1. 18/02/2014 CH.6.6 Properties of Kites and Trapezoids

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

3. Warm Up
Solve for x.
1. x = 3x2 – 12
x = 180 3.
4. Find FE.
5 or –5 43
156

4. Classwork/Homework
Classwork
(Pages 444 to 448 ) Exercises 1, 14 to 25, 26, 27 to 32, 34 to 36, 40, 42, 47, 48, 49.
Homework
Homework Booklet
Chapter: 6.6

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

6. Vocabulary kite trapezoid base of a trapezoid leg of a trapezoid
base angle of a trapezoid
isosceles trapezoid
midsegment of a trapezoid

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

9. 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?

10. Understand the Problem
Example 1 Continued 1
Understand the Problem
The answer will be the amount of wood Lucy has left after cutting the dowel. 2
Make a Plan
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 .

11. Example 1 Continued Solve N bisects JM. Pythagorean Thm.3
N bisects JM.
Pythagorean Thm.
Pythagorean Thm.

12. Example 1 Continued
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.

13. Example 1 Continued Look Back4
Look Back
To estimate the length of the diagonal, change the side length into decimals and round , and The length of the diagonal is approximately = 32. So the wood remaining is approximately 36 – 32 = 4. So 3.6 is a reasonable answer.

14. Check It Out! Example 1
What if...? Daryl is going to make a kite by doubling all the measures in the kite. What is the total amount of binding needed to cover the edges of his kite? How many packages of binding must Daryl buy?

15. Understand the Problem
Check It Out! Example 1 Continued 1
Understand the Problem
The answer has two parts.
• the total length of binding Daryl needs
• the number of packages of binding Daryl must buy

16. Check It Out! Example 1 Continued2
Make a Plan
The diagonals of a kite are perpendicular, so the four triangles are right triangles. Use the Pythagorean Theorem and the properties of kites to find the unknown side lengths. Add these lengths to find the perimeter of the kite.

17. Check It Out! Example 1 Continued
Solve 3
Pyth. Thm.
Pyth. Thm.
perimeter of PQRS =

18. Check It Out! Example 1 Continued
Daryl needs approximately inches of binding.
One package of binding contains 2 yards, or 72 inches.
packages of binding
In order to have enough, Daryl must buy 3 packages of binding.

19. Check It Out! Example 1 Continued4
Look Back
To estimate the perimeter, change the side lengths into decimals and round.
, and The perimeter of the kite is approximately
2(54) + 2 (41) = 190. So is a reasonable answer.

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

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

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

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

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

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

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

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

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

29. A trapezoid is a quadrilateral with exactly one pair of parallel sides
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.

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

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

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

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

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

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

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

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

39. Example 4A: Applying Conditions for Isosceles Trapezoids
Find the value of a so that PQRS is isosceles.
Trap. with pair base s   isosc. trap.
S  P
mS = mP
Def. of  s
Substitute 2a2 – 54 for mS and a for mP.
2a2 – 54 = a2 + 27
Subtract a2 from both sides and add 54 to both sides.
a2 = 81
a = 9 or a = –9
Find the square root of both sides.

40. 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.
AD = BC
Substitute 12x – 11 for AD and 9x – 2 for BC.
12x – 11 = 9x – 2
Subtract 9x from both sides and add 11 to both sides.
3x = 9
x = 3
Divide both sides by 3.

41. Check It Out! Example 4
Find the value of x so that PQST is isosceles.
Trap. with pair base s   isosc. trap.
Q  S
mQ = mS
Def. of  s
Substitute 2x for mQ and 4x2 – 13 for mS.
2x = 4x2 – 13
Subtract 2x2 and add 13 to both sides.
32 = 2x2
Divide by 2 and simplify.
x = 4 or x = –4

42. 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.

44. Example 5: Finding Lengths Using Midsegments
Find EF.
Trap. Midsegment Thm.
Substitute the given values.
EF = 10.75
Solve.

45. Substitute the given values.
Check It Out! Example 5
Find EH.
Trap. Midsegment Thm. 1
16.5 = (25 + EH) 2
Substitute the given values.
Simplify.
33 = 25 + EH
Multiply both sides by 2.
13 = EH
Subtract 25 from both sides.

46. Lesson Quiz: Part I
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LHJ 3. mPKL
about in.
81°
18°

47. Lesson Quiz: Part II
Use the diagram for Items 4 and 5.
4. mWZY = 61°. Find mWXY.
5. XV = 4.6, and WY = Find VZ.
6. Find LP.
119°
9.6 18

48. Warm Up 5 –1 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
 opp. s 
Vert. s Thm.

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

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

51. 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

52. Check It Out! Example 1
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 the frame is a rectangle.

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

54. In order to apply Theorems 6-5-1 through 6-5-5, the quadrilateral must be a parallelogram.
Caution
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.

55. You can also prove that a given quadrilateral is a
rectangle, rhombus, or square by using the definitions of the special quadrilaterals.
Remember!

56. 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.

57. Example 2B: 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 square.
Step 1 Determine if EFGH is a parallelogram.
Given
Quad. with diags. bisecting each other 
EFGH is a parallelogram.

58. Example 2B Continued
Step 2 Determine if EFGH is a rectangle.
Given.
EFGH is a rectangle.
with diags.   rect.
Step 3 Determine if EFGH is a rhombus.
with one pair of cons. sides   rhombus
EFGH is a rhombus.

59. Example 2B Continued
Step 4 Determine is EFGH is a square.
Since EFGH is a rectangle and a rhombus, it has four right angles and four congruent sides. So EFGH is a square by definition.
The conclusion is valid.

60. Check It Out! Example 2
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 .

61. 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)

62. Example 3A Continued
Step 1 Graph PQRS.

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

64. Example 3A Continued
Step 3 Determine if PQRS is a rhombus.
Since , 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.

65. Example 3B: 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.
W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3)
Step 1 Graph WXYZ.

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

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

68. Check It Out! Example 3A
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)

69. Check It Out! Example 3A Continued
Step 1 Graph KLMN.

70. Check It Out! Example 3A Continued
Step 2 Find KM and LN to determine if KLMN is a rectangle.
Since , KMLN is a rectangle.

71. Check It Out! Example 3A Continued
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.

72. Check It Out! Example 3A Continued
Step 4 Determine if KLMN is a square.
Since KLMN is a rectangle and a rhombus, it has four right angles and four congruent sides. So KLMN is a square by definition.

73. Check It Out! Example 3B
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)

74. Check It Out! Example 3B Continued
Step 1 Graph PQRS.

75. Check It Out! Example 3B Continued
Step 2 Find PR and QS to determine if PQRS is a rectangle.
Since , PQRS is not a rectangle. Thus PQRS is not a square.

76. Check It Out! Example 3B Continued
Step 3 Determine if PQRS is a rhombus.
Since (–1)(1) = –1, are perpendicular and congruent. PQRS is a rhombus.

77. Lesson Quiz: Part I
1. Given that AB = BC = CD = DA, what additional information is needed to conclude that ABCD is a square?

78. Lesson Quiz: Part II
2. 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

79. Lesson Quiz: Part III
3. 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.

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

81. Vocabulary
rectangle
rhombus
square

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

83. 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.

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

85. Check It Out! Example 1a
Carpentry The rectangular gate has diagonal braces.
Find HJ.
Rect.  diags. 
HJ = GK = 48
Def. of  segs.

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

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

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

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

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

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

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

94. Check It Out! Example 2a
CDFG is a rhombus. Find CD.
CG = GF
Def. of rhombus
5a = 3a + 17
Substitute
a = 8.5
Simplify
GF = 3a + 17 = 42.5
Substitute
CD = GF
Def. of rhombus
CD = 42.5
Substitute

95. Check It Out! 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°
Def. of rhombus
b b – 40 = 180°
Substitute.
7b = 217°
Simplify.
b = 31°
Divide both sides by 7.

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

97. 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.

98. Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms.
Helpful Hint

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

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

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

102. 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.
The diagonals are congruent perpendicular bisectors of each other.

103. SV ^ TW SV = TW = 122 so, SV @ TW . slope of TW = –11 1 slope of SV =
Check It Out! Example 3
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.
SV = TW = so, TW . 1 11
slope of SV =
slope of TW = –11
SV ^ TW

104. Check It Out! Example 3 Continued
Step 1 Show that SV and TW are congruent.
Since SV = TW,

105. Check It Out! Example 3 Continued
Step 2 Show that SV and TW are perpendicular.
Since

106. Check It Out! Example 3 Continued
Step 3 Show that SV and TW bisect each other.
Since SV and TW have the same midpoint, they bisect each other.
The diagonals are congruent perpendicular bisectors of each other.

107. Example 4: Using Properties of Special Parallelograms in Proofs
Given: ABCD is a rhombus. E is the midpoint of , and F is the midpoint of .
Prove: AEFD is a parallelogram.

108. Example 4 Continued ||

109. Check It Out! Example 4
Given: PQTS is a rhombus with diagonal
Prove:

110. Check It Out! Example 4 Continued
Statements
Reasons
1. PQTS is a rhombus.
1. Given.
2. Rhombus → each
diag. bisects opp. s 2.
3. QPR  SPR
3. Def. of  bisector. 4.
4. Def. of rhombus. 5.
5. Reflex. Prop. of  6.
6. SAS 7.
7. CPCTC

111. Lesson Quiz: Part I
A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length.
1. TR CE
35 ft
29 ft

112. Lesson Quiz: Part II
PQRS is a rhombus. Find each measure.
3. QP mQRP 42
51°

113. Lesson Quiz: Part III
5. 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.

114. Lesson Quiz: Part IV
6. Given: ABCD is a rhombus.
Prove: 