# holt mcdougal geometry 6-6 properties of kites and trapezoids 6-6 properties of kites and trapezoids...

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6-6Properties of Kites and Trapezoids

Holt McDougal GeometryWarm UpLesson PresentationLesson QuizHolt McDougal Geometry6-6Properties of Kites and TrapezoidsWarm UpSolve for x.

1. x2 + 38 = 3x2 122. 137 + x = 1803. 4. Find FE.

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsUse properties of kites to solve problems.

Use properties of trapezoids to solve problems.ObjectivesHolt McDougal Geometry6-6Properties of Kites and Trapezoidskitetrapezoidbase of a trapezoidleg of a trapezoidbase angle of a trapezoidisosceles trapezoidmidsegment of a trapezoid

VocabularyHolt McDougal Geometry6-6Properties of Kites and TrapezoidsA kite is a quadrilateral with exactly two pairs of congruent consecutive sides.

Holt McDougal Geometry6-6Properties of Kites and Trapezoids

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 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?

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 1 Continued

1 Understand the ProblemThe answer will be the amount of wood Lucy has left after cutting the dowel.

2Make a PlanThe 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 .

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsSolve

3N bisects JM.Pythagorean Thm. Pythagorean Thm.Example 1 Continued

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsLucy 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 cmLucy will have 3.6 cm of wood left over after the cut.Example 1 Continued

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsLook Back

4Example 1 ContinuedTo 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.

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck 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?

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck It Out! Example 1 Continued

1 Understand the ProblemThe answer has two parts. the total length of binding Daryl needs the number of packages of binding Daryl must buyHolt McDougal Geometry6-6Properties of Kites and Trapezoids

2Make a PlanThe 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.Check It Out! Example 1 Continued Holt McDougal Geometry6-6Properties of Kites and TrapezoidsSolve

3Pyth. Thm. Pyth. Thm. Check It Out! Example 1 Continued

perimeter of PQRS =

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsDaryl 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.Check It Out! Example 1 Continued packages of binding

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsLook Back

4Check It Out! Example 1 Continued To estimate the perimeter, change the side lengths into decimals and round. , and . The perimeter of the kite is approximately2(54) + 2 (41) = 190. So 191.3 is a reasonable answer.

Holt McDougal Geometry6-6Properties of Kites and Trapezoids Kite cons. sides

Example 2A: Using Properties of KitesIn 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 CDFmCBF = mCDFmBCD + mCBF + mCDF = 180Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 2A ContinuedSubstitute mCDF for mCBF.Substitute 52 for mCDF.Subtract 104 from both sides. mBCD + mCDF + mCDF = 180mBCD + 52 + 52 = 180mBCD = 76mBCD + mCBF + mCDF = 180Holt McDougal Geometry6-6Properties of Kites and TrapezoidsKite one pair opp. s

Example 2B: Using Properties of KitesDef. of s Polygon Sum Thm.In kite ABCD, mDAB = 54, and mCDF = 52. Find mABC.ADC ABCmADC = mABCmABC + mBCD + mADC + mDAB = 360mABC + mBCD + mABC + mDAB = 360Substitute mABC for mADC.Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 2B ContinuedSubstitute. Simplify. mABC + mBCD + mABC + mDAB = 360mABC + 76 + mABC + 54 = 3602mABC = 230mABC = 115Solve. Holt McDougal Geometry6-6Properties of Kites and TrapezoidsKite one pair opp. s

Example 2C: Using Properties of KitesDef. of s Add. Post. Substitute. Solve. In kite ABCD, mDAB = 54, and mCDF = 52. Find mFDA.CDA ABCmCDA = mABCmCDF + mFDA = mABC 52 + mFDA = 115 mFDA = 63 Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck 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 PRQmQPT = mQRT

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck It Out! Example 2a Continued Polygon Sum Thm.Substitute 78 for mPQR.mPQR + mQRP + mQPR = 18078 + mQRT + mQPT = 180Substitute. 78 + mQRT + mQRT = 18078 + 2mQRT = 1802mQRT = 102mQRT = 51Substitute. Subtract 78 from both sides. Divide by 2. Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck It Out! Example 2b

In kite PQRS, mPQR = 78, and mTRS = 59. Find mQPS.Kite one pair opp. s Add. Post. Substitute.Substitute. QPS QRSmQPS = mQRT + mTRSmQPS = mQRT + 59mQPS = 51 + 59mQPS = 110Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck It Out! Example 2c

Polygon Sum Thm.Def. of sSubstitute.Substitute.Simplify.In kite PQRS, mPQR = 78, and mTRS = 59. Find each mPSR.mSPT + mTRS + mRSP = 180mSPT = mTRSmTRS + mTRS + mRSP = 18059 + 59 + mRSP = 180mRSP = 62Holt McDougal Geometry6-6Properties of Kites and TrapezoidsA 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 McDougal Geometry6-6Properties of Kites and TrapezoidsIf the legs of a trapezoid are congruent, the trapezoid is an isosceles trapezoid. The following theorems state the properties of an isosceles trapezoid.Holt McDougal Geometry6-6Properties of Kites and Trapezoids

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsTheorem 6-6-5 is a biconditional statement. So it is true both forward and backward. Reading MathHolt McDougal Geometry6-6Properties of Kites and TrapezoidsIsos. 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 = 180100 + mB = 180mB = 80A B mA = mB mA = 80 Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 3B: Using Properties of Isosceles TrapezoidsKB = 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 = FMKJ = 32.7KB + BJ = KJ21.9 + BJ = 32.7BJ = 10.8

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 3B ContinuedSame line.Isos. trap. s base Isos. trap. legs SAS CPCTC Vert. s

KFJ MJF

BKF BMJFBK JBMFKJ JMFHolt McDougal Geometry6-6Properties of Kites and TrapezoidsIsos. trap. legs AASCPCTC Def. of segs. Substitute 10.8 for JB.Example 3B Continued

FBK JBM

FB = JBFB = 10.8Holt McDougal Geometry6-6Properties of Kites and TrapezoidsIsos. trap. s base Same-Side Int. s Thm.Def. of s Substitute 49 for mE.mF + mE = 180E H mE = mHmF = 131 mF + 49 = 180Simplify.Check It Out! Example 3a Find mF.

Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck It Out! Example 3b

JN = 10.6, and NL = 14.8. Find KM.Def. of segs. Segment Add Postulate Substitute.Substitute and simplify. Isos. trap. s base

KM = JLJL = JN + NLKM = JN + NLKM = 10.6 + 14.8 = 25.4Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 4A: Applying Conditions for Isosceles Trapezoids

Find the value of a so that PQRS is isosceles. a = 9 or a = 9Trap. 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 PmS = mP2a2 54 = a2 + 27 a2 = 81Holt McDougal Geometry6-6Properties of Kites and TrapezoidsExample 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 = BC12x 11 = 9x 23x = 9x = 3Holt McDougal Geometry6-6Properties of Kites and TrapezoidsCheck It Out! Example 4

Find the value of x so that PQST is isosceles.Subtract 2x2 and add 13 to both sides.x = 4 or x = 4Divide by 2

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