1. Properties of Kites 8-5,6 and Trapezoids Warm Up Lesson Presentation
Lesson Quiz
Holt Geometry

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

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

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

6. 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.

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

8. 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°

9. 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.

10. 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.

11. 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.

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

14. 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

15. Example 3B: Using Properties of Isosceles Trapezoids
KB = 21.9m and MF = 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.

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

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

18. 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.

19. 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.

21. Review: Triangle Midsegment Theorem
If a segment joins the midpoints of two sides of a triangle, then the segment is parallel to the third side, and is half of its length.
Midsegment x 2x

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

23. 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.
8 = EH
Subtract 25 from both sides.

24. 8-6 Identify Special Quadrilaterals Warm Up Lesson Presentation
Lesson Quiz
Holt Geometry

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

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

27. Example 1: Carpentry Application
A manufacturer 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

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

29. 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.

30. 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.

31. 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.

32. 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.

33. 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 .

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

35. 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