## Engage NY Eureka Math Geometry Module 3 Lesson 9 Answer Key

### Eureka Math Geometry Module 3 Lesson 9 Opening Exercise Answer Key

a. For each pair of similar figures, write the ratio of side lengths a: b or C: d that compares one pair of corresponding sides. Then, complete the third column by writing the ratio that compares the areas of the similar figures. Simplify ratios when possible.

Answer:

b.

i. State the relationship between the ratio of sides a: b and the ratio of the areas Area(A): Area(B).

Answer:

When the ratio of side lengths is a: b, then the ratio of the areas is a^{2} : b^{2}.

ii. Make a conjecture as to how the ratio of sides a: b will be related to the ratio of volumes Volume(S): Volume(T). Explain.

Answer:

When the ratio of side lengths is s: t, then the ratio of the volumes will probably be s^{3} : t^{3}. Area is twodimensionaI, and the comparison of areas was raised to the second power. Since volume is three-dimensional, I think the comparison of volumes will be raised to the third power.

c. What does it mean for two solids In three-dimensional space to be similar?

Answer:

It means that a sequence of basic rigid motions and dilations maps one figure onto the other.

### Eureka Math Geometry Module 3 Lesson 9 Exercise Answer Key

Exercise 1.

Each pair of solids shown below is similar. Write the ratio of side lengths a: b comparing one pair of corresponding sides. Then, complete the third column by writing the ratio that compares volumes of the similar figures. Simplify ratios when possible.

Answer:

Exercise 2.

Use the triangular prism shown below to answer the questions that follow.

a. Calculate the volume of the triangular prism.

Answer:

V = \(\frac{1}{2}\)(3)(3)(5)

V = \(\frac{45}{2}\)

V= 22.5

b. If one side of the triangular base is scaled by a factor of 2, the other side of the triangular base is scaled by a factor of 4, and the height of the prism is scaled by a factor of 3, what are the dimensions of the scaled triangular prism?

Answer:

The new dimensions of the base are 6 by 12, and the height is 15.

c. Calculate the volume of the scaled triangular prism.

Answer:

V = \(\frac{1}{2}\)6(12)(15)

V = \(\frac{1080}{2}\)

V = 540

d. Make a conjecture about the relationship between the volume of the original triangular prism and the scaled triangular prism.

Answer:

Answers vary. Accept any reasonable response. The correct response is that the volume of the scaled figure is equal to the volume of the original figure multiplied by the product of the scaled factors. For this specific problem, 540 = 2(3)(4)(22.5).

e. Do the volumes of the figures have the same relationship as was shown in the figures in Exercise 1? Explain.

Answer:

No, the figure was scaled differently in each perpendicular direction, so the volumes are related by the product of the scale factors 2 · 3 · 4 = 24.

Exercise 3.

Use the rectangular prism shown below to answer the questions that follow.

a. Calculate the volume of the rectangular prism.

Answer:

V = 1(8)(12)

V = 96

b. If one side of the rectangular base is scaled by a factor of \(\frac{1}{2}\), the other side of the rectangular base is scaled by a factor of 24, and the height of the prism is scaled by a factor of \(\frac{1}{3}\), what are the dimensions of the scaled rectangular prism?

Answer:

Note that some students may have selected different sides of the base to scale corn pared to the solution below. Regardless, the result in part (c) is the same.

The dimensions of the scaled rectangular prism are 4, 24, and 4.

OR

The dimensions of the scaled rectangular prism are 192, \(\frac{1}{2}\), and 4.

c. Calculate the volume of the scaled rectangular prism.

Answer:

V = 4(24)(4)

V = 384

d. Make a conjecture about the relationship between the volume of the original rectangular prism and the scaled rectangular prism.

Answer:

Answers vary. Accept any reasonable response. The correct response is that the volume of the scaled figure is equal to the volume of the original figure multiplied by the product of the scaled factors. For this specific problem, 384 = \(\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)\) (24)(96).

Exercise 4.

A manufacturing company needs boxes to ship their newest widget, which measures 2 × 4 × 5 in^{3}. Standard size boxes, 5-inch cubes, are inexpensive but require foam packaging so the widget is not damaged in transit. Foam packaging costs $0.03 per cubic inch. Specially designed boxes are more expensive but do not require foam packing. If the standard size box costs $0. 80 each, and the specially designedbox costs $3. 00 each, which kind of box should the company choose? Explain your answer.

Answer:

Volume of foam packaging needed in a standard box:

125 – (2 · 4 · 5) = 85; 85 in^{3} of foam packaging needed for standard-sized box.

Total cost of widget packed in a standard box:

85(0. 03) + 0.80 = 3.35;total cost is $3.35.

Therefore, the specially designed packages for $3.00 each are more cost effective.

### Eureka Math Geometry Module 3 Lesson 9 Problem Set Answer Key

Question 1.

Coffees sold at a deli come in similar-shaped cups. A small cup has a height of 4. 2’, and a large cup has a height of 5”. The large coffee holds 12 fluid ounces. How much coffee is in a small cup? Round your answer to the nearest tenth of an ounce.

Answer:

The scale factor of the smaller cup is \(\frac{4.2}{5}\), or 0.84. The cups are similar, so the scale factor is the same in all three perpendicular dimensions. Therefore, the volume of the small cup Is equal to the volume of the large cup times

(0. 84)^{3}, or 0.592 704.

Volume = (12)(0. 592 704)

Volume = 7.112448

The small coffee cup contains approximately 7.1 fluid ounces.

Question 2.

Right circular cylinder A has volume 2, 700 and radIus 3. Right circular cylinder B Is similar to cylinder A and has volume 6,400. Find the radius of cylinder B.

Answer:

Let r be the radius of cylinder B.

\(\frac{6400}{2700}=\frac{r^{3}}{3^{3}}\)

r^{3} = 63

r = 4

Question 3.

The Empire State Building is a 102-story skyscraper. Its height Is 1.250 ft. from the ground to the roof The length and width of the building are approximately 42411 and 187 ft, respectively. A manufacturing company plans to make a miniature version of the building and sell cases of them to souvenir shops.

a. The miniature version is just \(\frac{1}{2500}\) of the size of the original. What are the dimensions of the miniature Empire State Building?

Answer:

The height is 0.5 ft. the length is about 0. 17 ft. and the width is 0. 07 ft.

b. Determine the volume of the miniature building. Explain how you determined the volume.

Answer:

Answers wiiy since the Empire State Building has an irregular shape. Some students may model the shape of the building as o rectangular pyramid, rectangular prism, or combination of the two.

By modeling with a rectangular prism:

Volume = [(017)(0.07)] ‘(0.5)

Volume = 0.00595.

The volume of the miniature building is approximately 0.00595 ft^{3}.

By modeling with o rectangular pyramid:

Volume = \(\frac{1}{3}\)[(0. 17)(0.07) · (0.5)

Volume = 0.00198.

The volume of the miniature building is approximately 0.00198 ft^{3}.

By modeling with the composition of a rectangular prism and a rectangular pyramid:

Combination of prisms; answers vary. The solution that follows considers the highest 0.4 ft of the building to be a pyramid and the lower 0.4 ft. of the building to be rectangular

Volume = \(\frac{1}{3}\)(0. 17 × 0. 07 × 0. 5) + 0.17 × 0.07 × 0.5

Volume = \(\frac{4}{3}\)(0. 17 × 0.07 × 0.5)

Volume = 0.00793

The volume of the miniature building is approximately 0.00793 ft^{3}.

Question 4.

If a right square pyramid has a 2 × 2 square base and height 1, then its volume is \(\frac{4}{3}\). Use this information to find the volume of a right rectangular prism with base dimensions a × b and height h.

Answer:

Scale the right square pyramid by \(\frac{a}{2}\) and \(\frac{b}{2}\) the directions determined by the sides of the square base and by h in the direction perpendicular to the base. This turns the right square pyramid into a right rectangular pyramid with rectangular base of side lengths a and b and height h that has volume

\(\frac{4}{3} \cdot \frac{a}{2} \cdot \frac{b}{2}\) · h = \(\frac{1}{3}\)abh = \(\frac{1}{3}\) area of base × height.

Question 5.

The following solids are similar. The volume of the first solid is 100. Find the volume of the second solid.

Answer:

(1. 1)^{3}. 100 = 133. 1

Question 6.

A general cone has a height of 6. What fraction of the cone’s volume Is between a plane containing the base and a parallel plane halfway between the vertex of the cone and the base plane?

Answer:

The smaller top cone is similar to the whole cone and has volume \(\left(\frac{1}{2}\right)^{3}\) or \(\frac{1}{8}\) of the volume of the whole cone. So, the region between the two planes is the remaining part of the volume and, therefore, has \(\frac{7}{8}\) the volume of the whole cone.

Question 7.

A company uses rectangular boxes to package small electronic components for shipping. The box that is currently used can contain 500 of one type of component. The company wants to package twice as many pieces per box. Michael thinks that the box will hold twice as much if its dimensions are doubled. Shawn disagrees and says that Michael’s Idea provides a box that is much too large for 1,000 pieces. Explain why you agree or disagree with one or either of the boys. What would you recommend to the company?

Answer:

If the box’s dimensions were all doubled, the volume of the box should be enough to contain 2 or 8 times the number of components that the current box holds, which is far too large for 1,000 pieces. To double the volume of the box, the company could double the width, the height, or the length of the box, or extend all dimensions of the box by a scale factor of \(\sqrt[3]{2}\).

Question 8.

A dairy facility has bulk milk tanks that are shaped like right circular cylinders. They have replaced one of their bulk milk tanks with three smaller tanks that have the same height as the original but \(\frac{1}{3}\) the radius. Do the new tanks hold the same amount of milk as the original tank? If not, explain how the volumes compare.

Answer:

If the original tank had a radius of r and a height of h, then the volume of the original tank would be πr^{2}h. The radii of the new tanks would be r, and the volume of the new tanks combined would be 3 (π (\(\frac{1}{3}\)r)^{2} h), or \(\frac{1}{3}\)πr^{2} h.

The bases of the new tanks were scaled down by a scale factor of \(\frac{1}{3}\) in two directions, so the area of the base of each tank is \(\frac{1}{9}\) the area of the original tank. Combining the three smaller tanks provides a base area that is \(\frac{3}{9}\) or \(\frac{1}{3}\) the base area of the original tank. The ratio of the volume of the original tank to the three replacement tanks is (πr^{2} h): \(\frac{1}{3}\)(πr^{2} h) or 3: 1, which means the original tank holds three times as much milk as the replacements.

### Eureka Math Geometry Module 3 Lesson 9 Exit Ticket Answer Key

Question 1.

Two circular cylinders are similar. The ratio of the areas of their bases is 9:4. Find the ratio of the volumes of the similar solids.

Answer:

If the solids are similar, then their bases are similar as well. The areas of similar plane figures are related by the square of the scale factor relating the figures, and if the ratios of the areas of the bases is 9:4, then the scale factor of the two solids must be \(\sqrt{\frac{9}{4}}\), or \(\frac{3}{2}\). The ratio of lengths in the two solids is, therefore, 3:2.

By the scaling principle for volumes, the ratio of the volumes of the solids is the ratio 3^{3} : 2^{3}, or 27 : 8.

Question 2.

The volume of a rectangular pyramid is 60. The width of the base is then scaled by a factor of 3, the length of the base is scaled by a factor of \(\frac{5}{2}\), and the height of the pyramid is scaled such that the resulting image has the same volume as the original pyramid. Find the scale factor used for the height of the pyramid.

Answer:

The scaling principle for volumes says that the volume of a solid scaled in three perpendicular directions is equal to the area of the original solid times the product of the scale factors used in each direction. Since the volumes of the two solids are the same, it follows that the product of the scale factors in three perpendicular directions must be 1.

Lets represent the scale factor used for the height of the image:

Volume(A’) = (3 · \(\frac{5}{2}\) · s) (Area(A))

60 = \(\frac{15}{2}\) · s · (60)

s = \(\frac{2}{15}\)

The scale factor used to scale the height of the pyramid is \(\frac{2}{15}\).