Ch 5 Polygon Notebook Key

Transcription

1 hapter 5: iscovering and Proving Polygon Properties Lesson 5.1 Polygon Sum onjecture & Lesson 5.2 xterior ngles of a Polygon Warm up: efinition: xterior angle is an angle that forms a linear pair with one of the interior angles of a polygon. Measure the interior angles of QU to the nearest degree and put the measures into the diagram. raw one exterior angle at each vertex of QU. 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? ach interior angle forms a linear pair with an exterior angle re 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 at the right. ompare the different angle sums with the angle sums for the quadrilateral. re any of the angles equal? No What is the sum of the interior angles? 180 What is the sum of the exterior angles? 360 Q U m = m = m = m QU = m QU = m U = m Q = o you see a possible pattern? S. Stirling Page 1 of 11

2 Page nvestigation: s there a Polygon Sum Formula? Steps 1-2: eview your work from page 1 and examine the diagrams below. Step 3-4: omplete the sum of the interior angles column and drawing diagonals on the next page. Quadrilateral m = 113 m = 77 m = 56 m = 114 m +m +m +m = Pentagon FGH m F = 71 m GH = 157 m FG = 156 m H = 112 m FGH = 43 m F+m FG+m FGH+m GH+m H = F G H Hexagon JKLMNO m OJK = 112 m JKL = 159 m KLM = 108 m LMN = 105 m MNO = 140 m NOJ = 96 m OJK+m JKL+m KLM+m LMN+m MNO+m NOJ = J K L Octagon PQSUVW m WPQ = 119 m PQ = 130 m QS = 154 Q m S = 132 m SU = 131 m UV = 137 S m UVW = 131 m VWP = 147 M P U O N W V Page nvestigation: s there an xterior ngle Sum? Steps 1-5: eview your work from page 1 and examine the diagrams below. One exterior angle is drawn at each vertex. omplete the sum of the exterior angles column on the next page. F m H = 67 m = 103 m F = 84 m G = 106 H G H m H+m +m F+m G = H G G F K J m G = 63 m H = 66 m = 73 m J = 33 m KF = 54 m MF = 72 F m F = 59 m G = 80 m H = 104 m = 56 J m J = 61 m F+m G+m H+m +m J = 360 M m G+m H+m +m J+m KF+m MF = 360 S. Stirling Page 2 of 11

3 (nvestigation 5.1 Step 5: raw all possible diagonals from one vertex, which divides each polygon into triangles. Use these to develop a formula for the Polygon Sum onjecture. Quadrilateral Pentagon Hexagon iagonal forms 2 triangles, so 2 (180) = 360 iagonals form _3_ triangles: 3 (180) = 540 iagonals form _4_ triangles: 4 (180) = 720 Octagon ecagon iagonals form _6_ triangles: 6 (180) = 1080 iagonals form _8_ triangles: 8 (180) = 1440 nvestigation 5.1 and 5.2 Summary Use the results from Lesson 5.1 and 5.2 to fill in the table at the left. (he last column of the table should be completed after 5.2 nvestigation.) Number of sides of a polygon Sum of measures of interior angles n (n-2) Sum of measures of exterior angles (one at a vertex) Polygon Sum onjecture. he sum of the measures of the n angles of an n- gon is (n 2) 180 or 180n 360 xterior ngle Sum onjecture. he sum of the measures of a set of exterior angles of an n-gon is 360. S. Stirling Page 3 of 11

4 Page nvestigation: s there an xterior ngle 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. ry 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 = = 108 One exterior angle = = 72 What is the relationship between one interior and one exterior angle? Supplementary, = 180 quiangular Polygon onjecture You can find the measure of each interior angle of an equiangular n-gon by using either of these formulas: ( n 2)180 n or 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 = = 60 What is the relationship between one interior and one exterior angle? Supplementary Use this relationship to find the measure of one interior angle = 120 Use the formula to find the measure of one interior angle. (6 2) = = Same results? Yes Which method is easier? Finding one exterior angle first, because sum is always 360. S. Stirling Page 4 of 11

5 Lesson 5.3 Kite and rapezoid Properties efinition of kite quadrilateral with exactly two distinct pairs of congruent consecutive sides. Label vocab in the drawing.: K Measure then compare the opposite angles of the kite. Which pair will be congruent? K vertex angles (of a kite) he angles between the pairs of congruent sides. nonvertex angles (of a kite) he two angles between consecutive noncongruent sides of a kite. Kite ngles onjecture (o proof on h 5 WS page 2.) he nonvertex angles of a kite are congruent. K dditional measures: Label the diagram with the measures to help you write the conjectures. m = 90 he vertex angles: he diagonals of the kite (compared to eachother): m = 19 = 1.57 cm = 1.57 cm m = 19 = 1.68 cm = 4.57 cm m = 43 m = 43 he angles of the kite: m = 38 m = 118 he nonvertex angles: m = 86 m = 118 m = 71 m = 47 m = 47 m = 71 Kite ngle isector onjecture (o proof on h 5 WS page 2.) he vertex angles of a kite are bisected by a diagonal. and Kite iagonals onjecture (o proof on h 5 WS page 3.) he diagonals of a kite are perpendicular. Kite iagonal isector onjecture (o proof on h 5 WS page 3.) he diagonal connecting the vertex angles of a kite is the perpendicular bisector of the other diagonal. is the perpendicular bisector of. S. Stirling Page 5 of 11

6 efinition of trapezoid quadrilateral with exactly one pair of parallel sides. Label vocab in the drawing: efinition of isosceles trapezoid trapezoid whose legs are congruent. Measure the angles of the trapezoids below. Label the diagram with the measures to help you write the conjectures. m = 59 Q bases (of a trapezoid) he two parallel sides. base angles (of a trapezoid) pair of angles with a base of the trapezoid as a common side. legs are the two nonparallel sides. m = 121 m = 139 m = 41 S m Q = 46 m S = 134 rapezoid onsecutive ngles onj. he consecutive angles between the bases of a trapezoid are supplementary. m + m = 180 m + m = 180 sosceles rapezoid [ase ngles] onj. he base angles of an isosceles trapezoid are congruent. Q and S Measure diagonals of the trapezoids below. sosceles rapezoid iagonals onjecture (o proof on h 5 WS page 4.) he diagonals of an isosceles trapezoid are congruent. S O while S P P O S. Stirling Page 6 of 11

7 Lesson 5.4 Properties of Midsegments Page nvestigation 1: riangle Midsegment Properties Midsegment (of a triangle) is the line segment connecting the midpoints of the two sides. Steps 1 3: Z Y hree Midsegments onjecture he three midsegments of a triangle divide it into four congruent triangles. Steps 4 5: (eview orresponding ngles onjecture for parallel lines. he F shape!) lso label the drawing below. riangle Midsegment onjecture midsegment of a triangle is parallel to the third side and half the length of the third side. Y Y f, then corr. angles congruent. Y and Y Y = 1 2 Page nvestigation 2: rapezoid Midsegment Properties. Midsegment (of a trapezoid) is the line segment connecting the midpoints of the two nonparallel sides. Steps 1 8: f you do not have tracing paper, you may measure instead. Label angles with measures!! M P ll parallels: P MN ll congruent angles: f, then corresponding angles congruent. PMN and P NM and NM and MN N rapezoid Midsegment onjecture he 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: P+ MN 2 1 MN = P+ 2 = O ( ) S. Stirling Page 7 of 11

8 Lesson 5.5 Properties of Parallelograms Page nvestigation: Four Parallelogram Properties efinition of a Parallelogram: quadrilateral with two pairs of opposite sides parallel. Steps 1 4: ngles! Parallelogram Opposite ngles L onjecture (o proof on h 5 WS page 5.) he opposite angles of a parallelogram are congruent. J M K Parallelogram onsecutive ngles onjecture he consecutive angles of a parallelogram are supplementary. So to find all angles of a parallelogram: Make opposite angles equal Subtract an angle from 180 to get a consecutive angle. Steps 5 7: Side and diagonal lengths! P = 2.8 PL = 7.5 P = 6.2 L = 9.4 P ngle measures: f m L 22 = then m M = = 158 m K = 22 m J = 158 Parallelogram Opposite Sides onjecture (o proof on h 5 WS page 5.) he opposite sides of a parallelogram are congruent. Parallelogram iagonals onjecture (o proof on h 5 WS page 6.) he diagonals of a parallelogram bisect each other. L Side lengths: P= L PL = P L 1 P = = P = L = = L = S. Stirling Page 8 of 11

9 Lesson 5.5 Properties of Parallelograms: nvestigation nvestigate: f, then the quadrilateral is a parallelogram. re the converses of the previous conjectures true? Yes So what do you need to know to know that a quadrilateral is a parallelogram? oth pairs of opposite angles congruent onsecutive angles supplementary oth pairs of opposite sides congruent iagonals bisect each other Would the following conjecture work? ry to prove it deductively. f one pair of sides of a quadrilateral is both parallel and congruent, then the quadrilateral is a parallelogram. Proof Given: Quadrilateral PL P Prove: PL is a parallelogram various L, P L and diagonal P. P P L P LP Given f, cong. L P = P Same Segment. P L Given ΔP ΔLP SS ong. onj. PL is a parallelogram PL P oth pairs of opposite sides are congruent, then parallelogram S. Stirling Page 9 of 11

10 5.6 Properties of Special Parallelograms efinition of a hombus: quadrilateral with all sides congruent. Page 291 nvestigation 1: What an You raw with the ouble-dged Straightedge? Steps 1 3: omplete in the space below. Use your ruler! ouble-dged Straightedge onjecture f 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 nvestigation 2: o hombus iagonals Have Special Properties? Steps 1 3: hombus iagonals [Lengths] onjecture (o proof on h 5 WS page 7.) M H iagonal elationships: O HM m H = or 90 ecause a rhombus is a parallelogram: O M H he diagonals of a rhombus are perpendicular and they bisect each other (because diagonals of a parallelogram bisect each other.) O hombus iagonals ngles onjecture (o proof on h 5 WS page 6.) he diagonals of a rhombus bisect the angles of the rhombus. ngle elationships: HM bisects HO and MO ecause a rhombus is a parallelogram and opposite angles are congruent: H OH M OM S. Stirling Page 10 of 11

11 efinition of a ectangle: quadrilateral with all angles congruent. Page nvestigation 3: o ectangle iagonals Have Special Properties? Steps 1 2: ectangle iagonals onjecture he diagonals of a rectangle are congruent and bisect each other (because diagonals of a parallelogram bisect each other.) iagonal elationships: nd because a rectangle is a parallelogram and the diagonals bisect each other: s a result, you get many pairs of congruent triangles (and some congruent isosceles triangles). efinition of a Square: quadrilateral with all angles and sides congruent U Q S Square iagonals onjecture he diagonals of a square are congruent (a rectangle), perpendicular (a rhombus) and bisect each other (a parallelogram). Unique elationships? an you determine any measures without measuring? Since diagonals of a rhombus bisect opposite angles, m U= 45 n fact, all the acute angles are 45. You get many pairs of congruent right isosceles (45:45:90) triangles: ΔQU ΔUS ΔSQ Δ SQU as well as ΔU ΔS ΔSQ Δ QU S. Stirling Page 11 of 11