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Ch 5 Worksheets Key
Name ___________________________
Worksheet Chapter 5:
Discovering and Proving Polygon Properties
Lesson 5.1 Polygon Sum Conjecture & Lesson 5.2 Exterior Angles of a Polygon
Warm up:
Definition: Exterior angle is an angle that forms a linear pair with one of the interior angles
of a polygon.
Measure the interior angles of QUAD to the nearest degree and put the measures into the diagram.
Draw one exterior angle at each vertex of QUAD. Measure each exterior angle to the nearest degree and
put the measures into the diagram.
How could you have calculated the exterior angles if all you had was the interior angles?
Each interior angle forms a linear pair with an exterior angle (supplementary)
Are any of the angles equal? No
What is the sum of the interior angles? ≈ 360
What is the sum of the exterior angles? ≈ 360
Now repeat the above investigation for the triangle TRI at the right. Compare the different angle sums
with the angle sums for the quadrilateral.
Are any of the angles equal? No
What is the sum of the interior angles? 180
What is the sum of the exterior angles? 360
108
D
R
72
99
Q
81
120
142
38
60
I
U
60
86
120
94
A
T
39
141
m ∠DQU = 141.77 °
m ∠RT I = 38.64°
m ∠QUA = 59.70 °
m ∠TRI = 81.14°
m ∠UAD = 86.28 °
m ∠RIT = 60.21°
m ∠ADQ = 72.26 °
Do you see a possible pattern? Various conclusions
S. Stirling
Page 1 of 20
Ch 5 Worksheets Key
Name ___________________________
Page 258-259 5.1 Investigation: Is there a Polygon Sum Formula?
Steps 1-2: Review your work from page 1 and examine the diagrams below.
Step 3-4: Complete the sum of the interior angles column and drawing diagonals on the next page.
Pentagon EFGHI
m ∠GHI = 157°
m ∠IEF = 71°
m ∠EFG = 156° m ∠HIE = 112°
m ∠FGH = 43°
m ∠IEF+m ∠EFG+m ∠FGH+m ∠GHI+m ∠HIE = 540.00°
Quadrilateral ABCD
m ∠BCD = 113 °
m ∠ABC = 77°
m ∠DAB = 114°
m ∠CDA = 56°
m ∠DAB+m ∠ABC+m ∠BCD+m ∠CDA = 360.00°
B
F
A
77
114
E
113
156
157
C
56
G
43
71
112
D
H
I
Hexagon JKLMNO
Oc tagon PQRSTUVW
m ∠WPQ = 119 °
m ∠PQR = 130 °
m ∠QRS = 154 °
m ∠LMN = 105°
m ∠MNO = 140°
m ∠NOJ = 96°
m ∠OJK = 112°
m ∠JKL = 159°
m ∠KLM = 108°
m ∠OJK+m∠JKL+m ∠KLM+m ∠LMN+m ∠MNO+m ∠NOJ = 720.00 °
154 132
Q
108
159
S
R
L
K
m ∠RST = 132° m ∠UVW = 131 °
m ∠ST U = 131° m ∠VWP = 147 °
m ∠TUV = 137°
130
131
T
119
137
U
J
112
105
P
M
131
147
V
W
140
96
N
O
Page 262-263 5.2 Investigation: Is there an Exterior Angle Sum?
Steps 1-5: Review your work from page 1 and examine the diagrams below. One exterior angle is drawn
at each vertex. Complete the sum of the exterior angles column on the next page.
F
C
E
84
103
B
I
m∠HAB = 67°
D
m∠EBC = 103°
106
67
m∠FCD = 84°
m∠GDA = 106°
C
H
A
73
G
H
D
66
m∠HAB+m∠EBC+m∠FCD+m∠GDA = 360.00°
B
33
H
C
G
G
59
E
A
61
m∠FAB = 59°
m∠GBC = 80°
m∠HCD = 104°
m∠IDE = 56°
J
m∠JEA = 61°
m∠FAB+m ∠GBC+m∠HCD+m ∠IDE+m∠JEA = 360°
F
S. Stirling
54
A
80
56
63
K
72
I
m ∠GAB = 63 °
m ∠HBC = 66 °
m ∠ICD = 73 °
E
104
D
B
J
F
m ∠JDE = 33 °
m ∠KEF = 54 °
m ∠MFA = 72 °
M
m ∠GAB+m ∠HBC+m ∠ICD+m ∠JDE+m ∠KEF+m ∠MFA = 360°
Page 2 of 20
Ch 5 Worksheets Key
Name ___________________________
Page 262-263 5.2 Investigation: Is there an Exterior Angle Sum?
Steps 7-8: Use what you know about interior angle sums and exterior angle sums to calculate the measure
of each interior and each exterior angle of any equiangular polygon.
Try an example first. Use deductive reasoning.
Find the measure of an interior and an exterior angle of an equiangular pentagon. Show your calculations
below:
One interior angle = 540 ÷ 5 = 108
One exterior angle = 360 ÷ 5 = 72
What is the relationship between one interior
and one exterior angle?
108
72
Supplementary, 108 + 72 = 180
Equiangular Polygon Conjecture
You can find the measure of
each interior angle of an
equiangular n-gon by using
either of these formulas:
(n − 2)180
360
or 180 −
n
n
Or
You can find the measure of
each exterior angle of an
equiangular n-gon by using the
formula:
360
n
180n 360 180n − 360
−
=
n
n
n
More practice:
One exterior angle = 360 ÷ 6 = 60
What is the relationship between one interior
and one exterior angle?
60 120
Supplementary
Use this relationship to find the measure of one interior angle. 180 – 60 = 120
Use the formula to find the measure of one interior angle.
(6 − 2)180 720
=
= 120
6
6
Same results? Yes
Which method is easier? Finding one exterior angle
first, because sum is always 360.
S. Stirling
Page 3 of 20
Ch 5 Worksheets Key
Name ___________________________
5.1 EXERCISES Page 259-260 #3 – 11, 13, 14, 12, 16.
Show how you are finding your answers!
e = (5 – 2)180/5
= 540/5
= 108
110
90
72
112
a = 360 – 90 – 76 – 72 = 122
108
(6 – 2)180 = 720
b = (720 – 448)/2 = 136
72
180 – 108 = 72
f = 180 – 2 * 72
= 180 – 144 = 36
360 – 108 – 130
= 122
122
142
102
44
120
Triangle: d = 180 – 44 – 30 = 106
Quad: c = 360 – 252 = 108
Penta: g = (540 – 225)/3 = 105
Quad: h = 360 – 278 = 82
60 120
j = 720/6 = 120
k = 360 – 322 = 38
various
102 + 82 + 98 + 76 = 358
Interior sum should be 360.
various
135 + 3 * 131 + 26 = 554
Interior sum should be 540.
180 −
18 sides
360
= 160
n
26
102
360 – 200 = 160
60
82
76
98
82
131
140
( 9 − 2 )180 = 140
131
131
9
360
= 156
n
360
−
= −24
n
n = 15
180 −
( n − 2 )180 = 2700
n − 2 = 15
n = 17
S. Stirling
Page 4 of 20
Ch 5 Worksheets Key
Name ___________________________
5.1 Page 260 Exercise #12
a = 116, b = 64, c = 90, d = 82, e = 99, f = 88,
g = 150, h = 56, j = 106, k = 74, m = 136, n = 118, p = 99
5.1 Page 261 Exercise #16 You are building the window frame below. You will need to know the
measures the angles in order to cut the trapezoidal pieces. Show and explain how you would calculate the
measures of the angles of the trapezoids
67.5
112.5 112.5
135
The angles of the trapezoid
measure 67.5 and 112.5.
Each angle of the octagon:
(8 − 2)180
= 135
8
Around a point:
360 – 135 = 225
225 ÷ 2 = 112.5
Angles between the bases are
supplementary.
180 – 112.5 = 67.5
S. Stirling
Page 5 of 20
Ch 5 Worksheets Key
5.1 Page 260-261 Exercise #15, 18, 20, 21
Name ___________________________
360
= 160
n
360
−
= −20
n
n = 18
180 −
So twelfth century.
120
D
Draw a right isosceles triangle. The base
angles are both 45, so they are
complementary.
S. Stirling
Page 6 of 20
Ch 5 Worksheets Key
Name ___________________________
5.2 EXERCISES Page 263-265 #1 – 10
1. Sum exterior angles
decagon?
2. An exterior angle of
equiangular pentagon?
360
360
= 72
5
hexagon?
360
= 60
6
108
b = 45
40
3. How many sides
regular polygon, each
exterior angle 24º?
4. How many sides
polygon, sum interior
angles 7380º?
360
= 24
n
360 = 24n
n = 15
(n − 2)180 = 7380
n − 2 = 41
n = 43
c = 51
1
3
3
7
d = 115
5
7
112
Exterior angle sum is 360.
a = 360 – 252 = 108
Exterior angle sum is 360.
360 – 112 – 43 - 69 = 136
136/3 = 45.333
7-gon: (7 – 2)180/7 = 128.57
c = 180 – 128.57 = 51.43
d = (360 – 128.57)/2 = 115.715
39
30
106
108
108
135
135
30
55
44
136
129
97
162
44
51
102
129
83
97
83
Pentagon: (5 – 2)180/5 = 108
Octagon: (8 – 2)180/8 = 135
e = 180 – 108 = 72
f = 180 – 135 = 45
g = 360 – 108 – 135 = 117
h = 360 – 117 – 72 – 45 = 126
Triangle:
a = 180 – 56 – 94 = 30
b = 30 ||, alt. int. angles =
Triangle:
c = 180 – 44 – 30 = 106
d = 180 – 44 = 136
S. Stirling
a = 180 – 18 = 162
g = 180 – 86 – 39 = 55
d = 39 Isos. triangle
c = 180 – 39 * 2 = 102
e = (360 – 102)/2 = 129
f = 90 – 39 = 51
Large Pentagon:
540 – 94 – 90 – 162 = 194
h = 194/2 = 97
b = 180 – 97 = 83
Quad: k = 360 – 129 – 51 – 97 = 83
Page 7 of 20
Ch 5 Worksheets Key
Name ___________________________
Proof of the Kite Angles Conjecture
Conjecture: The nonvertex angles of a kite are congruent.
Given: Kite KITE with diagonal KT .
Prove: The nonvertex angles are congruent,
Kite KITE
K
E
I
∠E ≅ ∠I .
KT = KT
T
Same Segment.
Given
KE = KI and
ET = IT
∆KET ≅ ∆KIT
∠E ≅ ∠I
SSS Cong. Conj.
CPCTC or
Def. cong.
triangles
Def. of Kite
5.3 Page 272 Exercise #9 Proof of Kite Angle Bisector Conjecture
The vertex angles of a kite are bisected by a diagonal.
∠1 ≅ ∠2
∠3 ≅ ∠4
BN ≅ BN
Same Segment.
BE ≅ BY
YN ≅ EN
Given or
Def. of Kite
S. Stirling
∆BYN ≅ ∆BEN
CPCTC or Def.
cong. triangles
SSS Cong. Conj.
BN bisects ∠YBE
BN bisects ∠YNE
Def. angle bisector
Page 8 of 20
Ch 5 Worksheets Key
Name ___________________________
Proof of the Kite Diagonals Conjecture
Conjecture: The diagonals of a kite are perpendicular.
D
A
DB and AC .
Prove: The diagonals are perpendicular. DB ⊥ AC .
Given: Kite ABCD with diagonals
I
B
Kite ABCD
Given
∠DAI ≅ ∠BAI
C
m∠DIA + m∠BIA = 180
Diag. bisect vertex angles.
AD = AB
Def. of Kite
AI = AI
Same Segment.
Linear Pair Conj.
∆DAI ≅ ∆BAI
SAS Cong. Conj.
∠DIA ≅ ∠BIA
CPCTC
m∠DIA = m∠BIA = 90°
DB ⊥ AC
Algebra
Def. of Perpendicular
D
Proof of the Kite Diagonal Bisector Conjecture
Conjecture: The diagonal connecting the vertex angles of a kite
is the perpendicular bisector of the other diagonal.
A
I
Given: Kite ABCD with diagonals DB and AC .
Prove: AC is the perpendicular bisector of DB .
B
C
Kite ABCD
Given
DB ⊥ AC
∠DAI ≅ ∠BAI
Diag. of kite are
Perpendicular
Diag. kite bisect vertex
angles.
DI = IB
AD = AB
Def. of Kite
AI = AI
Same Segment.
S. Stirling
∆DAI ≅ ∆BAI
CPCTC
SAS Cong. Conj.
AC is the perpendicular bisector of DB
Def. of perp. bisector.
Page 9 of 20
Ch 5 Worksheets Key
Name ___________________________
5.3 Page 271 Proof of Isosceles Trapezoid Diagonals Conjecture
Conjecture: The diagonals of an isosceles trapezoid are congruent.
Given: Isosceles trapezoid TRAP with TP = RA.
Show: Diagonals are congruent, TA = RP.
TR = TR
PT = RA
Same Segment.
Given
Isosceles trapezoid TRAP
Given
m∠PTR = m∠TRA
∆PTR ≅ ∆ART
TA = RP
SAS Cong. Conj.
CPCTC
Isosceles Trap. base angles =
5.3 EXERCISES Page 271-274 #1 – 8, #14 – 15 construct, #19
Book Page 272-273 #11 – 13 sketch (Do on a separate sheet of paper.)
52
21
146
64 cm
128
Non-vertex angles =. y = 146
x = 360 – 47 – 146 * 2 = 21
20
21
146
12
Isos. so base angles =.
y = 128
Consecutive angles
supplementary.
x = 180 – 128 = 52
Perimeter:
20 * 2 + 12 * 2 = 64
15
Perimeter:
85 = 37 +18 + 2x
85 = 55 + 2x
30 = 2x
15 = x
72
99
61
38 cm
90
81
15
S. Stirling
15
Small Right triangle:
x = 180 – 90 – 18 = 72
Large Right triangle:
x = 180 – 90 – 29 = 61
29
Perimeter:
164 = y + 2(y +12) + (y – 12)
164 = y + 2 y + 24 + y – 12
164 = 4 y + 12
152 = 4 y 38 = y
Page 10 of 20
99
128
52
Ch 5 Worksheets Key
Name ___________________________
Vertex angle
3.0 cm
45
1.6 cm
90
48
42
y = 180 – 90 – 48
= 42
30
30
S. Stirling
30
w = 180 – 2 * 30
= 120
Page 11 of 20
Ch 5 Worksheets Key
5.3 Page 274 Exercise #19
Name ___________________________
a = 80, b = 20, c = 160, d = 20, e = 80, f = 80,
g = 110, h = 70, m = 110, n = 100
5.4 EXERCISES Page 271-274 #1 – 7
three; one
65
60
140
28
Perimeter TOP =
8 + 2*10 = 28
y = 180 – 40
= 140
||, corr. angles = . x = 60
Corresponding
angles of
congruent
triangles.
129
35
73
23
42 cm
Corresponding sides of
congruent triangles.
Perimeter = 6 + 8 + 9 = 23
S. Stirling
m = 180 – 51 = 129
p=
1
( 36 + 48 ) = 42
2
1
(13 + q )
2
48 = 13 + q
24 =
35 = q
Page 12 of 20
Ch 5 Worksheets Key
Name ___________________________
Section 5.5 Proofs
Proof of the Parallelogram Opposite Angles Conjecture
L
Conjecture: The opposite angles of a parallelogram are congruent.
R
2
Given: Parallelogram PARL with diagonal AL .
Prove: ∠PAR ≅ ∠PLR and ∠R ≅ ∠P .
Parallelogram
PARL
1
m∠2 = m∠3
PA LR
4
If ||, AIA cong.
Def. of Parallelogram
3
Given
P
A
m∠1 = m∠4
LP AR
m∠2 + m∠1 = m∠3 + m∠4
If ||, AIA cong.
Addition
Def. of Parallelogram
m∠PLR = m∠PAR
m∠R = m∠P
Substitution
If 2 angles of one triangle =
2 angles of another, the 3rd
angles are =.
P
Proof of the Parallelogram Opposite Sides Conjecture
Conjecture: The opposite sides of a parallelogram are congruent.
Given: Parallelogram PARL with diagonal
Prove:
PL ≅ RA
Parallelogram
PARL
and
PR .
PA ≅ LR .
PA LR
Def. of Parallelogram
L
∠APR ≅ ∠LRP
R
If ||, AIA cong.
Given
∠L ≅ ∠A
Opposite angles of
Parallelogram =
PR = PR
Same Segment.
S. Stirling
∆PAR ≅ ∆RLP
AAS Cong. Conj.
PL ≅ RA
and
PA ≅ LR
CPCTC
Page 13 of 20
A
Ch 5 Worksheets Key
Name ___________________________
5.5 Page 284 Exercise #13 Proof of the Parallelogram Diagonals Conj.
The diagonals of a parallelogram bisect each other.
given
∠EAL ≅ ∠ALN
∆ETA ≅ ∆NTL
def parallelogram
LT ≅ TA
EA ≅ LN
EN & LA
5.5 EXERCISES Page 271-274 #1 – 6, #7 – 8 construct, #16, 18
34 cm
132
27 cm
48
a = 180 – 48
= 132
80
16 in
14 in
63
78
63 m
x – 3 = 17
x = 20
Perim = 18 + 24 + 21 = 63
S. Stirling
20 + 3 = 23
Perim = 2*17 + 2*23 = 80
Page 14 of 20
bisect eachother.
Ch 5 Worksheets Key
Name ___________________________
1. Copied ∡ L.
2. Measure & draw
LA.
3 Make ∡ A = 130,
since consecutive
angles supp.
4. Measure & draw
AS = LT. Opp. Sides
=.
5. Draw TS.
8. Construct parallelogram DROP, given side
D
DR and diagonals DO
PR .
R
O
D
P
and
R
1. Measure, draw & bisect DO.
2. Measure other diagonal PR and bisect. Using the
midpoint of DO construct a circle with radius ½ PR.
3. With compass, measure DR & mark locations for
R & P on the circle.
Diag. bisect each other & opposite sides = in a
parallelogram.
Hexa:
720 ÷ 6 = 120
d = 360 – 90 – 120 – 108 = 42
e = (180 – 42)/2 = 69
30 stones make a 30gon; each angle =
( 30 − 2 )180 = 168
30
Penta: 540 ÷ 5 = 108
S. Stirling
a = 120,
b = 108,
c = 90,
d = 42,
e = 69
About a point:
360 – 168 = 192
192 ÷ 2 = 96 = b
Consecutive angles
supp.
180 – 96 = 84 = a
Page 15 of 20
Ch 5 Worksheets Key
Section 5.6 Proofs
Name ___________________________
Proof of the Rhombus Diagonals Angles Conjecture
Conjecture: The diagonals of a rhombus bisect the angles of the rhombus.
HM .
∠RMO .
Given: Rhombus RHOM with diagonal
Prove:
HM
bisects ∠RHO and
H
R
O
M
Rhombus RHOM
Given
RH = HO = OM = MR
∆MRH ≅ ∆MOH
Def. of Rhombus
SSS Cong. Conj.
HM = HM
∠RHM ≅ ∠OHM
∠RMH ≅ ∠OMH
Same Segment.
HM bisects ∠RHO and
∠RMO
CPCTC
Def. of angle bisector
Proof of the Rhombus Diagonals Conjecture
Conjecture: The diagonals of a rhombus are perpendicular, and
they bisect each other.
Given: Rhombus RHOM with diagonals
Prove:
HM
and
HM
and
H
R
X
RO .
M
RO are perpendicular bisectors of each other.
and RO are
perpendicular bisectors of
each other.
HM
Rhombus RHOM
HX = XM
RX = XO
Def of Perp. Bisector
Given
RH = RM
Def of Rhombus
Diagonals of a parallelogram
bisect each other.
∆RXH ≅ ∆RXM
Def of Perp.
m∠RXH = m∠RXM = 90°
SSS Cong. Conj.
RX = RX
Same Segment
RO ⊥ HM
CPCTC and Algebra
m∠RXH + m∠RXM = 180
Linear Pair
S. Stirling
Page 16 of 20
O
Ch 5 Worksheets Key
Name ___________________________
5.6 EXERCISES Page 271-274 #1 – 11
Sometimes
Always
Always
Sometimes
Always
Sometimes
Always
Always
Always
Sometimes: only if the
parallelogram is a rectangle.
45
37
90
20
5.6 Page 297 Exercise #28
a = 54, b = 36, c = 72, d = 108, e = 36, f = 144, g = 18,
h = 48, i = 48, k = 84
S. Stirling
Page 17 of 20
Ch 5 Worksheets Key
Name ___________________________
1. Copy PS
2. Make perp. at P and S.
3. Measure PE and make
arc from P then repeat
from S.
4. Draw IE.
Diag. of rectangle are = &
a rectangle has 90º angles.
1. Copy LV
2. Make perp. bisector
3. Measure ½ of LV
4. Find O and E
Diag. of square are perp
bisectors and are =.
1. Measured ∡ B, took
half & made bisected ∡ B.
2. Measure & draw BK.
3 Make bisected ∡ B at K.
Where the sides intersect
is A and E.
Diag. bisect opposite
angles in a Rhombus.
20.
If the diagonals of
a parallelogram
are equal, then the
parallelogram is a
rectangle.
S. Stirling
If the diagonals of a quadrilateral are
congruent and bisect each other, then
the quadrilateral is a rectangle.
Can be proved true! The proofs will
vary, but should use congruent
triangles to show the angles are 90º
each.
Page 18 of 20
Ch 5 Worksheets Key
Ch 5 Review
Name ___________________________
x = 10, y = 40
x = 60 cm
a = 116, c = 64
x = 38 cm
y = 34, z = 51
100
Exercise #13
Opposite sides are
parallel
Opposite sides are
congruent
Opposite angles are
congruent
Diagonals bisect each
other
Diagonals are
perpendicular
Diagonals are
congruent
Exactly one line of
symmetry
Exactly two lines of
symmetry
S. Stirling
Kite
Isosceles
trapezoid
Parallelogram Rhombus
Rectangle
Square
No
One pair
Yes
Yes
Yes
Yes
No
One pair
Yes
Yes
Yes
Yes
Non-Vertex No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
Yes
No
No
Yes
No
Yes
No
Yes
No
No
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
Yes
Yes
Yes 4
Page 19 of 20
Ch 5 Worksheets Key
Name ___________________________
Exercise #14
Regular decagon; each angle =
(10 − 2 )180 = 144
10
About a point:
360 – 144 = 216
216 ÷ 2 = 108 = b
Each part of the frame must be an isosceles
trapezoid, so consecutive angles between the
bases are supp.
180 – 108 = 72 = a
72
a
2 in
108
108
b
144
5.R Page 305 Exercise #15
a = 120, b = 60, c = 60, d = 120, e = 60, f = 30, g = 108,
m = 24, p = 84
S. Stirling
Page 20 of 20