Core Connections, Course 2 Checkpoint Materials

Core Connections, Course 2 Checkpoint Materials

Notes to Students (and their Teachers)

Students master different skills at different speeds. No two students learn exactly the same way at the same time. At some point you will be expected to perform certain skills accurately. Most of the Checkpoint problems incorporate skills that you should have been developing in grades 5 and 6. If you have not mastered these skills yet it does not mean that you will not be successful in this class. However, you may need to do some work outside of class to get caught up on them.

Starting in Chapter 1 and finishing in Chapter 9, there are 9 problems designed as Checkpoint problems. Each one is marked with an icon like the one above and numbered according to the chapter that it is in. After you do each of the Checkpoint problems, check your answers by referring to this section. If your answers are incorrect, you may need some extra practice to develop that skill. The practice sets are keyed to each of the Checkpoint problems in the textbook. Each has the topic clearly labeled, followed by the answers to the corresponding Checkpoint problem and then some completed examples. Next, the complete solution to the Checkpoint problem from the text is given, and there are more problems for you to practice with answers included.

Remember, looking is not the same as doing! You will never become good at any sport by just watching it, and in the same way, reading through the worked examples and understanding the steps is not the same as being able to do the problems yourself. How many of the extra practice problems do you need to try? That is really up to you. Remember that your goal is to be able to do similar problems on your own confidently and accurately. This is your responsibility. You should not expect your teacher to spend time in class going over the solutions to the Checkpoint problem sets. If you are not confident after reading the examples and trying the problems, you should get help outside of class time or talk to your teacher about working with a tutor.

Checkpoint Topics

1. Area and Perimeter of Polygons 2. Multiple Representations of Portions 3. Multiplying Fractions and Decimals 5 Order of Operations 6. Writing and Evaluating Algebraic Expressions 7A. Simplifying Expressions 7B. Displays of Data: Histograms and Box Plots 8. Solving Multi-Step Equations 9. Unit Rates and Proportions

Checkpoints

? 2013 CPM Educational Program. All rights reserved.

1

Checkpoint 1

Problem 1-141

Area and Perimeter of Polygons

Answers to problem 1-141: a. 96 cm2, 40 cm; b. 22 in.2, 25.05 in.; c. 144 cm2, 52 cm; d. 696.67 m2, 114.67 m

Area is the number of square units in a flat region. The formulas to calculate the area of several kinds of polygons are:

RECTANGLE

PARALLELOGRAM

TRAPEZOID

TRIANGLE

h

b

A = bh

h

b

A = bh

b1 h

h

h

b2

A

=

1 2

( b1

+

b2

)

h

b

b

A

=

1 2

bh

Perimeter is the distance around a figure on a flat surface. To calculate the perimeter of a polygon, add together the length of each side.

Example 1: Compute the area and perimeter.

6 feet

5 feet

4 feet 5 feet

6 feet

parallelogram A = bh = 6 ! 4 = 24!feet2 P = 6 + 6 + 5 + 5 = 22!feet

Example 2: Compute the area and perimeter.

7 cm

9 cm 8 cm

6 cm

triangle

A

=

1 2

bh

=

1 2

!6!7

=

21!cm2

P = 6 + 8 + 9 = 23!cm

Now we can go back and solve the original problem.

a. Rectangle: A = bh = 12 ! 8 = 96!cm2 ; P = 8 + 8 + 12 + 12 = 40!cm

b.

Triangle:

A

=

1 2

bh

=

1 2

!11! 4

=

22!in.2 ;

P = 11+ 9.05 + 5 = 25.05!in.

c. Parallelogram: A = bh = 16 ! 9 = 144!cm2 ; P = 16 + 16 + 10 + 10 = 52!cm

d.

Trapezoid:

A P

= =

1221( b+1

2+5b+2

)h

24

= +

1 2

(

25

44.67

+ 44.67 )!10

= 114.67!m

=

696.67

!

m2

;

2

? 2013 CPM Educational Program. All rights reserved.

Core Connections, Course 2

Here are some more to try. Find the area and perimeter of each figure.

9 cm

1.

11 cm

2.

10 in.

12 in. 6 in.

7.2 in.

3. Trapezoid

35 feet

12.7 feet

12.5 feet

20 feet feetfeet

5. Parallelogram

17.8 feet

4. Parallelogram

10.8 cm

10 cm 10.8 cm 16 cm

6.

5 in. 4 in.

12 in.

9 feet 9 feet

4 feet

5 feet

7. Trapezoid

15 cm

8.1 cm

8 cm

9.3 cm

9.9 in.

8.

6.3 m 10 m

6 m

9 cm

6 m

42 feet

9. Parallelogram

18 in.

16 in. 18 in.

29 in.

10.

27 feet

24.5 feet

39.2 feet

Checkpoints

? 2013 CPM Educational Program. All rights reserved.

3

11.

12. Trapezoid

21 in.

2 feet

20 in.

13 in.

32.7 in.

13. Trapezoid

18 feet

19.8 feet

16 feet

14.

16.2 feet

24 in. 5.1 in. 5 in. 3 in. 3.2 in.

32 feet

15. Parallelogram

10 cm

9 cm 10 cm

10 cm

Answers:

1. 99 cm2, 40 cm

3. 343.75 feet2, 85.5 feet

5. 36 feet2, 28 feet

7. 96 cm2, 41.4 cm

9. 464 in.2, 94 in.

11.

14 4

=

3

1 2

feet2

,

30 4

=

7

1 2

feet

13. 400 feet2, 86 feet

15. 90 cm2, 40 cm

16. Trapezoid

2.7 m

0.8 m

0.5 m

4.4 m

1.2 m

2. 36 in.2, 29.2 in. 4. 160 cm2, 53.6 cm 6. 24 in.2, 26.9 in. 8. 18 m2, 22.3 m 10. 514.5 feet2, 108.2 feet 12. 457 in.2, 90.7 in. 14. 7.65 in.2, 13.3 in. 16. 1.78 m2, 9.1 m

4

? 2013 CPM Educational Program. All rights reserved.

Core Connections, Course 2

Checkpoint 2

Problem 2-120

Multiple Representations of Portions

Answers to problem 2-120:

a. 43%,

43 100

,

b.

9 10

,

0.9,

90%

,

c.

39 100

,

0.39

,

d. 64%, 0.64

Portions of a whole may be represented in various ways as represented by this web. Percent means "per hundred" and the place value of a decimal will determine its name. Change a fraction in an equivalent fraction with 100 parts to name it as a percent.

fraction

words or

pictures

decimal

percent

Representations of a Portion

Example 1: Name the given portion as a fraction and as a percent. 0.3

Solution:

The

digit

3

is

in

the

tenths

place

so

0.3

=

three

tenths

=

3 10

.

On a diagram or a hundreds grid, 3 parts out of 10 is equivalent to 30 parts out of 100

so

3 10

=

30 100

=

30% .

Example 2: Name the given portion as a fraction and as a decimal. 35%

Solution:

35% =

35 100

=

thirty-five hundredths = 0.35 ;

35 100

=

7 20

Now we can go back and solve the original problem.

a.

0.43 is forty-three hundredths or

43 100

= 43%

b.

nine tenths is

9 10

=

9 10

!

10 10

=

90 100

=

90% ;

9 10

=

0.9

c.

39%

=

39 100

= ! thirty-nine

hundredths

=

0.39

d.

16 25

=

16 25

!

4 4

=

64 100

=

0.64

=

64%

Checkpoints

? 2013 CPM Educational Program. All rights reserved.

5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download