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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