1. Properties of Kites 6-6 and Trapezoids Warm Up Lesson Presentation
Lesson Quiz
Holt McDougal Geometry

2. Warm Up
Solve for x.
1. x = 3x2 – 12
x = 180 3.
4. Find FE.

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

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

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

7. 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?

8. 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 .

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

10. 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.

11. Example 1 Continued 4
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.

12. 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?

13. 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

14. 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.

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

16. 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.

17. 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.

18. 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.

19. 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°

20. 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°

21. 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.

22. 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.

23. 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

24. 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.

25. 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°

26. 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°

27. 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.

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

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

31. 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

32. 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.

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

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

35. 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.

36. 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.

37. 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.

38. 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.

39. 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

40. 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.

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

43. 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.

44. 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

45. 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.