skip counting let’s skip count by 3s to 30. 3 6 9 12 15 18 21 24 27 30

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Module 4 Lesson 11 Demonstrate the possible whole number side lengths of rectangles with areas of 24, 36, 48, or 72 square units using the associative property.

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Page 1: Skip counting Let’s skip count by 3s to 30. 3 6 9 12 15 18 21 24 27 30

Module 4 Lesson 11Demonstrate the possible whole number

side lengths of rectangles with areas of 24, 36, 48, or 72 square units using the

associative property.

Page 2: Skip counting Let’s skip count by 3s to 30. 3 6 9 12 15 18 21 24 27 30

Skip countingLet’s skip count by 3s to 30.

36

9

1215

18

21

24

27

30

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Skip countingLet’s skip count by 6s to 60.

612

18

2430

36

42

48

54

60

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Skip countingLet’s skip count by 7s to 70.

714

21

2835

42

49

56

63

70

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Skip countingLet’s skip count by 8s to 80.

816

24

3240

48

56

64

72

80

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Skip countingLet’s skip count by 9s to 90.

918

27

3645

54

63

72

81

90

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Find the Unknown Factor

6 x _____ = 12

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

2

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Find the Unknown Factor

4 x _____ = 12

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

3

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Find the Unknown Factor

3 x _____ = 12

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

4

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Find the Unknown Factor

3 x _____ = 24

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

8

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Find the Unknown Factor

4 x _____ = 24

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

6

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Find the Unknown Factor

8 x _____ = 24

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

3

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Find the Unknown Factor

6 x _____ = 36

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

6

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Find the Unknown Factor

4 x _____ = 36

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

9

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Find the Unknown Factor

9 x _____ = 36

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

4

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Find the Unknown Factor

9 x _____ = 72

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

8

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Find the Unknown Factor

8 x _____ = 72

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

9

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Find the Unknown Factor

8 x _____ = 48

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

6

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Find the Unknown Factor

2 x _____ = 24

FIND THE UNKNOWN FACTOR AND SAY THE EQUATION.

12

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Find the Area

On your white boards writeAn expression that we couldUse to solve the area of the SHADED rectangle.

On your white boards writeAn expression that we couldUse to solve the area of the UNSHADED rectangle. (3 x 5) (3 x

3)

HOW CAN WE USE THESE EXPRESSIONS TOFIND THE AREA OF THE LARGE RECTANGLE?

ADD THEM!! SO WHAT IS THE AREA?

15 + 9 = 24 sq. units

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Find the Area

On your white boards writeAn expression that we couldUse to solve the area of the SHADED rectangle.

On your white boards writeAn expression that we couldUse to solve the area of the UNSHADED rectangle. (3 x 10) (3 x

7)

HOW CAN WE USE THESE EXPRESSIONS TOFIND THE AREA OF THE LARGE RECTANGLE?

ADD THEM!! SO WHAT IS THE AREA?

30 + 21 = 51 sq. units

710

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Problem of the Day

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Concept Development

3

12

Write an expression to show how to find thearea of the rectangle.

3 x 12In the problem of the day you found what 3 x 12 equals… 36 square units!

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3

12

3 x (2 x 6) Why is this expression equal to the one you just wrote?

Write this expression on your white boards with the parentheses in a different place. When I put my hands in the air show me your boards.

(3 x 2) x 6

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(3 x 2) x 6

SOLVE 3 x 2 and then write the new expression on your boards.

6 x 6 What new possible side lengths did we find for a rectangle with an area of 36 squareunits? 6 and 6

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Let’s look at our expression, (3 x 2) x 6 again. Use the commutative property and switch the order of the factors in the parentheses.

(2 x 3) x 6

Will you be able to find new sidelengths by moving the parentheses?

2 x (3 x 6) and yes… are two new side lengths are 2 and 18.

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How is 3 x (3 x 4) equal to our original expression of 3 x 12? WRITE THIS EXPRESSION WITH THE

PARENTHESES IN A DIFFERENT PLACE. AT MY SIGNAL (dog picture), SHOW ME YOUR BOARD.

(3 x 3) x 4

What new side lengths did we find for a rectangle with an area of 36 sq. Units?

9 and 4

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LET’S LOOK AT OUR EXPRESSION:(3 X 3) X 4 AGAIN.

If I use the commutative property and switch the order of the factors in the parentheses, will I be able to find new side lengths by moving the parentheses?

NO – BECAUSE BOTH FACTORS ARE THE SAME.

Do you think we found all the possible whole number side lengths for thisrectangle?

Do we have a side length of 1?

A side length of 2? of 4?

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Work with your partner to look at the rest of your side lengths to see if you have thenumbers 4 – 10. Which of these numbers 4 through 10, aren’t included in your side

lengths?

Which numbers aren’t included? 5, 7, 8, and 10WHY NOT?

NOW DO YOU THINK WE FOUND ALL THE POSSIBLE SIDE LENGTHS?

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PROBLEM SET