1. Polygons

4. Why a hexagon?

6. Polygon comes from Greek. poly- means "many" -gon means "angle" Polygons 1.2-dimensional shapes 2.made of straight lines 3.Shape is "closed" (all the lines connect up). Polygon – classified by sides and angles

8. Polygon (straight sides) Not a Polygon (has a curve) Not a Polygon (open, not closed) ConvexConcave If there are any internal angles greater than 180° then it is concave.

9. RegularIrregular Simple Polygon (this one's a Pentagon) Complex Polygon (also a Pentagon) complex polygon intersects itself Regular - all angles are equal and all sides are equal

10. Concave Convex

11. Not regular polygons Regular polygons

12. Complex Polygon (a "star polygon", in this case, a pentagram)pentagram Concave Octagon Irregular Hexagon

13. Common polygons

14. SidesName nN-gon 3Triangle 4Quadrilateral 5Pentagon 6Hexagon 7Heptagon 8Octagon 10Decagon 12Dodecagon SidesName 9Nonagon, Enneagon 11Undecagon, Hendecagon 13Tridecagon, Triskaidecagon 14Tetradecagon, Tetrakaidecagon 15Pentadecagon, Pentakaidecagon 16Hexadecagon, Hexakaidecagon 17Heptadecagon, Heptakaidecagon 18Octadecagon, Octakaidecagon 19Enneadecagon, Enneakaidecagon 20Icosagon 30Triacontagon 40Tetracontagon 50Pentacontagon 60Hexacontagon 70Heptacontagon 80Octacontagon 90Enneacontagon 100Hectogon, Hecatontagon 1,000Chiliagon 10,000Myriagon

15. SidesName 9Nonagon, Enneagon 11Undecagon, Hendecagon 13Tridecagon, Triskaidecagon 14Tetradecagon, Tetrakaidecagon 15Pentadecagon, Pentakaidecagon 16Hexadecagon, Hexakaidecagon 17Heptadecagon, Heptakaidecagon 18Octadecagon, Octakaidecagon 19Enneadecagon, Enneakaidecagon 20Icosagon

16. 30Triacontagon 40Tetracontagon 50Pentacontagon 60Hexacontagon 70Heptacontagon 80Octacontagon 90Enneacontagon 100Hectogon, Hecatontagon 1,000Chiliagon 10,000Myriagon

17. Each segment that forms a polygon is a side of the polygon. The common endpoint of two sides is a vertex of the polygon. A segment that connects any two nonconsecutive vertices is a diagonal.

18. The sum of the angles of a quadrilateral is 360 degrees Quadrilateral

19. A four-sided polygon with two pairs of parallel sides. The sum of the angles of a parallelogram is 360 degrees Parallelogram

20. A four-sided polygon having all right angles. The sum of the angles of a rectangle is 360 degrees. Rectangle

21. A four-sided polygon having equal-length sides meeting at right angles. The sum of the angles of a square is 360 degrees Square

22. Rhombus A four-sided polygon having all four sides of equal length. The sum of the angles of a rhombus is 360 degrees.

23. Trapezoid A four-sided polygon having exactly one pair of parallel sides. The two sides that are parallel are called the bases of the trapezoid. The sum of the angles of a trapezoid is 360 degrees.

24. Triangle A three-sided polygon. The sum of the angles of a triangle is 180 degrees.

25. Equilateral Triangle or Equiangular Triangle A triangle having all three sides of equal length. The angles of an equilateral triangle all measure 60 degrees.

26. Isosceles Triangle A triangle having two sides of equal length.

27. Scalene Triangle A triangle having three sides of different lengths.

28. Acute Triangle A triangle having three acute angles.

29. Obtuse Triangle A triangle having an obtuse angle. One of the angles of the triangle measures more than 90 degrees.

30. A triangle having a right angle. One of the angles of the triangle measures 90 degrees. The side opposite the right angle is called the hypotenuse. The two sides that form the right angle are called the legs. Right Triangle

31. A right triangle has the special property that the sum of the squares of the lengths of the legs equals the square of the length of the hypotenuse. Pythagorean Theorem

32. http://www.mathplayground.com/matching_shapes.html

33. Quadrilateral

34. Identifying Polygons If it is a polygon, name it by the number of sides. polygon, hexagon

35. Identifying Polygons If it is a polygon, name it by the number of sides. polygon, heptagon

36. Identifying Polygons If it is a polygon, name it by the number of sides. not a polygon

37. Identifying Polygons If it is a polygon, name it by the number of its sides. not a polygon

38. Identifying Polygons If it is a polygon, name it by the number of its sides. polygon, nonagon

39. Identifying Polygons If it is a polygon, name it by the number of its sides. not a polygon

40. A polygon is concave if any part of a diagonal contains points in the exterior of the polygon. If no diagonal contains points in the exterior, then the polygon is convex. A regular polygon is always convex.

41. Classifying Polygons Regular or irregular? Concave or convex? irregular, convex

42. Classifying Polygons Regular or irregular? Concave or convex? irregular, concave

43. Classifying Polygons Regular or irregular? Concave or convex? regular, convex

44. Classifying Polygons Regular or irregular? Concave or convex? regular, convex

45. Classifying Polygons Regular or irregular? Concave or convex? irregular, concave

46. To find the sum of the interior angle measures of a convex polygon, draw all possible diagonals from one vertex of the polygon. This creates a set of triangles. The sum of the angle measures of all the triangles equals the sum of the angle measures of the polygon.

47. By the Triangle Sum Theorem, the sum of the interior angle measures of a triangle is 180°.

50. Example 3A: Finding Interior Angle Measures and Sums in Polygons Find the sum of the interior angle measures of a convex heptagon. (n – 2)180° (7 – 2)180° 900° Polygon  Sum Thm. A heptagon has 7 sides, so substitute 7 for n. Simplify.

51. Example 3B: Finding Interior Angle Measures and Sums in Polygons Find the measure of each interior angle of a regular 16-gon. Step 1 Find the sum of the interior angle measures. Step 2 Find the measure of one interior angle. (n – 2)180° (16 – 2)180° = 2520° Polygon  Sum Thm. Substitute 16 for n and simplify. The int.  s are , so divide by 16.

52. Example 3C: Finding Interior Angle Measures and Sums in Polygons Find the measure of each interior angle of pentagon ABCDE. (5 – 2)180° = 540° Polygon  Sum Thm. m  A + m  B + m  C + m  D + m  E = 540° Polygon  Sum Thm. 35c + 18c + 32c + 32c + 18c = 540Substitute. 135c = 540Combine like terms. c = 4Divide both sides by 135.

53. Example 3C Continued m  A = 35(4°) = 140° m  B = m  E = 18(4°) = 72° m  C = m  D = 32(4°) = 128°

54. Example 4 Find the sum of the interior angle measures of a convex 15-gon. (n – 2)180° (15 – 2)180° 2340° Polygon  Sum Thm. A 15-gon has 15 sides, so substitute 15 for n. Simplify.

55. Find the measure of each interior angle of a regular decagon. Step 1 Find the sum of the interior angle measures. Step 2 Find the measure of one interior angle. Example 5 (n – 2)180° (10 – 2)180° = 1440° Polygon  Sum Thm. Substitute 10 for n and simplify. The int.  s are , so divide by 10.

56. In the polygons below, an exterior angle has been measured at each vertex. Notice that in each case, the sum of the exterior angle measures is 360°.

58. Example 6: Finding Interior Angle Measures and Sums in Polygons Find the measure of each exterior angle of a regular 20-gon. A 20-gon has 20 sides and 20 vertices. sum of ext.  s = 360°. A regular 20-gon has 20  ext.  s, so divide the sum by 20. The measure of each exterior angle of a regular 20-gon is 18°. Polygon  Sum Thm. measure of one ext.  =

59. Example 7: Finding Interior Angle Measures and Sums in Polygons Find the value of b in polygon FGHJKL. 15b° + 18b° + 33b° + 16b° + 10b° + 28b° = 360° Polygon Ext.  Sum Thm. 120b = 360Combine like terms. b = 3Divide both sides by 120.

60. Find the measure of each exterior angle of a regular dodecagon. Example 8 A dodecagon has 12 sides and 12 vertices. sum of ext.  s = 360°. A regular dodecagon has 12  ext.  s, so divide the sum by 12. The measure of each exterior angle of a regular dodecagon is 30°. Polygon  Sum Thm. measure of one ext.

61. Example 9 Find the value of r in polygon JKLM. 4r° + 7r° + 5r° + 8r° = 360° Polygon Ext.  Sum Thm. 24r = 360Combine like terms. r = 15Divide both sides by 24.

62. Example 10: Art Application Ann is making paper stars for party decorations. What is the measure of  1?  1 is an exterior angle of a regular pentagon. By the Polygon Exterior Angle Sum Theorem, the sum of the exterior angles measures is 360°. A regular pentagon has 5  ext. , so divide the sum by 5.

63. Find the value of each variable. 1. x2. y3. z 218 4

64. Prove and apply properties of parallelograms. Use properties of parallelograms to solve problems. Objectives

65. A quadrilateral with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you use the symbol.

67. If a quadrilateral is a parallelogram its opposite sides are congruent its opposite angles are congruent its consecutive angles are supplementary

68. If a quadrilateral is a parallelogram diagonals bisect each other each diagonal splits the parallelogram into two congruent triangles.

69. http://www.mathwarehouse.com/geometry/quadrilaterals/parallelograms/ http://www.ies.co.jp/math/products/geo1/applets/para/para.html

70. Example 1: Properties of Parallelograms Def. of  segs. Substitute 74 for DE. In CDEF, DE = 74 mm, DG = 31 mm, and m  FCD = 42°. Find CF. CF = DE CF = 74 mm  opp. sides 

71. Substitute 42 for m  FCD. Example 2: Properties of Parallelograms Subtract 42 from both sides. m  EFC + m  FCD = 180° m  EFC + 42 = 180 m  EFC = 138° In CDEF, DE = 74 mm, DG = 31 mm, and m  FCD = 42°. Find m  EFC.  cons.  s supp.

72. Example 3: Properties of Parallelograms Substitute 31 for DG. Simplify. In CDEF, DE = 74 mm, DG = 31 mm, and m  FCD = 42°. Find DF. DF = 2DG DF = 2(31) DF = 62  diags. bisect each other.

73. Example 4 In KLMN, LM = 28 in., LN = 26 in., and m  LKN = 74°. Find KN. Def. of  segs. Substitute 28 for DE. LM = KN LM = 28 in.  opp. sides 

74. Example 5 Def. of  angles. In KLMN, LM = 28 in., LN = 26 in., and m  LKN = 74°. Find m  NML. Def. of   s. Substitute 74° for m  LKN.  NML   LKN m  NML = m  LKN m  NML = 74°  opp.  s 

75. Example 6 In KLMN, LM = 28 in., LN = 26 in., and m  LKN = 74°. Find LO. Substitute 26 for LN. Simplify. LN = 2LO 26 = 2LO LO = 13 in.  diags. bisect each other.

76. Example 7: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find YZ. Def. of  segs. Substitute the given values. Subtract 6a from both sides and add 4 to both sides. Divide both sides by 2. YZ = XW 8a – 4 = 6a + 10 2a = 14 a = 7 YZ = 8a – 4 = 8(7) – 4 = 52  opp.  s 

77. Example 8: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find m  Z. Divide by 27. Add 9 to both sides. Combine like terms. Substitute the given values. m  Z + m  W = 180° (9b + 2) + (18b – 11) = 180 27b – 9 = 180 27b = 189 b = 7 m  Z = (9b + 2)° = [9(7) + 2]° = 65°  cons.  s supp.

78. Example 9 EFGH is a parallelogram. Find JG. Substitute. Simplify. EJ = JG 3w = w + 8 2w = 8 w = 4 Divide both sides by 2. JG = w + 8 = 4 + 8 = 12 Def. of  segs.  diags. bisect each other.

79. Example 10 EFGH is a parallelogram. Find FH. Substitute. Simplify. FJ = JH 4z – 9 = 2z 2z = 9 z = 4.5 Divide both sides by 2. Def. of  segs. FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18  diags. bisect each other.

80. When you are drawing a figure in the coordinate plane, the name ABCD gives the order of the vertices.

81. Example 11 In PNWL, NW = 12, PM = 9, and m  WLP = 144°. Find each measure. 1. PW 2. m  PNW 18144°

82. Example 12 QRST is a parallelogram. Find each measure. 2. TQ 3. m  T 28 71°

83. Example 13 Three vertices of ABCD are A (2, –6), B (–1, 2), and C(5, 3). Find the coordinates of vertex D. (8, –5)

84. Practice Justify each statement. 1. 2. Evaluate each expression for x = 12 and y = 8.5. 3. 2x + 7 4. 16x – 9 5. (8y + 5)° Reflex Prop. of  Conv. of Alt. Int.  s Thm. 31 183 73°

85. Practice Solve for x. 1. 16x – 3 = 12x + 13 2. 2x – 4 = 90 ABCD is a parallelogram. Find each measure. 3. CD4. m  C 4 47 14 104°

86. Prove and apply properties of rectangles, rhombuses, and squares. Use properties of rectangles, rhombuses, and squares to solve problems. Objectives

87. A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

88. Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

89. Example 1: Craft Application A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM. Rect.  diags.  Def. of  segs. Substitute and simplify. KM = JL = 86  diags. bisect each other

90. Example 1a Carpentry The rectangular gate has diagonal braces. Find HJ. Def. of  segs. Rect.  diags.  HJ = GK = 48

91. Example 1b Carpentry The rectangular gate has diagonal braces. Find HK. Def. of  segs. Rect.  diags.  JL = LG JG = 2JL = 2(30.8) = 61.6 Substitute and simplify. Rect.  diagonals bisect each other

92. A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

94. Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

95. Example 2A: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find TV. Def. of rhombus Substitute given values. Subtract 3b from both sides and add 9 to both sides. Divide both sides by 10. WV = XT 13b – 9 = 3b + 4 10b = 13 b = 1.3

96. Example 2A Continued Def. of rhombus Substitute 3b + 4 for XT. Substitute 1.3 for b and simplify. TV = XT TV = 3b + 4 TV = 3(1.3) + 4 = 7.9

97. Rhombus  diag.  Example 2B: Using Properties of Rhombuses to Find Measures TVWX is a rhombus. Find m  VTZ. Substitute 14a + 20 for m  VTZ. Subtract 20 from both sides and divide both sides by 14. m  VZT = 90° 14a + 20 = 90° a = 5a = 5

98. Example 2B Continued Rhombus  each diag. bisects opp.  s Substitute 5a – 5 for m  VTZ. Substitute 5 for a and simplify. m  VTZ = m  ZTX m  VTZ = (5a – 5)° m  VTZ = [5(5) – 5)]° = 20°

99. Example 2a CDFG is a rhombus. Find CD. Def. of rhombus Substitute Simplify Substitute Def. of rhombus Substitute CG = GF 5a = 3a + 17 a = 8.5 GF = 3a + 17 = 42.5 CD = GF CD = 42.5

100. Example 2b CDFG is a rhombus. Find the measure. m  GCH if m  GCD = (b + 3)° and m  CDF = (6b – 40)° m  GCD + m  CDF = 180° b + 3 + 6b – 40 = 180° 7b = 217° b = 31° Def. of rhombus Substitute. Simplify. Divide both sides by 7.

101. Example 2b Continued m  GCH + m  HCD = m  GCD 2m  GCH = m  GCD Rhombus  each diag. bisects opp.  s 2m  GCH = (b + 3) 2m  GCH = (31 + 3) m  GCH = 17° Substitute. Simplify and divide both sides by 2.

102. A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

103. Example 3: Verifying Properties of Squares Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

104. Example 3 Continued Step 1 Show that EG and FH are congruent. Since EG = FH,

105. Example 3 Continued Step 2 Show that EG and FH are perpendicular. Since,

106. The diagonals are congruent perpendicular bisectors of each other. Example 3 Continued Step 3 Show that EG and FH are bisect each other. Since EG and FH have the same midpoint, they bisect each other.

107. Example 4 The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3), and W(1, –9). Show that the diagonals of square STVW are congruent perpendicular bisectors of each other. 1 11 slope of SV = slope of TW = –11 SV  TW SV = TW = 122 so, SV  TW.

108. Step 1 Show that SV and TW are congruent. Example 4 Continued Since SV = TW,

109. Step 2 Show that SV and TW are perpendicular. Example 4 Continued Since

110. The diagonals are congruent perpendicular bisectors of each other. Step 3 Show that SV and TW bisect each other. Since SV and TW have the same midpoint, they bisect each other. Example 4 Continued

111. Example 5: Using Properties of Special Parallelograms in Proofs Prove: AEFD is a parallelogram. Given: ABCD is a rhombus. E is the midpoint of, and F is the midpoint of.

112. Example 5 Continued ||

113. Example 6 Given: PQTS is a rhombus with diagonal Prove:

114. Example 6 Continued StatementsReasons 1. PQTS is a rhombus. 1. Given. 2. Rhombus → each diag. bisects opp.  s 3.  QPR   SPR3. Def. of  bisector. 4. Def. of rhombus. 5. Reflex. Prop. of  6. SAS 7. CPCTC 2. 4. 5. 7. 6.

115. Example 7 A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length. 1. TR2. CE 35 ft29 ft

116. Example 8 PQRS is a rhombus. Find each measure. 3. QP4. m  QRP 4251°

117. Example 9 The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

118. Example 10  Given: ABCD is a rhombus. Prove:

119. Practice 1. Find AB for A (–3, 5) and B (1, 2). 2. Find the slope of JK for J(–4, 4) and K(3, –3). ABCD is a parallelogram. Justify each statement. 3.  ABC   CDA 4.  AEB   CED 5 –1 Vert.  s Thm.  opp.  s 

120. Prove that a given quadrilateral is a rectangle, rhombus, or square. Objective

121. When you are given a parallelogram with certain properties, you can use the theorems below to determine whether the parallelogram is a rectangle.

122. Example 1: Carpentry Application A manufacture builds a mold for a desktop so that,, and m  ABC = 90°. Why must ABCD be a rectangle? Both pairs of opposites sides of ABCD are congruent, so ABCD is a. Since m  ABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.

123. Example 1a A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle? Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one  of WXYZ is a right . If one angle is a right , then by Theorem 6-5-1 the frame is a rectangle.

124. Below are some conditions you can use to determine whether a parallelogram is a rhombus.

125. To prove that a given quadrilateral is a square, it is sufficient to show that the figure is both a rectangle and a rhombus. You will explain why this is true in Exercise 43. You can also prove that a given quadrilateral is a rectangle, rhombus, or square by using the definitions of the special quadrilaterals. In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram.

126. Example 2A: Applying Conditions for Special Parallelograms Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: Conclusion: EFGH is a rhombus. The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

127. Example 2b Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given:  ABC is a right angle. Conclusion: ABCD is a rectangle. The conclusion is not valid. By Theorem 6-5-1, if one angle of a parallelogram is a right angle, then the parallelogram is a rectangle. To apply this theorem, you need to know that ABCD is a parallelogram.

128. Example 3A: Identifying Special Parallelograms in the Coordinate Plane Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)

129. Example 3A Continued Step 1 Graph PQRS.

130. Step 2 Find PR and QS to determine is PQRS is a rectangle. Example 3A Continued Since, the diagonals are congruent. PQRS is a rectangle.

131. Step 3 Determine if PQRS is a rhombus. Step 4 Determine if PQRS is a square. Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition. Example 3A Continued Since, PQRS is a rhombus.

132. Example 3B: Identifying Special Parallelograms in the Coordinate Plane W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3) Step 1 Graph WXYZ. Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.

133. Step 2 Find WY and XZ to determine is WXYZ is a rectangle. Thus WXYZ is not a square. Example 3B Continued Since, WXYZ is not a rectangle.

134. Step 3 Determine if WXYZ is a rhombus. Example 3B Continued Since (–1)(1) = –1,, PQRS is a rhombus.

135. Example4 Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)

136. Step 1 Graph KLMN. Example 4 Continued

137. Step 2 Find KM and LN to determine is KLMN is a rectangle. Since, KMLN is a rectangle.

138. Step 3 Determine if KLMN is a rhombus. Since the product of the slopes is –1, the two lines are perpendicular. KLMN is a rhombus. Example 4 Continued

139. Step 4 Determine if PQRS is a square. Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition. Example 4 Continued

140. Example 5 Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply. P(–4, 6), Q(2, 5), R(3, –1), S(–3, 0)

141. Example 5 Continued Step 1 Graph PQRS.

142. Step 2 Find PR and QS to determine is PQRS is a rectangle. Example 5 Continued Since, PQRS is not a rectangle. Thus PQRS is not a square.

143. Step 3 Determine if KLMN is a rhombus. Example 5 Continued Since (–1)(1) = –1, are perpendicular and congruent. KLMN is a rhombus.

144. Example 6 Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

145. Example 7 Determine if the conclusion is valid. If not, tell what additional information is needed to make it valid. Given: PQRS and PQNM are parallelograms. Conclusion: MNRS is a rhombus. valid

146. Example 8 Use the diagonals to determine whether a parallelogram with vertices A(2, 7), B(7, 9), C(5, 4), and D(0, 2) is a rectangle, rhombus, or square. Give all the names that apply. AC ≠ BD, so ABCD is not a rect. or a square. The slope of AC = –1, and the slope of BD = 1, so AC  BD. ABCD is a rhombus.

147. Practice Solve for x. 1. x 2 + 38 = 3x 2 – 12 2. 137 + x = 180 3. 4. Find FE. 5 or –5 43 156

148. Use properties of kites to solve problems. Use properties of trapezoids to solve problems. Objectives

149. A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

151. Example 1: Problem-Solving Application Lucy is framing a kite with wooden dowels. She uses two dowels that measure 18 cm, one dowel that measures 30 cm, and two dowels that measure 27 cm. To complete the kite, she needs a dowel to place along. She has a dowel that is 36 cm long. About how much wood will she have left after cutting the last dowel?

152. Example 1 Continued The answer will be the amount of wood Lucy has left after cutting the dowel. The diagonals of a kite are perpendicular, so the four triangles are right triangles. Let N represent the intersection of the diagonals. Use the Pythagorean Theorem and the properties of kites to find, and. Add these lengths to find the length of.

153. Solve N bisects JM. Pythagorean Thm. Example 1 Continued

154. Lucy needs to cut the dowel to be 32.4 cm long. The amount of wood that will remain after the cut is, 36 – 32.4  3.6 cm Lucy will have 3.6 cm of wood left over after the cut. Example 1 Continued

155. Kite  cons. sides  Example 2A: Using Properties of Kites In kite ABCD, m  DAB = 54°, and m  CDF = 52°. Find m  BCD. ∆BCD is isos. 2  sides  isos. ∆ isos. ∆  base  s  Def. of   s Polygon  Sum Thm.  CBF   CDF m  CBF = m  CDF m  BCD + m  CBF + m  CDF = 180°

156. Example 2A Continued Substitute m  CDF for m  CBF. Substitute 52 for m  CBF. Subtract 104 from both sides. m  BCD + m  CBF + m  CDF = 180° m  BCD + 52° + 52° = 180° m  BCD = 76° m  BCD + m  CBF + m  CDF = 180°

157. Kite  one pair opp.  s  Example 2B: Using Properties of Kites Def. of   s Polygon  Sum Thm. In kite ABCD, m  DAB = 54°, and m  CDF = 52°. Find m  ABC.  ADC   ABC m  ADC = m  ABC m  ABC + m  BCD + m  ADC + m  DAB = 360° m  ABC + m  BCD + m  ABC + m  DAB = 360° Substitute m  ABC for m  ADC.

158. Example 2B Continued Substitute. Simplify. m  ABC + m  BCD + m  ABC + m  DAB = 360° m  ABC + 76° + m  ABC + 54° = 360° 2m  ABC = 230° m  ABC = 115° Solve.

159. Kite  one pair opp.  s  Example 2C: Using Properties of Kites Def. of   s  Add. Post. Substitute. Solve. In kite ABCD, m  DAB = 54°, and m  CDF = 52°. Find m  FDA.  CDA   ABC m  CDA = m  ABC m  CDF + m  FDA = m  ABC 52° + m  FDA = 115° m  FDA = 63°

160. Example 2a In kite PQRS, m  PQR = 78°, and m  TRS = 59°. Find m  QRT. Kite  cons. sides  ∆PQR is isos. 2  sides  isos. ∆ isos. ∆  base  s  Def. of   s  RPQ   PRQ m  QPT = m  QRT

161. Example 2a Continued Polygon  Sum Thm. Substitute 78 for m  PQR. m  PQR + m  QRP + m  QPR = 180° 78° + m  QRT + m  QPT = 180° Substitute. 78° + m  QRT + m  QRT = 180° 78° + 2m  QRT = 180° 2m  QRT = 102° m  QRT = 51° Substitute. Subtract 78 from both sides. Divide by 2.

162. Example 2b In kite PQRS, m  PQR = 78°, and m  TRS = 59°. Find m  QPS. Kite  one pair opp.  s   Add. Post. Substitute.  QPS   QRS m  QPS = m  QRT + m  TRS m  QPS = m  QRT + 59° m  QPS = 51° + 59° m  QPS = 110°

163. Example 2c Polygon  Sum Thm. Def. of   s Substitute. Simplify. In kite PQRS, m  PQR = 78°, and m  TRS = 59°. Find each m  PSR. m  SPT + m  TRS + m  RSP = 180° m  SPT = m  TRS m  TRS + m  TRS + m  RSP = 180° 59° + 59° + m  RSP = 180° m  RSP = 62°

164. A trapezoid is a quadrilateral with exactly one pair of parallel sides. Each of the parallel sides is called a base. The nonparallel sides are called legs. Base angles of a trapezoid are two consecutive angles whose common side is a base.

165. If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.

167. Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.”

168. Isos.  trap.  s base  Example 3A: Using Properties of Isosceles Trapezoids Find m  A. Same-Side Int.  s Thm. Substitute 100 for m  C. Subtract 100 from both sides. Def. of   s Substitute 80 for m  B m  C + m  B = 180° 100 + m  B = 180 m  B = 80°  A   B m  A = m  B m  A = 80°

169. Example 3B: Using Properties of Isosceles Trapezoids KB = 21.9m and MF = 32.7. Find FB. Isos.  trap.  s base  Def. of  segs. Substitute 32.7 for FM. Seg. Add. Post. Substitute 21.9 for KB and 32.7 for KJ. Subtract 21.9 from both sides. KJ = FM KJ = 32.7 KB + BJ = KJ 21.9 + BJ = 32.7 BJ = 10.8

170. Example 3B Continued Same line. Isos. trap.   s base  Isos. trap.  legs  SAS CPCTC Vert.   s   KFJ   MJF  BKF   BMJ  FBK   JBM ∆FKJ  ∆JMF

171. Isos. trap.  legs  AAS CPCTC Def. of  segs. Substitute 10.8 for JB. Example 3B Continued ∆FBK  ∆JBM FB = JB FB = 10.8

172. Isos.  trap.  s base  Same-Side Int.  s Thm. Def. of   s Substitute 49 for m  E. m  F + m  E = 180°  E   H m  E = m  H m  F = 131° m  F + 49° = 180° Simplify. Check It Out! Example 3a Find m  F.

173. Check It Out! Example 3b JN = 10.6, and NL = 14.8. Find KM. Def. of  segs. Segment Add Postulate Substitute. Substitute and simplify. Isos.  trap.  s base  KM = JL JL = JN + NL KM = JN + NL KM = 10.6 + 14.8 = 25.4

174. Example 4A: Applying Conditions for Isosceles Trapezoids Find the value of a so that PQRS is isosceles. a = 9 or a = –9 Trap. with pair base  s   isosc. trap. Def. of   s Substitute 2a 2 – 54 for m  S and a 2 + 27 for m  P. Subtract a 2 from both sides and add 54 to both sides. Find the square root of both sides. S  PS  P m  S = m  P 2a 2 – 54 = a 2 + 27 a 2 = 81

175. Example 4B: Applying Conditions for Isosceles Trapezoids AD = 12x – 11, and BC = 9x – 2. Find the value of x so that ABCD is isosceles. Diags.   isosc. trap. Def. of  segs. Substitute 12x – 11 for AD and 9x – 2 for BC. Subtract 9x from both sides and add 11 to both sides. Divide both sides by 3. AD = BC 12x – 11 = 9x – 2 3x = 9 x = 3

176. Example 5 Find the value of x so that PQST is isosceles. Subtract 2x 2 and add 13 to both sides. x = 4 or x = –4 Divide by 2 and simplify. Trap. with pair base  s   isosc. trap. Q  SQ  S Def. of   s Substitute 2x 2 + 19 for m  Q and 4x 2 – 13 for m  S. m  Q = m  S 2x 2 + 19 = 4x 2 – 13 32 = 2x 2

177. The midsegment of a trapezoid is the segment whose endpoints are the midpoints of the legs. In Lesson 5-1, you studied the Triangle Midsegment Theorem. The Trapezoid Midsegment Theorem is similar to it.

179. Example 6: Finding Lengths Using Midsegments Find EF. Trap. Midsegment Thm. Substitute the given values. Solve. EF = 10.75

180. Example 7 Find EH. Trap. Midsegment Thm. Substitute the given values. Simplify. Multiply both sides by 2. 33 = 25 + EH Subtract 25 from both sides. 13 = EH 1 16.5 = ( 25 + EH ) 2

181. Practice 1. Erin is making a kite based on the pattern below. About how much binding does Erin need to cover the edges of the kite? In kite HJKL, m  KLP = 72°, and m  HJP = 49.5°. Find each measure. 2. m  LHJ3. m  PKL about 191.2 in. 81°18°

182. Practice Use the diagram for Items 4 and 5. 4. m  WZY = 61°. Find m  WXY. 5. XV = 4.6, and WY = 14.2. Find VZ. 6. Find LP. 119° 9.6 18