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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.
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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