NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
146
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Lesson 12: Decimal Expansions of Fractions, Part 2
Student Outcomes
Students develop an alternative method for computing the decimal expansion of a rational number.
Lesson Notes
In this lesson, students use the ideas developed in the previous lesson for finding decimal approximations to quantities
and apply them to computing the decimal expansion of rational numbers. This produces a method that allows one to
compute decimal expansions of fractions without resorting to the long-division algorithm. The general strategy is to
compare a rational number, written as a mixed number, with a decimal: 35
11= 3
211
= 3.1 + something smaller than a
tenth. The process continues until a repeating pattern emerges, as it must.
This lesson also includes a Rapid White Board Exchange fluency activity on the side topic of volume. It takes
approximately 10 minutes to complete and can be found at the end of this lesson.
Classwork
Discussion (20 minutes)
Example 1
Find the decimal expansion of ππ
ππ.
For fun, letβs see if we can find the decimal expansion of 35
11 without using long division.
To start, can we say between which two integers this number lies?
The number 35
11 would lie between 3 and 4 on the number line because
35
11=
33
11+
2
11= 3 +
2
11.
Could we say in which tenth between 3 and 4 the number 3 +2
11 lies? Is this tricky?
Provide time for students to discuss strategies in small groups; then, share their ideas with the class. Encourage students
to critique the reasoning of their classmates.
MP.3
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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We know that 35
11 has a decimal expansion beginning with 3 in the ones place because
35
11= 3 +
2
11. Now we
want to determine the tenths digit, the hundredths digit, and then the thousandths digit.
π. Ones Tenths Hundredths Thousandths
To figure out the tenths digit, we will use an inequality based on tenths. We are
looking for the consecutive integers, π and π + 1, so that
3 +π10
<3511
< 3 +π + 1
10 .
We can rewrite the middle term:
3 +π10
< 3 +2
11< 3 +
π + 110
.
This means weβre looking at
π
10<
2
11<
π + 1
10 .
Give students time to make sense of the inequalities 3 +π10
<3511
< 3 +π + 1
10 and
π
10<
2
11<
π + 1
10.
Since the intervals of tenths are represented by π
10 and
π + 1
10, consider using concrete numbers. The chart below may
help students make sense of the intervals and the inequalities.
Scaffolding:
An alternative way of asking this question is βIn which interval could we place the
fraction 2
11?β Show students
the number line labeled with tenths.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Multiplying through by 10, we get
π < 10 (2
11) < π + 1
π <20
11< π + 1.
So now we are asking the following: Between which two integers does 20
11 lie?
We have 20
11 is between
11
11= 1 and
22
11= 2. That is,
20
11 lies between 1 and 2.
(Some students might observe 20
11= 1 +
9
11, which again shows that
20
11 lies between 1 and 2.)
So we have that 1 <2011
< 2. This means that 1
10<
2
11<
2
10, and consequently 3 +
110
< 3 +2
11< 3 +
210
,
which is what we were first looking for.
So what does this say about the location of 35
11= 3 +
2
11 on the number line?
It means that 35
11 lies between 3.1 and 3.2 and that we now know the decimal expansion of
35
11 has a 1
in the tenths place.
π. π Ones Tenths Hundredths Thousandths
Now can we pin down in which hundredth interval 35
11 lies?
This time we are looking for the integer π with
3 +1
10+
π
100<
35
11< 3 +
1
10+
π + 1
100.
Provide time for students to make sense of this.
Subtracting 3 and 1
10 throughout leaves us with
π
100<
35
11β 3 β
1
10<
π + 1
100.
So we need to compute 35
11β 3 β
1
10, which is equal to 3 +
211
β 3 β1
10, which is equal to
2
11β
1
10. And we
can do that:
2
11β
1
10=
20
110β
11
110=
9
110.
So we are left thinking about
π
100<
9
110<
π + 1
100.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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We are now looking for consecutive integers π and π + 1 so that
π
100<
9
110<
π + 1
100.
Letβs multiply through by some number to make the integers π and π + 1 easier to access. Which number
should we multiply through by?
Multiplying through by 100 will eliminate the fractions at the beginning and at the end of the
inequality.
Multiplying through by 100, we get
π <900
110< π + 1.
This is now asking the following: Between which two integers, π and π + 1, does 900
110 lie?
900
110 or
90
11 is between
88
11, which is 8 and
99
11, which is 9. (Or students might observe
90
11= 8 +
2
11.)
Now we know that π = 8. What was the original inequality we were looking at?
3 +1
10+
π100
<3511
< 3 +1
10+
π + 1100
So we have 3 +1
10+
8100
<3511
< 3 +1
10+
9100
, telling us that 3511
lies between 3.18 and 3.19 and so has an 8
in the hundredthβs place of its decimal expansion.
π. π π Ones Tenths Hundredths Thousandths
Now we wonder in which thousandth 35
11 or 3 +
211
lies. We now seek the integer π where
3 +1
10+
8
100+
π
1000< 3 +
2
11< 3 +
1
10+
8
100+
π + 1
1000.
Subtracting 3 and 1
10 and
8
100 throughout gives
π
1000< 3 +
2
11β 3 β
1
10β
8
100<
π + 1
1000
π
1000<
2
11β
1
10β
8
100<
π + 1
1000.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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We need to work out 2
11β
1
10β
8
100:
2
11β (
1
10+
8
100) =
2
11β
18
100
=200
1100β
198
1100
=2
1100.
So we are looking for the integer π that fits the inequality
π
1000<
2
1100<
π + 1
1000.
Multiplying through by 1000 gives
π <2011
< π + 1.
Now we are asking the following: Between which two integers does 20
11 lie? What value of π do we need?
We have that 20
11 lies between
11
11, which is 1 and
22
11, which is 2. We also see this by writing
20
11= 1 +
9
11. We need π = 1.
Have we asked and answered this very question before?
Yes. It came up when we were looking for the tenths decimal digit.
Back to our original equation, substituting π = 1 now shows
3 +1
10+
8
100+
1
1000< 3 +
2
11< 3 +
1
10+
8
100+
2
1000.
Therefore, the next digit in the decimal expansion of 35
11 is a 1:
π. π π π Ones Tenths Hundredths Thousandths
We seem to be repeating our work now, so it is natural to ask the following: Have we reached the repeating
part of the decimal? That is, does 35
11= 3.18181818β¦ ? There are two ways we could think about this. Letβs
work out which fraction has a decimal expansion of 3.181818β¦ and see if it is 35
11.
Have students compute 0.181818β¦ =1899
=2
11 and so observe that 3.181818β¦ = 3 +
211
=3511
.
We do indeed now have the correct repeating decimal expansion.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Or we could try to argue directly that we must indeed be repeating our work. We do this as follows:
We first saw the fraction 2
11 when we asked in which interval of a tenth our decimal lies,
π
10<
2
11<
π + 1
10.
We next saw the fraction 2
11 when asking about which thousandth the decimal lies,
π
1000<
2
1100<
π + 1
1000,
which after multiplying through by 100 gives π
10<
2
11<
π + 1
10. In each scenario, we are two-elevenths along
an interval, one at the tenths scale and one at the thousandths scale. The situations are the same, just at
different scales, and so the same work applies. We are indeed in a repeating pattern of work.
(This argument is very subtle. Reassure students that they can always check any repeating decimal expansion
they suspect is correct by computing the fraction that goes with that expansion.)
Of course, we can also compute the decimal expansion of a fraction with the long division algorithm, too.
Exercises 1β3 (5 minutes)
Students work independently or in pairs to complete Exercises 1β3.
Exercises 1β3
1. Find the decimal expansion of π
π without using long division.
π
π=
π
π+
π
π
= π +π
π
The decimal expansion begins with the integer π.
Among the intervals of tenths, we are looking for integers π and π + π so that
π +πππ
< π +ππ
< π +π + π
ππ ,
which is the same as
π
ππ<
π
π<
π + π
ππ
π <ππ
π< π + π
and
ππ
π=
ππ
π+
π
π
= π +π
π.
The tenths digit is π.
Among the intervals of hundredths we are looking for integers π and π + π so that
π +π
ππ+
ππππ
<ππ
< π +π
ππ+
π + ππππ
,
which is equivalent to
π
πππ<
π
πβ
π
ππ<
π + π
πππ .
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Now
π
πβ
π
ππ=
π
ππ .
So we are looking for integers π and π + π where
π
πππ<
π
ππ<
π + π
πππ ,
which is the same as
π <πππ
< π + π .
But we already know that ππ
π= π +
π
π; therefore, the hundredths digit is π. We feel like we are repeating our work,
so we suspect π
π= π. πππβ¦. To check: π. ππππβ¦ =
ππ
=ππ
and π. ππππβ¦ = π +ππ
=ππ
. We are correct.
2. Find the decimal expansion of π
ππ without using long division.
Its decimal expansion begins with the integer π.
In the intervals of tenths, we are looking for integers π and π + π so that
π
ππ<
π
ππ<
π + π
ππ ,
which is the same as
π <ππ
ππ< π + π
ππ
ππ=
ππ
ππ+
π
ππ
= π +π
ππ
The tenths digit is π.
In the intervals of hundredths, we are looking for integers π and π + π so that
π
ππ+
π
πππ<
π
ππ<
π
ππ+
π + π
πππ .
This is equivalent to
π
πππ<
π
ππβ
π
ππ<
π + π
πππ .
Now
π
ππβ
π
ππ=
π
πππ ,
so we are looking for integers π and π + π where
π
πππ<
π
πππ<
π + π
πππ ,
which is the same as
π <ππππ
< π + π.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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As
ππ
ππ=
ππ
ππ+
π
ππ
= π +π
ππ
we see that the hundredths digit is π.
The fraction π
ππ has reappeared, which makes us suspect we are in a repeating pattern and we have
π
ππ= π. ππππππβ¦. To check: π. ππππππβ¦ =
ππππ
=π
ππ. We are correct.
3. Find the decimal expansion of the number ππ
ππ first without using long division and then again using long division.
The decimal expansion begins with the integer π.
In the interval of tenths, we are looking for integers π and π + π so that
π
ππ<
ππ
ππ<
π + π
ππ,
which is the same as
π <πππππ
< π + π.
Now
πππ
ππ=
πππ
ππ+
ππ
ππ
= π +ππ
ππ
showing that the tenths digit is π.
In the interval of hundredths, we are looking for integers π and π + π so that
π
ππ+
π
πππ<
ππ
ππ<
π
ππ+
π + π
πππ,
which is equivalent to
π
πππ<
ππ
ππβ
π
ππ<
π + π
πππ.
Now
ππ
ππβ
π
ππ=
ππ
πππ,
so we want
π
πππ<
ππ
πππ<
π + π
πππ,
which is the same as
π <πππππ
< π + π.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Now
πππ
ππ=
πππ
ππ+
ππ
ππ
= π +ππ
ππ
The hundredths digit is π. The reappearance of ππ
ππ makes us suspect that weβre in a repeating pattern and
ππ
ππ= π. ππππππβ¦. We check that π. ππππππβ¦ does indeed equal
ππ
ππ , and we are correct.
Fluency Exercise (10 minutes): Area and Volume I
Refer to the Rapid White Board Exchanges section in the Module 1 Module Overview for directions to administer a Rapid
White Board Exchange.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson.
We have an alternative method for computing the decimal expansions of rational numbers.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Exit Ticket (5 minutes)
Lesson Summary
For rational numbers, there is no need to guess and check in which interval of tenths, hundredths, or thousandths
the number will lie.
For example, to determine where the fraction π
π lies in the interval of tenths, compute using the following
inequality:
π
ππ<
π
π<
π + π
ππ
π <ππ
π< π + π
π < ππ
π< π + π
Use the denominator of ππ because we need to find the tenths digit of π
π.
Multiply through by ππ.
Simplify the fraction ππ
π.
The last inequality implies that π = π and π + π = π because π < πππ
< π. Then, the tenths digit of the decimal
expansion of π
π is π.
To find in which interval of hundredths π
π lies, we seek consecutive integers π and π + π so that
π
ππ+
π
πππ<
π
π<
π
ππ+
π + π
πππ.
This is equivalent to
π
πππ<
π
πβ
π
ππ<
π + π
πππ,
so we compute π
πβ
π
ππ=
π
ππ=
π
ππ. We have
π
πππ<
π
ππ<
π + π
πππ.
Multiplying through by πππ gives
π <πππ
< π + π.
This inequality implies that π = π and π + π = π because π < πππ
< π. Then, the hundredths digit of the decimal
expansion of π
π is π.
We can continue the process until the decimal expansion is complete or until we suspect a repeating pattern that
we can verify.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Name Date
Lesson 12: Decimal Expansions of Fractions, Part 2
Exit Ticket
Find the decimal expansion of 41
6 without using long division.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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Exit Ticket Sample Solutions
Find the decimal expansion of ππ
π without using long division.
ππ
π=
ππ
π+
π
π
= π +π
π
The ones digit is π.
To determine in which interval of tenths the fraction lies, we look for integers π and π + π so that
π +πππ
< π +ππ
< π +π + π
ππ,
which is the same as
π
ππ<
π
π<
π + π
ππ
π <πππ
< π + π.
We compute
ππ
π=
ππ
π+
π
π
= π +π
π
The tenths digit is π.
To determine in which interval of hundredths the fraction lies, we look for integers π and π + π so that
π +π
ππ+
ππππ
< π +ππ
< π +π
ππ+
π + ππππ
,
which is the same as
π
πππ<
π
πβ
π
ππ<
π + π
πππ.
Now
π
πβ
π
ππ=
π
ππ,
so we have
π
πππ<
π
ππ<
π + π
πππ.
This is the same as
π <πππ
< π + π.
Since
ππ
π= π +
π
π,
the hundredths digit is π.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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We are a third over a whole number of tenths and a third over a whole number of hundredths. We suspect we are in a
repeating pattern and that ππ
π= π. πππππβ¦.
To check:
π = π. ππππβ¦
πππ = ππ. ππππβ¦
πππ = ππ + π. ππππβ¦
πππ = ππ +π
π
πππ =πππ
π+
π
π
πππ =πππ
π
π =πππ
ππ
π =ππ
π
We are correct.
Problem Set Sample Solutions
1. Without using long division, explain why the tenths digit of π
ππ is a π.
In the interval of tenths, we are looking for integers π and π + π so that
π
ππ<
π
ππ<
π + π
ππ,
which is the same as
π <ππ
ππ< π + π
ππ
ππ=
ππ
ππ+
π
ππ
= π +π
ππ
In looking at the interval of tenths, we see that the number π
ππ must be between
π
ππ and
π
ππ because
π
ππ<
π
ππ<
π
ππ.
For this reason, the tenths digit of the decimal expansion of π
ππ must be π.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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2. Find the decimal expansion of ππ
π without using long division.
ππ
π=
ππ
π+
π
π
= π +π
π
The ones digit is π. In the interval of tenths, we are looking for integers π and π + π so that
π
ππ<
π
π<
π+π
ππ,
which is the same as
π <ππ
π< π + π
ππ
π=
ππ
π+
π
π
= π +π
π
The tenths digit is π. The difference between π
π and
π
ππ is
π
πβ
π
ππ=
π
ππ.
In the interval of hundredths, we are looking for integers π and π + π so that
π
πππ<
π
ππ<
π + π
πππ,
which is the same as
π <πππ
< π + π.
However, we already know that ππ
π= π +
π
π; therefore, the hundredths digit is π. Because we keep getting
π
π, we
can assume the digit of π will continue to repeat. Therefore, the decimal expansion of ππ
π is π. πππβ¦.
3. Find the decimal expansion of ππ
ππ to at least π digits without using long division.
In the interval of tenths, we are looking for integers π and π + π so that
π
ππ<
ππ
ππ<
π + π
ππ,
which is the same as
π <πππ
ππ< π + π
πππ
ππ=
ππ
ππ+
ππ
ππ= π +
ππ
ππ.
The tenths digit is π. The difference between ππ
ππ and
π
ππ is
ππ
ππβ
π
ππ=
ππ
πππ.
In the interval of hundredths, we are looking for integers π and π + π so that
π
πππ<
ππ
πππ<
π + π
πππ,
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
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which is the same as
π <πππ
ππ< π + π
πππ
ππ=
πππ
ππ+
ππ
ππ= π +
ππ
ππ.
The hundredths digit is π. The difference between ππ
ππ and (
πππ
+π
πππ) is
ππ
ππβ (
π
ππ+
π
πππ) =
ππ
ππβ
ππ
πππ=
ππ
ππππ.
In the interval of thousandths, we are looking for integers π and π + π so that
π
ππππ<
ππ
ππππ<
π + π
ππππ,
which is the same as
π <πππ
ππ< π + π
πππ
ππ=
πππ
ππ+
ππ
ππ= π +
ππ
ππ.
The thousandths digit is π. The difference between ππ
ππ and (
πππ
+π
πππ+
πππππ
) is
ππ
ππβ (
π
ππ+
π
πππ+
π
ππππ) =
ππ
ππβ
πππ
ππππ=
ππ
πππππ.
In the interval of ten-thousandths, we are looking for integers π and π + π so that
π
πππππ<
ππ
πππππ<
π + π
πππππ,
which is the same as
π <πππ
ππ< π + π
πππ
ππ=
ππ
ππ+
ππ
ππ= π +
ππ
ππ.
The ten-thousandths digit is π. The difference between ππ
ππ and (
πππ
+π
πππ+
πππππ
+π
πππππ) is
ππ
ππβ (
π
ππ+
π
πππ+
π
ππππ+
π
πππππ) =
ππ
ππβ
ππππ
πππππ=
ππ
ππππππ.
In the interval of hundred-thousandths, we are looking for integers π and π + π so that
π
ππππππ<
ππ
ππππππ<
π + π
ππππππ,
which is the same as
π <πππ
ππ< π + π
πππ
ππ=
πππ
ππ+
ππ
ππ= π +
ππ
ππ.
The hundred-thousandths digit is π. We see again the fraction ππ
ππ, so we can expect the decimal digits to repeat at
this point. Therefore, the decimal approximation of ππ
ππ is π. ππππππππππβ¦.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
161
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
4. Which number is larger, βππ or ππ
π? Answer this question without using long division.
The number βππ is between π and π. In the sequence of tenths, βππ is between π. π and π. π because
π. ππ < (βππ)π
< π. ππ. In the sequence of hundredths, βππ is between π. ππ and π. ππ because
π. πππ < (βππ)π
< π. πππ. In the sequence of thousandths, βππ is between π. πππ and π. πππ because
π. ππππ < (βππ)π
< π. ππππ. The decimal expansion of βππ is approximately π. πππβ¦.
ππ
π=
ππ
π+
π
π
= π +π
π
In the interval of tenths, we are looking for the integers π and π + π so that
π
ππ<
π
π<
π + π
ππ,
which is the same as
π <ππ
π< π + π
ππ
π=
π
π+
π
π
= π +π
π
The tenths digit is π. Since the fraction π
π has reappeared, we can assume that the next digit is also π, and the work
will continue to repeat. Therefore, the decimal expansion of ππ
π is π. ππππβ¦.
Therefore, ππ
π< βππ.
Alternatively: (βππ)π
= ππ and (ππ
π)
π=
πππ
ππ, which is less than
πππ
ππ or ππ. Thus,
ππ
π is the smaller number.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
162
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This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
5. Sam says that π
ππ= π. ππ, and Jaylen says that
π
ππ= π. πππ. Who is correct? Why?
In the interval of tenths, we are looking for integers π and π + π so that
π
ππ<
π
ππ<
π + π
ππ,
which is the same as
π <ππ
ππ<
π + π
ππ
ππ
ππ=
ππ
ππ+
π
ππ
= π +π
ππ
The tenths digit is π. The difference between π
ππ and
π
ππ is
π
ππβ
π
ππ=
π
πππ.
In the interval of hundredths, we are looking for integers π and π + π so that
π
πππ<
π
πππ<
π + π
πππ,
which is the same as
π <ππ
ππ< π + π
ππ
ππ=
ππ
ππ+
π
ππ
= π +π
ππ
The hundredths digit is π. Again, we see the fraction π
ππ, which means the next decimal digit will be π, as it was in
the tenths place. This means we will again see the fraction π
ππ, meaning we will have another digit of π. Therefore,
the decimal expansion of π
ππ is π. ππππβ¦.
Technically, Sam and Jaylen are incorrect because the fraction π
ππ is an infinite decimal. However, Sam is correct to
the first two decimal digits of the number, and Jaylen is correct to the first three decimal digits of the number.
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
163
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from G8-M7-TE-1.3.0-10.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
Area and Volume I
1. Find the area of the square shown below.
π¨ = (π π¦)π
= ππ π¦π
2. Find the volume of the cube shown below.
π½ = (π π¦)π
= ππ π¦π
3. Find the area of the rectangle shown below.
π¨ = (π ππ¦)(π ππ¦)
= ππ ππ¦π
4. Find the volume of the rectangular prism shown below.
π½ = (ππ ππ¦π)(π ππ¦)
= πππ ππ¦π
5. Find the area of the circle shown below.
π¨ = (π π¦)ππ
= πππ π¦π
NYS COMMON CORE MATHEMATICS CURRICULUM 8β’7 Lesson 12
Lesson 12: Decimal Expansions of Fractions, Part 2
164
This work is derived from Eureka Math β’ and licensed by Great Minds. Β©2015 Great Minds. eureka-math.org This file derived from G8-M7-TE-1.3.0-10.2015
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
6. Find the volume of the cylinder shown below.
π½ = (πππ π¦π)(ππ π¦)
= ππππ π¦π
7. Find the area of the circle shown below.
π¨ = (π π’π§. )ππ
= πππ π’π§π
8. Find the volume of the cone shown below.
π½ = (π
π) (πππ π’π§π)(ππ π’π§. )
= ππππ π’π§π
9. Find the area of the circle shown below.
π¨ = (π π¦π¦)ππ
= πππ π¦π¦π
10. Find the volume of the sphere shown below.
π½ = (π
π) π (ππ π¦π¦π)(π π¦π¦)
=ππππ
ππ π¦π¦π
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