1. Special Parallelograms
GEOMETRY LESSON 6-4
Pages Exercises
1. 38, 38, 38, 38
2. 26, 128, 128
3. 118, 31, 31
, 33.5, 113, 33.5
5. 32, 90, 58, 32
6. 90, 60, 60, 30
7. 55, 35, 55, 90
8. 60, 90, 30
9. 90, 55, 90
10. 4; LN = MP = 4
11. 3; LN = MP= 7
12. 1; LN = MP = 4
13. 9; LN = MP = 67
; LN = MP = = 9
; LN = MP = 12
16. Impossible; if the diag. of a are ,
16. (continued)
then it would have to be a rectangle and have right .
17. Yes; diag. in a mean it can be a rectangle with 2 opp. sides 2 cm long.
18. Impossible; in a , consecutive must be supp., so all must be right This would make it a rectangle. s 5 3 29 3 2 3 s 5 2 1 2 s s
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2. Special Parallelograms
GEOMETRY LESSON 6-4
19. Impossible; if the
figure is a , then the opp. the bisected is also bisected, and the figure is a rhombus. But the sides are not .
20. Yes; the are bisected so it could be a rhombus which is a .
21. Yes; the diag. are so it could be a square which is a .
22. The pairs of opp. sides of the frame remain , so the frame remains a
23. After measuring the sides, she can measure the diag. If the diag. are , then the figure is a rectangle by Thm
24. Square; a square is both a rectangle and a rhombus, so its diag. have the same
24. (continued)
properties of both.
Symbols may vary. Samples are given:
parallelogram:
rhombus:
rectangle:
square: ,
, , ,
, , , s R S
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3. Special Parallelograms
GEOMETRY LESSON 6-4
, , , ,
, , ,
, , , , , ,
35.
Diag. are , diag. are .
36.
Diag. are and .
37.
38. a. Opp. sides are and ; diag. bis. each other; opp are ; consec are suppl.
b. All sides are ; diag. are .
c. All are rt. ; diag. are bis. of each other; each diag. bis two . R S s
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4. Special Parallelograms
GEOMETRY LESSON 6-4
Answers may vary. Samples are given.
39. Draw diag. 1, and construct its midpt. Draw a line through the mdpt. Construct segments of length diag. 2 in opp. directions from mdpt. Then, bisect these segments. Connect these mdpts. with the endpts. of diag. 1.
40. Construct a rt. , and draw diag. 1 from its vertex. Construct the from the opp. end of diag. 1 to a side of the rt.
. Repeat to other side.
41. Same as 39, but construct a line at the midpt. of diag. 1.
42. Same as 41 except make the diag. =.
43. Draw diag. 1.
43. (continued)
Construct a at a pt. different than the mdpt. Construct segments on the line of length diag. 2 in opp. directions from the pt. Then, bisect these segments. Connect these midpts. to the endpts. of diag. 1.
44. Draw an acute with the smaller diag. as a side. Construct the line to the other
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5. Special Parallelograms
GEOMETRY LESSON 6-4
44. (continued)
side through the non-vertex endpt. of the smaller diag. Draw an arc with compass set to the length of the larger diag. from the non-diag. side of the , passing through the line. Draw the larger diag., and then draw the non- sides of the trapezoid.
45. Yes; since all right are , the opp. are
45. (continued)
and it is a Since it has all right
, it is a rectangle.
46. Yes; 4 sides are , so the opp. sides are
making it a Since it has sides it is also a rhombus.
47. Yes; a quad. with 4 sides is a and a with sides and 4 right is a square.
48. 30
49. x = 5, y = 32, z = 7.5
50. x = 7.5, y = 3
Drawings may vary. Samples are given.
51. Square, rectangle, isosceles trapezoid, kite. s s s s
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6. Special Parallelograms
GEOMETRY LESSON 6-4
51. (continued)
52. Rhombus, , trapezoid, kite.
53. For a < b: trapezoid, isosc.
trapezoid (a > b), , rhombus , kite.
For a > b: trapezoid, isos. trapezoid,
, rhombus (a < 2b), kite, rectangle, square (if a = 2b) 1 2
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7. Special Parallelograms
GEOMETRY LESSON 6-4
55. Answers may vary. Sample: only one diag. is needed.
56. Given ABCD with diag. AC. Let AC bisect BAD. Because ABC
DAC, AB = DA by CPCTC. But
since opp. sides of a
are , AB = CD and BC = DA.
So AB = BC = CD = DA, and ABCD is a rhombus. The new statement is true.
53. (continued)
54. a. Def. of a rhombus
b. Diagonals of a bisect each other.
c. AE AE
54. (continued)
d. Reflexive Prop. of .
e. ABC ADE
f. CPCTC
g Add. Post.
h. AEB and AED are rt
i suppl are rt Thm.
j. Def. of s
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8. Special Parallelograms
GEOMETRY LESSON 6-4
57. 16, 16
58. 2, 2
59. 1, 1
60. 1, 1
ABC ADC (ASA)
5. AB AD (CPCTC)
6. AB DC, AD BC (Opp. sides of a are .)
7. AB BC CD AD (Trans. Prop. of )
62. Answers may vary. Sample: The diagonals of a bisect each other so
AE CE. Both
AED and CED are right because
AC BD, and since DE DE by the Reflexive Prop.,
AED CED by SAS. By CPCTC AD
CD , and since opp. sides of a are , AB BC AD. s
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9. Special Parallelograms
GEOMETRY LESSON 6-4
62. (continued)
So ABCD is a rhombus because it
has sides.
63.
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10. Special Parallelograms
GEOMETRY LESSON 6-4
64. D
65. I
66. [2] Since diag. of a rhombus bisect each other, QS = 9 cm. Also, since all sides are , RS = 9 cm. So QRS is an equilateral and each interior
is or 60º. QTS is also an
equilateral , so its are 60º. By add. (m PST + m PSR = m RST), m RST = or 120.
[1] no work shown OR a response that is only partially correct
67. Yes; both pairs of opp. sides are .
68. No; there are not 2 necess. opp. sides that are both and .
69. Yes; the diag. bisect each other.
70. 6
71. 16
72. 5
73. RQ
74. RP
75. ST
76. 89
180 3 s
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