Download Document

Ch 5 Polygon Notebook 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.
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 ∠RT I = 38.64°
m ∠DQU = 141.77 °
m ∠TRI = 81.14°
m ∠QUA = 59.70°
m ∠RIT = 60.21°
m ∠UAD = 86.28 °
m ∠ADQ = 72.26°
Do you see a possible pattern?
S. Stirling
Page 1 of 11
Ch 5 Polygon Notebook Key
Name ___________________________
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.
Quadrilateral ABCD
Pentagon EFGHI
m ∠BCD = 113°
m ∠ABC = 77 °
m ∠IEF = 71°
m ∠CDA = 56 °
m ∠DAB = 114°
m ∠GHI = 157°
m ∠HIE = 112°
m ∠EFG = 156°
m ∠FGH = 43°
m ∠DAB+m ∠ABC+m ∠BCD+m ∠CDA = 360.00°
m ∠IEF+m ∠EFG+m ∠FGH+m ∠GHI+m ∠HIE = 540.00°
B
A
F
G
E
H
C
I
D
Oc tagon PQRSTUV W
Hexagon JKLMNO
m ∠WPQ = 119 °
m ∠PQR = 130°
m ∠QRS = 154°
m ∠LMN = 105°
m ∠MNO = 140°
m ∠NOJ = 96°
m ∠OJK = 112°
m ∠JKL = 159°
m ∠KLM = 108°
S
R
m ∠OJK+m∠JKL+m ∠KLM+m ∠LMN+m ∠MNO+m ∠NOJ = 720.00°
L
K
m ∠RST = 132° m ∠UVW = 131°
m ∠ST U = 131° m ∠VWP = 147°
m ∠TUV = 137°
Q
T
J
M
U
P
V
W
N
O
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.
F
C
E
B
I
m ∠HAB = 67°
D
m ∠EBC = 103 °
C
H
m ∠FCD = 84 °
A
m ∠GDA = 106 °
G
H
D
m ∠HAB+m ∠EBC+m ∠FCD+m∠GDA = 360.00°
B
J
H
E
C
G
G
A
D
K
F
m ∠FAB = 59 °
m ∠GBC = 80 °
m ∠MFA = 72 °
E
M
A
m ∠GAB+m ∠HBC+m ∠ICD+m ∠JDE+m ∠KEF+m ∠MFA = 360°
m ∠HCD = 104°
m ∠IDE = 56 °
J
m ∠JEA = 61 °
m ∠FAB+m ∠GBC+m ∠HCD+m ∠IDE+m ∠JEA = 360°
S. Stirling
m ∠ICD = 73 °
m ∠JDE = 33 °
m ∠KEF = 54 °
I
B
F
m ∠GAB = 63 °
m ∠HBC = 66 °
Page 2 of 11
Ch 5 Polygon Notebook 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 3 of 11
Ch 5 Polygon Notebook 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. 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
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
Which method is easier? Finding one exterior angle first, because sum is always 360.
S. Stirling
Page 4 of 11
Ch 5 Polygon Notebook Key
Name ___________________________
Lesson 5.3 Kite and Trapezoid Properties
Definition of kite A quadrilateral with exactly two
Label vocab in the drawing.: K
distinct pairs of congruent consecutive sides.
E
Measure then compare the opposite angles of the kite.
Which pair will be congruent?
I
I
T
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.
K
T
E
Kite Angles Conjecture
(Do proof on Ch 5 WS page 2.)
∠K ≅ ∠T
The nonvertex angles of a kite are congruent.
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°
A
m ∠IAD = 43°
I
The angles of the kite:
m ∠DCB = 38°
m ∠CBA = 118°
m ∠BAD = 86°
m ∠ADC = 118°
The nonvertex angles:
m ∠CDI = 71°
m ∠IDA = 47°
B
m ∠IBA = 47°
m ∠IBC = 71°
C
Kite Angle Bisector Conjecture (Do proof on Ch 5 WS page 2.)
The vertex angles of a kite are bisected by a diagonal.
Kite Diagonals Conjecture
∠DAI ≅ ∠BAI and
∠DCI ≅ ∠BCI
(Do proof on Ch 5 WS page 3.)
The diagonals of a kite are perpendicular.
DB ⊥ CA
Kite Diagonal Bisector Conjecture (Do proof on Ch 5 WS page 3.)
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 5 of 11
Ch 5 Polygon Notebook Key
Name ___________________________
Definition of trapezoid A quadrilateral with exactly
Label vocab in
the drawing:
one pair of parallel sides.
D
C
B
Definition of isosceles trapezoid A trapezoid whose
legs are congruent.
A
Measure the angles of the trapezoids below. Label the diagram
with the measures to help you write the conjectures.
B
Q
D
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.
m ∠DC A = 59°
m ∠CAB = 121°
R
m ∠BDC = 139°
m ∠ABD = 41°
S
m ∠RQT = 46°
A
m ∠RST = 134°
T
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 (Do proof on Ch 5 WS page 4.)
The diagonals of an isosceles trapezoid are congruent.
I
C
S
IO ≅ SC
A
while
RP ≠ AT
P
T
O
S. Stirling
Page 6 of 11
Ch 5 Polygon Notebook Key
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:
Y
A
C
Z
X
Three Midsegments Conjecture
The three midsegments of a triangle divide it into
four congruent triangles.
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 is
parallel to the third side
and half the length of the third side.
R
X
XY & RI
T
I
Y
If || , then corr. angles congruent.
∠TXY ≅ ∠R and
∠I ≅ ∠TYX
XY = 1 RI
2
Page 276-277 Investigation 2: Trapezoid Midsegment Properties.
Midsegment (of a trapezoid) is the line segment connecting the midpoints of the two
nonparallel sides.
Steps 1 – 8: If you do not have tracing paper, you may measure instead. Label angles with measures!!
P
M
T
A
N
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
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.
Side lengths:
PA + TR
OR
2
1
MN = ( PA + TR )
2
MN =
Page 7 of 11
Ch 5 Polygon Notebook Key
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 on Ch 5 WS page 5.)
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.
Steps 5 – 7: Side and diagonal lengths!
PA = 2.8
PL = 7.5
PR = 6.2
LA = 9.4
P
A
Parallelogram Diagonals Conjecture
(Do proof on Ch 5 WS page 6.)
X
The diagonals of a parallelogram bisect
each other.
Side lengths:
L
S. Stirling
Parallelogram Opposite Sides
Conjecture (Do proof on Ch 5 WS page 5.)
The opposite sides of a parallelogram
are congruent.
R
PA = LR
PL = AR
PR ≠ AL
1
PR = 3.1
2
1
LX = XA = LA = 4.7
2
PX = XR =
Page 8 of 11
Ch 5 Polygon Notebook Key
Name ___________________________
Lesson 5.5 Properties of Parallelograms: Investigation
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
Given
LR , PA ≅ LR and diagonal PR .
P
∠APR ≅ ∠LRP
If ||, AIA cong.
PR = PR
ΔPAR ≅ ΔRLP
Same Segment.
SAS Cong. Conj.
L
R
PL ≅ RA
PA ≅ LR
CPCTC
Given
PARL is a
parallelogram
Both pairs of opposite sides are
congruent, then parallelogram
S. Stirling
Page 9 of 11
A
Ch 5 Polygon Notebook Key
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
rhombus.
Page 292 Investigation 2: Do Rhombus Diagonals Have Special Properties?
Steps 1 – 3:
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
(Do proof on Ch 5 WS page 7.)
The diagonals of a rhombus are perpendicular
and they bisect each other (because diagonals of a parallelogram bisect each other.)
Rhombus Diagonals Angles Conjecture
(Do proof on Ch 5 WS page 6.)
The diagonals of a rhombus
bisect the angles of the rhombus.
S. Stirling
Angle Relationships:
HM bisects ∠RHO and ∠RMO
Because a rhombus is a parallelogram and
opposite angles are congruent:
∠RHX ≅ ∠ OHX ≅ ∠RMX ≅ ∠OMX
Page 10 of 11
Ch 5 Polygon Notebook Key
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:
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
Q
Since diagonals of a rhombus bisect opposite
angles, m∠ XUA = 45°
In fact, all the acute angles are 45.
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 11 of 11