Lesson 5.5 • Properties of Parallelograms
Name
Period
Date
In Exercises 1–7, ABCD is a parallelogram.
1. Perimeter ABCD _____
26 cm
D
2. AO 11, and BO 7.
3. Perimeter ABCD 46.
AC _____, BD _____
C
C
D
15 cm
AB _____, BC _____
C
D
B
A
A
4. a _____, b _____,
B
6. a _____, b _____,
BC 24. AB _____
D
C
c _____
110°
19°
C
D
C
b
b
c
B
x9
A
5. Perimeter ABCD 119, and
c _____
D
2x 7
O
27°
a
c
68°
A
B
26°
A
a
62°
B
B
A
7. Perimeter ABCD 16x 12. AD _____
D
C
4x 3
8. If the diagonals of a quadrilateral are 15 cm and
9 cm, what is the perimeter of the quadrilateral
formed by connecting the midpoints of the sides?
A
B
63
and BD
.
9. Construct a parallelogram with diagonals AC
Is your parallelogram unique? If not, construct a different
A
(noncongruent) parallelogram.
C
D
B
10. Ball B is struck at the same instant by two forces, F1 and F2.
Copy the figure and show the resultant force on the ball.
F2
11. Find each lettered angle measure.
B
G
F1
81°
h
26°
D
g
A
e
a
d
i
E
f
c
38° j
b
B
38°
F
k
C
12. If the perimeter of a parallelogram is 132 cm, the longest possible
length of a diagonal is less than _____.
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CHAPTER 5
35
5.
D
7. MN 56; perimeter ABCD 166; MP 12;
mDON 90°; DP cannot be determined;
mA mB cannot be determined;
mB mC 180°
C
A
B
6. T(13, 5)
8. AMNO is a parallelogram. By the Triangle Midseg AM
and MN
AO
.
ment Conjecture, ON
Flowchart Proof
7. N(1, 1)
9. a 11
8. PS 33
10.
D
C
A
OC _1 AC
2
MN _1 AC
2
Definition of
midpoint
Midsegment
Conjecture
B
11. Possible answer:
Given: Trapezoid TRAP with T R
RA
Show: TP
P
T
A
OC MN
Both congruent to _1 AC
2
MB _1 AB
2
ON _1 AB
2
Definition of
midpoint
Midsegment
Conjecture
ON MB
R
E
PT
. TEAP is a paralParagraph proof: Draw AE
lelogram. T AER because they are corresponding angles of parallel lines. T R because it
is given. Therefore, AER R, because both
are congruent to T, so AER is isosceles by the
Converse of the Isosceles Triangle Conjecture.
EA
because they are opposite sides of a paralTP
EA
because AER is isosceles.
lelogram and AR
RA
because both are congruent
Therefore, TP
.
to EA
Both congruent to _1 AB
2
CON A
NMB A
CA Conjecture
CA Conjecture
CON NMB
Both congruent to A
ONC MBN
SAS Conjecture
LESSON 5.4 • Properties of Midsegments
1. a 89°, b 54°, c 91°
2. x 21, y 7, z 32
3. x 17, y 11, z 6.5
4. Perimeter XYZ 66, PQ 37, ZX 27.5
1.6,
5. M(12, 6), N(14.5, 2); slope AB
1.6
slope MN
6. Pick a point P from which A and B can be viewed
over land. Measure AP and BP and find the
midpoints M and N. AB 2MN.
B
A
FG
by
9. Paragraph proof: Looking at FGR, HI
the Triangle Midsegment Conjecture. Looking at
PQ
for the same reason. Because
PQR, FG
PQ
, quadrilateral FGQP is a trapezoid and
FG
is the midsegment, so it is parallel to FG
DE
. Therefore, HI
FG
DE
PQ
.
and PQ
LESSON 5.5 • Properties of Parallelograms
1. Perimeter ABCD 82 cm
2. AC 22, BD 14
3. AB 16, BC 7
4. a 51°, b 48°, c 70°
5. AB 35.5
6. a 41°, b 86°, c 53°
M
N
7. AD 75
8. 24 cm
P
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ANSWERS
99
9. No
9. c, d, f, g
D
C
A
B
10. c, d, f, g
11. d, g
12. d, g
13. b
14. Rectangle
15. Trapezoid (isosceles)
16. None
17. Parallelogram
18. R(6, 9), S(9, 3)
D
LESSON 5.7 • Proving Quadrilateral Properties
C
A
1. (See flowchart proof at bottom of page.)
2. Flowchart Proof
B
AB CB
10.
Given
F2
Resultant
vector
F1 F2
F1
11. a 38°, b 142°, c 142°, d 38°, e 142°,
f 38°, g 52°, h 12°, i 61°, j 81°, k 61°
12. 66 cm
LESSON 5.6 • Properties of Special Parallelograms
1. OQ 16, mQRS 90°, SQ 32
2. mOKL 45°, mMOL 90°,
perimeter KLMN 32
AD CD
ABD CBD
A C
Given
SSS Conjecture
CPCTC
DB DB
Same segment
3. Possible answer:
Given: Rhombus ABCD
Show: ABO CBO CDO ADO
Flowchart Proof
O
A
B
Diagonals of parallelogram
bisect each other
4.
C
A
C
AO CO
3. OB 6, BC 11, mAOD 90°
D
D
AB CB CD AD
BO DO
Definition of rhombus
Diagonals of a parallelogram
bisect each other
B
ABO CBO CDO ADO
SSS Conjecture
5. c, d, f, g
6. d, e, g
7. f, g
8. h
Lesson 5.7, Exercise 1
APR CQR
BP DQ
AIA Conjecture
Given
AP CQ
APR CQR
CPCTC
AC and QP
bisect each other
AB CD
Subtraction of
congruent segments
ASA Conjecture
PR QR
Definition of bisect
Opposite sides of
parallelogram
congruent
AR CR
CPCTC
PAR QCR
AIA Conjecture
100
ANSWERS
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©2003 Key Curriculum Press