezoid, so ma M ma J 45; by the Same-Side
Interior Angles Theorem, ma K 180 45 135;
a K and a L are a pair of base angles of an isosceles
trapezoid, so ma L ma K 135.
30. Midpoint of AD
**** :
冢
Midpoint of BC
**** :
冢
y A
trapezoid, so ma K ma L 60; by the Same-Side
Interior Angles Theorem, ma M 180 60 120;
a M and a J are a pair of base angles of an isosceles
trapezoid, so ma J ma M 120.
F
D
1
C
1
x
EF
**** is the midsegment of ABCD.
31. Corresponding Angles Postulate
32. The quadrilateral on the right is a parallelogram; from
Exercise 3, a 1 ⬵ a 3 and a 2 ⬵ a 4, so both pairs of
opposite angles are congruent.
ma T 180 132 48
18. ma R 180 78 102;
33. The opposite sides of a parallelogram are congruent, so
ma S 180 110 70
mm95
19. ma Q 180 90 90;
2m 9 5
(9 5)
m 2
ma S 180 150 30
1
1
2
2
1
1
21. MN (14 16) (30) 15
2
2
1
1
22. MN (15 9) (24) 12
2
2
1
1
23. 9 (x 12)
24. 10 (16 x)
2
2
18 x 12
20 16 x
20. MN (9 7) (16) 8
6x
冣
B
E
16. a J and a K are a pair of base angles of an isosceles
17. ma R 180 78 102;
冣 冢
88 62
16 8
F , , (8, 4);
2
2
2 2
15. a L and a K are a pair of base angles of an isosceles
trapezoid, so ma K ma J 128; by the Same-Side
Interior Angles Theorem, ma M 180 128 52;
a M and a L are a pair of base angles of an isosceles
trapezoid, so ma L ma M 52.
冣 冢 冣
24 62
6 8
E , , (3, 4);
2
2
2 2
4x
In words, the length of the midsegment is half the sum of
the lengths of the bases.
6.5 Standardized Test Practice (p. 336)
1
2
30 13 x
34. C; 15 (13 x)
35. H
17 x
6.5 Mixed Review (p. 336)
1
2
46 27 x
36. The quadrilateral is a parallelogram because both pairs of
19 x
37. No; you can tell that one pair of sides must be parallel,
25. 23 (27 x)
opposite angles are congruent.
26. The diameter of the middle layer is
1
1
(10 22) (32) 16 inches.
2
2
27.
y A
B
but not both.
38. The quadrilateral is a parallelogram because the diagonals
bisect each other.
6.5 Algebra Skills (p. 336)
39. 10
D
1
1
43. x
AB and DC
28. ****
****
94
2
7
C
40. 13
1
8
44. 41. 17
5
18
45. 42. 45
1
12
46. 29. AD
**** and BC
****
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
Chapter 6 continued
Lesson 6.6
10. Opposite sides of the quadrilateral are congruent, so it’s a
parallelogram. One angle is right. The consecutive angles
of a parallelogram are supplementary, so each of the other
angles is also right. A shape with four congruent sides
and four right angles is a square.
6.6 Checkpoint (p. 338)
1. No; the diagram shows that all four angles are right
angles. By The Rectangle Corollary, you can conclude
that the quadrilateral is a rectangle. But you cannot conclude that PQRS is a square because no information is
given about the sides.
11. The diagram shows that all four angles are right angles so
you can conclude that PQRS is a rectangle. You are not
given any information about the sides so you cannot conclude that PQRS is a square.
2. No; the diagram shows that both pairs of opposite angles
are congruent. By Theorem 6.7, you can conclude that
the quadrilateral is a parallelogram. But you cannot conclude that WXYZ is a rhombus because no information is
given about the sides.
3. Yes; a L and a M are supplementary, so KL
**** 储 JM
**** by the
Same-Side Interior Angles Converse. Because a J and
a M are not supplementary, KJ
**** 储 LM
**&. The figure has
exactly one pair of parallel sides, so it is a trapezoid.
6.6 Guided Practice (p. 339)
1.
2.
3.
4.
12. No; both pairs of opposite angles are congruent, so the
figure is a parallelogram. There is no information about
the sides, so you cannot conclude that the figure is a
rhombus.
13. Yes; since 140 40 180, the top and bottom sides
are parallel by the Same-Side Interior Angles Converse.
Since 60 40 180, the other two sides are not
parallel. The figure has exactly one pair of parallel sides,
so it is a trapezoid.
14. No; there is not enough information to determine if the
Property g Rectangle Rhombus Square Trapezoid
Both pairs
of opposite ✓
✓
✓
✓
sides are 储.
Exactly
one pair of
✓
opposite
sides are 储.
Diagonals
✓
✓
are ∏ .
Diagonals
✓
✓
are ⬵.
figure has any parallel sides, so you cannot conclude that
the figure is a trapezoid.
15. Yes; the four angles are congruent, so the four angles are
right angles by the Quadrilateral Interior Angles Theorem
(4x 360, so x 90). A quadrilateral with four right
angles is a rectangle.
16. No; there is no information about the sides or the other 3
angles, so you cannot conclude that the figure is a square.
17. Yes; both pairs of opposite sides are congruent, so the
figure is a parallelogram.
18. rectangle
20. false
✓
✓
✓
✓
✓
✓
25.
23. true
y
P
1
S
R
1
✓
22. true
measure angles and slopes, you will see that BC
**** 储 DF
****
and ****
AB 储 FE
**** , so the quadrilateral has two pairs of
opposite sides parallel.
g Rectangle Rhombus Square Trapezoid
5. Both pairs
of opposite
✓
sides are
congruent.
6. Diagonals
bisect each ✓
other.
7. Both pairs
of opposite
✓
angles are
congruent.
8. All sides
are ⬵.
21. true
24. BEFD is a parallelogram; using the geometry software to
6.6 Practice and Applications (pp. 339–341)
Property
19. isosceles trapezoid
✓
✓
✓
✓
BA 储 CD
9. The diagram shows ****
**** . Because no information is
given to prove BC
**** 储 AD
**** , exactly one pair of opposite
sides is parallel. So, ABCD is a trapezoid.
x
PQ
**** and SR
**** are horizontal lines, so they are parallel to
52
3
each other. The slope of PS
**** 3 and the
21
1
3
52
slope of QR
3.
and
QR
****
PS
****
**** have
1
56
different slopes so they are not parallel. The length of
苶苶
苶1苶
)2苶
苶(5
苶苶
苶2苶
)2 兹(1
苶苶
)2苶
苶(3
苶苶
)2 兹1苶0苶
PS
**** 兹(2
and the length of QR
苶苶
苶6苶
)2苶
苶(5
苶苶
苶2苶
)2
**** 兹(5
兹(
苶1苶苶
)2苶
苶(3
苶苶
)2 兹1
苶0
苶, so PS
**** ⬵ QR
**** . Because the
quadrilateral has exactly one pair of parallel sides and
its legs are congruent, the quadrilateral is an isosceles
trapezoid.
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
95
Chapter 6 continued
6. A trapezoid is a quadrilateral with exactly one pair of
6.6 Standardized Test Practice (p. 341)
sides parallel.
26. B
7. Each endpoint of a polygon is called a vertex.
8. A segment that joins two nonconsecutive vertices of a
6.6 Mixed Review (p. 341)
27.
x
4
3
12
12x 12
polygon is called a diagonal.
x
4
7
21
7x 84
28. x1
9. The nonparallel sides of a trapezoid are called legs and
the parallel sides are called its bases.
10. No; the figure is not a polygon because one of its sides is
x 12
not a segment.
3
24
10
x
3x 240
10
5
x
8
5x 80
30. x 16
x 80
29. x
5
24
12
12x 120
11. Yes; the figure is a polygon formed by six straight sides,
so it is a hexagon.
12. Yes; the figure is a polygon formed by eight straight
sides, so it is an octagon.
x
3
20
5
5x 60
31. 32. x 10
x 12
8
1
33. x
2
x 16
13. Sample answer:
3
21
7
x
3x 147
34. 14. Sample answer:
x 49
6.6 Algebra Skills (p. 341)
35. 0.2
36. 0.375
37. 0.8333. . . ⬇ 0.83
38. 0.35
15. By Theorem 6.2, the opposite sides of a parallelogram
are congruent. So, BC AD 4, and DC AB 8.
Quiz 2 (p. 341)
16. By Theorem 6.4, the consecutive angles of a parallelo-
1. ABCD is a rhombus, so its diagonals are perpendicular.
The two acute angles of a triangle are complementary, so
x 90 55 35.
2. FGHJ is a rectangle, so its diagonals are congruent. So,
x 6.
gram are supplementary, so ma F 180 125 55.
By Theorem 6.3, opposite angles of a parallelogram are
congruent, so ma G ma J 125.
17. By Theorem 6.5, the diagonals of a parallelogram bisect
each other, so PQ QM 8;
PM PQ QM 8 8 16
3. PQRS is a square, so it has four right angles.
ma PQR ma PQS ma SQR 90
x x 90
2x 90
x 45
4. trapezoid
5. isosceles trapezoid
6. rhombus, square, rectangle, or parallelogram
Chapter 6 Summary and Review (pp. 342–345)
1. A parallelogram is a quadrilateral with both pairs of
opposite sides parallel.
2. A polygon is a plane figure that is formed by three or
more segments called sides.
3. A rhombus is a parallelogram with four congruent sides.
4. A rectangle is parallelogram with four right angles.
18. The quadrilateral is not a parallelogram because the
opposite angles are not congruent.
19. The quadrilateral is a parallelogram because the opposite
sides of the quadrilateral are congruent.
20. The quadrilateral is a parallelogram because its diagonals
bisect each other.
21. x 8;
y58
y 13
22. x 1 5
x 6;
23. y 10;
x 4 10
y45
x6
y1
24. false
25. true
26. false
27. true
5. A parallelogram with four congruent sides and four right
angles is a square.
96
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
Copyright © McDougal Littell Inc.
All rights reserved.
Chapter 6 continued
28. By the Same-Side Interior Angles Theorem,
7. Yes; by Theorem 6.6, if both pairs of opposite sides of a
quadrilateral are congruent, then it is a parallelogram.
ma P ma L 180
8. No; its opposite angles are not congruent.
ma P 111 180
9. Yes; by Theorem 6.9, if the diagonals of a quadrilateral
ma P 69.
29. ABCD is an isosceles trapezoid, so the base angles are
congruent. Therefore, ma C ma D 71.
1
1
1
30. TU (QP RS) (13 9) (22) 11
2
2
2
31. No; the opposite angles are congruent, so you know the
figure is a parallelogram. However, with no information
given about the sides, you cannot conclude that it is a
rhombus.
32. 84 98 182
bisect each other, then it is a parallelogram.
10. Yes; by Theorem 6.8, if an angle of a quadrilateral is
supplementary to both of its consecutive angles, then it
is a parallelogram.
11. To conclude that the figure is a rectangle, you need to
know that all four angles are right angles.
12. a 11;
3b 1 11
3b 12
84 79 163
b4
None of the consecutive angles are supplementary, so
none of the sides are parallel. Therefore, the figure is not
a trapezoid.
33. Yes; the diagonals bisect each other, so you know the
figure is a parallelogram. Furthermore, the diagonals are
congruent, so you can determine that the parallelogram is
a rectangle.
13. By the Same-Side Interior Angles Theorem,
ma S 180 68 112 and
ma Q 180 122 58.
1
2
1
2
1
2
14. WX (TU SV) (17 11) (28) 14
15. square
16. trapezoid
17. Yes; by definition, EFGH is a parallelogram. By using
the Same-Side Interior Angles Theorem three times, you
can prove that all four angles are right angles. Therefore,
EFGH is a rectangle.
Chapter 6 Test (p. 346)
1. Sample answer:
C
D
Chapter 6 Standardized Test (p. 347)
1. A; the figure has five sides, so it is a pentagon.
E
B
A
2. J; x 105 60 150 360
F
x 315 360
2. The figure is a polygon with seven sides, so it is a
x 45
heptagon.
3. No, the figure is not a polygon because two of the sides
intersect only one other side.
4. F; opposite sides of a rectangle are congruent, so
4. By the Interior Angles of a Quadrilateral Theorem, the sum
of the measures of the interior angles of a quadrilateral is
360. So, ma A ma B ma C ma D 360
90 80 70 ma D 360
240 ma D 360
ma D 120
5. By Theorem 6.2, opposite sides of a parallelogram are
congruent, so y 7; by Theorem 6.3, opposite angles of
a parallelogram are congruent, so 2x 122, x 61.
6. x 70 180
x 110;
(y 10) 70 180
3. C; Theorem 6.5
FG JH 3.
5. A; by the Same-Side Interior Angles Theorem,
ma C 180 124 56.
6. J
7. D; x 180 85 95 and ma X ma W, so y 85.
8. H; x 7 13
x 6;
4y 1 13
4y 12
y3
1
1
9. D; (14 8) (22) 11
2
2
y 80 180
y 100
Copyright © McDougal Littell Inc.
All rights reserved.
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
97
Chapter 6 continued
Chapter 6 Algebra Review (p. 349)
3
t
4
40
4t 120
1. t 30
4
x
9
63
9x 252
C (3, 6)
A (4, 0)
2
y2
CD ⏐2 6⏐ ⏐8⏐ 8
AB CD, so ****
AB is not congruent to CD
**** .
x1 x2 y1 y2
冢 2 2 冣
2 0 5 (7)
冢, 冣
2
2
2 2
冢, 冣
2
2
3. M , 35
5
x
14
5x 490
6. x 98
d
11
4
14
14d 44
44
d 14
22
1
3
7
7
54
9
8. h
10
9h 540
7.
(1, 1)
x1 x2 y1 y2
冢 2 2 冣
1 7 4 (2)
冢, 冣
2
2
8 2
冢, 冣
2 2
4. M , (4, 1)
x1 x2 y1 y2
冢 2 2 冣
4 6 2 4
冢, 冣
2
2
2 2
冢, 冣
2 2
5. M , h 60
3z
7
10
20
60z 70
70
z 60
7
1
1
6
6
4
c2
10.
7
21
7(c 2) 84
x
D (3, 2)
AB ⏐5 (4)⏐ ⏐9⏐ 9
w 15
8
5
4. 15
b
8b 75
75
b 9.375
8
7
7
5. 2
y
7y 14
B (5, 0)
2
x 28
2
6
5
w
2w 30
(1, 1)
9. 6. 2x 44
x 22;
ma ABC 44 44 88, so it is acute.
7. (5x 3) 48
5x 45
8
40
11.
15
r1
8(r 1) 600
c 2 12
r 1 75
c 10
r 74
4
1
v5
6
v 5 24
v 19
Chapters 1–6 Cumulative Practice (pp. 350 –351)
1. Each number is 9 more than the previous number. The
next two numbers in the pattern are 29 9 38 and
38 9 47.
98
y
2. 3. 12.
2.
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
x 9;
ma ABC 48 48 96, so it is obtuse.
8. (4x 7) 41
4x 48
x 12;
ma ABC 41 41 82, so it is acute.
9. ma Q 90 28 62
10. ma Z 180 146 34
11. a 1 and a 4 are same-side interior angles.
12. a 3 and a 5 are alternate exterior angles.
13. a 1 and a 7 are alternate interior angles.
14. a 4 and a 6 are vertical angles.
Copyright © McDougal Littell Inc.
All rights reserved.
Chapter 6 continued
15. ma 4 105 by the Corresponding Angles Postulate.
33.
16. ma 7 180 ma 4 by the Linear Pair Postulate. So,
ma 7 180 105 75.
17. ma 5 ma 7 by the Vertical Angles Theorem, so
ma 5 75.
18. ma 2 ma 4 by the Alternate Interior Angles Theorem,
so ma 2 105.
A square has four lines of symmetry.
19. x 54 68 180
20. 5x 2 13
x 122 180
5x 15
x 58
21. x 42 90
34. y 109;
35. x 6 12
x 109 180
x3
x 18;
x 71
2y 5
5
y 2
22. x 49 83
x 48
x 132
23. AB 兹(1
苶苶
苶(
苶3苶))
苶苶 苶苶(4苶苶
苶
1苶
)
2
36. x 6 10
x 4;
2
兹4苶2苶
苶苶32苶
2y 5 7
2y 12
兹1苶6苶苶
苶9
y6
兹2苶5苶
37. E
5
24. AB 兹(
苶2苶苶
苶
3苶
)2苶
苶(1苶苶
苶13苶苶
)2
F
H
兹(
苶5苶苶
)2苶
苶(
苶1苶2苶苶
)2
67
G
兹2苶5苶苶
苶4
1苶4苶
兹1苶6苶9苶
13
25. AB 兹(3
苶苶
苶
9苶
)2苶
苶(
苶4苶苶
苶4苶
)2
兹(
苶6苶苶
)2苶
苶(
苶8苶苶
)2
(ma C ma F 65), and ****
AC ⬵ DF
**** .
Interior Angles Converse, ****
AB 储 DE
**** .
兹1苶0苶0苶
40. ABED is a rectangle. Sample answer: It is given that
10
26. yes; 6 7 8, 6 8 7, and 7 8 6
28. no; 4 10 14
29. No; you can show that the three pairs of corresponding
angles are congruent, but no information is given about
any of the sides, so you cannot conclude that the
triangles are congruent.
30. No; you can show that a pair of consecutive sides are
congruent, but no information is given about their included
angles, so you cannot conclude that the triangles are
congruent.
31. yes; SAS Congruence Postulate
32. yes; HL Congruence Theorem
Copyright © McDougal Littell Inc.
All rights reserved.
38. AAS Congruence Theorem; a B ⬵ a E, a C ⬵ a F
39. Yes; a A is supplementary to a D so by the Same-Side
兹3苶6苶苶
苶4
6苶
27. no; 3 6 12
EFGH must be a trapezoid. a E is supplementary to a F,
so EH
**& 储 FG
**** . a F is not supplementary to a G so EF
**** is
not parallel to HG
**** . EFGH has exactly one pair of parallel
sides, so it is a trapezoid.
AB ⊥ AD
AB is perpendicular to
****
**** and that DE
**** ⊥ AD
**** . ****
the level ground, so ****
AB is perpendicular to every line in
the plane of the ground that intersects ****
AB ; therefore,
AB ⊥ BE
****
****, and so a ABE is a right angle. Similarly,
a DEB is a right angle. Each angle of quadrilateral ADEB
is a right angle, so ADEB must be a rectangle by the
Rectangle Corollary.
Chapters 5–6 Project (pp. 352–353)
1. octagon
2. Sample answer: Seeing the lines of symmetry is helpful
when making a symmetric design.
3. Sample answer: triangles; hexagons
Geometry, Concepts and Skills
Chapter 6 Worked-Out Solution Key
99