Notebook Chapter 5: Discovering and Proving Polygon Properties

Ch 5 Polygon Notebook Key L2
Name ___________________________
Notebook Chapter 5:
Discovering and Proving Polygon Properties
(5.1 Investigation 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
Octagon
Diagonal forms 2
triangles, so
Diagonals form _3_
triangles:
Diagonals form _4_
triangles:
Diagonals form _6_
triangles:
2 (180) = 360
3 (180) = 540
4 (180) = 720
6 (180) = 1080
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
n
180
360
540
720
900
1080
1260
1440
(n-2) 180
360
360
360
360
360
360
360
360
360
Polygon (Interior Angle) Sum Conjecture.
The sum of the measures of the n angles of an
n-gon 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.
The interior and exterior angle in any polygon are supplementary.
Equiangular Polygon Conjectures
The measure of each interior angle
of an equiangular n-gon is:
( n − 2)180
360
180
−
or
n
n
S. Stirling
The measure of each exterior
angle of an equiangular n-gon is:
360
n
Page 1 of 8
Ch 5 Polygon Notebook Key L2
Name ___________________________
Lesson 5.3 Kite and Trapezoid Properties
Definition of kite A quadrilateral with exactly two
distinct pairs of congruent consecutive sides.
Measure then compare the opposite angles of the kite.
Which pair will be congruent?
I
Vertex angle
Label vocab. in the drawing below:
vertex angles (of a kite) The angles
between the pairs of congruent sides.
non-vertex angles (of a kite) The two
angles between consecutive noncongruent sides of a kite.
Non-vertex
angle K
T
Non-vertex
angle
E
Vertex angle
Kite Angles Conjecture
(Do Investigations Kites Ch 5 WS page 8 and top of page 9.)
The nonvertex angles of a kite are congruent.
∠K ≅ ∠T
D
Label the diagram with the measures to help you write the conjectures.
m ∠DIA = 90°
The vertex angles:
The diagonals of the kite (c ompared to eachother):
DI = 1.57 cm IB = 1.57 cm
IA = 1.68 cm CI = 4.57 cm
The angles of the kite:
m ∠DCB = 38°
m ∠CBA = 118°
m ∠BAD = 86°
m ∠ADC = 118°
The nonvertex angles:
m ∠CDI = 71°
A
m ∠DCI = 19°
m ∠ICB = 19°
I
m ∠BAI = 43°
m ∠IAD = 43°
m ∠IDA = 47°
m ∠IBA = 47°
B
m ∠IBC = 71°
C
Kite Angle Bisector Conjecture
The vertex angles of a kite are bisected by a diagonal.
∠DAI ≅ ∠BAI and
∠DCI ≅ ∠BCI
Kite Diagonals Conjecture
The diagonals of a kite are perpendicular.
DB ⊥ CA
Kite Diagonal Bisector Conjecture
The diagonal connecting the vertex angles of a kite is the perpendicular bisector
of the other diagonal.
S. Stirling
CA is the perpendicular bisector of DB .
Page 2 of 8
Ch 5 Polygon Notebook Key L2
Name ___________________________
(Do Investigations Trapezoids Ch 5 WS bottom of page 9 and top of page 10.)
Definition of trapezoid A quadrilateral with exactly
one pair of parallel sides.
Definition of isosceles trapezoid A trapezoid whose
legs are congruent.
Measure the angles of the trapezoids below. Label the diagram
with the measures to help you write the conjectures.
Leg
legs are the two nonparallel sides.
Isosceles Trapezoid QRST
B
Trapezoid ABDC
Label vocab. in the drawing below:
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.
m ∠DC A = 59°
Q
m ∠CAB = 121°
41
m ∠BDC = 139°
m ∠RQT = 46°
46
m ∠ABD = 41°
D
139
Base
Angles
m ∠RST = 134°
134 R
Base Base
Angles
Base
134
S
46
121
T
A
59
Leg
C
Trapezoid Consecutive Angles Conj.
The consecutive angles between the
bases of a trapezoid are supplementary.
Isosceles Trapezoid [Base Angles] Conj.
The base angles of an isosceles trapezoid
are congruent.
m∠C + m∠A = 180
m∠D + m∠B = 180
∠Q ≅ ∠T and ∠R ≅ ∠S
Measure diagonals of the trapezoids below.
R
Isosceles Trapezoid Diagonals Conjecture
The diagonals of an isosceles trapezoid are congruent.
I
C
S
IO ≅ SC
A
while
RP ≠ AT
P
T
O
S. Stirling
Page 3 of 8
Ch 5 Polygon Notebook Key L2
Name ___________________________
Lesson 5.4 Properties of Midsegments
(Do Investigation 1: Triangle Midsegment Properties Ch 5 WS page 12.)
Midsegment (of a triangle) is the line segment connecting the midpoints of the two sides.
Y
A
C
Z
X
B
ΔAZY ≅ ΔZBX ≅ ΔYXC ≅ ΔXYZ
Triangle Midsegment Conjecture
A midsegment of a triangle is
parallel to the third side
and half the length of the third side.
R
X
T
Three Midsegments Conjecture
The three midsegments of a triangle divide it
into four congruent triangles.
Y
I
XY & RI
If || , then corr. angles congruent.
XY = 1 RI
2
∠TXY ≅ ∠R and
∠I ≅ ∠TYX
(Do Investigation 2: Triapezoid Midsegment Properties Ch 5 WS page 13.)
Midsegment (of a trapezoid) is the line segment connecting the midpoints of the two
nonparallel sides.
Trapezoid Midsegment Conjecture
A
The midsegment of a trapezoid is
P
parallel to the bases and
N
M
is equal in length to the average of the
lengths of the bases.
T
R
All parallels:
PA & TR & MN
All congruent angles: If || , then corresponding
angles congruent.
∠PMN ≅ ∠T and ∠P ≅ ∠NMT and
∠ANM ≅ ∠R and ∠A ≅ ∠MNR
S. Stirling
Side lengths:
PA + TR
OR
2
1
MN = ( PA + TR )
2
MN =
Page 4 of 8
Ch 5 Polygon Notebook Key L2
Lesson 5.5 Properties of Parallelograms
Name ___________________________
(Do Investigation: Four Parallelogram Properties Ch 5 WS pages 15 – 16.)
Definition of a Parallelogram: A quadrilateral with two pairs of opposite sides parallel.
Steps 1 – 4: Angles!
Parallelogram Opposite Angles
Conjecture
The opposite angles of a
parallelogram are congruent.
L
M
Parallelogram Consecutive Angles
Conjecture
The consecutive angles of a
parallelogram are supplementary.
J
K
Angle measures:
So to find all angles of a parallelogram:
If m∠L = 22° then
• Make opposite angles equal
m∠M = 180 − 22 = 158
m∠K = 22
m∠J = 158
• Subtract an angle from 180 to get a
consecutive angle.
PA = 2.8
P
PL = 7.5
PR = 6.2
A
LA = 9.4
Parallelogram Diagonals Conjecture
The diagonals of a parallelogram bisect
each other.
X
Side lengths:
PA = LR
PL = AR
L
S. Stirling
R
Parallelogram Opposite Sides
Conjecture
The opposite sides of a parallelogram
are congruent.
PR ≠ AL
1
PR = 3.1
2
1
LX = XA = LA = 4.7
2
PX = XR =
Page 5 of 8
Ch 5 Polygon Notebook Key L2
Name ___________________________
Lesson 5.5 Properties of Parallelograms (continued):
Investigate: “If ___, then the quadrilateral is a parallelogram.”
Are the converses of the previous conjectures true? Yes
So what do you need to know to know that a quadrilateral is a parallelogram?
Both pairs of opposite angles congruent
Consecutive angles supplementary
Both pairs of opposite sides congruent
Diagonals bisect each other
Would the following conjecture work? Try to prove it deductively.
If one pair of sides of a quadrilateral is both parallel and congruent, then the quadrilateral
is a parallelogram.
Proof
Given: Quadrilateral PARL PA &
Prove: PARL is a parallelogram
various
PA & LR
LR , PA ≅ LR and diagonal PR .
P
∠APR ≅ ∠LRP
If ||, AIA cong.
L
Given
PR = PR
ΔPAR ≅ ΔRLP
Same Segment.
SAS Cong. Conj.
R
PL ≅ RA
PA ≅ LR
CPCTC
Given
PARL is a
parallelogram
Both pairs of opposite sides are
congruent, then parallelogram
S. Stirling
Page 6 of 8
A
Ch 5 Polygon Notebook Key L2
Name ___________________________
5.6 Properties of Special Parallelograms
(Do Investigation 2: Do Rhombus Diagonals Have Special Properties? Ch 5 WS page 19.)
Definition of a Rhombus: A quadrilateral with all sides congruent.
Diagonal Relationships:
H
R
RO ⊥ HM
or m∠RXH = 90°
X
M
O
Because a rhombus is
a parallelogram:
RX ≅ XO
MX ≅ XH
Rhombus Diagonals [Lengths] Conjecture
The diagonals of a rhombus are perpendicular
and they bisect each other (because diagonals of a parallelogram bisect each other.)
Rhombus Diagonals Angles Conjecture
The diagonals of a rhombus
bisect the angles of the rhombus.
Angle Relationships:
HM bisects ∠RHO and ∠RMO
Because a rhombus is a parallelogram and
opposite angles are congruent:
∠RHX ≅ ∠OHX ≅ ∠RMX ≅ ∠OMX
S. Stirling
Page 7 of 8
Ch 5 Polygon Notebook Key L2
Name ___________________________
(Do Investigation 3: Do Rectangle Diagonals Have Special Properties? Ch 5 WS page 20.)
Definition of a Rectangle: A quadrilateral with all angles congruent.
R
Rectangle Diagonals Conjecture
The diagonals of a rectangle are congruent
E
and bisect each other (because diagonals of a
parallelogram bisect each other.)
X
Diagonal Relationships:
RC ≅ TE
And because a rectangle is a parallelogram and the
diagonals bisect each other:
RX ≅ XC ≅ TX ≅ XE
T
C
As a result, you get many pairs of congruent
triangles (and some congruent isosceles triangles).
Definition of a Square: A quadrilateral with all angles and sides congruent
Square Diagonals Conjecture
The diagonals of a square are
congruent (a rectangle), perpendicular (a
rhombus) and bisect each other (a parallelogram).
A
U
X
Unique Relationships? Can you determine any measures
without measuring?
S
Since diagonals of a rhombus bisect opposite
angles, m∠XUA = 45°
In fact, all the acute angles are 45.
Q
You get many pairs of congruent right isosceles
(45:45:90) triangles:
ΔQUA ≅ ΔUAS ≅ ΔASQ ≅ ΔSQU
as well as
ΔUXA ≅ ΔAXS ≅ ΔSXQ ≅ ΔQXU
S. Stirling
Page 8 of 8

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