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

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

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?

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.

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.

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?
(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).

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.

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

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.

b. If the scale factor from the picture to the poster is 500%, determine the dimensions of the poster.
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?
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.
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 𝟐,𝟓𝟎𝟎% 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.
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?
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?
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.
Scale factors:
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:
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.
Radius of large disk = 4 units
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?
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?

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?

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?
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?
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?
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?

$$\frac{6}{15}$$ = $$\frac{2}{5}$$ = $$\frac{4}{10}$$ = 0.4
($$\frac{4}{10}$$)2 A = $$\frac{16}{100}$$ A.
As a percent, $$\frac{16}{100}$$ A = 0.16A .