## Engage NY Eureka Math 8th Grade Module 7 Lesson 13 Answer Key

### Eureka Math Grade 8 Module 7 Lesson 13 Exploratory Challenge/Exercise Answer Key

Exploratory Challenge/Exercises 1–11

Exercise 1.
Rodney thinks that $$\sqrt [ 3 ]{ 64 }$$ is greater than $$\frac{17}{4}$$. Sam thinks that $$\frac{17}{4}$$ is greater. Who is right and why?
$$\sqrt [ 3 ]{ 64 }$$ = $$\sqrt{4^{3}}$$ = 4
and
$$\frac{17}{4}$$ = $$\frac{16}{4}$$ + $$\frac{1}{4}$$
= 4 + $$\frac{1}{4}$$
= 4 $$\frac{1}{4}$$
We see that $$\sqrt [ 3 ]{ 64 }$$ < $$\frac{17}{4}$$. Sam is correct. Exercise 2.
Which number is smaller, $$\sqrt [ 3 ]{ 27 }$$ or 2.89? Explain.
$$\sqrt [ 3 ]{ 27 }$$ = $$\sqrt{3^{3}}$$ = 3
We see that 2.89 is smaller than $$\sqrt [ 3 ]{ 27 }$$.

Exercise 3.
Which number is smaller, $$\sqrt{121}$$ or $$\sqrt [ 3 ]{ 125 }$$? Explain.
$$\sqrt{121}$$ = $$\sqrt{11^{2}}$$ = 11
$$\sqrt [ 3 ]{ 125 }$$ = $$\sqrt{5^{3}}$$ = 5
We see that $$\sqrt [ 3 ]{ 125 }$$ is smaller than $$\sqrt{121}$$.

Exercise 4.
Which number is smaller, $$\sqrt{49}$$ or $$\sqrt [ 3 ]{ 216 }$$? Explain.
$$\sqrt{49}$$ = $$\sqrt{7^{2}}$$ = 7
$$\sqrt [ 3 ]{ 216 }$$ = $$\sqrt{6^{3}}$$ = 6
We see that $$\sqrt [ 3 ]{ 216 }$$ is smaller than $$\sqrt{49}$$.

Exercise 5.
Which number is greater, $$\sqrt{50}$$ or $$\frac{319}{45}$$? Explain.
Students may use any method to determine the decimal expansion of the fraction.
The number $$\frac{319}{45}$$ is equal to$$7.0 \overline{8}$$.
The number $$\sqrt{50}$$ is between 7 and 8 because 72 < 50 < 82. The number $$\sqrt{50}$$ is between 7.0 and 7.1 because 72 < 50 < 7.12. The number $$\sqrt{50}$$ is between 7.07 and 7.08 because 7.072 < 50 < 7.082. The approximate decimal value of $$\sqrt{50}$$ is 7.07. Since 7.07 < $$7.0 \overline{8}$$, then $$\sqrt{50}$$ < $$\frac{319}{45}$$; therefore, the fraction $$\frac{319}{45}$$ is greater than $$\sqrt{50}$$ .
Alternately: ($$\sqrt{50}$$)2 = 50 and ($$\frac{319}{45}$$)2 = 101761/2025>101250/2025 = 50. So, 319/45 must be larger.

Exercise 6.
Which number is greater, $$\frac{5}{11}$$ or $$0. \overline{4}$$? Explain.
Students may use any method to determine the decimal expansion of the fraction.
The number $$\frac{5}{11}$$ is equal to $$0. \overline{45}$$. Since 0.44444… < 0.454545…, then $$0. \overline{4}$$ < $$\frac{5}{11}$$; therefore, the fraction $$\frac{5}{11}$$ is greater than 0.4 ̅.
Alternately: 0.444… = $$\frac{4}{9}$$, and we can compare the fractions $$\frac{4}{9}$$ and $$\frac{5}{11}$$ using their equivalents, $$\frac{44}{99}$$ and $$\frac{45}{99}$$ to see that $$\frac{5}{11}$$ is larger.

Exercise 7.
Which number is greater, $$\sqrt{38}$$ or $$\frac{154}{25}$$? Explain.
Students may use any method to determine the decimal expansion of the fraction.
$$\frac{154}{25}$$ = $$\frac{154 \times 4}{25 \times 4} = \frac{616}{10^{2}}$$ = 6.16
The number $$\sqrt{38}$$ is between 6 and 7 because 62 < 38 < 72. The number $$\sqrt{38}$$ is between 6.1 and 6.2 because
6.12 < 38 < 6.22. The number $$\sqrt{38}$$ is between 6.16 and 6.17 because 6.162 < 38 < 6.172. Since $$\sqrt{38}$$ is greater than 6.16, then $$\sqrt{38}$$ is greater than 154/25.
Alternately: ($$\sqrt{38}$$)2 = 38 and ($$\frac{154}{25}$$)2 = $$\frac{23716}{625}$$ < $$\frac{23750}{625}$$ = 38. So, $$\sqrt{38}$$ must be larger.

Exercise 8.
Which number is greater, $$\sqrt{2}$$ or $$\frac{15}{9}$$? Explain.
Students may use any method to determine the decimal expansion of the fraction.
The number $$\frac{15}{9}$$ is equal to$$1 . \overline{6}$$.

The number $$\sqrt{2}$$ is between 1 and 2 because 12 < 2 < 22. The number $$\sqrt{2}$$ is between 1.4 and 1.5 because
1.42 < 2 < 1.52. Therefore, $$\sqrt{2}$$ < $$\frac{15}{9}$$; the fraction $$\frac{15}{9}$$ is greater.
Alternately: ($$\sqrt{2}$$)2 = 2 and ($$\frac{15}{9}$$)2 = ($$\frac{5}{3}$$)2 = $$\frac{28}{9}$$>2. So, $$\frac{15}{9}$$ must be larger.

Exercise 9.
Place each of the following numbers at its approximate location on the number line: $$\sqrt{25}$$, $$\sqrt{28}$$, $$\sqrt{30}$$, $$\sqrt{32}$$, $$\sqrt{35}$$, and $$\sqrt{36}$$.
The solutions are shown in red: The number $$\sqrt{25}$$ = $$\sqrt{5^{2}}$$ = 5.
The numbers $$\sqrt{28}$$,$$\sqrt{30}$$,$$\sqrt{32}$$, and $$\sqrt{35}$$ are between 5 and 6. The number $$\sqrt{28}$$ is between 5.2 and 5.3 because 5.22 < 28 < 5.32. The number $$\sqrt{30}$$ is between 5.4 and 5.5 because 5.42 < 30 < 5.52. The number $$\sqrt{32}$$ is between 5.6 and 5.7 because 5.62 < 32 < 5.72. The number $$\sqrt{35}$$ is between 5.9 and 6.0 because
5.92 < 35 < 62.
The number$$\sqrt{36}$$ = $$\sqrt{6^{2}}$$ = 6.

Exercise 10.
Challenge: Which number is larger, $$\sqrt{5}$$ or $$\sqrt [ 3 ]{ 11 }$$?
The number $$\sqrt{5}$$ is between 2 and 3 because 22 < 5 < 32. The number $$\sqrt{5}$$ is between 2.2 and 2.3 because
2.22 < 5 < 2.32. The number $$\sqrt{5}$$ is between 2.23 and 2.24 because 2.232 < 5 < 2.242. The number $$\sqrt{5}$$ is between 2.236 and 2.237 because 2.2362 < 5 < 2.2372. The decimal expansion of $$\sqrt{5}$$ is approximately 2.236….
The number $$\sqrt [ 3 ]{ 11 }$$ is between 2 and 3 because 2^3 < 11 < 3^3. The number $$\sqrt [ 3 ]{ 11 }$$ is between 2.2 and 2.3 because 2.2^3 < 11 < 2.3^3. The number $$\sqrt [ 3 ]{ 11 }$$ is between 2.22 and 2.23 because 2.22^3 < 11 < 2.23^3. The decimal expansion of $$\sqrt [ 3 ]{ 11 }$$ is approximately 2.22…. Since 2.22″…” < 2.236″…” , then $$\sqrt [ 3 ]{ 11 }$$ < $$\sqrt{5}$$; therefore, $$\sqrt{5}$$ is larger.

Alternately:
($$\sqrt [ 3 ]{ 11 }$$)^6 = 112 = 121
($$\sqrt{5}$$)^6 = 5^3 = 125
We see that $$\sqrt{5}$$ must be larger.

Exercise 11.
A certain chessboard is being designed so that each square has an area of 3 in2. What is the length of one edge of the board rounded to the tenths place? (A chessboard is composed of 64 squares as shown.) The area of one square is 3 in2. So, if x is the length of one side of one square,
x2 = 3
$$\sqrt{x^{2}}$$ = $$\sqrt{3}$$
x = $$\sqrt{3}$$.
There are 8 squares along one edge of the board, so the length of one edge is 8×$$\sqrt{3}$$. The number $$\sqrt{3}$$ is between 1 and 2 because
12 < 3 < 22. The number $$\sqrt{3}$$ is between 1.7 and 1.8 because 1.72 < 3 < 1.82. The number $$\sqrt{3}$$ is between 1.73 and 1.74 because 1.732 < 3 < 1.742. The number $$\sqrt{3}$$ is approximately 1.73. So, the length of one edge of the chessboard is about
8×1.73 inches, which is approximately 13.8 in.
Note: Some students may determine the total area of the board, 64×3 = 192, and then determine the approximate value of $$\sqrt{192}$$ as 13.8 to answer the question.

### Eureka Math Grade 8 Module 7 Lesson 13 Problem Set Answer Key

Question 1.
Which number is smaller, $$\sqrt [ 3 ]{ 343 }$$ or $$\sqrt{48}$$ ? Explain.
√(3&343) = $$\sqrt{7^{3}}$$ = 7
The number $$\sqrt{48}$$ is between 6 and 7 but definitely less than 7. Therefore, $$\sqrt{48}$$ < $$\sqrt [ 3 ]{ 343 }$$, and √$$\sqrt{48}$$ is smaller.

Question 2.
Which number is smaller, $$\sqrt{100}$$ or $$\sqrt [ 3 ]{ 1000 }$$? Explain.
$$\sqrt{100}$$ = $$\sqrt{10^{2}}$$ = 10
$$\sqrt [ 3 ]{ 1000 }$$ = $$\sqrt{10^{3}}$$ = 10
The numbers $$\sqrt{100}$$ and $$\sqrt [ 3 ]{ 1000 }$$ are equal because both are equal to 10.

Question 3.
Which number is larger, $$\sqrt{87}$$ or $$\frac{929}{99}$$? Explain.
Students may use any method to compute the first few decimal places of a fraction.
The number $$\frac{929}{99}$$ is equal to $$9 . \overline{38}$$.

The number $$\sqrt{87}$$ is between 9 and 10 because 92 < 87 < 102. The number $$\sqrt{87}$$ is between 9.3 and 9.4 because 9.32 < 87 < 9.42. The number $$\sqrt{87}$$ is between 9.32 and 9.33 because 9.322 < 87 < 9.332. Since $$\sqrt{87}$$ < 9.3, then $$\sqrt{87}$$ < $$\frac{929}{99}$$. The fraction $$\frac{929}{99}$$ is larger.

Question 4.
Which number is larger, $$\frac{9}{13}$$ or $$0 . \overline{692}$$? Explain.
Students may use any method to compute the first few decimal places of a fraction.
The number $$\frac{9}{13}$$ is equal to $$0 . \overline{692307}$$. Since 0.692307… < 0.692692…, then we see that 9/13 < $$0 . \overline{692}$$.
The decimal $$0 . \overline{692}$$ is larger.

Question 5.
Which number is larger, 9.1 or $$\sqrt{82}$$? Explain.
The number $$\sqrt{82}$$ is between 9 and 10 because 92 < 82 < 102. The number $$\sqrt{82}$$ is between 9.0 and 9.1 because 9.02 < 82 < 9.12. Since $$\sqrt{82}$$ < 9.1, then the number 9.1 is larger than the number $$\sqrt{82}$$.

Question 6.
Place each of the following numbers at its approximate location on the number line: $$\sqrt{144}$$, $$\sqrt [ 3 ]{ 1000 }$$, $$\sqrt{130}$$, $$\sqrt{110}$$, $$\sqrt{120}$$, $$\sqrt{115}$$, and $$\sqrt{133}$$. Explain how you knew where to place the numbers. The solutions are shown in red: The number $$\sqrt{144}$$ = $$\sqrt{12^{2}}$$ = 12.
The number $$\sqrt [ 3 ]{ 1000 }$$ = $$\sqrt{10^{3}}$$ = 10.

The numbers $$\sqrt{110}$$, $$\sqrt{115}$$, and $$\sqrt{120}$$ are all between 10 and 11 because when squared, their value falls between 102 and 112. The number $$\sqrt{110}$$ is between 10.4 and 10.5 because 10.42 < 110 < 10.52. The number $$\sqrt{115}$$ is between 10.7 and 10.8 because 10.72 < 115 < 10.82. The number $$\sqrt{120}$$ is between 10.9 and 11 because 10.92 < 120 < 112.
The numbers √130 and $$\sqrt{133}$$ are between 11 and 12 because when squared, their value falls between 112 and 122. The number √130 is between 11.4 and 11.5 because 11.42 < 130 < 11.52. The number $$\sqrt{133}$$ is between 11.5 and 11.6 because 11.52 < 133 < 11.62.

Question 7.
Which of the two right triangles shown below, measured in units, has the longer hypotenuse? Approximately how much longer is it? Let x represent the length of the hypotenuse of the triangle on the left.
72 + 22 = x2
49 + 4 = x2
53 = x2
$$\sqrt{53}$$ = x
The number $$\sqrt{53}$$ is between 7 and 8 because 72 < 53 < 82. The number $$\sqrt{53}$$ is between 7.2 and 7.3 because 7.22 < 53 < 7.32. The number $$\sqrt{53}$$ is between 7.28 and 7.29 because 7.282 < 53 < 7.292. The approximate decimal value of $$\sqrt{53}$$ is 7.28.
Let y represent the length of the hypotenuse of the triangle on the right.
52 + 52 = y2
25 + 25 = y2
50 = y2
$$\sqrt{50}$$ = y
The number $$\sqrt{50}$$ is between 7 and 8 because 72 < 50 < 82. The number $$\sqrt{50}$$ is between 7.0 and 7.1 because 7.02 < 50 < 7.12. The number $$\sqrt{50}$$ is between 7.07 and 7.08 because 7.072 < 50 < 7.082. The approximate decimal value of $$\sqrt{50}$$ is 7.07.

The triangle on the left has the longer hypotenuse. It is approximately 0.21 units longer than the hypotenuse of the triangle on the right.
Note: Based on their experience, some students may reason that $$\sqrt{50}$$ < $$\sqrt{53}$$. To answer completely, students must determine the decimal expansion to approximate how much longer one hypotenuse is than the other.

### Eureka Math Grade 8 Module 7 Lesson 13 Exit Ticket Answer Key

Question 1.
Place each of the following numbers at its approximate location on the number line: $$\sqrt{12}$$, $$\sqrt{16}$$, $$\frac{20}{6}$$, $$3 . \overline{53}$$, and $$\sqrt [ 3 ]{ 27 }$$. The number $$\sqrt{12}$$ is between 3.4 and 3.5 since 3.42 < 12 < 3.52.
The number $$\sqrt{16}$$ = $$\sqrt{4^{2}}$$ = 4.
The number $$\frac{20}{6}$$ is equal to $$3 . \overline{3}$$.
The number $$\sqrt [ 3 ]{ 27 }$$ = $$\sqrt{3^{3}}$$ = 3. 