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Properties of Kites and Trapezoids. 6-6. Holt Geometry. Warm Up Solve for x. 1. x 2 + 38 = 3 x 2 – 12 2. 137 + x = 180 3. 4. Find FE. Objectives. Use properties of kites to solve problems. Use properties of trapezoids to solve problems. Vocabulary. kite trapezoid - PowerPoint PPT Presentation
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Holt Geometry
6-6 Properties of Kites and Trapezoids6-6 Properties of Kites and Trapezoids
Holt Geometry
Holt Geometry
6-6 Properties of Kites and Trapezoids
Warm UpSolve for x.
1. x2 + 38 = 3x2 – 12
2. 137 + x = 180
3.
4. Find FE.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Use properties of kites to solve problems.
Use properties of trapezoids to solve problems.
Objectives
Holt Geometry
6-6 Properties of Kites and Trapezoids
kitetrapezoidbase of a trapezoidleg of a trapezoidbase angle of a trapezoidisosceles trapezoidmidsegment of a trapezoid
Vocabulary
Holt Geometry
6-6 Properties of Kites and Trapezoids
A kite is a quadrilateral with exactly two pairs of congruent consecutive sides.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Holt Geometry
6-6 Properties of Kites and Trapezoids
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?
Holt Geometry
6-6 Properties of Kites and Trapezoids
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 .
Holt Geometry
6-6 Properties of Kites and Trapezoids
Solve33
N bisects JM.
Pythagorean Thm.
Pythagorean Thm.
Example 1 Continued
Holt Geometry
6-6 Properties of Kites and Trapezoids
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
Holt Geometry
6-6 Properties of Kites and Trapezoids
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.
Holt Geometry
6-6 Properties of Kites and Trapezoids
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?
Holt Geometry
6-6 Properties of Kites and Trapezoids
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
Holt Geometry
6-6 Properties of Kites and Trapezoids
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.
Check It Out! Example 1 Continued
Holt Geometry
6-6 Properties of Kites and Trapezoids
Solve33
Pyth. Thm.
Pyth. Thm.
Check It Out! Example 1 Continued
perimeter of PQRS =
Holt Geometry
6-6 Properties of Kites and Trapezoids
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.
Check It Out! Example 1 Continued
packages of binding
Holt Geometry
6-6 Properties of Kites and Trapezoids
Look Back44
Check It Out! Example 1 Continued
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
6-6 Properties 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
6-6 Properties 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
6-6 Properties 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
6-6 Properties 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
6-6 Properties 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
6-6 Properties of Kites and Trapezoids
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
Holt Geometry
6-6 Properties of Kites and Trapezoids
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.
Subtract 78 from both sides.
Divide by 2.
Holt Geometry
6-6 Properties of Kites and Trapezoids
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°
Holt Geometry
6-6 Properties of Kites and Trapezoids
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°
Holt Geometry
6-6 Properties 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
6-6 Properties 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
6-6 Properties of Kites and Trapezoids
Holt Geometry
6-6 Properties of Kites and Trapezoids
Theorem 6-6-5 is a biconditional statement. So it is true both “forward” and “backward.”
Reading Math
Holt Geometry
6-6 Properties 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
6-6 Properties 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
6-6 Properties 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
6-6 Properties 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
6-6 Properties 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
6-6 Properties of Kites and Trapezoids
Check 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 = JL
JL = JN + NL
KM = JN + NL
KM = 10.6 + 14.8 = 25.4
Holt Geometry
6-6 Properties of Kites and Trapezoids
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
Holt Geometry
6-6 Properties of Kites and Trapezoids
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
Holt Geometry
6-6 Properties of Kites and Trapezoids
Check 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 = –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
6-6 Properties 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
6-6 Properties of Kites and Trapezoids
Holt Geometry
6-6 Properties 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
6-6 Properties 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.13 = EH
116.5 = (25 + EH)2
Holt Geometry
6-6 Properties of Kites and Trapezoids
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°
Holt Geometry
6-6 Properties of Kites and Trapezoids
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