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