TEKS/STAAR-BASED LESSONS - Having a HOOT in Fourth Grade

TEKSING TOWARD STAAR © 2014

TEKS/STAAR-BASED

LESSONS

PARENT GUIDESix Weeks 2

®MATHEMATICS

TEKSING TOWARD STAAR

TEKSING TOWARD STAAR2014

TEKSING TOWARD STAARSix Weeks 1 Scope and Sequence

Grade 4 Mathematics

Lesson TEKS/Lesson Content

Lesson 1

4.3A/represent a fraction a/b as a sum of fractions 1/b, where a and b are whole numbers and b> 0, including when a > b

4.3B/decompose a fraction in more than one way into a sum of fractions with the samedenominator using concrete and pictorial models and recording results with symbolicrepresentations

Lesson 24.3C/determine if two given fractions are equivalent using a variety of methods

4.3D/compare two fractions with different numerators and different denominators and representthe comparison using the symbols >, =, or <

Lesson 34.3E/represent and solve addition and subtraction of fractions with equal denominators usingobjects and pictorial models that build to the number line and properties of operations

4.3F/evaluate the reasonableness of sums and differences of fractions using benchmarkfractions 0, 1/4, 1/2, 3/4, and 1, referring to the same whole

Lesson 4

4.4B/determine products of a number and 10 or 100 using properties of operations and placevalue understandings

4.4G/round to the nearest 10, 100, or 1,000 or use compatible numbers to estimate solutionsinvolving whole numbers

Lesson 5

4.4D/use strategies and algorithms, including the standard algorithm, to multiply up to a four-digit number by a one-digit number and to multiply a two-digit number by a two-digit number.Strategies may include mental math, partial products, and the commutative, associative, anddistributive properties

4.4C/represent the product of 2 two-digit numbers using arrays, area models, or equations,including perfect squares through 15 by 15;

4.4H/solve with fluency one- and two-step problems involving multiplication…

Lesson 6

4.5A/represent multi-step problems involving the four operations (multiplication only in lesson)with whole numbers using strip diagrams and equations with a letter standing for the unknownquantity

4.5B/represent problems using an input-output table and numerical expressions to generate anumber pattern that follows a given rule representing the relationship of the values in theresulting sequence and their position in the sequence (multiplication only in lesson)

Lesson 74.5C/use models to determine the formulas for the perimeter of a rectangle (l + w + l + w or 2l +2w), including the special form for perimeter of a square (4s) and the area of a rectangle (l x w)

4.5D/solve problems related to perimeter and area of rectangles where dimensions are wholenumbers

Lesson 8

4.6A/identify …rays, angles…and perpendicular…lines

4.6C/apply knowledge of right angles to identify acute, right, and obtuse triangles

4.6D/ classify two-dimensional figures based on the presence or absence of parallel orperpendicular lines or the presence or absence of angles of a specified size

Review

Assessment

NOTES:

STAAR Category 1 GRADE 4 TEKS 4.3A/4.3B

TEKSING TOWARD STAAR 2014 Page 1

LESSON 1 - 4.3A & 4.3B

Lesson Focus

For TEKS 4.3A students are expected to represent a fraction a/b as a sum of fractions1/b, where a and b are whole numbers and b > 0, including when a > b.

For TEKS 4.3B students are expected to decompose a fraction in more than one wayinto a sum of fractions with the same denominator using concrete and pictorial modelsand recording results with symbolic representations.

For these TEKS students should be able to apply mathematical process standards torepresent and generate fractions to solve problems.

For STAAR Category 1 students should be able to demonstrate an understanding ofhow to represent and manipulate numbers and expressions.

Process Standards Incorporated Into Lesson

4.1.A Apply mathematics to problems arising in everyday life, society, and theworkplace.

4.1.B Use a problem-solving model that incorporates analyzing given information,formulating a plan or strategy, determining a solution, justifying the solution,and evaluating the problem-solving process and the reasonableness of asolution.

4.1.C Select tools, including real objects, manipulatives, paper and pencil, andtechnology as appropriate, and techniques, including mental math, estimation,

4.1.E Create and use representations to organize, record, and communicatemathematical ideas

Vocabulary for Lesson

PART I PART II PART IIIfraction fraction greater than 1 decomposenumerator sum of unit fractions symbolic representationdenominatorunit fractionsum of unit fractionsimproper fractionmixed number



Math Background Part I - Fractions

A fraction is a number that names part of a whole or part of a group. The numeratoris the top number in a fraction and represents how many equal parts are described bythe fraction. The denominator is the bottom number in a fraction and represents thetotal number of equal parts in the whole.

EXAMPLE: Five eighths is a fraction. The fraction represents 5 out of 8 equal partsof a whole object or a whole group.

58

Unit Fractions

A unit fraction names 1 equal part of a whole. The numerator of a unit fraction isalways 1.

EXAMPLE: One-eighth is a unit fraction. The fraction represents 1 out of 8 equalparts of a whole.

NOTE: The quantity 1b

represents one part of a whole that has been divided into

equal parts where a and b are whole numbers and b > 0, including when a > b.

Part of a Whole

Braden is hiking on a nature trail at a state park. He has hiked46

of the length of the

trail. The diagram represents the length of the trail that Braden has already hiked.

The length of the trail is divided in to 6 equal sections. The diagram shows 4 out of 6equal parts shaded to represent the length of the trail Braden has already hiked.

The distance Braden has already hiked can be represented by a sum of unit fractions.16

+ 16

+ 16

+ 16

= 46

The numerator shows the number of addends in the sum of unit fractions. So, 46

is

the distance Braden has already hiked on the trail.

The model also shows 2 out of 6 equal parts are not shaded to represent the length ofthe trail Braden still has left to hike.

numerator: 1 part of the whole

denominator: 8 equal parts in the whole18

numerator: Braden has already hiked 4 equal parts of the trail.

denominator: The trail is divided into 6 equal parts.

46

numerator: Braden has 2 equal parts of the trail left to hike.

denominator: The trail is divided into 6 equal parts.26

4616

4616

4616

4616

116

16



The distance Braden has left to hike can also be represented by a sum of unit fractions.16

+16

=26

The numerator shows the number of addends in the sum of unit fractions. So, Braden

has 26

of the distance of the trail left to hike.

Part of a Group

Jackson has a collection of marbles. Find the fraction of his collection that is blue.

The model shows a total of 12 marbles. The model shows 8 out of 12 marbles are notblue.

The number of marbles that are not blue can be represented by a sum of unit fractions.

112

+112

+112

+112

+112

+112

+112

+112

=8

12


of

the marbles in Jackson's collection are not blue.

The model also shows 4 out of 12 of the marbles are blue.

The number of blue marbles Jackson has can also be represented by a sum of unitfractions.

112

+ 112

+ 112

+ 112

= 412


of

the marbles in Jackson's collection are blue.

redred

blueblue yellow

green redredblue

blueyellow

green

numerator: Jackson has 8 marbles that are not green.

denominator: Jackson has a total of 12 marbles.812

numerator: Jackson has 4 blue marbles.

denominator: Jackson has a total of 12 marbles.

412



Math Background Part II - Fractions Greater than 1

A fraction can be used to represent a number that is greater than 1. A fraction withits numerator greater than its denominator is sometimes called an improper

fraction. The fraction 32

is an example of a fraction that is greater than 1. There is

nothing wrong with the fraction32

, but it has a numerator that is greater than the

denominator, so the value of the fraction is greater than 1.

EXAMPLE 1: Kendra has a piece of lace that is 43

feet long. She needs a piece of

lace that is 1 foot long to decorate a hair clip. Does she have enough lace?

The fraction43

can be written as a sum of unit fractions.

13

+13

+13

+13

=43

The fraction43

can be represented by a diagram.

The diagram shows that 43

is equal to 1 and 13

, so 43

= 113

.

Kendra has a length of lace 113

feet long. She has enough lace to decorate the hair

clip because she has more than 1 foot of lace.

EXAMPLE 2: Leisa needs a length of leather string 52

yards long to weave a bracelet.

She has a length of string 2 yards long. Does she have enough string to weave thebracelet?

The fraction 52

can be written as a sum of unit fractions.

12

+ 12

+ 12

+ 12

+ 12

= 52

The fraction52

can be represented by a diagram.

113

13

13

13

1 112

12

12

12

12



The diagram shows that 52

is equal to 2 and 12

, so 52

= 122

.

So, Leisa does not have enough leather string to weave the bracelet because she onlyhas 2 yards of leather string.

EXAMPLE 3: Ricardo baked two pans of gingerbread. He cut each pan of gingerbreadinto 6 equal pieces. Ricardo gave Alissa and George 1 piece each of the gingerbread.The amount of gingerbread left is shown below.

Use fraction strips to represent the amount of gingerbread left.

Represent the amount of gingerbread left as a sum of unit fractions.

16

+ 16

+ 16

+ 16

+ 16

+ 16

+ 16

+ 16

+ 16

+ 16

= 106

The model shows that106

=4

16

EXAMPLE 4: Celine cut each of 2 yards of lace into 8 equal pieces. She used 3pieces of the lace for a craft project.

Use a number line to represent the amount of lace Celine has left.

Represent the amount of lace left as a sum of unit fractions.

18

+18

+18

+18

+18

+18

+18

+18

+18

+18

+18

+18

+18

=138

The model shows that 138

= 518

.

4616

4616

4616

4616

14616

4616

4616

4616

14616

4616

0 1 2

18

18

18

18

18

18

18

18

18

18

18

18

18

STAAR Category 1 GRADE 4 TEKS 4.3C/4.3D


LESSON 2 - 4.3C & 4.3D

Lesson Focus

For TEKS 4.3C students are expected to determine if two given fractions are equivalentusing a variety of methods.

For TEKS 4.3D students are expected to compare two fractions with differentnumerators and different denominators and represent the comparison using thesymbols >, =, or <.


For STAAR Category 1 students should be able to demonstrate an understanding ofhow to represent and manipulate numbers and expressions.




4.1.D Communicate mathematical ideas, reasoning, and their implications usingmultiple representations, including symbols, diagrams, graphs, and languageas appropriate.

4.1.E Create and use representations to organize, record, and communicatemathematical ideas.

4.1.G Display, explain, and justify mathematical ideas and arguments using precisemathematical language in written or oral communication.


PART I PART I PART IIequivalent fraction common factors multiplesimplest form simplest form least common multiplenumerator factor common denominatordenominator greatest common factor



Math Background Part I - Equivalent Fractions

Sometimes two fractions are written differently but actually name equal parts. Theseare called equivalent fractions. Equivalent fractions name the same part of awhole in different ways.

Model Equivalent Fractions

EXAMPLE 1: This model shows that 12

, 24

, and 48

are equivalent fractions.

The model shows48

,24

, and12

are equivalent fractions because they are equal

distances from 0 on the number line.

EXAMPLE 2: This model also shows 12

, 24

, and 48

are equivalent fractions.

Rectangle 1, Rectangle 2, and Rectangle 3 are the same size.Rectangle 1 is divided into 8 equal parts, and 4 of the parts are shaded.

The fraction 48

represents the shaded part of the whole.

Rectangle 2 is divided into 4 equal parts, and 2 of the parts are shaded.

The fraction 24


Rectangle 3 is divided into 2 equal parts, and 1 of the parts is shaded.

The fraction 12


The model shows an equal amount is shaded in all three rectangles. So 48

, 24

, and 12

are equivalent fractions because they represent equal amounts of a whole.

0 114

24

34

0 118

28

38

48

58

68

78

120 1

Rectangle 1 Rectangle 2 Rectangle 3



EXAMPLE 3: This group shows 8 circles arranged in 2 rows. There are 4 circles ineach row.

There are two ways to look at what part of the group is shaded.

You can see that 6 of the 8 circles in the group are shaded.

The fraction 68

names this part of the group that is shaded.

You can see that 3 of the 4 columns of circles are shaded.

The fraction34

names this part of the group that is shaded.

So, the fractions68

and34

are equivalent fractions because they describe an equal

part of the group.

NOTE: Rows are horizontal. Columns are vertical.

EXAMPLE 4: This model represents two equivalent fractions.

The model shows 6 rectangles arranged in two rows, with 3 rectangles in each row.

In the model, 2 of the 6 rectangles are shaded.

The fraction 26

names this part of the model that is shaded.

In the model, 1 of the 3 columns is shaded.

The fraction13

names this part of the model that is shaded.

Because the fractions 26

and 13

represent the same part of the model, they are

equivalent fractions.

column

row



Find Equivalent Fractions

To find an equivalent fraction, you can multiply or divide the numerator and thedenominator by the same number. This is the same as multiplying or dividing by afraction that is equal to 1.

EXAMPLE 1: Multiplying or dividing by a fraction that is equal to 1 does not changethe value of the fraction because you are multiplying or dividing by 1.

1 1 3 32 2 3 6

2 2 2 14 4 2 2

EXAMPLE 2: A fraction is in simplest form when its numerator and denominatorhave no common factors other than 1. A factor is a whole number that dividesexactly into another number.

The fraction23

is in simplest form because 2 and 3 have no common factors other

than 1. To go from a fraction in simplest form to an equivalent fraction, multiply by afraction equal to 1.

The table below shows some fractions that are equivalent to23

.

Original Fraction Fraction Equal to 1 Multiply Equivalent Fraction

23

22

2 2 43 2 6

46

23

55

2 5 103 5 15

1015

23

1111

2 11 223 11 33

2233

EXAMPLE 3: When you write an equivalent fraction with a smaller denominator thanthe one in the original fraction, it is called simplifying the fraction or stating thefraction in simplest form.

To go from a smaller denominator to alarger denominator, multiply by a

fraction equal to 1.

3 times as many parts,so 3 times as manyparts are shaded.

To go from a larger denominator toa smaller denominator, divide by a

fraction equal to 1.

Half as many parts,so half as many parts

are shaded.



To simplify a fraction, divide the numerator and denominator by the same number.

To obtain a fraction in the simplest form, divide by the greatest common factor ofthe numerator and denominator.

The table below shows some examples of simplified fractions.

OriginalFraction

Factors GreatestCommon Factor

Division by GreatestCommon Factor

SimplestForm

46

4 2 26 2 3

24 2 26 2 3

23

1525

15 3 525 5 5

515 5 325 5 5

35

812

8 2 412 3 4

48 4 212 4 3

23

1228

12 3 428 7 4

412 4 328 4 7

37

EXAMPLE 4: What fraction with a denominator of 40 is equivalent to410

?

4 ?10 40

Look at the two denominators. Think of a multiplication fact that relates 10 and 40.

4 10 = 40

To go from a smaller denominator to a larger denominator, multiply the numeratorand the denominator by the same number.

Since 4 10 = 40, multiply by 4.

Multiply both the numerator and the denominator of4

10by 4.

4 4 1610 4 40

The fraction1640

is equivalent to410

.



Math Background Part II - Comparing Two Fractions

Models to Compare Two Fractions

Models can be used to compare two fractions. The models will show a comparisonthat can be represented using a >, <, or = symbol. fraction that2

EXAMPLE 1: Compare the two fractions represented by the models.

Look at the shaded areas. The shaded area of the bottom model is greater than theshaded area of the top model.

The models show that 58

is less than 46

, or 58

< 46

.

The models also show that 46

is greater than 58

, or 46

> 58

.

EXAMPLE 2: Compare the amounts of brown sugar in the measuring cups.

Use the pictures to compare the fractions.

The amount shaded for23

is greater than12

, so23

>12

.

EXAMPLE 3: Compare the fractions represented by the models.

Write: 24

= 36

Say: two fourths is equal to three sixths

EXAMPLE 4: Compare the fractions represented by the models

23

1 cup

12

1 cup

58

46



Write:48

>24

Say: four eighths is greater than two fourths

Number Lines to Compare Fractions

Number lines can be used to compare fractions using benchmarks on the number lines.

EXAMPLE: Compare 34

and 710

.

The number lines show that 34

is a greater distance from 0 than 710

.

So, 34

> 710

and 710

< 34

.

Equivalent Fractions to Compare Fractions

To compare fractions that have unlike denominators, rewrite them as equivalentfractions with a common denominator. To find a common denominator for twofractions, use the least common multiple of the two different denominators. Amultiple of a number is the product of the number and another whole number.

The least common multiple of two numbers is the smallest number that is in bothlists of multiples. To find the least common multiple of two numbers, list the multiplesof both numbers. Identify the smallest number found in both lists.

EXAMPLE: Find the least common multiple of 3 and 5.

Multiples of 3 are: 3, 6, 9, 12, 15 , 18, …

Multiples of 5 are: 5, 10, 15 , 20, …

The smallest number in both lists of multiples is 15.

The least common multiple of 3 and 5 is 15.

Guidelines for Comparing Two Fractions That Have Different Denominators

Step 1: Find the least common multiple for the denominators of the two fractions.

Step 2: Rewrite each fraction as an equivalent fraction using the least commonmultiple as the denominator.

Step 3: Compare the numerators of the two rewritten fractions.

0 12

34

1

0 12

710

1



EXAMPLE: Compare 34

and 56

.

Find the least common multiple for the denominators.Find the multiples of 4 and 6.

Multiples of 4 are: 4, 8, 12, 16, …

Multiples of 6 are: 6, 12, 18, 24, …

The smallest number in both lists of multiples is 12, so the least common multiple of4 and 6 is 12.

Rewrite each fraction as an equivalent fraction using the least common multiple asthe denominator.

Rewrite 34

and 56

using 12 as the denominators.

3 ?4 12 3 3 9

4 3 12

3 94 12

5 ?6 12

5 2 106 2 12

5 106 12

Compare the numerators of the two rewritten fractions.

10 > 9 and 9 < 10

Since 10 > 9, then56

>34

and since 9 < 10, then34

<56

.

Confirm the comparison of the fractions using models.

The model shows that 34

= 912

and 56

= 1012

.

The model also shows that34

<56

and9

12<

1012

.

So, the model confirms 34

< 56

and 56

> 34

.

4 3 12

6 2 12

=34

=912

=56

=1012

STAAR Category 2 GRADE 4 TEKS 4.3E/4.3F


LESSON 3 - 4.3E & 4.3F

Lesson Focus

For TEKS 4.3E students are expected to represent and solve addition and subtractionof fractions with equal denominators using objects and pictorial models that build tothe number line and properties of operations. This lesson focuses on the use offraction strips, fraction bars, and other linear models, as well as sketches of the linearfraction models and strip diagrams.

For TEKS 4.3F students are expected to evaluate the reasonableness of sums anddifferences of fractions using benchmark fractions 0, 1/4, 1/2, 3/4, and 1, referring tothe same whole.


For STAAR Category 2 students should be able to demonstrate how to performoperations and represent algebraic relationships.



4.1.D Communicate mathematical ideas, reasoning, and their implications usingmultiple representations, including symbols, diagrams, graphs, and languageas appropriate


4.1.F Analyze mathematical relationships to connect and communicate mathematicalideas.



PART I PART IIequivalent fractions equivalent fractionsmixed numbers mixed numbersequal denominators equal denominatorsnumerators numeratorssimplest form simplest formproperties of addition



Math Background Part I - Addition of Fractions

Remembering what you know about adding whole numbers can be used to help youadd fractions. You will also use what you know about equivalent fractions andmixed numbers.

Only parts that refer to the same whole can be joined to add fractions.

EXAMPLE: Jackson had 1 lizard in his reptile cage. He caught 2 more lizards to putinto the cage. When Jackson adds 1 lizard and 2 lizards, he has 3 lizards. Thenumber of lizards changes, but they are still lizards.

Jackson puts 1 fourth of a container of meal worms in his reptile cage to feed 1 lizard.He needs to add another 2 fourths of a container to feed the 2 new lizards he put inhis cage. When Jackson adds 1 fourth and 2 fourths, he finds he needs to put 3fourths of a container of meal worms in his cage to feed 3 lizards. The number offourths changes, but they are still fourths.

Adding Fractions with Equal Denominators

To add fractions, check the denominators. If the fractions have equal denominators,add the numerators. The denominator stays the same because you are adding thesame kind of thing.

EXAMPLE:

Model Addition of Fractions with Equal Denominators

Addition of fractions can be represented using models, pictures, words, and numbers.A model shows the relationship between the fractions in the problem.

EXAMPLE 1: Emily is baking blueberry cupcakes. She needs 38

of a stick of butter

for the cupcake batter and 38

of a stick of butter for the icing. Find the amount of

butter she needs in all.

Use the operation of addition to solve the problem.

Model the problem to find the sum of 38

and 38

.

The model shows that Emily needs68

of a stick of butter.

+ =+ Add the numerators to get the new numerator

Use the same denominator

18

18

18

18

18

38

38

+ = 68

18



Use a model to find the simplest form of 68

. A fraction is in simplest form when

the numerator and denominator have no common factor greater than 1.

The model shows the simplest form of68

is34

.

So, the model shows that Emily needs34


Add without using a model.

Step 1Write the sum of thefractions as an expression.

Step 2Add the numeratorsto find the sum.

Step 3Write the sum in simplest form.

38

+ 38

38

+ 38

= 68

68

6 28 2

= 34

Either way, Emily needs 34


EXAMPLE 2: Stella has these pieces of blueberry pie left over from 2 different piesshe baked for a holiday dinner.

She has decided to combine the pieces so they are in the same pie pan. Find theamount of pie she will have in the pie pan.



and 36

.

18

18

18

18

18

38

38

34

14

14

14

18



The model shows 16

+ 36

= 46

.

Use a model to find the simplest form of46

.


is23

.

So, the model shows Stella will have23

of a pie in the pie pan after she moves the

pieces into one pie pan.


Step 1Write the sum of the fractionsas an addition expression.

Step 2Add the numeratorsto find the sum.

Step 3Write the sum in simplest form.

16

+36

16

+36

=46

46

4 26 2

=23

Either way, Stella will have 23

of a pie in the pie pan after she moves the pieces into

one pie pan.

06

16

26

36

46

56

66

06

16

26

36

46

56

66

03

13

23

33



EXAMPLE 3: Find the sum of 26

and 46

.

Use a model to represent the problem.

The model shows 26

+ 46

= 66

= 1.


To add fractions with the same denominators, add the numerators. Then write thesum over the denominator.

2 4 2 4 6 16 6 6 6

Either way, the sum is 66

or 1.

EXAMPLE 4: Find the sum of 78

and 68

.

Use the Commutative Property of Addition to find the sum.

78

+ 68

= 78

+ 1 58 8

= 7 18 8

+ 58

= 88

+ 58

= 518

Evaluate Reasonableness of Sums of Fractions with Equal Denominators

Ryan walked 26

mile to the city library. Then he walked another 56

mile to the city

park. Find the distance Ryan walked to the library and the park.



and 56

.

The model shows Ryan traveled 26

+ 56

= 76

miles.

The model also shows Ryan traveled 1 + 16

, or 116

miles.

26

116

16

16

16

16

16

+56

116

16

16

16

16

16

+ =



Evaluate the reasonableness of the sum using benchmark fractions.

Compare the addends to the benchmarks 0,12

, and 1.

The sum is less than 12

+ 1 = 112

.

So, 76

, or 116

is a reasonable sum.

116

16

16

16

16

16

0 12

1

116

16

16

16

16

16

0 12

1

26

is closer to12

than 0.56

is closer to 1 than 12

.



Math Background Part II - Subtraction of Fractions

Remembering what you know about adding fractions can be used to help you subtractfractions. You will also use what you know about equivalent fractions and mixednumbers.

Subtracting Fractions with Equal Denominators

To subtract fractions, check the denominators. If the fractions have equaldenominators, subtract the numerators. The denominator stays the same becauseyou are subtracting the same kind of thing.

EXAMPLE:

Model Subtraction of Fractions with Equal Denominators

Subtraction of fractions can be represented using models, pictures, words, andnumbers. A model shows the relationship between the fractions in the problem.

EXAMPLE 1: Reagan lives 56

mile from school. Marissa lives 26

mile from school.

Find how much farther Reagan lives from school than Marissa.

Use the operation of subtraction to solve the problem. Find the difference between

56

and 26

.

Model the problem to find the difference.

The model shows Reagan lives36

mile farther from school than Marissa.


.

— =— Subtract the numerators to get the new numerator

Use the same denominator

06

16

26

36

46

56

66

02

12

22

06

16

26

36

46

56

66

Distance toMarissa's house

Difference is 36

mile

Distance toReagan's house




is12

.

So, the model shows Reagan lives12


Subtract without using a model.

Step 1Write the difference of thefractions as an expression.

Step 2Subtract the numerators

to find the difference.

Step 3Write the difference in simplest form.

56

‒ 26

56

‒ 26

= 36

36

3 36 3

= 12

Either way, Reagan lives 12


EXAMPLE 2: Stella made a poster for her history project. She has decided to fill 78

of the space on the poster with pictures and written information. She has already

filled 38

of the space with pictures. Find the space on the poster she has left to fill

with written information.

Use the operation of subtraction to solve the problem. Find the difference between78

and 38

.

Model the problem to find the difference.

The model shows 78

‒ 38

= 48

.

So, the model shows Stella has 48

of the space to fill with written information.


.

The model shows the simplest form of 48

is 12

.

18

18

18

18

18

18

18

118

18

18

18

18

18

18

18

18

12

12

1



So, the model shows Stella has12

of the space on the poster to fill with written

information.

Subtract without using a model.

Step 1Write the difference of thefractions as an expression.




78

‒ 38

78

‒ 38

= 48

48

4 48 4

= 12

Either way, Emily lives 12


EXAMPLE 3: Gabriel needs to walk 58

mile from his house to get to the city park. He

has already walked38

mile. Find the distance he has left to walk to the city park.

Use a model to find the difference.

The model shows 58

‒ 38

= 28

.

So, the model shows Gabriel has 28

mile left to walk to the city park.

Subtract without using a model.Step 1

Write the difference of thefractions as an expression.




58

‒ 38

58

‒ 38

= 28

28

2 28 2

= 14

Either way, Gabriel has 28

or 14

mile left to walk to the city park.

EXAMPLE 4: Find the difference between37

and57

.

Use a model to represent the problem.

The model shows 57

‒ 37

= 27

.

Subtract without using a model.To subtract fractions with the same denominators, subtract the numerators. Thenwrite the difference over the denominator.

5 3 5 – 3 2–

7 7 7 7

Either way, the difference between57

and37

is27

.

Total distance to city parkDistance already walked



Evaluate Reasonableness of Differences Between Fractionswith Equal Denominators

Yasmin lives910

mile from Melissa. She decides to ride her bicycle to Melissa's house

to work on their homework together. Yasmin has ridden 410

of a mile when she gets

to the city library. Find how much farther she needs to ride to reach Melissa's house.

Use the operation of subtraction to solve the problem.

Model the problem to find the difference between 910

and 410

.

The model shows 910

‒ 410

= 510

.

So, the model shows Yasmin has another510

mile to ride to get to Melissa's house.

Evaluate the reasonableness of the difference using benchmark fractions.

Compare the fractions to 0, 12

, and 1.

Estimate 910

as 1. Estimate 410

as 12

. 1 ‒ 12

= 12

The difference is 510

. The simplest form of 510

is 12

.

So 510

mile is a reasonable distance Yasmin has left to get to Melissa's house.

1110

110

110

110

110

110

110

110

110

110

110

110

110

110

110

110

110

110

110

110

1110

110

110

110

110

110

110

110

110

110

0 12

1

STAAR Category 2 GRADE 4 TEKS 4.4B/4.4G


LESSON 4 - 4.4B & 4.4G

Lesson Focus

For TEKS 4.4B students are expected to determine products of a number and 10 or100 using properties of operations and place value understandings.

For TEKS 4.4G students are expected to round to the nearest 10, 100, or 1,000 or usecompatible numbers to estimate solutions involving whole numbers.

For these TEKS students should be able to apply mathematical process standards todevelop and use strategies and methods for whole number computations and decimalsums and differences in order to solve problems with efficiency and accuracy.

For STAAR Category 2 students should be able to perform operations and representalgebraic relationships.




4.1.C Select tools, including real objects, manipulatives, paper and pencil, andtechnology as appropriate, and techniques, including mental math, estimation,and number sense as appropriate, to solve problems.





PART I PART IInumber sentences estimationequation overestimateexpanded notation underestimate

roundingcompatible numbers



Math Background Part I - Product of a Whole Number and 10 or 100

There are patterns when you multiply numbers by 1, 10, and 100.

Look at these number sentences. Notice the patterns.

5 1 = 5 5 10 = 50 5 100 = 500

6 1 = 6 6 10 = 60 6 100 = 600

7 1 = 7 7 10 = 70 7 100 = 700

8 1 = 8 8 10 = 80 8 100 = 800

When you multiply a number by 1, the product equals the number you started with.

5 1 = 5

When you multiply a number by 10, put a zero at the end of the number you startedwith to give you the product.

5 10 = 50

When you multiply a number by 100, put two zeros at the end of the number youstarted with to give you the product.

5 100 = 500

EXAMPLE 1: What pair of numbers could complete this equation?

10 =

When a number is multiplied by 10, put a zero at the end of that number to get theproduct.

If 15 was in the , the number in the would be 150, so 15 10 = 150.



EXAMPLE 2: Jaime arranged his collection of pennies in rows of 10.



Find the number of pennies in his collection.

There are 12 rows of pennies, with 10 pennies in each row.

When you multiply a number by 10, put a zero at the end of the number you startedwith to give you the product.

12 10 = 120

So, Jamie has 120 pennies in his collection.

EXAMPLE 3: Each section in an auditorium has 200 seats. The auditorium has 8sections. Find the number of seats in the auditorium.

Make a quick sketch to find the solution to the problem.

So, the number of seats in the auditorium is 1,600.

Use place value to find the solution to the problem.

8 200 = 8 2 hundreds

= 16 hundreds

= 1,600

EXAMPLE 4: Fill in the missing numbers.

____ 100 = 200

The missing number is 2 because when you multiply a number by 100 you put twozeros at the end of the number you started with to give you the product.

2 100 = 200

3 ____ = 300

The missing number is 100 because when you multiply a number by 100 you put twozeros at the end of the number you started with to give you the product.

3 100 = 300

4 10 = ____

The missing number is 40 because when you multiply a number by 10 you put onezero at the end of the number you started with to give you the product.

4 10 = 40

14 100 = _______

The missing number is 1,400 because when you multiply a number by 100 you puttwo zeros at the end of the number you started with to give you the product.

14 100 = 1,400

H H H H H H H H H H T

H H H H H H

10 hundreds = 1,000

6 hundreds = 600



EXAMPLE 5: Decide if four different pairs of numbers will complete this equation.

100 =

Will and complete the equation?

These two numbers will NOT work because the problem asks for multiplying anumber by 100. These two numbers would work for multiplying 345 by 10 because345 10 = 3,450.


These two numbers will work because when you multiply 345 by 100, you place twozeros to the right of 234, so 345 100 = 34,500.


These two numbers will NOT work because the problem asks for multiplying anumber by 100. These two numbers would work for multiplying 35 by 1,000because 35 1,000 = 35,000.


These two numbers will NOT work because the problem asks for multiplying anumber by 100. These numbers do not show multiplying by 100.

EXAMPLE 6: Find the product of 3,000 and 10.

Use place value.

The product is 10 times larger than 3,000, so the place value of the 3 increases from1,000 to 10,000.

3,000 10 = 30,000

EXAMPLE 7: Find the product of 436 and 10.

Use the properties of operations and place value.

Multiplying 436 by 10 means that each value from expanded notation is beingmultiplied by 10.

436 = (400 + 30 + 6)

10(436) =

10(400 + 30 + 6) =

10(400) + 10(30) + 10(6)=

4,000 + 300 + 60 = 4,360

345 3,450

345 34,500

35 35,000

345 34,000



Math Background Part II - Estimating Products

Sometimes an exact answer to a problem is not needed. Estimation can be used tofind an answer that is close to the exact answer. Estimation can be used to solvethese problems that ask about how many or approximately how much.

One way to estimate an answer to a problem is to round the numbers before workingthe problem. You can round numbers to the nearest ten, nearest hundred, or nearestthousand. A number line or a set of rounding rules can be used to round numbers.

Rounding Rules

When rounding a number to the nearest ten, look at the ones place.If the digit in the ones place is 0 to 4, the digit in the tens place stays the same.

Change the digit in the ones place to a zero.If the digit in the ones place is 5 to 9, the digit in the tens place rounds to the next

higher ten. Change the digit in the ones place to a zero.

When rounding a number to the nearest hundred, look at the tens place.If the digit in the tens place is 0 to 4, the digit in the hundreds place stays the same.

Change the digits in the ones and tens place to zeros.If the digit in the tens place is 5 to 9, the digit in the hundreds place rounds to the

next higher hundred. Change the digits in the ones and tens places to zeros.

When rounding a number to the nearest thousand, look at the hundreds place.If the digit in the hundreds place is 0 to 4, the digit in the thousands place stays the

same. Change the digits in the ones, tens and hundreds place to zeros.If the digit in the tens place is 5 to 9, the digit in the thousands place rounds to the

next higher thousand. Change the digits in the ones, tens and hundreds places tozeros.

Rounding to Estimate Solutions to Multiplication Problems

One way to estimate an answer to a multiplication problem is to round the numbersbefore working the problem.

Rounding One Factor to Estimate a Product

If one factor is a one-digit number and the other factor is a two-digit number, onlyround the two-digit factor.

EXAMPLE 1: During 4 months Jamie earned $93 each month. About how much didJamie earn during these 4 months?

Since the problem says about how much, estimate the answer.Round 93 to the nearest ten.

On a number line, 93 is closer to 90 than to 100.

The number 93 rounds to 90.

Multiply the estimated amount Jamie earned each month by the number of months.

90 4 360

Jamie earned about $360 during the 4 months.

90 93 100



EXAMPLE 2: A ticket to a school play costs $5. Lucinda sold 74 tickets.Approximately how much did Lucinda collect for the tickets she sold?Estimate the product of 74 $5 to solve the problem.

74 5

70 5 350

Lucinda collected about $350 for the tickets she sold.NOTE: Since 70 is less than 74, 70 5 is less than 74 5. The estimate of 350 isless than the actual product. This is an underestimate. So, Lucinda collected morethan $350 for the tickets she sold.

EXAMPLE 3: The fourth grade play was performed on 4 different days. Each day, all389 tickets were sold. About how many tickets were sold for the 4 days?Estimate the product of 4 389 to solve the problem.

4 389

4 400 1,600

About 1,600 tickets were sold for the 4 days.NOTE: Since 400 is more than 389, 4 400 is more than 4 389. The estimate of1,600 is more than the actual product. This is an overestimate. Less than 1,600tickets were sold for the 4 days.

Rounding Each Factor to Estimate a Product

If both factors are two-digit numbers, round both factors.EXAMPLE: The school auditorium has 38 rows of 53 seats. About how many seatsare in the auditorium?Estimate the product of 38 53 to solve the problem.

38 53

40 50 2,000About 2,000 seats are in the auditorium.NOTE: This estimate is close to the actual product because one factor was roundedup 2 and one factor was rounded down 3.

Front-End Estimation to Estimate a Product

The front digits of factors can be multiplied to estimate a product. The estimatedproduct will always be an underestimate if the front digit of one or both of thefactors is rounded down.EXAMPLE 1: About how much will 8 gallons of paint cost if each gallon costs $33?Estimate the product of 8 $33 to solve the problem.

8 33

8 30 2408 gallons of paint will cost about $240.

NOTE: This is an underestimate because one of the factors was rounded down.

Only 74 was rounded.

Only 389 was rounded.

Both 38 and 53 were rounded.



EXAMPLE 2: About how much will 12 cords of fire wood cost if each cord costs $219?

Estimate the product of 12 $219 to solve the problem.

12 219

10 200 2,000

The 12 cords of wood will cost about $2,000.

This is an underestimate because both factors were rounded down.

Compatible Numbers to Estimate a Product

Another way to estimate is by using compatible numbers. Compatible numbers arenumbers that are easy to add, subtract, multiply, or divide. Using compatiblenumbers makes the computation easier.

Compatible numbers can be helpful when estimating the answer to a multiplication ordivision problem. Changing the numbers to other numbers that form a basic fact canhelp you solve the problem in your head.

EXAMPLE: Estimate the product of 19 32.

Think:19 is close to 2032 is close to 30

Use the basic fact 2 3 6 to help solve the problem in your head: 20 30 600

The product of 19 32 is approximately 600.

Any factor is compatible with a multiple of 10, because there are shortcutsfor multiplying by multiples of 10.

EXAMPLE 1: Kevitt can make 42 skateboards in a week. About how manyskateboards can Kevitt make in 9 weeks?

Estimate the product of 8 42 to solve the problem.

9 42

10 42 420

Kevitt can make about 420 skateboards in 9 weeks.

NOTE: Estimating only one factor gives a product closer to the actual product thanestimating both factors would give.

EXAMPLE 2: Each of the 43 sections of a rodeo arena has 98 seats. About howmany seats are in the rodeo arena?

Estimate the product of 43 98 to solve the problem.

43 98

40 100 4,000

The rodeo arena has about 4,000 seats.

NOTE: This estimate is close to the actual product because one factor is roundeddown 3 and the other factor is rounded up 2.

Only 9 was rounded.

Both 43 and 98 were rounded.



Range Estimates

An estimate can be a range of numbers.

EXAMPLE 1: Lance took pictures at the holiday parade. He used 6 rolls of film. Thenumber of pictures that were taken with each roll of film was from 12 pictures to 24pictures. What could be the total number of pictures Lance took?

Find the least number of pictures Lance could have taken with the 6 rolls.

He could have taken up to 5 rolls with 12 pictures and 1 roll with 24 pictures.

Find the product of 5 12 and then add 24 to find the least number of picturesLance could have taken.

5 12 = 60

60 24 = 84

The least number of pictures Lance could have taken is 84.

Find the greatest number of pictures Lance could have taken with the 6 rolls.

He could have taken up to 5 rolls with 24 pictures and 1 roll with 12 pictures.

Find the product of 5 24 and then add 12 to find the greatest number of picturesLance could have taken.

5 24 = 120

120 12 = 132

The greatest number of pictures Lance could have taken is 132.

Lance took at least 84 pictures, but not more than 132 pictures.

The total number of pictures Lance took is between 84 and 132.

NOTE: A reasonable estimate of the number of pictures Lance took is between 80and 130.



LESSON 5 - 4.4C & 4.4D

Lesson Focus

For TEKS 4.4C students are expected to represent the product of 2 two-digit numbersusing arrays, area models, or equations, including perfect squares through 15 by 15.

For TEKS 4.4D students are expected to use strategies and algorithms, including thestandard algorithm, to multiply up to a four-digit number by a one-digit number andto multiply a two-digit number by a two-digit number. Strategies may include mentalmath, partial products, and the commutative, associative, and distributive properties.

For these TEKS students should be able to apply mathematical process standards todevelop and use strategies and methods for whole number computations and decimalsums and differences in order to solve problems with efficiency and accuracy.

For STAAR Category 2 students should be able to perform operations and representalgebraic relationships.










PART I and PART II PART I and PART IImultiplication partial productsCommutative Property expanded formAssociative Property standard algorithmDistributive Property mental matharray area model



Math Background Part I - Multiplication by a One-Digit Number

Use multiplication to combine two or more groups that are equal in value. You cannot memorize the answer to every possible multiplication problem up to four digitstimes one digit, but you do not have to. What you know about place value andmultiplication facts can be used to multiply greater numbers.

Strategies that can be used to multiply greater numbers include using a model, usingthe Associative Property, using the Commutative Property, using the DistributiveProperty, using place value and expanded form, using partial products, using mentalmath, and using the standard algorithm.

Multiply a Two-Digit Number by a One-Digit Number

Several strategies can be used to multiply a two-digit number by a one-digit number.

EXAMPLE 1: Cowboy Stadium gave tours to 3 groups of 23 children on Wednesday.How many children took the tour on Wednesday?

To solve the problem, multiply 3 23 =

There is more than one way to find the answer to the problem.

Use a Model

Step 1 Step 2 Step 3Use base ten blocksto model 3 23.

Break the model intotens and ones.

Add the tens and the onesto find the product.

3 groups of 2 tens 3 ones (6 1 ten) (3 3 ones)(6 10) (3 3)

60 9

(6 10) + (3 3)60 + 9

69

The model shows 3 23 = 69.

So, 69 children took the tour on Wednesday.

NOTE: In Step 2, the model is broken into two parts.Each part shows a partial product. The partial products are 60 and 9.

Use the Distributive Property

The Distributive Property states that multiplying a sum by a number is the same asmultiplying each addend by the number and then adding the products.

The model shows 3 23.

Think of 20 as 10 + 13. Break apart the model to show 3 (10 + 13).



Use the Distributive Property. Find the product each smaller rectangle represents.Then find the sum of the products.

3 10 = 303 13 = 39

30 + 39 = 69

Think of 23 as a different sum. Think of 23 as 11 + 12. Break apart the model toshow 3 (11 + 12).

Use the Distributive Property. Find the product each smaller rectangle represents.Then find the sum of the products.

3 11 = 333 12 = 36

33 + 36 = 69


Use Place Value and Expanded form

Expanded form is a way to write numbers that shows the place value of each digit.Place value and expanded form can help to multiply.

3 23 = 3 (20 + 3)= (3 20) + (3 3)= 60 + (3 3)= 60 + 9= 69


Use Partial Products

Multiply by finding and listing all the partial products, and then adding. Theproducts you add to get the final product are called partial products.NOTE: Place the two-digit number first when multiplying a one-digit number by a

two-digit number.

23

39

6069

Multiply the ones. 3 3 ones = 9 ones

Multiply the tens. 3 2 tens = 6 tens

T O

Add the partial products.

3

10 13

3

11 12

Write 23 in expanded form.

Use the Distributive Property.

Multiply the tens.

Multiply the ones.





Use the Standard Algorithm

Multiply without listing the partial products.


No matter which way you use to find the product, 69 children took the tour onWednesday.

EXAMPLE 2: Film comes in rolls of 24 pictures. Elliott used 4 rolls of film during afield trip to the zoo. How many pictures did Elliott take?

To solve the problem, multiply 4 24 = .


Use a Model

4 24 = 96

So, Elliott took 96 pictures on the field trip to the zoo.



4 24 = 4 (20 + 4)

= (4 20) + (4 4)

= 80 + 16

= 96


Multiply the ones. 3 3 ones = 9 onesWrite 9 in the ones place.

23

369

T O

Multiply the tens. 3 2 tens = 6 tensWrite 6 in the tens place.

Regroup 10 ones = 1 ten.




Multiply by finding all the partial products, and then adding.

Elliott took 96 pictures on the field trip to the zoo.


Use the standard algorithm to multiply without listing the partial products.


No matter which way you use to find the answer, Elliott took 96 pictures on the fieldtrip to the zoo.

CAUTION: When you regroup, do not multiply the regrouped tens again.

244

168096



T O


T O

Multiply the ones. 4 4 ones = 16 onesSince 16 is 1 ten+ 6 ones, write 6 in the ones placeand write 1 above the tens place.

24

496

Multiply the tens. 4 2 tens = 8 tensAdd the 8 tens to the 1 ten 8 tens 1 ten = 9 tensWrite 9 in the tens place.

1

This 1 ten comes from multiplying the ones by 4.( 4 4 ones = 16 ones)Do not multiply it again.Multiply tens and then add on the 1 ten.

24

496

1



Multiply a Three-Digit Number by a One-Digit Number

If you know how to multiply 1-digit numbers such as 6 7, you can also multiplylarger numbers such as 6 777.

EXAMPLE 1: A box of special order pencils with a school name and logo contains 777pencils. How many pencils are in 6 boxes?

To solve the problem, multiply 777 by 6.

Use a Model

Model 6 777.

6 777 = 4200 + 420 + 42= 4,662

So, there are 4,662 logo pencils in 6 boxes.



6 777 = 6 (700 + 70 + 7)

= (6 700) + (6 70) + (6 7)

= (4,200) + (420) + (42)

= 4,662

So, there are 4,662 logo pencils in 6 boxes.

6

700 70 7+ +

Write 777 in expanded form.

6 4200

700 70 7+ +

Multiply the hundreds.

6 4200 420

700 70 7+ +

Multiply the tens.

6 4200 420 42

700 70 7+ +

Multiply the ones.




Multiply by listing all the partial products, then adding the partial products.777

642

4204,200

4,662

So, there are 4,662 special order pencils in 6 boxes.


Multiply without listing the partial products. Use regrouping.

7776

4,662

So, there are 4,662 special order pencils in 6 boxes.

Using any of the strategies to solve the problem, there are 4,662 special order pencilsin 6 boxes.

Multiply the ones. 6 7 ones = 42

Multiply the tens. 6 7 tens = 420

Multiply the hundreds. 6 7 hundreds = 4,200

Add the partial products. 42 + 420 + 4,200 = 4,662

Multiply the ones 6 7 ones = 42 onesRegroup: 42 ones = 4 tens + 2 onesWrite 2 in the ones place. Write 4 above the tens place.

Multiply the tens 6 7 tens = 42 tensRegroup: 42 tens = 420 = 4 hundreds + 2 tensAdd the 2 tens to the 4 tens you already have 2 tens + 4 tens = 6 tensWrite 6 in the tens place. Write 4 above the hundreds place.

Multiply the hundreds 6 7 hundreds = 42 hundredsRegroup: 42 hundreds = 4,200 = 4 thousands + 2 hundredsAdd the 2 hundreds to the 4 hundreds you already have 2 hundreds + 4 hundreds = 6 hundredsWrite 6 in the hundreds place. Write 4 in the thousands place.

44



Multiply a Four-Digit Number by a One-Digit Number

EXAMPLE: The fire department in a Texas city responded to an average of 5,555 callsper year during 7 years. Find the number of calls they responded to in the 7 years.


To solve the problem, multiply 5,555 by 7. Multiply the value of each digit in the4-digit number by the value of the 1-digit number, one at a time. List the partialproducts and then add.

So, the fire department responded to 38,885 calls during the 7 years.


Multiply without listing the partial products.

So, the fire department responded to 38,885 calls during the 7 years.

Either way, the fire department responded to 38,885 calls during the 7 years.

Multiply the ones.Since 35 ones is 3 tens and 5 ones,Write 5 in the ones place.Write 3 above the tens place so you won't forget it.

55557

38885

Multiply the tens.Since 35 tens is 3 hundreds and 5 tens,add the 5 tens to the 3 tens you already have.Write 8 in the tens place.Write 3 above the hundreds place so you won't forget it.

Multiply the hundreds.Since 35 hundreds is 3 thousands and 5 hundreds,add the 5 hundreds to the 3 hundreds you already have.Write 8 in the hundreds place.Write 3 above the thousands place so you won't forget it.

33 3

Multiply the thousands.Since 35 thousands is 3 ten thousands and 5 thousands,add the 5 thousands to the 3 thousands you already have.Write 8 in the thousands place.Write 3 in the ten thousands place.


Add the partial products. 35 + 350 + 3500 + 35000 = 38885


Multiply the hundreds. 7 5 hundreds = 3500 hundreds

Multiply the thousands. 7 5 thousands = 35000 thousands

55557

35350

35003500038885



Math Background Part II - Multiplication of Two-Digit Numbers

You can not memorize the answer to every possible two-digit by two-digit multiplicationproblem, but you do not have to. What you know about place value and multiplicationfacts, and how to break apart numbers can be used to multiply a two-digit number bya two-digit number.

EXAMPLE 1: For basketball games a professional team sets up seats on the floor.Sections 11 and 12 together have 15 rows with 28 seats in each row. Find thenumber of seats in these two sections.



Using Arrays

An array can be used to solve the problem 15 28.Draw an array to represent the factors.Break apart the array into smaller arrays to show factors broken into tens and ones.Label the smaller arrays.Find the product of each smaller array.Then find the sum of the partial products.

40 + 100 + 80 + 200 = 420

So, the total number of seats in the two sections is 420.

Using Area Models

An area model can be used to solve the problem 15 28.Draw a rectangle to represent the factors.Break apart the model into tens and ones to show the partial products.

20 8

10

28

5

10 20=200

5 20= 100 5 8 =40

10 8= 80

15

10

20

5

8



Write the product for each of the smaller rectangles.

(10 2 tens) (10 8 ones) (5 2 tens) (5 8 ones)

(10 20) (10 8) (5 20) (5 8)

200 80 100 40

Add to find the product for the whole model.

200 + 80 + 100 + 40 = 420


Using Partial Products

The problem 15 28 can be solved by finding and listing partial products, then findingthe sum of the partial products.


Using the Standard Algorithm

The problem 15 28 can be solved without listing partial products.


No matter which way you use to solve the problem, the total number of seats in thesetwo sections is 420.

HTO


281540

10080

200420

Multiply by the ones.5 8 = 405 20 = 100

Multiply by the tens.10 8 = 8010 20 = 200

Multiply by the ones. 5 28 =5 8 = 40 0 ones with 4 tens to regroup5 20 = 100 10 tens + 4 tens = 14 tens

14 tens = 1 hundred 4 tensSo, 5 28 = 140

HTO

2815

140

4


HTOT O2815

140280420

Multiply by the tens. 10 28 =10 8 = 80 8 tens + 0 ones10 20 = 200 2 hundreds

So, 10 28 = 280

1



EXAMPLE 2: A tree farm planted 34 rows of trees with 29 trees in each row. Howmany trees were planted at the tree farm?





So, the total number of trees planted at the tree farm is 986.


The problem 34 29 can be solved without listing partial products.

So, the total number of trees planted at the tree farm is 986.

No matter which way you use to find the answer, the total number of trees planted atthe tree farm is 986.

CAUTION: Be careful with regrouped tens.

HTOT O


342936

27080

600986

Multiply by the ones.9 4 = 369 30 = 270

Multiply by the tens.20 4 = 8020 30 = 600

1

It helps to cross out the regrouped 3 tens onceyou add them to the multiplied tens from 4 2tens. That way you will not add them again bymistake.

2934

116870986

33

When you multiply 4 9,you regroup 3 tens.

When you multiply 3 9,you regroup again. Now youhave 2 tens.

21

HTOT O3429

306

Multiply by the ones. 9 34 =9 4 = 36 6 ones with 3 tens to regroup9 30 = 270 27 tens + 3 tens = 30 tens = 3 hundreds

3 hundreds + 0 tens + 6 onesSo, 9 34 = 300 + 6 = 306

3


HTOT O3429

306680986

Multiply by the tens. 20 34 =20 4 = 80 8 tens + 0 ones20 30 = 600 6 hundreds

6 hundreds + 8 tens + 6 onesSo, 20 34 = 600 + 80 = 680



EXAMPLE 3: A necklace has 14 strings of beads. Each string has 28 beads. Find thenumber of beads in the necklace.

To solve the problem, multiply 14 by 28.



So, there are 392 beads in the necklace.


The problem 15 28 can be solved without listing partial products and using what youknow about regrouping.


Using the Distributive Property

What you know about the Distributive Property of Multiplication can be used to findthe solution to 14 28. Break apart one of the factors before multiplying.

Break apart one factor into numbers that are easy to multiply. 14 x 28 = (10 + 4) x 28

Multiply. 10 x 28 = 2804 x 28 = 112

Add the two products.112280392


Multiply by the ones.8 4 ones = 328 10 ones = 80

Multiply the tens.20 4 tens = 8020 10 tens = 200Add the partial products 32 + 80 + 80 + 200 = 392

TOTens Ones1428328080

200392

TO

1428

3Multiply by the ones 8 x 14 ones = ?

8 x 4 ones = 32 2 ones with 3 tens to regroup8 x 10 ones = 80 8 tens + 3 tens = 11 tens

So, 8 x 14 = 112112

Multiply by the tens 20 x 14 tens = ?20 x 4 ones = 80 8 tens + 0 ones20 x 10 ones = 200 2 hundreds + 0 tens + 0 ones

So, 20 x 14 = 280.

Add the partial products. 112 + 280 = 392

3

112280

392

1428



Using any of these multiplication procedures for multiplying two-digit numbers, thereare 392 beads in the necklace.

CAUTION: Zeros may seem like “nothing” in a factor or product, but they are veryimportant.

EXAMPLE: The website http://users/htcomp.net receives an average of 305 visitsper week. At this rate, about how many visits would the website receive in 4 weeks?

To find the solution to the problem, multiply 305 by 4.

At this rate, the website would receive about 1,220 visits in 4 weeks.

Checking Multiplication

Always check multi-digit multiplication because so many steps are involved that it iseasy to make a mistake.

These are 2 different methods that can be used to check multiplication.Reverse the factors.Use the lattice method.

Jerissa found the product of 24 x 38 = 912. Now she needs to check to make sureher multiplication is correct.

EXAMPLE 1: Jerissa can reverse the factors.

2438

192720912

3824

152760912

Reversing the factors shows her work is correct.

EXAMPLE 2: Jerissa could also use the lattice method.

Step 1: Draw a grid.

Write one factor on top.

Write the other factor on the right.

HTO2

4 x 0 = 0 tens0 tens + 2 tens = 2 tens

4 x 5 = 20 2 tens + 0 onesThere are no tens in 305, but that does not mean wecan forget about the tens.

4 x 300 = 1,200 tens 1 thousand + 2 hundreds

3054

1220

31

1

31

1

If reversing the factors gives the same product,then the multiplication is correct.

If reversing the factors does not give the same product,then one of the products is not correct.

2 4

3

8



Step 2: In each square, write a product.

Multiply the digit at the top of the column by the digitto the right of the row.

Note: Use a diagonal line to separate the digits in each product.

If the product is 1-digit, write the product as 0__.

Write 2 x 3 as .

If the product is 2-digits, write the tens digit in the top leftand write the ones digit in the bottom right.

Write 4 x 3 as .

Step 3: Add along the diagonals.

Begin at the lower right.

For 2-digit sums, add the tens digit to the digits in thenext diagonal.

Step 4: Read the product.

Begin reading the product at the top left and end at the bottom right.24 x 38 = 912

The lattice method shows her work is correct.

Using either method to check her multiplication, Jerissa's product is correct.

06

12

2 4

3

8

12

06

16

32

21

1

9

0

3

8

12

06

16

32

2 4

STAAR Category 2 GRADE 4 TEKS 4.5A/4.4H


LESSON 6 - 4.5A & 4.4H

Lesson Focus

For TEKS 4.5A students are expected to represent multi-step problems involving thefour operations with whole numbers using strip diagrams and equations with a letterstanding for the unknown quantity. Focus for this lesson is multiplication.

For TEKS 4.4H students are expected to solve with fluency one- and two-stepproblems involving multiplication and division, including interpreting remainders.Focus for this lesson is multiplication.

For these TEKS students should be able to apply mathematical process standards todevelop concepts of expressions and equations.

For STAAR Category 2 students should be able to how to perform operations andrepresent algebraic relationships.








PART Isymbolconstantvariableexpressionequationstrip diagram



Math Background Part I - Multi-Step Multiplication Problems

More than one step is often needed to solve math problems. Making a plan can helpto solve multi-step problems. Multi-step multiplication problems can be representedusing strip diagrams and equations.

Describe Relationships Mathematically

Relationships can be described mathematically by replacing words and sentences withnumbers, symbols, expressions, and equations. Describing relationships with numbers,symbols, expressions, and equations can help to solve problems.

A symbol is something that represents something else.

EXAMPLES:The symbol + means add.The symbol – means subtract.The symbol < means less than.

A constant is a quantity that stays the same.EXAMPLES:25 (number of cents in a quarter)12 (number of inches in a foot)12 (number of months is a year)

A variable is a quantity that can have different values. A letter is a symbol that canstand for a variable. Any letter can be chosen to stand for a variable. Often the firstletters of the important words make the meaning of the variables easy to remember.EXAMPLES:n (number of inches tall you are)t (amount of time you spend on homework)c (number of cents in your pocket)

Writing an Expression

An expression names an amount. An expression is a variable or combination ofvariables, numbers, and symbols that represents a mathematical relationship.

An expression can be a number. An expression can be a variable.

EXAMPLE: 6 EXAMPLE: w

An expression can be a combination of numbers, variables, and operations.

EXAMPLES:6 + n6 n

To write an expression that describes what is going on in a word problem, think abouta word expression. Use numbers when you know what they are. Use variables whenyou do not know the numbers.

Problem WordExpression

AlgebraicExpression

You have 3 full boxes of pencils.Write an expression to represent the total number ofpencils.

full box 3 b 3



Writing an Equation

An equation is a mathematical sentence with an equals sign. An equation alwayssays two expressions are equal.EXAMPLES:5 + 3 = 8x + 7 = 15

In an equation with only one variable there may be only one number that will makethe sentence true. In an equation with more than one variable, there is often morethan one way to make the equation true.EXAMPLES:b + 3 = 11P = 4 s

To write an equation, think about which two quantities are equal to each other. Thenwrite an expression for each quantity.

Problem WordExpression

AlgebraicExpression

You have 6 reams of paper.Write an equation to represent the totalnumber of pieces of paper.

ream 6 = pieces r 6 = p



Representing Multi-Step Problems

Multi-step multiplication problems can be represented using strip diagrams andequations.

EXAMPLE 1: A fourth grade class is planning a field trip to a museum for a specialdinosaur exhibit. Tickets will need to be purchased for students and teachers.

How much will it cost to purchase tickets for 117 students and 8 teachers?

Use strip diagrams and equations to solve the problem step-by-step.

It will cost $649 for 117 student and 8 teacher tickets to the dinosaur exhibit.

Dinosaur ExhibitADMISSION

Student $5Teacher $8

STEP 1: Find the total cost of the tickets for the students.

Total cost for 117 students

117 $5 each for 117 students.

s5 117 = s

585 = s

117 117117117

STEP 2: Find the total cost of tickets for the teachers.

Total cost for 117 students

$8 each for 8 teachers.

t8 8 = t

64 = t

8 8 8 8 8 8 8 8

STEP 3: Find the total cost for students and teachers.

c Total cost for students and teachers

Total costfor teachers

64585

Total cost for students

585 64 = c

649 = c



EXAMPLE 2: Shonda's computer has 3 hard drives with 32 gigabytes of space each,and 2 hard drives with 64 gigabytes of space each. The files on her computer use 83gigabytes of space. How much hard drive space does her computer have left?

Use strip diagrams and equations to solve the problem step-by-step.

Shonda has 141 gigabytes of hard drive space left on her computer.

2 hard drives with 64 gigabytes

Total space on 2 hard drives with 64 gigabytesg

64 64

2 64 = g

128 = g

STEP 2: Find how much hard drive space is on 2 harddrives with 64 gigabytes of space each.

Total space on 3 hard drives with 32 gigabytes

323232 3 hard drives with 32 gigabytes

n3 32 = n

96 = n

STEP 1: Find how much hard drive space is on 3 harddrives with 32 gigabytes of space each.Find how much hard drive space is on 3 harddrives with 32 gigabytes of space each.

STEP 3: Find the total hard drive space on the computer.

c Total hard drive space on computer

Total space on 64gigabyte hard drives

12896

Total space on 32gigabyte hard drives

96 128 = c

224 = c

STEP 4: The files use 83 gigabytes of space. Find howmuch hard drive space the computer has left.

224 83 = a

141 = a

224 Total hard drive space on computer

83

Space left Space used

a



LESSON 7 - 4.5C & 4.5D

Lesson Focus

For TEKS 4.5C students are expected to use models to determine the formulas for theperimeter of a rectangle (l + w + l + w or 2l + 2w), including the special form forperimeter of a square (4s) and the area of a rectangle (l x w). This TEKS is notassessed on STAAR.

For TEKS 4.5D students are expected to solve problems related to perimeter and areaof rectangles where dimensions are whole numbers.

For this TEKS students should be able to apply mathematical process standards todevelop concepts of expressions and equations.

For STAAR Category 3 students should be able to demonstrate an understanding ofhow to represent and apply geometry and measurement concepts.







PART I PART I PART IIlength inch areacentimeter perimetermillimeter formula

variable



Math Background Part I - Perimeter

Measure Length

The Grade 4 Reference Materials includes two rulers. One is a metric ruler and one isa customary ruler. To find the length of an object, use a ruler.

EXAMPLE 1: Use a ruler to measure the length of the chain.

Notice that the ruler is numbered in centimeters. The marks between each numberindicate a space that is 1 millimeter in length from mark to mark. Each centimeterequals the length of 10 millimeters.

The 0-centimeter mark on the ruler is placed at the left end of the chain.

The right end of the chain aligns with the mark on the ruler that is 8 marks to theright of the 10, between 10 centimeters and 11 centimeters.

So, the length of the chain is 10 810

centimeters.

EXAMPLE 2: Use the ruler to measure the length of the nail.

Notice that the ruler is marked in inches. The marks between each number indicate a

space that is18

inch in length from mark to mark. Each inch equals 8 times the

length of18

inch.

The 0-inch mark on the ruler is placed at the left end of the nail.

The right end of the nail aligns with the mark on the ruler that is 6 marks to theright of the 4, between 4 inches and 5 inches.

The length of the nail is 468

inches or 4 34

inches.

Perimeter of a Rectangle

Two Greek words join together to form the word perimeter. Peri means around andmetron means measure. Perimeter is the measure of the distance around a figure.To find the perimeter (distance around) a rectangle, add the lengths of all its sides.

0 1 2 3 4 5 6 7 8 9 10 11Centimeters

0 1 2 3 4 5Inches



EXAMPLE 1: A model for a square concrete patio was built using 1-centimeter tiles.A sketch of the model is shown. What is the perimeter of the model of the patio?

The perimeter of the model is the sum of the lengths of all its sides.

A square is a special rectangle. All four of a square’s sides are the same length.

A formula can be used to find the perimeter of a square.

Perimeter (square) is length of side + length of side + length of side + length of side

Perimeter (square) is 4 times length of a side

P = 4s

This formula can be used to find the perimeter of the model of the patio.

P = 4s

P = 4 4

P = 16

Using the formula, the perimeter of the model of the patio is 16 centimeters.

EXAMPLE 2: A model for a rectangular concrete patio was built using 1-inch colortiles. A sketch of the model is shown. What is the perimeter of the model of thepatio?

The perimeter of a rectangle is the sum of the lengths of all its sides.

In a rectangle, opposite sides are the same length.

A formula can be used to find the perimeter of a rectangle.

Perimeter (rectangle) is length + width + length + width

P = l + w + l + w

This formula can be used to find the perimeter of the model.

P = l + w + l + w

P = 10 + 4 + 10 + 4 = 28

Using this formula, the perimeter of the model of the patio is 28 inches.

4 cm

4 cm 4 cm

4 cm

10 in.

4 in. 4 in.

10 in.



Another formula can also be used to find the perimeter of a rectangle.

Perimeter (rectangle) is 2 times length + 2 times width

P = (2l) + (2w)

This formula can also be used to find the perimeter of the model.

P = (2l) + (2w)

P = (2 10) + (2 4)

P = 20 + 8

P = 28

Using this formula, the perimeter of the model of the patio is 28 inches.

So, using either formula, the perimeter of the model of the patio is 28 inches.

EXAMPLE 3: The length of this rectangle is 10 centimeters and the width of thisrectangle is 30 centimeters. Find the perimeter of this rectangle.

Two opposite sides have a length of 10 centimeters and 2 opposite sides have a lengthof 30 centimeters.

Find the perimeter by finding the sum of the lengths of the sides of the rectangle.

30 + 10 + 30 + 10 = 80

The perimeter of the rectangle is 80 centimeters.

Find the perimeter using the formula P = l + w + l + w.

P = l + w + l + w

P = 10 + 30 + 10 + 30

P = 80

Using this formula, the perimeter of the rectangle is 80 centimeters.

Find the perimeter using the formula P = 2l + 2w.

P = (2l) + (2w)

P = (2 10) + (2 30)

P = 20 + 60

P = 80

Using this formula, the perimeter of the rectangle is 80 centimeters.

So, using any of these methods the perimeter of the rectangle is 80 inches.

10 cm

30 cm



Grade 4 Reference Materials

Formulas for the perimeter of a square and a the perimeter of a rectangle are shownon the STAAR Grade 4 Mathematics Reference Materials. These formulas usealgebraic symbols to represent the dimensions of the figures.

The formula given for the perimeter of a square is P = 4s, in which P represents"perimeter" and s represents "length of side". Read the formula as perimeter equals4 times length of side.

The first formula given for the perimeter of a rectangle is P = l + w + l + w, in whichP represents "perimeter", l represents "length" and w represents "width". Read theformula as perimeter equals length plus width plus length plus width.

The second formula given for the perimeter of a rectangle is P = (2l) + (2w), inwhich P represents "perimeter", l represents "length" and w represents "width".Read the formula as perimeter equals two times length plus two times width.

NOTE: The information above is included on STAAR Grade 4 Mathematics ReferenceMaterials. Students are NOT expected to memorize the information on thechart. Students ARE expected to be able to utilize the information

When using a formula to solve a problem, follow these steps:Identify what is being asked in the problem.Identify the type of figure in the problem.Identify the quantities in the problem.Identify the formula needed to solve the problem.Substitute the variables in the formula with the quantities from the problem.Solve the problem.

PerimeterSquare P = 4s

RectangleP = l + w + l + w

orP = (2l) + (2w)



Math Background Part II - Area

Area is the measure of how many unit squares are needed to cover a flat surface.The units used to measure area are based on units of length.

EXAMPLE: The length of each side of the square below is 1 inch. The perimeter ofthe square is 4 inches. The area of the square is measured in square inches.

A square inch is the size of a square that is exactly 1 inch on each side.

Write: 1 square inch

Say: one inch square or one square inch

Area is measured in square units, such as square feet or square meters. The unitused to measure area is the square of the unit used to measure the lengths of thesides.

Here are some units for measuring area:

Customary Units Metric Unitssquare inch square millimetersquare foot square centimetersquare yard square metersquare mile square kilometer

Area of a Rectangle

To find the area of a rectangle, find the number of square units needed to cover therectangle.

EXAMPLE 1: A square patio was built using 1-meter sections of natural stone. Asketch of the patio is shown. Find the area of the patio.

The area of a square can be found by counting the number of squares inside thesquare.

1 in.

1 in.

1 in. 1 in.

1

5

9

13

2

6

10

14

3

7

11

15

4

8

12

16



So, the area of the patio is 16 square meters.

The area of a square can be found by multiplying the number of squares in thelength by the number of squares in the width.

Area = number of squares in the length number of squares in the widthArea = 4 4Area = 16

So, the area of the patio is 16 square meters.

The area of a square can also be found using a formula. The formula for the areaof a square given on the Grade 4 Reference Materials is A = s s

The formula for area of a square can be used to find the area of the patio.A = s sA = 4 4A = 16

So, the area of the patio using the formula for the area of a square is 16 squaremeters.

The area of the patio is 16 square meters using any of these methods to find the areaof a square.

EXAMPLE 2: A rectangular patio was built using 1-foot square concrete tiles. Asketch of the patio is shown. Find the area of the patio.

The area of a rectangle can be found by counting the number of squares inside therectangle.

1

11

21

31

41

5151

2

12

22

32

42

5251

3

13

23

33

43

5351

4

14

24

34

44

5451

5

15

25

35

45

5551

6

16

26

36

46

5651

7

17

27

37

47

5751

8

18

28

38

48

5851

9

19

29

39

49

59

10

20

30

40

50

60

4

4



So, the area of the patio is 60 square feet.

The area of a rectangle can be found by multiplying the number of squares in thelength by the number of squares in the width.

Area = number of squares in the length number of squares in the width

Area = 10 6

Area = 60

So, the area of the patio is 60 square feet.

The area of a rectangle can also be found using a formula. The formula for the areaof a rectangle given on the Grade 4 Reference Materials is A = l w.

The formula for the area of a rectangle can be used to find the area of the patio.

A = l w

A = 10 6

A = 60

So, the area of the patio using the formula for the area of a rectangle is 60 squarefeet.

The area of the patio is 60 square feet using any of these methods to find the area ofa rectangle.

EXAMPLE 3: The length of this rectangle is 10 centimeters and the width of thisrectangle is 30 centimeters. Find the area of this rectangle

The area of a rectangle can be found by multiplying the length of the rectangle bythe width of the rectangle.

10

6

10 cm

30 cm



Area = length width

Area = 10 30

Area = 300

So, the area of the rectangle is 300 square centimeters.

A formula can be used to find the area of the rectangle. The formula for the areaof a rectangle given on the Grade 4 Reference Materials is A = l w.

A = l w

A = 10 30

A = 300

So, the area of the rectangle using the formula for the area of a rectangle is 300square centimeters.

The area of the rectangle is 300 square centimeters using either of these methods tofind the area of a rectangle.

Grade 4 Reference Materials

Formulas for the area of a square and a the area of a rectangle are shown on theSTAAR Grade 4 Mathematics Reference Materials. These formulas use algebraicsymbols to represent the dimensions of the figures.

The formula given for the area of a square is A = s s, in which A represents "area"and s represents "length of side". Read the formula as area equals length of sidetimes length of side.

The formula given for the area of a rectangle is A = l w, in which A represents"area", l represents "length" and w represents "width". Read the formula as areaequals length times width.

NOTE: The information above is included on STAAR Grade 4 Mathematics ReferenceMaterials. Students are NOT expected to memorize the information on the chart.Students ARE expected to be able to utilize the information.

When using a formula to solve a problem, follow these steps:Identify what is being asked in the problem.Identify the type of figure in the problem.Identify the quantities in the problem.Identify the formula needed to solve the problem.Substitute the variables in the formula with the quantities from the problem.Solve the problem.

AreaSquare A = s s

Rectangle A = l w

STAAR Category 3 GRADE 4 TEKS 4.6A/4.6C/4.6D


LESSON 8 - 4.6A, 4.6C & 4.6D

Lesson Focus

For TEKS 4.6A students are expected to identify points, lines, line segments, rays,angles, and perpendicular and parallel lines. Focus for this lesson is rays, angles, andperpendicular lines and parallel lines.

For TEKS 4.6C students are expected to apply knowledge of right angles to identifyacute, right, and obtuse triangles.

For TEKS 4.6D students are expected to classify two-dimensional figures based on thepresence or absence of parallel or perpendicular lines or the presence or absence ofangles of a specified size.

For these TEKS students should be able to apply mathematical process standards toanalyze geometric attributes in order to develop generalizations about their properties.

For STAAR Category 3 students should be able to demonstrate an understanding ofhow to represent and apply geometry and measurement concepts.





PART I PART IIray two-dimensionalendpoint triangleline segment right triangleangle acute triangle symbol obtuse trianglevertex quadrilateralright angle rectangle° symbol squareclassify trapezoidacute angle parallelogramobtuse angle rhombusintersecting linespointperpendicular lines symbol



Math Background Part I - Rays, Angles, and Perpendicular Lines

Rays

A ray is part of a line. A ray has one endpoint and goes on forever in one direction.The symbol

indicates a ray.

Write: AB

Say: ray AB

Angles

An angle is formed when two rays meet at an endpoint. The endpoint where therays meet is called the vertex of the angle. The symbol indicates an angle.

This angle can be named by writing the point on one ray, then writing the vertex,and then writing the point on the other ray. The vertex is always the point named inthe middle.

Write: BAC or CAB Say: angle BAC or angle CAB

This angle can also be named by writing just the letter of the vertex.

Write: A Say: angle A

Types of Angles

Angles are measured in degrees. The ° symbol stands for degrees. The measureof an angle tells the size of the opening between the two rays.

Right Angles

This angle is called a right angle.

The measure of a right angle is 90°.

A small square is placed at the vertexof an angle to show that it is a right angle.

A right angle is used as a benchmark for classifying other angles.

Acute Angles

These angles are called acute angles.

The opening between the rays of an acute angleis smaller than the opening between the rays of a right angle.

So, an acute angle is smaller than a right angle.

The measure of an acute angle is less than 90°.

A B

A

B

C

90°



Obtuse Angles

These angles are called obtuse angles.

The opening between the rays of an obtuse angleis larger than the opening between the rays of a right angle.

So, an obtuse angle is larger than a right angle.

The measure of an obtuse angle is greater than 90°.

Straight Angles

These angles are called straight angles because a straight angle forms a line.

The opening between the rays of a straight angle islarger than the opening between the rays of a right angle.

So, a straight angle is larger than a right angle.

The measure of a straight angle is greater than 90°.

Angle Classification Chart

Grade 4 students should know and be able to use the information in this chart.

Classification of Angles

Type of Angle Example Description

Right angle

Angles in squares and rectangles are rightangles.

The small box at the vertex is a symbol thatshows that the angle is a right angle.

Acute angleAn acute angle is smaller than a right angle.

Compared to a right angle, an acute angleseems more closed.

Obtuse angle

An obtuse angle is larger than a right angle.

Compared to a right angle, an obtuse angleseems more open.

Straight angle

A straight angle is larger than a right angle.

Compared to a right angle, a straight angleis much more open and forms a straight line.

TEACHER NOTE: Make a copy of the Angles Classification chart for each student.Copy the master in the Cardstock folder for Six Weeks 2 - Lesson 8.

S

M

B

P



Intersecting Lines

Intersecting lines are lines that cross. The lines cross at a point.

EXAMPLE: Think about streets that meet and cross at intersections.

Write: AB

intersects CD

at E.

Say: Line AB intersects line CD at point E.

Perpendicular Lines

Two intersecting lines that meet to form square corners are called perpendicularlines. A square corner forms a right angle.

EXAMPLE 1: Line DE is perpendicular to line FG. A small is placed at thevertex of an angle to show that it is a right angle. Angle X is a right angle.

Write: FG

DE

Say: Line FG is perpendicular to line DE.

D E

F

G

X

GAS

SUPERMARKET

Community Center

A

C

E

D

B

The symbol means“is perpendicular to”.



EXAMPLE 2: A piece of paper can be folded in half and in half again. When the pieceof paper is opened up, the folds form perpendicular lines.

Your teacher will give you a sheet of white copy paper. Follow the directions to foldthe sheet of paper to form perpendicular lines. Use a ruler and pencil to trace thefold lines. Put a small square in the corner to show the angles are right angles.

EXAMPLE 3: The bottom and side of a dry erase board form perpendicular lines.

EXAMPLE 4: The rails of a fence are usually perpendicular to the posts of the fence.

EXAMPLE 5: Line segments are perpendicular in a figure if they intersect to formright angles.

This figure shows a right angle at each of the vertices: A, B, C, and D.

The two sides that form each vertex form a right angle and are perpendicular.

Side AB and side BC are perpendicular.

Side BC and side CD are perpendicular.

Side CD and side AD are perpendicular.

Side AD and side AB are perpendicular.

A B

C D

Homework: Read pages 35-43



Math Background Part II - Classifying Two-Dimensional Figures

Two-dimensional figures can be classified based on the presence or absence of anglesof a specified type or the presence or absence of parallel or perpendicular lines.

Classifying Triangles

Triangles are plane, two-dimensional figures with three sides. What you know aboutangles can be used to classify triangles. A triangle can be classified by the measure ofits angles.

Right Triangle

A right triangle is a triangle with one right angle. One angle in a right trianglemeasures exactly 90°. A small square indicates the right angle. These are examplesof right triangles.

Acute Triangle

An acute triangle is a triangle with three acute angles. All three angles measure lessthan a right angle. These are examples of acute triangles.

Obtuse Triangle

An obtuse triangle is a triangle with one obtuse angle. One angle in an obtusetriangle measures greater than a right angle. These are examples of obtuse triangles.

Classifying Quadrilaterals

Quadrilaterals are plane, two-dimensional figures with four sides. Quadrilaterals canbe classified based on the presence or absence of parallel or perpendicular lines andthe presence or absence of angles of a specific type.

Rectangle

A rectangle is a quadrilateral with 2 pairs of parallel sides that are congruent and 4right angles.



The first pair of parallel sides is marked with two tick marks on each side to showthey are congruent. The second pair of parallel sides is marked with two tick markson each side to show they are congruent.All four corners are marked with a small square to show they are right angles.

Square

A square is a quadrilateral with 2 pairs of parallel sides, 4 congruent sides, and 4right angles.

All four sides are marked with one tick mark to show they are congruent.All four corners are marked with a small square to show they are right angles.

Trapezoid

A trapezoid is a quadrilateral with 1 pair of parallel sides.

The arrow marks on the two lines show the lines are parallel.

Parallelogram

A parallelogram is a quadrilateral with 2 pairs of parallel sides.

The first pair of parallel sides is marked with two tick marks on each side to showthey are congruent. The second pair of parallel sides is marked with two tick markson each side to show they are congruent.

Rhombus

A rhombus is a quadrilateral with 2 pairs of parallel sides and 4 congruent sides. Thisis a rhombus.

The first pair of parallel sides is marked with two tick marks on each side to showthey are congruent. The second pair of parallel sides is marked with two tick markson each side to show they are congruent.



Two-Dimensional Figures Classification Chart

Grade 4 students should know and be able to use the information in this chart.

Classification of Triangles

Type of Triangle Examples Description

Right triangle Exactly one angle is a right triangle.

A small square indicates the right angle.

Right triangle Exactly three angles are acute angles.

The angles are smaller than right angles.

Obtuse triangle Exactly one angle is an obtuse angle.

The obtuse angle is larger than a right angle.

Classification of Quadrilaterals

Quadrilateral Example Properties

Rectangle2 pairs of parallel sides2 pairs of congruent sides4 right angles

Square2 pairs of parallel sides4 congruent sides4 right angles

Trapezoid 1 pair of parallel sides

Parallelogram2 pairs of parallel sides2 pairs of congruent sides

Rhombus2 pairs of parallel sides4 congruent sides

Documents

TEKS/STAAR-BASED LESSONS - Having a HOOT in Fourth Grade