Transcript Slide 1

Properties
of
Kites
6-6
of Kites and Trapezoids
6-6 Properties
and Trapezoids
Warm Up
Lesson Presentation
Lesson Quiz
Holt
Holt
Geometry
Geometry
6-6 Properties of Kites and Trapezoids
Warm Up
Solve for x.
1. x2 + 38 = 3x2 – 12 5 or –5
2. 137 + x = 180
43
3.
156
4. Find FE.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Objectives
Use properties of kites to solve
problems.
Use properties of trapezoids to solve
problems.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Vocabulary
kite
trapezoid
base of a trapezoid
leg of a trapezoid
base angle of a trapezoid
isosceles trapezoid
midsegment of a trapezoid
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
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
.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 1 Continued
3
Solve
N bisects JM.
Pythagorean Thm.
Pythagorean Thm.
Holt Geometry
6-6 Properties of Kites and Trapezoids
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.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 1 Continued
4
Look Back
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
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
Holt Geometry
6-6 Properties of Kites and Trapezoids
Check It Out! Example 1 Continued
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.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Check It Out! Example 1 Continued
3
Solve
Pyth. Thm.
Pyth. Thm.
perimeter of PQRS =
Holt Geometry
6-6 Properties of Kites and Trapezoids
Check It Out! Example 1 Continued
Daryl needs approximately 191.3 inches of binding.
One package of binding contains 2 yards, or 72 inches.
packages of binding
In order to have enough, Daryl must buy 3 packages
of binding.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Check It Out! Example 1 Continued
4
Look Back
To estimate the perimeter, change the side lengths
into decimals and round.
, and
kite is approximately
. The perimeter of the
2(54) + 2 (41) = 190. So 191.3 is a reasonable
answer.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 2A: Using Properties of Kites
In kite ABCD, mDAB = 54°, and
mCDF = 52°. Find mBCD.
Kite  cons. sides 
∆BCD is isos.
2  sides isos. ∆
CBF  CDF
isos. ∆ base s 
mCBF = mCDF
Def. of   s
mBCD + mCBF + mCDF = 180° Polygon  Sum Thm.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 2A Continued
mBCD + mCBF + mCDF = 180°
Substitute mCDF
mBCD + mCBF + mCDF = 180°
for mCBF.
mBCD + 52° + 52° = 180°
mBCD = 76°
Holt Geometry
Substitute 52 for
mCBF.
Subtract 104
from both sides.
6-6 Properties of Kites and Trapezoids
Example 2B: Using Properties of Kites
In kite ABCD, mDAB = 54°, and
mCDF = 52°. Find mABC.
ADC  ABC
Kite  one pair opp. s 
Def. of  s
Polygon  Sum Thm.
mABC + mBCD + mADC + mDAB = 360°
mADC = mABC
Substitute mABC for mADC.
mABC + mBCD + mABC + mDAB = 360°
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 2B Continued
mABC + mBCD + mABC + mDAB = 360°
mABC + 76° + mABC + 54° = 360°
2mABC = 230°
mABC = 115°
Holt Geometry
Substitute.
Simplify.
Solve.
6-6 Properties of Kites and Trapezoids
Example 2C: Using Properties of Kites
In kite ABCD, mDAB = 54°, and
mCDF = 52°. Find mFDA.
CDA  ABC
Kite  one pair opp. s 
mCDA = mABC Def. of  s
mCDF + mFDA = mABC  Add. Post.
52° + mFDA = 115°
mFDA = 63°
Holt Geometry
Substitute.
Solve.
6-6 Properties of Kites and Trapezoids
Check It Out! Example 2a
In kite PQRS, mPQR = 78°,
and mTRS = 59°. Find
mQRT.
Kite  cons. sides 
∆PQR is isos.
RPQ  PRQ
mQPT = mQRT
Holt Geometry
2  sides  isos. ∆
isos. ∆  base s 
Def. of  s
6-6 Properties of Kites and Trapezoids
Check It Out! Example 2a Continued
mPQR + mQRP + mQPR = 180° Polygon  Sum Thm.
78° + mQRT + mQPT = 180° Substitute 78 for
mPQR.
78° + mQRT + mQRT = 180° Substitute.
78° + 2mQRT = 180° Substitute.
2mQRT = 102° Subtract 78 from
both sides.
mQRT = 51°
Holt Geometry
Divide by 2.
6-6 Properties of Kites and Trapezoids
Check It Out! Example 2b
In kite PQRS, mPQR = 78°,
and mTRS = 59°. Find
mQPS.
QPS  QRS
Kite  one pair opp. s 
mQPS = mQRT + mTRS  Add. Post.
mQPS = mQRT + 59°
Substitute.
mQPS = 51° + 59°
mQPS = 110°
Substitute.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Check It Out! Example 2c
In kite PQRS, mPQR = 78°,
and mTRS = 59°. Find each
mPSR.
mSPT + mTRS + mRSP = 180° Polygon  Sum Thm.
mSPT = mTRS
Def. of  s
mTRS + mTRS + mRSP = 180° Substitute.
59° + 59° + mRSP = 180° Substitute.
Simplify.
mRSP = 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
Reading Math
Theorem 6-6-5 is a biconditional statement. So it
is true both “forward” and “backward.”
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 3A: Using Properties of Isosceles
Trapezoids
Find mA.
mC + mB = 180°
100 + mB = 180
Same-Side Int. s Thm.
Substitute 100 for mC.
mB = 80°
A  B
Subtract 100 from both sides.
Isos. trap. s base 
mA = mB
Def. of  s
mA = 80°
Substitute 80 for mB
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 
KJ = FM
Def. of  segs.
KJ = 32.7 Substitute 32.7 for FM.
KB + BJ = KJ
Seg. Add. Post.
21.9 + BJ = 32.7 Substitute 21.9 for KB and 32.7 for KJ.
BJ = 10.8 Subtract 21.9 from both sides.
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 3B Continued
Same line.
KFJ  MJF
Isos. trap.  s base 
Isos. trap.  legs 
∆FKJ  ∆JMF
SAS
BKF  BMJ
CPCTC
FBK  JBM
Vert. s 
Holt Geometry
6-6 Properties of Kites and Trapezoids
Example 3B Continued
Isos. trap.  legs 
∆FBK  ∆JBM
AAS
CPCTC
Holt Geometry
FB = JB
Def. of  segs.
FB = 10.8
Substitute 10.8 for JB.
6-6 Properties of Kites and Trapezoids
Check It Out! Example 3a
Find mF.
mF + mE = 180°
E  H
mE = mH
mF + 49° = 180°
mF = 131°
Holt Geometry
Same-Side Int. s Thm.
Isos. trap. s base 
Def. of  s
Substitute 49 for mE.
Simplify.
6-6 Properties of Kites and Trapezoids
Check It Out! Example 3b
JN = 10.6, and NL = 14.8.
Find KM.
Isos. trap. s base 
KM = JL
JL = JN + NL
Def. of  segs.
KM = JN + NL
Substitute.
Segment Add Postulate
KM = 10.6 + 14.8 = 25.4 Substitute and simplify.
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.
Trap. with pair base
s   isosc. trap.
S  P
mS = mP
2a2
– 54 =
a2
a2
Substitute 2a2 – 54 for mS and
+ 27 2
a + 27 for mP.
= 81
a = 9 or a = –9
Holt Geometry
Def. of  s
Subtract a2 from both sides and add
54 to both sides.
Find the square root of both sides.
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.
AD = BC
Def. of  segs.
Substitute 12x – 11 for AD and
12x – 11 = 9x – 2 9x – 2 for BC.
3x = 9
x=3
Holt Geometry
Subtract 9x from both sides and add
11 to both sides.
Divide both sides by 3.
6-6 Properties of Kites and Trapezoids
Check It Out! Example 4
Find the value of x so that
PQST is isosceles.
Q  S
mQ = mS
Trap. with pair base
s   isosc. trap.
Def. of  s
2 + 19 for mQ
Substitute
2x
2x2 + 19 = 4x2 – 13 and 4x2 – 13 for mS.
32 = 2x2
x = 4 or x = –4
Holt Geometry
Subtract 2x2 and add
13 to both sides.
Divide by 2 and simplify.
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.
EF = 10.75
Holt Geometry
Solve.
6-6 Properties of Kites and Trapezoids
Check It Out! Example 5
Find EH.
Trap. Midsegment Thm.
1
16.5 = 2 (25 + EH) Substitute the given values.
Simplify.
33 = 25 + EH
Multiply both sides by 2.
13 = EH
Subtract 25 from both sides.
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?
about 191.2 in.
In kite HJKL, mKLP = 72°,
and mHJP = 49.5°. Find each
measure.
2. mLHJ
Holt Geometry
81°
3. mPKL
18°
6-6 Properties of Kites and Trapezoids
Lesson Quiz: Part II
Use the diagram for Items 4 and 5.
4. mWZY = 61°. Find mWXY.
119°
5. XV = 4.6, and WY = 14.2. Find VZ.
9.6
6. Find LP.
18
Holt Geometry