Download Ch 5 Polygon Notebook L1

Ch 5 Note Sheets
Name ___________________________
Note Sheets Chapter 5:
Discovering and Proving Polygon Properties
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
Polygon (interior Angle) Sum Conjecture.
The sum of the measures of the n angles of an
n-gon is
Exterior Angle Sum Conjecture.
The sum of the measures of a set of exterior
angles of an n-gon is
The interior and exterior angle in any polygon are
(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 ____
triangles:
Diagonals form ____
triangles:
Diagonals form ____
triangles:
____ (180) = ______
____ (180) = ______
____ (180) = ______
____ (180) = ______
Equiangular Polygon Conjectures
The measure of each interior angle
of an equiangular n-gon is:
S. Stirling
The measure of each exterior
angle of an equiangular n-gon is:
Page 1 of 9
Ch 5 Note Sheets
Lesson 5.3 Kite and Trapezoid Properties
Definition of kite
Name ___________________________
Label vocab in the drawing.:
K
Measure then compare the opposite angles of the kite.
Which pair will be congruent?
E
I
I
T
K
T
E
Kite Angles Conjecture (Do proof on Ch 5 WS page 8.)
angles of a kite are
The
vertex angles (of a kite) The angles
between the pairs of congruent sides.
nonvertex angles (of a kite) The two
angles between consecutive noncongruent
sides of a kite.
Additional measures: Label the diagram with the measures to help you write the conjectures.
D
m ∠DIA = 90°
The vertex angles:
The diagonals of the kite (c ompared to eachother):
m ∠DCI = 19°
DI = 1.57 cm IB = 1.57 cm
m ∠ICB = 19°
IA = 1.68 cm CI = 4.57 cm
m ∠BAI = 43°
m ∠IAD = 43°
The angles of the kite:
m ∠DCB = 38°
m ∠CBA = 118°
m ∠BAD = 86°
m ∠ADC = 118°
A
I
The nonvertex angles:
m ∠CDI = 71°
m ∠IDA = 47°
B
m ∠IBA = 47°
m ∠IBC = 71°
C
Kite Angle Bisector Conjecture
The vertex angles of a kite are
Kite Diagonals Conjecture
The diagonals of a kite are
(Do proof on Ch 5 WS page 8.)
(Do proof on Ch 5 WS page 9.)
Kite Diagonal Bisector Conjecture (Do proof on Ch 5 WS page 9.)
The diagonal connecting the vertex angles of a kite is the
of the other diagonal.
S. Stirling
Page 2 of 9
Ch 5 Note Sheets
Name ___________________________
Definition of trapezoid
Label vocab in
the drawing:
D
Definition of isosceles trapezoid
C
B
Measure the angles of the trapezoids below. Label the diagram
with the measures to help you write the conjectures.
B
A
Q
m∠RQT = 46°
m∠RST = 134°
D
m∠DC A = 59°
R
m∠CAB = 121°
m∠BDC = 139°
m∠ABD = 41°
S
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.
legs are the two nonparallel sides.
A
T
C
Trapezoid Consecutive Angles Conj.
The consecutive angles between the
Isosceles Trapezoid [Base Angles] Conj.
The base angles of an isosceles trapezoid are
bases of a trapezoid are
Measure diagonals of the trapezoids below.
Isosceles Trapezoid Diagonals Conjecture
The diagonals of an isosceles trapezoid are
R
(Do proof on Ch 5 WS page 10.)
I
C
A
S
P
T
O
S. Stirling
Page 3 of 9
Ch 5 Note Sheets
Name ___________________________
Lesson 5.4 Properties of Midsegments
Page 275-276 Investigation 1: Triangle Midsegment Properties
Midsegment (of a triangle) is the line segment connecting the midpoints of the two sides.
Steps 1 – 3:
C
A
Three Midsegments Conjecture
The three midsegments of a triangle divide it into
B
Steps 4 – 5: (Review Corresponding Angles Conjecture for parallel lines. The F shape!) Also label the
drawing below.
Triangle Midsegment Conjecture
A midsegment of a triangle
R
I
T
Page 276-277 Investigation 2: Trapezoid Midsegment Properties.
Midsegment (of a trapezoid) is the line segment connecting the midpoints of the two
nonparallel sides.
P
M
T
Steps 1 – 8: If you do not have tracing paper, you
may measure instead. Label angles with
measures!!
A
Trapezoid Midsegment Conjecture
The midsegment of a trapezoid is
N
R
Side lengths:
S. Stirling
Page 4 of 9
Ch 5 Note Sheets
S. Stirling
Name ___________________________
Page 5 of 9
Ch 5 Note Sheets
Lesson 5.5 Properties of Parallelograms
Name ___________________________
Page 281-282 Investigation: Four Parallelogram Properties
Definition of a Parallelogram: A quadrilateral with two pairs of opposite sides parallel.
Steps 1 – 4: Angles!
Parallelogram Opposite Angles
Conjecture (Do proof Ch 5 WS page 13.)
The opposite angles of a
parallelogram
L
M
Parallelogram Consecutive
Angles Conjecture
The consecutive angles of a
parallelogram
J
K
Angle measures:
If m∠L = 22° then
So to find all angles of a parallelogram:
Steps 5 – 7: Side and diagonal lengths!
P
A
Parallelogram Opposite Sides
Conjecture (Do proof on Ch 5 WS page 13.)
The opposite sides of a parallelogram
Parallelogram Diagonals Conjecture
(Do proof on Ch 5 WS page 14.)
The diagonals of a parallelogram
Side lengths:
L
S. Stirling
R
Page 6 of 9
Ch 5 Note Sheets
Name ___________________________
Lesson 5.5 Properties of Parallelograms: Investigation
Investigate: “If ___, then the quadrilateral is a parallelogram.”
Are the converses of the previous conjectures true?
So what do you need to know to know that a quadrilateral is a parallelogram?
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
LR , PA ≅ LR and diagonal PR .
P
L
S. Stirling
R
Page 7 of 9
A
Ch 5 Note Sheets
5.6 Properties of Special Parallelograms
Name ___________________________
Definition of a Rhombus: A quadrilateral with all sides congruent.
Page 291 Investigation 1: What Can You Draw with the Double-Edged Straightedge?
Steps 1 – 3: Complete in the space below. Use your ruler!
Double-Edged Straightedge
Conjecture
If two parallel lines are intersected by a
second pair of parallel lines that are the
same distance apart as the first pair,
then the parallelogram formed is a
Page 292 Investigation 2: Do Rhombus Diagonals Have Special Properties?
Steps 1 – 3:
H
R
M
Diagonal Relationships:
O
Rhombus Diagonals [Lengths] Conjecture
(Do proof on Ch 5 WS page 16.)
The diagonals of a rhombus are
and they
Rhombus Diagonals Angles Conjecture
Angle Relationships:
(Do proof on Ch 5 WS page 16.)
The diagonals of a rhombus
S. Stirling
Page 8 of 9
Ch 5 Note Sheets
Name ___________________________
Definition of a Rectangle: A quadrilateral with all angles congruent.
Page 293-4 Investigation 3: Do Rectangle Diagonals Have Special Properties?
Steps 1 – 2:
Rectangle Diagonals Conjecture
The diagonals of a rectangle are
and
R
E
Diagonal Relationships:
T
C
Definition of a Square: A quadrilateral with all angles and sides congruent
A
Square Diagonals Conjecture
The diagonals of a square are
U
X
S
Unique Relationships? Can you determine any measures
without measuring?
Q
S. Stirling
Page 9 of 9