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```20/19/02/2014 CH.7.1
Factors and Greatest
Common Factors
Warm Up
Solve for x.
1. 16x – 3 = 12x + 13 4
2. 2x – 4 = 90 47
ABCD is a parallelogram. Find each
measure.
3. CD
14
4.
mC
104&deg;
Warm Up
Tell whether the second number is a factor
of the first number
1. 50, 6
no
2. 105, 7
yes
3. List the factors of 28. &plusmn;1, &plusmn;2, &plusmn;4, &plusmn;7,
&plusmn;14, &plusmn;28
Tell whether each number is prime or
composite. If the number is composite, write
it as the product of two numbers.
4. 11 prime
5. 98 composite; 49  2
Classwork and
Homework
Classwork
Homework
7.1
(Pages 459
to 461) Exercises
1, 17 to 30, 31,
32 to 35, 36, 37,
38, 47 to 55, 57,
58, 60 to 68.
Homework booklet
Ch. 7.1
Objectives
Write the prime factorization of
numbers.
Find the GCF of monomials.
Vocabulary
prime factorization
greatest common factor
The whole numbers that are multiplied to find a
product are called factors of that product. A
number is divisible by its factors.
You can use the factors of a number to write the
number as a product. The number 12 can be
factored several ways.
Factorizations of 12







The order of factors does not change the product,
but there is only one example below that cannot
be factored further. The circled factorization is
the prime factorization because all the factors
are prime numbers. The prime factors can be
written in any order, and except for changes in
the order, there is only one way to write the
prime factorization of a number.
Factorizations of 12







Remember!
A prime number has exactly two factors, itself
and 1. The number 1 is not prime because it only
has one factor.
Example 1: Writing Prime Factorizations
Write the prime factorization of 98.
Method 1 Factor tree
Method 2 Ladder diagram
Choose any two factors
Choose a prime factor of 98
of 98 to begin. Keep finding
to begin. Keep dividing by
factors until each branch
prime factors until the
ends in a prime factor.
quotient is 1.
98
2 98
7 49
2  49
7 7

7
7
1
98 = 2  7  7
98 = 2  7  7
The prime factorization of 98 is 2  7  7 or 2  72.
Check It Out! Example 1
Write the prime factorization of each number.
a. 40
40
2  20
2  10
2  5
40 = 23  5
The prime factorization
of 40 is 2  2  2  5 or
23  5.
b. 33
11 33
3
33 = 3  11
The prime factorization
of 33 is 3  11.
Check It Out! Example 1
Write the prime factorization of each number.
c. 49
d. 19
49
7  7
49 = 7  7
The prime factorization
of 49 is 7  7 or 72.
1 19
19
19 = 1  19
The prime factorization
of 19 is 1  19.
Factors that are shared by two or more whole
numbers are called common factors. The greatest
of these common factors is called the greatest
common factor, or GCF.
Factors of 12: 1, 2, 3, 4, 6, 12
Factors of 32: 1, 2, 4, 8, 16, 32
Common factors: 1, 2, 4
The greatest of the common factors is 4.
Example 2A: Finding the GCF of Numbers
Find the GCF of each pair of numbers.
100 and 60
Method 1 List the factors.
factors of 100: 1, 2, 4,
5, 10, 20, 25, 50, 100
List all the factors.
factors of 60: 1, 2, 3, 4, 5,
6, 10, 12, 15, 20, 30, 60
Circle the GCF.
The GCF of 100 and 60 is 20.
Example 2B: Finding the GCF of Numbers
Find the GCF of each pair of numbers.
26 and 52
Method 2 Prime factorization.
26 =
2  13
52 = 2  2  13
2  13 = 26
Write the prime
factorization of each
number.
Align the common
factors.
The GCF of 26 and 52 is 26.
Check It Out! Example 2a
Find the GCF of each pair of numbers.
12 and 16
Method 1 List the factors.
factors of 12: 1, 2, 3, 4, 6, 12
List all the factors.
factors of 16: 1, 2, 4, 8, 16
Circle the GCF.
The GCF of 12 and 16 is 4.
Check It Out! Example 2b
Find the GCF of each pair of numbers.
15 and 25
Method 2 Prime factorization.
15 = 1  3  5
25 = 1  5  5
1
5=5
Write the prime
factorization of each
number.
Align the common
factors.
The GCF of 15 and 25 is 5.
You can also find the GCF of monomials that
include variables. To find the GCF of monomials,
write the prime factorization of each coefficient
and write all powers of variables as products.
Then find the product of the common factors.
Example 3A: Finding the GCF of Monomials
Find the GCF of each pair of monomials.
15x3 and 9x2
15x3 = 3  5  x  x  x
9x2 = 3  3  x  x
3
Write the prime factorization of
each coefficient and write
powers as products.
Align the common factors.
x  x = 3x2 Find the product of the common
factors.
The GCF of 15x3 and 9x2 is 3x2.
Example 3B: Finding the GCF of Monomials
Find the GCF of each pair of monomials.
8x2 and 7y3
Write the prime
factorization of each
8x2 = 2  2  2 
xx
coefficient and write
7y3 =
7
y  y  y powers as products.
Align the common
factors.
The GCF 8x2 and 7y3 is 1.
There are no
common factors
other than 1.
Helpful Hint
If two terms contain the same variable raised to
different powers, the GCF will contain that
variable raised to the lower power.
Check It Out! Example 3a
Find the GCF of each pair of monomials.
18g2 and 27g3
18g2 = 2  3  3 
27g3 =
gg
Write the prime factorization
of each coefficient and
write powers as products.
3  3  3  g  g  g Align the common factors.
33
gg
Find the product of the
common factors.
The GCF of 18g2 and 27g3 is 9g2.
Check It Out! Example 3b
Find the GCF of each pair of monomials.
Write the prime
factorization of
each coefficient
and write powers
as products.
16a6 and 9b
16a6 = 2  2  2  2  a  a  a  a  a  a
9b =
The GCF of 16a6 and 9b is 1.
33b
Align the common
factors.
There are no common factors
other than 1.
Check It Out! Example 3c
Find the GCF of each pair of monomials.
8x and 7v2
8x = 2  2  2  x
7v2 =
7vv
The GCF of 8x and 7v2 is 1.
Write the prime factorization
of each coefficient and
write powers as products.
Align the common factors.
There are no common
factors other than 1.
Example 4: Application
A cafeteria has 18 chocolate-milk cartons and
24 regular-milk cartons. The cook wants to
arrange the cartons with the same number of
cartons in each row. Chocolate and regular
milk will not be in the same row. How many
rows will there be if the cook puts the greatest
possible number of cartons in each row?
The 18 chocolate and 24 regular milk cartons must
be divided into groups of equal size. The number of
cartons in each row must be a common factor of 18
and 24.
Example 4 Continued
Factors of 18: 1, 2, 3, 6, 9, 18
Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24
Find the common
factors of 18
and 24.
The GCF of 18 and 24 is 6.
The greatest possible number of milk cartons in
each row is 6. Find the number of rows of each type
of milk when the cook puts the greatest number of
cartons in each row.
Example 4 Continued
18 chocolate milk cartons
= 3 rows
6 containers per row
24 regular milk cartons
6 containers per row
= 4 rows
When the greatest possible number of types of
milk is in each row, there are 7 rows in total.
Check It Out! Example 4
Adrianne is shopping for a CD storage unit.
She has 36 CDs by pop music artists and 48
CDs by country music artists. She wants to put
the same number of CDs on each shelf without
putting pop music and country music CDs on
the same shelf. If Adrianne puts the greatest
possible number of CDs on each shelf, how
many shelves does her storage unit need?
The 36 pop and 48 country CDs must be divided into
groups of equal size. The number of CDs in each row
must be a common factor of 36 and 48.
Check It Out! Example 4 Continued
Find the common
Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36 factors of 36
and 48.
Factors of 48: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
The GCF of 36 and 48 is 12.
The greatest possible number of CDs on each shelf
is 12. Find the number of shelves of each type of
CDs when Adrianne puts the greatest number of
CDs on each shelf.
36 pop CDs
12 CDs per shelf
= 3 shelves
48 country CDs
12 CDs per shelf
= 4 shelves
When the greatest possible number of CD types
are on each shelf, there are 7 shelves in total.
Lesson Quiz: Part I
Write the prime factorization of each number.
1. 50
2  52
2. 84
22  3  7
Find the GCF of each pair of numbers.
3. 18 and 75 3
4. 20 and 36 4
Lesson Quiz: Part II
Find the GCF of each pair of monomials.
5. 12x and 28x3 4x
6. 27x2 and 45x3y2 9x2
7. Cindi is planting a rectangular flower bed with 40
orange flower and 28 yellow flowers. She wants
to plant them so that each row will have the
same number of plants but of only one color. How
many rows will Cindi need if she puts the greatest
possible number of plants in each row?
17
Warm Up
Solve for x.
1. x2 + 38 = 3x2 – 12 5 or –5
2. 137 + x = 180
43
3.
156
4. Find FE.
Classwork/Homework
Classwork
(Pages 444 to
448 ) Exercises
1, 14 to 25,
26, 27 to 32, 34
to 36, 40, 42, 47,
48, 49
.
Homework
Homework
Booklet
Chapter: 6.6
Objectives
Use properties of kites to solve
problems.
Use properties of trapezoids to solve
problems.
Vocabulary
kite
trapezoid
base of a trapezoid
leg of a trapezoid
base angle of a trapezoid
isosceles trapezoid
midsegment of a trapezoid
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
.
Example 1 Continued
3
Solve
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
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.
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
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.
Check It Out! Example 1 Continued
3
Solve
perimeter of PQRS =
Pyth. Thm.
Pyth. Thm.
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.
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.
Example 2A: Using Properties of Kites
In kite ABCD, mDAB = 54&deg;, and
mCDF = 52&deg;. 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&deg;Polygon  Sum Thm.
Example 2A Continued
mBCD + mCBF + mCDF = 180&deg;
Substitute mCDF
mBCD + mCDF + mCDF = 180&deg;
for mCBF.
Substitute 52 for
mBCD + 52&deg; + 52&deg; = 180&deg;
mCDF.
mBCD = 76&deg;
Subtract 104
from both sides.
Example 2B: Using Properties of Kites
In kite ABCD, mDAB = 54&deg;, and
mCDF = 52&deg;. Find mABC.
ADC  ABC
mADC = mABC
Kite  one pair opp. s 
Def. of  s
Polygon  Sum Thm.
mABC + mBCD + mADC + mDAB = 360&deg;
Substitute mABC for mADC.
mABC + mBCD + mABC + mDAB = 360&deg;
Example 2B Continued
mABC + mBCD + mABC + mDAB = 360&deg;
mABC + 76&deg; + mABC + 54&deg; = 360&deg; Substitute.
2mABC = 230&deg; Simplify.
mABC = 115&deg;
Solve.
Example 2C: Using Properties of Kites
In kite ABCD, mDAB = 54&deg;, and
mCDF = 52&deg;. 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&deg; + mFDA = 115&deg;
mFDA = 63&deg;
Substitute.
Solve.
Check It Out! Example 2a
In kite PQRS, mPQR = 78&deg;,
and mTRS = 59&deg;. Find
mQRT.
Kite  cons. sides 
∆PQR is isos.
RPQ  PRQ
mQPT = mQRT
2  sides  isos. ∆
isos. ∆  base s 
Def. of  s
Check It Out! Example 2a Continued
mPQR + mQRP + mQPR = 180&deg; Polygon  Sum Thm.
78&deg; + mQRT + mQPT = 180&deg;Substitute 78 for
mPQR.
78&deg; + mQRT + mQRT = 180&deg;Substitute.
78&deg; + 2mQRT = 180&deg; Substitute.
2mQRT = 102&deg; Subtract 78 from
both sides.
mQRT = 51&deg; Divide by 2.
Check It Out! Example 2b
In kite PQRS, mPQR = 78&deg;,
and mTRS = 59&deg;. Find
mQPS.
QPS  QRS
Kite  one pair opp. s 
mQPS = mQRT + mTRS  Add. Post.
mQPS = mQRT + 59&deg;
Substitute.
mQPS = 51&deg; + 59&deg;
mQPS = 110&deg;
Substitute.
Check It Out! Example 2c
In kite PQRS, mPQR = 78&deg;,
and mTRS = 59&deg;. Find each
mPSR.
mSPT + mTRS + mRSP = 180&deg;Polygon  Sum Thm.
mSPT = mTRS
Def. of  s
mTRS + mTRS + mRSP = 180&deg; Substitute.
59&deg; + 59&deg; + mRSP = 180&deg;Substitute.
mRSP = 62&deg; Simplify.
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.
Reading Math
Theorem 6-6-5 is a biconditional statement. So it
is true both “forward” and “backward.”
Example 3A: Using Properties of Isosceles
Trapezoids
Find mA.
mC + mB = 180&deg;
100 + mB = 180
Same-Side Int. s Thm.
Substitute 100 for mC.
mB = 80&deg;
A  B
Subtract 100 from both sides.
Isos. trap. s base 
mA = mB
Def. of  s
mA = 80&deg;
Substitute 80 for mB
Example 3B: Using Properties of Isosceles
Trapezoids
KB = 21.9 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.
Example 3B Continued
Same line.
KFJ  MJF
Isos. trap.  s base 
Isos. trap.  legs 
∆FKJ  ∆JMF
SAS
BKF  BMJ
CPCTC
FBK  JBM
Vert. s 
Example 3B Continued
Isos. trap.  legs 
∆FBK  ∆JBM
AAS
CPCTC
FB = JB
Def. of  segs.
FB = 10.8
Substitute 10.8 for JB.
Check It Out! Example 3a
Find mF.
mF + mE = 180&deg;
E  H
mE = mH
mF + 49&deg; =
180&deg; mF = 131&deg;
Same-Side Int. s Thm.
Isos. trap. s base 
Def. of  s
Substitute 49 for mE.
Simplify.
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.
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
Def. of  s
Substitute 2a2 – 54 for mS and
+ 27 2
a + 27 for mP.
= 81
a = 9 or a = –9
Subtract a2 from both sides and add
54 to both sides.
Find the square root of both sides.
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
Subtract 9x from both sides and add
11 to both sides.
Divide both sides by 3.
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
Subtract 2x2 and add
13 to both sides.
Divide by 2 and simplify.
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.
EF = 10.75
Solve.
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.
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&deg;,
and mHJP = 49.5&deg;. Find each
measure.
2. mLHJ
81&deg;
3. mPKL
18&deg;
Lesson Quiz: Part II
Use the diagram for Items 4 and 5.
4. mWZY = 61&deg;. Find mWXY.
119&deg;
5. XV = 4.6, and WY = 14.2. Find VZ.
9.6
6. Find LP.
18
Warm Up
1. Find AB for A (–3, 5) and B (1, 2). 5
2. Find the slope of JK for J(–4, 4) and K(3, –3). –1
ABCD is a parallelogram. Justify each
statement.
3. ABC  CDA
 opp. s 
4. AEB  CED Vert. s Thm.
Objective
Prove that a given quadrilateral is a
rectangle, rhombus, or square.
When you are given a parallelogram with certain
properties, 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
mABC = 90&deg;. Why must
ABCD be a rectangle?
Both pairs of opposites sides of ABCD are
congruent, so ABCD is a . Since mABC = 90&deg;,
one angle
ABCD is a right angle. ABCD is a
rectangle by Theorem 6-5-1.
Check It Out! Example 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.
Caution
In order to apply Theorems 6-5-1 through 6-5-5,
the quadrilateral must be a parallelogram.
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.
Remember!
You can also prove that a given quadrilateral is a
rectangle, rhombus, or square by using the
definitions of the special quadrilaterals.
Example 2A: Applying Conditions for Special
Parallelograms
Determine 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
Parallelograms
Determine 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
Quad. with diags.
EFGH is a parallelogram. bisecting each other 
Example 2B Continued
Step 2 Determine if EFGH is a rectangle.
Given.
EFGH is a rectangle.
with diags.   rect.
Step 3 Determine if EFGH is a rhombus.
EFGH is a rhombus.
with one pair of cons. sides
  rhombus
Example 2B Continued
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.
Check It Out! Example 2
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)
Example 3A Continued
Step 1 Graph
PQRS.
Example 3A Continued
Step 2 Find PR and QS to determine if PQRS is a
rectangle.
Since
, the diagonals are congruent.
PQRS is a rectangle.
Example 3A Continued
Step 3 Determine if PQRS is a rhombus.
Since
, 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.
Example 3B: 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.
W(0, 1), X(4, 2), Y(3, –2),
Z(–1, –3)
Step 1 Graph
WXYZ.
Example 3B Continued
Step 2 Find WY and XZ to determine if WXYZ is a
rectangle.
Since
, WXYZ is not a rectangle.
Thus WXYZ is not a square.
Example 3B Continued
Step 3 Determine if WXYZ is a rhombus.
Since (–1)(1) = –1,
rhombus.
, WXYZ is a
Check It Out! Example 3A
Use the diagonals to determine whether a
parallelogram with the given vertices is a
rectangle, rhombus, or square. Give all the
names that apply.
K(–5, –1), L(–2, 4), M(3, 1), N(0, –4)
Check It Out! Example 3A Continued
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.
Check It Out! Example 3A Continued
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.
Check It Out! Example 3A Continued
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
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(–4, 6) , Q(2, 5) , R(3, –1) , S(–3, 0)
Check It Out! Example 3B Continued
Step 1 Graph
PQRS.
Check It Out! Example 3B Continued
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.
Check It Out! Example 3B Continued
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?
Lesson Quiz: Part II
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.
Objectives
Prove and apply properties of
rectangles, rhombuses, and squares.
Use properties of rectangles,
rhombuses, and squares to solve
problems.
Vocabulary
rectangle
rhombus
square
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. 
KM = JL = 86
Def. of  segs.
 diags. bisect each other
Substitute and simplify.
Check It Out! Example 1a
Carpentry The rectangular gate has diagonal braces.
Find HJ.
Rect.  diags. 
HJ = GK = 48
Def. of  segs.
Check It Out! Example 1b
Carpentry The rectangular gate has diagonal braces.
Find HK.
Rect.  diags. 
Rect.  diagonals bisect each other
JL = LG
JG = 2JL = 2(30.8) = 61.6
Def. of  segs.
Substitute and simplify.
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.
WV = XT
13b – 9 = 3b + 4
10b = 13
b = 1.3
Def. of rhombus
Substitute given values.
Subtract 3b from both sides and add 9 to both sides.
Divide both sides by 10.
Example 2A Continued
TV = XT
Def. of rhombus
TV = 3b + 4
Substitute 3b + 4 for XT.
TV = 3(1.3) + 4 = 7.9
Substitute 1.3 for b and simplify.
Example 2B: Using Properties of Rhombuses to Find Measures
TVWX is a rhombus. Find
mVTZ.
mVZT = 90&deg;
14a + 20 = 90&deg;
a=5
Rhombus  diag. 
Substitute 14a + 20 for mVTZ.
Subtract 20 from both sides and divide both
sides by 14.
Example 2B Continued
mVTZ = mZTX
Rhombus  each diag. bisects opp. s
mVTZ = (5a – 5)&deg;
Substitute 5a – 5 for mVTZ.
mVTZ = [5(5) – 5)]&deg;
= 20&deg;
Substitute 5 for a and simplify.
Check It Out! Example 2a
CDFG is a rhombus. Find CD.
CG = GF
5a = 3a + 17
a = 8.5
GF = 3a + 17 = 42.5
Def. of rhombus
Substitute
Simplify
Substitute
CD = GF
Def. of rhombus
CD = 42.5
Substitute
Check It Out! Example 2b
CDFG is a rhombus.
Find the measure.
mGCH if mGCD = (b + 3)&deg;
and mCDF = (6b – 40)&deg;
mGCD + mCDF = 180&deg;
b + 3 + 6b – 40 = 180&deg;
7b = 217&deg;
b = 31&deg;
Def. of rhombus
Substitute.
Simplify.
Divide both sides by 7.
Check It Out! Example 2b Continued
mGCH + mHCD = mGCD
2mGCH = mGCD
Rhombus  each diag. bisects
opp. s
2mGCH = (b + 3)
Substitute.
2mGCH = (31 + 3)
Substitute.
mGCH = 17&deg;
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.
Helpful Hint
Rectangles, rhombuses, and squares are sometimes referred to as special
parallelograms.
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
,
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 diagonals are congruent perpendicular bisectors of each other.
Check It Out! Example 3
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.
SV = TW =
122 so, SV  TW .
1
slope of SV =
11
slope of TW = –11
SV  TW
Check It Out! Example 3 Continued
Step 1 Show that SV and TW are congruent.
Since SV = TW,
Check It Out! Example 3 Continued
Step 2 Show that SV and TW are perpendicular.
Since
Check It Out! Example 3 Continued
Step 3 Show that SV and TW bisect each other.
Since SV and TW have the same midpoint, they bisect each other.
The diagonals are congruent perpendicular bisectors of each other.
Example 4: Using Properties of Special Parallelograms in Proofs
Given: ABCD is a rhombus. E is
the midpoint of
, and F
is the midpoint of
.
Prove: AEFD is a parallelogram.
Example 4 Continued
||
Check It Out! Example 4
Given: PQTS is a rhombus with diagonal
Prove:
Check It Out! Example 4 Continued
Statements
Reasons
1. PQTS is a rhombus.
1. Given.
2.
2. Rhombus → each
diag. bisects opp. s
3. QPR  SPR
3. Def. of  bisector.
4.
4. Def. of rhombus.
5.
5. Reflex. Prop. of 
6.
6. SAS
7.
7. CPCTC
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
Lesson Quiz: Part II
PQRS is a rhombus. Find each measure.
3. QP
4. mQRP
42
51&deg;
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
6. Given: ABCD is a rhombus.
Prove:
ABE  CDF

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