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Ch 5 Polygon Notebook L2 Key
Name ___________________________
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
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. Do you see a pattern?
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
R
D
Q
I
T
U
A
m ∠DQU = 141.77 °
m ∠QUA = 59.70°
m ∠UAD = 86.28 °
m ∠RT I = 38.64°
m ∠TRI = 81.14°
m ∠RIT = 60.21°
m ∠ADQ = 72.26°
S. Stirling
Page 1 of 9
Ch 5 Polygon Notebook L2 Key
Name ___________________________
(Investigation 5.1 Step 5: Draw all possible diagonals from one vertex, which divides each polygon into triangles.
Use these to develop a formula for the Polygon Sum Conjecture.
Quadrilateral
Pentagon
Hexagon
Diagonal forms 2 triangles, so
Diagonals form _3_ triangles:
Diagonals form _4_ triangles:
__2__ (180) = ___360____
__3__ (180) = ___540____
__4__ (180) = ___720____
Octagon
Decagon
Diagonals form _6_ triangles:
Diagonals form _8_ triangles:
__6__ (180) = ___1080____
__8__ (180) = ___1440____
Investigation 5.1 and 5.2 Summary
Use the results from Lesson 5.1 and 5.2 to fill in the table at the left. (The last column of the table should be
completed after 5.2 Investigation.)
Number of
sides of a
polygon
Sum of
measures of
interior angles
Sum of measures
of exterior angles
(one at a vertex)
3
4
5
6
7
8
9
10
11
12
13
n
180
360
540
720
900
1080
1260
1440
1620
1800
1980
(n-2) 180
360
360
360
360
360
360
360
360
360
360
360
360
S. Stirling
Polygon Sum Conjecture.
The sum of the measures of the n angles of an ngon is (n – 2) 180 or 180n – 360
Exterior Angle Sum Conjecture.
The sum of the measures of a set of exterior angles
of an n-gon is 360.
Page 2 of 9
Ch 5 Polygon Notebook L2 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:
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
You can find the measure of
each interior angle of an
equiangular n-gon by using
either of these formulas:
( n − 2)180
360
180
−
or
n
n
You can find the measure of
each exterior angle of an
equiangular n-gon by using the
formula:
360
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
(6 − 2)180 720
=
= 120
Use the formula to find the measure of one interior angle.
6
6
Same results? Yes
S. Stirling
Page 3 of 9
Ch 5 Polygon Notebook L2 Key
Name ___________________________
Lesson 5.3 Kite and Trapezoid Properties
Definition of kite A quadrilateral with exactly two
K
Vocabulary:
distinct pairs of congruent consecutive sides.
E
I
(Do investigation on Ch 5 WS page 3.)
Measure then compare the opposite angles of the kite.
Which pair will be congruent?
I
T
vertex angles (of a kite) The angles
between the pairs of congruent sides.
K
∠K & ∠T
T
E
nonvertex angles (of a kite) The two
angles between consecutive noncongruent
sides of a kite.
∠E & ∠I
Kite Angles Conjecture
The nonvertex angles of a kite are congruent.
∠K ≅ ∠T but ∠I ≅ ∠E
D
Kite Angle Bisector Conjecture (Do investigation on Ch 5 WS page 3).)
The vertex angles of a kite are bisected by a diagonal.
∠DAI ≅ ∠BAI and
∠DCI ≅ ∠BCI
48
66
2.61
90
5.35
I
2.36
24
24
48
66
C
Kite Diagonals Conjecture
B
(Do investigation on Ch 5 WS page 3.)
The diagonals of a kite are perpendicular.
DB ⊥ CA
Kite Diagonal Bisector Conjecture (Do investigation on Ch 5 WS page 4.)
The diagonal connecting the vertex angles of a kite is the perpendicular bisector
of the other diagonal.
CA is the perpendicular bisector of DB , The diagonals do not bisect each other.
S. Stirling
A
42
2.36
Page 4 of 9
42
Ch 5 Polygon Notebook L2 Key
Name ___________________________
Vocabulary:
Definition of trapezoid A quadrilateral with exactly
D
one pair of parallel sides.
C
B
Definition of isosceles trapezoid A trapezoid whose
legs are congruent.
(Do investigation on Ch 5 WS page 4.)
A
Below are Trapezoid ABDC and Isosceles Trapezoid ISOE:
A
I
E
50
63
CD & AB
130
130
117
B
bases (of a trapezoid) The two
parallel sides.
base angles (of a trapezoid) A pair
of angles with a base of the
trapezoid as a common side.
S
C
∠A & ∠B and ∠C & ∠D
130
legs are the two nonparallel sides.
50
CA & DB
O
50
m∠C + m∠A = 180
m∠D + m∠B = 180
D
Trapezoid Consecutive Angles Conjecture
The consecutive angles between the bases of a
trapezoid are supplementary.
m∠S + m∠O = 180
Isosceles Trapezoid [Base Angles] Conjecture
The base angles of an isosceles trapezoid are congruent.
∠I ≅ ∠S and
∠E ≅ ∠O
m∠I + m∠E = 180
I
(Do investigation on Ch 5 WS page 5.)
Isosceles Trapezoid Diagonals Conjecture
The diagonals of an isosceles
IO ≅ SC but
trapezoid are congruent.
S
C
R
O
RP ≅ TA
A
T
S. Stirling
Page 5 of 9
P
Ch 5 Polygon Notebook L2 Key
Name ___________________________
Lesson 5.4 Properties of Midsegments
Page 275-276 Investigation 1: Triangle Midsegment Properties (Do investigation on Ch 5 WS page 6.)
Midsegment (of a triangle) is the line segment connecting the midpoints of the two sides.
Three Midsegments Conjecture
The three midsegments of a triangle divide it into
four congruent triangles.
Y
A
C
Z
X
B
ΔAYZ ≅ ΔYCX ≅ ΔZXB ≅ ΔXZY
R
Triangle Midsegment Conjecture
A midsegment of a triangle is
parallel to the third side
and half the length of the third side.
X
T
∠TXY ≅ ∠R and ∠I ≅ ∠TYX
XY & RI If corr. angles congruent, then ||.
I
Y
XY = 1 RI
2
Page 276-277 Investigation 2: Trapezoid Midsegment Properties. (Do investigation on Ch 5 WS page 7.)
Midsegment (of a trapezoid) is the line segment connecting the midpoints of the two
nonparallel sides.
A
P
Trapezoid Midsegment Conjecture
The midsegment of a trapezoid is
parallel to the bases and
is equal in length to the average of the
lengths of the bases.
T
∠PMN ≅ ∠T
MN & PA & TR If corr. angles congruent, then ||.
S. Stirling
N
M
R
Side lengths:
PA + TR
OR
2
1
MN = ( PA + TR )
2
MN =
Page 6 of 9
Ch 5 Polygon Notebook L2 Key
Lesson 5.5 Properties of Parallelograms
Name ___________________________
Page 281-282 Investigation: Four Parallelogram Properties (Do investigation on Ch 5 WS page 8.)
Definition of a Parallelogram: A quadrilateral with two pairs of opposite sides parallel.
Angles:
L
∠M ≅ ∠J and ∠L ≅ ∠K
various: m∠L + m∠ J = 180
Parallelogram Opposite Angles
Conjecture
The opposite angles of a
parallelogram are congruent.
M
Parallelogram Consecutive Angles
Conjecture
The consecutive angles of a
parallelogram are supplementary.
J
K
Sides and diagonal lengths: (Do investigation on Ch 5 WS page 9.)
P
A
Parallelogram Diagonals Conjecture
The diagonals of a parallelogram bisect each
other.
N
L
S. Stirling
Parallelogram Opposite Sides Conjecture
The opposite sides of a parallelogram are
congruent.
R
PA ≅ LR and PL ≅ AR
N is the midpoint of PR and LA
PR and LA bisect each other.
Page 7 of 9
Ch 5 Polygon Notebook L2 Key
5.6 Properties of Special Parallelograms
Name ___________________________
Definition of a Rhombus: A quadrilateral with all sides congruent.
Page 292 Investigation 2: Do Rhombus Diagonals Have Special Properties?
(Do investigation on Ch 5 WS page 10.)
Rhombus Diagonals Angles
Conjecture
The diagonals of a rhombus bisect the
angles of the rhombus.
H
R
O
M
∠MRO ≅ ∠ORH ≅ ∠ HOR ≅ ∠ROM and
∠RHM ≅ ∠MHO ≅ ∠OMH ≅ ∠HMR
RO ⊥ HM
Rhombus Diagonals Conjecture
RO and HM bisect each other.
The diagonals of a rhombus are perpendicular.
and they bisect each other (because diagonals of a parallelogram bisect each other.)
Definition of a Rectangle: A quadrilateral with all angles congruent.
Page 293-4 Investigation 3: Do Rectangle Diagonals Have Special Properties?
(Do investigation on Ch 5 WS page 11.)
Rectangle Diagonals Conjecture
The diagonals of a rectangle are congruent
and bisect each other (because diagonals of a parallelogram
bisect each other.)
R
RC ≅ ET
RN ≅ NC ≅ NT ≅ NE
E
N
T
S. Stirling
C
Page 8 of 9
Ch 5 Polygon Notebook L2 Key
Name ___________________________
Definition of a Square: A quadrilateral with all angles and sides congruent
Square Diagonals Conjecture
The diagonals of a square are
congruent, (because diagonals of a rectangle are
congruent.)
A
U
perpendicular (because diagonals of a rrhombus are
perpendicular.)
S
and bisect each other (because diagonals of a
parallelogram bisect each other.)
Q
The small triangles formed will always have angle measures of 45 – 45 – 90
S. Stirling
Page 9 of 9