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Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 1 of 20
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
Do you see a possible pattern? Various conclusions
Q
UA
D
mADQ = 72.26
mUAD = 86.28
mQUA = 59.70
mDQU = 141.77
T
R
I
mRIT = 60.21
mTRI = 81.14
mRTI = 38.64
60
72
86
142
120
94
108
38
39
81
60
120
141
99
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 2 of 20
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.
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.
mDAB+mABC+mBCD+mCDA = 360.00
mDAB = 114
mABC = 77mBCD = 113
mCDA = 56
Quadrilateral ABCD
D
C
BA
mIEF+mEFG+mFGH+mGHI+mHIE = 540.00
mIEF = 71
mEFG = 156
mFGH = 43
mGHI = 157
mHIE = 112
Pentagon EFGHI
I
H
G
F
E
mOJK+mJKL+mKLM+mLMN+mMNO+mNOJ = 720.00
mOJK = 112
mJKL = 159
mKLM = 108
mLMN = 105
mMNO = 140
mNOJ = 96
Hexagon JKLMNO
ON
M
LK
J
mHAB+mEBC+mFCD+mGDA = 360.00
mHAB = 67
mEBC = 103
mFCD = 84
mGDA = 106
D
C
B
A
HG
F
E
mFAB+mGBC+mHCD+mIDE+mJEA = 360
mJEA = 61
mIDE = 56
mHCD = 104
mGBC = 80
mFAB = 59A
G
H
I
J
B
C
D
EF
B
I
J
K
M
G
C
D
E
F
A
H
mMFA = 72
mKEF = 54
mJDE = 33
mICD = 73
mHBC = 66
mGAB = 63
mGAB+mHBC+mICD+mJDE+mKEF+mMFA = 360
67
103
84
106
80
104
59 61
56
63
73
66
33
54
72
77
113 56
114
mIEF+mEFG+mFGH+mGHI+mHIE = 540.00
mIEF = 71
mEFG = 156
mFGH = 43
mGHI = 157
mHIE = 112
Pentagon EFGHI
I
H
G
F
E 43 71
157
156
112
mOJK+mJKL+mKLM+mLMN+mMNO+mNOJ = 720.00
mOJK = 112
mJKL = 159
mKLM = 108
mLMN = 105
mMNO = 140
mNOJ = 96
Hexagon JKLMNO
ON
M
LK
J
96
105
108
112
140
159
mWPQ = 119
mPQR = 130
mQRS = 154
mRST = 132
mSTU = 131
mTUV = 137
mUVW = 131
mVWP = 147
Octagon PQRSTUVW
W V
U
T
SR
Q
P
132
119
130
137
131
131
147
154
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 3 of 20
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?
Supplementary, 108 + 72 = 180
Equiangular Polygon Conjecture
Or 180 360 180 360n n
n n n
More practice:
One exterior angle = 360 ÷ 6 = 60
What is the relationship between one interior
and one exterior angle?
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
1206 6
Same results? Yes
Which method is easier? Finding one exterior angle
first, because sum is always 360.
You can find the measure of
each interior angle of an
equiangular n-gon by using
either of these formulas:
( 2)180n
n
or
360180
n
You can find the measure of
each exterior angle of an
equiangular n-gon by using the
formula:
360
n
108 72
120 60
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 4 of 20
5.1 EXERCISES Page 259-260 #3 – 11, 13, 14, 12, 16.
Show how you are finding your answers!
various
76
82
82 102 98
102 + 82 + 98 + 76 = 358
Interior sum should be 360.
131
various
60
26
135 + 3 * 131 + 26 = 554
Interior sum should be 540.
131
131
18 sides
9 2 180140
9
140 360 – 200 = 160
360180 160
n
2 180 2700
2 15
17
n
n
n
360180 156
36024
15
n
n
n
a = 360 – 90 – 76 – 72 = 122
90
110
112
(6 – 2)180 = 720
b = (720 – 448)/2 = 136
e = (5 – 2)180/5
= 540/5
= 108
180 – 108 = 72
f = 180 – 2 * 72
= 180 – 144 = 36
108 72
72
44 102
Triangle: d = 180 – 44 – 30 = 106
Quad: c = 360 – 252 = 108
Penta: g = (540 – 225)/3 = 105
Quad: h = 360 – 278 = 82
122
360 – 108 – 130
= 122
j = 720/6 = 120
k = 360 – 322 = 38
60 120
120
142
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 5 of 20
5.1 Page 260 Exercise #12
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
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
The angles of the trapezoid
measure 67.5 and 112.5.
Each angle of the octagon:
(8 2)180135
8
Around a point:
360 – 135 = 225
225 2 = 112.5
Angles between the bases are
supplementary.
180 – 112.5 = 67.5
135
112.5 112.5
67.5
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 6 of 20
5.1 Page 260-261 Exercise #15, 18, 20, 21
120
360180 160
36020
18
n
n
n
So twelfth century.
D
Draw a right isosceles triangle. The base
angles are both 45, so they are
complementary.
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 7 of 20
5.2 EXERCISES Page 263-265 #1 – 10
108
1. Sum exterior angles
decagon?
360
2. An exterior angle of
equiangular pentagon?
36072
5
hexagon?
36060
6
3. How many sides
regular polygon, each
exterior angle 24º?
36024
360 24
15
n
n
n
4. How many sides
polygon, sum interior
angles 7380º?
( 2)180 7380
2 41
43
n
n
n
145
3b
351
7c 5
1157
d
Exterior angle sum is 360.
a = 360 – 252 = 108
112
40
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
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
108 108
135
135
136
44
30
44 106
30
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
Triangle:
a = 180 – 56 – 94 = 30
b = 30 ||, alt. int. angles =
Triangle:
c = 180 – 44 – 30 = 106
d = 180 – 44 = 136
162
55
39
102
129 129
51
97 97
83
83
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 8 of 20
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, E I .
5.3 Page 272 Exercise #9 Proof of Kite Angle Bisector Conjecture
The vertex angles of a kite are bisected by a diagonal.
Same Segment.
KT KT
SSS Cong. Conj.
IKET K T
Given
Kite KITE
Def. of Kite
KE KI and
ET IT CPCTC or
Def. cong.
triangles
E I
T
E I
K
Same Segment.
BN BN
SSS Cong. Conj.
BYN BEN
Given or
Def. of Kite
BE BY
YN EN
1 2
3 4
BN bisects YBE
BN bisects YNE
Def. angle bisector
CPCTC or Def.
cong. triangles
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 9 of 20
Proof of the Kite Diagonals Conjecture
Conjecture: The diagonals of a kite are perpendicular.
Given: Kite ABCD with diagonals DB and AC .
Prove: The diagonals are perpendicular. DB AC .
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.
Given: Kite ABCD with diagonals DB and AC .
Prove: AC is the perpendicular bisector of DB .
Same Segment.
AI AI
SAS Cong. Conj.
ADAI B I
Given
Kite ABCD
Def. of Kite
AD AB
DIA BIA
I
D
C
B
A
Diag. bisect vertex angles.
DAI BAI
180m DIA m BIA
Linear Pair Conj.
DB AC
Def. of Perpendicular
90m DIA m BIA
Algebra
Same Segment.
AI AI
SAS Cong. Conj.
ADAI B I
Given
Kite ABCD
Def. of Kite
AD AB
CPCTC
DI IB
Diag. kite bisect vertex
angles.
ADAI B I DB AC
Diag. of kite are
Perpendicular
AC is the perpendicular bisector of DB
Def. of perp. bisector.
I
D
C
B
A
CPCTC
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 10 of 20
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.
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.)
Given
PT RA Same Segment.
TR TR
SAS Cong. Conj.
RPTR A T
Given
Isosceles trapezoid TRAP
Isosceles Trap. base angles =
m PTR m TRA CPCTC
TA RP
21
146 64 cm
52
128
15 72
61
99
38 cm
Perimeter:
20 * 2 + 12 * 2 = 64
Non-vertex angles =. y = 146
x = 360 – 47 – 146 * 2 = 21 Isos. so base angles =.
y = 128
Consecutive angles
supplementary.
x = 180 – 128 = 52 12
20
146
128
52
21
Perimeter:
85 = 37 +18 + 2x
85 = 55 + 2x
30 = 2x
15 = x
15 15 Small Right triangle:
x = 180 – 90 – 18 = 72
Large Right triangle:
x = 180 – 90 – 29 = 61
29
90 81 99
Perimeter:
164 = y + 2(y +12) + (y – 12)
164 = y + 2 y + 24 + y – 12 164 = 4 y + 12
152 = 4 y 38 = y
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 11 of 20
30 30
45
30
w = 180 – 2 * 30
= 120
3.0 cm
1.6 cm
48 90
y = 180 – 90 – 48
= 42 42
Vertex angle
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 12 of 20
5.3 Page 274 Exercise #19
5.4 EXERCISES Page 271-274 #1 – 7
a = 80, b = 20, c = 160, d = 20, e = 80, f = 80,
g = 110, h = 70, m = 110, n = 100
28
42 cm
three; one
60
140
65
23
129
73
35
Perimeter TOP =
8 + 2*10 = 28
y = 180 – 40
= 140
||, corr. angles = . x = 60
Corresponding
angles of
congruent
triangles.
Corresponding sides of
congruent triangles.
Perimeter = 6 + 8 + 9 = 23
m = 180 – 51 = 129
1
36 48 422
p
1
24 132
48 13
35
q
q
q
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 13 of 20
Section 5.5 Proofs Proof of the Parallelogram Opposite Angles Conjecture
Conjecture: The opposite angles of a parallelogram are congruent.
Given: Parallelogram PARL with diagonal AL .
Prove: PAR PLR and R P .
Proof of the Parallelogram Opposite Sides Conjecture
Conjecture: The opposite sides of a parallelogram are congruent.
Given: Parallelogram PARL with diagonal PR .
Prove: PL RA and PA LR .
Same Segment.
PR PR
AAS Cong. Conj.
PAR RLP
Given
Parallelogram
PARL Def. of Parallelogram
PA LR
If ||, AIA cong.
APR LRP
PL RA and PA LR
CPCTC
AP
RL
Opposite angles of
Parallelogram =
AL
Substitution
Given
Parallelogram
PARL Def. of Parallelogram
PA LR
If ||, AIA cong.
2 3m m
Addition Def. of Parallelogram
LP AR
If ||, AIA cong.
1 4m m 3 42 1 m mm m
m PLR m PAR
4
2
3
1
LR
P
A
m R m P
If 2 angles of one triangle =
2 angles of another, the 3rd
angles are =.
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 14 of 20
5.5 Page 284 Exercise #13 Proof of the Parallelogram Diagonals Conj.
The diagonals of a parallelogram bisect each other.
5.5 EXERCISES Page 271-274 #1 – 6, #7 – 8 construct, #16, 18
def parallelogram
given
EAL ALN
EA LN LT TA
ETA NTL
&EN LA bisect eachother.
34 cm 132
27 cm 48
16 in
14 in
63 m
80 63
78
a = 180 – 48
= 132
Perim = 18 + 24 + 21 = 63
x – 3 = 17
x = 20
20 + 3 = 23
Perim = 2*17 + 2*23 = 80
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 15 of 20
8. Construct parallelogram DROP, given side DR and diagonals DO and PR .
D
D
R
P
O
R
a = 120,
b = 108,
c = 90,
d = 42,
e = 69
30 stones make a 30-
gon; each angle =
30 2 180168
30
About a point:
360 – 168 = 192
192 ÷ 2 = 96 = b
Consecutive angles
supp.
180 – 96 = 84 = a
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.
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
Penta: 540 ÷ 5 = 108
d = 360 – 90 – 120 – 108 = 42
e = (180 – 42)/2 = 69
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 16 of 20
Section 5.6 Proofs
Proof of the Rhombus Diagonals Angles Conjecture
Conjecture: The diagonals of a rhombus bisect the angles of the rhombus.
Given: Rhombus RHOM with diagonal HM .
Prove: HM bisects RHO and RMO .
Proof of the Rhombus Diagonals Conjecture
Conjecture: The diagonals of a rhombus are perpendicular, and
they bisect each other.
Given: Rhombus RHOM with diagonals HM and RO .
Prove: HM and RO are perpendicular bisectors of each other.
HR
MO
Same Segment.
HM HM
SSS Cong. Conj.
MRH MOH
Given
Rhombus RHOM
Def. of Rhombus
RHM OHM
RMH OMH
RH HO OM MR
CPCTC & Opposite angles
of a parallelogram are =.
HM bisects RHO and
RMO Def. of angle bisector
X
HR
MO
Given
Rhombus RHOM
Def of Rhombus
RH RM Def of Perp.
RX HM
Diagonals of a parallelogram
bisect each other.
RX XOHX XM
Same Segment
RX RX
SSS Cong. Conj.
RXH RXM
90m RXH m RXM
CPCTC and Algebra
180m RXH m RXM
Linear Pair supp.
HM and RO are
perpendicular bisectors of
each other.
Def of Perp. Bisector
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 17 of 20
5.6 EXERCISES Page 271-274 #1 – 11
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
Sometimes
Always
Always
Sometimes
Always
Sometimes
Always
Always
Always
Sometimes: only if the
parallelogram is a rectangle.
20
37 45
90
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 18 of 20
20. On page 295. 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.
If the diagonals of
a parallelogram
are equal, then the
parallelogram is a
rectangle.
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. 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. 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.
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 19 of 20
Ch 5 Review
Exercise #13
Kite Isosceles
trapezoid
Parallelogram Rhombus Rectangle Square
Opposite sides are
parallel
No One pair Yes Yes Yes Yes
Opposite sides are
congruent
No One pair Yes Yes Yes Yes
Opposite angles are
congruent
Non-Vertex No Yes Yes Yes Yes
Diagonals bisect each
other
No No Yes Yes Yes Yes
Diagonals are
perpendicular
Yes No No Yes No Yes
Diagonals are
congruent
No Yes No No Yes Yes
Exactly one line of
symmetry
Yes Yes No No No No
Exactly two lines of
symmetry
No No No Yes Yes Yes 4
x = 10, y = 40 x = 60 cm
a = 116, c = 64
100
x = 38 cm
y = 34, z = 51
Ch 5 Worksheets Key Name ___________________________
S. Stirling Page 20 of 20
Exercise #14 On page 305. Show all work!!
5.R Page 305 Exercise #15
a = 120, b = 60, c = 60, d = 120, e = 60, f = 30, g = 108,
m = 24, p = 84
Regular decagon; each angle =
10 2 180144
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
144
108
a 72
108 b
2 in
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