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Ch 5 Worksheets Key Name ___________________________ 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? 360 108 D R 72 99 Q 81 120 142 38 60 I U 60 86 120 94 A T 39 141 m ∠DQU = 141.77 ° m ∠RT I = 38.64° m ∠QUA = 59.70 ° m ∠TRI = 81.14° m ∠UAD = 86.28 ° m ∠RIT = 60.21° m ∠ADQ = 72.26 ° Do you see a possible pattern? Various conclusions S. Stirling Page 1 of 20 Ch 5 Worksheets 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. Pentagon EFGHI m ∠GHI = 157° m ∠IEF = 71° m ∠EFG = 156° m ∠HIE = 112° m ∠FGH = 43° m ∠IEF+m ∠EFG+m ∠FGH+m ∠GHI+m ∠HIE = 540.00° Quadrilateral ABCD m ∠BCD = 113 ° m ∠ABC = 77° m ∠DAB = 114° m ∠CDA = 56° m ∠DAB+m ∠ABC+m ∠BCD+m ∠CDA = 360.00° B F A 77 114 E 113 156 157 C 56 G 43 71 112 D H I Hexagon JKLMNO Oc tagon PQRSTUVW 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° m ∠OJK+m∠JKL+m ∠KLM+m ∠LMN+m ∠MNO+m ∠NOJ = 720.00 ° 154 132 Q 108 159 S R L K m ∠RST = 132° m ∠UVW = 131 ° m ∠ST U = 131° m ∠VWP = 147 ° m ∠TUV = 137° 130 131 T 119 137 U J 112 105 P M 131 147 V W 140 96 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 84 103 B I m∠HAB = 67° D m∠EBC = 103° 106 67 m∠FCD = 84° m∠GDA = 106° C H A 73 G H D 66 m∠HAB+m∠EBC+m∠FCD+m∠GDA = 360.00° B 33 H C G G 59 E A 61 m∠FAB = 59° m∠GBC = 80° m∠HCD = 104° m∠IDE = 56° J m∠JEA = 61° m∠FAB+m ∠GBC+m∠HCD+m ∠IDE+m∠JEA = 360° F S. Stirling 54 A 80 56 63 K 72 I m ∠GAB = 63 ° m ∠HBC = 66 ° m ∠ICD = 73 ° E 104 D B J F m ∠JDE = 33 ° m ∠KEF = 54 ° m ∠MFA = 72 ° M m ∠GAB+m ∠HBC+m ∠ICD+m ∠JDE+m ∠KEF+m ∠MFA = 360° Page 2 of 20 Ch 5 Worksheets 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? 108 72 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 or 180 − n n Or You can find the measure of each exterior angle of an equiangular n-gon by using the formula: 360 n 180n 360 180n − 360 − = n n n More practice: One exterior angle = 360 ÷ 6 = 60 What is the relationship between one interior and one exterior angle? 60 120 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 = = 120 6 6 Same results? Yes Which method is easier? Finding one exterior angle first, because sum is always 360. S. Stirling Page 3 of 20 Ch 5 Worksheets Key Name ___________________________ 5.1 EXERCISES Page 259-260 #3 – 11, 13, 14, 12, 16. Show how you are finding your answers! e = (5 – 2)180/5 = 540/5 = 108 110 90 72 112 a = 360 – 90 – 76 – 72 = 122 108 (6 – 2)180 = 720 b = (720 – 448)/2 = 136 72 180 – 108 = 72 f = 180 – 2 * 72 = 180 – 144 = 36 360 – 108 – 130 = 122 122 142 102 44 120 Triangle: d = 180 – 44 – 30 = 106 Quad: c = 360 – 252 = 108 Penta: g = (540 – 225)/3 = 105 Quad: h = 360 – 278 = 82 60 120 j = 720/6 = 120 k = 360 – 322 = 38 various 102 + 82 + 98 + 76 = 358 Interior sum should be 360. various 135 + 3 * 131 + 26 = 554 Interior sum should be 540. 180 − 18 sides 360 = 160 n 26 102 360 – 200 = 160 60 82 76 98 82 131 140 ( 9 − 2 )180 = 140 131 131 9 360 = 156 n 360 − = −24 n n = 15 180 − ( n − 2 )180 = 2700 n − 2 = 15 n = 17 S. Stirling Page 4 of 20 Ch 5 Worksheets Key Name ___________________________ 5.1 Page 260 Exercise #12 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 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 67.5 112.5 112.5 135 The angles of the trapezoid measure 67.5 and 112.5. Each angle of the octagon: (8 − 2)180 = 135 8 Around a point: 360 – 135 = 225 225 ÷ 2 = 112.5 Angles between the bases are supplementary. 180 – 112.5 = 67.5 S. Stirling Page 5 of 20 Ch 5 Worksheets Key 5.1 Page 260-261 Exercise #15, 18, 20, 21 Name ___________________________ 360 = 160 n 360 − = −20 n n = 18 180 − So twelfth century. 120 D Draw a right isosceles triangle. The base angles are both 45, so they are complementary. S. Stirling Page 6 of 20 Ch 5 Worksheets Key Name ___________________________ 5.2 EXERCISES Page 263-265 #1 – 10 1. Sum exterior angles decagon? 2. An exterior angle of equiangular pentagon? 360 360 = 72 5 hexagon? 360 = 60 6 108 b = 45 40 3. How many sides regular polygon, each exterior angle 24º? 4. How many sides polygon, sum interior angles 7380º? 360 = 24 n 360 = 24n n = 15 (n − 2)180 = 7380 n − 2 = 41 n = 43 c = 51 1 3 3 7 d = 115 5 7 112 Exterior angle sum is 360. a = 360 – 252 = 108 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 39 30 106 108 108 135 135 30 55 44 136 129 97 162 44 51 102 129 83 97 83 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 Triangle: a = 180 – 56 – 94 = 30 b = 30 ||, alt. int. angles = Triangle: c = 180 – 44 – 30 = 106 d = 180 – 44 = 136 S. Stirling 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 Page 7 of 20 Ch 5 Worksheets Key Name ___________________________ 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, Kite KITE K E I ∠E ≅ ∠I . KT = KT T Same Segment. Given KE = KI and ET = IT ∆KET ≅ ∆KIT ∠E ≅ ∠I SSS Cong. Conj. CPCTC or Def. cong. triangles Def. of Kite 5.3 Page 272 Exercise #9 Proof of Kite Angle Bisector Conjecture The vertex angles of a kite are bisected by a diagonal. ∠1 ≅ ∠2 ∠3 ≅ ∠4 BN ≅ BN Same Segment. BE ≅ BY YN ≅ EN Given or Def. of Kite S. Stirling ∆BYN ≅ ∆BEN CPCTC or Def. cong. triangles SSS Cong. Conj. BN bisects ∠YBE BN bisects ∠YNE Def. angle bisector Page 8 of 20 Ch 5 Worksheets Key Name ___________________________ Proof of the Kite Diagonals Conjecture Conjecture: The diagonals of a kite are perpendicular. D A DB and AC . Prove: The diagonals are perpendicular. DB ⊥ AC . Given: Kite ABCD with diagonals I B Kite ABCD Given ∠DAI ≅ ∠BAI C m∠DIA + m∠BIA = 180 Diag. bisect vertex angles. AD = AB Def. of Kite AI = AI Same Segment. Linear Pair Conj. ∆DAI ≅ ∆BAI SAS Cong. Conj. ∠DIA ≅ ∠BIA CPCTC m∠DIA = m∠BIA = 90° DB ⊥ AC Algebra Def. of Perpendicular D 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. A I Given: Kite ABCD with diagonals DB and AC . Prove: AC is the perpendicular bisector of DB . B C Kite ABCD Given DB ⊥ AC ∠DAI ≅ ∠BAI Diag. of kite are Perpendicular Diag. kite bisect vertex angles. DI = IB AD = AB Def. of Kite AI = AI Same Segment. S. Stirling ∆DAI ≅ ∆BAI CPCTC SAS Cong. Conj. AC is the perpendicular bisector of DB Def. of perp. bisector. Page 9 of 20 Ch 5 Worksheets Key Name ___________________________ 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. TR = TR PT = RA Same Segment. Given Isosceles trapezoid TRAP Given m∠PTR = m∠TRA ∆PTR ≅ ∆ART TA = RP SAS Cong. Conj. CPCTC Isosceles Trap. base angles = 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.) 52 21 146 64 cm 128 Non-vertex angles =. y = 146 x = 360 – 47 – 146 * 2 = 21 20 21 146 12 Isos. so base angles =. y = 128 Consecutive angles supplementary. x = 180 – 128 = 52 Perimeter: 20 * 2 + 12 * 2 = 64 15 Perimeter: 85 = 37 +18 + 2x 85 = 55 + 2x 30 = 2x 15 = x 72 99 61 38 cm 90 81 15 S. Stirling 15 Small Right triangle: x = 180 – 90 – 18 = 72 Large Right triangle: x = 180 – 90 – 29 = 61 29 Perimeter: 164 = y + 2(y +12) + (y – 12) 164 = y + 2 y + 24 + y – 12 164 = 4 y + 12 152 = 4 y 38 = y Page 10 of 20 99 128 52 Ch 5 Worksheets Key Name ___________________________ Vertex angle 3.0 cm 45 1.6 cm 90 48 42 y = 180 – 90 – 48 = 42 30 30 S. Stirling 30 w = 180 – 2 * 30 = 120 Page 11 of 20 Ch 5 Worksheets Key 5.3 Page 274 Exercise #19 Name ___________________________ a = 80, b = 20, c = 160, d = 20, e = 80, f = 80, g = 110, h = 70, m = 110, n = 100 5.4 EXERCISES Page 271-274 #1 – 7 three; one 65 60 140 28 Perimeter TOP = 8 + 2*10 = 28 y = 180 – 40 = 140 ||, corr. angles = . x = 60 Corresponding angles of congruent triangles. 129 35 73 23 42 cm Corresponding sides of congruent triangles. Perimeter = 6 + 8 + 9 = 23 S. Stirling m = 180 – 51 = 129 p= 1 ( 36 + 48 ) = 42 2 1 (13 + q ) 2 48 = 13 + q 24 = 35 = q Page 12 of 20 Ch 5 Worksheets Key Name ___________________________ Section 5.5 Proofs Proof of the Parallelogram Opposite Angles Conjecture L Conjecture: The opposite angles of a parallelogram are congruent. R 2 Given: Parallelogram PARL with diagonal AL . Prove: ∠PAR ≅ ∠PLR and ∠R ≅ ∠P . Parallelogram PARL 1 m∠2 = m∠3 PA LR 4 If ||, AIA cong. Def. of Parallelogram 3 Given P A m∠1 = m∠4 LP AR m∠2 + m∠1 = m∠3 + m∠4 If ||, AIA cong. Addition Def. of Parallelogram m∠PLR = m∠PAR m∠R = m∠P Substitution If 2 angles of one triangle = 2 angles of another, the 3rd angles are =. P Proof of the Parallelogram Opposite Sides Conjecture Conjecture: The opposite sides of a parallelogram are congruent. Given: Parallelogram PARL with diagonal Prove: PL ≅ RA Parallelogram PARL and PR . PA ≅ LR . PA LR Def. of Parallelogram L ∠APR ≅ ∠LRP R If ||, AIA cong. Given ∠L ≅ ∠A Opposite angles of Parallelogram = PR = PR Same Segment. S. Stirling ∆PAR ≅ ∆RLP AAS Cong. Conj. PL ≅ RA and PA ≅ LR CPCTC Page 13 of 20 A Ch 5 Worksheets Key Name ___________________________ 5.5 Page 284 Exercise #13 Proof of the Parallelogram Diagonals Conj. The diagonals of a parallelogram bisect each other. given ∠EAL ≅ ∠ALN ∆ETA ≅ ∆NTL def parallelogram LT ≅ TA EA ≅ LN EN & LA 5.5 EXERCISES Page 271-274 #1 – 6, #7 – 8 construct, #16, 18 34 cm 132 27 cm 48 a = 180 – 48 = 132 80 16 in 14 in 63 78 63 m x – 3 = 17 x = 20 Perim = 18 + 24 + 21 = 63 S. Stirling 20 + 3 = 23 Perim = 2*17 + 2*23 = 80 Page 14 of 20 bisect eachother. Ch 5 Worksheets Key Name ___________________________ 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. 8. Construct parallelogram DROP, given side D DR and diagonals DO PR . R O D P and R 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 d = 360 – 90 – 120 – 108 = 42 e = (180 – 42)/2 = 69 30 stones make a 30gon; each angle = ( 30 − 2 )180 = 168 30 Penta: 540 ÷ 5 = 108 S. Stirling a = 120, b = 108, c = 90, d = 42, e = 69 About a point: 360 – 168 = 192 192 ÷ 2 = 96 = b Consecutive angles supp. 180 – 96 = 84 = a Page 15 of 20 Ch 5 Worksheets Key Section 5.6 Proofs Name ___________________________ Proof of the Rhombus Diagonals Angles Conjecture Conjecture: The diagonals of a rhombus bisect the angles of the rhombus. HM . ∠RMO . Given: Rhombus RHOM with diagonal Prove: HM bisects ∠RHO and H R O M Rhombus RHOM Given RH = HO = OM = MR ∆MRH ≅ ∆MOH Def. of Rhombus SSS Cong. Conj. HM = HM ∠RHM ≅ ∠OHM ∠RMH ≅ ∠OMH Same Segment. HM bisects ∠RHO and ∠RMO CPCTC Def. of angle bisector Proof of the Rhombus Diagonals Conjecture Conjecture: The diagonals of a rhombus are perpendicular, and they bisect each other. Given: Rhombus RHOM with diagonals Prove: HM and HM and H R X RO . M RO are perpendicular bisectors of each other. and RO are perpendicular bisectors of each other. HM Rhombus RHOM HX = XM RX = XO Def of Perp. Bisector Given RH = RM Def of Rhombus Diagonals of a parallelogram bisect each other. ∆RXH ≅ ∆RXM Def of Perp. m∠RXH = m∠RXM = 90° SSS Cong. Conj. RX = RX Same Segment RO ⊥ HM CPCTC and Algebra m∠RXH + m∠RXM = 180 Linear Pair S. Stirling Page 16 of 20 O Ch 5 Worksheets Key Name ___________________________ 5.6 EXERCISES Page 271-274 #1 – 11 Sometimes Always Always Sometimes Always Sometimes Always Always Always Sometimes: only if the parallelogram is a rectangle. 45 37 90 20 5.6 Page 297 Exercise #28 a = 54, b = 36, c = 72, d = 108, e = 36, f = 144, g = 18, h = 48, i = 48, k = 84 S. Stirling Page 17 of 20 Ch 5 Worksheets Key Name ___________________________ 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. 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. 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. 20. If the diagonals of a parallelogram are equal, then the parallelogram is a rectangle. S. Stirling 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. Page 18 of 20 Ch 5 Worksheets Key Ch 5 Review Name ___________________________ x = 10, y = 40 x = 60 cm a = 116, c = 64 x = 38 cm y = 34, z = 51 100 Exercise #13 Opposite sides are parallel Opposite sides are congruent Opposite angles are congruent Diagonals bisect each other Diagonals are perpendicular Diagonals are congruent Exactly one line of symmetry Exactly two lines of symmetry S. Stirling Kite Isosceles trapezoid Parallelogram Rhombus Rectangle Square No One pair Yes Yes Yes Yes No One pair Yes Yes Yes Yes Non-Vertex No Yes Yes Yes Yes No No Yes Yes Yes Yes Yes No No Yes No Yes No Yes No No Yes Yes Yes Yes No No No No No No No Yes Yes Yes 4 Page 19 of 20 Ch 5 Worksheets Key Name ___________________________ Exercise #14 Regular decagon; each angle = (10 − 2 )180 = 144 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 72 a 2 in 108 108 b 144 5.R Page 305 Exercise #15 a = 120, b = 60, c = 60, d = 120, e = 60, f = 30, g = 108, m = 24, p = 84 S. Stirling Page 20 of 20