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Engage NY Eureka Math 7th Grade Module 3 Lesson 4 Answer Key

Eureka Math Grade 7 Module 3 Lesson 4 Example Answer Key

Example 1.

Answer:

→ What is happening when you factor and write equivalent expressions for parts (e), (f), (g), and (h)?
→ In the same way that dividing is the opposite or inverse operation of multiplying, factoring is the opposite of expanding.
→ What are the terms being divided by?
→ They are being divided by a common factor.
Have students write an expression that is equivalent to 8x+4.
→ Would it be incorrect to factor out a 2 instead of a 4?

Example 2
Let the variables x and y stand for positive integers, and let 2x, 12y, and 8 represent the area of three regions in the array. Determine the length and width of each rectangle if the width is the same for each rectangle.

Answer:
→ What does 2x represent in the first region of the array?
→ The region has an area of 2x or can be covered by 2x square units.
→ What does the entire array represent?
→ The entire array represents 2x+12y+8 square units.
→ What is the common factor of 2x, 12y, and 8?
→ The common factor of 2x, 12y, and 8 is 2.
→ What are the missing values, and how do you know?
→ The missing values are x, 6y, and 4. If the products are given in the area of the regions, divide the regions by 2 to determine the missing values.

Example 3.
A new miniature golf and arcade opened up in town. For convenient ordering, a play package is available to purchase. It includes two rounds of golf and 20 arcade tokens, plus $3.00 off the regular price. There is a group of six friends purchasing this package. Let g represent the cost of a round of golf, and let t represent the cost of a token. Write two different expressions that represent the total amount this group spent. Explain how each expression describes the situation in a different way.
Answer:
→ What two equivalent expressions could be used to represent the situation?
→ 6(2g+20t-3)
Each person will pay for two rounds of golf and 20 tokens and will be discounted $3.00. This expression is six times the quantity of each friend’s cost.
→ 12g+120t-18
The total cost is equal to 12 games of golf plus 120 tokens, minus $18.00 off the entire bill.

→ What does it mean to take the opposite of a number?
→ You can determine the additive inverse of a number or a multiplicative inverse.
→ What is the opposite of 2?
→ -2
→ What is (-1)(2)?
→ -2
→ What is (-1)(n)?
→ -n
→ What are two mathematical expressions that represent the opposite of (2a+3b)?
→ (-1)(2a+3b) or -(2a+3b)
→ Use the distributive property to write (-1)(2a+3b) as an equivalent expression.
→ -2a-3b or -2a+(-3b)
→ To go from -2a-3b to -(2a+3b), what process occurs?
→ The terms -2a and -3b are written as (-1)(2a) and (-1)(3b), and the -1 is factored out of their sum.

Example 4.
→ What does it mean to take the opposite of a number?
→ You can determine the additive inverse of a number or a multiplicative inverse.
→ What is the opposite of 2?
→ -2
→ What is (-1)(2)?
→ -2
→ What is (-1)(n)?
→ -n
→ What are two mathematical expressions that represent the opposite of (2a+3b)?
→ (-1)(2a+3b) or -(2a+3b)
→ Use the distributive property to write (-1)(2a+3b) as an equivalent expression.
→ -2a-3b or -2a+(-3b)
→ To go from -2a-3b to -(2a+3b), what process occurs?
→ The terms -2a and -3b are written as (-1)(2a) and (-1)(3b), and the -1 is factored out of their sum.

Example 5
Rewrite 5a-(a-3b) in standard form. Justify each step, applying the rules for subtracting and the distributive property.
Answer:
5a+(-(a+-3b)) Subtraction as adding the inverse
5a+(-1)(a+-3b) The opposite of a number is the same as multiplying by –1.
5a+(-1)(a)+(-1)(-3b) Distributive property
5a+-a+3b Multiplying by -1 is the same as the opposite of the number.
4a+3b Collect like terms

Eureka Math Grade 7 Module 3 Lesson 4 Exercise Answer Key

Exercise 1.
Rewrite the expressions as a product of two factors.
a. 72t+8
Answer:
8(9t+1)

b. 55a+11
Answer:
11(5a+1)

c. 36z+72
Answer:
36(z+2)

d. 144q-15
Answer:
3(48q-5)

e. 3r+3s
Answer:
3(r+s)

Exercise 2.
Write the product and sum of the expressions being represented in the rectangular array.
Answer:

2(12d + 4e + 3); 24d + 8e + 6

b. Factor 48j+60k+24 by finding the greatest common factor of the terms.
Answer:
12(4j+5k+2)

Exercise 3.
For each expression, write each sum as a product of two factors. Emphasize the importance of the distributive property. Use various equivalent expressions to justify equivalency.
a. 2∙3+5∙3
Both have a common factor of 3, so the two factors would be 3(2+5). Demonstrate that 3(7) is equivalent to 6+15, or 21.

b. (2+5)+(2+5)+(2+5)
Answer:
This expression is 3 groups of (2+5) or 3(2)+3(5), which is 3(2+5).

c. 2∙2+(5+2)+(5∙2)
Answer:
Rewrite the expression as
2∙2+(5∙2)+(2+5), so 2(2+5)+(2+5), which equals 3(2+5).

d. x∙3+5∙3
Answer:
The greatest common factor is 3, so factor out the 3:

e. 3(x+5). (x+5)+(x+5)+(x+5)
Answer:
Similar to part (b), this is 3 groups of (x+5), so

f. 3(x+5). 2x+(5+x)+5∙2
Answer:
Combine like terms and then identify the common factor. 3x+15, where 3 is the common factor: 3(x+5). Or,
2x+2∙5+(x+5), so that 2(x+5)+(x+5)=3(x+5). Or, use the associative property and write:
2x+(5∙2)+(5+x)
2(x+5)+(5+x)
3(x+5).

g. x∙3+y∙3
Answer:
The greatest common factor is 3, so 3(x+y). (x+y)+(x+y)+(x+y)
There are 3 groups of

h. (x + y) + (x+ y) + (x+y)
Answer:
There are 3 groups of (x+y), so 3(x+y)

i. 2x + (y + x) + 2y
Answer:
Combine like terms and then identify the common factor. 3x+3y, where 3 is the common factor. 3(x+y). Or,
2x+2y+(x+y), so that 2(x+y)+(x+y) is equivalent to 3(x+y). Or, use the associative property, and write:
2x+2y+(y+x)
2(x+y)+(x+y)
3(x+y).

Exercise 4.
a. What is the opposite of (-6v+1)?
Answer:
-(-6v+1)

b. Using the distributive property, write an equivalent expression for part (a).
Answer:
6v-1

Exercise 5.
Expand each expression and collect like terms.
a. -3(2p-3q)
Answer:
-3(2p+(-3q)) Subtraction as adding the inverse
-3∙2p+(-3)∙(-3q) Distributive property
-6p+9q Apply integer rules

b. -a-(a-b)
Answer:
-a+(-(a+-b)) Subtraction as adding the inverse
-1a+(-1(a+-1b)) The opposite of a number is the same as multiplying by –1.
-1a+(-1a)+1b Distributive property
-2a+b Apply integer addition rules

Eureka Math Grade 7 Module 3 Lesson 4 Problem Set Answer Key

Question 1.
Write each expression as the product of two factors.
a. 1∙3+7∙3
Answer:
3(1+7)

b. (1+7)+(1+7)+(1+7)
Answer:
3(1+7)

c. 2∙1+(1+7)+(7∙2)
Answer:
3(1+7)

d. h∙3+6∙3
Answer:
3(h+6)

e. (h+6)+(h+6)+(h+6)
Answer:
3(h+6)

f. 2h+(6+h)+6∙2
Answer:
3(h+6)

g. j∙3+k∙3
Answer:
3(j+k)

h. (j+k)+(j+k)+(j+k)
Answer:
3(j+k)

i. 3(j+k) 2j+(k+j)+2k
Answer:
3(j+k)

Question 2.
Write each sum as a product of two factors.
a. 6∙7+3∙7
Answer:
7(6+3)

b. (8+9)+(8+9)+(8+9)
Answer:
3(8+9)

c. 4+(12+4)+(5∙4)
Answer:
4(1+4+5)

d. 2y∙3+4∙3
Answer:
3(2y+4)

e. (x+5)+(x+5)
Answer:
2(x+5)

f. 3x+(2+x)+5∙2
Answer:
4(x+3)

g. f∙6+g∙6
Answer:
6(f+g)

h. (c+d)+(c+d)+(c+d)+(c+d)
Answer:
4(c+d)

i. 2r+r+s+2s
Answer:
3(r+s)

Question 3.
Use the following rectangular array to answer the questions below.

a. Fill in the missing information.

b. Write the sum represented in the rectangular array.
Answer:
15f + 5g + 45

c. Use the missing information from part (a) to write the sum from part (b) as a product of two factors.
Answer:
5(3f+g+9)

Question 4.
Write the sum as a product of two factors.
a. 81w+48
Answer:
3(27w+16)

b. 10-25t
Answer:
5(2-5t)

c. 12a+16b+8
Answer:
4(3a+4b+2)

Question 5.
Xander goes to the movies with his family. Each family member buys a ticket and two boxes of popcorn. If there are five members of his family, let t represent the cost of a ticket and p represent the cost of a box of popcorn. Write two different expressions that represent the total amount his family spent. Explain how each expression describes the situation in a different way.
Answer:
5(t+2p)
Five people each buy a ticket and two boxes of popcorn, so the cost is five times the quantity of a ticket and two boxes of popcorn.
5t+10p
There are five tickets and 10 boxes of popcorn total. The total cost will be five times the cost of the tickets, plus 10 times the cost of the popcorn.

Question 6.
Write each expression in standard form.
a. -3(1-8m-2n)
Answer:
-3(1+(-8m)+(-2n))
-3+24m+6n

b. 5-7(-4q+5)
Answer:
5+-7(-4q+5)
5+28q+(-35)
28q-35+5
28q-30

c. -(2h-9)-4h
Answer:
-(2h-9)-4h
-(2h+(-9))+(-4h)
-2h+9+(-4h)
-6h+9

d. 6(-5r-4)-2(r-7s-3)
Answer:
6(-5r-4)-2(r-7s-3)
6(-5r+-4)+-2(r-7s+-3)
-30r+-24+-2r+14s+6
-30r+-2r+14s+-24+6
-32r+14s-18

Question 7.
Combine like terms to write each expression in standard form.
a. (r-s)+(s-r)
Answer:
0

b. (-r+s)+(s-r)
Answer:
-2r+2s

c. (-r-s)-(-s-r)
Answer:
0

d. (r-s)+(s-t)+(t-r)
Answer:
0

e. (r-s)-(s-t)-(t-r)
Answer:
2r-2s

Eureka Math Grade 7 Module 3 Lesson 4 Exit Ticket Answer Key

Question 1.
Write the expression below in standard form.
3h-2(1+4h)
Answer:
3h-2(1+4h)
3h+(-2(1+4h)) Subtraction as adding the inverse
3h+(-2∙1)+(-2h∙4) Distributive property
3h+(-2)+(-8h) Apply integer rules
-5h-2 Collect like terms

Question 2.
Write the expression below as a product of two factors.
6m+8n+4
Answer:
The GCF for the terms is 2. Therefore, the factors are 2(3m+4n+2).

Engage NY Eureka Math 6th Grade Module 1 Lesson 15 Answer Key

Eureka Math Grade 6 Module 1 Lesson 15 Exercise Answer Key

Exercise 1.
Create a table to determine how many views the website probably had one hour after the end of the broadcast based on how many views it had two and three hours after the end of the broadcast. Using this relationship, predict how many views the website will have 4, 5, and 6 hours after the end of the broadcast.
Answer:

Exercise 2.
What is the constant number, c, that makes these ratios equivalent?
Answer:
12

Using an equation, represent the relationship between the number of views, y, the website received and the number of hours, h, after this morning’s news broadcast.
Answer:
v = 12h

Exercise 3.
Use the table created In Exercise 1 to Identify sets of ordered pairs that can be graphed.
Answer:
(1, 12), (2,24), (3, 36), (4, 48), (5, 60), (6, 72)

Exercise 4.
Use the ordered pairs you created to depict the relationship between hours and number of views on a coordinate plane.
Label your axes and create a title for the graph. Do the points you plotted lie on a line?

Answer:

Exercise 5
Predict how many views the website will have after twelve hours. Use at least two representations (e.g., tape diagram, table, double number line diagram) to justify your answer.
Answer:

Exercise 6
Also on the news broadcast, a chef from a local Italian restaurant demonstrated how he makes fresh pasta daily for his restaurant. The recipe for his pasta is below:
3 eggs, beaten
1 teaspoon salt
2 cups all-purpose flour
2 tablespoons water
2 tablespoons vegetable oil
Determine the ratio of the number of tablespoons of water to the number of eggs.
Answer:
2: 3

Provided the information in the table below, complete the table to determine ordered pairs. Use the ordered pairs to graph the relationship of the number of tablespoons of water to the number of eggs.

Answer:

Answer:

What would you have to do to the graph in order to find how many eggs would be needed If the recipe was larger and called for 16 tablespoons of water?
Answer:
Extend the graph.

Demonstrate on your graph.

How many eggs would be needed if the recipe called for 16 tablespoons of water?
Answer:
24

Exercise 7.
Determine how many tablespoons of water will be needed If the chef is making a large batch of pasta and the recipe increases to 36 eggs. Support your reasoning using at least one diagram you find applies best to the situation, and explain why that tool is the best to use.
Answer:
Answers may vary but should include reasoning for each tool. For example, extending the table/double number line diagram because values were already given to find the pattern or using a tape diagram to determine the equivalent ratios.

Eureka Math Grade 6 Module 1 Lesson 15 Problem Set Answer Key

Question 1.
The producer of the news station posted an article about the high school’s football championship ceremony on a new website. The website had 500 views after four hours. Create a table to show how many views the website would have had after the first, second, and third hours after posting, if the website receives views at the same rate. How many views would the website receive after 5 hours?
Answer:

Question 2.
Write an equation that represents the relationship from Problem 1. Do you see any connections between the equations you wrote and the ratio of the number of views to the number of hours?
Answer:
125h = v

Question 3.
Use the table in Problem 1 to make a list of ordered pairs that you could plot on a coordinate plane.
Answer:
(1, 125), (2, 250), (3, 375), (4, 500), (5, 625)

Question 4.
Graph the ordered pairs on a coordinate plane. Label your axes and create a title for the graph.
Answer:

Question 5.
Use multiple tools to predict how many views the website would have after 12 hours.
Answer:
Answers may vary but could include all representations from the module. The correct answer is 1,500 views.

Eureka Math Grade 6 Module 1 Lesson 15 Exit Ticket Answer Key

Question 1.
Jen and Nikki are making bracelets to sell at the local market. They determined that each bracelet would have eight beads and two charms.

Complete the table below to show the ratio of the number of charms to the number of beads.

Answer:

Create ordered pairs from the table, and plot the pairs on the graph below. Label the axes of the graph, and provide a title.

Answer:

Eureka Math Grade 6 Module 1 Lesson 15 Exploratory Challenge Answer Key

Question 1.
At the end of this morning’s news segment, the local television station highlighted area pets that need to be adopted. The station posted a specific website on the screen for viewers to find more information on the pets shown and the adoption process. The station producer checked the website two hours after the end of the broadcast and saw that the website had 24 views. One hour after that, the website had 36 views.
Answer:
At the end of this morning’s news segment, the local television station highlighted area pets that need to be adopted. The station posted a specific website on the screen for viewers to find more information on the pets shown and the adoption process. The station producer checked the website two hours after the end of the broadcast and saw that the website had 24 views. One hour after that, the website had 36 views.

Engage NY Eureka Math 7th Grade Module 4 Lesson 16 Answer Key

Eureka Math Grade 7 Module 4 Lesson 16 Example Answer Key

Example 1.
A school has 60% girls and 40% boys. If 20% of the girls wear glasses and 40% of the boys wear glasses, what percent of all students wears glasses?
Answer:
Let n represent the number of students in the school.
The number of girls is 0.6n. The number of boys is 0.4n.

The number of girls wearing glasses is as follows: 0.2(0.6n) = 0.12n.

The number of boys wearing glasses is as follows: 0.4(0.4n) = 0.16n.

The total number of students wearing glasses is 0.12n + 0.16n = 0.28n.
0.28 = 28%, so 28% of the students wear glasses.

Example 2.
The weight of the first of three containers is 12% more than the second, and the third container is 20% lighter than the second. By what percent is the first container heavier than the third container?
Answer:
Let n represent the weight of the second container. (The tape diagram representation for the second container is divided into five equal parts to show 20%. This will be useful when drawing a representation for the third container and also when sketching a 12% portion for the first container since it will be slightly bigger than half of the 20% portion created.)

The weight of the first container is (1.12)n.

The weight of the third container is (0.80)n.

The following represents the difference in weight between the first and third container:
1.12n – 0.80n = 0.32n
Recall that the weight of the third container is 0.8n
0.32n ÷ 0.8n = 0.4. The first container is 40% heavier than the third container.
Or 1.4 × 100% = 140%, which also shows that the first container is 40% heavier than the third container.

Example 3.
In one year’s time, 20% of Ms. McElroy’s investments increased by 5%, 30% of her investments decreased by 5%, and 50% of her investments increased by 3%. By what percent did the total of her investments increase?
Answer:
Let n represent the dollar amount of Ms. McElroy’s investments before the changes occurred during the year.

After the changes, the following represents the dollar amount of her investments:
0.2n(1.05) + 0.3n(0.95) + 0.5n(1.03)
= 0.21n + 0.285n + 0.515n
= 1.01n.

Since 1.01 = 101%, Ms. McElroy’s total investments increased by 1%.

Eureka Math Grade 7 Module 4 Lesson 16 Exercise Answer Key

Exercise 1.
How does the percent of students who wear glasses change if the percent of girls and boys remains the same (that is, 60% girls and 40% boys), but 20% of the boys wear glasses and 40% of the girls wear glasses?
Answer:
Let n represent the number of students in the school.
The number of girls is 0.6n. The number of boys is 0.4n.

Girls who wear glasses:
40% of 60% of n = 0.4 × 0.6n = 0.24n

Boys who wear glasses:

Students who wear glasses:

32% of students wear glasses.

Exercise 2.
How would the percent of students who wear glasses change if the percent of girls is 40% of the school and the percent of boys is 60% of the school, and 40% of the girls wear glasses and 20% of the boys wear glasses? Why?
Answer:
The number of students wearing glasses would be equal to the answer for Example 1 because all of the percents remain the same except that a swap is made between the boys and girls. So, the number of boys wearing glasses is swapped with the number of girls, and the number of girls wearing glasses is swapped with the number of boys, but the total number of students wearing glasses is the same.

Let n represent the number of students in the school.
The number of boys is 0.6n. The number of girls is 0.4n.

Boys who wear glasses:

Girls who wear glasses:

Students who wear glasses:

Exercise 3.
Matthew’s pet dog is 7% heavier than Harrison’s pet dog, and Janice’s pet dog is 20% lighter than Harrison’s. By what percent is Matthew’s dog heavier than Janice’s?
Answer:
Let h represent the weight of Harrison’s dog.
Matthew’s dog is 1.07h, and Janice’s dog is 0.8h.
Since 1.07 ÷ 0.8 = \(\frac{107}{80}\) = 1.3375, Mathew’s dog is 33.75% heavier than Janice’s dog.

Exercise 4.
A concert had 6,000 audience members in attendance on the first night and the same on the second night. On the first night, the concert exceeded expected attendance by 20%, while the second night was below the expected attendance by 20%. What was the difference in percent of concert attendees and expected attendees for both nights combined?
Answer:
Let x represent the expected number of attendees on the first night and y represent the number expected on the second night.

1.2x = 6,000
x = 5,000
6,000 – 5,000 = 1,000
The first night was attended by 1,000 more people than expected.

0.8y = 6,000
y = 7,500
7,500 – 6,000 = 1,500
The second night was attended by 1,500 less people than expected.
5,000 + 7,500 = 12,500
12,500 people were expected in total on both nights.
1,500 – 1,000 = 500. \(\frac{500}{12,500}\) × 100% = 4%. The concert missed its expected attendance by 4%.

Eureka Math Grade 7 Module 4 Lesson 16 Problem Set Answer Key

Question 1.
One container is filled with a mixture that is 30% acid. A second container is filled with a mixture that is 50% acid. The second container is 50% larger than the first, and the two containers are emptied into a third container. What percent of acid is the third container?
Answer:
Let t be the amount of mixture in the first container. Then the second container has 1.5t, and the third container has 2.5t.
The amount of acid in the first container is 0.3t, the amount of acid in the second container is 0.5(1.5t) = 0.75t, and the amount of acid in the third container is 1.05t. The percent of acid in the third container is
\(\frac{1.05}{2.5}\) × 100% = 42%.

Question 2.
The store’s markup on a wholesale item is 40%. The store is currently having a sale, and the item sells for 25% off the retail price. What is the percent of profit made by the store?
Answer:
Let w represent the wholesale price of an item.
Retail price: 1.4w
Sale price: 1.4w – (1.4w × 0.25) = 1.05w
The store still makes a 5% profit on a retail item that is on sale.

Question 3.
During lunch hour at a local restaurant, 90% of the customers order a meat entrée and 10% order a vegetarian entrée. Of the customers who order a meat entrée, 80% order a drink. Of the customers who order a vegetarian entrée, 40% order a drink. What is the percent of customers who order a drink with their entrée?
Answer:
Let e represent lunch entrées.
Meat entrées: 0.9e
Vegetarian entrées: 0.1e
Meat entrées with drinks: 0.9e × 0.8 = 0.72e
Vegetarian entrées with drinks: 0.1e × 0.4 = 0.04e
Entrées with drinks: 0.72e + 0.04e = 0.76e. Therefore, 76% of lunch entrées are ordered with a drink.

Question 4.
Last year’s spell – a – thon spelling test for a first grade class had 15% more words with four or more letters than this year’s spelling test. Next year, there will be 5% less than this year. What percent more words have four or more letters in last year’s test than next year’s?
Answer:
Let t represent this year’s amount of spell – a – thon words with four letters or more.
Last year: 1.15t
Next year: 0.95t
1.15 t ÷ 0.95t × 100% ≈ 121%. There were about 21% more words with four or more letters last year than there will be next year.

Question 5.
An ice cream shop sells 75% less ice cream in December than in June. Twenty percent more ice cream is sold in July than in June. By what percent did ice cream sales increase from December to July?
Answer:
Let j represent sales in June.
December: 0.25j
July: 1.20j
1.20 ÷ 0.25 = 4.8 × 100% = 480%. Ice cream sales in July increase by 380% from ice cream sales in December.

Question 6.
The livestock on a small farm the prior year consisted of 40% goats, 10% cows, and 50% chickens. This year, there is a 5% decrease in goats, 9% increase in cows, and 15% increase in chickens. What is the percent increase or decrease of livestock this year?
Answer:
Let l represent the number of livestock the prior year.
Goats decrease: 0.4l – (0.4l × 0.05) = 0.38l or 0.95(0.4l) = 0.38l
Cows increase: 0.1 l + (0.1l × 0.09 ) = 0.109l or 1.09(0.1l) = 0.109l
Chickens increase: 0.5k + ( 0.5k × 0.15) = 0.575l or 1.15(0.5l) = 0.575l
0.38l + 0.109l + 0.575l = 1.064l. There is an increase of 6.4% in livestock.

Question 7.
In a pet shelter that is occupied by 55% dogs and 45% cats, 60% of the animals are brought in by concerned people who found these animals in the streets. If 90% of the dogs are brought in by concerned people, what is the percent of cats that are brought in by concerned people?
Answer:
Let c represent the percent of cats brought in by concerned people.
0.55(0.9) + (0.45)(c) = 1(0.6)
0.495 + 0.45c = 0.6
0.495 – 0.495 + 0.45c = 0.6 – 0.495
0.45c = 0.105
0.45c ÷ 0.45 = 0.105 ÷ 0.45
c ≈ 0.233
About 23% of the cats brought into the shelter are brought in by concerned people.

Question 8.
An artist wants to make a particular teal color paint by mixing a 75% blue hue and 25% yellow hue. He mixes a blue hue that has 85% pure blue pigment and a yellow hue that has 60% of pure yellow pigment. What is the percent of pure pigment that is in the resulting teal color paint?
Answer:
Let p represent the teal color paint.
(0.75 × 0.85p) + (0.25 × 0.6p) = 0.7875p
78.75% of pure pigment is in the resulting teal color paint.

Question 9.
On Mina’s block, 65% of her neighbors do not have any pets, and 35% of her neighbors own at least one pet. If 25% of the neighbors have children but no pets, and 60% of the neighbors who have pets also have children, what percent of the neighbors have children?
Answer:
Let n represent the number of Mina’s neighbors.
Neighbors who do not have pets: 0.65n
Neighbors who own at least one pet: 0.35n
Neighbors who have children but no pets: 0.25 × 0.65n = 0.1625n
Neighbors who have children and pets: 0.6 × 0.35n = 0.21n
Percent of neighbors who have children: 0.1625n + 0.21n = 0.3725n
37.25% of Mina’s neighbors have children.

Eureka Math Grade 7 Module 4 Lesson 16 Exit Ticket Answer Key

Question 1.
Jodie spent 25% less buying her English reading book than Claudia. Gianna spent 9% less than Claudia. Gianna spent more than Jodie by what percent?
Answer:
Let c represent the amount Claudia spent, in dollars. The number of dollars Jodie spent was 0.75c, and the number of dollars Gianna spent was 0.91c. 0.91c ÷ 0.75c = \(\frac{91}{75}\) × 100% = 121 \(\frac{1}{3}\)%. Gianna spent 21 \(\frac{1}{3}\)% more than Jodie.

Question 2.
Mr. Ellis is a teacher who tutors students after school. Of the students he tutors, 30% need help in computer science and the rest need assistance in math. Of the students who need help in computer science, 40% are enrolled in Mr. Ellis’s class during the school day. Of the students who need help in math, 25% are enrolled in his class during the school day. What percent of the after – school students are enrolled in Mr. Ellis’s classes?
Answer:
Let t represent the after – school students tutored by Mr. Ellis.
Computer science after – school students: 0.3t
Math after – school students: 0.7t

After – school computer science students who are also Mr. Ellis’s students: 0.4 × 0.3t = 0.12t
After – school math students who are also Mr. Ellis’s students: 0.25 × 0.7t = 0.175t

Number of after – school students who are enrolled in Mr. Ellis’s classes: 0.12t + 0.175t = 0.295t
Out of all the students Mr. Ellis tutors, 29.5% of the tutees are enrolled in his classes.

Engage NY Eureka Math 7th Grade Module 3 Lesson 5 Answer Key

Eureka Math Grade 7 Module 3 Lesson 5 Example Answer Key

Example 1.
Write the sum, and then write an equivalent expression by collecting like terms and removing parentheses.
a. 2x and – 2x + 3
Answer:
2x + (- 2x + 3)
(2x + (- 2x)) + 3 Associative property, collect like terms
0 + 3 Additive inverse
3 Additive identity property of zero

b. 2x – 7 and the opposite of 2x
Answer:
2x + (- 7) + (- 2x)
2x + (- 2x) + (- 7) Commutative property, associative property
0 + (- 7) Additive inverse
– 7 Additive identity property of zero

c. The opposite of (5x – 1) and 5x
Answer:
– (5x – 1) + 5x
– 1(5x – 1) + 5x Taking the opposite is equivalent to multiplying by –1.
– 5x + 1 + 5x Distributive property
(- 5x + 5x) + 1 Commutative property, any order property
0 + 1 Additive inverse
1 Additive identity property of zero

Example 2.
• (\(\frac{3}{4}\))×(\(\frac{4}{3}\) )=
Answer:
1

• 4×\(\frac{1}{4}\) =
Answer:
1

• \(\frac{1}{9}\) ×9=
Answer:
1

• (- \(\frac{1}{3}\) )× – 3=
Answer:
1

• (- \(\frac{6}{5}\) )×(- 5/6)=
Answer:
1

→ What are these pairs of numbers called?
→ Reciprocals
→ What is another term for reciprocal?
→ The multiplicative inverse
→ What happens to the sign of the expression when converting it to its multiplicative inverse?
→ There is no change to the sign For example, the multiplicative inverse of – 2 is (- \(\frac{1}{2}\) ) The negative sign remains the same.
→ What can you conclude from the pattern in the answers?
→ The product of a number and its multiplicative inverse is equal to 1.
→ Earlier, we saw that 0 is a special number because it is the only number that when added to another number, results in that number again Can you explain why the number 1 is also special?
→ One is the only number that when multiplied with another number, results in that number again.
→ This property makes 1 special among all the numbers Mathematicians have a special name for 1, the multiplicative identity; they call the property the multiplicative identity property of one.
As an extension, ask students if there are any other special numbers that they have learned Students should respond: Yes; – 1 has the property that multiplying a number by it is the same as taking the opposite of the number Share with students that they are going to learn later in this module about another special number called pi.
As a class, write the product, and then write an equivalent expression in standard form State the properties for each step After discussing questions, review the properties and definitions in the Lesson Summary emphasizing the multiplicative identity property of one and the multiplicative inverse.

Write the product, and then write the expression in standard form by removing parentheses and combining like terms Justify each step.
a. The multiplicative inverse of \(\frac{1}{5}\) and (2x – \(\frac{1}{5}\) )
Answer:
5(2x – \(\frac{1}{5}\) )
5(2x) – 5\(\frac{1}{5}\) Distributive property
10x – 1 Multiplicative inverses

b. The multiplicative inverse of 2 and (2x + 4)
Answer:
(\(\frac{1}{2}\) )(2x + 4)
(\(\frac{1}{2}\) )(2x) + (\(\frac{1}{2}\) )(4) Distributive property
1x + 2 Multiplicative inverses, multiplication
x + 2 Multiplicative identity property of one

c. The multiplicative inverse of (\(\frac{1}{3x + 5}\) ) and \(\frac{1}{3}\)
Answer:
(3x + 5)∙\(\frac{1}{3}\)
3x(\(\frac{1}{3}\) ) + 5(\(\frac{1}{3}\) ) Distributive property
1x + \(\frac{5}{3}\) Multiplicative inverse
x + \(\frac{5}{3}\) Multiplicative identity property of one

Eureka Math Grade 7 Module 3 Lesson 5 Opening Exercise Answer Key

a. In the morning, Harrison checked the temperature outside to find that it was – 12°F Later in the afternoon, the temperature rose 12°F Write an expression representing the temperature change What was the afternoon temperature?
Answer:
– 12 + 12; the afternoon temperature was 0°F.

b. Rewrite subtraction as adding the inverse for the following problems and find the sum.
i. 2 – 2
Answer:
2 + (- 2)=0

ii – 4 – (- 4)
Answer:
(- 4) + 4=0

iii. The difference of 5 and 5
Answer:
5 – 5=5 + (- 5)=0

iv. g – g
g + (- g)=0

c. What pattern do you notice in part (a) and (b)?
Answer:
The sum of a number and its additive inverse is equal to zero.

d. Add or subtract.
i. 16 + 0
Answer:
16

ii. 0 – 7
Answer:
0 + (- 7)= – 7

iii – 4 + 0
– 4

iv. 0 + d
Answer:
d

v. What pattern do you notice in parts (i) through (iv)?
Answer:
The sum of any quantity and zero is equal to the value of the quantity.

e. Your younger sibling runs up to you and excitedly exclaims, “I’m thinking of a number If I add it to the number 2 ten times, that is, 2 + my number + my number + my number, and so on, then the answer is 2 What is my number?” You almost immediately answer, “zero,” but are you sure? Can you find a different number (other than zero) that has the same property? If not, can you justify that your answer is the only correct answer?
Answer:
No, there is no other number On a number line, 2 can be represented as a directed line segment that starts at 0, ends at 2, and has length 2 Adding any other (positive or negative) number v to 2 is equivalent to attaching another directed line segment with length |v| to the end of the first line segment for 2:

If v is any number other than 0, then the directed line segment that represents v will have to have some length, so 2 + v will have to be a different number on the number line Adding v again just takes the new sum further away from the point 2 on the number line.

Eureka Math Grade 7 Module 3 Lesson 5 Exercise Answer Key

Exercise 1
With a partner, take turns alternating roles as writer and speaker The speaker verbalizes how to rewrite the sum and properties that justify each step as the writer writes what is being spoken without any input At the end of each problem, discuss in pairs the resulting equivalent expressions.
Write the sum, and then write an equivalent expression by collecting like terms and removing parentheses whenever possible.
a – 4 and 4b + 4
Answer:
– 4 + (4b + 4)
(- 4 + 4) + 4b Any order, any grouping
0 + 4b Additive inverse
4b Additive identity property of zero

b. 3x and 1 – 3x
Answer:
3x + (1 – 3x)
3x + (1 + (- 3x)) Subtraction as adding the inverse
(3x + (- 3x)) + 1 Any order, any grouping
0 + 1 Additive inverse
1 Additive identity property of zero

c. The opposite of 4x and – 5 + 4x
Answer:
– 4x + (- 5 + 4x)
(- 4x + 4x) + (- 5) Any order, any grouping
0 + (- 5) Additive inverse
– 5 Additive identity property of zero

d. The opposite of – 10t and t – 10t
Answer:
10t + (t – 10t)
(10t + (- 10t)) + t Any order, any grouping
0 + t Additive inverse
t Additive identity property of zero

e. The opposite of (- 7 – 4v) and – 4v
Answer:
– (- 7 – 4v) + (- 4v)
– 1(- 7 – 4v) + (- 4v) Taking the opposite is equivalent to multiplying by –1.
7 + 4v + (- 4v) Distributive property
7 + 0 Any grouping, additive inverse
7 Additive identity property of zero

Exercise 2.
Write the product, and then write the expression in standard form by removing parentheses and combining like terms Justify each step.
a. The reciprocal of 3 and – 6y – 3x
Answer:
(\(\frac{1}{3}\) )(- 6y + (- 3x)) Rewrite subtraction as an addition problem
(\(\frac{1}{3}\) )(- 6y) + (\(\frac{1}{3}\) )(- 3x) Distributive property
– 2y – 1x Multiplicative inverse
– 2y – x Multiplicative identity property of one

b. The multiplicative inverse of 4 and 4h – 20
Answer:
(\(\frac{1}{4}\) )(4h + (- 20)) Rewrite subtraction as an addition problem
(\(\frac{1}{4}\) )(4h) + (\(\frac{1}{4}\) )(- 20) Distributive property
1h + (- 5) Multiplicative inverse
h – 5 Multiplicative identity property of one

c. The multiplicative inverse of – \(\frac{1}{6}\) and 2 – \(\frac{1}{6}\) j
(- 6)(2 + (- \(\frac{1}{6}\) j)) Rewrite subtraction as an addition problem
(- 6)(2) + (- 6)(- \(\frac{1}{6}\) j) Distributive property
– 12 + 1j Multiplicative inverse
– 12 + j Multiplicative identity property of one

Eureka Math Grade 7 Module 3 Lesson 5 Problem Set Answer Key

Question 1.
Fill in the missing parts.
a. The sum of 6c – 5 and the opposite of 6c
(6c – 5) + (- 6c)
_____ Rewrite subtraction as addition
Answer:
(6c + (- 5)) + (- 6c)
6c + (- 6c) + (- 5)
Regrouping/any order (or commutative property of addition)
0 + (- 5) ___
Answer:
Additive inverse
__ Additive identity property of zero
Answer:
– 5

b. The product of – 2c + 14 and the multiplicative inverse of – 2
(- 2c + 14)(- \(\frac{1}{2}\) )
(- 2c)(- \(\frac{1}{2}\) ) + (14)(- \(\frac{1}{2}\) )
Distributive property
__ Multiplicative inverse, multiplication
1c + (- 7)
1c – 7 Adding the additive inverse is the same as subtraction
c – 7 ____
Answer:
Multiplicative identity property of one

Question 2.
Write the sum, and then rewrite the expression in standard form by removing parentheses and collecting like terms.
a. 6 and p – 6
Answer:
6 + (p – 6)
6 + (- 6) + p
0 + p
p

b. 10w + 3 and – 3
Answer:
(10w + 3) + (- 3)
10w + (3 + (- 3))
10w + 0
10w

c – x – 11 and the opposite of – 11
Answer:
(- x + (- 11)) + 11
– x + ((- 11) + (11))
– x + 0
– x

d. The opposite of 4x and 3 + 4x
Answer:
(- 4x) + (3 + 4x)
((- 4x) + 4x) + 3
0 + 3
3

e. 2g and the opposite of (1 – 2g)
Answer:
2g + (- (1 – 2g))
2g + (- 1) + 2g
2g + 2g + (- 1)
4g + (- 1)
4g – 1

Question 3.
Write the product, and then rewrite the expression in standard form by removing parentheses and collecting like terms.
a. 7h – 1 and the multiplicative inverse of 7
Answer:
(7h + (- 1))(\(\frac{1}{7}\) )
(\(\frac{1}{7}\) )(7h) + (\(\frac{1}{7}\) )(- 1)
h – \(\frac{1}{7}\)

b. The multiplicative inverse of – 5 and 10v – 5
(- \(\frac{1}{5}\) )(10v – 5)
(- \(\frac{1}{5}\) )(10v) + (- \(\frac{1}{5}\) )(- 5)
– 2v + 1

Eureka Math Grade 7 Module 3 Lesson 5 Exit Ticket Answer Key

Question 1.
Find the sum of 5x + 20 and the opposite of 20 Write an equivalent expression in standard form Justify each step.
Answer:
(5x + 20) + (- 20)
5x + (20 + (- 20)) Associative property of addition
5x + 0 Additive inverse
5x Additive identity property of zero

Question 2.
For 5x + 20 and the multiplicative inverse of 5, write the product and then write the expression in standard form, if possible Justify each step.
Answer:
(5x + 20)(\(\frac{1}{5}\) )
(5x)(\(\frac{1}{5}\) ) + 20(\(\frac{1}{5}\) ) Distributive property
1x + 4 Multiplicative inverses, multiplication
x + 4 Multiplicative identity property of one

Engage NY Eureka Math 7th Grade Module 3 Lesson 19 Answer Key

Eureka Math Grade 7 Module 3 Lesson 19 Example Answer Key

Example: Area of a Parallelogram
The coordinate plane below contains figure P, parallelogram ABCD.

Answer:

a. Write the ordered pairs of each of the vertices next to the vertex points.
Answer:
See figure.

b. Draw a rectangle surrounding figure P that has vertex points of A and C. Label the two triangles in the figure as S and T.
Answer:
See figure.

c. Find the area of the rectangle.
Answer:
Base = 8 units
Height = 6 units
Area = 8 units × 6 units = 48 sq. units

d. Find the area of each triangle.
Answer:
Figure S
Base = 3 units
Height = 6 units
Area = \(\frac{1}{2}\) × 3 units × 6 units
= 9 sq. units
Figure T
Base = 3 units
Height = 6 units
Area = \(\frac{1}{2}\) × 3 units × 6 units
= 9 sq. units

e. Use these areas to find the area of parallelogram ABCD.
Answer:
Area P = Area of rectangle – Area S – Area T
= 48 sq. units – 9 sq. units – 9 sq. units = 30 sq. units

The coordinate plane below contains figure R, a rectangle with the same base as the parallelogram above.

Answer:

f. Draw triangles S and T and connect to figure R so that you create a rectangle that is the same size as the rectangle you created on the first coordinate plane.
Answer:
See figure.

g. Find the area of rectangle R.
Answer:
Base = 5 units
Height = 6 units
Area = 30 sq. units

h. What do figures R and P have in common?
Answer:
They have the same area. They share the same base and have the same height.

Eureka Math Grade 7 Module 3 Lesson 19 Exercise Answer Key

Exercise 1.
Find the area of triangle ABC.

Answer:
A = \(\frac{1}{2}\) × 7 units × 4 units = 14 sq.units

Exercise 2.
Find the area of quadrilateral ABCD two different ways.

Answer:

\(\frac{1}{2}\) × 2 × 5 + 2 × 5 + \(\frac{1}{2}\) × 1 × 5 = 5 + 10 + 2.5 = 17.5
The area is 17.5 sq.units.

\(\frac{1}{2}\) × (5 + 2) × 5 = 17.5
The area is 17.5 sq.units.

Exercise 3.
The area of quadrilateral ABCD is 12 sq. units. Find x.

Answer:
Area = base×height
12 sq.units = 2x
6 units = x

Exercise 4.
The area of triangle ABC is 14 sq. units. Find the length of side \(\overline{B C}\).

Answer:
Area = \(\frac{1}{2}\) × base×height
14 sq.units = \(\frac{1}{2}\) × BC × (7 units)
BC = 4 units

Exercise 5.
Find the area of triangle ABC.

Answer:
Area of rectangle ARST = 11 units × 10 units = 110 sq.units
Area of triangle ARB = \(\frac{1}{2}\) × 7 units×10 units = 35 sq.units
Area of triangle BSC = \(\frac{1}{2}\) × 4 units×5 units = 10 sq.units
Area of triangle ATC = \(\frac{1}{2}\) × 11 units×5 units = 27.5 sq.units
Area of triangle ABC = Area of ARST – Area of ARB – Area of BSC – Area of ATC = 37.5 sq.units

Eureka Math Grade 7 Module 3 Lesson 19 Problem Set Answer Key

Find the area of each figure.
Question 1.

Answer:
Area = 13.5 sq. units

Question 2.

Answer:
Area = 4.5π sq. units ≈14.13 sq. units

Question 3.

Answer:
Area = 48 sq. units

Question 4.

Answer:
Area = (2π + 16) sq. units ≈ 22.28 sq. units

Question 5.

Answer:
Area = 68 sq. units

Question 6.

Answer:
Area = 46 sq. units

For Problems 7–9, draw a figure in the coordinate plane that matches each description.
Question 7.
A rectangle with an area of 18 sq. units

Answer:

Question 8.
A parallelogram with an area of 50 sq. units

Answer:

Question 9.
A triangle with an area of 25 sq. units

Answer:

Find the unknown value labeled as x on each figure.
Question 10.
The rectangle has an area of 80 sq. units.

Answer:
x = 8

Question 11.
The trapezoid has an area of 115 sq. units.

Answer:
x = 10

Question 12.
Find the area of triangle ABC.

Answer:
Area = 6.5 sq. units

Question 13.
Find the area of the quadrilateral using two different methods. Describe the methods used, and explain why they result in the same area.

Answer:
Area = 15 sq. units
One method is by drawing a rectangle around the figure. The area of the parallelogram is equal to the area of the rectangle minus the area of the two triangles. A second method is to use the area formula for a parallelogram (Area = base × height).

Question 14.
Find the area of the quadrilateral using two different methods. What are the advantages or disadvantages of each method?

Answer:
Area = 60 sq. units
One method is to use the area formula for a trapezoid, A = \(\frac{1}{2}\) (base 1 + base 2)×height. The second method is to split the figure into a rectangle and a triangle. The second method requires more calculations. The first method requires first recognizing the figure as a trapezoid and recalling the formula for the area of a trapezoid.

Eureka Math Grade 7 Module 3 Lesson 19 Exit Ticket Answer Key

The figure ABCD is a rectangle. AB = 2 units, AD = 4 units, and AE = FC = 1 unit.

Question 1.
Find the area of rectangle ABCD.
Answer:
Area = 4 units × 2 units = 8 sq. units

Question 2.
Find the area of triangle ABE.
Answer:
Area = \(\frac{1}{2}\) × 1 unit × 2 units = 1 sq. unit

Question 3.
Find the area of triangle DCF.
Answer:
Area = \(\frac{1}{2}\) × 1 unit × 2 units = 1 sq. unit

Question 4.
Find the area of the parallelogram BEDF two different ways.
Answer:
Area = Area of ABCD – Area of ABE – Area of DCF
= (8 – 1 – 1) sq. units = 6 sq. units

Area = base × height
= 3 units × 2 units = 6 sq. units

Engage NY Eureka Math 7th Grade Module 3 Lesson 18 Answer Key

Eureka Math Grade 7 Module 3 Lesson 18 Example Answer Key

Example 1.
Find the area of the following semicircle. Use π ≈ \(\frac{22}{7}\) .

Answer:
If the diameter of the circle is 14 cm, then the
radius is 7 cm. The area of the semicircle is half
of the area of the circular region.
A ≈ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ (7 cm)²
A ≈ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ 49 cm²
A ≈ 77 cm²

What is the area of the quarter circle? Use π ≈ \(\frac{22}{7}\) .

Answer:
A ≈ \(\frac{1}{4}\) ∙ \(\frac{22}{7}\) (6 cm)²
A ≈ \(\frac{1}{4}\) ∙ \(\frac{22}{7}\) ∙ 36 cm²
A ≈ \(\frac{198}{7}\) cm²

Example 2.
Marjorie is designing a new set of placemats for her dining room table. She sketched a drawing of the placement on graph paper. The diagram represents the area of the placemat consisting of a rectangle and two semicircles at either end. Each square on the grid measures 4 inches in length.
Find the area of the entire placemat. Explain your thinking regarding the solution to this problem.

Answer:
The length of one side of the rectangular section is 12 inches in length, while the width is 8 inches. The radius of the semicircular region is 4 inches. The area of the rectangular part is (8 in) ∙ (12 in) = 96 in². The total area must include the two semicircles on either end of the placemat. The area of the two semicircular regions is the same as the area of one circle with the same radius. The area of the circular region is A = π ∙ (4 in)² = 16π in². In this problem, using π ≈ 3.14 makes more sense because there are no fractions in the problem. The area of the semicircular regions is approximately 50.24 in². The total area for the placemat is the sum of the areas of the rectangular region and the two semicircular regions, which is approximately (96+50.24) in² = 146.24 in².

If Marjorie wants to make six placemats, how many square inches of fabric will she need? Assume there is no waste.
Answer:
There are 6 placemats that are each 146.24 in², so the fabric needed for all is 6 ∙ 146.24 in² = 877.44 in².

Marjorie decides that she wants to sew on a contrasting band of material around the edge of the placemats. How much band material will Marjorie need?
Answer:
The length of the band material needed will be the sum of the lengths of the two sides of the rectangular region and the circumference of the two semicircles (which is the same as the circumference of one circle with the same radius).
P = (l+l+2πr)
P = (12+12+2 ∙ π ∙ 4) = 49.12
The perimeter is 49.12 in².

Example 3.
The circumference of a circle is 24π cm. What is the exact area of the circle?
Draw a diagram to assist you in solving the problem.

What information is needed to solve the problem?
Answer:
The radius is needed to find the area of the circle. Let the radius be r cm. Find the radius by using the circumference formula.
C = 2πr
24π = 2πr
(\(\frac{1}{2}\) π)24π = (\(\frac{1}{2}\) π)2πr
12 = r
The radius is 12 cm.

Next, find the area.
Answer:
A = π r²
A = π(12)²
A = 144π
The exact area of the circle is 144π cm².

Eureka Math Grade 7 Module 3 Lesson 18 Exercise Answer Key

Opening Exercise
Draw a circle with a diameter of 12 cm and a square with a side length of 12 cm on grid paper. Determine the area of the square and the circle.

Answer:
Area of square: A = (12 cm)² = 144 cm²; Area of circle: A = π ∙ (6 cm)² = 36π cm²

Brainstorm some methods for finding half the area of the square and half the area of the circle.
Answer:
Some methods include folding in half and counting the grid squares and cutting each in half and counting the squares.

Find the area of half of the square and half of the circle, and explain to a partner how you arrived at the area.
Answer:
The area of half of the square is 72 cm². The area of half of the circle is 18π cm². Some students may count the squares; others may realize that half of the square is a rectangle with side lengths of 12 cm and 6 cm and use A = l ∙ w to determine the area. Some students may fold the square vertically, and some may fold it horizontally. Some students will try to count the grid squares in the semicircle and find that it is easiest to take half of the area of the circle.

What is the ratio of the new area to the original area for the square and for the circle?
Answer:
The ratio of the areas of the rectangle (half of the square) to the square is 72:144 or 1:2. The ratio for the areas of the circles is 18π:36π or 1:2.

Find the area of one – fourth of the square and one – fourth of the circle, first by folding and then by another method. What is the ratio of the new area to the original area for the square and for the circle?
Answer:
Folding the square in half and then in half again will result in one – fourth of the original square. The resulting shape is a square with a side length of 6 cm and an area of 36 cm². Repeating the same process for the circle will result in an area of 9π cm². The ratio for the areas of the squares is 36:144 or 1:4. The ratio for the areas of the circles is 9π:36π or 1:4.

Write an algebraic expression that expresses the area of a semicircle and the area of a quarter circle.
Answer:
Semicircle: A = \(\frac{1}{2}\) πr²; Quarter circle: A = \(\frac{1}{4}\) πr²

Exercise 1.
Find the area of a circle with a diameter of 42 cm. Use π ≈ \(\frac{22}{7}\) .
Answer:
If the diameter of the circle is 42 cm, then the radius is 21 cm .
A = πr²
A ≈ \(\frac{22}{7}\) (21 cm)²
A ≈ 1386 cm²

Exercise 2.
The circumference of a circle is 9π cm.
a. What is the diameter?
Answer:
If C = πd, then 9π cm = πd.
Solving the equation for the diameter, d, \(\frac{1}{\pi}\) ∙ 9π cm = \(\frac{1}{\pi}\) π ∙ d.
So, 9 cm = d.

b. What is the radius?
Answer:
If the diameter is 9 cm, then the radius is half of that or \(\frac{9}{2}\) cm.

c. What is the area?
Answer:
The area of the circle is A = π ∙ (\(\frac{9}{2}\) cm)², so A = \(\frac{81}{4}\) π cm².

Exercise 3.
If students only know the radius of a circle, what other measures could they determine? Explain how students would use the radius to find the other parts.
Answer:
If students know the radius, then they can find the diameter. The diameter is twice as long as the radius. The circumference can be found by doubling the radius and multiplying the result by π. The area can be found by multiplying the radius times itself and then multiplying that product by π.

Exercise 4.
Find the area in the rectangle between the two quarter circles if AF = 7 ft, FB = 9 ft, and HD = 7 ft. Use π ≈ \(\frac{22}{7}\) . Each quarter circle in the top – left and lower – right corners have the same radius.

Answer:
The area between the quarter circles can be found by subtracting the area of the two quarter circles from the area of the rectangle. The area of the rectangle is the product of the length and the width. Side AB has a length of 16 ft and Side AD has a length of 14 ft. The area of the rectangle is
A = 16 ft ∙ 14 ft = 224 ft². The area of the two quarter circles is the same as the area of a semicircle, which is half the area of a circle. A = \(\frac{1}{2}\) πr².
A ≈ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ (7 ft)²
A ≈ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ 49 ft²
A ≈ 77 ft²
The area between the two quarter circles is 224 ft² – 77 ft² = 147 ft².

Eureka Math Grade 7 Module 3 Lesson 18 Problem Set Answer Key

Question 1.
Mark created a flower bed that is semicircular in shape. The diameter of the flower bed is 5 m .
a. What is the perimeter of the flower bed? (Approximate π to be 3.14.)

Answer:
The perimeter of this flower bed is the sum of the diameter and one – half the circumference of a circle with the same diameter.
P = diameter+\(\frac{1}{2}\) π ∙ diameter
P ≈ 5 m+\(\frac{1}{2}\) ∙ 3.14 ∙ 5 m
P ≈ 12.85 m

b. What is the area of the flower bed? (Approximate π to be 3.14.)
Answer:
A = \(\frac{1}{2}\) π (2.5 m)²
A = \(\frac{1}{2}\) π (6.25 m²)
A ≈ 0.5 ∙ 3.14 ∙ 6.25 m²
A ≈ 9.8 m²

Question 2.
A landscape designer wants to include a semicircular patio at the end of a square sandbox. She knows that the area of the semicircular patio is 25.12 cm².
a. Draw a picture to represent this situation.
Answer:

b. What is the length of the side of the square?
Answer:
If the area of the patio is 25.12 cm², then we can find the radius by solving the equation A = \(\frac{1}{2}\) πr² and substituting the information that we know. If we approximate π to be 3.14 and solve for the radius, r, then
25.12 cm² ≈ \(\frac{1}{2}\) πr²
\(\frac{2}{1}\) ∙ 25.12 cm² ≈ \(\frac{2}{1}\) ∙ \(\frac{1}{2}\) πr²
50.24 cm² ≈ 3.14r²
\(\frac{1}{3.14}\) ∙ 50.24 cm² ≈ \(\frac{1}{3.14}\) ∙ 3.14r²
16 cm² ≈ r²
4 cm ≈ r
The length of the diameter is 8 cm; therefore, the length of the side of the square is 8 cm.

Question 3.
A window manufacturer designed a set of windows for the top of a two – story wall. If the window is comprised of 2 squares and 2 quarter circles on each end, and if the length of the span of windows across the bottom is 12 feet, approximately how much glass will be needed to complete the set of windows?

Answer:
The area of the windows is the sum of the areas of the two quarter circles and the two squares that make up the bank of windows. If the span of windows is 12 feet across the bottom, then each window is 3 feet wide on the bottom. The radius of the quarter circles is 3 feet, so the area for one quarter circle window is A = \(\frac{1}{4}\) π ∙ (3 ft)², or A ≈ 7.065 ft². The area of one square window is A = (3 ft)², or 9 ft². The total area is A = 2(area of quarter circle)+2(area of square), or A ≈ (2 ∙ 7.065 ft² )+(2 ∙ 9 ft² ) ≈ 32.13 ft².

Question 4.
Find the area of the shaded region. (Approximate π to be \(\frac{22}{7}\) .)

Answer:
A = \(\frac{1}{4}\) π(12 in)²
A = \(\frac{1}{4}\) π ∙ 144 in²
A ≈ \(\frac{1}{4}\) ∙ \(\frac{22}{7}\) ∙ 144 in²
A ≈ 792/7 in² or 113.1 in²

Question 5.
The figure below shows a circle inside of a square. If the radius of the circle is 8 cm, find the following and explain your solution.

a. The circumference of the circle
Answer:
C = 2π ∙ 8 cm
C = 16π cm

b. The area of the circle
Answer:
A = π ∙ (8 cm)²
A = 64 π cm²

c. The area of the square
Answer:
A = 16 cm ∙ 16 cm
A = 256 cm²

Question 6.
Michael wants to create a tile pattern out of three quarter circles for his kitchen backsplash. He will repeat the three quarter circles throughout the pattern. Find the area of the tile pattern that Michael will use. Approximate π as 3.14.

Answer:
There are three quarter circles in the tile design. The area of one quarter circle multiplied by 3 will result in the total area.
A = \(\frac{1}{4}\) π ∙ (16 cm)²
A ≈ \(\frac{1}{4}\) ∙ 3.14 ∙ 256 cm²
A ≈ 200.96 cm²

A ≈ 3 ∙ 200.96 cm²
A ≈ 602.88 cm²
The area of the tile pattern is approximately 602.88 cm².

Question 7.
A machine shop has a square metal plate with sides that measure 4 cm each. A machinist must cut four semicircles with a radius of \(\frac{1}{2}\) cm and four quarter circles with a radius of 1 cm from its sides and corners. What is the area of the plate formed? Use \(\frac{22}{7}\) to approximate π.

Answer:
The area of the metal plate is determined by subtracting the four quarter circles (corners) and the four half – circles (on each side) from the area of the square. Area of the square: A = (4 cm)² = 16 cm².
The area of four quarter circles is the same as the area of a circle with a radius of
1 cm: A ≈ \(\frac{22}{7}\) (1 cm)² ≈ \(\frac{22}{7}\) cm².
The area of the four semicircles with radius \(\frac{1}{2}\) cm is
A ≈ 4 ∙ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ (\(\frac{1}{2}\) cm)²
A ≈ 4 ∙ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ \(\frac{1}{4}\) cm² ≈ 11/7 cm².
The area of the metal plate is
A ≈ 16 cm² – \(\frac{22}{7}\) cm² – \(\frac{11}{7}\) cm² ≈ \(\frac{79}{7}\) cm²

Question 8.
A graphic artist is designing a company logo with two concentric circles (two circles that share the same center but have different radii). The artist needs to know the area of the shaded band between the two concentric circles. Explain to the artist how he would go about finding the area of the shaded region.

Answer:
The artist should find the areas of both the larger and smaller circles. Then, the artist should subtract the area of the smaller circle from the area of the larger circle to find the area between the two circles. The area of the larger circle is
A = π ∙ (9 cm)\(\frac{22}{7}\) or 81π cm\(\frac{22}{7}\) .
The area of the smaller circle is
A = π(5 cm)\(\frac{22}{7}\) or 25π cm\(\frac{22}{7}\) .
The area of the region between the circles is 81π cm\(\frac{22}{7}\) – 25π cm\(\frac{22}{7}\) = 56π cm\(\frac{22}{7}\) . If we approximate π to be 3.14, then A ≈ 175.84 cm\(\frac{22}{7}\) .

Question 9.
Create your own shape made up of rectangles, squares, circles, or semicircles, and determine the area and perimeter.
Answer:
Student answers may vary.

Eureka Math Grade 7 Module 3 Lesson 18 Exit Ticket Answer Key

Question 1.
Ken’s landscape gardening business creates odd – shaped lawns that include semicircles. Find the area of this semicircular section of the lawn in this design. Use \(\frac{22}{7}\) for π.

Answer:
If the diameter is 5 m, then the radius is \(\frac{5}{2}\) m. Using the formula for area of a semicircle, A = \(\frac{1}{2}\) πr², A ≈ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ (\(\frac{5}{2}\) m)². Using the order of operations,
A ≈ \(\frac{1}{2}\) ∙ \(\frac{22}{7}\) ∙ \(\frac{25}{4}\) m² ≈ \(\frac{550}{56}\) m² ≈ 9.8 m².

Question 2.
In the figure below, Ken’s company has placed sprinkler heads at the center of the two small semicircles. The radius of the sprinklers is 5 ft. If the area in the larger semicircular area is the shape of the entire lawn, how much of the lawn will not be watered? Give your answer in terms of π and to the nearest tenth. Explain your thinking.

Answer:
The area not covered by the sprinklers would be the area between the larger semicircle and the two smaller ones. The area for the two semicircles is the same as the area of one circle with the same radius of 5 ft. The area not covered by the sprinklers can be found by subtracting the area of the two smaller semicircles from the area of the large semicircle.

Area Not Covered = Area of large semicircle – Area of two smaller semicircles
A = \(\frac{1}{2}\) π ∙ (10 ft)² – (2 ∙ (\(\frac{1}{2}\) (π ∙ (5 ft)² )))
A = \(\frac{1}{2}\) π ∙ 100 ft² – π ∙ 25 ft²
A = 50π ft² – 25π ft² = 25π ft²
Let π ≈ 3.14
A ≈ 78.5 ft²
The sprinklers will not cover 25π ft² or 78.5 ft² of the lawn.

Engage NY Eureka Math 7th Grade Module 4 Lesson 18 Answer Key

Eureka Math Grade 7 Module 4 Lesson 18 Example Answer Key

Example 1.
All of the 3-letter passwords that can be formed using the letters A and B are as follows: AAA, AAB, ABA, ABB, BAA, BAB, BBA, BBB.
a. What percent of passwords contain at least two B’s?
Answer:
There are four passwords that contain at least two B’s: ABB, BAB, BBA, and BBB. There are eight passwords total.
\(\frac{4}{8}\) = \(\frac{1}{2}\) = 50%, so 50% of the passwords contain at least two B’s.

b. What percent of passwords contain no A’s?
Answer:
There is one password that contains no A’s. There are eight passwords total.
\(\frac{1}{8}\) = 0.125 = 12.5%, so 12.5% of the passwords contain no A’s.

Example 2.
In a set of 3-letter passwords, 40% of the passwords contain the letter B and two of another letter. Which of the two sets below meets the criteria? Explain how you arrived at your answer.

Answer:
For each set, I counted how many passwords have the letter B and two of another letter. Then, I checked to see if that quantity equaled 40% of the total number of passwords in the set.
In Set 1, CBC, AAB, ABA, CCB, BAA, and BCC are the passwords that contain a B and two of another letter. Set 1 meets the criteria since there are 15 passwords total and 40% of 15 is 6.
Quantity = Percent × Whole
6 = 0.4(15)
6 = 6 → True

In Set 2, EBE, EEB, CCB, and CBC are the only passwords that contain a B and two others of the same letter. Set 2 meets the criteria since there are 10 passwords total and 40% of 10 is 4.
Quantity = Percent × Whole
4 = 0.4(10)
4 = 4 → True
So, both Sets 1 and 2 meet the criteria.

Example 3.
Look at the 36 points on the coordinate plane with whole number coordinates between 1 and 6, inclusive.

a. Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 7.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 6.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 5.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 4.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 3.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 2.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 8.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 9.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 10.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 11.
Draw a line through each of the points which have an x-coordinate and y-coordinate sum of 12.
Answer:

b. What percent of the 36 points have a coordinate sum of 7?
Answer:
\(\frac{6}{36}\) = \(\frac{1}{6}\) = 16 \(\frac{2}{3}\)%

c. Write a numerical expression that could be used to determine the percent of the 36 points that have a coordinate sum of 7.
Answer:
There are six coordinate points in which the sum of the x-coordinate and the y-coordinate is 7. So,
\(\frac{6}{36}\) × 100%.

d. What percent of the 36 points have a coordinate sum of 5 or less?
Answer:
\(\frac{10}{36}\) × 100% = 27 \(\frac{7}{9}\)%

e. What percent of the 36 points have a coordinate sum of 4 or 10?
Answer:
\(\frac{6}{36}\) × 100% = 16 \(\frac{2}{3}\)%

Eureka Math Grade 7 Module 4 Lesson 18 Exercise Answer Key

Opening Exercise
You are about to switch out your books from your locker during passing period but forget the order of your locker combination. You know that there are the numbers 3, 16, and 21 in some order. What is the percent of locker combinations that start with 3?
Locker Combination Possibilities:
3, 16, 21
21, 16, 3
16, 21, 3
21, 3, 16
16, 3, 21
3, 21, 16
Answer:
\(\frac{2}{6}\) = \(\frac{1}{3}\) = \(0.33 \overline{3}\) = \(33 . \overline{3} \%\)

Exercises 1–2

Exercise 1.
How many 4-letter passwords can be formed using the letters A and B?
Answer:
16: AAAA, AAAB, AABB, ABBB, AABA, ABAA, ABAB, ABBA,
BBBB, BBBA, BBAA, BAAA, BBAB, BABB, BABA, BAAB

Exercise 2.
What percent of the 4-letter passwords contain
a. No A’s?
Answer:
\(\frac{1}{16}\) = 0.0625 = 6.25%

b. Exactly one A?
Answer:
\(\frac{4}{16}\) = \(\frac{1}{4}\) = 25%

c. Exactly two A’s?
Answer:
\(\frac{6}{16}\) = 0.375 = 37.5%

d. Exactly three A’s?
Answer:
\(\frac{4}{16}\) = \(\frac{1}{4}\) = 25%

e. Four A’s?
Answer:
\(\frac{1}{16}\) = 0.0625 = 6.25%

f. The same number of A’s and B’s?
Answer:
\(\frac{6}{16}\) = 0.375 = 37.5%

→ Which categories have percents that are equal?
No A’s and four A’s have the same percents.
Exactly one A and exactly three A’s have the same percents.
Exactly two A’s and the same number of A’s and B’s also have the same percents.

→ Why do you think they are equal?
Four A’s is the same as saying no B’s, and since there are only two letters, no B’s is the same as no A’s.
The same reasoning can be used for exactly one A and exactly three A’s. If there are exactly three A’s, then this would mean that there is exactly one B, and since there are only two letters, exactly one B is the same as exactly one A.
Finally, exactly two A’s and the same number of A’s and B’s are the same because the same amount of A’s and B’s would be two of each.

Exercises 3–4

Exercise 3.
Shana read the following problem:
“How many letter arrangements can be formed from the word triangle that have two vowels and two consonants (order does not matter)?”
She answered that there are 30 letter arrangements.
Twenty percent of the letter arrangements that began with a vowel actually had an English definition. How many letter arrangements that begin with a vowel have an English definition?
Answer:
0.20 × 30 = 6
Six have a formal English definition.

Exercise 4.
Using three different keys on a piano, a songwriter makes the beginning of his melody with three notes, C, E, and G: CCE, EEE, EGC, GCE, CEG, GEE, CGE, GGE, EGG, EGE, GCG, EEC, ECC, ECG, GGG, GEC, CCG, CEE, CCC, GEG, CGC.
a. From the list above, what is the percent of melodies with all three notes that are different?
Answer:
\(\frac{6}{21}\) ≈ 28.6%

b. From the list above, what is the percent of melodies that have three of the same notes?
Answer:
\(\frac{3}{21}\) ≈ 14.3%

Eureka Math Grade 7 Module 4 Lesson 18 Problem Set Answer Key

Question 1.
A six-sided die (singular for dice) is thrown twice. The different rolls are as follows:
1 and 1, 1 and 2, 1 and 3, 1 and 4, 1 and 5, 1 and 6,
2 and 1, 2 and 2, 2 and 3, 2 and 4, 2 and 5, 2 and 6,
3 and 1, 3 and 2, 3 and 3, 3 and 4, 3 and 5, 3 and 6,
4 and 1, 4 and 2, 4 and 3, 4 and 4, 4 and 5, 4 and 6,
5 and 1, 5 and 2, 5 and 3, 5 and 4, 5 and 5, 5 and 6,
6 and 1, 6 and 2, 6 and 3, 6 and 4, 6 and 5, 6 and 6.
a. What is the percent that both throws will be even numbers?
Answer:
\(\frac{9}{36}\) = 25%

b. What is the percent that the second throw is a 5?
Answer:
\(\frac{6}{36}\) = 16 \(\frac{2}{3}\)%

c. What is the percent that the first throw is lower than a 6?
Answer:
\(\frac{30}{36}\) = 83 \(\frac{1}{3}\)%

Question 2.
You have the ability to choose three of your own classes, art, language, and physical education. There are three art classes (A1, A2, A3), two language classes (L1, L2), and two P.E. classes (P1, P2) to choose from. The order does not matter and you must choose one from each subject.

Compare the percent of possibilities with A1 in your schedule to the percent of possibilities with L1 in your schedule.
Answer:
A1: \(\frac{4}{12}\) = 33 \(\frac{1}{3}\)% L1: \(\frac{6}{12}\) = 50%
There is a greater percent with L1 in my schedule.

Question 3.
Fridays are selected to show your school pride. The colors of your school are orange, blue, and white, and you can show your spirit by wearing a top, a bottom, and an accessory with the colors of your school. During lunch, 11 students are chosen to play for a prize on stage. The table charts what the students wore.

a. What is the percent of outfits that are one color?
Answer:
\(\frac{2}{11}\) = 18 \(\frac{2}{11}\)%

b. What is the percent of outfits that include orange accessories?
Answer:
\(\frac{5}{11}\) = 45 \(\frac{5}{11}\)%

Question 4.
Shana wears two rings (G represents gold, and S represents silver) at all times on her hand. She likes fiddling with them and places them on different fingers (pinky, ring, middle, index) when she gets restless. The chart is tracking the movement of her rings.

a. What percent of the positions shows the gold ring on her pinky finger?
Answer:
\(\frac{4}{14}\) ≈ 28.57%

b. What percent of the positions shows both rings on the same finger?
Answer:
\(\frac{4}{14}\) = 28 \(\frac{4}{7}\)%

Question 5.
Use the coordinate plane below to answer the following questions.

a. What is the percent of the 36 points whose quotient of \(\frac{x \text { -coordinate }}{y-\text { coordinate }}\) is greater than one?
Answer:
\(\frac{15}{36}\) = 41 \(\frac{2}{3}\)%

b. What is the percent of the 36 points whose coordinate quotient is equal to one?
Answer:
\(\frac{6}{36}\) = 16\(\frac{2}{3}\)%

Eureka Math Grade 7 Module 4 Lesson 18 Exit Ticket Answer Key

There are a van and a bus transporting students on a student camping trip. Arriving at the site, there are 3 parking spots. Let v represent the van and b represent the bus. The chart shows the different ways the vehicles can park.

a. In what percent of the arrangements are the vehicles separated by an empty parking space?
Answer:
\(\frac{2}{6}\) = 33 \(\frac{1}{3}\)%

b. In what percent of the arrangements are the vehicles parked next to each other?
Answer:
\(\frac{4}{6}\) = 66 \(\frac{2}{3}\)%

c. In what percent of the arrangements does the left or right parking space remain vacant?
Answer:
\(\frac{4}{6}\) = 66 \(\frac{2}{3}\)%

Engage NY Eureka Math 7th Grade Module 3 Lesson 22 Answer Key

Eureka Math Grade 7 Module 3 Lesson 22 Example Answer Key

Example 1.
The pyramid in the picture has a square base, and its lateral faces are triangles that are exact copies of one another. Find the surface area of the pyramid.

Answer:
The surface area of the pyramid consists of one square base and four lateral triangular faces.
B = s²
B = (6 cm)²
B = 36 cm²
The pyramid’s base area is 36 cm².

LA = 4(\(\frac{1}{2}\) bh)
LA = 4 ∙ \(\frac{1}{2}\) (6 cm ∙ 7 cm)
LA = 2(6 cm ∙ 7 cm)
LA = 2(42 cm² )
LA = 84 cm²
The pyramid’s lateral area is 84 cm².
SA = LA + B
SA = 84 cm² + 36 cm²
SA = 120 cm²
The surface area of the pyramid is 120 cm².

Example 2: Using Cubes
There are 13 cubes glued together forming the solid in the diagram. The edges of each cube are \(\frac{1}{4}\) inch in length. Find the surface area of the solid.

Answer:
The surface area of the solid consists of 46 square faces, all having side lengths of \(\frac{1}{4}\) inch. The area of a square having sides of length \(\frac{1}{4}\) inch is \(\frac{1}{16}\) in² .
SA = 46 ∙ A_square
SA = 46 ∙ \(\frac{1}{16}\) in²
SA = \(\frac{46}{16}\) in²
SA = 2 \(\frac{14}{16}\) in²
SA = 2 \(\frac{7}{8}\) in² The surface are of the solid is 2 \(\frac{7}{8}\) in².

Example 3.
Find the total surface area of the wooden jewelry box. The sides and bottom of the box are all \(\frac{1}{4}\) inch thick.
What are the faces that make up this box?

Answer:
The box has a rectangular bottom, rectangular lateral faces, and a rectangular top that has a smaller rectangle removed from it. There are also rectangular faces that make up the inner lateral faces and the inner bottom of the box.

How does this box compare to other objects that you have found the surface area of?
Answer:
The box is a rectangular prism with a smaller rectangular prism removed from its inside. The total surface area is equal to the surface area of the larger right rectangular prism plus the lateral area of the smaller right rectangular prism.

Large Prism: The surface area of the large right rectangular prism makes up the outside faces of the box, the rim of the box, and the inside bottom face of the box.

SA = LA + 2B
LA = P ∙ h
LA = 32 in. ∙ 4 in.
LA = 128 in²

B = lw
B = 10 in. ∙ 6 in.
B = 60 in²
The lateral area is 128 in².
The base area is 60 in².

SA = LA + 2B
SA = 128 in² + 2(60 in²)
SA = 128 in² + 120 in²
SA = 248 in²
The surface area of the larger prism is 248 in².

Small Prism: The smaller prism is \(\frac{1}{2}\) in. smaller in length and width and \(\frac{1}{4}\) in. smaller in height due to the thickness of the sides of the box.
SA = LA + 1B
LA = P ∙ h
LA = 2(9 \(\frac{1}{2}\) in. + 5 \(\frac{1}{2}\) in.) ∙ 3 \(\frac{3}{4}\) in.
LA = 2(14 in. + 1 in.) ∙ 3 \(\frac{3}{4}\) in.
LA = 2(15 in.) ∙ 3 \(\frac{3}{4}\) in.
LA = 30 in. ∙ 3 \(\frac{3}{4}\) in.
LA = 90 in² + \(\frac{90}{4}\) in²
LA = 90 in² + 22 \(\frac{1}{2}\) in²
LA = 112 \(\frac{1}{2}\) in²
The lateral area is 112 \(\frac{1}{2}\) in².

Surface Area of the Box
SA_box = SA + LA
SA_box = 248 in² + 112 \(\frac{1}{2}\) in²
SA_box = 360 \(\frac{1}{2}\) in²
The total surface area of the box is 360 \(\frac{1}{2}\) in².

Eureka Math Grade 7 Module 3 Lesson 22 Exercise Answer Key

Opening Exercise
What is the area of the composite figure in the diagram? Is the diagram a net for a three-dimensional image? If so, sketch the image. If not, explain why.

Answer:
There are four unit squares in each square of the figure. There are 18 total squares that make up the figure, so the total area of the composite figure is
A = 18 ∙ 4 units² = 72 units².
The composite figure does represent the net of a three-dimensional figure. The figure is shown below.

Eureka Math Grade 7 Module 3 Lesson 22 Problem Set Answer Key

Question 1.
For each of the following nets, draw (or describe) the solid represented by the net and find its surface area.
a. The equilateral triangles are exact copies.

Answer:
The net represents a triangular pyramid where the three lateral faces are identical to each other and the triangular base.
SA = 4B since the faces are all the same size and shape.

SA = 4B
SA = 4(35 \(\frac{1}{10}\) mm² )
SA = 140 mm² + \(\frac{4}{10}\) mm²
SA = 140 \(\frac{2}{5}\) mm²

B = \(\frac{1}{2}\) bh
B = \(\frac{1}{2}\) ∙ 9 mm ∙ 7 \(\frac{4}{5}\) mm
B = \(\frac{9}{2}\) mm ∙ 7 \(\frac{4}{5}\) mm
B = \(\frac{63}{2}\) mm² + \(\frac{36}{10}\) mm²
B = \(\frac{315}{10}\) mm² + \(\frac{36}{10}\) mm²
B = \(\frac{351}{10}\) mm² The surface area of the triangular pyramid is 140 \(\frac{2}{5}\) mm².
B = 35 \(\frac{1}{10}\) mm²

b.

Answer:
The net represents a square pyramid that has four identical lateral faces that are triangles. The base is a square.
B = s²
B = (12 in.)²
B = 144 in²

LA = 2(12 in. ∙ 14 \(\frac{3}{4}\) in.)
SA = LA + B
LA = 4 ∙ \(\frac{1}{2}\) (bh)
LA = 4 ∙ \(\frac{1}{2}\) (12 in. ∙ 14 \(\frac{3}{4}\) in.)
LA = 2(168 in² + 9 in² )
LA = 336 in² + 18 in²
LA = 354 in²
SA = LA + B
SA = 354 in² + 144 in²
SA = 498 in²
The surface area of the square pyramid is 498 in².

Question 2.
Find the surface area of the following prism.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (4 cm + 6 \(\frac{1}{2}\) cm + 4 \(\frac{1}{5}\) cm + 5 \(\frac{1}{4}\) cm) ∙ 9 cm
LA = (19 cm + \(\frac{1}{2}\) cm + \(\frac{1}{5}\) cm + \(\frac{1}{4}\) cm) ∙ 9 cm
LA = (19 cm + \(\frac{10}{20}\) cm + \(\frac{4}{20}\) cm + \(\frac{5}{20}\) cm) ∙ 9 cm
LA = (19 cm + \(\frac{19}{20}\) cm) ∙ 9 cm
LA = 171 cm² + \(\frac{171}{20}\) cm²
LA = 171 cm² + 8 \(\frac{11}{20}\) cm²
LA = 179 \(\frac{11}{20}\) cm²

B = A_rectangle + A_triangle
B = (5 \(\frac{1}{4}\) cm ∙ 4 cm) + \(\frac{1}{2}\)(4 cm ∙ 1 \(\frac{1}{4}\) cm)
B = (20 cm² + 1 cm² ) + (2 cm ∙ 1 \(\frac{1}{4}\) cm)
B = 21 cm² + 2 \(\frac{1}{2}\) cm²
B = 23 \(\frac{1}{2}\) cm²

SA = LA + 2B
SA = 179 \(\frac{11}{20}\) cm² + 2(23 \(\frac{1}{2}\) cm² )
SA = 179 \(\frac{11}{20}\) cm² + 47 cm²
SA = 226 \(\frac{11}{20}\) cm²
The surface area of the prism is 226 \(\frac{11}{20}\) cm².

Question 3.
The net below is for a specific object. The measurements shown are in meters. Sketch (or describe) the object, and then find its surface area.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = 6 cm ∙ \(\frac{1}{2}\) cm
LA = 3 cm²

B = (\(\frac{1}{2}\) cm ∙ \(\frac{1}{2}\) cm) + (\(\frac{1}{2}\) cm ∙ 1 cm) + (\(\frac{1}{2}\) cm ∙ 1 \(\frac{1}{2}\) cm)
B = (\(\frac{1}{4}\) cm²) + (\(\frac{1}{2}\) cm² ) + (\(\frac{3}{4}\) cm² )
B = (\(\frac{1}{4}\)cm²) + (\(\frac{2}{4}\) cm²) + (\(\frac{3}{4}\) cm²)
B = \(\frac{6}{4}\) cm²
B = 1 \(\frac{1}{2}\) cm²

SA = LA + 2B
SA = 3 cm² + 2(1 \(\frac{1}{2}\) cm² )
SA = 3 cm² + 3 cm²
SA = 6 cm²

The surface area of the object is 6 cm².

Question 4.
In the diagram, there are 14 cubes glued together to form a solid. Each cube has a volume of \(\frac{1}{8}\) in³. Find the surface area of the solid.

Answer:
The volume of a cube is s³, and 1/8 in³ is the same as (\(\frac{1}{2}\) in.)³, so the cubes have edges that are \(\frac{1}{2}\) in. long. The cube faces have area s², or (\(\frac{1}{2}\) in.)², or \(\frac{1}{4}\) in². There are 42 cube faces that make up the surface of the solid.
SA = \(\frac{1}{4}\) in² ∙ 42
SA = 10 \(\frac{1}{2}\) in²
The surface area of the solid is 10 \(\frac{1}{2}\) in².

Question 5.
The nets below represent three solids. Sketch (or describe) each solid, and find its surface area.
a.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = 12 ∙ 3
LA = 36 cm²

B = s²
B = (3 cm)²
B = 9 cm²

SA =
36 cm² + 2(9 cm²)
SA = 36 cm² + 18 cm²
SA = 54 cm²

b.

Answer:
SA = 3A_square + 3A_{rt triangle} + A_{equ triangle}
A_square = s²
A_square = (3 cm)²
A_square = 9 cm²
A_{rt triangle} = \(\frac{1}{2}\) bh
A_{rt triangle} = \(\frac{1}{2}\) ∙ 3 cm ∙ 3 cm
A_{rt triangle} = \(\frac{9}{2}\)
A_{rt triangle} = 4 \(\frac{1}{2}\) cm²
A_{equ triangle} = \(\frac{1}{2}\) bh
A_{equ triangle} = \(\frac{1}{2}\) ∙ (4 \(\frac{1}{5}\) cm) ∙ (3 \(\frac{7}{10}\) cm)
A_{equ triangle} = 2 \(\frac{1}{10}\) cm ∙ 3 \(\frac{7}{10}\) cm
A_{equ triangle} = \(\frac{21}{10}\) cm ∙ \(\frac{37}{10}\) cm
A_{equ triangle} = 7 \(\frac{77}{100}\) cm²
A_{equ triangle} = 7 \(\frac{77}{100}\) cm²
SA = 3(9 cm² ) + 3(4 \(\frac{1}{2}\) cm² ) + 7 \(\frac{77}{100}\) cm²
SA = 27 cm² + (12 + \(\frac{3}{2}\)) cm² + 7 \(\frac{77}{100}\) cm²
SA = 47 cm² + \(\frac{1}{2}\) cm² + \(\frac{77}{100}\) cm²
SA = 47 cm² + \(\frac{50}{100}\) cm² + \(\frac{77}{100}\) cm²
SA = 47 cm² + \(\frac{127}{100}\) cm²
SA = 47 cm² + 1 cm² + \(\frac{27}{100}\) cm²
SA = 48 \(\frac{27}{100}\) cm²

c.

Answer:
SA = 3A_{rt triangle} + A_{equ triangle}
SA = 3(4 \(\frac{1}{2}\) ) cm² + 7 \(\frac{77}{100}\) cm²
SA = 12 cm² + \(\frac{3}{2}\) cm² + 7 cm² + \(\frac{77}{100}\) cm²
SA = 20 cm² + \(\frac{1}{2}\) cm² + \(\frac{77}{100}\) cm²
SA = 20 cm² + 1 cm² + \(\frac{27}{100}\) cm²
SA = 21 \(\frac{27}{100}\) cm²

d. How are figures (b) and (c) related to figure (a)?
Answer:
If the equilateral triangular faces of figures (b) and (c) were matched together, they would form the cube in part (a).

Question 6.
Find the surface area of the solid shown in the diagram. The solid is a right triangular prism (with right triangular bases) with a smaller right triangular prism removed from it.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (4 in. + 4 in. + 5 \(\frac{13}{20}\) in.) ∙ 2 in.
LA = (13 \(\frac{13}{20}\) in.) ∙ 2 in.
LA = 26 in² + \(\frac{13}{10}\) in²
LA = 26 in² + 1 in² + \(\frac{3}{10}\) in²
LA = 27 \(\frac{3}{10}\) in²
The \(\frac{1}{4}\)in. by 4 \(\frac{19}{20}\) in. rectangle has to be taken away from the lateral area:
A = lw
A = 4 \(\frac{19}{20}\) in ∙ \(\frac{1}{4}\) in
A = 1 in² + 19/80 in²
A = 1 19/80 in²

LA = 27 \(\frac{3}{10}\) in²-1 \(\frac{19}{80}\) in²
LA = 27 \(\frac{24}{80}\) in²-1 \(\frac{19}{80}\) in²
LA = 26 \(\frac{5}{80}\) in²
LA = 26 \(\frac{1}{16}\) in²

Two bases of the larger and smaller triangular prisms must be added:
SA = 26 \(\frac{1}{16}\) in² + 2(3 \(\frac{1}{2}\) in ∙ \(\frac{1}{4}\) in) + 2(\(\frac{1}{2}\) ∙ 4 in ∙ 4 in)
SA = 26 \(\frac{1}{16}\) in² + 2 ∙ \(\frac{1}{4}\) in ∙ 3 \(\frac{1}{2}\) in + 16 in²
SA = 26 \(\frac{1}{16}\) in² + \(\frac{1}{2}\) in ∙ 3 \(\frac{1}{2}\) in + 16 in²
SA = 26 \(\frac{1}{16}\) in² + (\(\frac{3}{2}\) in² + \(\frac{1}{4}\) in² ) + 16 in²
SA = 26 \(\frac{1}{16}\) in² + 1 in² + \(\frac{8}{16}\) in² + \(\frac{4}{16}\) in² + 16 in²
SA = 43 \(\frac{13}{16}\) in²
The surface area of the solid is 43 \(\frac{13}{16}\) in².

Question 7.
The diagram shows a cubic meter that has had three square holes punched completely through the cube on three perpendicular axes. Find the surface area of the remaining solid.

Answer:
Exterior surfaces of the cube (SA₁):
SA₁ = 6(1 m)²-6(\(\frac{1}{2}\) m)²
SA₁ = 6(1 m² )-6(\(\frac{1}{4}\) m²)
SA₁ = 6 m²–\(\frac{6}{4}\) m²
SA₁ = 6 m²-(1 \(\frac{1}{2}\) m²)
SA₁ = 4 \(\frac{1}{2}\) m²
Just inside each square hole are four intermediate surfaces that can be treated as the lateral area of a rectangular prism. Each has a height of \(\frac{1}{4}\) m and perimeter of \(\frac{1}{2}\) m + \(\frac{1}{2}\) m + \(\frac{1}{2}\) m + \(\frac{1}{2}\) m or 2 m.
SA₂ = 6(LA)
SA₂ = 6(2 m ∙ \(\frac{1}{4}\) m)
SA₂ = 6 ∙ \(\frac{1}{2}\) m²
SA₂ = 3 m²

The total surface area of the remaining solid is the sum of these two areas:
SA_T = SA₁ + SA₂.
SA_T = 4 \(\frac{1}{2}\) m² + 3 m²
SA_T = 7 \(\frac{1}{2}\) m²
The surface area of the remaining solid is 7 \(\frac{1}{2}\) m².

Eureka Math Grade 7 Module 3 Lesson 22 Exit Ticket Answer Key

Question 1.
The right hexagonal pyramid has a hexagon base with equal-length sides. The lateral faces of the pyramid are all triangles (that are exact copies of one another) with heights of 15 ft. Find the surface area of the pyramid.

Answer:
SA = LA + 1B
LA = 6 ∙ \(\frac{1}{2}\) (bh)
LA = 6 ∙ \(\frac{1}{2}\) (5 ft. ∙ 15 ft.)
LA = 3 ∙ 75 ft²
LA = 225 ft²

B = A_rectangle + 2A_triangle
B = (8 ft. ∙ 5 ft.) + 2 ∙ \(\frac{1}{2}\) (8 ft. ∙ 3 ft.)
B = 40 ft² + (8 ft. ∙ 3 ft.)
B = 40 ft² + 24 ft²

B = 64 ft²
SA = LA + 1B
SA = 225 ft² + 64 ft²
SA = 289 ft²
The surface area of the pyramid is 289 ft².

Question 2.
Six cubes are glued together to form the solid shown in the diagram. If the edges of each cube measure 1 \(\frac{1}{2}\) inches in length, what is the surface area of the solid?

Answer:
There are 26 square cube faces showing on the surface area of the solid (5 each from the top and bottom view, 4 each from the front and back view, 3 each from the left and right side views, and 2 from the “inside” of the front).
A = s²
A = (1 \(\frac{1}{2}\) in.)²
A = (1 \(\frac{1}{2}\) in.)(1 \(\frac{1}{2}\) in.)
A = 1 \(\frac{1}{2}\) in.(1 in. + \(\frac{1}{2}\) in.)
A = (1 \(\frac{1}{2}\) in. ∙ 1 in.) + (1 \(\frac{1}{2}\) in. ∙ \(\frac{1}{2}\) in.)
A = 1 \(\frac{1}{2}\) in² + \(\frac{3}{4}\) in²
A = 1 \(\frac{2}{4}\) in² + \(\frac{3}{4}\) in²
A = 1 \(\frac{5}{4}\) in² = 2 \(\frac{1}{4}\) in²

SA = 26 ∙ (2 \(\frac{1}{4}\) in²)
SA = 52 in² + \(\frac{26}{4}\) in²
SA = 52 in² + 6 in² + \(\frac{1}{2}\) in²
SA = 58 \(\frac{1}{2}\) in²
The surface area of the solid is 58 \(\frac{1}{2}\) in².

Engage NY Eureka Math 7th Grade Module 3 Lesson 21 Answer Key

Eureka Math Grade 7 Module 3 Lesson 21 Example Answer Key

Example 1: Lateral Area of a Right Prism
A right triangular prism, a right rectangular prism, and a right pentagonal prism are pictured below, and all have equal heights of h.

a. Write an expression that represents the lateral area of the right triangular prism as the sum of the areas of its lateral faces.
Answer:
a ∙ h + b ∙ h + c ∙ h

b. Write an expression that represents the lateral area of the right rectangular prism as the sum of the areas of its lateral faces.
Answer:
a ∙ h + b ∙ h + a ∙ h + b ∙ h

c. Write an expression that represents the lateral area of the right pentagonal prism as the sum of the areas of its lateral faces.
Answer:
a ∙ h + b ∙ h + c ∙ h + d ∙ h + e ∙ h

d. What value appears often in each expression and why?
Answer:
h; Each prism has a height of h; therefore, each lateral face has a height of h.

e. Rewrite each expression in factored form using the distributive property and the height of each lateral face.
Answer:
h(a + b + c)
h(a + b + a + b)
h(a + b + c + d + e)

f. What do the parentheses in each case represent with respect to the right prisms?
Answer:

The perimeter of the base of the corresponding prism.

g. How can we generalize the lateral area of a right prism into a formula that applies to all right prisms?
Answer:
If LA represents the lateral area of a right prism, P represents the perimeter of the right prism’s base, and h represents the distance between the right prism’s bases, then:
LA = P_base ∙ h.

Relevant Vocabulary
RIGHT PRISM: Let E and E’ be two parallel planes. Let B be a triangular or rectangular region or a region that is the union of such regions in the plane E. At each point P of B, consider the segment PP’ perpendicular to E, joining P to a point P’ of the plane E’. The union of all these segments is a solid called a right prism.

There is a region B’ in E’ that is an exact copy of the region B. The regions B and B’ are called the base faces (or just bases) of the prism. The rectangular regions between two corresponding sides of the bases are called lateral faces of the prism. In all, the boundary of a right rectangular prism has 6 faces: 2 base faces and 4 lateral faces. All adjacent faces intersect along segments called edges (base edges and lateral edges).

CUBE: A cube is a right rectangular prism all of whose edges are of equal length.
SURFACE: The surface of a prism is the union of all of its faces (the base faces and lateral faces).
NET: A net is a two-dimensional diagram of the surface of a prism.

1. Why are the lateral faces of right prisms always rectangular regions?
Answer:
Because along a base edge, the line segments PP’ are always perpendicular to the edge, forming a rectangular region.

2. What is the name of the right prism whose bases are rectangles?
Answer:
Right rectangular prism

3. How does this definition of right prism include the interior of the prism?
Answer:
The union of all the line segments fills out the interior.

Eureka Math Grade 7 Module 3 Lesson 21 Exercise Answer Key

Opening Exercise: Surface Area of a Right Rectangular Prism
On the provided grid, draw a net representing the surfaces of the right rectangular prism (assume each grid line represents 1 inch). Then, find the surface area of the prism by finding the area of the net.

Answer:
There are six rectangular faces that make up the net.
The four rectangles in the center form one long rectangle that is 20 in. by 3 in.
Area = lw
Area = 3 in ∙ 20 in
Area = 60 in²

Two rectangles form the wings, both 6 in by 4 in.
Area = lw
Area = 6 in ∙ 4 in
Area = 24 in²
The area of both wings is 2(24 in² ) = 48 in².

The total area of the net is
A = 60 in² + 48 in² = 108 in²

The net represents all the surfaces of the rectangular prism, so its area is equal to the surface area of the prism. The surface area of
the right rectangular prism is 108 in².

Exercise 1.
Marcus thinks that the surface area of the right triangular prism will be half that of the right rectangular prism and wants to use the modified formula SA = \(\frac{1}{2}\) (2lw + 2lh + 2wh). Do you agree or disagree with Marcus? Use nets of the prisms to support your argument.

Answer:
The surface area of the right rectangular prism is 108 in², so Marcus believes the surface areas of each right triangular prism is 54 in².

The net of the right triangular prism has one less face than the right rectangular prism. Two of the rectangular faces on the right triangular prism (rectangular regions 1 and 2 in the diagram) are the same faces from the right rectangular prism, so they are the same size. The areas of the triangular bases (triangular regions 3 and 4 in the diagram) are half the area of their corresponding rectangular faces of the right rectangular prism. These four faces of the right triangular prism make up half the surface area of the right rectangular prism before considering the fifth face; no, Marcus is incorrect.

The areas of rectangular faces 1 and 2, plus the areas of the triangular regions 3 and 4 is 54 in². The last rectangular region has an area of 30 in². The total area of the net is 54 in² + 30 in² or 84 in², which is far more than half the surface area of the right rectangular prism.

Eureka Math Grade 7 Module 3 Lesson 21 Problem Set Answer Key

Question 1.
For each of the following nets, highlight the perimeter of the lateral area, draw the solid represented by the net, indicate the type of solid, and then find the solid’s surface area.
a. Right rectangular prism

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (2 \(\frac{1}{2}\) cm + 7 \(\frac{1}{2}\) cm + 2 \(\frac{1}{2}\) cm + 7 \(\frac{1}{2}\) cm) ∙ 5 cm
LA = 20 cm ∙ 5 cm
LA = 100 cm²

B = lw
B = 2 \(\frac{1}{2}\) cm ∙ 7 \(\frac{1}{2}\) cm
B = \(\frac{5}{2}\) cm ∙ \(\frac{15}{2}\) cm
B = \(\frac{75}{4}\) cm²

SA = 100 cm² + 2(\(\frac{75}{4}\) cm² )
SA = 100 cm² + 37.5 cm²
SA = 137.5 cm²
The surface area of the right rectangular prism is 137.5 cm²

b. Right triangular prism

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (10 in. + 8 in. + 10 in.) ∙ 12 in.
LA = 28 in. ∙ 12 in.
LA = 336 in²

B = \(\frac{1}{2}\) bh
B = \(\frac{1}{2}\) (8 in.)(9 \(\frac{1}{5}\) in.)
B = 4 in.(9 \(\frac{1}{5}\) in.)
B = (36 + \(\frac{4}{5}\) )in²
B = 36 \(\frac{4}{5}\) in²

SA = 336 in² + 2(36 \(\frac{4}{5}\) in²)
SA = 336 in² + (72 + \(\frac{8}{5}\) )in²
SA = 408 in² + 1 \(\frac{3}{5}\) in²
SA = 409 \(\frac{3}{5}\) in²
The surface area of the right triangular prism is 409 3/5 in²2.

Question 2.
Given a cube with edges that are \(\frac{3}{4}\) inch long:
a. Find the surface area of the cube.
Answer:
SA = 6s²2
SA = 6(\(\frac{3}{4}\) in.)²
SA = 6 (\(\frac{3}{4}\) in.) ∙ (\(\frac{3}{4}\) in.)
SA = 6(\(\frac{9}{16}\) in²)
SA = \(\frac{27}{8}\) in² or 3 \(\frac{3}{8}\) in²

b. Joshua makes a scale drawing of the cube using a scale factor of 4. Find the surface area of the cube that Joshua drew.
Answer:
\(\frac{3}{4}\) in. ∙ 4 = 3 in.; The edge lengths of Joshua’s drawing would be 3 inches.
SA = 6(3 in.)²
SA = 6(9 in²)
SA = 54 in²

c. What is the ratio of the surface area of the scale drawing to the surface area of the actual cube, and how does the value of the ratio compare to the scale factor?
Answer:
54 ÷ 3 \(\frac{3}{8}\)
54 ÷ \(\frac{27}{8}\)
54 ∙ \(\frac{8}{27}\)
2 ∙ 8 = 16. The ratios of the surface area of the scale drawing to the surface area of the actual cube is 16:1. The value of the ratio is 16. The scale factor of the drawing is 4, and the value of the ratio of the surface area of the drawing to the surface area of the actual cube is 4² or 16.

Question 3.
Find the surface area of each of the following right prisms using the formula SA = LA + 2B.
a.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (12 \(\frac{1}{2}\) mm + 10 mm + 7 \(\frac{1}{2}\) mm) ∙ 15 mm
LA = 30 mm ∙ 15 mm
LA = 450 mm²

B = \(\frac{1}{2}\) bh
B = \(\frac{1}{2}\) ∙ (7 \(\frac{1}{2}\) mm) ∙ (10 mm)
B = \(\frac{1}{2}\) ∙ (70 + 5) mm²
B = \(\frac{1}{2}\) ∙ 75 mm²
B = 75/2 mm²

SA = 450 mm² + 2(\(\frac{75}{2}\) mm²)
SA = 450 mm² + 75 mm²
SA = 525 mm²

The surface area of the prism is 525 mm².

b.

SA = LA + 2B
LA = P ∙ h
LA = (9 \(\frac{3}{25}\) in. + 6 \(\frac{1}{2}\) in. + 4 in.) ∙ 5 in in. ∙ 2 \(\frac{1}{2}\) in.
LA = (\(\frac{228}{25}\) in. + \(\frac{13}{2}\) in. + 4 in.) ∙ 5 in
LA = (\(\frac{456}{50}\) in. + \(\frac{325}{50}\) in. + \(\frac{200}{50}\) in.) ∙ 5 in.
LA = (\(\frac{981}{50}\) in.) ∙ 5 in.
LA = \(\frac{49,050}{50}\) in²
LA = 98 \(\frac{1}{10}\) in²

B = \(\frac{1}{2}\) bh
B = \(\frac{1}{2}\) ∙ 9 \(\frac{3}{25}\)
B = \(\frac{1}{2}\) ∙ \(\frac{228}{25}\) in. ∙ \(\frac{5}{2}\) in.
B = \(\frac{1,140}{100}\) in²
B = 11 \(\frac{2}{5}\) in²
2B = 2 ∙ 11 \(\frac{2}{5}\) in²
2B = 22 \(\frac{4}{5}\) in²

SA = LA + 2B
SA = 98 \(\frac{1}{10}\) in² + 22 \(\frac{4}{5}\) in²
SA = 120 \(\frac{9}{10}\) in²
The surface area of the prism is 120 \(\frac{9}{10}\) in².

c.

SA = LA + 2B
LA = P ∙ h
LA = (\(\frac{1}{8}\) in. + \(\frac{1}{2}\) in. + \(\frac{1}{8}\) in. + \(\frac{1}{4}\) in. + \(\frac{1}{2}\) in. + \(\frac{1}{4}\) in.) ∙ 2 in.
LA = (1 \(\frac{3}{4}\) in.) ∙ 2 in.
LA = 2 in² + 1 \(\frac{1}{2}\) in²
LA = 3 \(\frac{1}{2}\) in²

B = A_rectangle + 2A_triangle
B = (\(\frac{1}{2}\) in. ∙ \(\frac{1}{5}\) in.) + 2 ∙ \(\frac{1}{2}\) (\(\frac{1}{8}\) in. ∙ \(\frac{1}{5}\) in.)
B = (\(\frac{1}{10}\) in² ) + (\(\frac{1}{40}\) in² )
B = \(\frac{1}{10}\) in² + \(\frac{1}{40}\)
B = \(\frac{4}{40}\) in² + \(\frac{1}{40}\) in²
B = \(\frac{5}{40}\) in²
B = \(\frac{1}{8}\) in²

SA = 3 \(\frac{1}{2}\) in² + 2(\(\frac{1}{8}\) in²) in²
SA = 3 \(\frac{1}{2}\) in² + \(\frac{1}{4}\) in²
SA = 3 \(\frac{2}{4}\) in² + \(\frac{1}{4}\) in²
SA = 3 \(\frac{3}{4}\) in²
The surface area of the prism is 3 \(\frac{3}{4}\) in².

d.

SA = LA + 2B
LA = P ∙ h
LA = (13 cm + 13 cm + 8.6 cm + 8.6 cm) ∙ 2 \(\frac{1}{4}\) cm
LA = (26 cm + 17.2 cm) ∙ 2 \(\frac{1}{4}\) cm
LA = (43.2)cm ∙ 2 \(\frac{1}{4}\) cm
LA = (86.4 cm² + 10.8 cm²)
LA = 97.2 cm²

SA = LA + 2B
SA = 97.2 cm² + 2(95 cm²)
SA = 97.2 cm² + 190 cm²
SA = 287.2 cm²

B = \(\frac{1}{2}\) (10 cm ∙ 7 cm ) + \(\frac{1}{2}\) (12 cm ∙ 10 cm)
B = \(\frac{1}{2}\) (190 cm²)
B = \(\frac{1}{2}\) (70 cm² + 120 cm²)
B = 95 cm²
The surface area of the prism is 287.2 cm².

Question 4.
A cube has a volume of 64 m³. What is the cube’s surface area?
Answer:
A cube’s length, width, and height must be equal. 64 = 4 ∙ 4 ∙ 4 = 4³, so the length, width, and height of the cube are all 4 m.
SA = 6s²
SA = 6(4 m)²
SA = 6(16 m²)
SA = 96 m²

Question 5.
The height of a right rectangular prism is 4 \(\frac{1}{2}\) ft. The length and width of the prism’s base are 2 ft. and 1 \(\frac{1}{2}\) ft. Use the formula SA = LA + 2B to find the surface area of the right rectangular prism.
Answer:
SA = LA + 2B
LA = P ∙ h
LA = (2 ft. + 2 ft. + 1 \(\frac{1}{2}\) ft. + 1 \(\frac{1}{2}\) ft.) ∙ 4 \(\frac{1}{2}\) ft.
LA = (2 ft + 2 ft. + 3 ft.) ∙ 4 \(\frac{1}{2}\) ft.
LA = 7 ft. ∙ 4 \(\frac{1}{2}\) ft.
LA = 28 ft² + 3 \(\frac{1}{2}\) ft²
LA = 31 \(\frac{1}{2}\) ft²

B = lw
B = 2 ft. ∙ 1 \(\frac{1}{2}\) ft.
B = 3 ft²

SA = LA + 2b
SA = 31 \(\frac{1}{2}\) ft² + 2(3 ft²)
SA = 31 \(\frac{1}{2}\) ft² + 6 ft²
SA = 37 \(\frac{1}{2}\) ft²
The surface area of the right rectangular prism is 37 \(\frac{1}{2}\) ft².

Question 6.
The surface area of a right rectangular prism is 68 \(\frac{2}{3}\) in². The dimensions of its base are 3 in and 7 in Use the formula SA = LA + 2B and LA = Ph to find the unknown height h of the prism.
Answer:
SA = LA + 2B
SA = P ∙ h + 2B
68 \(\frac{2}{3}\) in² = 20 in. ∙ (h) + 2(21 in²)
68 \(\frac{2}{3}\) in² = 20 in. ∙ (h) + 42 in²
68 \(\frac{2}{3}\) in² – 42 in² = 20 in. ∙ (h) + 42 in² – 42 in²
26 \(\frac{2}{3}\) in² = 20 in. ∙ (h) + 0 in²
26 \(\frac{2}{3}\) in² ∙ \(\frac{1}{20 \mathrm{in.}}\) = 20 in ∙ \(\frac{1}{20 \mathrm{in.}}\) ∙ (h)
\(\frac{80}{3}\) in² ∙ \(\frac{1}{20 \mathrm{in.}}\) = 1 ∙ h
\(\frac{4}{3}\) in. = h
h = \(\frac{4}{3}\) in. or 1 \(\frac{1}{3}\) in.
The height of the prism is 1 \(\frac{1}{3}\) in.

Question 7.
A given right triangular prism has an equilateral triangular base. The height of that equilateral triangle is approximately 7.1 cm. The distance between the bases is 9 cm. The surface area of the prism is 319 \(\frac{1}{2}\) cm². Find the approximate lengths of the sides of the base.
Answer:
SA = LA + 2B
LA = P ∙ h
LA = 3(x cm) ∙ 9 cm
LA = 27x cm²

Let x represent the number of centimeters in each side of the equilateral triangle.
B = \(\frac{1}{2}\) lw
B = \(\frac{1}{2}\) ∙ (x cm) ∙ 7.1 cm
B = 3.55x cm²

319 \(\frac{1}{2}\) cm² = LA + 2B
319 \(\frac{1}{2}\) cm² = 27x cm² + 2(3.55x cm²)
319 \(\frac{1}{2}\) cm² = 27x cm² + 7.1x cm²
319 \(\frac{1}{2}\) cm² = 34.1x cm²
319 \(\frac{1}{2}\) cm² = 34 \(\frac{1}{10}\) x cm²
\(\frac{639}{2}\) cm² = \(\frac{341}{10}\) x cm²
\(\frac{639}{2}\) cm² ∙ \(\frac{10}{341 \mathrm{~cm}}\) = \(\frac{341}{10}\) x cm² ∙ \(\frac{10}{341 \mathrm{~cm}}\)
\(\frac{3195}{341}\) cm = x
x = \(\frac{3195}{341}\) cm
x ≈ 9.4 cm
The lengths of the sides of the equilateral triangles are approximately 9.4 cm each.

Eureka Math Grade 7 Module 3 Lesson 21 Exit Ticket Answer Key

Question 1.
Find the surface area of the right trapezoidal prism. Show all necessary work.

Answer:
SA = LA + 2B
LA = P ∙ h
LA = (3 cm + 7 cm + 5 + 11 cm ) ∙ 6 cm
LA = 26 cm ∙ 6 cm
LA = 156 cm²
Each base consists of a 3 cm by 7 cm rectangle and right triangle with a base of 3 cm and a height of 4 cm. Therefore, the area of each base:
B = A_r + A_t
B = lw + \(\frac{1}{2}\) bh
B = (7 cm ∙ 3 cm) + (\(\frac{1}{2}\) ∙ 3 cm ∙ 4 cm)
B = 21 cm² + 6 cm²
B = 27 cm²

SA = LA + 2B
SA = 156 cm² + 2(27 cm² )
SA = 156 cm² + 54 cm²
SA = 210 cm²
The surface of the right trapezoidal prism is 210 cm².

Engage NY Eureka Math 7th Grade Module 3 Lesson 23 Answer Key

Eureka Math Grade 7 Module 3 Lesson 23 Exploratory Challenge Answer Key

Exploratory Challenge: The Volume of a Right Prism
What is the volume of the right prism pictured on the right? Explain.

Answer:

The volume of the right prism is 36 units³ because the prism is filled with 36 cubes that are 1 unit long, 1 unit wide, and 1 unit high, or 1 unit³.

Draw the same diagonal on the square base as done above; then, darken the grid lines on the lower right triangular prism. What is the volume of that right triangular prism? Explain.
Answer:
The volume of the right triangular prism is 18 units³. There are 15 cubes from the original right prism and 6 right triangular prisms that are each half of a cube. The 6 right triangular prisms can be paired together to form 3 cubes, or 3 units³. Altogether the area of the right triangular prism is (15+3) units³, or 18 units³.

How could we create a right triangular prism with five times the volume of the right triangular prism pictured to the right, without changing the base? Draw your solution on the diagram, give the volume of the solid, and explain why your solution has five times the volume of the triangular prism.

Answer:

If we stack five exact copies of the base (or bottom floor), the prism then has five times the number of unit cubes as the original, which means it has five times the volume, or 90 units³.

What could we do to cut the volume of the right triangular prism pictured on the right in half without changing the base? Draw your solution on the diagram, give the volume of the solid, and explain why your solution has half the volume of the given triangular prism.

Answer:

If we slice the height of the prism in half, each of the unit cubes that make up the triangular prism will have half the volume as in the original right triangular prism. The volume of the new right triangular prism is 9 units³.

To find the volume (V) of any right prism …
Answer:
Multiply the area of the right prism’s base (B) times the height of the right prism (h), V = Bh.

Eureka Math Grade 7 Module 3 Lesson 23 Example Answer Key

Example: The Volume of a Right Triangular Prism
Find the volume of the right triangular prism shown in the diagram using V = Bh.

Answer:
V = Bh
V = (\(\frac{1}{2}\) lw)h
V = (\(\frac{1}{2}\) ∙ 4 m ∙ \(\frac{1}{2}\) m) ∙ 6 \(\frac{1}{2}\) m
V = (2 m ∙ \(\frac{1}{2}\) m) ∙ 6 \(\frac{1}{2}\) m
V = 1 m² ∙ 6 \(\frac{1}{2}\) m
V = 6 \(\frac{1}{2}\) m³ The volume of the triangular prism is 6 \(\frac{1}{2}\) m³.

Eureka Math Grade 7 Module 3 Lesson 23 Exercise Answer Key

Opening Exercise
The volume of a solid is a quantity given by the number of unit cubes needed to fill the solid. Most solids—rocks, baseballs, people—cannot be filled with unit cubes or assembled from cubes. Yet such solids still have volume. Fortunately, we do not need to assemble solids from unit cubes in order to calculate their volume. One of the first interesting examples of a solid that cannot be assembled from cubes, but whose volume can still be calculated from a formula, is a right triangular prism.

What is the area of the square pictured on the right? Explain.

Answer:

The area of the square is 36 units² because the region is filled with 36 square regions that are 1 unit by 1 unit, or 1 unit².

Draw the diagonal joining the two given points; then, darken the grid lines within the lower triangular region. What is the area of that triangular region? Explain.
Answer:
The area of the triangular region is 18 units². There are 15 unit squares from the original square and 6 triangular regions that are \(\frac{1}{2}\) unit². The 6 triangles can be paired together to form 3 units². Altogether the area of the triangular region is (15+3) units², or 18 units².

Exercise: Multiple Volume Representations
The right pentagonal prism is composed of a right rectangular prism joined with a right triangular prism. Find the volume of the right pentagonal prism shown in the diagram using two different strategies.

Answer:
Strategy #1
The volume of the pentagonal prism is equal to the sum of the volumes of the rectangular and triangular prisms.
V = V_{rectangular prism} + V_{triangular prism}
V = Bh
V = (lw)h
V = (4 m ∙ 6 \(\frac{1}{2}\) m) ∙ 6 \(\frac{1}{2}\) m
V = (24 m²+2 m²) ∙ 6 \(\frac{1}{2}\) m
V = 26 m² ∙ 6 \(\frac{1}{2}\) m
V = 156 m³+13 m³
V = 169 m³

V = Bh
V = (\(\frac{1}{2}\) lw)h
V = (\(\frac{1}{2}\) ∙ 4 m ∙ \(\frac{1}{2}\) m) ∙ 6 \(\frac{1}{2}\) m
V = (2 m ∙ \(\frac{1}{2}\) m) ∙ 6 \(\frac{1}{2}\) m
V = (1 m² ) ∙ 6 \(\frac{1}{2}\) m
V = 6 \(\frac{1}{2}\) m³
So the total volume of the pentagonal prism is 169 m³+6 \(\frac{1}{2}\) m³, or 175 \(\frac{1}{2}\) m³ .

Strategy #2
The volume of a right prism is equal to the area of its base times its height. The base is a rectangle and a triangle.
V = Bh
B = A_rectangle + A_triangle
A_rectangle = 4 m ∙ 6 \(\frac{1}{2}\) m
A_rectangle = 24 m²+2 m²
A_rectangle = 26 m²

A_triangle = \(\frac{1}{2}\) ∙ 4 m ∙ \(\frac{1}{2}\) m
A_triangle = 2 m ∙ \(\frac{1}{2}\) m
A_triangle = 1 m²

V = Bh
V = 27 m² ∙ 6 \(\frac{1}{2}\) m
V = 162 m³+13 \(\frac{1}{2}\) m³
V = 175 \(\frac{1}{2}\) m³

B = 26 m²+1 m²
B = 27 m²
The volume of the right pentagonal prism is 175 \(\frac{1}{2}\) m³.

Eureka Math Grade 7 Module 3 Lesson 23 Problem Set Answer Key

Question 1.
Calculate the volume of each solid using the formula V = Bh (all angles are 90 degrees).
a.

Answer:
V = Bh
V = (8 cm ∙ 7 cm) ∙ 12 \(\frac{1}{2}\) cm
V = (56 ∙ 12 \(\frac{1}{2}\)) cm³
V = 672 cm³+28 cm³
V = 700 cm³
The volume of the solid is 700 cm³.

b.

Answer:
V = Bh
V = (\(\frac{3}{4}\) in. ∙ \(\frac{3}{4}\) in.) ∙ \(\frac{3}{4}\) in.
V = (9/16) ∙ \(\frac{3}{4}\) in³
V = \(\frac{27}{64}\) in³
The volume of the cube is \(\frac{27}{64}\) in³.

c.

Answer:
V = Bh
B = A_rectangle+A_square
B = lw+s²
B = (2 \(\frac{1}{2}\) in. ∙ 4 \(\frac{1}{2}\) in.)+(1 \(\frac{1}{2}\) in.)²
B = (10 in²+1 \(\frac{1}{4}\) in² )+(1 \(\frac{1}{2}\) in. ∙ 1 \(\frac{1}{2}\) in.)
B = 11 \(\frac{1}{4}\) in²+(1 \(\frac{1}{2}\) in²+\(\frac{3}{4}\) in² )
B = 11 \(\frac{1}{4}\) in²+\(\frac{3}{4}\) in²+1 \(\frac{1}{2}\) in²
B = 12 in²+1 \(\frac{1}{2}\) in²
B = 13 \(\frac{1}{2}\) in²

V = Bh
V = 13 \(\frac{1}{2}\) in² ∙ \(\frac{1}{2}\) in.
V = \(\frac{13}{2}\) in³+\(\frac{1}{4}\) in³
V = 6 in³+\(\frac{1}{2}\) in³+\(\frac{1}{4}\) in³
V = 6 \(\frac{3}{4}\) in³
The volume of the solid is 6 \(\frac{3}{4}\) in³.

d.

Answer:
V = Bh
B = (A_{(lg rectangle})-(A_{sm rectangle})
B = (lw)₁– (lw)₂
B = (6 yd. ∙ 4 yd.)-(1 \(\frac{1}{3}\) yd. ∙ 2 yd.)
B = 24 yd²-(2 yd²+\(\frac{2}{3}\) yd² )
B = 24 yd²-2 yd²–\(\frac{2}{3}\) yd²
B = 22 yd²–\(\frac{2}{3}\) yd²
B = 21 \(\frac{1}{3}\) yd²

V = Bh
V = (21 \(\frac{1}{3}\) yd²) ∙ \(\frac{2}{3}\) yd.
V = 14 yd³+(\(\frac{1}{3}\) yd² ∙ \(\frac{2}{3}\) yd.)
V = 14 yd³+\(\frac{2}{9}\) yd³
V = 14 \(\frac{2}{9}\) yd³
The volume of the solid is 14 \(\frac{2}{9}\) yd³.

e.

Answer:
V = Bh_prism
B = \(\frac{1}{2}\) bh_triangle
B = \(\frac{1}{2}\) ∙ 4 cm ∙ 4 cm
B = 2 ∙ 4 cm²
B = 8 cm²

V = Bh
V = 8 cm² ∙ 6 \(\frac{7}{10}\) cm
V = 48 cm³+\(\frac{56}{10}\) cm³
V = 48 cm³+5 cm³+6/10 cm³
V = 53 cm³+\(\frac{3}{5}\) cm³
V = 53 \(\frac{3}{5}\) cm³
The volume of the solid is 53 \(\frac{3}{5}\) cm³.

f.

Answer:
V = Bh_prism
B = \(\frac{1}{2}\) bh_triangle
B = \(\frac{1}{2}\) ∙ 9 \(\frac{3}{25}\) in. ∙ 2 \(\frac{1}{2}\) in.
B = \(\frac{1}{2}\) ∙ 2 \(\frac{1}{2}\) in. ∙ 9 3/25 in.
B = (1 \(\frac{1}{4}\) ) ∙ (9 \(\frac{3}{25}\) )in²
B = (\(\frac{5}{4}\) ∙ \(\frac{228}{25}\) )in²
B = \(\frac{57}{5}\) in²

V = Bh
V = (\(\frac{57}{5}\) in² ) ∙ 5 in
V = 57 in³
The volume of the solid is 57 in³.

g.

Answer:
V = Bh
B = A_rectangle + A_triangle
B = lw+\(\frac{1}{2}\) bh
B = (5 \(\frac{1}{4}\) cm ∙ 4 cm)+\(\frac{1}{2}\) (4 cm ∙ 1 \(\frac{1}{4}\) cm)
B = (20 cm²+1 cm² )+(2 cm ∙ 1 \(\frac{1}{4}\) cm)
B = 21 cm²+2 cm²+\(\frac{1}{2}\) cm²
B = 23 cm²+\(\frac{1}{2}\) cm²
B = 23 \(\frac{1}{2}\) cm²

V = Bh
V = 23 \(\frac{1}{2}\) cm² ∙ 9 cm
V = 207 cm³+\(\frac{9}{2}\) cm³
V = 207 cm³+4 cm³+\(\frac{1}{2}\) cm³
V = 211 \(\frac{1}{2}\) cm³
The volume of the solid is 211 \(\frac{1}{2}\) cm³.

h.

Answer:
V = Bh
V = \(\frac{1}{8}\) in² ∙ 2 in.
V = \(\frac{1}{4}\) in³

B = A_rectangle+2A_triangle
B = lw+2 ∙ \(\frac{1}{2}\) bh
B = (\(\frac{1}{2}\) in. ∙ \(\frac{1}{5}\) in.)+(1 ∙ \(\frac{1}{8}\) in. ∙ \(\frac{1}{5}\) in.)
B = \(\frac{1}{10}\) in²+\(\frac{1}{40}\) in²
B = \(\frac{4}{40}\) in²+\(\frac{1}{40}\) in²
The volume of the solid is \(\frac{1}{4}\) in³.
B = \(\frac{5}{40}\) in²
B = \(\frac{1}{8}\) in²

Question 2.
Let l represent the length, w the width, and h the height of a right rectangular prism. Find the volume of the prism when
a. l = 3 cm, w = 2 \(\frac{1}{2}\) cm, and h = 7 cm.
Answer:
V = lwh
V = 3 cm ∙ 2 \(\frac{1}{2}\) cm ∙ 7 cm
V = 21 ∙ (2 \(\frac{1}{2}\) ) cm³
V = 52 \(\frac{1}{2}\) cm³ The volume of the prism is 52 \(\frac{1}{2}\) cm³.

l = \(\frac{1}{4}\) cm, w = 4 cm, and h = 1 \(\frac{1}{2}\) cm.
V = lwh
V = \(\frac{1}{4}\) cm ∙ 4 cm ∙ 1 \(\frac{1}{2}\) cm
V = 1 \(\frac{1}{2}\) cm³ The volume of the prism is 1 \(\frac{1}{2}\) cm³.

Question 3.
Find the length of the edge indicated in each diagram.
a. V = Bh
Let h represent the number of inches in the height of the prism.

93 \(\frac{1}{2}\) in³ = 22 in² ∙ h
93 \(\frac{1}{2}\) in³ = 22h in²
22h = 93.5 in
h = 4.25 in
The height of the right rectangular prism is 4 \(\frac{1}{4}\) in.

What are possible dimensions of the base?
11 in by 2 in, or 22 in by 1 in

b. V = Bh
Let h represent the number of meters in the height of the triangular base of the prism.

V = (\(\frac{1}{2}\) bh_triangle) ∙ h_prism
4 \(\frac{1}{2}\) m³ = (\(\frac{1}{2}\) ∙ 3 m ∙ h) ∙ 6 m
4 \(\frac{1}{2}\) m³ = \(\frac{1}{2}\) ∙ 18 m² ∙ h
4 \(\frac{1}{2}\) m³ = 9h m²
9h = 4.5 m
h = 0.5 m
The height of the triangle is \(\frac{1}{2}\) m.

Question 4.
The volume of a cube is 3 \(\frac{3}{8}\) in³. Find the length of each edge of the cube.
Answer:
V = s³, and since the volume is a fraction, the edge length must also be fractional.
3 \(\frac{3}{8}\) in³ = \(\frac{27}{8}\) in³
3 \(\frac{3}{8}\) in³ = \(\frac{3}{2}\) in. ∙ \(\frac{3}{2}\) in. ∙ \(\frac{3}{2}\) in.
3 \(\frac{3}{8}\) in³ = (\(\frac{3}{2}\) in.)³
The lengths of the edges of the cube are \(\frac{3}{2}\) in., or 1 \(\frac{1}{2}\) in.

Question 5.
Given a right rectangular prism with a volume of 7 \(\frac{1}{2}\) ft³, a length of 5 ft., and a width of 2 ft., find the height of the prism.
Answer:
V = Bh
V = (lw)h Let h represent the number of feet in the height of the prism.
7 \(\frac{1}{2}\) ft³ = (5ft. ∙ 2ft.) ∙ h
7 \(\frac{1}{2}\) ft³ = 10 ft² ∙ h
7.5 ft³ = 10h ft²
h = 0.75 ft.
The height of the right rectangular prism is \(\frac{3}{4}\) ft. (or 9 in.).

Eureka Math Grade 7 Module 3 Lesson 23 Exit Ticket Answer Key

Question 1.
The base of the right prism is a hexagon composed of a rectangle and two triangles. Find the volume of the right hexagonal prism using the formula V = Bh.

Answer:
The area of the base is the sum of the areas of the rectangle and the two triangles.
B = A_rectangle+2 ∙ A_triangle
A_rectangle = lw
A_rectangle = 2 \(\frac{1}{4}\) in. ∙ 1 \(\frac{1}{2}\) in.
A_rectangle = (\(\frac{9}{4}\) ∙ \(\frac{3}{2}\)) in²
A_rectangle = \(\frac{27}{8}\) in²

A_triangle = \(\frac{1}{2}\) lw
A_triangle = \(\frac{1}{2}\) (1 \(\frac{1}{2}\) in. ∙ \(\frac{3}{4}\) in.)
A_triangle = (\(\frac{1}{2}\) ∙ \(\frac{3}{2}\) ∙ \(\frac{3}{4}\) )in²
A_triangle = \(\frac{9}{16}\) in²

B = \(\frac{27}{8}\) in²+2(\(\frac{9}{16}\) in² )
B = \(\frac{27}{8}\) in²+\(\frac{9}{8}\) in²
B = \(\frac{36}{8}\) in²
B = \(\frac{9}{2}\) in²

V = Bh
V = (\(\frac{9}{2}\) in² ) ∙ 3 in.
V = \(\frac{27}{2}\) in³
V = 13 \(\frac{1}{2}\) in³

The volume of the hexagonal prism is 13 \(\frac{1}{2}\) in³.

Engage NY Eureka Math 7th Grade Module 3 Lesson 24 Answer Key

Eureka Math Grade 7 Module 3 Lesson 24 Exploratory Challenge Answer Key

Exploratory Challenge: Measuring a Container’s Capacity
A box in the shape of a right rectangular prism has a length of 12 in, a width of 6 in., and a height of 8 in. The base and the walls of the container are \(\frac{1}{4}\) in thick, and its top is open. What is the capacity of the right rectangular prism?
(Hint: The capacity is equal to the volume of water needed to fill the prism to the top.)

If the prism is filled with water, the water will take the shape of a right rectangular prism slightly smaller than the container. The dimensions of the smaller prism are a length of 11 \(\frac{1}{2}\) in, a width of 5 \(\frac{1}{2}\) in, and a height of 7 \(\frac{3}{4}\) in.
V = Bh
V = (lw)h
V = (11 \(\frac{1}{2}\) in ∙ 5 \(\frac{1}{2}\) in) ∙ 7 \(\frac{3}{4}\) in
V = (\(\frac{23}{2}\) in ∙ \(\frac{11}{2}\) in) ∙ \(\frac{31}{4}\) in
V = (25\(\frac{3}{4}\) in²) ∙ \(\frac{31}{4}\) in
V = \(\frac{7843}{16}\) in³
V = 490 \(\frac{3}{16}\) in³
The capacity of the right rectangular prism is 490 \(\frac{3}{16}\) in³.

Eureka Math Grade 7 Module 3 Lesson 24 Example Answer Key

Example 1: Measuring Liquid in a Container in Three Dimensions
A glass container is in the form of a right rectangular prism. The container is 10 cm long, 8 cm wide, and 30 cm high. The top of the container is open, and the base and walls of the container are 3 mm (or 0.3 cm) thick. The water in the container is 6 cm from the top of the container. What is the volume of the water in the container?

Answer:
Because of the walls and base of the container, the water in the container forms a right rectangular prism that is 9.4 cm long, 7.4 cm wide, and 23.7 cm tall.
V = Bh
V = (lw)h
V = (9.4 cm ∙ 7.4 cm) ∙ 23.7 cm
V = (\(\frac{94}{10}\) cm ∙ \(\frac{74}{10}\) cm) ∙ \(\frac{237}{10}\) cm
V = (\(\frac{6,956}{100}\) cm² ) ∙ \(\frac{237}{10}\) cm
V = \(\frac{1,648,572}{1,000}\) cm³
V = 1,648.572 cm³
The volume of the water in the container is 1,648.6 cm³.

Example 2.
7.2 L of water are poured into a container in the shape of a right rectangular prism. The inside of the container is 50 cm long, 20 cm wide, and 25 cm tall. How far from the top of the container is the surface of the water? (1 L = 1,000 cm³)

Answer:
7.2 L = 7,200 cm³
V = Bh
V = (lw)h
7,200 cm³ = (50 cm)(20 cm)h
7,200 cm³ = 1,000 cm² ∙ h
7,200 cm³ ∙ \(\frac{1}{1,000 \mathrm{~cm}^{2}}\) = 1,000 cm² ∙ \(\frac{1}{1,000 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{7,200}{1,000}\) cm = 1 ∙ h
7.2 cm = h
The depth of the water is 7.2 cm. The height of the container is 25 cm.
25 cm – 7.2 cm = 17.8 cm
The surface of the water is 17.8 cm from the top of the container.

Example 3.
A fuel tank is the shape of a right rectangular prism and has 27 L of fuel in it. It is determined that the tank is \(\frac{3}{4}\) full. The inside dimensions of the base of the tank are 90 cm by 50 cm. What is the height of the fuel in the tank? How deep is the tank? (1 L = 1,000 cm³)
Answer:
Let the height of the fuel in the tank be h cm.
27 L = 27,000 cm³
V = Bh
V = (lw)h
27,000 cm³ = (90 cm ∙ 50 cm) ∙ h
27,000 cm³ = (4,500 cm² ) ∙ h
27,000 cm³ ∙ \(\frac{1}{4,500 \mathrm{~cm}^{2}}\) = 4,500 cm² ∙ \(\frac{1}{4,500 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{27,000}{4,500}\) cm = 1 ∙ h
6 cm = h
The height of the fuel in the tank is 6 cm. The height of the fuel is \(\frac{3}{4}\) the depth of the tank. Let d represent the depth of the tank in centimeters.
6 cm = \(\frac{3}{4}\) d
6 cm ∙ \(\frac{4}{3}\) = \(\frac{3}{4}\) ∙ \(\frac{4}{3}\) ∙ d
8 cm = d
The depth of the fuel tank is 8 cm.

Eureka Math Grade 7 Module 3 Lesson 24 Problem Set Answer Key

Question 1.
Mark wants to put some fish and decorative rocks in his new glass fish tank. He measured the outside dimensions of the right rectangular prism and recorded a length of 55 cm, width of 42 cm, and height of 38 cm. He calculates that the tank will hold 87.78 L of water. Why is Mark’s calculation of volume incorrect? What is the correct volume? Mark also failed to take into account the fish and decorative rocks he plans to add. How will this affect the volume of water in the tank? Explain.
Answer:
V = Bh = (lw)h
V = 55 cm ∙ 42 cm ∙ 38 cm
V = 2,310 cm² ∙ 38 cm
V = 87,780 cm³
87,780 cm³ = 87.78 L
Mark measured only the outside dimensions of the fish tank and did not account for the thickness of the sides of the tank. If he fills the tank with 87.78 L of water, the water will overflow the sides. Mark also plans to put fish and rocks in the tank, which will force water out of the tank if it is filled to capacity.

Question 2.
Leondra bought an aquarium that is a right rectangular prism. The inside dimensions of the aquarium are 90 cm long, by 48 cm wide, by 60 cm deep. She plans to put water in the aquarium before purchasing any pet fish. How many liters of water does she need to put in the aquarium so that the water level is 5 cm below the top?
Answer:
If the aquarium is 60 cm deep, then she wants the water to be 55 cm deep. Water takes on the shape of its container, so the water will form a right rectangular prism with a length of 90 cm, a width of 48 cm, and a height of 55 cm.
V = Bh = (lw)h
V = (90 cm ∙ 48 cm) ∙ 55 cm
V = 4,320 cm² ∙ 55 cm
V = 237,600 cm³
237,600 cm³ = 237.6 L
The volume of water needed is 237.6 L.

Question 3.
The inside space of two different water tanks are shown below. Which tank has a greater capacity? Justify your answer.

Answer:
V₁ = Bh = (lw)h
V₁ = (6 in. ∙ 1 \(\frac{1}{2}\) in.) ∙ 3 in.
V₁ = (6
V₁ = 9 in² ∙ 3 in.
V₁ = 27 in³

V₂ = Bh = (lw)h
V₂ = (1 \(\frac{1}{2}\) in. ∙ 2 in.) ∙ 9 in.
V₂ = (2 in² ÷ 1 in²) ∙ 9 in .
V₂ = 3 in² ∙ 9 in.
V₂ = 27 in³
The tanks have the same volume, 27 in³. Each prism has a face with an area of 18 in²(base) and a height that is 1 \(\frac{1}{2}\) in.

Question 4.
The inside of a tank is in the shape of a right rectangular prism. The base of that prism is 85 cm by 64 cm. What is the minimum height inside the tank if the volume of the liquid in the tank is 92 L ?
Answer:
V = Bh = (lw)h
92,000 cm³ = (85 cm ∙ 64 cm) ∙ h
92,000 cm³ = 5,440 cm² ∙ h
92,000 cm³ ∙ \(\frac{1}{5,440 \mathrm{~cm}^{2}}\) = 5,440 cm² ∙ \(\frac{1}{5,440 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{92,000}{5,440}\) cm = 1 ∙ h
16 \(\frac{31}{34}\) cm = h
The minimum height of the inside of the tank is 16 \(\frac{31}{34}\) cm.

Question 5.
An oil tank is the shape of a right rectangular prism. The inside of the tank is 36.5 cm long, 52 cm wide, and 29 cm high. If 45 liters of oil have been removed from the tank since it was full, what is the current depth of oil left in the tank?
Answer:
V = Bh = (lw)h
V = (36.5 cm ∙ 52 cm) ∙ 29 cm
V = 1,898 cm² ∙ 29 cm
V = 55,042 cm³
The tank has a capacity of 55,042 cm³, or 55.042 L.
55.042 L – 45 L = 10.042 L

If 45 L of oil have been removed from the tank, then 10.042 L are left in the tank.
V = Bh = (lw)h
10,042 cm³ = (36.5 cm ∙ 52 cm) ∙ h
10,042 cm³ = 1,898 cm² ∙ h
10,042 cm³ ∙ \(\frac{1}{1,898 \mathrm{~cm}^{2}}\) = 1,898 cm² ∙ \(\frac{1}{1,898 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{10,042}{1,898}\) cm = 1 ∙ h
5.29 cm ≈ h
The depth of oil left in the tank is approximately 5.29 cm.

Question 6.
The inside of a right rectangular prism – shaped tank has a base that is 14 cm by 24 cm and a height of 60 cm. The tank is filled to its capacity with water, and then 10.92 L of water is removed. How far did the water level drop?
Answer:
V = Bh = (lw)h
V = (14 cm ∙ 24 cm) ∙ 60 cm
V = 336 cm² ∙ 60 cm
V = 20,160 cm³
The capacity of the tank is 20,160 cm³ or 20.16 L.
20,160 cm³ – 10,920 cm³ = 9,240 cm³
When 10.92 L or 10,920 cm³ of water is removed from the tank, there remains 9,240 cm³ of water in the tank.

V = Bh = (lw)h
9,240 cm³ = (14 cm ∙ 24 cm) ∙ h
9,240 cm³ = 336 cm² ∙ h
9,240 cm³ ∙ \(\frac{1}{336 \mathrm{~cm}^{2}}\) = 336 cm² ∙ \(\frac{1}{336 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{9,240}{336}\) cm = 1 ∙ h
27 \(\frac{1}{2}\) cm = h
The depth of the water left in the tank is 27 \(\frac{1}{2}\) cm.
60 cm – 27 \(\frac{1}{2}\) cm = 32 \(\frac{1}{2}\) cm
This means that the water level has dropped 32 \(\frac{1}{2}\) cm.

Question 7.
A right rectangular prism – shaped container has inside dimensions of 7 \(\frac{1}{2}\) cm long and 4 \(\frac{3}{5}\) cm wide. The tank is \(\frac{3}{5}\) full of vegetable oil. It contains 0.414 L of oil. Find the height of the container.
Answer:
V = Bh = (lw)h
414 cm³ = (7 \(\frac{1}{2}\) cm ∙ 4 \(\frac{3}{5}\) cm) ∙ h
414 cm³ = 34 \(\frac{1}{2}\) cm² ∙ h
414 cm³ = \(\frac{69}{2}\) cm² ∙ h
414 cm³ ∙ \(\frac{2}{69 \mathrm{~cm}^{2}}\) = \(\frac{69}{2}\) cm² ∙ \(\frac{2}{69 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{828}{69}\) cm = 1 ∙ h
12 cm = h
The vegetable oil in the container is 12 cm deep, but this is only \(\frac{3}{5}\) of the container’s depth. Let d represent the depth of the container in centimeters.
12 cm = \(\frac{3}{5}\) ∙ d
12 cm ∙ \(\frac{5}{3}\) = \(\frac{3}{5}\) ∙ \(\frac{5}{3}\) ∙ d
\(\frac{60}{3}\) cm = 1 ∙ d
20 cm = d
The depth of the container is 20 cm.

Question 8.
A right rectangular prism with length of 10 in, width of 16 in, and height of 12 in is \(\frac{2}{3}\) filled with water. If the water is emptied into another right rectangular prism with a length of 12 in, a width of 12 in, and height of 9 in, will the second container hold all of the water? Explain why or why not. Determine how far (above or below) the water level would be from the top of the container.
Answer:
\(\frac{2}{3}\) ∙ 12 in = \(\frac{24}{3}\) in = 8 in The height of the water in the first prism is 8 in.
V = Bh = (lw)h
V = (10 in ∙ 16 in) ∙ 8 in
V = 160 in² ∙ 8 in
V = 1,280 in³
The volume of water is 1,280 in³.

V = Bh = (lw)h
V = (12 in ∙ 12 in) ∙ 9 in
V = 144 in² ∙ 9 in
V = 1,296 in³
The capacity of the second prism is 1,296 in³, which is greater than the volume of water, so the water will fit in the second prism.

V = Bh = (lw)h Let h represent the depth of the water in the second prism in inches.
1,280 in³ = (12 in ∙ 12 in) ∙ h
1,280 in³ = (144 in² ) ∙ h
1,280 in³ ∙ \(\frac{1}{144 \mathrm{in}^{2}}\) = 144 in² ∙ \(\frac{1}{144 \mathrm{in}^{2}}\) ∙ h
\(\frac{1,280}{144}\) in = 1 ∙ h
8 \(\frac{128}{144}\) in = h
8 \(\frac{8}{9}\) in = h
The depth of the water in the second prism is 8 \(\frac{8}{9}\) in.
9 in – 8 \(\frac{8}{9}\) in = \(\frac{1}{9}\) in
The water level will be \(\frac{1}{9}\) in from the top of the second prism.

Eureka Math Grade 7 Module 3 Lesson 24 Exit Ticket Answer Key

Lawrence poured 27.328 L of water into a right rectangular prism – shaped tank. The base of the tank is 40 cm by 28 cm. When he finished pouring the water, the tank was \(\frac{2}{3}\) full. (1 L = 1,000 cm³)
a. How deep is the water in the tank?
Answer:
27.328 L = 27,328 cm³
V = Bh
V = (lw)h
27,328 cm³ = (40 cm ∙ 28 cm) ∙ h
27,328 cm³ = 1,120 cm² ∙ h
27,328 cm³ ∙ \(\frac{1}{1,120 \mathrm{~cm}^{2}}\) = 1,120 cm² ∙ \(\frac{1}{1,120 \mathrm{~cm}^{2}}\) ∙ h
\(\frac{27,328}{1,120}\) cm = 1 ∙ h
24\(\frac{280}{1,120}\) cm = h
24 \(\frac{2}{5}\) cm = h
The depth of the water is 24 \(\frac{2}{5}\) cm.

b. How deep is the tank?
Answer:
The depth of the water is \(\frac{2}{3}\) the depth of the tank. Let d represent the depth of the tank in centimeters.
24 \(\frac{2}{5}\) cm = \(\frac{2}{3}\) ∙ d
24 \(\frac{2}{5}\) cm ∙ \(\frac{3}{2}\) = \(\frac{2}{3}\) ∙ \(\frac{3}{2}\) ∙ d
36 cm ÷ \(\frac{3}{5}\) cm = 1d
36 \(\frac{3}{5}\) cm = d
The depth of the tank is 36 \(\frac{3}{5}\) cm.

c. How many liters of water can the tank hold in total?
Answer:
V = Bh
V = (lw)h
V = (40 cm ∙ 28 cm) ∙ 36 \(\frac{3}{5}\) cm
V = 1,120 cm² ∙ 36 \(\frac{3}{5}\) cm
V = 40,320 cm³ ÷ 672 cm³
V = 40,992 cm³
40,992 cm³ = 40.992 L The tank can hold up to 41.0 L of water.

Engage NY Eureka Math 7th Grade Module 3 Lesson 25 Answer Key

Eureka Math Grade 7 Module 3 Lesson 25 Example Answer Key

Example 1: Volume of a Fish Tank
Jay has a small fish tank. It is the same shape and size as the right rectangular prism shown in the Opening Exercise.
a. The box it came in says that it is a 3-gallon tank. Is this claim true? Explain your reasoning. Recall that 1 gal = 231 in³.
Answer:
The volume of the tank is 715 in³. To convert cubic inches to gallons, divide by 231.
715 in³ ∙ \(\frac{1 \text { gallon }}{231 \mathrm{in}^{3}}\) = 3.09 gallons
The claim is true if you round to the nearest whole gallon.

b. The pet store recommends filling the tank to within 1.5 in of the top. How many gallons of water will the tank hold if it is filled to the recommended level?
Answer:
Use 8.5 in. instead of 10 in. to calculate the volume. V = 11 in ∙ 6.5 in ∙ 8.5 in = 607.75 in³.
607.75 in³ ∙ \(\frac{1 \text { gallon }}{231 \mathrm{in}^{3}}\) = 2.63 gallons

c. Jay wants to cover the back, left, and right sides of the tank with a background picture. How many square inches will be covered by the picture?
Answer:
Back side area = 10 in ∙ 11 in = 110 in²
Left and right side area = 2(6.5 in)(10 in) = 130 in²
The total area to be covered with the background picture is 240 in².

d. Water in the tank evaporates each day, causing the water level to drop. How many gallons of water have evaporated by the time the water in the tank is four inches deep? Assume the tank was filled to within
Answer:
1.5 in. of the top to start.
Volume when water is 4 in. deep:
11 in. ∙ 6.5 in. ∙ 4 in = 286in³
Difference in the two volumes:
607.75 in³-286 in³ = 321.75 in³
When the water is filled to within 1.5 in of the top, the volume is 607.75 in³. When the water is 4 in deep, the volume is 286 in³. The difference in the two volumes is 321.75 in³. Converting cubic inches to gallons by dividing by 231 gives a difference of 1.39 gal., which means 1.39 gal. of water have evaporated.

Eureka Math Grade 7 Module 3 Lesson 25 Exercise Answer Key

Opening Exercise
What is the surface area and volume of the right rectangular prism?

Answer:
Surface Area = 2(11 in)(6.5 in) + 2(10 in)(6.5 in) + 2(11 in)(10 in) = 493 in²
Volume = 11 in ∙ 6.5 in ∙ 10 in = 715 in³

Exercise 1: Fish Tank Designs
Two fish tanks are shown below, one in the shape of a right rectangular prism (R) and one in the shape of a right trapezoidal prism (T).

a. Which tank holds the most water? Let Vol(R) represent the volume of the right rectangular prism and Vol(T) represent the volume of the right trapezoidal prism. Use your answer to fill in the blanks with Vol(R) and Vol(T).
__________________ < __________________
Answer:
Volume of the right rectangular prism: (25 in × 10 in) × 15 in = 3,750 in³
Volume of the right trapezoidal prism: (31 in × 8 in) × 15 in = 3,720 in³
The right rectangular prism holds the most water.
Vol(T) < Vol(R)

b. Which tank has the most surface area? Let SA(R) represent the surface area of the right rectangular prism and SA(T) represent the surface area of the right trapezoidal prism. Use your answer to fill in the blanks with SA(R) and SA(T).
__________________ < __________________
Answer:
The surface area of the right rectangular prism:
2(25 in × 10 in) + 2(25 in × 15 in) + 2(10 in × 15 in) = 500 in² + 750 in² + 300 in² = 1,550 in²
The surface area of the right trapezoidal prism:
2(31 in × 8 in) + 2(10 in × 15 in) + (25 in × 15 in) + (31 in × 15 in) = 496 in² + 300 in² + 375 in² + 555 in² = 1726 in²
The right trapezoidal prism has the most surface area.
SA(R) < SA(T)

c. Water evaporates from each aquarium. After the water level has dropped \(\frac{1}{2}\) inch in each aquarium, how many cubic inches of water are required to fill up each aquarium? Show work to support your answers.
Answer:
The right rectangular prism will need 125 in³ of water. The right trapezoidal prism will need 124 in³ of water. First, decrease the height of each prism by a half inch and recalculate the volumes. Then, subtract each answer from the original volume of each prism.
NewVol(R) = (25 in)(10 in)(14.5 in) = 3,625 in³
NewVol(T) = (31 in)(8 in)(14.5 in) = 3,596 in³
3,750 in³-3,625 in³ = 125 in³
3,720 in³-3,596 in³ = 124 in³

Exercise 2: Design Your Own Fish Tank
Design at least three fish tanks that will hold approximately 10 gallons of water. All of the tanks should be shaped like right prisms. Make at least one tank have a base that is not a rectangle. For each tank, make a sketch, and calculate the volume in gallons to the nearest hundredth.
Answer:
Three possible designs are shown below.

10 gal. is 2,310 in³
Rectangular Base: Volume = 2,304 in³ or 9.97 gal.
Triangular Base: Volume = 2,240 in³ or 9.70 gal.
Hexagonal Base: Volume = 2,325 in³ or 10.06 gal.

Challenge: Each tank is to be constructed from glass that is \(\frac{1}{4}\) in. thick. Select one tank that you designed, and determine the difference between the volume of the total tank (including the glass) and the volume inside the tank. Do not include a glass top on your tank.
Answer:
Height = 12 in-\(\frac{1}{4}\) in = 11.75 in
Length = 24 in-\(\frac{1}{2}\) in = 23.5 in
Width = 8 in-\(\frac{1}{2}\) in = 7.5 in
Inside Volume = 2,070.9 in³
The difference between the two volumes is 233.1 in³, which is approximately 1 gal.

Eureka Math Grade 7 Module 3 Lesson 25 Problem Set Answer Key

Question 1.
The dimensions of several right rectangular fish tanks are listed below. Find the volume in cubic centimeters, the capacity in liters (1 L = 1000 cm³), and the surface area in square centimeters for each tank. What do you observe about the change in volume compared with the change in surface area between the small tank and the extra-large tank?

Answer:

While the volume of the extra-large tank is about five times the volume of the small tank, its surface area is less than three times that of the small tank.

Question 2.
A rectangular container 15 cm long by 25 cm wide contains 2.5 L of water.

a. Find the height of the water level in the container. (1 L = 1000 cm³)
2.5 L = 2,500 cm³
To find the height of the water level, divide the volume in cubic centimeters by the area of the base.
\(\frac{2,500 \mathrm{~cm}^{3}}{25 \mathrm{~cm} \cdot 15 \mathrm{~cm}}\) = 6 \(\frac{2}{3}\) cm

b. If the height of the container is 18 cm, how many more liters of water would it take to completely fill the container?
Answer:
Volume of tank: (25 cm × 15 cm) × 18 cm = 6,750 cm³
Capacity of tank: 6.75 L
Difference: 6.75 L- 2.5 L = 4.25 L

c. What percentage of the tank is filled when it contains 2.5 L of water?
Answer:
\(\frac{2.5 L}{6.75 L}\) = 0.37 = 37%

Question 3.
A rectangular container measuring 20 cm by 14.5 cm by 10.5 cm is filled with water to its brim. If 300 cm³ are drained out of the container, what will be the height of the water level? If necessary, round to the nearest tenth.

Answer:
Volume: (20 cm × 14.5 cm) × 10.5 cm = 3,045 cm³
Volume after draining: 2,745 cm³
Height (divide the volume by the area of the base):
\(\frac{2745 \mathrm{~cm}^{3}}{20 \mathrm{~cm} \times 14.5 \mathrm{~cm}}\) ≈ 9.5 cm

Question 4.
Two tanks are shown below. Both are filled to capacity, but the owner decides to drain them. Tank 1 is draining at a rate of 8 liters per minute. Tank 2 is draining at a rate of 10 liters per minute. Which tank empties first?

Answer:
Tank 1 Volume: 75 cm × 60 cm × 60 cm = 270,000 cm³
Tank 2 Volume: 90 cm × 40 cm × 85 cm = 306,000 cm³
Tank 1 Capacity: 270 L
Tank 2 Capacity: 306 L

To find the time to drain each tank, divide the capacity by the rate (liters per minute).
Time to drain tank 1: \(\frac{270 \mathrm{~L}}{8 \frac{\mathrm{L}}{\mathrm{min}}}\) = 33.75 min.
Time to drain tank 2: \(\frac{360 \mathrm{~L}}{10 \frac{\mathrm{L}}{\mathrm{min}}}\) = 30.6 min.
Tank 2 empties first.

Question 5.
Two tanks are shown below. One tank is draining at a rate of 8 liters per minute into the other one, which is empty. After 10 minutes, what will be the height of the water level in the second tank? If necessary, round to the nearest minute.

Answer:
Volume of the top tank: 45 cm × 50 cm × 55 cm = 123,750 cm³
Capacity of the top tank: 123.75 L
At 8 L/min for 10 minutes, 80 L will have drained into the bottom tank after 10 minutes.
That is 80,000 cm³. To find the height, divide the volume by the area of the base.
\(\frac{80,000 \mathrm{~cm}^{3}}{100 \mathrm{~cm} \cdot 35 \mathrm{~cm}}\) ≈ 22.9 cm
After 10 minutes, the height of the water in the bottom tank will be about 23 cm.

Question 6.
Two tanks with equal volumes are shown below. The tops are open. The owner wants to cover one tank with a glass top. The cost of glass is $0.05 per square inch. Which tank would be less expensive to cover? How much less?

Answer:
Dimensions: 12 in. long by 8 in. wide by 10 in. high
Surface area: 96 in²
Cost: \(\frac{\$ 0.05}{\mathrm{in}^{2}}\) ∙ 96 in² = $4.80

Dimensions: 15 in. long by 8 in. wide by 8 in. high
Surface area: 120 in²
Cost: \(\frac{\$ 0.05}{\mathrm{in}^{2}}\) ∙ 120 in² = $6.00

The first tank is less expensive. It is $1.20 cheaper.

Question 7.
Each prism below is a gift box sold at the craft store.

a. What is the volume of each prism?
Answer:
(a) Volume = 336 cm³
(b) Volume = 750 cm³
(c) Volume = 990 cm³
(d) Volume = 1130.5 cm³

b. Jenny wants to fill each box with jelly beans. If one ounce of jelly beans is approximately 30 cm^3, estimate how many ounces of jelly beans Jenny will need to fill all four boxes? Explain your estimates.
Answer:
Divide each volume in cubic centimeters by 30.
(a) 11.2 ounces
(b) 25 ounces
(c) 33 ounces
(d) 37.7 ounces
Jenny would need a total of 106.9 ounces.

Question 8.
Two rectangular tanks are filled at a rate of 0.5 cubic inches per minute. How long will it take each tank to be half-full?
a. Tank 1 Dimensions: 15 in. by 10 in. by 12.5 in.
Answer:
Volume: 1,875 in³
Half of the volume is 937.5 in³.
To find the number of minutes, divide the volume by the rate in cubic inches per minute.
Time: 1,875 minutes.

b. Tank 2 Dimensions: 2 \(\frac{1}{2}\) in. by 3 \(\frac{3}{4}\) in. by 4 \(\frac{3}{8}\) in.
Answer:
Volume: \(\frac{2625}{64}\) in³
Half of the volume is \(\frac{2625}{128}\) in³.
To find the number of minutes, divide the volume by the rate in cubic inches per minute.
Time: 41 minutes

Eureka Math Grade 7 Module 3 Lesson 25 Exit Ticket Answer Key

Melody is planning a raised bed for her vegetable garden.

a. How many square feet of wood does she need to create the bed?
Answer:
2(4 ft)(1.25 ft) + 2(2.5 ft)(1.25 ft) = 16.25 ft²
The dimensions in feet are 4 ft. by 1.25 ft. by 2.5 ft. The lateral area is 16.25 ft².

b. She needs to add soil. Each bag contains 1.5 cubic feet. How many bags will she need to fill the vegetable garden?
Answer:
V = 4 ft ∙ 1.25 ft ∙ 2.5 ft = 12.5 ft³
The volume is 12.5 ft³. Divide the total cubic feet by 1.5 ft³ to determine the number of bags.
12.5 ft³ ÷ 1.5 ft³ = 8 \(\frac{1}{3}\)
Melody will need to purchase 9 bags of soil to fill the garden bed.

Engage NY Eureka Math 7th Grade Module 3 Lesson 26 Answer Key

Eureka Math Grade 7 Module 3 Lesson 26 Example Answer Key

Example 1.

The insulated box shown is made from a large cube with a hollow inside that is a right rectangular prism with a square base. The figure on the right is what the box looks like from above.
a. Calculate the volume of the outer box.
Answer:
24 cm × 24 cm × 24 cm = 13,824 cm³

b. Calculate the volume of the inner prism.
Answer:
18 cm × 18 cm × 21 \(\frac{1}{4}\) cm = 6,885 cm³

c. Describe in words how you would find the volume of the insulation.
Answer:
Find the volume of the outer cube and the inner right rectangular prism, and then subtract the two volumes.

d. Calculate the volume of the insulation in cubic centimeters.
Answer:
13,824 cm³ – 6,885 cm³ = 6,939cm³

e. Calculate the amount of water the box can hold in liters.
Answer:
6939 cm³ = 6939 mL = \(\frac{(6939 \mathrm{~mL})}{1000 \frac{\mathrm{mL}}{\mathrm{L}}}\) = 6.939 L

Eureka Math Grade 7 Module 3 Lesson 26 Exercise Answer Key

Opening Exercise
Explain to your partner how you would calculate the area of the shaded region. Then, calculate the area.

Answer:
Find the area of the outer rectangle, and subtract the area of the inner rectangle.
6 cm × 3 cm – 5 cm × 2 cm = 8 cm²

Exercise 1: Brick Planter Design
You have been asked by your school to design a brick planter that will be used by classes to plant flowers. The planter will be built in the shape of a right rectangular prism with no bottom so water and roots can access the ground beneath. The exterior dimensions are to be 12 ft. × 9 ft. × 2 \(\frac{1}{2}\) ft. The bricks used to construct the planter are 6 in. long, 3 \(\frac{1}{2}\) in. wide, and 2 in. high.
a. What are the interior dimensions of the planter if the thickness of the planter’s walls is equal to the length of the bricks?
Answer:

6 in = \(\frac{1}{2}\) ft.
Interior length:
12 ft. – \(\frac{1}{2}\) ft. – \(\frac{1}{2}\) ft. = 11 ft.
Interior width:
9 ft. – \(\frac{1}{2}\) ft. – \(\frac{1}{2}\) ft. = 8 ft.
Interior dimensions:
11 ft. × 8 ft. × 2 \(\frac{1}{2}\) ft.

b. What is the volume of the bricks that form the planter?
Answer:
Solution 1
Subtract the volume of the smaller interior prism V_S from the volume of the large exterior prism V_L.
V_Brick = V_L – V_S
V_Brick = (12 ft. × 9 ft. × 2 \(\frac{1}{2}\) ft.) – (11 ft. × 8 ft. × 2 \(\frac{1}{2}\) ft.)
V_Brick = 270 ft³ – 220 ft³
V_Brick = 50 ft³

Solution 2
The volume of the brick is equal to the area of the base times the height.

B = \(\frac{1}{2}\) ft. × (8 \(\frac{1}{2}\) ft. + 11 \(\frac{1}{2}\) ft. + 8 \(\frac{1}{2}\) ft. + 11 \(\frac{1}{2}\) ft.)
B = \(\frac{1}{2}\) ft. × (40 ft.) = 20 ft²
V = Bh
V = (20 ft² )(2 \(\frac{1}{2}\) ft.) = 50 ft³

c. If you are going to fill the planter \(\frac{3}{4}\) full of soil, how much soil will you need to purchase, and what will be the height of the soil?
Answer:
The height of the soil will be \(\frac{3}{4}\) of 2 \(\frac{1}{2}\)feet.
\(\frac{3}{4}\) (\(\frac{5}{2}\) ft.) = \(\frac{15}{8}\) ft.; The height of the soil will be \(\frac{15}{8}\) ft. (or 1 \(\frac{7}{8}\) ft.).
The volume of the soil in the planter:
V = (11 ft × 8 ft. × \(\frac{15}{8}\) ft.)
V = (11 ft. × 15 ft² ) = 165 ft³

d. How many bricks are needed to construct the planter?
Answer:
P = 2(8 \(\frac{1}{2}\)ft.) + 2(11 \(\frac{1}{2}\)ft.)
P = 17 ft. + 23 ft. = 40 ft.

3 \(\frac{1}{2}\) in. = \(\frac{7}{24}\) ft.

We can then divide the perimeter by the width of each brick in order to determine the number of bricks needed for each layer of the planter.
40 ÷ \(\frac{7}{24}\) = \(\frac{960}{7}\)
40 × \(\frac{24}{7}\) = \(\frac{960}{7}\) ≈ 137.1

Each layer of the planter requires approximately 137.1 bricks.
2 in. = \(\frac{1}{6}\) ft.
The height of the planter, 2 \(\frac{1}{2}\) ft., is equal to the product of the number of layers of brick, n, and the height of each brick, \(\frac{1}{6}\) ft.
2\(\frac{1}{2}\) = l(\(\frac{1}{6}\))
6(2 \(\frac{1}{2}\)) = l
15 = l
There are 15 layers of bricks in the planter.

The total number of bricks, b, is equal to the product of the number of bricks in each layer (\(\frac{960}{7}\)) and the number of layers (15).
b = \(\frac{960}{7}\))(15)
b = \(\frac{14400}{7}\) ≈ 2057.1
It is not reasonable to purchase 0.1 brick; we must round up to the next whole brick, which is 2,058 bricks. Therefore, 2,058 bricks are needed to construct the planter.

e. Each brick used in this project costs $0.82 and weighs 4.5 lb. The supply company charges a delivery fee of $15 per whole ton (2000 lb) over 4000 lb How much will your school pay for the bricks (including delivery) to construct the planter?
Answer:
If the school purchases 2058 bricks, the total weight of the bricks for the planter,
2058(4.5 lb) = 9261 lb
The number of whole tons over 4,000 pounds,
9261 – 4000 = 5261
Since 1 ton = 2000 lb., there are 2 whole tons (4000 lb.) in 5,261 lb.
Total cost = cost of bricks + cost of delivery
Total cost = 0.82(2058) + 2(15)
Total cost = 1687.56 + 30 = 1717.56
The cost for bricks and delivery will be $1,717.56.

f. A cubic foot of topsoil weighs between 75 and 100 lb. How much will the soil in the planter weigh?
Answer:
The volume of the soil in the planter is 165 ft³.
Minimum weight:
Minimum weight = 75 lb(165)
Minimum weight = 12375 lb.

Maximum weight:
Maximum weight = 100 lb(165)
Maximum weight = 16500 lb.
The soil in the planter will weigh between 12,375 lb. and 16,500 lb.

g. If the topsoil costs $0.88 per cubic foot, calculate the total cost of materials that will be used to construct the planter.
Answer:
The total cost of the top soil:
Cost = 0.88(165) = 145.2; The cost of the top soil will be $145.20.
The total cost of materials for the brick planter project:
Cost = (cost of bricks) + (cost of soil)
Cost = $1,717.56 + $145.20
Cost = $1,862.76
The total cost of materials for the brick planter project will be $1,862.76.

Exercise 2: Design a Feeder
You did such a good job designing the planter that a local farmer has asked you to design a feeder for the animals on his farm. Your feeder must be able to contain at least 100,000 cubic centimeters, but not more than 200,000 cubic centimeters of grain when it is full. The feeder is to be built of stainless steel and must be in the shape of a right prism but not a right rectangular prism. Sketch your design below including dimensions. Calculate the volume of grain that it can hold and the amount of metal needed to construct the feeder.
The farmer needs a cost estimate. Calculate the cost of constructing the feeder if \(\frac{1}{2}\) cm thick stainless steel sells for $93.25 per square meter.
Answer:
Answers will vary. Below is an example using a right trapezoidal prism.
This feeder design consists of an open – top container in the shape of a right trapezoidal prism. The trapezoidal sides of the feeder will allow animals easier access to feed at its bottom. The dimensions of the feeder are shown in the diagram.

B = \(\frac{1}{2}\)(b₁ + b₂ )h
B = \(\frac{1}{2}\) (100 cm + 80 cm)∙30 cm
B = \(\frac{1}{2}\) (180 cm)∙30 cm
B = 90 cm∙30 cm
B = 2700 cm²

V = Bh
V = (2,700 cm² )(60 cm)
V = 162,000 cm³

The volume of the solid prism is 162,000 cm³, so the volume that the feeder can contain is slightly less, depending on the thickness of the metal used.

The exterior surface area of the feeder tells us the area of metal required to build the feeder.
SA = (LA – A_top) + 2B
SA = 60 cm∙(40 cm + 80 cm + 40 cm) + 2(2,700 cm² )
SA = 60 cm(160 cm) + 5,400 cm²
SA = 9,600 cm² + 5,400 cm²
SA = 15,000 cm²
The feeder will require 15,000 cm² of metal.
1 m² = 10,000 cm², so 15,000 cm² = 1.5 m²
Cost = 93.25(1.5) = 139.875
Since this is a measure of money, the cost must be rounded to the nearest cent, which is $139.88.

Eureka Math Grade 7 Module 3 Lesson 26 Problem Set Answer Key

Question 1.
A child’s toy is constructed by cutting a right triangular prism out of a right rectangular prism.

a. Calculate the volume of the rectangular prism.
Answer:
10 cm × 10 cm × 12 \(\frac{1}{2}\) cm = 1250 cm³

b. Calculate the volume of the triangular prism.
Answer:
\(\frac{1}{2}\) (5 cm × 2 \(\frac{1}{2}\) cm) × 12 \(\frac{1}{2}\) cm = 78 \(\frac{1}{8}\) cm³

c. Calculate the volume of the material remaining in the rectangular prism.
Answer:
1250 cm³ – 78 \(\frac{1}{8}\) cm³ = 1171 \(\frac{7}{8}\) cm³

d.
What is the largest number of triangular prisms that can be cut from the rectangular prism?
Answer:
\(\frac{1250 \mathrm{~cm}^{3}}{78 \frac{1}{8} \mathrm{~cm}^{3}}\) = 16

e. What is the surface area of the triangular prism (assume there is no top or bottom)?
Answer:
5.6 cm × 12 \(\frac{1}{2}\) cm + 2 \(\frac{1}{2}\) cm × 12 \(\frac{1}{2}\) cm + 5 cm × 12 \(\frac{1}{2}\) cm = 163 \(\frac{3}{4}\) cm²

Question 2.
A landscape designer is constructing a flower bed in the shape of a right trapezoidal prism. He needs to run three identical square prisms through the bed for drainage.

a. What is the volume of the bed without the drainage pipes?
Answer:
\(\frac{1}{2}\) (14 ft. + 12 ft.) × 3 ft. × 16 ft. = 624 ft³

b. What is the total volume of the three drainage pipes?
Answer:
3(\(\frac{1}{4}\) ft² × 16 ft.) = 12 ft³

c. What is the volume of soil if the planter is filled to 3/4 of its total capacity with the pipes in place?
Answer:
\(\frac{3}{4}\) (624 ft³ ) – 12 ft³ = 456 ft³

d. What is the height of the soil? If necessary, round to the nearest tenth.
Answer:
\(\frac{456 \mathrm{ft}^{3}}{\frac{1}{2}(14 \mathrm{ft} + 12 \mathrm{ft}) \times 16 \mathrm{ft}}\)≈ 2.2 ft.

e. If the bed is made of 8 ft. × 4 ft. pieces of plywood, how many pieces of plywood will the landscape designer need to construct the bed without the drainage pipes?
Answer:
2(3 \(\frac{1}{4}\) ft. × 16 ft.) + 12 ft. × 16 ft. + 2(\(\frac{1}{2}\) (12 ft. + 14 ft.) × 3 ft.) = 374 ft²
374 ft² ÷ \(\frac{(8 \mathrm{ft} \times 4 \mathrm{ft})}{\text { piece of plywood }}\) = 11.7, or 12 pieces of plywood

f. If the plywood needed to construct the bed costs $35 per 8 ft. × 4 ft. piece, the drainage pipes cost $125 each, and the soil costs $1.25/cubic foot, how much does it cost to construct and fill the bed?
Answer:
\(\frac{\$ 35}{\text { piece of plywood }}\)(12 pieces of plywood) + \(\frac{\$ 125}{\text { pipe }}\) (3 pipes) + \(\frac{\$ 1.25}{f t^{3} \text { soil }}\)(456 ft³ soil) = $1,365.00

Eureka Math Grade 7 Module 3 Lesson 26 Exit Ticket Answer Key

Lawrence is designing a cooling tank that is a square prism. A pipe in the shape of a smaller 2 ft × 2 ft square prism passes through the center of the tank as shown in the diagram, through which a coolant will flow.

a. What is the volume of the tank including the cooling pipe?
Answer:
7 ft. × 3 ft. × 3 ft. = 63 ft³

b. What is the volume of coolant that fits inside the cooling pipe?
Answer:
2 ft. × 2 ft. × 7 ft. = 28 ft³

c. What is the volume of the shell (the tank not including the cooling pipe)?
Answer:
63 ft³ – 28 ft³ = 35 ft³

d. Find the surface area of the cooling pipe.
Answer:
2 ft. × 7 ft. × 4 = 56 ft²

Engage NY Eureka Math 7th Grade Module 4 Lesson 17 Answer Key

Eureka Math Grade 7 Module 4 Lesson 17 Example Answer Key

Example 1.
A 5 – gallon container of trail mix is 20% nuts. Another trail mix is added to it, resulting in a 12 – gallon container of trail mix that is 40% nuts.
a. Write an equation to describe the relationships in this situation.
Answer:
Let j represent the percent of nuts in the second trail mix that is added to the first trail mix to create the resulting 12 – gallon container of trail mix.
0.4(12) = 0.2(5) + j(12 – 5)

b. Explain in words how each part of the equation relates to the situation.
Answer:
Quantity = Percent×Whole
(Resulting gallons of trail mix)(Resulting % of nuts) = (1^st trail mix in gallons)(% of nuts) + (2^nd trail mix in gallons)(% of nuts)

c. What percent of the second trail mix is nuts?
Answer:
4.8 = 1 + 7j
4.8 – 1 = 1 – 1 + 7j
3.8 = 7j
j ≈ 0.5429
About 54% of the second trail mix is nuts.

Example 2.
Soil that contains 30% clay is added to soil that contains 70% clay to create 10 gallons of soil containing 50% clay. How much of each of the soils was combined?
Answer:
Let x be the amount of soil with 30% clay.
(1^st soil amount)(% of clay) + (2^nd soil amount)(% of clay) = (resulting amount)(resulting % of clay)
(0.3)(x) + (0.7)(10 – x) = (0.5)(10)
0.3x + 7 – 0.7x = 5
– 0.4x + 7 – 7 = 5 – 7
– 0.4x = – 2
x = 5
5 gallons of the 30% clay soil and 10 – 5 = 5, so 5 gallons of the 70% clay soil must be mixed to make 10 gallons of 50% clay soil.

Eureka Math Grade 7 Module 4 Lesson 17 Exercise Answer Key

Opening Exercise
Imagine you have two equally – sized containers. One is pure water, and the other is 50% water and 50% juice. If you combined them, what percent of juice would be the result?

Answer:

25% of the resulting mixture is juice because \(\frac{0.5}{2}\) = \(\frac{1}{4}\).

If a 2 – gallon container of pure juice is added to 3 gallons of water, what percent of the mixture is pure juice?

Answer:
Let x represent the percent of pure juice in the resulting juice mixture.

If a 2 – gallon container of juice mixture that is 40% pure juice is added to 3 gallons of water, what percent of the mixture is pure juice?

Answer:

→ How many gallons of the juice mixture is pure juice?
(2 gallons)(0.40) = 0.8 gallons

→ What percent is pure juice out of the resulting mixture?
16%

→ Does this make sense relative to the prior problem?
Yes, because the mixture should have less juice than in the prior problem

If a 2 – gallon juice cocktail that is 40% pure juice is added to 3 gallons of pure juice, what percent of the resulting mixture is pure juice?

Answer:

What is the difference between this problem and the previous one?
Instead of adding water to the two gallons of juice mixture, pure juice is added, so the resulting liquid contains 3.8 gallons of pure juice.
What percent is pure juice out of the resulting mixture?
Let x represent the percent of pure juice in the resulting mixture.
x(5) = 40%(2) + 100%(3)
5x = 0.8 + 3
5x = 3.8
x = 0.76
The mixture is 76% pure juice.

Exercise 1.
Represent each situation using an equation, and show all steps in the solution process.
a. A 6 – pint mixture that is 25% oil is added to a 3 – pint mixture that is 40% oil. What percent of the resulting mixture is oil?
Answer:
Let x represent the percent of oil in the resulting mixture.
0.25(6) + 0.40(3) = x(9)
1.5 + 1.2 = 9x
2.7 = 9x
x = 0.3
The resulting 9 – pint mixture is 30% oil.

b. An 11 – ounce gold chain of 24% gold was made from a melted down 4 – ounce charm of 50% gold and a golden locket. What percent of the locket was pure gold?
Let x represent the percent of pure gold in the locket.
0.5(4) + (x)(7) = 0.24(11)
2 + 7x = 2.64
2 – 2 + 7x = 2.64 – 2
\(\frac{7x}{7}\) = \(\frac{0.64}{7}\)
x≈0.0914
The locket was about 9% gold.

c. In a science lab, two containers are filled with mixtures. The first container is filled with a mixture that is 30% acid. The second container is filled with a mixture that is 50% acid, and the second container is 50% larger than the first. The first and second containers are then emptied into a third container. What percent of acid is in the third container?
Answer:
Let m represent the total amount of mixture in the first container.
0.3m is the amount of acid in the first container.
0.5(m + 0.5m) is the amount of acid in the second container.
0.3m + 0.5(m + 0.5m) = 0.3m + 0.5(1.5m) = 1.05m is the amount of acid in the mixture in the third container.
m + 1.5m = 2.5m is the amount of mixture in the third container. So, \(\frac{1.05 m}{2.5 m}\) = 0.42 = 42% is the percent of acid in the third container.

Exercise 2.
The equation (0.2)(x) + (0.8)(6 – x) = (0.4)(6) is used to model a mixture problem.
a. How many units are in the total mixture?
Answer:
6 units

b. What percents relate to the two solutions that are combined to make the final mixture?
Answer:
20% and 80%

c. The two solutions combine to make 6 units of what percent solution?
Answer:
40%

d. When the amount of a resulting solution is given (for instance, 4 gallons) but the amounts of the mixing solutions are unknown, how are the amounts of the mixing solutions represented?
Answer:
If the amount of gallons of the first mixing solution is represented by the variable x, then the amount of gallons of the second mixing solution is 4 – x.

Eureka Math Grade 7 Module 4 Lesson 17 Problem Set Answer Key

Question 1.
A 5 – liter cleaning solution contains 30% bleach. A 3 – liter cleaning solution contains 50% bleach. What percent of bleach is obtained by putting the two mixtures together?
Answer:
Let x represent the percent of bleach in the resulting mixture.
0.3(5) + 0.5(3) = x(8)
1.5 + 1.5 = 8x
3 ÷ 8 = 8x ÷ 8
x = 0.375
The percent of bleach in the resulting cleaning solution is 37.5%.

Question 2.
A container is filled with 100 grams of bird feed that is 80% seed. How many grams of bird feed containing 5% seed must be added to get bird feed that is 40% seed?
Answer:
Let x represent the amount of bird feed, in grams, to be added.
0.8(100) + 0.05x = 0.4(100 + x)
80 + 0.05x = 40 + 0.4x
80 – 40 + 0.05x = 40 – 40 + 0.4x
40 + 0.05x = 0.4x
40 + 0.05x – 0.05x = 0.4x – 0.05x
40 ÷ 0.35 = 0.35x ÷ 0.35
x ≈ 114.3
About 114.3 grams of the bird seed containing 5% seed must be added.

Question 3.
A container is filled with 100 grams of bird feed that is 80% seed. Tom and Sally want to mix the 100 grams with bird feed that is 5% seed to get a mixture that is 40% seed. Tom wants to add 114 grams of the 5% seed, and Sally wants to add 115 grams of the 5% seed mix. What will be the percent of seed if Tom adds 114 grams? What will be the percent of seed if Sally adds 115 grams? How much do you think should be added to get 40% seed?
Answer:
If Tom adds 114 grams, then let x be the percent of seed in his new mixture. 214x = 0.8(100) + 0.05(114). Solving, we get the following:
x = \(\frac{80 + 5.7}{214}\) = \(\frac{85.7}{214}\) ≈ 0.4005 = 40.05%.
If Sally adds 115 grams, then let y be the percent of seed in her new mixture. 215y = 0.8(100) + 0.05(115). Solving, we get the following:
y = \(\frac{80 + 5.75}{215}\) = \(\frac{85.75}{215}\) ≈ 0.3988 = 39.88%.
The amount to be added should be between 114 and 115 grams. It should probably be closer to 114 because 40.05% is closer to 40% than 39.88%.

Question 4.
Jeanie likes mixing leftover salad dressings together to make new dressings. She combined 0.55 L of a 90% vinegar salad dressing with 0.45 L of another dressing to make 1 L of salad dressing that is 60% vinegar. What percent of the second salad dressing was vinegar?
Answer:
Let c represent the percent of vinegar in the second salad dressing.
0.55(0.9) + (0.45)(c) = 1(0.6)
0.495 + 0.45c = 0.6
0.495 – 0.495 + 0.45c = 0.6 – 0.495
0.45c = 0.105
0.45c ÷ 0.45 = 0.105 ÷ 0.45
c ≈ 0.233
The second salad dressing was around 23% vinegar.

Question 5.
Anna wants to make 30 mL of a 60% salt solution by mixing together a 72% salt solution and a 54% salt solution. How much of each solution must she use?
Answer:
Let s represent the amount, in milliliters, of the first salt solution.
0.72(s) + 0.54(30 – s) = 0.60(30)
0.72s + 16.2 – 0.54s = 18
0.18s + 16.2 = 18
0.18s + 16.2 – 16.2 = 18 – 16.2
0.18s = 1.8
s = 10
Anna needs 10 mL of the 72% solution and 20 mL of the 54% solution.

Question 6.
A mixed bag of candy is 25% chocolate bars and 75% other filler candy. Of the chocolate bars, 50% of them contain caramel. Of the other filler candy, 10% of them contain caramel. What percent of candy contains caramel?
Answer:
Let c represent the percent of candy containing caramel in the mixed bag of candy.
0.25(0.50) + (0.75)(0.10) = 1(c)
0.125 + 0.075 = c
0.2 = c
In the mixed bag of candy, 20% of the candy contains caramel.

Question 7.
A local fish market receives the daily catch of two local fishermen. The first fisherman’s catch was 84% fish while the rest was other non – fish items. The second fisherman’s catch was 76% fish while the rest was other non – fish items. If the fish market receives 75% of its catch from the first fisherman and 25% from the second, what was the percent of other non – fish items the local fish market bought from the fishermen altogether?
Answer:
Let n represent the percent of non – fish items of the total market items.
0.75(0.16) + 0.25(0.24) = n
0.12 + 0.06 = n
0.18 = n
The percent of non – fish items in the local fish market is 18%.

Eureka Math Grade 7 Module 4 Lesson 17 Exit Ticket Answer Key

A 25% vinegar solution is combined with triple the amount of a 45% vinegar solution and a 5% vinegar solution resulting in 20 milliliters of a 30% vinegar solution.
Question 1.
Determine an equation that models this situation, and explain what each part represents in the situation.
Answer:
Let s represent the number of milliliters of the first vinegar solution.
(0.25)(s) + (0.45)(3s) + (0.05)(20 – 4s) = (0.3)(20)
(0.25)(s) represents the amount of the 25% vinegar solution.
(0.45)(3s) represents the amount of the 45% vinegar solution, which is triple the amount of the 25% vinegar solution.
(0.05)(20 – 4s) represents the amount of the 5% vinegar solution, which is the amount of the remainder of the solution.
(0.3)(20) represents the result of the mixture, which is 20 mL of a 30% vinegar solution.

Question 2.
Solve the equation and find the amount of each of the solutions that were combined.
Answer:
0.25s + 1.35s + 1 – 0.2s = 6
1.6s – 0.2s + 1 = 6
1.4s + 1 – 1 = 6 – 1
1.4s ÷ 1.4 = 5 ÷ 1.4
s ≈ 3.57
3s ≈ 3(3.57) = 10.71
20 – 4s ≈ 20 – 4(3.57) = 5.72
Around 3.57 mL of the 25% vinegar solution, 10.71 mL of the 45% vinegar solution and 5.72 mL of the
5% vinegar solution were combined to make 20 mL of the 30% vinegar solution.

Engage NY Eureka Math 7th Grade Module 4 Lesson 15 Answer Key

Eureka Math Grade 7 Module 4 Lesson 15 Example Answer Key

Example 1.
What percent of the area of the large square is the area of the small square?

Answer:
Scale factor of the large square to the small square: \(\frac{1}{5}\)
Area of the large square to the small square: (\(\frac{1}{5}\))² = \(\frac{1}{25}\) = \(\frac{4}{100}\) = 0.04 = 4%
The area of the small square is only 4% of the area of the large square.

Example 2.
What percent of the area of the large disk lies outside the shaded disk?

Answer:
Radius of the shaded disk = 2
Radius of large disk = 4
Scale factor of the large disk to the shaded disk: \(\frac{2}{4}\) = \(\frac{1}{2}\)
Area of the large disk to the shaded disk:
(\(\frac{1}{2}\))² = \(\frac{1}{4}\) = 25%
Area outside shaded disk: \(\frac{3}{4}\) = 75%

Example 3.
If the area of the shaded region in the larger figure is approximately 21.5 square inches, write an equation that relates the areas using scale factor and explain what each quantity represents. Determine the area of the shaded region in the smaller scale drawing.

Answer:
Scale factor of corresponding sides:
\(\frac{6}{10}\) = \(\frac{3}{5}\) = 60%

Area of shaded region of smaller figure: Assume A is the area of the shaded region of the larger figure.
(\(\frac{3}{5}\))² A = \(\frac{9}{25}\) A
= \(\frac{9}{25}\)(21.5)
= 7.74
In this equation, the square of the scale factor, (\(\frac{3}{5}\))², multiplied by the area of the shaded region in the larger figure, 21.5 sq.in., is equal to the area of the shaded region of the smaller figure, 7.74 sq.in.
The area of shaded region of the smaller scale drawing is about 7.74 sq.in.

Example 4.
Use Figure 1 below and the enlarged scale drawing to justify why the area of the scale drawing is k² times the area of the original figure.

Answer:
Area of Figure 1: Area of scale drawing:
Area = lw Area = lw
Area = (kl)(kw)
Area = k² lw
Since the area of Figure 1 is lw, the area of the scale drawing is k² multiplied by the area of Figure 1.

Explain why the expressions (kl)(kw) and k² lw are equivalent. How do the expressions reveal different information about this situation?
Answer:
(kl)(kw) is equivalent to klkw by the associative property, which can be written kklw using the commutative property. This is sometimes known as “any order, any grouping.” kklw is equal to k² lw because k × k = k². (kl)(kw) shows the area as the product of each scaled dimension, while k² lw shows the area as the scale factor squared, times the original area (lw).

Eureka Math Grade 7 Module 4 Lesson 15 Exercise Answer Key

Opening Exercise
For each diagram, Drawing 2 is a scale drawing of Drawing 1. Complete the accompanying charts. For each drawing, identify the side lengths, determine the area, and compute the scale factor. Convert each scale factor into a fraction and percent, examine the results, and write a conclusion relating scale factors to area.

Answer:

Answer:

The length of each side in Drawing 1 is 12 units, and the length of each side in Drawing 2 is 6 units.


Answer:

Scale factor: \(\frac{1}{2}\)
Quotient of areas: \(\frac{1}{4}\)
Conclusion: (\(\frac{1}{2}\))(\(\frac{1}{2}\)) = (\(\frac{1}{2}\))² = \(\frac{1}{4}\)
The quotient of the areas is equal to the square of the scale factor.

Exercise 1.
The Lake Smith basketball team had a team picture taken of the players, the coaches, and the trophies from the season. The picture was 4 inches by 6 inches. The team decided to have the picture enlarged to a poster and then enlarged again to a banner measuring 48 inches by 72 inches.
a. Sketch drawings to illustrate the original picture and enlargements.
Answer:

b. If the scale factor from the picture to the poster is 500%, determine the dimensions of the poster.
Answer:
Quantity = Percent × Whole
Poster height = Percent × Picture height
Poster height = 500% × 4 in.
Poster height = (5.00)(4 in.)
Poster height = 20 in.

Quantity = Percent × Whole
Poster width = Percent × Picture width
Poster width = 500% × 6 in.
Poster width = (5.00)(6 in.)
Poster width = 30 in.

The dimensions of the poster are 20 in. by 30 in.

c. What scale factor is used to create the banner from the picture?
Answer:
Quantity = Percent × Whole
Banner width = Percent × Picture width
72 = Percent × 6
\(\frac{72}{6}\) = Percent
12 = 1,200%

Quantity = Percent × Whole
Banner height = Percent × Picture height
48 = Percent × 4
\(\frac{48}{4}\) = Percent
12 = 1,200%

The scale factor used to create the banner from the picture is 1,200%.

d. What percent of the area of the picture is the area of the poster? Justify your answer using the scale factor and by finding the actual areas.
Answer:
Area of picture:
A = lw
A = (4)(6)
A = 24
Area = 24 sq.in.

Area of poster:
A = lw
A = (20)(30)
A = 600
Area = 600 sq.in.

Quantity = Percent × Whole
Area of Poster = Percent × Area of Picture
600 = Percent × 24
\(\frac{600}{24}\) = Percent
25 = 2,500%

Using scale factor:
Scale factor from picture to poster was given earlier in the problem as 500% = \(\frac{500}{100}\) = 5.
The area of the poster is the square of the scale factor times the corresponding area of the picture. So, the area of the poster is ????,????????????% the area of the original picture.

e. Write an equation involving the scale factor that relates the area of the poster to the area of the picture.
Answer:
Quantity = Percent × Whole
Area of Poster = Percent × Area of Picture
A = 2,500% p
A = 25p

f. Assume you started with the banner and wanted to reduce it to the size of the poster. What would the scale factor as a percent be?
Answer:
Banner dimensions: 48 in. × 72 in.
Poster dimensions: 20 in. × 30 in.
Quantity = Percent × Whole
Poster = Percent × Banner
30 = Percent × 72
\(\frac{30}{72}\) = \(\frac{5}{12}\) = \(\frac{5}{12}\) × 100% = 41 \(\frac{2}{3}\)%

g. What scale factor would be used to reduce the poster to the size of the picture?
Answer:
Poster dimensions: 20 in. × 30 in.
Picture dimensions: 4 in. × 6 in.
Quantity = Percent × Whole
Picture width = Percent × Poster width
6 = Percent × 30
\(\frac{6}{30}\) = \(\frac{1}{5}\) = 0.2 = 20%

Eureka Math Grade 7 Module 4 Lesson 15 Problem Set Answer Key

Question 1.
What percent of the area of the larger circle is shaded?

a. Solve this problem using scale factors.
Answer:
Scale factors:
Shaded small circle: radius = 1 unit
Shaded medium circle: radius = 2 units
Large circle: radius = 3 units, area = A
Area of small circle: (\(\frac{1}{3}\))² A = \(\frac{1}{9}\) A
Area of medium circle: (\(\frac{2}{3}\))² A = \(\frac{4}{9}\) A
Area of shaded region: \(\frac{1}{9}\) A + 4/9 A = \(\frac{5}{9}\) A = \(\frac{5}{9}\) A × 100% = 55 \(\frac{5}{9}\)%A
The area of the shaded region is 55 \(\frac{5}{9}\)% of the area of the entire circle.

b. Verify your work in part (a) by finding the actual areas.
Areas:
Answer:
Small circle: A = πr²
A = π(1 unit)²
A = 1π unit²
Medium circle: A = πr²
A = π(2 units)²
A = 4π units²
Area of shaded circles: 1π unit² + 4π units² = 5π units²
Large circle: A = πr²
A = π(3 units)²
A = 9π units²
Percent of shaded to large circle: \(\frac{5 \pi \text { units }^{2}}{9 \pi \text { units }^{2}}\) = \(\frac{5}{9}\) = \(\frac{5}{9}\) × 100% = 55 \(\frac{5}{9}\)%

Question 2.
The area of the large disk is 50.24 units².

a. Find the area of the shaded region using scale factors. Use 3.14 as an estimate for π.
Answer:
Radius of small shaded circles = 1 unit
Radius of larger shaded circle = 2 units
Radius of large disk = 4 units
Scale factor of shaded region:
Small shaded circles: \(\frac{1}{4}\)
Large shaded circle: \(\frac{2}{4}\)
If A represents the area of the large disk, then the total shaded area:
(\(\frac{1}{4}\))² A + (\(\frac{1}{4}\))² A + (\(\frac{2}{4}\))² A
= \(\frac{1}{16}\) A + \(\frac{1}{16}\) A + \(\frac{4}{16}\) A
= \(\frac{6}{16}\) A
= \(\frac{6}{16}\)(50.24 units²)
The area of the shaded region is 18.84 units².

b. What percent of the large circular region is unshaded?
Answer:
Area of the shaded region is 18.84 square units. Area of total is 50.24 square units. Area of the unshaded region is 31.40 square units. Percent of large circular region that is unshaded is
\(\frac{31.4}{50.24}\) = \(\frac{5}{8}\) = 0.625 = 62.5%.

Question 3.
Ben cut the following rockets out of cardboard. The height from the base to the tip of the smaller rocket is 20 cm. The height from the base to the tip of the larger rocket is 120 cm. What percent of the area of the smaller rocket is the area of the larger rocket?

Answer:
Height of smaller rocket: 20 cm
Height of larger rocket: 120 cm
Scale factor:
Quantity = Percent × Whole
Actual height of larger rocket = Percent × height of smaller rocket
120 = Percent × 20
6 = Percent
600%
Area of larger rocket:
(scale factor)² (area of smaller rocket)
(6)² (area of smaller rocket)
36A
36 = 36 × 100% = 3,600%
The area of the larger rocket is 3,600% the area of the smaller rocket.

Question 4.
In the photo frame depicted below, three 5 inch by 5 inch squares are cut out for photographs. If these cut-out regions make up 3/16 of the area of the entire photo frame, what are the dimensions of the photo frame?

Answer:
Since the cut-out regions make up \(\frac{3}{16}\) of the entire photo frame, then each cut-out region makes up (\(\frac{\frac{3}{16}}{3}\) = \(\frac{1}{16}\) of the entire photo frame.
The relationship between the area of the scale drawing is
(square factor)² × area of original drawing.

The area of each cut-out is \(\frac{1}{16}\) of the area of the original photo frame. Therefore, the square of the scale factor is \(\frac{1}{16}\). Since (\(\frac{1}{4}\))² = \(\frac{1}{16}\), the scale factor that relates the cut-out to the entire photo frame is \(\frac{1}{4}\), or 25%.
To find the dimensions of the square photo frame:
Quantity = Percent × Whole
Small square side length = Percent × Photo frame side length
5 in. = 25% × Photo frame side length
5 in. = \(\frac{1}{4}\) × Photo frame side length
4(5) in. = 4(\(\frac{1}{4}\)) × Photo frame side length
20 in. = Photo frame side length
The dimensions of the square photo frame are 20 in. by 20 in.

Question 5.
Kelly was online shopping for envelopes for party invitations and saw these images on a website.

The website listed the dimensions of the small envelope as 6 in. by 8 in. and the medium envelope as 10 in. by 13 \(\frac{1}{3}\) in.
a. Compare the dimensions of the small and medium envelopes. If the medium envelope is a scale drawing of the small envelope, what is the scale factor?
Answer:
To find the scale factor,
Quantity = Percent × Whole
Medium height = Percent × small height
10 = Percent × 6
\(\frac{10}{6}\) = \(\frac{5}{3}\) = \(\frac{5}{3}\) × 100% = 166 \(\frac{2}{3}\)%

Quantity = Percent × Whole
Medium width = Percent × Small width
13 \(\frac{1}{3}\) = Percent × 8
\(\frac{13 \frac{1}{3}}{8}\) = \(\frac{5}{3}\) = \(\frac{5}{3}\) × 100% = 166 \(\frac{2}{3}\)%

b. If the large envelope was created based on the dimensions of the small envelope using a scale factor of 250%, find the dimensions of the large envelope.
ans;:
Scale factor is 250%, so multiply each dimension of the small envelope by 2.50.
Large envelope dimensions are as follows:
(6 in.)(2.5) = 15 in.
(8 in.)(2.5) = 20 in.

c. If the medium envelope was created based on the dimensions of the large envelope, what scale factor was used to create the medium envelope?
Answer:
Scale factor:
Quantity = Percent × Whole
Medium = Percent × Large
10 = Percent × 15
\(\frac{10}{15}\) = Percent
\(\frac{2}{3}\) = \(\frac{2}{3}\) × 100% = 66 \(\frac{2}{3}\)%

Quantity = Percent × Whole
Medium = Percent × Large
13 \(\frac{1}{3}\) = Percent × 20
\(\frac{13 \frac{1}{3}}{20}\) = Percent
\(\frac{2}{3}\) = \(\frac{2}{3}\) × 100% = 66 \(\frac{2}{3}\)%

d. What percent of the area of the larger envelope is the area of the medium envelope?
Answer:
Scale factor of larger to medium: 66 \(\frac{2}{3}\)% = \(\frac{2}{3}\)
Area: (\(\frac{2}{3}\))² = \(\frac{4}{9}\) = \(\frac{4}{9}\) × 100% = 44 \(\frac{4}{9}\)%
The area of the medium envelope is 44 \(\frac{4}{9}\)% of the larger envelope.

Eureka Math Grade 7 Module 4 Lesson 15 Exit Ticket Answer Key

Question 1.
Write an equation relating the area of the original (larger) drawing to its smaller scale drawing. Explain how you determined the equation. What percent of the area of the larger drawing is the smaller scale drawing?

Answer:
Scale factor:
Quantity = Percent × Whole
Scale Drawing Length = Percent × Original Length
6 = Percent × 15
\(\frac{6}{15}\) = \(\frac{2}{5}\) = \(\frac{4}{10}\) = 0.4
The area of the scale drawing is equal to the square of the scale factor times the area of the original drawing. Using A to represent the area of the original drawing, then the area of the scale is
(\(\frac{4}{10}\))² A = \(\frac{16}{100}\) A.
As a percent, \(\frac{16}{100}\) A = 0.16A .
Therefore, the area of the scale drawing is 16% of the area of the original drawing.