Worksheet Chapter 5: Discovering and Proving Polygon ... 5 Worksheets Key Name _____ S. Stirling Page 4 of 20 5.1 EXERCISES Page 259-260 #3 – 11, 13, 14, 12, 16. Show how you are

Ch 5 Worksheets Key Name ___________________________

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

m∠ADQ = 72.26°

m∠UAD = 86.28°

m∠QUA = 59.70°

m∠DQU = 141.77°

T

R

I

m∠RIT = 60.21°

m∠TRI = 81.14°

m∠RT I = 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.

m∠DAB+m∠ABC+m∠BCD+m∠CDA = 360.00°

m∠DAB = 114°m∠ABC = 77°m∠BCD = 113°

m∠CDA = 56°

Quadrilateral ABCD

D

C

BA

m∠IEF+m∠EFG+m∠FGH+m∠GHI+m∠HIE = 540.00°

m∠IEF = 71°m∠EFG = 156°m∠FGH = 43°

m∠GHI = 157°m∠HIE = 112°

Pentagon EFGHI

ON

M

LK

J

m∠HAB+m∠EBC+m∠FCD+m∠GDA = 360.00°

m∠HAB = 67°

m∠EBC = 103°

m∠FCD = 84°

m∠GDA = 106°

D

C

B

A

HG

F

E

m∠FAB+m∠GBC+m∠HCD+m∠IDE+m∠JEA = 360°m∠JEA = 61°m∠IDE = 56°m∠HCD = 104°m∠GBC = 80°m∠FAB = 59°

A

G

H

I

J

B

C

D

EF

B

I

J

K

M

G

C

D

E

F

A

H

m∠MFA = 72°m∠KEF = 54°m∠JDE = 33°m∠ICD = 73°m∠HBC = 66°m∠GAB = 63°

m∠GAB+m∠HBC+m∠ICD+m∠JDE+m∠KEF+m∠MFA = 360°

67

103 84

106

80

104

59 61

56

63

73

66

33

54

72

77

113 56

114

I

H

G

F

E 43 71

157

156

112

m∠OJK+m∠JKL+m∠KLM+m∠LMN+m∠MNO+m∠NOJ = 720.00°

m∠OJK = 112°m∠JKL = 159°m∠KLM = 108°

m∠LMN = 105°m∠MNO = 140°m∠NOJ = 96°

Hexagon JKLMNO

96

105

108

112

140

159

m∠WPQ = 119°m∠PQR = 130°m∠QRS = 154°

m∠RST = 132°m∠ST U = 131°m∠TUV = 137°

m∠UVW = 131°m∠VWP = 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

nn

− =

− = −

=

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)180

1358

− =

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

nn

− =

− = −

=

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? 360

725

=

hexagon? 360

606

=

3. How many sides regular polygon, each exterior angle 24º? 360

24

360 24

15

nn

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

( )136 48 42

2p = + =

( )124 13

248 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

PA

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 2 180

16830

−=

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 HM bisects RHO∠ and RMO∠

Def. of angle bisector

X

HR

MO

Given

Rhombus RHOM

Def of Rhombus

RH RM= Def of Perp.

RO HM⊥ Diagonals of a parallelogram bisect each other.

HX XM= RX XO=

Same Segment

RX RX=

SSS Cong. Conj.

RXH RXM∆ ≅ ∆

90m RXH m RXM∠ = ∠ = °

CPCTC and Algebra

180m RXH m RXM∠ + ∠ =

Linear Pair

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. 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 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 2 180

14410

−=

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

Documents

Worksheet Chapter 5: Discovering and Proving Polygon ... 5 Worksheets Key Name _____ S. Stirling Page 4 of 20 5.1 EXERCISES Page 259-260 #3 – 11, 13, 14, 12, 16. Show how you are