Geometry Geometry Chapter 6Chapter 6

QuadrilateralsKites and Trapezoids

Warm UpSolve for x.

1. x2 + 38 = 3x2 – 12

2. 137 + x = 180

3.

4. Find FE.

5 or –5

43

156

Use properties of kites to solve problems.

Use properties of trapezoids to solve problems.

Your Math Goal Today is…

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

Vocabulary

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

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?

Example 1 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 .

Solve33

N bisects JM.

Pythagorean Thm.

Pythagorean Thm.

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.

Example 1 Continued

Look Back44

Example 1 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.

In Your Notes

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?

In Your Notes

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

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.

In Your Notes

Solve33

Pyth. Thm.

Pyth. Thm.

In Your Notes

perimeter of PQRS =

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.

In Your Notes

packages of binding

Look Back44

In Your Notes

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.

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°

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°

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.

Example 2B Continued

Substitute.

Simplify.

mABC + mBCD + mABC + mDAB = 360°

mABC + 76° + mABC + 54° = 360°

2mABC = 230°

mABC = 115° Solve.

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°

In Your Notes

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

In Your Notes

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.

Subtract 78 from both sides.

Divide by 2.

In Your Notes

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

Kite one pair opp. s

Add. Post.

Substitute.

Substitute.

QPS QRS

mQPS = mQRT + mTRS

mQPS = mQRT + 59°

mQPS = 51° + 59°

mQPS = 110°

In Your Notes

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°

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.

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

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

Reading Math

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°

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

Example 3B Continued

Same line.

Isos. trap. s base

Isos. trap. legs

SAS

CPCTC

Vert. s

KFJ MJF

BKF BMJ

FBK JBM

∆FKJ ∆JMF

Isos. trap. legs

AAS

CPCTC

Def. of segs.

Substitute 10.8 for JB.

Example 3B Continued

∆FBK ∆JBM

FB = JB

FB = 10.8

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.

In Your Notes

Find mF.

In Your Notes

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

Example 4A: 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

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.

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

In Your Notes

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

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.

Example 5: Finding Lengths Using Midsegments

Find EF.

Trap. Midsegment Thm.

Substitute the given values.

Solve.EF = 10.75

In Your Notes

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