Download Note Sheets Chapter 5: Discovering and Proving Polygon Properties

Ch 5 Polygon Notebook Key L2
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
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
# triangles times angle sum one triangle
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.
(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
Equiangular Polygon Conjectures
The measure of each interior angle of an
equiangular n-gon is:
 n  2 180 or
n
S. Stirling
180n  360
360
or 180 
n
n
The measure of each exterior
angle of an equiangular n-gon is:
360
n
Page 1 of 7
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? The non-vertex
I
Vertex angle
Label the vocab. in the drawing of KITE:
106
Non-vertex
angle K
vertex angles (of a kite) The angles between
the pairs of congruent sides.
53
53
148
T
non-vertex angles (of a kite) The two angles
between consecutive non-congruent sides of a
kite.
Non-vertex
angle
E
Vertex angle
D
(Do 5.3 Investigations Kites on Ch 5 WS page 8 and top of page 9.)
71 47
Use the diagram at the right to help you write the conjectures.
Illustrate how to apply the conjectures with that diagram.
1.57 cm
90
Kite Angles Conjecture
The non-vertex angles of a kite are congruent.
43
1.68 cm 43
I
1.57 cm
4.57 cm
CDA  ABC
47
71
Kite Angle Bisector Conjecture
The vertex angles of a kite are
bisected by a diagonal.
DAI  BAI and DCI  BCI
B
19
19
C
Also notice the isosceles
triangles DAB and DCB .
As well as the many pairs of
congruent triangles:
DCI  BCI , etc…
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
A
CA is the perpendicular bisector of DB .
Page 2 of 7
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.
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.
legs are the two nonparallel sides.
Measure the angles of the trapezoids below. Label the diagram
with the measures to help you write the conjectures.
Isosceles Trapezoid QRST
B
Trapezoid ABDC
Q
41
Leg
46
D
139
Base
Angles
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 7
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
AZY  ZBX  YXC  XYZ
Triangle Midsegment Conjecture
A midsegment of a triangle is
parallel to the third side
and half the length of the third side.
R
87
X
3.2 cm
93
87
1.6 cm
52
T
Three Midsegments Conjecture
The three midsegments of a triangle divide it
into four congruent triangles.
128
Y
52
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
The midsegment of a trapezoid is
A
P 20 mm
parallel to the bases and
N
M
is equal in length to the average of the
49 mm
lengths of the bases.
T
78 mm
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 7
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
22
M
158
Parallelogram Consecutive Angles
Conjecture
The consecutive angles of a
parallelogram are supplementary.
158
J
22
K
mL  22 then
mM  180  22  158 consecutive angles supp.
mK  22 opposite angles equal
mJ  158 opposite angles equal or consecutive angles supp.
Find the angle measures, If
P
PA  2.8
A
2.8 cm
PL  7.5
PR  6.2
LA  9.4
3.1
4.7
X
7.5 cm
7.5 cm
Side lengths:
4.7
L
S. Stirling
3.1
R
2.8 cm
Parallelogram Opposite Sides
Conjecture
The opposite sides of a parallelogram
are congruent.
PA  LR
PL  AR
PR  AL
Parallelogram Diagonals Conjecture
The diagonals of a parallelogram bisect
each other.
1
PR  3.1
2
1
LX  XA  LA  4.7
2
PX  XR 
Page 5 of 7
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
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.
R
H
1
2
Diagonal Relationships:
RO  HM
or mRXH  90
8 7
X
6
3 4
M
5
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
And because opposite angles are congruent:
8  7  3  4
Also RO bisects HRM and HOM
And because opposite angles are congruent:
1  2  5  6
S. Stirling
Page 6 of 7
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:
QUA  UAS  ASQ  SQU
as well as
UXA  AXS  SXQ  QXU
S. Stirling
Page 7 of 7