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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 mDQU = 141.77 mRT I = 38.64 mQUA = 59.70 mTRI = 81.14 mUAD = 86.28 mRIT = 60.21 mADQ = 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 anglesPentagon column and drawing diagonals on the next page. EFGHI mIEFPentagon = 71 EFGHImGHI = 157 mGHI = = 157 mIEF = 71 mHIE 112 mEFG = 156 mEFG = 156 mHIE = 112 mFGH = 43 mFGH = 43 mIEF+mEFG+mFGH+mGHI+mHIE = 540.00 = 540.00 mIEF+mEFG+mFGH+mGHI+mHIE Quadrilateral ABCD mBCD = 113 mABC = 77 mDAB = 114 mCDA = 56 mDAB+mABC+mBCD+mCDA = 360.00 F B A 77 114 E 113 156 71 H I mLMN = 105 mJKL = 159 mMNO = 140 mLMN = 105 mOJK = 112 mJKL = 159 mMNO = 140 mNOJ = 96 mKLM = 108 mOJK = 112 Hexagon JKLMNO Oc tagon PQRSTUVW mWPQ = 119 mPQR = 130 mQRS = 154 mNOJ = 96 mKLM = 108 mRST = 132 mUVW = 131 mST U = 131 mVWP = 147 mTUV = 137 mOJK+mJKL+mKLM+mLMN+mMNO+mNOJ = 720.00 mOJK+mJKL+mKLM+mLMN+mMNO+mNOJ = 720.00 L K J 154 132 Q 108 159 J S R L K G 157 112 Hexagon JKLMNO D 43 H I C 56 F G E 130 131 T 119 137 U M 112 105 P M 131 147 N O 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 mHAB = 67 D mEBC = 103 106 67 mFCD = 84 mGDA = 106 C G 33 G 56 59 A 61 E mGBC = 80 mJEA = 61 63 m HBC = 66 m ICD = 73 54 A K 72 I F m JDE = 33 m KEF = 54 m MFA = 72 M J mFAB+mGBC+mHCD+mIDE+mJEA = 360 S. Stirling m GAB = 63 m GAB+m HBC+m ICD+m JDE+m KEF+m MFA = 360 mHCD = 104 mIDE = 56 J E 104 80 B D 66 B D F 73 G H mHAB+mEBC+mFCD+mGDA = 360.00 H mFAB = 59 C H A 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: 360 (n 2)180 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 S. Stirling Triangle: a = 180 – 56 – 94 = 30 b = 30 ||, alt. int. angles = Triangle: c = 180 – 44 – 30 = 106 d = 180 – 44 = 136 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 E I . K E KT KT I 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 mDIA mBIA 180 Diag. bisect vertex angles. AD AB Def. of Kite AI AI Linear Pair Conj. DAI BAI SAS Cong. Conj. Same Segment. DIA BIA CPCTC mDIA mBIA 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. Given: Kite ABCD with diagonals Prove: DB and AC is the perpendicular bisector of A I AC . 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 mPTR mTRA 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 35 q 24 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 m2 m3 PA LR 4 If ||, AIA cong. Def. of Parallelogram 3 Given m1 m4 LP AR P A m2 m1 m3 m4 If ||, AIA cong. Addition Def. of Parallelogram mPLR mPAR mR mP 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: PR . PL RA and PA LR . Parallelogram PARL 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 RH HO OM MR Def. of Rhombus Given MRH MOH HM HM Same Segment. RHM OHM RMH OMH SSS Cong. Conj. CPCTC & Opposite angles of a parallelogram are =. HM bisects RHO and RMO 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 RX XO HX XM Def of Perp. Bisector Given RH RM Def of Rhombus Diagonals of a parallelogram bisect each other. RXH RXM SSS Cong. Conj. Def of Perp. mRXH mRXM 90 RX RX Same Segment RX HM CPCTC and Algebra mRXH mRXM 180 Linear Pair supp. 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. On page 295. 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 On page 305. Show all work!! 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