Worksheet Chapter 5: Discovering and Proving Polygon ...€¦ · S. Stirling Page 1 of 20 Worksheet Chapter 5: Discovering and Proving Polygon Properties ... Finding one exterior

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S. Stirling Page 1 of 20

Worksheet 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 (supplementary)

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

Do you see a possible pattern? Various conclusions

Q

UA

D

mADQ = 72.26

mUAD = 86.28

mQUA = 59.70

mDQU = 141.77

T

R

I

mRIT = 60.21

mTRI = 81.14

mRTI = 38.64

60

72

86

142

120

94

108

38

39

81

60

120

141

99



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.

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.

mDAB+mABC+mBCD+mCDA = 360.00

mDAB = 114

mABC = 77mBCD = 113

mCDA = 56

Quadrilateral ABCD

D

C

BA

mIEF+mEFG+mFGH+mGHI+mHIE = 540.00

mIEF = 71

mEFG = 156

mFGH = 43

mGHI = 157

mHIE = 112

Pentagon EFGHI

I

H

G

F

E

mOJK+mJKL+mKLM+mLMN+mMNO+mNOJ = 720.00

mOJK = 112

mJKL = 159

mKLM = 108

mLMN = 105

mMNO = 140

mNOJ = 96

Hexagon JKLMNO

ON

M

LK

J

mHAB+mEBC+mFCD+mGDA = 360.00

mHAB = 67

mEBC = 103

mFCD = 84

mGDA = 106

D

C

B

A

HG

F

E

mFAB+mGBC+mHCD+mIDE+mJEA = 360

mJEA = 61

mIDE = 56

mHCD = 104

mGBC = 80

mFAB = 59A

G

H

I

J

B

C

D

EF

B

I

J

K

M

G

C

D

E

F

A

H

mMFA = 72

mKEF = 54

mJDE = 33

mICD = 73

mHBC = 66

mGAB = 63

mGAB+mHBC+mICD+mJDE+mKEF+mMFA = 360

67

103

84

106

80

104

59 61

56

63

73

66

33

54

72

77

113 56

114

mIEF+mEFG+mFGH+mGHI+mHIE = 540.00

mIEF = 71

mEFG = 156

mFGH = 43

mGHI = 157

mHIE = 112

Pentagon EFGHI

I

H

G

F

E 43 71

157

156

112

mOJK+mJKL+mKLM+mLMN+mMNO+mNOJ = 720.00

mOJK = 112

mJKL = 159

mKLM = 108

mLMN = 105

mMNO = 140

mNOJ = 96

Hexagon JKLMNO

ON

M

LK

J

96

105

108

112

140

159

mWPQ = 119

mPQR = 130

mQRS = 154

mRST = 132

mSTU = 131

mTUV = 137

mUVW = 131

mVWP = 147

Octagon PQRSTUVW

W V

U

T

SR

Q

P

132

119

130

137

131

131

147

154



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

Or 180 360 180 360n n

n n 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

Use the formula to find the measure of one interior angle. (6 2)180 720

1206 6

Same results? Yes

Which method is easier? Finding one exterior angle

first, because sum is always 360.

You can find the measure of

each interior angle of an

equiangular n-gon by using

either of these formulas:

( 2)180n

n

or

360180

n

You can find the measure of

each exterior angle of an

equiangular n-gon by using the

formula:

360

n

108 72

120 60



5.1 EXERCISES Page 259-260 #3 – 11, 13, 14, 12, 16.

Show how you are finding your answers!

various

76

82

82 102 98

102 + 82 + 98 + 76 = 358

Interior sum should be 360.

131

various

60

26

135 + 3 * 131 + 26 = 554

Interior sum should be 540.

131

131

18 sides

9 2 180140

9

140 360 – 200 = 160

360180 160

n

2 180 2700

2 15

17

n

n

n

360180 156

36024

15

n

n

n

a = 360 – 90 – 76 – 72 = 122

90

110

112

(6 – 2)180 = 720

b = (720 – 448)/2 = 136

e = (5 – 2)180/5

= 540/5

= 108

180 – 108 = 72

f = 180 – 2 * 72

= 180 – 144 = 36

108 72

72

44 102

Triangle: d = 180 – 44 – 30 = 106

Quad: c = 360 – 252 = 108

Penta: g = (540 – 225)/3 = 105

Quad: h = 360 – 278 = 82

122

360 – 108 – 130

= 122

j = 720/6 = 120

k = 360 – 322 = 38

60 120

120

142



5.1 Page 260 Exercise #12

5.1 Page 261 Exercise #16 You are building the window frame below. You will need to know the

measures the angles in order to cut the trapezoidal pieces. Show and explain how you would calculate the

measures of the angles of the trapezoids

a = 116, b = 64, c = 90, d = 82, e = 99, f = 88,

g = 150, h = 56, j = 106, k = 74, m = 136, n = 118, p = 99

The angles of the trapezoid

measure 67.5 and 112.5.

Each angle of the octagon:

(8 2)180135

8

Around a point:

360 – 135 = 225

225 2 = 112.5

Angles between the bases are

supplementary.

180 – 112.5 = 67.5

135

112.5 112.5

67.5



5.1 Page 260-261 Exercise #15, 18, 20, 21

120

360180 160

36020

18

n

n

n

So twelfth century.

D

Draw a right isosceles triangle. The base

angles are both 45, so they are

complementary.



5.2 EXERCISES Page 263-265 #1 – 10

108

1. Sum exterior angles

decagon?

360

2. An exterior angle of

equiangular pentagon?

36072

5

hexagon?

36060

6

3. How many sides

regular polygon, each

exterior angle 24º?

36024

360 24

15

n

n

n

4. How many sides

polygon, sum interior

angles 7380º?

( 2)180 7380

2 41

43

n

n

n

145

3b

351

7c 5

1157

d

Exterior angle sum is 360.

a = 360 – 252 = 108

112

40

Exterior angle sum is 360.

360 – 112 – 43 - 69 = 136

136/3 = 45.333

7-gon: (7 – 2)180/7 = 128.57

c = 180 – 128.57 = 51.43

d = (360 – 128.57)/2 = 115.715

Pentagon: (5 – 2)180/5 = 108

Octagon: (8 – 2)180/8 = 135

e = 180 – 108 = 72

f = 180 – 135 = 45

g = 360 – 108 – 135 = 117

h = 360 – 117 – 72 – 45 = 126

108 108

135

135

136

44

30

44 106

30

a = 180 – 18 = 162

g = 180 – 86 – 39 = 55

d = 39 Isos. triangle

c = 180 – 39 * 2 = 102

e = (360 – 102)/2 = 129

f = 90 – 39 = 51

Large Pentagon:

540 – 94 – 90 – 162 = 194

h = 194/2 = 97

b = 180 – 97 = 83

Quad: k = 360 – 129 – 51 – 97 = 83

Triangle:

a = 180 – 56 – 94 = 30

b = 30 ||, alt. int. angles =

Triangle:

c = 180 – 44 – 30 = 106

d = 180 – 44 = 136

162

55

39

102

129 129

51

97 97

83

83



Proof of the Kite Angles Conjecture

Conjecture: The nonvertex angles of a kite are congruent.

Given: Kite KITE with diagonal KT .

Prove: The nonvertex angles are congruent, E I .

5.3 Page 272 Exercise #9 Proof of Kite Angle Bisector Conjecture

The vertex angles of a kite are bisected by a diagonal.

Same Segment.

KT KT

SSS Cong. Conj.

IKET K T

Given

Kite KITE

Def. of Kite

KE KI and

ET IT CPCTC or

Def. cong.

triangles

E I

T

E I

K

Same Segment.

BN BN

SSS Cong. Conj.

BYN BEN

Given or

Def. of Kite

BE BY

YN EN

1 2

3 4

BN bisects YBE

BN bisects YNE

Def. angle bisector

CPCTC or Def.

cong. triangles



Proof of the Kite Diagonals Conjecture

Conjecture: The diagonals of a kite are perpendicular.

Given: Kite ABCD with diagonals DB and AC .

Prove: The diagonals are perpendicular. DB AC .

Proof of the Kite Diagonal Bisector Conjecture

Conjecture: The diagonal connecting the vertex angles of a kite

is the perpendicular bisector of the other diagonal.

Given: Kite ABCD with diagonals DB and AC .

Prove: AC is the perpendicular bisector of DB .

Same Segment.

AI AI

SAS Cong. Conj.

ADAI B I

Given

Kite ABCD

Def. of Kite

AD AB

DIA BIA

I

D

C

B

A

Diag. bisect vertex angles.

DAI BAI

180m DIA m BIA

Linear Pair Conj.

DB AC

Def. of Perpendicular

90m DIA m BIA

Algebra

Same Segment.

AI AI

SAS Cong. Conj.

ADAI B I

Given

Kite ABCD

Def. of Kite

AD AB

CPCTC

DI IB

Diag. kite bisect vertex

angles.

ADAI B I DB AC

Diag. of kite are

Perpendicular

AC is the perpendicular bisector of DB

Def. of perp. bisector.

I

D

C

B

A

CPCTC



5.3 Page 271 Proof of Isosceles Trapezoid Diagonals Conjecture

Conjecture: The diagonals of an isosceles trapezoid are congruent.

Given: Isosceles trapezoid TRAP with TP = RA.

Show: Diagonals are congruent, TA = RP.

5.3 EXERCISES Page 271-274 #1 – 8, #14 – 15 construct, #19

Book Page 272-273 #11 – 13 sketch (Do on a separate sheet of paper.)

Given

PT RA Same Segment.

TR TR

SAS Cong. Conj.

RPTR A T

Given

Isosceles trapezoid TRAP

Isosceles Trap. base angles =

m PTR m TRA CPCTC

TA RP

21

146 64 cm

52

128

15 72

61

99

38 cm

Perimeter:

20 * 2 + 12 * 2 = 64

Non-vertex angles =. y = 146

x = 360 – 47 – 146 * 2 = 21 Isos. so base angles =.

y = 128

Consecutive angles

supplementary.

x = 180 – 128 = 52 12

20

146

128

52

21

Perimeter:

85 = 37 +18 + 2x

85 = 55 + 2x

30 = 2x

15 = x

15 15 Small Right triangle:

x = 180 – 90 – 18 = 72

Large Right triangle:

x = 180 – 90 – 29 = 61

29

90 81 99

Perimeter:

164 = y + 2(y +12) + (y – 12)

164 = y + 2 y + 24 + y – 12 164 = 4 y + 12

152 = 4 y 38 = y



30 30

45

30

w = 180 – 2 * 30

= 120

3.0 cm

1.6 cm

48 90

y = 180 – 90 – 48

= 42 42

Vertex angle




5.4 EXERCISES Page 271-274 #1 – 7

a = 80, b = 20, c = 160, d = 20, e = 80, f = 80,

g = 110, h = 70, m = 110, n = 100

28

42 cm

three; one

60

140

65

23

129

73

35

Perimeter TOP =

8 + 2*10 = 28

y = 180 – 40

= 140

||, corr. angles = . x = 60

Corresponding

angles of

congruent

triangles.

Corresponding sides of

congruent triangles.

Perimeter = 6 + 8 + 9 = 23

m = 180 – 51 = 129

1

36 48 422

p

1

24 132

48 13

35

q

q

q



Section 5.5 Proofs Proof of the Parallelogram Opposite Angles Conjecture

Conjecture: The opposite angles of a parallelogram are congruent.

Given: Parallelogram PARL with diagonal AL .

Prove: PAR PLR and R P .

Proof of the Parallelogram Opposite Sides Conjecture

Conjecture: The opposite sides of a parallelogram are congruent.

Given: Parallelogram PARL with diagonal PR .

Prove: PL RA and PA LR .

Same Segment.

PR PR

AAS Cong. Conj.

PAR RLP

Given

Parallelogram

PARL Def. of Parallelogram

PA LR

If ||, AIA cong.

APR LRP

PL RA and PA LR

CPCTC

AP

RL

Opposite angles of

Parallelogram =

AL

Substitution

Given

Parallelogram

PARL Def. of Parallelogram

PA LR

If ||, AIA cong.

2 3m m

Addition Def. of Parallelogram

LP AR

If ||, AIA cong.

1 4m m 3 42 1 m mm m

m PLR m PAR

4

2

3

1

LR

P

A

m R m P

If 2 angles of one triangle =

2 angles of another, the 3rd

angles are =.



5.5 Page 284 Exercise #13 Proof of the Parallelogram Diagonals Conj.

The diagonals of a parallelogram bisect each other.

5.5 EXERCISES Page 271-274 #1 – 6, #7 – 8 construct, #16, 18

def parallelogram

given

EAL ALN

EA LN LT TA

ETA NTL

&EN LA bisect eachother.

34 cm 132

27 cm 48

16 in

14 in

63 m

80 63

78

a = 180 – 48

= 132

Perim = 18 + 24 + 21 = 63

x – 3 = 17

x = 20

20 + 3 = 23

Perim = 2*17 + 2*23 = 80



8. Construct parallelogram DROP, given side DR and diagonals DO and PR .

D

D

R

P

O

R

a = 120,

b = 108,

c = 90,

d = 42,

e = 69

30 stones make a 30-

gon; each angle =

30 2 180168

30

About a point:

360 – 168 = 192

192 ÷ 2 = 96 = b

Consecutive angles

supp.

180 – 96 = 84 = a

1. Copied L.

2. Measure & draw

LA.

3 Make A = 130,

since consecutive

angles supp.

4. Measure & draw

AS = LT. Opp. Sides

=.

5. Draw TS.

1. Measure, draw & bisect DO.

2. Measure other diagonal PR and bisect. Using the

midpoint of DO construct a circle with radius ½ PR.

3. With compass, measure DR & mark locations for

R & P on the circle.

Diag. bisect each other & opposite sides = in a

parallelogram.

Hexa:

720 ÷ 6 = 120

Penta: 540 ÷ 5 = 108

d = 360 – 90 – 120 – 108 = 42

e = (180 – 42)/2 = 69



Section 5.6 Proofs

Proof of the Rhombus Diagonals Angles Conjecture

Conjecture: The diagonals of a rhombus bisect the angles of the rhombus.

Given: Rhombus RHOM with diagonal HM .

Prove: HM bisects RHO and RMO .

Proof of the Rhombus Diagonals Conjecture

Conjecture: The diagonals of a rhombus are perpendicular, and

they bisect each other.

Given: Rhombus RHOM with diagonals HM and RO .

Prove: HM and RO are perpendicular bisectors of each other.

HR

MO

Same Segment.

HM HM

SSS Cong. Conj.

MRH MOH

Given

Rhombus RHOM

Def. of Rhombus

RHM OHM

RMH OMH

RH HO OM MR

CPCTC & Opposite angles

of a parallelogram are =.

HM bisects RHO and

RMO Def. of angle bisector

X

HR

MO

Given

Rhombus RHOM

Def of Rhombus

RH RM Def of Perp.

RX HM

Diagonals of a parallelogram

bisect each other.

RX XOHX XM

Same Segment

RX RX

SSS Cong. Conj.

RXH RXM

90m RXH m RXM

CPCTC and Algebra

180m RXH m RXM

Linear Pair supp.

HM and RO are

perpendicular bisectors of

each other.

Def of Perp. Bisector



5.6 EXERCISES Page 271-274 #1 – 11


a = 54, b = 36, c = 72, d = 108, e = 36, f = 144, g = 18,

h = 48, i = 48, k = 84

Sometimes

Always

Always

Sometimes

Always

Sometimes

Always

Always

Always

Sometimes: only if the

parallelogram is a rectangle.

20

37 45

90



20. On page 295. If the diagonals of a quadrilateral are

congruent and bisect each other, then

the quadrilateral is a rectangle.

Can be proved true! The proofs will

vary, but should use congruent

triangles to show the angles are 90º

each.

If the diagonals of

a parallelogram

are equal, then the

parallelogram is a

rectangle.

1. Copy LV

2. Make perp. bisector

3. Measure ½ of LV

4. Find O and E

Diag. of square are perp

bisectors and are =.

1. Copy PS

2. Make perp. at P and S.

3. Measure PE and make

arc from P then repeat

from S.

4. Draw IE.

Diag. of rectangle are = &

a rectangle has 90º angles.

1. Measured B, took

half & made bisected B.

2. Measure & draw BK.

3 Make bisected B at K.

Where the sides intersect

is A and E.

Diag. bisect opposite

angles in a Rhombus.



Ch 5 Review

Exercise #13

Kite Isosceles

trapezoid

Parallelogram Rhombus Rectangle Square

Opposite sides are

parallel

No One pair Yes Yes Yes Yes

Opposite sides are

congruent

No One pair Yes Yes Yes Yes

Opposite angles are

congruent

Non-Vertex No Yes Yes Yes Yes

Diagonals bisect each

other

No No Yes Yes Yes Yes

Diagonals are

perpendicular

Yes No No Yes No Yes

Diagonals are

congruent

No Yes No No Yes Yes

Exactly one line of

symmetry

Yes Yes No No No No

Exactly two lines of

symmetry

No No No Yes Yes Yes 4

x = 10, y = 40 x = 60 cm

a = 116, c = 64

100

x = 38 cm

y = 34, z = 51



Exercise #14 On page 305. Show all work!!

5.R Page 305 Exercise #15

a = 120, b = 60, c = 60, d = 120, e = 60, f = 30, g = 108,

m = 24, p = 84

Regular decagon; each angle =

10 2 180144

10

About a point:

360 – 144 = 216

216 ÷ 2 = 108 = b

Each part of the frame must be an isosceles

trapezoid, so consecutive angles between the

bases are supp.

180 – 108 = 72 = a

144

108

a 72

108 b

2 in

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Worksheet Chapter 5: Discovering and Proving Polygon ...€¦ · S. Stirling Page 1 of 20 Worksheet Chapter 5: Discovering and Proving Polygon Properties ... Finding one exterior