18/02/2014 CH.6.6 Properties of Kites and Trapezoids

18/02/2014 CH.6.6Properties of Kites

and Trapezoids

Warm UpSolve for x.

1. 16x – 3 = 12x + 13

2. 2x – 4 = 90

ABCD is a parallelogram. Find each measure.

3. CD 4. mC

4

47

14 104°

Warm UpSolve for x.

1. x2 + 38 = 3x2 – 12

2. 137 + x = 180

3.

4. Find FE.

5 or –5

43

156

Classwork/Homework

Classwork (Pages 444 to 448 ) Exercises

1, 14 to 25, 26, 27 to 32, 34 to 36, 40, 42, 47,

48, 49.

HomeworkHomework Booklet Chapter: 6.6

Use properties of kites to solve problems.

Use properties of trapezoids to solve problems.

Objectives

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

1 Understand the Problem

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

2 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 .

Solve3

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 Back4

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.

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?

Check It Out! Example 1 Continued

1 Understand the Problem

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

2 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.


Solve3

Pyth. Thm.

Pyth. Thm.


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.


packages of binding

Look Back4


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 mCDF.

Subtract 104 from both sides.

mBCD + mCDF + 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°

Check It Out! Example 2a

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

Check It Out! Example 2a Continued

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.

Check It Out! Example 2b

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°

Check It Out! Example 2c

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.


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 = 21.9 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


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.


∆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.


Find mF.


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


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


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

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°

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

Warm Up

1. Find AB for A (–3, 5) and B (1, 2).

2. Find the slope of JK for J(–4, 4) and K(3, –3).

ABCD is a parallelogram. Justify each statement.

3. ABC CDA

4. AEB CED

5

–1

Vert. s Thm.

opp. s

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

Objective

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

Example 1: Carpentry Application

A manufacture 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.


A carpenter’s square can be used to test that an angle is a right angle. How could the contractor use a carpenter’s square to check that the frame is a rectangle?

Both pairs of opp. sides of WXYZ are , so WXYZ is a parallelogram. The contractor can use the carpenter’s square to see if one of WXYZ is a right . If one angle is a right , then by Theorem 6-5-1 the frame is a rectangle.

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

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.

You can also prove that a given quadrilateral is arectangle, rhombus, or square by using the definitions of the special quadrilaterals.

Remember!

Example 2A: 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 rhombus.

The conclusion is not valid. By Theorem 6-5-3, if one pair of consecutive sides of a parallelogram are congruent, then the parallelogram is a rhombus. By Theorem 6-5-4, if the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. To apply either theorem, you must first know that ABCD is a parallelogram.

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


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


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.


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 .

Example 3A: Identifying Special Parallelograms in the Coordinate Plane

Use the diagonals to determine whether a parallelogram with the given vertices is a rectangle, rhombus, or square. Give all the names that apply.

P(–1, 4), Q(2, 6), R(4, 3), S(1, 1)


Step 1 Graph PQRS.

Step 2 Find PR and QS to determine if PQRS is a rectangle.


Since , the diagonals are congruent. PQRS is a rectangle.

Step 3 Determine if PQRS is a rhombus.

Step 4 Determine if PQRS is a square.

Since PQRS is a rectangle and a rhombus, it has four right angles and four congruent sides. So PQRS is a square by definition.


Since , PQRS is a rhombus.

Example 3B: Identifying Special Parallelograms in the Coordinate Plane

W(0, 1), X(4, 2), Y(3, –2), Z(–1, –3)

Step 1 Graph WXYZ.


Step 2 Find WY and XZ to determine if WXYZ is a rectangle.

Thus WXYZ is not a square.


Since , WXYZ is not a rectangle.

Step 3 Determine if WXYZ is a rhombus.


Since (–1)(1) = –1, , WXYZ is a rhombus.

Check It Out! Example 3A


K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)

Step 1 Graph KLMN.

Check It Out! Example 3A Continued


Step 2 Find KM and LN to determine if KLMN is a rectangle.

Since , KMLN is a rectangle.

Step 3 Determine if KLMN is a rhombus.

Since the product of the slopes is –1, the two lines are perpendicular. KLMN is a rhombus.


Step 4 Determine if KLMN is a square.

Since KLMN is a rectangle and a rhombus, it has four right angles and four congruent sides. So KLMN is a square by definition.


Check It Out! Example 3B


P(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)

Check It Out! Example 3B Continued

Step 1 Graph PQRS.

Step 2 Find PR and QS to determine if PQRS is a rectangle.


Since , PQRS is not a rectangle. Thus PQRS is not a square.

Step 3 Determine if PQRS is a rhombus.


Since (–1)(1) = –1, are perpendicular and congruent. PQRS is a rhombus.

Lesson Quiz: Part I

1. Given that AB = BC = CD = DA, what additional

information is needed to conclude that ABCD is a

square?


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

Lesson Quiz: Part III

3. Use the diagonals to determine whether a parallelogram with vertices A(2, 7), B(7, 9), C(5, 4), and D(0, 2) is a rectangle, rhombus, or square. Give all the names that apply.

AC ≠ BD, so ABCD is not a rect. or a square. The slope of AC = –1, and the slope of BD= 1, so AC BD. ABCD is a rhombus.

Prove and apply properties of rectangles, rhombuses, and squares.

Use properties of rectangles, rhombuses, and squares to solve problems.

Objectives

rectanglerhombussquare

Vocabulary

A second type of special quadrilateral is a rectangle. A rectangle is a quadrilateral with four right angles.

Since a rectangle is a parallelogram by Theorem 6-4-1, a rectangle “inherits” all the properties of parallelograms that you learned in Lesson 6-2.

Example 1: Craft Application

A woodworker constructs a rectangular picture frame so that JK = 50 cm and JL = 86 cm. Find HM.

Rect. diags.

Def. of segs.

Substitute and simplify.

KM = JL = 86

diags. bisect each other


Carpentry The rectangular gate has diagonal braces. Find HJ.

Def. of segs.

Rect. diags.

HJ = GK = 48


Carpentry The rectangular gate has diagonal braces. Find HK.

Def. of segs.

Rect. diags.

JL = LG

JG = 2JL = 2(30.8) = 61.6 Substitute and simplify.

Rect. diagonals bisect each other

A rhombus is another special quadrilateral. A rhombus is a quadrilateral with four congruent sides.

Like a rectangle, a rhombus is a parallelogram. So you can apply the properties of parallelograms to rhombuses.

Example 2A: Using Properties of Rhombuses to Find Measures

TVWX is a rhombus. Find TV.

Def. of rhombus

Substitute given values.

Subtract 3b from both sides and add 9 to both sides.


WV = XT

13b – 9 = 3b + 4

10b = 13

b = 1.3


Def. of rhombus

Substitute 3b + 4 for XT.

Substitute 1.3 for b and simplify.

TV = XT

TV = 3b + 4

TV = 3(1.3) + 4 = 7.9

Rhombus diag.

Example 2B: Using Properties of Rhombuses to Find Measures

TVWX is a rhombus. Find mVTZ.

Substitute 14a + 20 for mVTZ.

Subtract 20 from both sides and divide both sides by 14.

mVZT = 90°

14a + 20 = 90°

a = 5


Rhombus each diag. bisects opp. s

Substitute 5a – 5 for mVTZ.

Substitute 5 for a and simplify.

mVTZ = mZTX

mVTZ = (5a – 5)°

mVTZ = [5(5) – 5)]° = 20°


CDFG is a rhombus. Find CD.

Def. of rhombus

Substitute

Simplify

Substitute

Def. of rhombus

Substitute

CG = GF

5a = 3a + 17

a = 8.5

GF = 3a + 17 = 42.5

CD = GF

CD = 42.5


CDFG is a rhombus. Find the measure.

mGCH if mGCD = (b + 3)°and mCDF = (6b – 40)°

mGCD + mCDF = 180°

b + 3 + 6b – 40 = 180°

7b = 217°

b = 31°

Def. of rhombus

Substitute.

Simplify.


Check It Out! Example 2b Continued

mGCH + mHCD = mGCD

2mGCH = mGCDRhombus each diag. bisects opp. s

2mGCH = (b + 3)

2mGCH = (31 + 3)

mGCH = 17°

Substitute.

Substitute.

Simplify and divide both sides by 2.

A square is a quadrilateral with four right angles and four congruent sides. In the exercises, you will show that a square is a parallelogram, a rectangle, and a rhombus. So a square has the properties of all three.

Rectangles, rhombuses, and squares are sometimes referred to as special parallelograms.

Helpful Hint

Example 3: Verifying Properties of Squares

Show that the diagonals of square EFGH are congruent perpendicular bisectors of each other.

Example 3 Continued

Step 1 Show that EG and FH are congruent.

Since EG = FH,

Example 3 Continued

Step 2 Show that EG and FH are perpendicular.

Since ,

The diagonals are congruent perpendicular bisectors of each other.

Example 3 Continued

Step 3 Show that EG and FH are bisect each other.

Since EG and FH have the same midpoint, they bisect each other.


The vertices of square STVW are S(–5, –4), T(0, 2), V(6, –3) , and W(1, –9) . Show that the diagonals of square STVW are congruent perpendicular bisectors of each other.

111

slope of SV =

slope of TW = –11

SV ^ TW

SV = TW = 122 so, SV @ TW .

Step 1 Show that SV and TW are congruent.


Since SV = TW,

Step 2 Show that SV and TW are perpendicular.


Since

The diagonals are congruent perpendicular bisectors of each other.

Step 3 Show that SV and TW bisect each other.

Since SV and TW have the same midpoint, they bisect each other.


Example 4: Using Properties of Special Parallelograms in Proofs

Prove: AEFD is a parallelogram.

Given: ABCD is a rhombus. E is the midpoint of , and F is the midpoint of .

Example 4 Continued

||


Given: PQTS is a rhombus with diagonal

Prove:


Statements Reasons

1. PQTS is a rhombus. 1. Given.

2. Rhombus → eachdiag. bisects opp. s

3. QPR SPR 3. Def. of bisector.

4. Def. of rhombus.

5. Reflex. Prop. of

6. SAS

7. CPCTC

2.

4.

5.

7.

6.

Lesson Quiz: Part I

A slab of concrete is poured with diagonal spacers. In rectangle CNRT, CN = 35 ft, and NT = 58 ft. Find each length.

1. TR 2. CE

35 ft 29 ft


PQRS is a rhombus. Find each measure.

3. QP 4. mQRP

42 51°

Lesson Quiz: Part III

5. The vertices of square ABCD are A(1, 3), B(3, 2), C(4, 4), and D(2, 5). Show that its diagonals are congruent perpendicular bisectors of each other.

Lesson Quiz: Part IV

ABE CDF

6. Given: ABCD is a rhombus. Prove:

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18/02/2014 CH.6.6 Properties of Kites and Trapezoids

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