6-6

Holt Geometry

6-6 Properties of Kites and Trapezoids6-6 Properties of Kites and Trapezoids

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Geometry

6-6 Properties of Kites and Trapezoids

Warm UpSolve for x.

1. 137 + x = 180

2.

3. Find FE.

43

156

Holt Geometry


Use properties of kites to solve problems.

Use properties of trapezoids to solve problems.

Objectives

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kitetrapezoidbase of a trapezoidleg of a trapezoidbase angle of a trapezoidisosceles trapezoidmidsegment of a trapezoid

Vocabulary

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

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Holt Geometry


Class Example: 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?

Holt Geometry


Class Example Continued

11 Understand the Problem

The answer will be the amount of wood Lucy has left after cutting the dowel.

22 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 .

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Solve33

N bisects JM.

Pythagorean Thm.

Pythagorean Thm.


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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.


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Look Back44

Class Examples Continued

To estimate the length of the diagonal, change the side length into decimals and round. , and . The length of the diagonal is approximately 10 + 22 = 32. So the wood remaining is approximately 36 – 32 = 4. So 3.6 is a reasonable answer.

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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?

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Check It Out! Example 1 Continued

11 Understand the Problem

The answer has two parts.• the total length of binding Daryl needs• the number of packages of binding Daryl must buy

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22 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.


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Solve33

Pyth. Thm.

Pyth. Thm.


perimeter of PQRS =

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Daryl needs approximately 191.3 inches of binding.One package of binding contains 2 yards, or 72 inches.

In order to have enough, Daryl must buy 3 packages of binding.


packages of binding

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Look Back44


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 191.3 is a reasonable answer.

Holt Geometry


Kite cons. sides

Example 2: 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


Example 2 Continued

Substitute mCDF for mCBF.

Substitute 52 for mCBF.

Subtract 104 from both sides.


mBCD + 52° + 52° = 180°

mBCD = 76°


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Kite one pair opp. s


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.

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Example 3 Continued

Substitute.

Simplify.

mABC + mBCD + mABC + mDAB = 360°

mABC + 76° + mABC + 54° = 360°

2mABC = 230°

mABC = 115° Solve.

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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



In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT.

Kite cons. sides

∆PQR is isos. 2 sides isos. ∆

isos. ∆ base s

Def. of s

RPQ PRQ

mQPT = mQRT

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Polygon Sum Thm.

Substitute 78 for mPQR.

mPQR + mQRP + mQPR = 180°

78° + mQRT + mQPT = 180°

Substitute. 78° + mQRT + mQRT = 180°

78° + 2mQRT = 180°

2mQRT = 102°

mQRT = 51°

Substitute.


Divide by 2.

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In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQPS.


Add. Post.

Substitute.

Substitute.

QPS QRS

mQPS = mQRT + mTRS

mQPS = mQRT + 59°

mQPS = 51° + 59°

mQPS = 110°

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Polygon Sum Thm.

Def. of s

Substitute.

Substitute.Simplify.

In kite PQRS, mPQR = 78°, and mTRS = 59°. Find each mPSR.

mSPT + mTRS + mRSP = 180°

mSPT = mTRS

mTRS + mTRS + mRSP = 180°

59° + 59° + mRSP = 180°

mRSP = 62°

Holt Geometry


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.

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

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Holt Geometry


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

Reading Math

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Isos. trap. s base

Example 8: Using Properties of Isosceles Trapezoids

Find mA.

Same-Side Int. s Thm.

Substitute 100 for mC.


Def. of s

Substitute 80 for mB

mC + mB = 180°

100 + mB = 180

mB = 80°

A B

mA = mB

mA = 80°

Holt Geometry


Example 9: 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


Example 9 Continued

Same line.

Isos. trap. s base

Isos. trap. legs

SAS

CPCTC

Vert. s

KFJ MJF

BKF BMJ

FBK JBM

∆FKJ ∆JMF

Holt Geometry


Isos. trap. legs

AAS

CPCTC

Def. of segs.

Substitute 10.8 for JB.

Example 9 Continued

∆FBK ∆JBM

FB = JB

FB = 10.8

Holt Geometry


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.


Find mF.

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JN = 10.6, and NL = 14.8. Find KM.

Def. of segs.

Segment Add Postulate

Substitute.

Substitute and simplify.

Isos. trap. s base

KM = JL

JL = JN + NL

KM = JN + NL

KM = 10.6 + 14.8 = 25.4

Holt Geometry


Example 12: Applying Conditions for Isosceles Trapezoids

Find the value of a so that PQRS is isosceles.

a = 9 or a = –9

Trap. with pair base s isosc. trap.

Def. of s

Substitute 2a2 – 54 for mS and a2

+ 27 for mP.

Subtract a2 from both sides and add 54 to both sides.

Find the square root of both sides.

S P

mS = mP

2a2 – 54 = a2 + 27

a2 = 81

Holt Geometry


Example 13: 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.

Substitute 12x – 11 for AD and 9x – 2 for BC.

Subtract 9x from both sides and add 11 to both sides.

Divide both sides by 3.

AD = BC

12x – 11 = 9x – 2

3x = 9

x = 3

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Find the value of x so that PQST is isosceles.

Subtract 2x2 and add 13 to both sides.

x = 4 or x = –4 Divide by 2 and simplify.

Trap. with pair base s isosc. trap.Q S

Def. of s

Substitute 2x2 + 19 for mQ and 4x2 – 13 for mS.

mQ = mS

2x2 + 19 = 4x2 – 13

32 = 2x2

Holt Geometry


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.

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Holt Geometry


Example 15: Finding Lengths Using Midsegments

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.EF = 10.75

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Find EH.

Trap. Midsegment Thm.

Substitute the given values.

Simplify.

Multiply both sides by 2.33 = 25 + EH

Subtract 25 from both sides.13 = EH

116.5 = (25 + EH)2

Holt Geometry


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, mKLP = 72°,and mHJP = 49.5°. Find eachmeasure.

2. mLHJ 3. mPKL

about 191.2 in.

81° 18°

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Lesson Quiz: Part II

Use the diagram for Items 4 and 5.

4. mWZY = 61°. Find mWXY.

5. XV = 4.6, and WY = 14.2. Find VZ.

6. Find LP.

119°

9.6

18

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