Holt Geometry
8-5,6Properties of Kites and Trapezoids8-5,6 Properties of Kites and Trapezoids
Holt Geometry
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Holt Geometry
8-5,6Properties of Kites and Trapezoids
Use properties of kites to solve problems.
Use properties of trapezoids to solve problems.
Objectives
Holt Geometry
8-5,6Properties of Kites and Trapezoids
kitetrapezoidbase of a trapezoidleg of a trapezoidbase angle of a trapezoidisosceles trapezoidmidsegment of a trapezoid
Vocabulary
Holt Geometry
8-5,6Properties of Kites and Trapezoids
A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Kite cons. sides
Example 2A: Using Properties of Kites
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD.
∆BCD is isos. 2 sides isos. ∆
isos. ∆ base s
Def. of s
Polygon Sum Thm.
CBF CDF
mCBF = mCDF
mBCD + mCBF + mCDF = 180°
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 2A Continued
Substitute mCDF for mCBF.
Substitute 52 for mCBF.
Subtract 104 from both sides.
mBCD + mCBF + mCDF = 180°
mBCD + 52° + 52° = 180°
mBCD = 76°
mBCD + mCBF + mCDF = 180°
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Kite one pair opp. s
Example 2B: Using Properties of Kites
Def. of s Polygon Sum Thm.
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC.
ADC ABC
mADC = mABC
mABC + mBCD + mADC + mDAB = 360°
mABC + mBCD + mABC + mDAB = 360°
Substitute mABC for mADC.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 2B Continued
Substitute.
Simplify.
mABC + mBCD + mABC + mDAB = 360°
mABC + 76° + mABC + 54° = 360°
2mABC = 230°
mABC = 115° Solve.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Kite one pair opp. s
Example 2C: Using Properties of Kites
Def. of s
Add. Post.
Substitute.
Solve.
In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mFDA.
CDA ABC
mCDA = mABC
mCDF + mFDA = mABC
52° + mFDA = 115°
mFDA = 63°
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
If the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Isos. trap. s base
Example 3A: Using Properties of Isosceles Trapezoids
Find mA.
Same-Side Int. s Thm.
Substitute 100 for mC.
Subtract 100 from both sides.
Def. of s
Substitute 80 for mB
mC + mB = 180°
100 + mB = 180
mB = 80°
A B
mA = mB
mA = 80°
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 3B Continued
Same line.
Isos. trap. s base
Isos. trap. legs
SAS
CPCTC
Vert. s
KFJ MJF
BKF BMJ
FBK JBM
∆FKJ ∆JMF
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Isos. trap. legs
AAS
CPCTC
Def. of segs.
Substitute 10.8 for JB.
Example 3B Continued
∆FBK ∆JBM
FB = JB
FB = 10.8
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Isos. trap. s base
Same-Side Int. s Thm.
Def. of s
Substitute 49 for mE.
mF + mE = 180°
E H
mE = mH
mF = 131°
mF + 49° = 180°
Simplify.
Check It Out! Example 3a
Find mF.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Review: Triangle Midsegment TheoremIf 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
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 5: Finding Lengths Using Midsegments
Find EF.
Trap. Midsegment Thm.
Substitute the given values.
Solve.EF = 10.75
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Check It Out! Example 5
Find EH.
Trap. Midsegment Thm.
Substitute the given values.
Simplify.
Multiply both sides by 2.33 = 25 + EH
Subtract 25 from both sides.8 = EH
116.5 = (25 + EH)2
Holt Geometry
8-5,6Properties of Kites and Trapezoids8-6 Identify Special Quadrilaterals
Holt Geometry
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Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Prove that a given quadrilateral is a rectangle, rhombus, or square.
Objective
Holt Geometry
8-5,6Properties of Kites and Trapezoids
When you are given a parallelogram with certainproperties, you can use the theorems below to determine whether the parallelogram is a rectangle.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 1: Carpentry Application
A manufacturer builds a mold for a desktop so that , , and mABC = 90°. Why must ABCD be a rectangle?
Both pairs of opposites sides of ABCD are congruent, so ABCD is a . Since mABC = 90°, one angle ABCD is a right angle. ABCD is a rectangle by Theorem 6-5-1.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Below are some conditions you can use to determine whether a parallelogram is a rhombus.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 2B: Applying Conditions for Special ParallelogramsDetermine 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
EFGH is a parallelogram. Quad. with diags. bisecting each other
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Example 2B Continued
Step 2 Determine if EFGH is a rectangle.
Given.
EFGH is a rectangle.
Step 3 Determine if EFGH is a rhombus.
EFGH is a rhombus.
with diags. rect.
with one pair of cons. sides rhombus
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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.
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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 .
Holt Geometry
8-5,6Properties of Kites and Trapezoids
Lesson Quiz: Part I
1. Given that AB = BC = CD = DA, what additional
information is needed to conclude that ABCD is a
square?
Holt Geometry
8-5,6Properties of Kites and Trapezoids
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