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Engage NY Eureka Math 7th Grade Module 6 Mid Module Assessment Answer Key

Eureka Math Grade 7 Module 6 Mid Module Assessment Task Answer Key

Question 1.
In each problem, set up and solve an equation for the unknown angles.
a. Four lines meet at a point. Find the measures m° and n°.

b. Two lines meet at the vertex of two rays. Find the measures m° and n°.

c. Two lines meet at a point that is the vertex of two rays. Find the measures m° and n°.

d. Three rays have a common vertex on a line. Find the measures m° and n°.

Answer:

a. n° = 90° , vertical angles

25° + (90°) + 40° + m° = 180°
155° + m° = 180°
155° – 155° + m° = 180° – 155°
m° = 25°

b. 50° + 90° + n° = 180°
140° + n° = 180°
140° – 140° + n° = 180° – 140°
n° = 40°

m° + 50° = 90°
m° + 50° – 50° = 90° – 50°
m° = 40°

c. m° + 52° = 90°
m° + 52° – 52° = 90° – 52°
m° = 38°

40 + 52 + (38) + n° = 180
130 + n° = 180
130 – 130 + n° = 180 – 130
n° = 50°

d. n° + 62° = 90°
n° + 62° – 62° = 90° – 62°
n° = 28°”

m° + 62° + (28°) + 27° = 180°
m° + 117° = 180
m° + 117° – 117° = 180° – 117°
m° = 63°

Question 2.
Use tools to construct a triangle based on the following given conditions.
a. If possible, use your tools to construct a triangle with angle measurements 20°, 55°, and 105°, and leave evidence of your construction. If it is not possible, explain why.
b. Is it possible to construct two different triangles that have the same angle measurements? If it is, construct examples that demonstrate this condition, and label all angle and length measurements. If it is not possible, explain why.
Answer:
a. Solutions will vary. An example of a correctly constructed triangle is shown here.

b. Solutions will vary; refer to the rubric.

Question 3.
In each of the following problems, two triangles are given. For each: (1) state if there are sufficient or insufficient conditions to show the triangles are identical, and (2) explain your reasoning.


Answer:

a. The triangles are identical by the two angles and included side condition. The marked side is between the given angles.
△ABC ↔ △YXZ

b. There is insufficient evidence to determine that the triangles are identical. In △DEF , the marked side is between the marked angles, but in △ABC , the marked side is not between the marked angles.

c. The triangles are identical by the two sides and included angle condition. △DEF ↔ △GIH

d. The triangles are not identical. In △ABC , the marked side is opposite ∠B . In △WXY , the marked side is opposite ∠W . ∠B and ∠W are not necessarily equal in measure.

Question 4.
Use tools to draw rectangle ABCD with AB = 2 cm and BC = 6 cm. Label all vertices and measurements.
Answer:

Question 5.
The measures of two complementary angles have a ratio of 3:7. Set up and solve an equation to determine the measurements of the two angles.
Answer:

3x + 7x = 90
10x = 90
(\(\frac{1}{10}\))10x = (\(\frac{1}{10}\))90
x = 9

Measure of Angle 1: 3(9) = 27 . The measure of the first angle is 27° .
Measure of Angle 2: 7(9) = 63 . The measure of the second angle is 63° .

Question 6.
The measure of the supplement of an angle is 12° less than the measure of the angle. Set up and solve an equation to determine the measurements of the angle and its supplement.
Answer:

Let y° be the number of degrees in the angle.

y + (y – 12) = 180
2y – 12 = 180
2y – 12 + 12 = 180 + 12
2y = 192
(\(\frac{1}{2}\))2y = (\(\frac{1}{2}\))192
y = 96

Measure of the angle: 96°
Measure of its supplement: (96)° – 12° = 84°

Question 7.
Three angles are at a point. The ratio of two of the angles is 2:3, and the remaining angle is 32° more than the larger of the first two angles. Set up and solve an equation to determine the measures of all three angles.
Answer:

2x + 3x + (3x + 32) = 360
8x + 32 = 360
8x + 32 – 32 = 360 – 32
8x = 328
(\(\frac{1}{8}\))8x = (\(\frac{1}{8}\))328
x = 41

Measure of Angle 1: 2(41)° = 82°
Measure of Angle 2: 3(41)° = 123°
Measure of Angle 3: 3(41)° + 32° = 155°

Question 8.
Draw a right triangle according to the following conditions, and label the provided information. If it is not possible to draw the triangle according to the conditions, explain why. Include a description of the kind of figure the current measurements allow. Provide a change to the conditions that makes the drawing feasible.
a. Construct a right triangle ABC so that AB = 3 cm, BC = 4 cm, and CA = 5 cm; the measure of angle B is 90°.
b. Construct triangle DEF so that DE = 4 cm, EF = 5 cm, and FD = 11 cm; the measure of angle D is 50°.
Answer:
a.

b. It is not possible to draw this triangle because the lengths of the two shorter sides do not sum to be greater than the longest side. In this situation, the total lengths of \(\overline{D E}\) and \(\overline{E F}\) are less than the length of \(\overline{F D}\); there is no way to arrange \(\overline{D E}\) and \(\overline{E F}\) so that they meet. If they do not meet, there is no arrangement of three non-collinear vertices of a triangle; therefore, a triangle cannot be formed. I would change \(\overline{E F}\)to 9 cm instead of 5 cm so that the three sides would form a triangle.

Engage NY Eureka Math 7th Grade Module 6 Lesson 27 Answer Key

Eureka Math Grade 7 Module 6 Lesson 27 Example Answer Key

Example 1.
A swimming pool holds 10,000 ft³ of water when filled. Jon and Anne want to fill the pool with a garden hose. The garden hose can fill a five – gallon bucket in 30 seconds. If each cubic foot is about 7.5 gallons, find the flow rate of the garden hose in gallons per minute and in cubic feet per minute. About how long will it take to fill the pool with a garden hose? If the hose is turned on Monday morning at 8:00 a.m., approximately when will the pool be filled?
Answer:
→ If the hose fills a 5 – gallon bucket in 30 seconds, how much would it fill in 1 minute? Find the flow rate in gallons per minute.
→ It would fill 10 gallons in 1 minute; therefore, the flow rate is 10 \(\frac{\mathrm{gal}}{\mathrm{min}}\).
→ Find the flow rate in cubic feet per minute.
Convert gallons to cubic feet: (10 gal.) \(\frac{1}{7.5} \frac{\mathrm{ft}^{3}}{\mathrm{gal}}\) = 1 \(\frac{1}{3}\)ft³
Therefore, the flow rate of the garden hose in cubic feet per minute is 1 \(\frac{1}{3} \frac{\mathrm{ft}^{3}}{\mathrm{~min}}\)

→ How many minutes would it take to fill the 10,000 ft³ pool?
\(\frac{10,000 \mathrm{ft}^{3}}{1 \frac{1}{3}\left(\frac{\mathrm{ft}^{3}}{1 \mathrm{~min}}\right)}\) = 7,500 min.
→ How many days and hours is 7,500 minutes?
(7,500 min.) \(\frac{1 \mathrm{~h}}{60 \mathrm{~min}}\) = 125 h, or 5 days and 5 hours

→ At what time will the pool be filled?
The pool begins to fill at 8:00 a.m. on Monday, so 5 days and 5 hours later on Saturday at 1:00 p.m., the pool will be filled.

Example 2.
A square pipe (a rectangular prism – shaped pipe) with inside dimensions of 2 in.×2 in. has water flowing through it at a flow speed of 3 \(\frac{\mathrm{ft}}{\mathrm{s}}\). The water flows into a pool in the shape of a right triangular prism, with a base in the shape of a right isosceles triangle and with legs that are each 5 feet in length. How long will it take for the water to reach a depth of
4 feet?
Answer:
→ This problem is slightly different than the previous example. In this example, we are given a flow speed (also known as linear flow speed) instead of a flow rate.

→ What do you think the term flow speed means based on what you know about flow rate and what you know about the units of each?
Flow speed in this problem is measured in feet per second. Flow rate in Example 1 is measured in gallons per minute (or cubic feet per minute).
Flow speed is the distance that the liquid moves in one unit of time.

→ Now, let’s go back to our example. The water is traveling at a flow speed of 3 \(\frac{\mathrm{ft}}{\mathrm{s}}\). This means that for each second that the water is flowing out of the pipe, the water travels a distance of 3 ft. Now, we need to determine the volume of water that passes per second; in other words, we need to find the flow rate.

→ Each second, the water in a cross – section of the pipe will travel 3 ft. This is the same as the volume of a right rectangular prism with dimensions 2 in.×2 in.×3 ft.

→ The volume of this prism in cubic feet is \(\frac{1}{6}\) ft.×\(\frac{1}{6}\) ft.×3 ft. = \(\frac{1}{12}\) ft³, and the volume of water flowing out of the pipe every second is \(\frac{1}{12}\) ft³. So, the flow rate is \(\frac{1}{12}\) \(\frac{\mathrm{ft}^{3}}{\mathrm{~s}}\).

→ Seconds is a very small unit of time when we think about filling up a pool. What is the flow rate in cubic feet per minute?
\(\frac{\frac{1}{12} \mathrm{ft}^{3}}{1 \mathrm{~s}} \cdot \frac{60 \mathrm{~s}}{1 \mathrm{~min}}\) = 5 \(\frac{\mathrm{ft}^{3}}{\min }\)

→ What is the volume of water that will be in the pool once the water reaches a depth of 4 ft.?
The volume of water in the pool will be \(\frac{1}{2}\)(5 ft.)(5 ft.)(4 ft.) = 50 ft³.

→ Now that we know our flow rate and the total volume of water in the pool, how long will it take for the pool to fill to a depth of 4 ft.?
\(\frac{50 \mathrm{ft}^{3}}{5 \frac{\mathrm{ft}^{3}}{\mathrm{~min}}}\) = 10 min.
It will take 10 minutes to fill the pool to a depth of 4 ft

Eureka Math Grade 7 Module 6 Lesson 27 Exercise Answer Key

Exercise 1.
A park fountain is about to be turned on in the spring after having been off all winter long. The fountain flows out of the top level and into the bottom level until both are full, at which point the water is just recycled from top to bottom through an internal pipe. The outer wall of the top level, a right square prism, is five feet in length; the thickness of the stone between outer and inner wall is 1 ft.; and the depth is 1 ft. The bottom level, also a right square prism, has an outer wall that is 11 ft. long with a 2 ft. thickness between the outer and inner wall and a depth of 2 ft. Water flows through a 3 in.×3 in. square pipe into the top level of the fountain at a flow speed of 4 \(\frac{\mathrm{ft}}{\mathrm{s}}\). Approximately how long will it take for both levels of the fountain to fill completely?

Answer:
Volume of top:
3 ft.×3 ft.×1 ft. = 9 ft³
Volume of bottom:
(7 ft.×7 ft.×2 ft.) – (5 ft.×5 ft.×2 ft.) = 48 ft³
Combined volume of both levels:
9 ft³ + 48 ft³ = 57 ft³

With a flow speed of 4 \(\frac{\mathrm{ft}}{\mathrm{s}}\) through a 3 in.×3 in. square pipe, the volume of water moving through the pipe in one second is equivalent to the volume of a right rectangular prism with dimensions 3 in.×3 in.×4 ft. The volume in feet is
\(\frac{1}{4}\) ft.×\(\frac{1}{4}\) ft.×4 ft. = \(\frac{1}{4}\) ft³. Therefore, the flow rate is \(\frac{1}{2}\) because \(\frac{1}{4}\) ft³ of water flows every second.

Volume of water that will flow in one minute:
\(\frac{\frac{1}{4} \mathrm{ft}^{3}}{1 \mathrm{~s}} \cdot \frac{60 \mathrm{~s}}{1 \mathrm{~min}}\) = 15 \(\frac{\mathrm{ft}^{3}}{\mathrm{~min}}\)
Time needed to fill both fountain levels: \(\frac{57 \mathrm{ft}^{3}}{15 \frac{\mathrm{ft}^{3}}{\min }}\) = 3.8 min.; it will take 3.8 minutes to fill both fountain levels.

Exercise 2.
A decorative bathroom faucet has a 3 in.×3 in. square pipe that flows into a basin in the shape of an isosceles trapezoid prism like the one shown in the diagram. If it takes one minute and twenty seconds to fill the basin completely, what is the approximate speed of water flowing from the faucet in feet per second?

Answer:
Volume of the basin in cubic inches:
\(\frac{1}{2}\) (3 in. + 15 in.)(10 in.)×4.5 in. = 405 in³
Approximate volume of the basin in cubic feet:
(405 in³ )(\(\frac{1 \mathrm{ft}^{3}}{1.728 \mathrm{in}^{3}}\)) = 0.234375 ft³
Based on the rate of water flowing out the faucet, the volume of water can also be calculated as follows:
Let s represent the distance that the water is traveling in 1 second.
3 in. ∙ 3 in. ∙ s = \(\frac{1}{4}\) ft. ∙ \(\frac{1}{4}\) ft. ∙ s ft. = 0.234375 ft³
Therefore, the speed of the water flowing from the faucet is 3.75 \(\frac{\mathrm{ft}}{\mathrm{s}}\).

Eureka Math Grade 7 Module 6 Lesson 27 Problem Set Answer Key

Question 1.
Harvey puts a container in the shape of a right rectangular prism under a spot in the roof that is leaking. Rainwater is dripping into the container at an average rate of 12 drops a minute. The container Harvey places under the leak has a length and width of 5 cm and a height of 10 cm. Assuming each raindrop is roughly 1 cm³, approximately how long does Harvey have before the container overflows?
Answer:
Volume of the container in cubic centimeters:
5 cm×5 cm×10 cm = 250 cm³
Number of minutes until the container is filled with rainwater:
(250 cm³)(\(\frac{1 \mathrm{~min}}{12 \mathrm{~cm}^{3}}\))≈20.8 min.

Question 2.
A large square pipe has inside dimensions 3 in.×3 in., and a small square pipe has inside dimensions 1 in.×1 in. Water travels through each of the pipes at the same constant flow speed. If the large pipe can fill a pool in 2 hours, how long will it take the small pipe to fill the same pool?
Answer:
If s is the length that the water travels in one minute, then in one minute the large pipe provides \(\frac{1}{4}\) ft. ∙ \(\frac{1}{4}\) ft. ∙ s ft. of water. In one minute, the small pipe provides one – ninth as much, \(\frac{1}{12}\) ft. ∙ \(\frac{1}{12}\) ft. ∙ s ft. of water. Therefore, it will take the small pipe nine times as long. It will take the small pipe 18 hours to fill the pool.

Question 3.
A pool contains 12,000 ft³ of water and needs to be drained. At 8:00 a.m., a pump is turned on that drains water at a flow rate of 10 ft³ per minute. Two hours later, at 10:00 a.m., a second pump is activated that drains water at a flow rate of 8 ft³ per minute. At what time will the pool be empty?
Answer:
Water drained in the first two hours: \(\frac{10 \mathrm{ft}^{3}}{1 \mathrm{~min}}\) ∙ 120 min. = 1,200 ft³
Volume of water that still needs to be drained: 12,000 ft³ – 1,200 ft³ = 10,800 ft³
Amount of time needed to drain remaining water with both pumps working:
10,800 ft³ (\(\frac{1 \mathrm{~min}}{10 \mathrm{ft}^{3}} + \frac{1 \mathrm{~min}}{8 \mathrm{ft}^{3}}\)) = 600 min., or 10 h.
The total time needed to drain the pool is 12 hours, so the pool will drain completely at 8:00 p.m.

Question 4.
In the previous problem, if water starts flowing into the pool at noon at a flow rate of 3 ft³ per minute, how much longer will it take to drain the pool?
Answer:
At noon, the first pump will have been on for four hours, and the second pump will have been on for two hours. The cubic feet of water drained by the two pumps together at noon is
240 min.(\(\frac{10 \mathrm{ft}^{3}}{1 \mathrm{~min}}\)) + 120 min.(\(\frac{8 \mathrm{ft}^{3}}{1 \mathrm{~min}}\)) = 3,360 ft³
Volume of water that still needs to be drained:
12,000 ft³ – 3,360 ft³ = 8,640 ft²
If water is entering the pool at 3 \(\frac{\mathrm{ft}^{3}}{\mathrm{~min}}\) but leaving it at 18 \(\frac{\mathrm{ft}^{3}}{\mathrm{~min}}\), the net effect is that water is leaving the pool at 15 \(\frac{\mathrm{ft}^{3}}{\mathrm{~min}}\)
8,640 ft³ (\(\frac{1 \mathrm{~min}}{15 \mathrm{ft}^{3}}\)) = 576 min., or 9 h. and 36 min.
The pool will finish draining at 9:36 p.m. the same day. It will take an additional 1 hour and 36 minutes to drain the pool.

Question 5.
A pool contains 6,000 ft³ of water. Pump A can drain the pool in 15 hours, Pump B can drain it in 12 hours, and Pump C can drain it in 10 hours. How long will it take all three pumps working together to drain the pool?
Answer:
Rate at which Pump A drains the pool: \(\frac{1}{15}\) pool per hour
Rate at which Pump B drains the pool: \(\frac{1}{12}\) pool per hour
Rate at which Pump C drains the pool: \(\frac{1}{10}\) pool per hour
Together, the pumps drain the pool at (\(\frac{1}{15}\) + \(\frac{1}{12}\) + \(\frac{1}{10}\)) pool per hour, or \(\frac{1}{4}\) pool per hour. Therefore, it will take
4 hours to drain the pool when all three pumps are working together.

Question 6.
A 2,000 – gallon fish aquarium can be filled by water flowing at a constant rate in 10 hours. When a decorative rock is placed in the aquarium, it can be filled in 9.5 hours. Find the volume of the rock in cubic feet (1 ft³ = 7.5 gal.)
Answer:
Rate of water flow into aquarium:
\(\frac{2,000 \text { gal }}{10 \mathrm{~h}} = \frac{200 \mathrm{gal}}{1 \mathrm{~h}}\)
Since it takes half an hour less time to fill the aquarium with the rock inside, the volume of the rock is
\(\frac{200 \text { gal }}{1 \mathrm{~h}}\) = 100 gal.
Volume of the rock:
100 gal.(\(\frac{1 \mathrm{ft}^{3}}{7.5 \mathrm{gal}}\))≈13.3 ft³; the volume of the rock is approximately 13.3 ft³.

Eureka Math Grade 7 Module 6 Lesson 27 Exit Ticket Answer Key

Question 1.
Jim wants to know how much his family spends on water for showers. Water costs $1.50 for 1,000 gallons. His family averages 4 showers per day. The average length of a shower is 10 minutes. He places a bucket in his shower and turns on the water. After one minute, the bucket has 2.5 gallons of water. About how much money does his family spend on water for showers in a 30 – day month?
Answer:
Number of gallons of water in one day of showering (four ten – minute showers): 4(10 min.)(\(\frac{2.5 \mathrm{gal}}{1 \mathrm{~min}}\)) = 100 gal.
Number of gallons of water in 30 days: (30 days)(\(\frac{100 \mathrm{gal}}{1 \mathrm{~min}}\)) = 3,000 gal.
Cost of showering for 30 days: (3,000 gal.)(\(\frac{\$ 1.50}{1,000 \mathrm{gal}}\)) = $4.50
The family spends $4.50 in a 30 – day month on water for showers.

Engage NY Eureka Math 7th Grade Module 6 Lesson 26 Answer Key

Eureka Math Grade 7 Module 6 Lesson 26 Example Answer Key

Example 1.
Find the volume of the following three – dimensional object composed of two right rectangular prisms.

Answer:
Volume of object = Volume of top prism + Volume of bottom prism
Volume of top prism: Volume of bottom prism:
Volume_{top prism} = (4 m)(5 m)(5 m)
= 100 m³

Volume_{bottom prism} = (10 m)(10 m)(3 m)
= 300 m³
The volume of the object is 100 m³ + 300 m³ = 400 m³.

There are different ways the volume of a composite figure may be calculated. If the figure is like the one shown in Example 1, where the figure can be decomposed into separate prisms and it would be impossible for the prisms to share any one dimension, the individual volumes of the decomposed prisms can be determined and then summed. If, however, the figure is similar to the figure in Exercise 1, there are two possible strategies. In Exercise 1, the figure can be decomposed into two individual prisms, but a dimension is shared between the two prisms—in this case, the height. Instead of calculating the volume of each prism and then taking the sum, we can calculate the area of the entire base by decomposing it into shapes we know and then multiplying the area of the base by the height.

Example 2.
Find the volume of the right prism shown in the diagram whose base is the region between two right triangles. Use two different strategies.

Answer:
Strategy 1
The volume of the right prism is equal to the difference of the volumes of the two triangular prisms.
Volume of object = Volume_{large prism} – Volume_{small prism}
Volume of large prism:
Volume_{large prism} = (\(\frac{1}{2}\) ∙ 3 cm ∙ 4 cm)4 \(\frac{1}{2}\)cm
= 27 cm³

Volume of small prism:
Volume_{small prism} = (\(\frac{1}{2}\) ∙ 1 \(\frac{1}{2}\)cm ∙ 2 cm) 4\(\frac{1}{2}\)cm
= 6 \(\frac{3}{4}\)cm³

The volume of the object is 20 \(\frac{1}{4}\)cm³.

Strategy 2
The volume of a right prism is equal to the area of its base times its height. The base is the region between two right triangles.
Volume of object = Bh
B = Area_{large triangle} – Area_{small triangle}
Area_{large triangle} = \(\frac{1}{2}\) ∙ 3 cm ∙ 4 cm = 6 cm²
Area_{small triangle} = \(\frac{1}{2}\) ∙ 1 1\(\frac{1}{2}\)cm ∙ 2 cm = 1 \(\frac{1}{2}\)cm²
B = 6 cm² – 1 \(\frac{1}{2}\)cm² = 4 \(\frac{1}{2}\)cm²

Volume of object:
Volume_object = Bh
= (4 \(\frac{1}{2}\)cm² ∙ 4 \(\frac{1}{2}\)cm)
= 20 \(\frac{1}{4}\)cm³

The volume of the object is 20 \(\frac{1}{4}\)cm³.

→ Write a numeric expression to represent the volume of the figure in Strategy 1.
(\(\frac{1}{2}\) ∙ 3 cm ∙ 4 cm)4 \(\frac{1}{2}\)cm – (\(\frac{1}{2}\) ∙ 1 \(\frac{1}{2}\)cm ∙ 2 cm)4 \(\frac{1}{2}\)cm

→ Write a numeric expression to represent the volume of the figure in Strategy 2.
4 \(\frac{1}{2}\)cm

→ How do the numeric expressions represent the problem differently?
The first expression is appropriate to use when the volume of the smaller prism is being subtracted away from the volume of the larger prism, whereas the second expression is used when the area of the base of the composite figure is found and then multiplied by the height to determine the volume.

→ What property allows us to show that these representations are equivalent?
The distributive property

Example 3.
A box with a length of 2 ft., a width of 1.5 ft., and a height of 1.25 ft. contains fragile electronic equipment that is packed inside a larger box with three inches of styrofoam cushioning material on each side (above, below, left side, right side, front, and back).
a. Give the dimensions of the larger box.
b. Design styrofoam right rectangular prisms that could be placed around the box to provide the cushioning (i.e., give the dimensions and how many of each size are needed).
c. Find the volume of the styrofoam cushioning material by adding the volumes of the right rectangular prisms in the previous question.
d. Find the volume of the styrofoam cushioning material by computing the difference between the volume of the larger box and the volume of the smaller box.
Answer:
a. Length 2.5 ft., width 2 ft., and height 1.75 ft.

b. Possible answer: Two pieces with dimensions 2.5 ft. ∙ 2 ft. ∙ 3 in. and four pieces with dimensions 2 ft. ∙ 1.25 ft. ∙ 3 in.

c. V₁ = 2(2.5 ft. ∙ 2 ft. ∙ 0.25 ft.) = 2.5 ft³
V₂ = 4(2 ft. ∙ 1.25 ft. ∙ 0.25 ft.) = 2.5 ft³
V₁ + V₂ = 2.5 ft³ + 2.5 ft³ = 5 ft³

d. (2.5 ft. ∙ 2 ft. ∙ 1.75 ft.) – (2 ft. ∙ 1.5 ft. ∙ 1.25 ft.) = 8.75 ft³ – 3.75 ft³ = 5 ft³

Eureka Math Grade 7 Module 6 Lesson 26 Exercise Answer Key

Exercise 1.
Find the volume of the following three – dimensional figure composed of two right rectangular prisms.

Answer:
Area of Base_{back prism} = (2 in.)( 10 in.)
= 20 in.²

Area of Base_{front prism} = (6 in.)(2 in.)
= 12 in²
Area of Base = 20 in² + 12 in² = 32 in²
The volume of the object is (9 in.)(32 in² ) = 288 in³.

Exercise 2.
The right trapezoidal prism is composed of a right rectangular prism joined with a right triangular prism. Find the volume of the right trapezoidal prism shown in the diagram using two different strategies.

Answer:
Strategy 1
The volume of the trapezoidal prism is equal to the sum of the volumes of the rectangular and triangular prisms.
Volume of object = Volume of rectangular prism + Volume of triangular prism
Volume of rectangular prism: Volume of triangular prism:
Volume_{rectangular prism} = Bh
= (lw)h
= (3 cm ∙ 2 cm) ∙ 1 \(\frac{1}{2}\)cm
= 9 cm³

Volume_{triangular prism} = Bh = (\(\frac{1}{2}\) lw)h
= (\(\frac{1}{2}\) ∙ 3 cm ∙ 2\(\frac{1}{4}\)cm) ∙ 1 \(\frac{1}{2}\)cm
= 5 \(\frac{1}{16}\) cm³
The volume of the object is 9 cm³ + 5 \(\frac{1}{16}\) cm³ = 14 \(\frac{1}{16}\) cm³.

Strategy 2
The volume of a right prism is equal to the area of its base times its height. The base consists of a rectangle and a triangle.
Volume of object = Bh
B = Area_rectangle + Area_triangle
Area_rectangle = 3 cm ∙ 2 cm = 6 cm²
Area_triangle = \(\frac{1}{2}\) ∙ 3 cm ∙ 2 \(\frac{1}{4}\)cm = 3 \(\frac{3}{8}\)cm²
B = 6 cm² + 3 \(\frac{3}{8}\)cm² = 9 \(\frac{3}{8}\)cm²

Volume of object:
Volume_object = Bh
= (9 \(\frac{3}{8}\)cm² )(1\(\frac{1}{2}\)cm)
= 14 \(\frac{1}{16}\)cm³
The volume of the object is 14\(\frac{1}{16}\)cm³.

→ Write a numeric expression to represent the volume of the figure in Strategy 1.
(3 cm ∙ 2 cm) ∙ 1 \(\frac{1}{2}\)cm + (\(\frac{1}{2}\) ∙ 3 cm ∙ 2 \(\frac{1}{4}\)cm) ∙ 1 \(\frac{1}{2}\)cm

→ Write a numeric expression to represent the volume of the figure in Strategy 2.
(3 cm ∙ 2 cm + \(\frac{1}{2}\) ∙ 3 cm ∙ 2 \(\frac{1}{4}\)cm )(1 \(\frac{1}{2}\)cm)

→ How do the numeric expressions represent the problem differently?
The first expression is appropriate to use when individual volumes of the decomposed figure are being added together, whereas the second expression is used when the area of the base of the composite figure is found and then multiplied by the height to determine the volume.

→ What property allows us to show that these representations are equivalent?
The distributive property.

Eureka Math Grade 7 Module 6 Lesson 26 Problem Set Answer Key

Question 1.
Find the volume of the three – dimensional object composed of right rectangular prisms.

Answer:
Volume_object = Volume_{top and bottom prisms} + Volume_{middle prism}
Volume of top and bottom prisms:
V = 2(12 in. ∙ 12 in. ∙ 3 in.)
= 864 in³

Volume of middle prism:
= 4 in. ∙ 4 in. ∙ 8 in.
= 128 in³
The volume of the object is 864 in³ + 128 in³ = 992 in³.

Question 2.
A smaller cube is stacked on top of a larger cube. An edge of the smaller cube measures \(\frac{1}{2}\) cm in length, while the larger cube has an edge length three times as long. What is the total volume of the object?

Answer:
Volume_object = Volume_{small cube} + Volume_{large cube}
Volume_{small cube} = (\(\frac{1}{2}\) cm)³
= \(\frac{1}{8}\) cm³

Volume_{large cube} = (\(\frac{3}{2}\) cm)³
= \(\frac{27}{8}\) cm³

V = \(\frac{1}{8}\) cm³ + \(\frac{27}{8}\) cm³
= 3 \(\frac{1}{2}\) cm³

The total volume of the object is 3\(\frac{1}{2}\) cm³.

Question 3.
Two students are finding the volume of a prism with a rhombus base but are provided different information regarding the prism. One student receives Figure 1, while the other receives Figure 2.

a. Find the expression that represents the volume in each case; show that the volumes are equal.
b. How does each calculation differ in the context of how the prism is viewed?
Answer:
a. Figure 1
2(\(\frac{1}{2}\) ∙ 14.6 mm ∙ 3.6 mm) ∙ 9 mm
473.04 mm³

Figure 2
((8 mm ∙ 6.57 mm) ∙ 9 mm)
473.04 mm³

b. In Figure 1, the prism is treated as two triangular prisms joined together. The volume of each triangular prism is found and then doubled, whereas in Figure 2, the prism has a base in the shape of a rhombus, and the volume is found by calculating the area of the rhomboid base and then multiplying by the height.

Question 4.
Find the volume of wood needed to construct the following side table composed of right rectangular prisms.

Answer:
Volume of bottom legs: V = 2(8 in. ∙ 1 in. ∙ 0.75 in.)
= 12 in³
Volume of vertical legs: V = 2(1 in. ∙ 9.5 in. ∙ 0.75 in.)
= 14.25 in³
Volume of tabletop: V = 8 in. ∙ 6 in. ∙ 1.5 in.
= 72 in³
The volume of the table is
12 in³ + 14.25 in³ + 72 in³ = 98.25 in³.

Question 5.
A plastic die (singular for dice) for a game has an edge length of 1.5 cm. Each face of the cube has the number of cubic cutouts as its marker is supposed to indicate (i.e., the face marked 3 has 3 cutouts). What is the volume of the die?

Answer:
Number of cubic cutouts:
1 + 2 + 3 + 4 + 5 + 6 = 21
Volume of cutout cubes: Volume of large cube:
V = 21(2 mm)³
V = 168 mm³ = 0.168 cm³ V = (1.5 cm)³
V = 3.375 cm³
The total volume of the die is
3.375 cm³ – 0.168 cm³ = 3.207 cm³.

Question 6.
A wooden cube with an edge length of 6 inches has square holes (holes in the shape of right rectangular prisms) cut through the centers of each of the three sides as shown in the figure. Find the volume of the resulting solid if the square for the holes has an edge length of 1 inch.

Answer:
Think of making the square holes between opposite sides by cutting three times: The first cut removes 6 in³, and the second and third cuts each remove 5 in³. The resulting solid has a volume of (6 in.)³ – 6 in³ – 5 in³ – 5 in³ = 200 in³.

Question 7.
A right rectangular prism has each of its dimensions (length, width, and height) increased by 50%. By what percent is its volume increased?
Answer:
V = l ∙ w ∙ h
V’ = 1.5l ∙ 1.5w ∙ 1.5h
V’ = 3.375lwh
The larger volume is 337.5% of the smaller volume. The volume has increased by 237.5%.

Question 8.
A solid is created by putting together right rectangular prisms. If each of the side lengths is increase by 40%, by what percent is the volume increased?
Answer:
If each of the side lengths is increased by 40%, then the volume of each right rectangular prism is multiplied by 1.4³ = 2.744. Since this is true for each right rectangular prism, the volume of the larger solid, V’, can be found by multiplying the volume of the smaller solid, V, by 2.744 = 274.4% (i.e., V’ = 2.744V). This is an increase of 174.4%.

Eureka Math Grade 7 Module 6 Lesson 26 Exit Ticket Answer Key

Question 1.
A triangular prism has a rectangular prism cut out of it from one base to the opposite base, as shown in the figure. Determine the volume of the figure, provided all dimensions are in millimeters.
Is there any other way to determine the volume of the figure? If so, please explain.

Answer:
Possible response:
Volume of the triangular prism: (\(\frac{1}{2}\) ∙ 16 mm ∙ 13 mm)(14 mm) = 1,456 mm³
Volume of the rectangular prism: (6 mm ∙ 3 mm ∙ 14 mm) = 252 mm³
Volume of the composite prism: 1,456 mm³ – 252 mm³ = 1,204 mm³
The calculations above subtract the volume of the cutout prism from the volume of the main prism. Another strategy would be to find the area of the base of the figure, which is the area of the triangle less the area of the rectangle, and then multiply by the height to find the volume of the prism.

Engage NY Eureka Math 7th Grade Module 6 Lesson 25 Answer Key

Eureka Math Grade 7 Module 6 Lesson 25 Example Answer Key

Example 1.
Calculate the volume of the following prism.

Answer:
→ What is the initial difficulty of determining the volume of this prism?
The base is not in the shape of a rectangle or triangle.

→ Do we have a way of finding the area of a kite in one step? If not, how can we find the area of a kite?
We do not have a way to find the area of a kite in one step. We can break the kite up into smaller shapes. One way would be to break it into two triangles, but it can also be broken into four triangles.

Once students understand that the base must be decomposed into triangles, allow them time to solve the problem.
→ Provide a numeric expression that determines the area of the kite – shaped base.
Area of the base: \(\frac{1}{2}\)(20 in.∙18 in.) + \(\frac{1}{2}\)(4 in.∙18 in.)

→ Find the volume of the prism.
Volume of the prism: (\(\frac{1}{2}\)(20 in.∙18 in.) + \(\frac{1}{2}\)(4 in.∙18 in.))(3 in.) = 648 in³

Example 2.
A container is shaped like a right pentagonal prism with an open top. When a cubic foot of water is dumped into the container, the depth of the water is 8 inches. Find the area of the pentagonal base.
Answer:
→ How can we use volume to solve this problem?
We know that the formula for the volume of a prism is V = Bh, where B represents the area of the pentagonal base and h is the height of the prism.

→ What information do we know from reading the problem?
The volume is 1 cubic foot, and the height of the water is 8 inches.

→ Knowing the volume formula, can we use this information to solve the problem?
We can use this information to solve the problem, but the information is given using two different dimensions.

→ How can we fix this problem?
We have to change inches into feet so that we have the same units.

→ How can we change inches into feet?
Because 12 inches make 1 foot, we have to divide the 8 inches by 12 to get the height in feet.

→ Convert 8 inches into feet.
\(\frac{8}{12}\) ft. = \(\frac{2}{3}\) ft.

→ Now that we know the volume of the water is 1 cubic foot and the height of the water is \(\frac{2}{3}\) feet, how can we determine the area of the pentagonal base?
Use the volume formula.
Use the information we know to find the area of the base.
1 ft³ = B(\(\frac{2}{3}\)ft.) Therefore, the area of the pentagonal base is \(\frac{3}{2}\) ft², or 1\(\frac{1}{2}\) ft².

Example 3.
Two containers are shaped like right triangular prisms, each with the same height. The base area of the larger container is 200% more than the base area of the smaller container. How many times must the smaller container be filled with water and poured into the larger container in order to fill the larger container?
Answer:
Solution by manipulating the equation of the volume of the smaller prism:
→ Let us call the area of the base of the smaller prism, B. Write an expression for the area of the larger base, and explain how it models the situation.
The base area of the larger prism is 3B because 200% more means that its area is 300% of B. Both prisms have the same height, h.

→ Compute the volume of the smaller prism.
The volume of the smaller prism is V_S = Bh.

→ What is the volume of the larger prism?
The volume of the larger prism is V_L = 3Bh.

→ How many times greater is the volume of the larger prism relative to the smaller prism?
\(\frac{3Bh}{Bh}\) = 3 The smaller container must be filled three times in order to fill the larger container.

Solution by substituting values for the smaller prism’s dimensions:
→ To solve this problem, create two right triangular prisms. What dimensions should we use for the smaller container?
Answers will vary, but for this example we will use a triangle that has a base of 10 inches and a height of 5 inches. The prism will have a height of 2 inches.

→ What is the area of the base for the smaller container? Explain.
The area of the base of the smaller container is 25 in² because A = \(\frac{1}{2}\)(10 in.)(5 in.) = 25 in².

→ What is the volume of the smaller container? Explain.
The volume of the smaller container is 50 in³ because V = 25 in² × 2 in. = 50 in³.

→ What do we know about the larger container?
The area of the larger container’s base is 200% more than the area of the smaller container’s base.
The height of the larger container is the same as the height of the smaller container.

→ If the area of the larger container’s base is 200% more than the area of the smaller container’s base, what is the area of the larger container’s base?
The area of the larger container’s base would be 25 in² + 2(25 in²) = 75 in².

→ What is the volume of the larger container? Explain.
The volume of the larger container is 150 in³ because V = 75 in² × 2 in. = 150 in³.

→ How many times must the smaller container be filled with water and poured into the larger container in order to fill the larger container? Explain.
The smaller container would have to be filled three times in order to fill the larger container. Each time you fill the smaller container, you will have 50 in³; therefore, you will need to fill the smaller container three times to get a volume of 150 in³.

→ Would your answer be different if we used different dimensions for the containers? Why or why not?
Our answer would not change if we used different dimensions. Because the area of the base of the larger container is triple the area of the base of the smaller container—and because the two heights are the same—the volume of the larger container is triple that of the smaller container.

Eureka Math Grade 7 Module 6 Lesson 25 Exercise Answer Key

Exercise 1.
a. Show that the following figures have equal volumes.

b. How can it be shown that the prisms will have equal volumes without completing the entire calculation?
Answer:
a. Volume of triangular prism:
\(\frac{1}{2}\)(8 cm × 9 cm) × 8 cm = 288 cm³

Volume of rectangular prism:
8 cm × 8 cm × 4.5 cm = 288 cm³

b. If one base can be cut up and rearranged to form the other base, then the bases have the same area. If the prisms have the same height, then the cutting and rearranging of the bases can show how to slice and rearrange one prism so that it looks like the other prism. Since slicing and rearranging does not change volume, the volumes are the same.

Exercise 2.
Two aquariums are shaped like right rectangular prisms. The ratio of the dimensions of the larger aquarium to the dimensions of the smaller aquarium is 3:2.
Addie says the larger aquarium holds 50% more water than the smaller aquarium.
Berry says that the larger aquarium holds 150% more water.
Cathy says that the larger aquarium holds over 200% more water.
Are any of the girls correct? Explain your reasoning.
Answer:
Cathy is correct. If the ratio of the dimensions of the larger aquarium to the dimensions of the smaller aquarium is 3:2, then the volume must be 1.5³, or 3.375 times greater than the smaller aquarium. Therefore, the larger aquarium’s capacity is 337.5% times as much water, which is 237.5% more than the smaller aquarium. Cathy said that the larger aquarium holds over 200% more water than the smaller aquarium, so she is correct.

Eureka Math Grade 7 Module 6 Lesson 25 Problem Set Answer Key

Question 1.
The pieces in Figure 1 are rearranged and put together to form Figure 2.

a. Use the information in Figure 1 to determine the volume of the prism.
b. Use the information in Figure 2 to determine the volume of the prism.
c. If we were not told that the pieces of Figure 1 were rearranged to create Figure 2, would it be possible to determine whether the volumes of the prisms were equal without completing the entire calculation for each?
Answer:
a. Volume: \(\frac{1}{2}\)(6.9 cm∙6 cm)(5 cm) = 103.5 cm³
b. Volume: (\(\frac{1}{2}\)(6.9 cm + 13.8 cm) ∙ 2 cm)(5 cm) = 103.5 cm³
c. Both prisms have the same height, so as long as it can be shown that both bases have the same area, both prisms must have equal volumes. We could calculate the area of the triangle base and the trapezoid base and find that they are equal in area and be sure that both volumes are equal.

Question 2.
Two right prism containers each hold 75 gallons of water. The height of the first container is 20 inches. The of the second container is 30 inches. If the area of the base in the first container is 6 ft², find the area of the base in the second container. Explain your reasoning.
Answer:
We know that the volume of each of the two containers is 75 gallons; therefore, the volumes must be equal.
In order to find the volume of the first container, we could multiply the area of the base (6 ft²) by the height
(20 inches). To find the volume of the second container, we would also multiply the area of its base, which we will call A (area in square feet), and the height (30 inches). These two expressions must equal each other since both containers have the same volume.
6 × 20 = A × 30
120 = 30A
120 ÷ 30 = 30A ÷ 30
4 = A
Therefore, the area of the base in the second container is 4 ft². Note: The units for the volume are 1 ft. × 1 ft. × 1 in. in this computation. Converting the inches to feet would make the computation in cubic feet, but it would not change the answer for A.

Question 3.
Two containers are shaped like right rectangular prisms. Each has the same height, but the base of the larger container is 50% more in each direction. If the smaller container holds 8 gallons when full, how many gallons does the larger container hold? Explain your reasoning.
Answer:
The larger container holds 18 gallons because each side length of the base is 1.5 times larger than the smaller container’s dimensions. Therefore, the area of the larger container’s base is 1.5², or 2.25 times larger than the smaller container. Because the height is the same in both containers, the volume of the larger container must be 2.25 times larger than the smaller container. 8 gal. × 2.25 = 18 gal.

Question 4.
A right prism container with the base area of 4 ft² and height of 5 ft. is filled with water until it is 3 ft. deep. If a solid cube with edge length 1 ft. is dropped to the bottom of the container, how much will the water rise?
Answer:
The volume of the cube is 1 ft³. Let the number of feet the water will rise be x. Then, the volume of the water over the 3 ft. mark is 4 x ft³ because this represents the area of the base (4 ft²) times the height (x). Because the volume of the cube is 1 ft³, 4 x ft³ must equal 1 ft³.
4x = 1
4x ÷ 4 = 1 ÷ 4
x = \(\frac{1}{4}\)
Therefore, the water will rise \(\frac{1}{4}\) ft., or 3 inches.

Question 5.
A right prism container with a base area of 10 ft² and height 9 ft. is filled with water until it is 6 ft. deep. A large boulder is dropped to the bottom of the container, and the water rises to the top, completely submerging the boulder without causing overflow. Find the volume of the boulder.
Answer:
The increase in volume is the same as the volume of the boulder. The height of the water increases 3 ft. Therefore, the increase in volume is 10 ft² (area of the base) multiplied by 3 ft. (i.e., the change in height).
V = 10 ft² × 3 ft. = 30 ft³
Because the increase in volume is 30 ft³, the volume of the boulder is 30 ft³.

Question 6.
A right prism container with a base area of 8 ft² and height 6 ft. is filled with water until it is 5 ft. deep. A solid cube is dropped to the bottom of the container, and the water rises to the top. Find the length of the cube.
Answer:
When the cube is dropped into the container, the water rises 1 foot, which means the volume increases 8 cubic feet. Therefore, the volume of the cube must be 8 cubic feet. We know that the length, width, and height of a cube are equal, so the length of the cube is 2 feet because 2 ft. × 2 ft. × 2 ft. = 8 ft³, which is the volume of the cube.

Question 7.
A rectangular swimming pool is 30 feet wide and 50 feet long. The pool is 3 feet deep at one end, and 10 feet deep at the other.
a. Sketch the swimming pool as a right prism.
b. What kind of right prism is the swimming pool?
c. What is the volume of the swimming pool in cubic feet?
d. How many gallons will the swimming pool hold if each cubic feet of water is about 7.5 gallons?
Answer:
a.

b. The swimming pool is a right trapezoidal prism.

c. Area of base = \(\frac{50 \mathrm{ft} \cdot(10 \mathrm{ft} + 3 \mathrm{ft})}{2}\) = 325 ft²
Volume of pool = 325 ft² × 30 ft. = 9,750 ft³

d. (9,750 ft³ ) \(\frac{(7.5 \mathrm{gal})}{1 \mathrm{ft}^{3}}\) = 73,125 gal. The pool will hold 73,125 gal.

Question 8.
A milliliter (mL) has a volume of 1 cm³. A 250 mL measuring cup is filled to 200 mL. A small stone is placed in the measuring cup. The stone is completely submerged, and the water level rises to 250 mL.
a. What is the volume of the stone in cm³?
b. Describe a right rectangular prism that has the same volume as the stone.
Answer:
a. When the stone is dropped into the measuring cup, the increase in volume is 250 mL – 200 mL = 50 mL. We know that 1 mL has a volume of 1 cm³; therefore, the stone has a volume of 50 cm³.

b. Answers will vary. Possible answers are listed below.
1 cm × 1 cm × 50 cm
1 cm × 2 cm × 25 cm
1 cm × 5 cm × 10 cm
2 cm × 5 cm × 5 cm

Eureka Math Grade 7 Module 6 Lesson 25 Exit Ticket Answer Key

Question 1.
Determine the volume of the following prism. Explain how you found the volume.

Answer:
To find the volume of the prism, the base must be decomposed into triangles and rectangles since there is no way to find the area of the base as is. The base can be decomposed into two triangles and a rectangle, and their areas must be summed to find the area of the base. Once the area of the base is determined, it should be multiplied by the height to find the volume of the entire prism.
Area of both triangles: 2(\(\frac{1}{2}\)(5 in. × 6 in.)) = 30 in²
Area of the rectangle: 15 in. × 5 in. = 75 in²
Total area of the base: (30 + 75) in² = 105 in²
Volume of the prism: (105 in²)(1.5 in.) = 157.5 in³

Engage NY Eureka Math 7th Grade Module 6 Lesson 22 Answer Key

Eureka Math Grade 7 Module 6 Lesson 22 Example Answer Key

Example 1.
a. The circle to the right has a diameter of 12 cm. Calculate the area of the shaded region.

b. Sasha, Barry, and Kyra wrote three different expressions for the area of the shaded region. Describe what each student was thinking about the problem based on his or her expression.
Sasha’s expression: \(\frac{1}{4}\) π(6²)
Barry’s expression: π(6² ) – \(\frac{3}{4}\) π(6²)
Kyra’s expression: \(\frac{1}{2}\)(\(\frac{1}{2}\) π(6²))
Answer:
a.
→ What information do we need to calculate the area?
We need the radius of the circle and the shaded fraction of the circle.

→ What is the radius of the circle? How do you know?
The radius of the circle is 6 cm since the length of the radius is half the length of the diameter.

→ Now that we know the radius, how can we find the area of the shaded region?
Find the area of one quarter of the circle because the circle is divided into four identical parts, and only one part is shaded.

→ Choose a method discussed, and calculate the area of the shaded region.
A_{quarter circle} = \(\frac{1}{4}\)πr²
The area in squared centimeters:
\(\frac{1}{4}\)π(6)² = 9π ≈ 28.27
The area of the shaded region is about 28.27 cm².

b. Sasha’s expression: \(\frac{1}{4}\) π(6²)
Sasha’s expression gets directly to the shaded area as it represents a quarter of the area of the whole circle.

Barry’s expression: π(6² ) – \(\frac{3}{4}\) π(6²)
Barry’s expression shows the area of the whole circle minus the unshaded area of the circle or three – quarters of the area of the circle.

Kyra’s expression: \(\frac{1}{2}\)(\(\frac{1}{2}\) π(6²))
Kyra’s expression arrives at the shaded area by taking half the area of the whole circle, which is half of the circle, and taking half of that area, which leaves a quarter of the area of the whole circle.

Example 2.
The square in this figure has a side length of 14 inches. The radius of the quarter circle is 7 inches.
a. Estimate the shaded area.

b. What is the exact area of the shaded region?
c. What is the approximate area using π ≈ \(\frac{22}{7}\)?
Answer:
→ Describe a strategy to find the area of the shaded region.
Find the area of the entire square, and subtract the area of the unshaded region because the unshaded region is four quarter circles of equal radius or a whole circle.

→ What is the difference between parts (b) and (c) in Example 2?
Part (b) asks for the exact area, which means the answer must be left in terms of pi; part (c) asks for the approximate area, which means the answer must be rounded off.

→ How would you estimate the shaded area?
Responses will vary. One possible response might be that the shaded area looks to be approximately one – quarter the area of the entire square, roughly \(\frac{1}{4}\)(14 in.)² = 49 in².

→ What is the area of the square?
A = 14 in. ∙ 14 in. = 196 in²
The area of the square is 196 in².

→ What is the area of each quarter circle?
A = \(\frac{\pi}{4}\) r² = \(\frac{\pi}{4}\)(7 in.)² = \(\frac{49 \pi}{4}\) in²
The area of each quarter circle is \(\frac{49 \pi}{4}\) in².
The area of the entire unshaded region, or all four quarter circles, is 49π in².

→ What is the exact area of the shaded region?
A = 196 in² – 49π in²
The area of the shaded region is (196 – 49π) in².

→ What is the approximate area using π ≈ \(\frac{22}{7}\)?
A ≈ 196 in² – (\(\frac{22}{7}\)) (7 in.)²
≈ 196 in² – 154 in²
≈ 42 in²
The area of the shaded region is approximately 42 in².

Eureka Math Grade 7 Module 6 Lesson 22 Exercise Answer Key

Exercise 1.
a. Find the area of the shaded region of the circle to the right.

b. Explain how the expression you used represents the area of the shaded region.
Answer:
a. A = \(\frac{3}{8}\)πr²
A = \(\frac{3}{8}\)(π) (12 ft)²
A = 54π ft²
A≈169.65 ft²
The shaded area of the circle is approximately 169.65 ft².

b. The expression \(\frac{3}{8}\)(π) (12)² takes the area of a whole circle with a radius of 12 ft., which is just the portion that reads (π) (12)², and then multiplies that by \(\frac{3}{8}\). The shaded region is just three out of eight equal pieces of the circle.

Exercise 2.
Calculate the area of the figure below that consists of a rectangle and two quarter circles, each with the same radius. Leave your answer in terms of pi.

Answer:
A_rectangle = l ∙ w
A = 6 in. ∙ 4 in.
A = 24 in²
The area of the rectangle is 24 in².

A_{half circle} = \(\frac{1}{2}\)πr²
A = \(\frac{1}{2}\)(π)(4 in.)²
A = 8π in²

The area of the two quarter circles, or one semicircle, is 8π in².
The area of the entire figure is A = (24 + 8π) in².

Exercise 3.
The vertices A and B of rectangle ABCD are centers of circles each with a radius of 5 inches.
a. Find the exact area of the shaded region.

b. Find the approximate area using π ≈ \(\frac{22}{7}\).
c. Find the area to the nearest hundredth using the π key on your calculator.
Answer:
a. A_rectangle = 10 in. ∙ 5 in. = 50 in²
A_semicircle = \(\frac{1}{2}\) π(5 in.)²
A_semicircle = 25\(\frac{\pi}{2}\) in²
A_{shaded area} = (50 – 25\(\frac{\pi}{2}\)) in²

b. A_{shaded area} = (50 – 25\(\frac{\pi}{2}\) in²
A_{shaded area} ≈ (50 – \(\frac{25}{2}\)(\(\frac{22}{7}\))) in²
A_{shaded area} ≈ (50 – \(\frac{275}{7}\)) in²
A_{shaded area} ≈ 10 \(\frac{5}{7}\)in²
The area of the shaded region in the figure is approximately 10\(\frac{5}{7}\) in².

A_{shaded area} = (50 – \(\frac{25 \pi}{2}\)) in²
A_{shaded area} ≈ 10.73 in²
The area of the shaded region in the figure is approximately 10.73 in².

Exercise 4.
The diameter of the circle is 12 in. Write and explain a numerical expression that represents the area of the shaded region.

Answer:
A_shaded = A_{quarter circle} – A_triangle
A_shaded = \(\frac{\pi}{4}\)(6in.)² – \(\frac{1}{2}\)(6 in. ∙ 6 in.)
A_shaded = (9π – 18) in²
The expression represents the area of one quarter of the entire circle less the area of the right triangle, whose legs are formed by radii of the circle.

Eureka Math Grade 7 Module 6 Lesson 22 Problem Set Answer Key

Question 1.
A circle with center O has an area of 96 in². Find the area of the shaded region.

Peyton’s Solution
A = \(\frac{1}{3}\)(96 in² ) = 32 in²

Monte’s Solution
A = \(\frac{96}{120}\) in² = 0.8 in²
Answer:
Peyton solved the problem correctly because he correctly identified the shaded region as one – third of the area of the entire circle. The shaded region represents \(\frac{1}{3}\) of the circle because 120° is one third of 360°. To find the area of the shaded region, one – third of the area of the entire circle, 96 in², must be calculated, which is what Peyton did to get his solution.

Question 2.
The following region is bounded by the arcs of two quarter circles, each with a radius of 4 cm, and by line segments 6 cm in length. The region on the right shows a rectangle with dimensions 4 cm by 6 cm. Show that both shaded regions have equal areas.

Answer:
A = ((4 cm ∙ 6 cm) – \(\frac{1}{4}\)π(4 cm)² ) + \(\frac{1}{4}\)π(4 cm)²
A = 24 cm²

A = 4 cm ∙ 6 cm
A = 24 cm²

Question 3.
A square is inscribed in a paper disc (i.e., a circular piece of paper) with a radius of 8 cm. The paper disc is red on the front and white on the back. Two edges of the circle are folded over. Write and explain a numerical expression that represents the area of the figure. Then, find the area of the figure.

Answer:

Numeric expression for the area: 4(\(\frac{1}{2}\) ∙ 8 cm ∙ 8 cm)
The shaded (red) area is the same as the area of the square. The radius is 8 cm, which is the length of one leg of each of the four equal – sized right triangles within the square. Thus, we find the area of one triangle and multiply by 4.
The area of the shaded region is 128 cm².

Question 4.
The diameters of four half circles are sides of a square with a side length of 7 cm.

a. Find the exact area of the shaded region.
b. Find the approximate area using π ≈ \(\frac{22}{7}\).
c. Find the area using the π button on your calculator and rounding to the nearest thousandth.
Answer:

a. Figure 2 isolates one quarter of Figure 1. The shaded area in Figure 2 can be found as follows:
Shaded area = Area of the quarter circle – Area of the isosceles right triangle.
Shaded area:
(\(\frac{ \pi}{4}\) (\(\frac{7}{2}\) cm)² ) – (\(\frac{1}{2}\) ∙ \(\frac{7}{2}\) cm ∙ \(\frac{7}{2}\) cm) = \(\frac{49 \pi}{16}\) cm² – \(\frac{49}{8}\) cm²
= \(\frac{49}{16}\) (π – 2) cm²
The area of the shaded region is \(\frac{49}{16}\)(π – 2) cm². There are 8 such regions in the figure, so we multiply this answer by 8.
Total shaded area:
8(\(\frac{49}{16}\) (π – 2)) = \(\frac{49}{2}\) (π – 2) = \(\frac{49 \pi}{2}\) – 49
The exact area of the shaded region is (\(\frac{49 \pi}{2}\) – 49) cm².

b. A_{total shaded} ≈ \(\frac{49}{2}\) (\(\frac{22}{7}\) – 2) cm²
A_{total shaded} ≈ (77 – 49) cm²
A_{total shaded} ≈ 28 cm²
The approximate area of the shaded region is 28 cm².

c. A_{total shaded} = \(\frac{49}{2}\) (π – 2) cm²
A_{total shaded} ≈ 27.969 cm²
The approximate area of the shaded region is 27.969 cm².

Question 5.

A square with a side length of 14 inches is shown below, along with a quarter circle (with a side of the square as its radius) and two half circles (with diameters that are sides of the square). Write and explain a numerical expression that represents the area of the figure.
Answer:

Numeric expression for the area: 1\(\frac{1}{4}\)π(14 in.)² – (\(\frac{1}{2}\) ∙ 14 in. ∙ 14 in.)
The shaded area in Figure 1 is the same as the shaded area in Figure 2. This area can be found by subtracting the area of the right triangle with leg lengths of 14 in. from the area of the quarter circle with a radius of 14 in.
\(\frac{1}{4}\) π(14 in.)² – (\(\frac{1}{2}\) ∙ 14 in. ∙ 14 in.) = (49π – 98) in²

Question 6.
Three circles have centers on segment AB. The diameters of the circles are in the ratio 3:2:1. If the area of the largest circle is 36 ft², find the area inside the largest circle but outside the smaller two circles.

Answer:
Since all three circles are scale drawings of each other, the ratio of the areas of the circles is 9:4:1. This ratio provides a means to find the areas of the two smaller circles.
Area of medium – sized circle in ft²:
\(\frac{9}{4}\) = \(\frac{36}{x}\)
x = 16
The area of the medium – sized circle is 16 ft².

Area of small – sized circle in ft²:
\(\frac{9}{1}\) = \(\frac{36}{y}\)
y = 4
The area of the small – sized circle is 4 ft².

The area inside the largest circle but outside the smaller two circles is
A = 36 ft² – 16 ft² – 4 ft²
A = 16 ft².
The area inside the largest circle but outside the smaller two circles is 16 ft².

Question 7.
A square with a side length of 4 ft. is shown, along with a diagonal, a quarter circle (with a side of the square as its radius), and a half circle (with a side of the square as its diameter). Find the exact, combined area of regions I and II.

Answer:
The area of I is the same as the area of III in the following diagram.

Since the area of I is the same as the area of III, we need to find the combined area of II and III. The combined area of II and III is half the area of II, III, and IV. The area of II, III, and IV is the area of the quarter circle minus the area of the triangle.
A_{II and III} = \(\frac{1}{2}\)(\(\frac{1}{4}\)π(4 ft.)² – \(\frac{1}{2}\) ∙ 4 ft ∙ 4 ft.)
A_{II and III} = \(\frac{1}{2}\)(\(\frac{16 \pi}{4}\) ft² – (\(\)\frac{4 \mathrm{ft} \cdot 4 \mathrm{ft}}{2}/\(\))
A_{II and III} = (2π – 4) ft²
The combined area of I and II is (2π – 4) ft².

Eureka Math Grade 7 Module 6 Lesson 22 Exit Ticket Answer Key

Question 1.
A circle with a 10 cm radius is cut into a half circle and two quarter circles. The three circular arcs bound the region below.
a. Write and explain a numerical expression that represents the area.
b. Then, find the area of the figure.

Answer:
a. Numeric expression 1 for the area:
10 cm ∙ 20 cm
The expression for the area represents the region when it is cut into three pieces and rearranged to make a complete rectangle as shown.

Numeric expression 2 for the area: (50π) cm² + (200 – 50π) cm²
The expression for the area is calculated as is; in other words, by finding the area of the semicircle in the top portion of the figure and then the area of the carved – out regions in the bottom portion of the figure.

b. The area of the figure is 200 cm².

Engage NY Eureka Math 7th Grade Module 6 Lesson 24 Answer Key

Eureka Math Grade 7 Module 6 Lesson 24 Example Answer Key

Example 1.
Determine the surface area of the image.

Answer:
Surface area of top or bottom prism:
Lateral sides = 4(12 in. × 3 in.) = 144 in²
Base face = 12 in. × 12 in. = 144 in²
Base face with hole = 12 in. × 12 in. – 4 in. × 4 in.
= 128 in²
There are two of these, making up 832 in².
Surface area of middle prism:
Lateral sides: = 4(4 in. × 8 in.) = 128 in²
Surface area: 832 in² + 128 in² = 960 in²

→ Describe the method you used to determine the surface area.
Answers will vary. I determined the surface area of each prism separately and added them together. Then, I subtracted the area of the sections that were covered by another prism.

→ If all three prisms were separate, would the sum of their surface areas be the same as the surface area you determined in this example?
No, if the prisms were separate, there would be more surfaces shown. The three separate prisms would have a greater surface area than this example. The area would be greater by the area of four 4 in. × 4 in. squares (64 in²).

Example 2.
a. Determine the surface area of the cube.

b. A square hole with a side length of 4 inches is cut through the cube. Determine the new surface area.

Answer:
a. Surface area = 6s²
SA = 6(12 in.)²
SA = 6(144 in²)
SA = 864 in²
→ Explain how 6(12 in.)² represents the surface area of the cube.
The area of one face, one square with a side length of 12 in., is (12 in.)², and so a total area of all six faces is 6(12 in.)².

b. Area of interior lateral sides
= 4(12 in. × 4 in.)
= 192 in²

Surface area of cube with holes
= 6 (12 in.)² – 2(4 in. × 4 in.) + 4(12 in. × 4 in.)
= 864 in² – 32 in² + 192 in²
= 1,024 in²
→ How does cutting a hole in the cube change the surface area?
We have to subtract the area of the square at the surface from each end.
We also have to add the area of each of the interior faces to the total surface area.

→ What happens to the surfaces that now show inside the cube?
These are now part of the surface area.

→ What is the shape of the piece that was removed from the cube?
A rectangular prism was cut out of the cube with the following dimensions: 4 in. × 4 in. × 12 in.

→ How can we use this to help us determine the new total surface area?
We can find the surface area of the cube and the surface area of the rectangular prism, but we will have to subtract the area of the square bases from the cube and also exclude these bases in the area of the rectangular prism.

→ Why is the surface area larger when holes have been cut into the cube?
There are more surfaces showing now. All of the surfaces need to be included in the surface area.

→ Explain how the expression 6(12 in.)² – 2(4 in. × 4 in.) + 4(12 in. × 4 in.) represents the surface area of the cube with the hole.
From the total surface area of a whole (uncut) cube, 6(12 in.)², the area of the bases (the cuts made to the surface of the cube) are subtracted: 6(12 in.)² – 2(4 in. × 4 in.). To this expression we add the area of the four lateral faces of the cutout prism, 4(12 in. × 4 in.). The complete expression then is 6(12 in.)² – 2(4 in. × 4 in.) + 4(12 in. × 4 in.)

Example 3.
A right rectangular pyramid has a square base with a side length of 10 inches. The surface area of the pyramid is 260 in². Find the height of the four lateral triangular faces.
Answer:
Area of base = 10 in. × 10 in. = 100 in²
Area of the four faces = 260 in² – 100 in² = 160 in²
The total area of the four faces is 160 in².
Therefore, the area of each triangular face is 40 in².
Area of lateral side = \(\frac{1}{2}\)bh
40 in² = \(\frac{1}{2}\)(10 in.)h
40 in² = (5 in)h
h = 8 in.
The height of each lateral triangular face is 8 inches.

→ What strategies could you use to help you solve this problem?
I could draw a picture of the pyramid and label the sides so that I can visualize what the problem is asking me to do.

→ What information have we been given? How can we use the information?
We know the total surface area, and we know the length of the sides of the square.
We can use the length of the sides of the square to give us the area of the square base.

→ How does the area of the base help us determine the height of the lateral triangular face?
First, we can subtract the area of the base from the total surface area in order to determine what is left for the lateral sides.
Next, we can divide the remaining area by 4 to get the area of just one triangular face.
Finally, we can work backward. We have the area of the triangle, and we know the base is 10″ in.” , so we can solve for the height.

Eureka Math Grade 7 Module 6 Lesson 24 Exercise Answer Key

Determine the surface area of each figure. Assume all faces are rectangles unless it is indicated otherwise.
Exercise 1.

Answer:
Top and bottom = 2(18 m × 5 m) = 180 m²
Extra interior sides = 2(5 m × 7 m) = 70 m²
Left and right sides = 2(5 m × 12 m) = 120 m²
Front and back sides = 2((18 m × 12 m) – (8 m × 7 m))
= 2(216 m² – 56 m² )
= 2(160 m² )
= 320 m²
Surface area = 180 m² + 70 m² + 120 m² + 320 m²
= 690 m²

Exercise 2.
In addition to your calculation, explain how the surface area of the following figure was determined.

Answer:
The surface area of the prism is found by taking the sum of the areas of the trapezoidal front and back and the areas of the back of the four different – sized rectangles that make up the lateral faces.
Area top = 25 cm × 8 cm
= 200 cm²
Area bottom = 24 cm × 8 cm
= 192 cm²
Area sides = (23 cm × 8 cm) + (16 cm × 8 cm)
= 312 cm²
Area front and back = 2(\(\frac{1}{2}\)(16 cm + 23 cm)(24 cm))
= 2(468 cm² )
= 936 cm²
Surface area = 200 cm² + 192 cm² + 312 cm² + 936 cm²
= 1,640 cm²

Exercise 3.

Answer:
Surface area of one of the prisms on the sides:
Area of front and back = 2(2 in. × 14 in.)
= 56 in²
Area of top and bottom = 2(2 in. × 10 in.)
= 40 in²
Area of side = 14 in. × 10 in. = 140 in²
Area of side with hole = 14 in. × 10 in. – 3 in. × 3 in.
= 131 in²

There are two such rectangular prisms, so the surface area of both is 734 in².
Surface area of middle prism:
Area of front and back = 2(3 in. × 1²in.) = 72 in²
Area of sides = 2(3 in. × 1² in.) = 72 in²
Surface area of middle prism = 72 in² + 72 in² = 144 in²
The total surface area of the figure is 734 in² + 144 in² = 878 in².

Exercise 4.
In addition to your calculation, explain how the surface area was determined.

Answer:
The surface area of the prism is found by taking the area of the base of the rectangular prism and the area of its four lateral faces and adding it to the area of the four lateral faces of the pyramid.
Area of base = 9 ft. × 9 ft.
= 81 ft²
Area of rectangular sides = 4(9 ft. × 5 ft.)
= 180 ft²
Area of triangular sides = 4(\(\frac{1}{2}\)(9 ft.)(6 ft.))
= 108 ft²
Surface area = 81 ft² + 180 ft² + 108 ft²
= 369 ft²

Exercise 5.
A hexagonal prism has the following base and has a height of 8 units. Determine the surface area of the prism.

Answer:
Area of bases = 2(48 + 6 + 20 + 10) = 168
Area of 4 unit sides = 4(5 × 8)
= 160
Area of other sides = (4 × 8) + (12 × 8)
= 128
Surface area = 168 + 160 + 128
= 456
The surface area of the hexagonal prism is 456 units².

Exercise 6.
Determine the surface area of each figure.
a.

b. A cube with a square hole with 3 m side lengths has been cut through the cube.

c. A second square hole with 3 m side lengths has been cut through the cube.

Answer:
a. SA = 6s²
= 6(9 m)²
= 6(81m²)
= 486 m²

b. Lateral sides of the hole = 4(9 m × 3 m) = 108 m²
Surface area of cube with holes = 486 m² – 2(3 m × 3 m) + 108 m²
= 576 m²

c. Surface area = 576 m² – 4(3 m × 3 m) + 2(4(3 m × 3 m))
= 612 m²

Exercise 7.
The figure below shows 28 cubes with an edge length of 1 unit. Determine the surface area.

Answer:
Area top and bottom = 24 units²
Area sides = 18 units²
Area front and back = 28 units²
Surface area
= 24 units² + 18 units ² + 28 units²
= 70 units²

Exercise 8.
The base rectangle of a right rectangular prism is 4 ft. × 6 ft. The surface area is 288 ft². Find the height. Let h be the height in feet.
Answer:
Area of one base: 4 ft. × 6 ft. = 24 ft²
Area of two bases: 2(24 ft²) = 48 ft²
Numeric area of four lateral faces: 288 ft² – 48 ft² = 240 ft²
Algebraic area of four lateral faces: 2(6h + 4h) ft²
Solve for h.
2(6h + 4h) = 240
10h = 120
h = 12
The height is 12 feet.

Eureka Math Grade 7 Module 6 Lesson 24 Problem Set Answer Key

Determine the surface area of each figure.
Question 1.
In addition to the calculation of the surface area, describe how you found the surface area.

Answer:
Area of top = 28 cm × 13 cm
= 364 cm²
Area of bottom = 28 cm × 12 cm
= 336 cm²
Area of left and right sides = 28 cm × 20 cm + 15 cm × 28 cm
= 980 cm²
Area of front and back sides = 2(12 cm × 15 cm) + \(\frac{1}{2}\)(5 cm × 12 cm)
= 2(180 cm² + 30 cm² )
= 2(210 cm² )
= 420 cm²
Surface area = 364 cm² + 336 cm² + 980 cm² + 420 cm²
= 2,100 cm²
Split the area of the two trapezoidal bases into triangles and rectangles, take the sum of the areas, and then add the areas of the four different – sized rectangles that make up the lateral faces.

Question 2.

Answer:
Area of front and back = 2(18.4 in. × 8.6 in.) = 316.48 in²
Area of sides = 2(8.6 in. × 24 in.) = 412.8 in²
Area of top and bottom = 2(18.4 in. × 24 in.) – (4.8 in. × 2.2 in.)
= 2(441.6 in² – 10.56 in²)
= 2(431.04 in²)
= 862.08 in²
Surface area = 316.48 in² + 412.8 in² + 862.08 in²
= 1,591.36 in²

Question 3.

Answer:
Area of front and back = 2(\(\frac{1}{2}\)(32 m + 16 m)15 m)
= 720 m²
Area of top = 16 m × 36 m
= 576 m²
Area of left and right sides = 2(17 m × 36 m)
= 2(612 m² )
= 1,224 m²
Area of bottom = 32 m × 36 m
= 1,152 m²
Surface area = 720 m² + 1,152 m² + 1,224 m² + 576 m²
= 3,672 m²

Question 4.
Determine the surface area after two square holes with a side length of 2 m are cut through the solid figure composed of two rectangular prisms.

Answer:
Surface area of top prism before the hole is drilled:
Area of top = 4 m × 5 m
= 20 m²
Area of front and back = 2(4 m × 5 m)
= 40 m²
Area of sides = 2(5 m × 5 m)
= 50 m²
Surface area of bottom prism before the hole is drilled:
Area of top = 10 m × 10 m – 20 m²
= 80 m²
Area of bottom = 10 m × 10 m
= 100 m²
Area of front and back = 2(10 m × 3 m)
= 60 m²
Area of sides = 2(10 m × 3 m)
= 60 m²

Surface area of interiors:
Area of interiors = 4(2 m × 4 m) + 4(2 m × 3 m)
= 56 m²
Surface Area = 110 m² + 300 m² + 56 m² – 16 m²
= 450 m²

Question 5.
The base of a right prism is shown below. Determine the surface area if the height of the prism is 10 cm. Explain how you determined the surface area.

Answer:
Take the sum of the areas of the two bases made up of two right triangles, and add to it the sum of the areas of the lateral faces made up of rectangles of different sizes.
Area of sides = (20 cm × 10 cm) + (15 cm × 10 cm) + (24 cm × 10 cm) + (7 cm × 10 cm)
= 200 cm² + 150 cm² + 240 cm² + 70 cm²
= 660 cm²
Area of bases = 2(\(\frac{1}{2}\)(7 cm × 24 cm) + 1/2 (20 cm × 15 cm))
= (7 cm × 24 cm) + (20 cm × 15 cm)
= 168 cm² + 300 cm²
= 468 cm²
Surface area = 660 cm² + 468 cm²
= 1,128 cm²

Eureka Math Grade 7 Module 6 Lesson 24 Exit Ticket Answer Key

Question 1.
Determine the surface area of the right rectangular prism after the two square holes have been cut. Explain how you determined the surface area.

Answer:
Area of top and bottom = 2(15 cm × 6 cm)
= 180 cm²
Area of sides = 2(6 cm × 8 cm)
= 96 cm²
Area of front and back = 2(15 cm × 8 cm) – 4(5 cm × 5 cm)
= 140 cm²
Area inside = 8(5 cm × 6 cm)
= 240 cm²
Surface area = 180 cm² + 96 cm² + 140 cm² + 240 cm²
= 656 cm²
Take the sum of the areas of the four lateral faces and the two bases of the main rectangular prism, and subtract the areas of the four square cuts from the area of the front and back of the main rectangular prism. Finally, add the lateral faces of the prisms that were cut out of the main prism.

Engage NY Eureka Math 7th Grade Module 6 Lesson 23 Answer Key

Eureka Math Grade 7 Module 6 Lesson 23 Example Answer Key

Example 1.
a. Calculate the surface area of the rectangular prism.

b. Imagine that a piece of the rectangular prism is removed. Determine the surface area of both pieces.

c. How is the surface area in part (a) related to the surface area in part (b)?
Answer:
a. Surface area:
2(3 in. × 6 in.) + 2(3 in. × 12 in.) + 2(6 in. × 12 in.)
= 2(18 in²) + 2(36 in²) + 2(72 in² )
= 36 in² + 72 in² + 144 in²
= 252 in²

b.
→ How did you determine the surface area of the shape on the left?
→ I was able to calculate the area of the sides that are rectangles using length times width. For the two bases that are C – shaped, I used the area of the original top and bottom and subtracted the piece that was taken off.

c. If I add the surface area of both figures, I will get more than the surface area of the original shape.
234 in² + 90 in² = 324 in²
324 in² – 252 in² = 72 in²
72 in² is twice the area of the region where the two pieces fit together.
There are 72 more square inches when the prisms are separated.

Eureka Math Grade 7 Module 6 Lesson 23 Exercise Answer Key

Opening Exercise
Calculate the surface area of the square pyramid.

Answer:
Area of the square base:
s² = (8 cm)²
= 64 cm²
Area of the triangular lateral sides:
\(\frac{1}{2}\) bh = \(\frac{1}{2}\)(8 cm)(5 cm)
= 20 cm²
There are four lateral sides. So, the area of all 4 triangles is 80 cm².

Surface area:
(8 cm ∙ 8 cm) + 4(1/2 (8 cm ∙ 5 cm)) = 64 cm² + 80 cm² = 144 cm²

→ Explain the process you used to determine the surface area of the pyramid.
Answers will vary. I drew a net to determine the area of each side and added the areas together.

→ The surface area of a pyramid is the union of its base region and all its lateral faces.
→ Explain how (8 cm ∙ 8 cm) + 4(\(\frac{1}{2}\)(8 cm∙5 cm)) represents the surface area.
The area of the square with side lengths of 8 cm is represented by (8 cm ∙ 8 cm), and the area of the four lateral faces with base lengths of 8 cm and heights of 5 cm is represented by 4(\(\frac{1}{2}\) (8 cm∙5 cm)).

→ Would this method work for any prism or pyramid?
Answers will vary. Calculating the area of each face, including bases, will determine the surface area even if the areas are determined in different orders, by using a formula or net, or any other method.

Determine the surface area of the right prisms.
Exercise 1.

Answer:
Area of top and bottom: 2(\(\frac{1}{2}\)(15 ft.  × 8 ft.) )
= 15 ft. × 8 ft. = 120 ft²
Area of front: 15 ft. × 20 ft. = 300 ft²
Area that can be seen from left: 17 ft. × 20 ft. = 340 ft²
Area that can be seen from right: 8 ft. × 20 ft. = 160 ft²
Surface area: 120 ft² + 300 ft² + 340 ft² + 160 ft² = 920 ft²

Exercise 2.

Answer:
Area of front and back: 2(\(\frac{1}{2}\)(10 yd. + 4 yd.)4 yd.)
= 14 yd. × 4 yd. = 56 yd²
Area of top: 4 yd. × 15 yd. = 60 yd²
Area that can be seen from left and right: 2(5 yd. × 15 yd.)
= 2(75 yd²) = 150 yd²
Area of bottom: 10 yd. × 15 yd. = 150 yd²
Surface area: 56 yd² + 60 yd² + 150 yd² + 150 yd² = 416 yd²

Exercise 3.

Answer:
Area of top and bottom: 2((8 ft. × 6 ft.) + (7 ft. × 2 ft.))
= 2(48 ft² + 14 ft² )
= 2(62 ft² ) = 124 ft²
Area for back: 8 ft. × 3 ft. = 24 ft²

Area for front: 7 ft. × 3 ft. = 21 ft²

Area of corner cutout: (2 ft. × 3 ft.) + (1 ft. × 3 ft.) = 9 ft²
Area of right side: 6 ft. × 3 ft. = 18 ft²
Area of left side: 8 ft. × 3 ft. = 24 ft²

Surface area: 124 ft² + 24 ft² + 21 ft² + 9 ft² + 18 ft² + 24 ft² = 220 ft²

Exercise 4.

Answer:
Surface area of top prism:
Area of top: 4 m × 5 m = 20 m²
Area of front and back sides: 2(4 m × 5 m) = 40 m²
Area of left and right sides: 2(5 m × 5 m) = 50 m²
Total surface area of top prism: 20 m² + 40 m² + 50 m² = 110 m²
Surface area of bottom prism:
Area of top: 10 m × 10 m – 20 m² = 80 m²
Area of bottom: 10 m × 10 m = 100 m²
Area of front and back sides: 2(10 m × 3 m) = 60 m²
Area of left and right sides: 2(10 m × 3 m) = 60 m²
Total surface area of bottom prism: 80 m² + 100 m² + 60 m² + 60 m²
= 300 m²

Surface area: 110 m² + 300 m² = 410 m²

Exercise 5.

Answer:
Area of top and bottom faces: 2(10 in. × 2 in.) + 2(6 in. × 2 in.)
= 40 in² + 24 in²
= 64 in²
Area of lateral faces: 2(9 in. × 2 in.) + 2(6 in. × 9 in.) + 2(9 in. × 10 in.)
= 36 in² + 108 in² + 180 in²
= 324 in²
Surface area:
64 in² + 324 in² = 388 in²

Eureka Math Grade 7 Module 6 Lesson 23 Problem Set Answer Key

Determine the surface area of the figures.
Question 1.

Answer:
Area of top and bottom: 2(9 cm × 4 cm) = 72 cm²
Area of left and right sides: 2(4 cm × 9 cm) = 72 cm²
Area of front and back: 2(9 cm × 4 cm) + 2(4.5 cm × 5 cm)
= 117 cm²
Surface area: 72 cm² + 72 cm² + 117 cm² = 261 cm²

Question 2.

Answer:
Area of front and back: 2(9 ft. × 2 ft.) = 36 ft²

Area of sides: 2(8 ft. × 2 ft.) = 32 ft²

Area of top and bottom: 2(9 ft. × 8 ft.) – 2(4 ft. × 3 ft.)
= 144 ft² – 24 ft²
= 120 ft²
Area of interior sides: 2(4 ft. × 2 ft.) = 16 ft²
Surface area: 36 ft² + 32 ft² + 120 ft² + 16 ft² = 204 ft²

Question 3.

Answer:
Surface area of top prism:
Area of top: 8 in. × 6 in. = 48 in²
Area of front and back sides: 2(6 in. × 8 in.) = 96 in²
Area of left and right sides: 2(8 in. × 8 in.) = 128 in²
Total surface area of top prism: 48 in² + 96 in² + 128 in² = 272 in²
Surface area of bottom prism:
Area of top: 16 in. × 16 in. – 48 in² = 208 in²
Area of bottom: 16 in. × 16 in. = 256 in²
Area of front and back sides: 2(16 in. × 4 in.) = 128 in²
Area of left and right sides: 2(16 in. × 4 in.) = 128 in²
Total surface area of bottom prism: 208 in² + 256 in² + 128 in² + 128 in²
= 720 in²

Surface area: 272 in² + 720 in² = 992 in²

Question 4.

Answer:
Area of the rectangle base: 14 ft. × 30 ft. = 420 ft²
Area of the triangular lateral sides:
Area of front and back: \(\frac{1}{2}\)bh
= 2(\(\frac{1}{2}\)(14 ft.)(24 ft.))
= 336 ft²

Area that can be seen from left and right: \(\frac{1}{2}\)bh
= 2(1\(\frac{1}{2}\)(30 ft.)(20 ft.))
= 600 ft²
Surface area: 420 ft² + 336 ft² + 600 ft² = 1,356 ft²

Question 5.

Answer:
Area of front and back: 2(\(\frac{1}{2}\)(8 cm × 15 cm))
= 8 cm × 15 cm
= 120 cm²
Area of bottom: 8 cm × 25 cm
= 200 cm²
Area that can be seen from left side: 25 cm × 15 cm
= 375 cm²
Area that can be seen from right side: 25 cm × 17 cm
= 425 cm²

Surface area: 120 cm² + 200 cm² + 375 cm² + 425 cm² = 1,120 cm²

Eureka Math Grade 7 Module 6 Lesson 23 Exit Ticket Answer Key

Determine and explain how to find the surface area of the following right prisms.
Question 1.

Answer:
Area of top and bottom: 2(\(\frac{1}{2}\)(12 ft. × 5 ft.))
= 12 ft. × 5 ft.
= 60 ft²
Area of front: 12 ft. × 15 ft.
= 180 ft²
Area seen from left: 13 ft. × 15 ft.
= 195 ft²
Area seen from right: 5 ft. × 15 ft.
= 75 ft²
Surface area: 60 ft² + 180 ft² + 195 ft² + 75 ft²
= 510 ft²
To find the surface area of the triangular prism, I must sum the areas of two triangles (the bases that are equal in area) and the areas of three different – sized rectangles.

Question 2.

Answer:
Area of front and back: 2(10 ft. × 1 ft.) = 20 ft²
Area of sides: 2(10 ft. × 1 ft.) = 20 ft²
Area of top and bottom: 2(10 ft. × 5 ft.) + 2(4 ft. × 5 ft.)
= 100 ft² + 40 ft²
= 140 ft²
Surface area: 20 ft² + 20 ft² + 140 ft²
= 180 ft²
To find the surface area of the prism, I must sum the composite area of the bases with rectangular areas of the sides of the prism.

Engage NY Eureka Math 7th Grade Module 6 Lesson 21 Answer Key

Eureka Math Grade 7 Module 6 Lesson 21 Example Answer Key

Example 1.
Examine the change in dimension and area of the following square as it increases by 2 units from a side length of 4 units to a new side length of 6 units. Observe the way the area is calculated for the new square. The lengths are given in units, and the areas of the rectangles and squares are given in units squared.

a. Based on the example above, draw a diagram for a square with a side length of 3 units that is increasing by 2 units. Show the area calculation for the larger square in the same way as in the example.
b. Draw a diagram for a square with a side length of 5 units that is increased by 3 units. Show the area calculation for the larger square in the same way as in the example.
c. Generalize the pattern for the area calculation of a square that has an increase in dimension. Let the length of the original square be a units and the increase in length be b units. Use the diagram below to guide your work.

Answer:
Area of the 6 by 6 square = (4 + 2)² units² = 4² units² + 2(2 ∙ 4) units² + 2² units² = 36 units²
The area of the 6 by 6 square can be calculated either by multiplying its sides represented by (4 + 2)(4 + 2) or by adding the areas of the subsections, represented by (4 ∙ 4), 2(2 ∙ 4), and (2 ∙ 2).
a.

Area of the 5 by 5 square = (3 + 2)² units² = 3² units² + 2(2 ∙ 3) units² + 2² units² = 25 units²
The area of the 5 by 5 square can be calculated either by multiplying its sides represented by (3 + 2)(3 + 2) or by adding the areas of the subsections, represented by (3 ∙ 3), 2(2 ∙ 3), and (2 ∙ 2).

b.

Area of the 8 by 8 square = (5 + 3)² units² = 5² units² + 2(3 ∙ 5) units² + 3² units² = 64 units²
The area of the 8 by 8 square can be calculated either by multiplying its sides represented by (5 + 3)(5 + 3) or by adding the areas of the subsections, represented by (5 ∙ 5), 2(3 ∙ 5), and (3 ∙ 3).

c.

Area of the (a + b) by (a + b) square = (a + b)² units² = (a² + 2ab + b² ) units²
The area of the square with side length (a + b) is equal to the sum of the areas of the subsections. Describing the area as (a + b)² units² is a way to state the area in terms of the dimensions of the figure; whereas describing the area as (a² + 2ab + b² ) units² is a way to state the area in terms of the sum of the areas of the sections formed by the extension of b units in each dimension.
Show how the distributive property can be used to rewrite the expression (a + b)²:

Example 2.
Bobby draws a square that is 10 units by 10 units. He increases the length by x units and the width by 2 units.
a. Draw a diagram that models this scenario.
b. Assume the area of the large rectangle is 156 unit². Find the value of x.
Answer:
a.

b. 156 = 100 + 20 + 10x + 2x
156 = 120 + 12x
156 – 120 = 120 + 12x – 120
36 = 12x
(\(\frac{1}{12}\))36 = (\(\frac{1}{12}\))12x
x = 3

Example 3.
The dimensions of a square with a side length of x units are increased. In this figure, the indicated lengths are given in units, and the indicated areas are given in units².

a. What are the dimensions of the large rectangle in the figure?
b. Use the expressions in your response from part (a) to write an equation for the area of the large rectangle, where A represents area.
c. Use the areas of the sections within the diagram to express the area of the large rectangle.
d. What can be concluded from parts (b) and (c)?
e. Explain how the expressions (x + 2)(x + 3) and x² + 3x + 2x + 6 differ within the context of the area of the figure.
Answer:
a. The length (or width) is (x + 3) units, and the width (or length) is (x + 2) units.
b. A = (x + 3)(x + 2) units²
c. A = (x² + 3x + 2x + 6) units²
d. (x + 2)(x + 3) = x² + 3x + 2x + 6
e. The expression (x + 2)(x + 3) shows the area is equal to the quantity of the width increased by 2 units times the quantity of the length increased by 3 units. The expression x² + 3x + 2x + 6 shows the area as the sum of four sections of the expanded rectangle.

Eureka Math Grade 7 Module 6 Lesson 21 Exercise Answer Key

Opening Exercise
Patty is interested in expanding her backyard garden. Currently, the garden plot has a length of 4 ft. and a width of 3 ft.
a. What is the current area of the garden?

Patty plans on extending the length of the plot by 3 ft. and the width by 2 ft.
b. What will the new dimensions of the garden be? What will the new area of the garden be?
c. Draw a diagram that shows the change in dimension and area of Patty’s garden as she expands it. The diagram should show the original garden as well as the expanded garden.
d. Based on your diagram, can the area of the garden be found in a way other than by multiplying the length by the width?
e. Based on your diagram, how would the area of the original garden change if only the length increased by 3 ft.? By how much would the area increase?
f. How would the area of the original garden change if only the width increased by 2 ft.? By how much would the area increase?
g. Complete the following table with the numeric expression, area, and increase in area for each change in the dimensions of the garden.

h. Will the increase in both the length and width by 3 ft. and 2 ft., respectively, mean that the original area will increase strictly by the areas found in parts (e) and (f)? If the area is increasing by more than the areas found in parts (e) and (f), explain what accounts for the additional increase.
Answer:

a. The garden has an area of 12 ft².
b. The new dimensions of the garden will be 7 ft. by 5 ft., and it will have an area of 35 ft².
Part (c) asks students to draw a plan of the expanded garden and quantify how the area increases. Allow students time to find a way to show this. Share out student responses; if none are able to produce a valid response, share the diagram shown on the next page.
c.

d. The area can be found by taking the sum of the smaller sections of areas.
e. The area of the garden would increase by 9 ft².
f. The area of the garden would increase by 8 ft².
g.

h. The area of the garden increases not only by 9 ft² and 8 ft², but also by an additional 6 ft². This additional 6 ft² is the corresponding area formed by the 3 ft. and 2 ft. extensions in both dimensions (length and width); that is, this area results from not just an extension in the length or just in the width but because the extensions occurred in both length and width.

Eureka Math Grade 7 Module 6 Lesson 21 Problem Set Answer Key

Question 1.
A square with a side length of a units is decreased by b units in both length and width.

Use the diagram to express(a – b)² in terms of the other a², ab, and b² by filling in the blanks below:
(a – b)² = a² – b(a – b) – b(a – b) – b²
= a² – _____ + _____ – _____ + _____ – b²
= a² – 2ab + _____ – b²
= _____________________________________
Answer:
(a – b)² = a² – b(a – b) – b(a – b) – b²
= a² – ba + b² – ba + b² – b²
= a² – 2ab + 2b² – b²
= a² – 2ab + b²

Question 2.
In Example 3, part (c), we generalized that (a + b)² = a² + 2ab + b². Use these results to evaluate the following expressions by writing 1,001 = 1,000 + 1.
a. Evaluate 101².
b. Evaluate 1,001².
c. Evaluate 21².
Answer:
a. (100 + 1)² = 100² + 2(100 ∙ 1) + 1²
= 10,000 + 200 + 1
= 10,201

b. (1,000 + 1)² = 1,000² + 2(1,000 ∙ 1) + 1²
= 1,000,000 + 2,000 + 1
= 1,002,001

c. (20 + 1)² = 20² + 2(20 ∙ 1) + 1²
= 400 + 40 + 1
= 441

Question 3.
Use the results of Problem 1 to evaluate 999². by writing 999 = 1,000 – 1.
Answer:
(1,000 – 1)² = 1,000² – 2(1,000 ∙ 1) + 1²
= 1,000,000 – 2,000 + 1
= 998,001

Question 4.
The figures below show that 8² – 5² is equal to (8 – 5)(8 + 5).

a. Create a drawing to show that a² – b² = (a – b)(a + b).
b. Use the result in part (a), a² – b² = (a – b)(a + b), to explain why:
i) 35² – 5² = (30)(40).
ii) 21² – 18² = (3)(39).
iii) 104² – 63² = (41)(167).
c. Use the fact that 35² = (30)(40) + 5² = 1,225 to create a way to mentally square any two – digit number ending in 5.
Answer:
a.

b. i) 35² – 5² = (35 – 5)(35 + 5) = (30)(40)
ii) 21² – 18² = (21 – 18)(21 + 18) = (3)(39)
iii) 104² – 63² = (104 – 63)(104 + 63) = (41)(167)

c. 15² = (10)(20) + 25 = 225
25² = (20)(30) + 25 = 625
35² = (30)(40) + 25 = 1,225
In general, if the first digit is n, then the first digit(s) are n(n + 1), and the last two digits are 25.

Question 5.
Create an area model for each product. Use the area model to write an equivalent expression that represents the area.
a. (x + 1)(x + 4) =
b. (x + 5)(x + 2) =
c. Based on the context of the area model, how do the expressions provided in parts (a) and (b) differ from the equivalent expression answers you found for each?
Answer:
a. (x + 1)(x + 4) = x² + x + 4x + 4

b. (x + 5)(x + 2) = x² + 5x + 2x + 10

c. The expression provided in each question shows the area as a length times width product or as the product of two expanded lengths. Each equivalent expression shows the area as the sum of the sections of the expanded rectangle.

Question 6.
Use the distributive property to multiply the following expressions.
a. (2 + 6)(2 + 4)
b. (x + 6)(x + 4); draw a figure that models this multiplication problem.
c. (10 + 7)(10 + 7)
d. (a + 7)(a + 7)
e. (5 – 3)(5 + 3)
f. (x – 3)(x + 3)
Answer:
a. (2 + 6)(2 + 4)
(2 + 6)(2 + 4) = (2 + 6) ∙ 2 + (2 + 6) ∙ 4
= (2 ∙ 2 + 6 ∙ 2) + (2 ∙ 4 + 6 ∙ 4)
= 2² + 10(2) + 24
= 48

b. (x + 6)(x + 4) = (x + 6) ∙ x + (x + 6) ∙ 4
= (x ∙ x + 6 ∙ x) + (x ∙ 4 + 6 ∙ 4)
= x² + 10x + 24

c. (10 + 7)(10 + 7) = (10 + 7) ∙ 10 + (10 + 7) ∙ 7
= (10 ∙ 10 + 7 ∙ 10) + (7 ∙ 10 + 7 ∙ 7)
= 10² + 2(7 ∙ 10) + 49
= 10² + 140 + 49
= 289

d. (a + 7)(a + 7) = (a + 7) ∙ a + (a + 7) ∙ 7
= (a ∙ a + 7 ∙ a) + (a ∙ 7 + 7 ∙ 7)
= a² + 2(7 ∙ a) + 49
= a² + 14a + 49

e. (5 – 3)(5 + 3) = (5 – 3) ∙ 5 + (5 – 3) ∙ 3
= (5 ∙ 5 – 3 ∙ 5) + (5 ∙ 3 – 3 ∙ 3)
= 5² – (3 ∙ 5) + (5 ∙ 3) – 3²
= 5² – 3²
= 16

f. (x – 3)(x + 3) = (x – 3) ∙ x + (x – 3) ∙ 3
= (x ∙ x – 3 ∙ x) + (x ∙ 3 – 3 ∙ 3)
= x² – (3 ∙ x) + (x ∙ 3) – 3²
= x² – 9

Eureka Math Grade 7 Module 6 Lesson 21 Exit Ticket Answer Key

Question 1.
Create an area model to represent this product: (x + 4)(x + 2).
Answer:

Question 2.
Write two different expressions that represent the area.
Answer:
(x + 2)(x + 4) and x² + 6x + 8

Question 3.
Explain how each expression represents different information about the situation.
Answer:
The expression (x + 2)(x + 4) shows the area is equal to the quantity of the length increased by 4 times the quantity of the width increased by 2. The expression x² + 4x + 2x + 8 shows the area as the sum of four sections of the expanded rectangle.

Question 4.
Show that the two expressions are equal using the distributive property.
Answer:
(x + 4)(x + 2) = (x + 4) ∙ x + (x + 2) ∙ 2
(x + 4) ∙ x + (x + 4) ∙ 2 = (x ∙ x + 4 ∙ x) + (x ∙ 2 + 4 ∙ 2)
(x ∙ x + 4 ∙ x) + (x ∙ 2 + 4 ∙ 2) = x² + 6x + 8

Engage NY Eureka Math 7th Grade Module 6 Lesson 20 Answer Key

Eureka Math Grade 7 Module 6 Lesson 20 Example Answer Key

Example 1.
A landscape company wants to plant lawn seed. A 20 lb. bag of lawn seed will cover up to 420 sq.ft. of grass and costs $49.98 plus the 8% sales tax. A scale drawing of a rectangular yard is given. The length of the longest side is 100 ft. The house, driveway, sidewalk, garden areas, and utility pad are shaded. The unshaded area has been prepared for planting grass. How many 20 lb. bags of lawn seed should be ordered, and what is the cost?

Answer:
The following calculations demonstrate how to find the area of the lawn by subtracting the area of the home from the area of the entire yard.
→ Find the non – grassy sections in the map of the yard and their areas.
A₁ = 4 units • 4 units = 16 units²
A₂ = 1 units • 13 units = 13 units²
A₃ = 7 units • 13 units = 91 units²
A₄ = 1 units • 6 units = 6 units²
A₅ = 6 units • 1 units = 6 units²
A₆ = 1 units • 6 units = 6 units²
A₇ = 2 units • 1 units = 2 units²

→ What is the total area of the non – grassy sections?
A₁ + A₂ + A₃ + A₄ + A₅ + A₆ + A₇ = 16 units² + 13 units² + 91 units² + 6 units² + 6 units² + 6 units² + 2 units² = 140 units²

→ What is the area of the grassy section of the yard?
Subtract the area of the non – grassy sections from the area of the yard.
A = (20 units • 18 units) – 140 units² = 220 units²

→ What is the scale of the map of the yard?
The scale of the map is 5 ft.
What is the grassy area in square feet?
220 units² • 25 (ft²)/(units² ) = 5,500 ft²
The area of the grassy space is 5,500 ft².

→ If one 20 lb. bag covers 420 square feet, write a numerical expression for the number of bags needed to cover the grass in the yard. Explain your expression.
→ Grassy area ÷ area that one bag of seed covers
5,500 ÷ 420
→ How many bags are needed to cover the grass in the yard?
5,500 ÷ 420≈13.1
It will take 14 bags to seed the yard.
→ What is the final cost of seeding the yard?
1.08 • 14 • $49.98≈$755.70
The final cost with sales tax is $755.70.

Example 2.
Ten dartboard targets are being painted as shown in the following figure. The radius of the smallest circle is 3 in., and each successive larger circle is 3 in. more in radius than the circle before it. A can of red paint and a can of white paint is purchased to paint the target. Each 8 oz. can of paint covers 16 ft². Is there enough paint of each color to create all ten targets?

Answer:
Let each circle be labeled as in the diagram.

Radius of C₁ is 3 in.; area of C₁ is 9π in².
Radius of C₂ is 6 in.; area of C₂ is 36π in².
Radius of C₃ is 9 in.; area of C₃ is 81π in².
Radius of C₄ is 12 in.; area of C₄ is 144π in².

→ Write a numerical expression that represents the area painted red. Explain how your expression represents the situation.
→ The area of red and white paint in square inches is found by finding the area between circles of the target board.
Red paint: (144π in² – 81π in²) + (36π in² – 9π in²)
White paint: (81π in² – 36π in²) + 9π in²
The following calculations demonstrate how to find the area of red and white paint in the target.
Target area painted red
The area between C₄ and C₃: 144π in² – 81π in² = 63π in²
The area between C₂ and C₁: 36π in² – 9π in² = 27π in²
Area painted red in one target: 63π in² + 27π in² = 90π in²; approximately 282.7 in²
Area of red paint for one target in square feet: 282.7 in² (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈1.96 ft²
Area to be painted red for ten targets in square feet: 1.96 ft²×10 = 19.6 ft²

Target area painted white
The area between C₃ and C₂: 81π in² – 36π in² = 45π in²
The area of C₁: 9π in²
Area painted white in one target: 45π in² + 9π in² = 54π in²; approximately 169.6 in²
Area of white paint for one target in square feet: 169.6 in² (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈1.18 ft²
Area of white paint for ten targets in square feet: 1.18 ft²×10 = 11.8 ft²
There is not enough red paint in one 8 oz. can of paint to complete all ten targets; however, there is enough white paint in one 8 oz. can of paint for all ten targets.

Eureka Math Grade 7 Module 6 Lesson 20 Exercise Answer Key

Opening Exercise
Find the area of each shape based on the provided measurements. Explain how you found each area.

Answer:

Triangular region: half base times height
area = \(\frac{1}{2}\) • 10 units • 7.5 units
= 37.5 units²

Parallelogram: base times height
area = 12 units • 9 units
= 108 units²

Regular hexagon: area of the shown triangle times six for the six triangles that fit into the hexagon
area = 6(\(\frac{1}{2}\)6 units • 5.2 units)
= 93.6 units²

Semicircle: half the area of a circle with the same radius
area = \(\frac{\pi}{2}\) (4.5 unit)²
= 10.125π units²
≈31.81 units²

Exercise 1.
A landscape contractor looks at a scale drawing of a yard and estimates that the area of the home and garage is the same as the area of a rectangle that is 100 ft. × 35 ft. The contractor comes up with 5,500 ft². How close is this estimate?
Answer:
The entire yard (home and garage) has an area of 100 ft.×35 ft. = 3,500 ft². The contractor’s estimate is 5,500 ft². He is 2,000 ft² over the actual area, which is quite a bit more (2,000 ft² is roughly 57% of the actual area).

Eureka Math Grade 7 Module 6 Lesson 20 Problem Set Answer Key

Question 1.
A farmer has four pieces of unfenced land as shown to the right in the scale drawing where the dimensions of one side are given. The farmer trades all of the land and $10,000 for 8 acres of similar land that is fenced. If one acre is equal to 43,560 ft², how much per square foot for the extra land did the farmer pay rounded to the nearest cent?

Answer:

A₁ = \(\frac{1}{2}\) (6 units∙4 units) = 12 units²
A₂ = \(\frac{1}{2}\) (6 units + 7 units)(4 units) = 26 units²
A₃ = (3 units • 6 units) + (3 units • 5 units) = 33 units²
A₄ = (4 units • 7 units) + (3 units • 3 units) + \(\frac{1}{2}\)(3 units∙4 units) = 43 units²

The sum of the farmer’s four pieces of land:
A₁ + A₂ + A₃ + A₄ = 12 units² + 26 units² + 33 units² + 43 units² = 114 units²

The sum of the farmer’s four pieces of land in square feet:
6 units = 300 ft.; divide each side by 6.
1 unit = 50 ft. and 1 unit² = 2,500 ft²
114 • 2,500 = 285,000
The total area of the farmer’s four pieces of land: 285,000 ft².

The sum of the farmer’s four pieces of land in acres:
285,000 ÷ 43,560≈6.54
The farmer’s four pieces of land total about 6.54 acres.
Extra land purchased with $10,000: 8 acres – 6.54 acres = 1.46 acres
Extra land in square feet:
1.46 acres(\(\frac{43,560 \mathrm{ft}^{2}}{1 \text { acre }}\))≈63597.6 ft²
Price per square foot for extra land:
(\(\frac{\$ 10,000}{63,597.6 \mathrm{ft}^{2}}\))≈$0.16

Question 2.
An ordinance was passed that required farmers to put a fence around their property. The least expensive fences cost $10 for each foot. Did the farmer save money by moving the farm?
Answer:
At $10 for each foot, $10,000 would purchase 1,000 feet of fencing. The perimeter of the third piece of land (labeled A₃) has a perimeter of 1,200 ft. So, it would have cost over $10,000 just to fence that piece of property. The farmer did save money by moving the farm.

Question 3.
A stop sign is an octagon (i.e., a polygon with eight sides) with eight equal sides and eight equal angles. The dimensions of the octagon are given. One side of the stop sign is to be painted red. If Timmy has enough paint to cover 500 ft², can he paint 100 stop signs? Explain your answer.

Answer:
Area of top trapezoid = \(\frac{1}{2}\)(12 in. + 29 in.)(8.5 in.) = 174.25 in²
Area of middle rectangle = 12 in. • 29 in. = 348 in²
Area of bottom trapezoid = \(\frac{1}{2}\)(12 in. + 29 in.)(8.5 in.) = 174.25 in²
Total area of stop sign in square inches:
A₁ + A₂ + A₃ = 174.25 in² + 348 in² + 174.25 in² = 696.5 in²
Total area of stop sign in square feet:
696.5 in ² (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈4.84 ft ²
Yes, the area of one stop sign is less than 5 ft ² (approximately 4.84 ft ²). Therefore, 100 stop signs would be less than 500 ft ².

Question 4.
The Smith family is renovating a few aspects of their home. The following diagram is of a new kitchen countertop. Approximately how many square feet of counter space is there?

Answer:

A₁ = (20 in. + 16 in.)(18 in. + 14 in.) = 1,152 in²
A₂ = (18 in. • 7 in.) + 1/2 (49π in² )
≈(126 in² + 77 in² )
≈203 in²
A₃ = (50 in. • 16 in.) – (17 in. • 16 in.) = 528 in²

Total area of counter space in square inches:
A₁ + A₂ + A₃ ≈ 1,152 in² + 203 in² + 528 in²
A₁ + A₂ + A₃ ≈ 1,883 in²

Total area of counter space in square feet:
1,883 in² (\(\frac{1 \mathrm{ft}^{2}}{144 \mathrm{in}^{2}}\))≈13.1 ft²
There is approximately 13.1 ft² of counter space.

Question 5.
In addition to the kitchen renovation, the Smiths are laying down new carpet. Everything but closets, bathrooms, and the kitchen will have new carpet. How much carpeting must be purchased for the home?

Answer:

A₁ = (9 units • 7 units) + 3 units² = 66 units²
A₂ = (6 units • 7 units) – 4 units² = 38 units²
A₃ = (6 units • 7 units) – 4 units² = 38 units²
A₄ = 2 units • 11 units = 22 units²
A₅ = (5 units • 3 units) + (4 units⋅6 units) = 39 〖units²
A₆ = 5 units • 8 units = 40 〖units²

Total area that needs carpeting:
A₁ + A₂ + A₃ + A₄ + A₅ + A₆ = 66 units² + 38 units² + 38 units² + 22 units² + 39 units² + 40 units²
= 243 units²

Scale: 1 unit = 2 ft.; 1 unit² = 4 ft²
Total area that needs carpeting in square feet:
243 units² (\(\frac{4 \mathrm{ft}^{2}}{1 \mathrm{unit}^{2}}\)) = 972 ft²

Question 6.
Jamie wants to wrap a rectangular sheet of paper completely around cans that are 8\(\frac{1}{2}\) in. high and 4 in. in diameter. She can buy a roll of paper that is 8\(\frac{1}{2}\) in. wide and 60 ft. long. How many cans will this much paper wrap?
Answer:
A can with a 4 – inch diameter has a circumference of 4π in. (i.e., approximately 12.57 in.). 60 ft. is the same as 720 in.; 720 in. ÷ 12.57 in. is approximately 57.3 in., so this paper will cover 57 cans.

Eureka Math Grade 7 Module 6 Lesson 20 Exit Ticket Answer Key

Question 1.
A homeowner called in a painter to paint the bedroom walls and ceiling. The bedroom is 18 ft. long, 12 ft. wide, and 8 ft. high. The room has two doors each 3 ft. by 7 ft. and three windows each 3 ft. by 5 ft. The doors and windows do not have to be painted. A gallon of paint can cover 300 ft². A hired painter claims he will need 4 gallons. Show that the estimate is too high.
Answer:
Area of 2 walls: 2(18 ft. • 8 ft.) = 288 ft²
Area of remaining 2 walls: 2(12 ft. • 8 ft.) = 192 ft²
Area of ceiling: 18 ft. • 12 ft. = 216 ft²
Area of 2 doors: 2(3 ft. • 7 ft.) = 42 ft²
Area of 3 windows: 3(3 ft. • 5 ft.) = 45 ft²
Area to be painted: (288 ft² + 192 ft² + 216 ft²) – (42 ft² + 45 ft²) = 609 ft²
Gallons of paint needed: 609 ÷ 300 = 2.03; The painter will need a little more than 2 gal.
The painter’s estimate for how much paint is necessary is too high.

Engage NY Eureka Math 7th Grade Module 6 Lesson 19 Answer Key

Eureka Math Grade 7 Module 6 Lesson 19 Example Answer Key

Example 1.
If slices parallel to the tabletop (with height a whole number of units from the tabletop) were taken of this figure, then what would each slice look like?

Answer:

Example 2.
If slices parallel to the tabletop were taken of this figure, then what would each slice look like?

Answer:

Example 3.
Given the level slices in the figure, how many unit cubes are in the figure?

Answer:
The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 18 cubes between Level 0 and Level 1.
Level 2: There are 13 cubes between Level 1 and Level 2.
Level 3: There are 9 cubes between Level 2 and Level 3.
Level 4: There are 5 cubes between Level 3 and Level 4.
The total number of cubes in the solid is 45.

Eureka Math Grade 7 Module 6 Lesson 19 Exercise Answer Key

Exercise 1.
Based on the level slices you determined in Example 2, how many unit cubes are in the figure?
Answer:
The number of unit cubes can be determined by counting the shaded squares in Levels 1–3.
Level 1: Five shaded squares; there are 5 cubes between Level 0 and Level 1.
Level 2: Two shaded squares; there are 2 cubes between Level 1 and Level 2.
Level 3: One shaded square; there is 1 cube between Level 2 and Level 3.
The total number of cubes in the solid is 8.

Exercise 2.
a. If slices parallel to the tabletop were taken of this figure, then what would each slice look like?

b. Given the level slices in the figure, how many unit cubes are in the figure?
Answer:
a.

b. The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 12 cubes between Level 0 and Level 1.
Level 2: There are 7 cubes between Level 1 and Level 2.
Level 3: There are 6 cubes between Level 2 and Level 3.
Level 4: There are 3 cubes between Level 3 and Level 4.
The total number of cubes in the solid is 28.

Exercise 3.
Sketch your own three-dimensional figure made from cubes and the slices of your figure. Explain how the slices relate to the figure.
Answer:
Responses will vary.

Eureka Math Grade 7 Module 6 Lesson 19 Problem Set Answer Key

In the given three-dimensional figures, unit cubes are stacked exactly on top of each other on a tabletop. Each block is either visible or below a visible block.
Question 1.
a. The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many cubes are in the figure?

Answer:
a.

b. Given the level slices in the figure, how many cubes are in the figure?
The number of unit cubes can be determined by counting the shaded squares in Levels 1–3.
Level 1: There are 9 cubes between Level 0 and Level 1.
Level 2: There are 5 cubes between Level 1 and Level 2.
Level 3: There are 3 cubes between Level 2 and Level 3.
The total number of cubes in the solid is 17.

Question 2.
a. The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many cubes are in the figure?

Answer:
a.

b. The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 15 cubes between Level 0 and Level 1.
Level 2: There are 6 cubes between Level 1 and Level 2.
Level 3: There are 2 cubes between Level 2 and Level 3.
Level 4: There is 1 cube between Level 3 and Level 4.
The total number of cubes in the solid is 24.

Question 3.
a. The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?
b. Given the level slices in the figure, how many cubes are in the figure?

Answer:
a.

b. The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 6 cubes between Level 0 and Level 1.
Level 2: There are 3 cubes between Level 1 and Level 2.
Level 3: There is 1 cube between Level 2 and Level 3.
Level 4: There is 1 cube between Level 3 and Level 4.
The total number of cubes in the solid is 11.

Question 4.
John says that we should be including the Level 0 slice when mapping slices. Naya disagrees, saying it is correct to start counting cubes from the Level 1 slice. Who is right?
Answer:
Naya is right because the Level 0 slice and Level 1 slice are the tops and bottoms of the same set of cubes; counting cubes in both slices would be double counting cubes.

Question 5.
Draw a three-dimensional figure made from cubes so that each successive layer farther away from the tabletop has one less cube than the layer below it. Use a minimum of three layers. Then draw the slices, and explain the connection between the two.
Answer:
Responses will vary.

Eureka Math Grade 7 Module 6 Lesson 19 Exit Ticket Answer Key

Question 1.
The following three-dimensional figure is built on a tabletop. If slices parallel to the tabletop are taken of this figure, then what would each slice look like?

Answer:

Question 2.
Given the level slices in the figure, how many cubes are in the figure?
Answer:
The number of unit cubes can be determined by counting the shaded squares in Levels 1–4.
Level 1: There are 11 cubes between Level 0 and Level 1.
Level 2: There are 6 cubes between Level 1 and Level 2.
Level 3: There are 5 cubes between Level 2 and Level 3.
Level 4: There is 1 cube between Level 3 and Level 4.
The total number of cubes in the solid is 23.

Engage NY Eureka Math 7th Grade Module 6 Lesson 18 Answer Key

Eureka Math Grade 7 Module 6 Lesson 18 Example Answer Key

Example 1.
With your group, discuss whether a right rectangular prism can be sliced at an angle so that the resulting slice looks like the figure in Figure 1. If it is possible, draw an example of such a slice into the following prism.

Answer:
Once students have had some time to attempt Example 1, pose the following questions:
→ Is it possible to make a triangular slice from this prism? Where would this slice have to be made? Students may choose their method of how to represent the slice (e.g., drawing a 2D or 3D sketch or describing the slice in words).
→ Yes, it is possible. It can be done by slicing off a corner of the right rectangular prism.

Here, and elsewhere, some students may be able to visualize this right away, while others may struggle.
Allow some time to see if any one group has a valid answer. If there is a valid answer, use it for the next item; otherwise, share the following triangular slice of a right rectangular prism.
→ Here is a slice that results in a triangular region. Does this triangle have three equal sides, two equal sides, or no equal sides?
Encourage students to use a ruler to verify their answers.

→ The slice is a triangle with no equal sides.
→ At how many points does the slice meet an edge of the right rectangular prism? What makes these points important with respect to the triangle?
→ The slice meets an edge of the right rectangular prism at three points, one on each of three edges; these three points are the vertices of the triangle.
→ Find another slice that will create another scalene triangular region. Mark the vertices on the edges of the prism.

Example 2.
With your group, discuss whether a right rectangular prism can be sliced at an angle so that the resulting slice looks like the figure in Figure 4. If it is possible, draw an example of such a slice into the following prism.

Answer:
Once students have had some time to attempt Example 2, pose the following questions:
→ Is it possible to slice this right rectangular prism to form a quadrilateral cross section? Remember, we are slicing at an angle now. Consider what you know about vertices and the edges they fall on from Example 1.

Again, some students may be able to visualize this right away, while some might not. Give students time to experiment with the solution. Use a valid response to the question to move forward, or simply share the image provided.

→ Here is one possible slice in the shape of a quadrilateral. Notice that the slice is, in fact, made by a plane.
A common error, especially when outlining a quadrilateral slice by its vertices, is to make a slice that is not a true slice because the figure could not be made by a single plane.
→ What must be true about the opposite sides of the quadrilateral?
→ The opposite sides are parallel.
The conclusion that the opposite sides of the quadrilateral region are parallel is based on the above image. This conclusion likely comes from their understanding that the opposite faces of the right rectangular prism are parallel. Therefore, since the opposite sides lie in these faces, they too must be parallel.
This might be an appropriate time to show them an image like the following:

→ Though the segments lie in planes that are a constant distance apart, the segments are not parallel.
→ Because the earlier slice is made by one plane, the segments that form the sides of the quadrilateral-shaped slice both lie in the same plane as each other and in opposite faces that are an equal distance apart. Together, this means the segments are parallel.
→ Not only is this slice a quadrilateral, but it is a special quadrilateral. What kind of special quadrilateral-shaped slice must it be?
→ Since we have determined the opposite sides of the quadrilateral to be parallel, the quadrilateral must be a parallelogram.
Have students draw another example of a slice through the right rectangular prism that results in a parallelogram shape.

Example 3.
a. If slicing a plane through a right rectangular prism so that the slice meets the three faces of the prism, the resulting slice is in the shape of a triangle; if the slice meets four faces, the resulting slice is in the shape of a quadrilateral. Is it possible to slice the prism in a way that the region formed is a pentagon (as in Figure 5)? A hexagon (as in Figure 6)? An octagon (as in Figure 7)?

b. Draw an example of a slice in a pentagon shape and a slice in a hexagon shape.
Answer:
a. Yes, it is possible to slice a right rectangular prism with a plane so that the resulting cross section is a pentagon; the slice would have to meet five of the six faces of the prism. Similarly, it is possible for the slice to take the shape of a hexagon if the slice meets all six faces. It is impossible to create a slice in the shape of an octagon because a right rectangular prism has six faces, and it is not possible for the shape of a slice to have more sides than the number of faces of the solid.

b.

Example 4.
a. With your group, discuss whether a right rectangular pyramid can be sliced at an angle so that the resulting slice looks like the figure in Figure 8. If it is possible, draw an example of such a slice into the following pyramid.

b. With your group, discuss whether a right rectangular pyramid can be sliced at an angle so that the resulting slice looks like the figure in Figure 9. If it is possible, draw an example of such a slice into the pyramid above.

Answer:
Allow students time to experiment with this slice. Remind students that marking the vertices of the slice on the edges of the pyramid facilitates the drawing of the slice.
If there is no valid response, share the figure below.

Ask students to find a second slice in the shape of a pentagon.

b. It is impossible to create a slice in the shape of a hexagon because a right rectangular pyramid has five faces, and it is not possible for the shape of a slice to have more sides than the number of faces of the solid.

Eureka Math Grade 7 Module 6 Lesson 18 Exercise Answer Key

Exercise 1.
a. With your group, discuss how to slice a right rectangular prism so that the resulting slice looks like the figure in Figure 2. Justify your reasoning.

b. With your group, discuss how to slice a right rectangular prism so that the resulting slice looks like the figure in Figure 3. Justify your reasoning.

Answer:
a. I would use a ruler to measure two segments of equal length on two edges that meet at a common vertex. Then, I would join these two endpoints with a third segment.
b. I would use a ruler to measure three segments of equal length on three edges that meet at a common vertex.

Exercise 2.
In Example 2, we discovered how to slice a right rectangular prism to makes the shapes of a rectangle and a parallelogram. Are there other ways to slice a right rectangular prism that result in other quadrilateral-shaped slices?
Answer:
Allow students more time to experiment with other possible slices that might result in another kind of quadrilateral.
If there is no valid response, share the figure below.

A slice can be made to a right rectangular prism at an angle so that the resulting cross section is a trapezoid
(as in the example shown above). In addition to slicing at an angle, it is also possible to slice perpendicular to a face or base to form a trapezoid-shaped slice.

Eureka Math Grade 7 Module 6 Lesson 18 Problem Set Answer Key

Question 1.
Draw a slice into the right rectangular prism at an angle in the form of the provided shape, and draw each slice as a 2D shape.

Answer:

Question 2.
Draw slices at an angle in the form of each given shape into each right rectangular pyramid, and draw each slice as a 2D shape.

Answer:

Question 3.
Why is it not possible to draw a slice in the shape of a hexagon for a right rectangular pyramid?
Answer:
It is not possible for the shape of a slice to have more sides than the number of faces of the solid.

Question 4.
If the slicing plane meets every face of a right rectangular prism, then the slice is a hexagonal region. What can you say about opposite sides of the hexagon?
Answer:
The opposite sides of the hexagon lie in opposite faces; therefore, they are parallel.

Question 5.
Draw a right rectangular prism so that rectangles ABCD and A’B’C’D’ are base faces. The line segments AA’, BB’, CC’, and DD’ are edges of the lateral faces.
a. A slicing plane meets the prism so that vertices A, B, C, and D lie on one side of the plane, and vertices A’, B’, C’, and D’ lie on the other side. Based on the slice’s position, what other information can be concluded about the slice?
b. A slicing plane meets the prism so that vertices A, B, C, and B’ are on one side of the plane, and vertices A’, D’, C’, and D are on the other side. What other information can be concluded about the slice based on its position?
Answer:
a. The slice misses the base faces ABCD and A’B’C’D’ since all the vertices of each face lie on the same side of the plane. The slice meets each of the lateral faces in an interval since each lateral face has two vertices on each side. The slice is a quadrilateral. In fact, the slice is a parallelogram because opposite faces of a right rectangular prism lie in parallel planes.

b. The slice meets each face in line segments because, in each case, three of the vertices of the face are on one side of the plane and the remaining vertex lies in the opposite side. The slice is a hexagon because it has six edges. Opposite sides of the hexagon are parallel since they lie in parallel planes.

Eureka Math Grade 7 Module 6 Lesson 18 Exit Ticket Answer Key

Question 1.
Draw a slice that has the maximum possible number of sides for each solid. Explain how you got your answer.

Answer:
The slice in the right rectangular prism should be hexagonal (diagrams will vary); the slice in the right rectangular pyramid should be pentagonal (again, diagrams will vary).
The edges of a slice are determined by the number of faces the slicing plane meets; there cannot be more sides to the polygon than there are faces of the solid.

Engage NY Eureka Math 7th Grade Module 6 Lesson 17 Answer Key

Eureka Math Grade 7 Module 6 Lesson 17 Example Answer Key

Example 1.
Use the models you built to assist in a sketch of a pyramid. Though you are sketching from a model that is opaque, use dotted lines to represent the edges that cannot be seen from your perspective.
Answer:
Sketches will vary; emphasize the distinction between the pyramids by asking how each must be drawn.

Students may struggle with this example; remind them that attempting these sketches is not a test of artistic ability but rather an exercise in becoming more familiar with the structure of a pyramid. They are working toward visualizing how a plane slices a rectangular pyramid perpendicular and parallel to its base.

Example 2.
Sketch a right rectangular pyramid from three vantage points: (1) from directly over the vertex, (2) from facing straight on to a lateral face, and (3) from the bottom of the pyramid. Explain how each drawing shows each view of the pyramid.
Answer:
Possible sketches:

1. From directly overhead, all four lateral faces are visible.
2. From facing a lateral face, the one entire lateral face is visible, as well as a bit of the adjacent lateral faces. If the pyramid were transparent, I would be able to see the entire base.
3. From the bottom, only the rectangular base is visible.

Example 3.
Assume the following figure is a top-down view of a rectangular pyramid. Make a reasonable sketch of any two adjacent lateral faces. What measurements must be the same between the two lateral faces? Mark the equal measurement. Justify your reasoning for your choice of equal measurements.

Answer:

The equal sides in each triangle are the same between both triangles.

Students may think that the heights of the triangles are equal in length, when in fact they are not, unless the base is a square. The triangle with the shorter base has a height greater than that of the triangle with the longer base.
An easy way of making this point is by looking at a right rectangular pyramid with a rectangular base of exaggerated dimensions: a very long length in contrast to a very short width. Though students do not yet have the Pythagorean theorem at their disposal to help them quantify the difference in heights of the lateral faces, an image should be sufficiently persuasive.

Example 4.
a. A slicing plane passes through segment a parallel to base B of the right rectangular pyramid below. Sketch what the slice will look like into the figure. Then sketch the resulting slice as a two-dimensional figure. Students may choose how to represent the slice (e.g., drawing a 2D or 3D sketch or describing the slice in words).

b. What shape does the slice make? What is the relationship between the slice and the rectangular base of the pyramid?
Answer:
a.
b. The slice is a rectangle; the slice looks a lot like the rectangular base but is smaller in size.

Example 5.
A slice is to be made along segment a perpendicular to base B of the right rectangular pyramid below.
a. Which of the following figures shows the correct slice? Justify why each of the following figures is or is not a correct diagram of the slice.

b. A slice is taken through the vertex of the pyramid perpendicular to the base. Sketch what the slice will look like into the figure. Then, sketch the resulting slice itself as a two-dimensional figure.

Answer:
a.
This is not a slice by a plane because there is space between the base and the lateral face. This could be a slice by a rectangle with the same width as in the figure but not a plane that extends in all directions.

This is not a slice by a plane perpendicular to the base because the marked rectangular region is in the same plane as the lateral face in which it lies.

This is a slice made by a plane meeting the rectangular pyramid perpendicular to its base. The slice shows all the possible points at which the slicing plane would make contact with the right rectangular pyramid. The slice is in the shape of an isosceles trapezoid.

b. A slice is taken through the vertex of the pyramid perpendicular to the base. Sketch what the slice will look like into the figure. Then, sketch the resulting slice itself as a two-dimensional figure.

Eureka Math Grade 7 Module 6 Lesson 17 Problem Set Answer Key

Question 1.
A side view of a right rectangular pyramid is given. The line segments lie in the lateral faces.

a. For segments n, s, and r, sketch the resulting slice from slicing the right rectangular pyramid with a slicing plane that contains the line segment and is perpendicular to the base.
b. For segment m, sketch the resulting slice from slicing the right rectangular pyramid with a slicing plane that contains the segment and is parallel to the base.
Note: To challenge yourself, you can try drawing the slice into the pyramid.
c. A top view of a right rectangular pyramid is given. The line segments lie in the base face. For each line segment, sketch the slice that results from slicing the right rectangular pyramid with a plane that contains the line segment and is perpendicular to the base.

Answer:


Note that the diagram for the slice made through s is from a perspective different from the one in the original pyramid. From the original perspective, the slice itself would not be visible and would appear as follows:

c.

Eureka Math Grade 7 Module 6 Lesson 17 Exit Ticket Answer Key

Question 1.
Two copies of the same right rectangular pyramid are shown below. Draw in the slice along segment c perpendicular to the base and the slice along segment c parallel to the base. Then, sketch the resulting slices as two-dimensional figures.

Slice Perpendicular to Base Slice Parallel to Base
Answer:

Engage NY Eureka Math 7th Grade Module 6 Lesson 16 Answer Key

Eureka Math Grade 7 Module 6 Lesson 16 Example Answer Key

Example 1.
Consider a ball B. Figure 3 shows one possible slice of B.
a. What figure does the slicing plane form? Students may choose their method of representation of the slice (e.g., drawing a 2D sketch, a 3D sketch, or describing the slice in words).

b. Will all slices that pass through B be the same size? Explain your reasoning.
c. How will the plane have to meet the ball so that the plane section consists of just one point?
Answer:
a. A circle (or disc)
b. No, different slices can result in circles of different sizes; it will depend on where the slicing plane meets the ball.
c. If you picture the ball and the plane as distinct but being brought toward each other, the plane section of just one point occurs when the plane just makes contact with the ball.

Example 2.
The right rectangular prism in Figure 4 has been sliced with a plane parallel to face ABCD. The resulting slice is a rectangular region that is identical to the parallel face.

a. Label the vertices of the rectangular region defined by the slice as WXYZ.
b. To which other face is the slice parallel and identical?
c. Based on what you know about right rectangular prisms, which faces must the slice be perpendicular to?
Answer:
a.
b. The slice is parallel and identical to the face EFGH.
c. Since the slice is parallel to two faces, it will be perpendicular to whichever sides those faces are perpendicular to. Therefore, the slice is perpendicular to faces ABHG, CDFE, BCEH, and ADFG.

Example 3.
The right rectangular prism in Figure 6 has been sliced with a plane perpendicular to BCEH. The resulting slice is a rectangular region with a height equal to the height of the prism.

a. Label the vertices of the rectangle defined by the slice as WXYZ.
b. To which other face is the slice perpendicular?
c. What is the height of rectangle WXYZ?
d. Joey looks at WXYZ and thinks that the slice may be a parallelogram that is not a rectangle. Based on what is known about how the slice is made, can he be right? Justify your reasoning.
Answer:
a.
b. The slice is perpendicular to the face ADFG.
c. 6 in.
d. The slice was made perpendicular to face BCEH. Then we know that the angles in the slice, ∠X and ∠Y, formed by the slicing plane and face BCEH, are right angles. If we focus on ∠X of the slice, since it is a right angle, we know that \(\overline{W X}\) must be perpendicular to face BCEH. \(\overline{W X}\) lies in face ABCD, which is perpendicular to both BCEH and to ADFG, so \(\overline{W X}\)is perpendicular to ABCD. This means that \(\overline{W X}\) must also be perpendicular to \(\overline{W Z}\). A similar argument can be made for ∠Y of the slice, making all four angles of WXYZ right angles and making WXYZ a rectangle.

Eureka Math Grade 7 Module 6 Lesson 16 Exercise Answer Key

Exercise 1.
Discuss the following questions with your group.
The right rectangular prism in Figure 5 has been sliced with a plane parallel to face LMON.

a. Label the vertices of the rectangle defined by the slice as RSTU.
b. What are the dimensions of the slice?
c. Based on what you know about right rectangular prisms, which faces must the slice be perpendicular to?
Answer:
a.
b. 10 cm × 11 cm
c. Since the slice is parallel to two faces, it will be perpendicular to whichever sides those faces are perpendicular to. Therefore, the slice is perpendicular to faces JKLM, JMOP, NOPQ, and KLNQ.

In the following exercises, the points at which a slicing plane meets the edges of the right rectangular prism have been marked. Each slice is either parallel or perpendicular to a face of the prism. Use a straightedge to join the points to outline the rectangular region defined by the slice, and shade in the rectangular slice.
Exercise 2.
A slice parallel to a face

Answer:

Exercise 3.
A slice perpendicular to a face

Answer:

Exercise 4.
A slice perpendicular to a face

Answer:

In Exercises 5–6, the dimensions of the prisms have been provided. Use the dimensions to sketch the slice from each prism, and provide the dimensions of each slice.
Exercise 5.
A slice parallel to a face

Answer:

Exercise 6.
A slice perpendicular to a face

Answer:

Eureka Math Grade 7 Module 6 Lesson 16 Problem Set Answer Key

Question 1.
A right rectangular prism is shown along with line segments that lie in a face. For each line segment, draw and give the approximate dimensions of the slice that results when the slicing plane contains the given line segment and is perpendicular to the face that contains the line segment.

a.

b.

c.

d.

e.

f.

g.

Answer:
a.

b.

c.

d.

e.

f.

g.

Eureka Math Grade 7 Module 6 Lesson 16 Exit Ticket Answer Key

In the following figures, use a straightedge to join the points where a slicing plane meets with a right rectangular prism to outline the slice.
i) Label the vertices of the rectangular slice WXYZ.
ii) State any known dimensions of the slice.
iii) Describe two relationships slice WXYZ has in relation to faces of the right rectangular prism.
Question 1.

Answer:

Sides \(\overline{W Z}\) and \(\overline{X Y}\) are 6 cm in length. Slice WXYZ is perpendicular to faces ABCD and EFGH.

Question 2.

Answer:

Sides \(\overline{W Z}\) and \(\overline{X Y}\) are 7 cm in length. Sides \(\overline{W X}\) and \(\overline{Z Y}\) are 8 cm in length. Slice WXYZ is parallel to faces ABCD and EFGH and perpendicular to faces CDEH, ADEF, ABGF, and BCHG.
Note: Students are only required to state two of the relationships the slice has with the faces of the prism.

Engage NY Eureka Math 7th Grade Module 6 Lesson 15 Answer Key

Eureka Math Grade 7 Module 6 Lesson 15 Example Answer Key

Example 1.
A triangular fence with two equal angles, ∠S = ∠T, is used to enclose some sheep. A fence is constructed inside the triangle that exactly cuts the other angle into two equal angles: ∠SRW = ∠TRW. Show that the gates, represented by \(\overline{S W}\) and \(\overline{W T}\), are the same width.

Answer:
There is a correspondence △SRW ↔ △TRW that matches two pairs of angles of equal measurement, ∠S = ∠T and ∠SRW = ∠TRW, and one pair of sides of equal length shared, side \(\overline{R W}\). The triangles satisfy the two angles and side opposite a given angle condition. From the correspondence, we can conclude that SW = WT, or that the gates are of equal width.

Example 2.
In △ABC, AC = BC, and △ABC ↔ △B’ A’ C’. John says that the triangle correspondence matches two sides and the included angle and shows that ∠A = ∠B’. Is John correct?

Answer:
We are told that AC = BC. The correspondence △ABC ↔ △B’A’C’ tells us that BC ↔ A’C’, CA ↔ C’B’, and ∠C↔∠C’, which means △ABC is identical to △B’A’C’ by the two sides and included angle condition. From the correspondence, we can conclude that ∠A = ∠B’; therefore, John is correct.

Eureka Math Grade 7 Module 6 Lesson 15 Exercise Answer Key

Exercise 1.
Mary puts the center of her compass at the vertex O of the angle and locates points A and B on the sides of the angle. Next, she centers her compass at each of A and B to locate point C. Finally, she constructs the ray \(\overrightarrow{O C}\). Explain why ∠BOC = ∠AOC.

Answer:
Since Mary uses one compass adjustment to determine points A and B, OA = OB. Mary also uses the same compass adjustment from B and A to find point C; this means BC = AC. Side \(\overrightarrow{O C}\) is common to both the triangles,
△OBC and △OAC. Therefore, there is a correspondence △OBC ↔ △OAC that matches three pairs of equal sides, and the triangles are identical by the three sides condition. From the correspondence, we conclude that
∠BOC = ∠AOC.

Exercise 2.
Quadrilateral ACBD is a model of a kite. The diagonals \(\overline{A B}\) and \(\overline{C D}\) represent the sticks that help keep the kite rigid.
a. John says that ∠ACD = ∠BCD. Can you use identical triangles to show that John is correct?

b. Jill says that the two sticks are perpendicular to each other. Use the fact that ∠ACD = ∠BCD and what you know about identical triangles to show ∠AEC = 90°.
c. John says that Jill’s triangle correspondence that shows the sticks are perpendicular to each other also shows that the sticks cross at the midpoint of the horizontal stick. Is John correct? Explain.
Answer:
a. From the diagram, we see that AC = BC, and AD = BD. \(\overline{C D}\) is a common side to both triangles, △ACD and △BCD. There is a correspondence △ACD ↔ △BCD that matches three pairs of equal sides; the two triangles are identical by the three sides condition. From the correspondence, we conclude that ∠ACD = ∠BCD. John is correct.

b. Since we know that AC = BC and ∠ACD = ∠BCD, and that △ACE and △BCE share a common side, \(\overline{C E}\), we can find a correspondence that matches two pairs of equal sides and a pair of equal, included angles. The triangles are identical by the two sides and included angle condition. We can then conclude that ∠AEC = ∠BEC. Since both angles are adjacent to each other on a straight line, we also know their measures must sum to 180°. We can then conclude that each angle measures 90°.

c. Since we have established that △ACE and △BCE are adjacent to each other, we know that AE = BE. This means that E is the midpoint of \(\overline{A B}\), by definition.

Exercise 3.
In △ABC, ∠A = ∠B, and △ABC ↔ △B’A’C’. Jill says that the triangle correspondence matches two angles and the included side and shows that AC = B’C’. Is Jill correct?

Answer:
We are told that ∠A = ∠B. The correspondence △ABC ↔ △B’A’C’ tells us that ∠A = ∠B’, ∠B = ∠A’, and AB = B’A’, which means △ABC is identical to △B’A’C’ by the two angles and included side condition. From the correspondence, we can conclude that AC = B’C’; therefore, Jill is correct.

Exercise 4.
Right triangular corner flags are used to mark a soccer field. The vinyl flags have a base of 40 cm and a height of 14 cm.
a. Mary says that the two flags can be obtained by cutting a rectangle that is 40 cm×14 cm on the diagonal. Will that create two identical flags? Explain.

b. Will measures the two non-right angles on a flag and adds the measurements together. Can you explain, without measuring the angles, why his answer is 90°?
Answer:
a. If the flag is to be cut from a rectangle, both triangles will have a side of length 40 cm, a length of 14 cm, and a right angle. There is a correspondence that matches two pairs of equal sides and an included pair of equal angles to the corner flag; the two triangles are identical to the corner flag as well as to each other.

b. The two non-right angles of the flags are adjacent angles that together form one angle of the four angles of the rectangle. We know that a rectangle has four right angles, so it must be that the two non-right angles of the flag together sum to 90°.

Eureka Math Grade 7 Module 6 Lesson 15 Problem Set Answer Key

Question 1.
Jack is asked to cut a cake into 8 equal pieces. He first cuts it into equal fourths in the shape of rectangles, and then he cuts each rectangle along a diagonal.
Did he cut the cake into 8 equal pieces? Explain.

Answer:

Yes, Jack cut the cake into 8 equal pieces. Since the first series of cuts divided the cake into equal fourths in the shape of rectangles, we know that the opposite sides of the rectangles are equal in length; that means all 8 triangles have two sides that are equal in length to each other. Each of the triangular pieces also has one right angle because we know that rectangles have four right angles. Therefore, there is a correspondence between all 8 triangles that matches two pairs of equal sides and an equal, 90° non-included angle, determining 8 identical pieces of cake.

Question 2.
The bridge below, which crosses a river, is built out of two triangular supports. The point M lies on \(\overline{B C}\) . The beams represented by \(\overline{A M}\) and \(\overline{D M}\) are equal in length, and the beams represented by \(\overline{A B}\) and \(\overline{D C}\) are equal in length. If the supports were constructed so that ∠A and ∠D are equal in measurement, is point M the midpoint of \(\overline{B C}\)? Explain.

Answer:
Yes, M is the midpoint of \(\overline{B C}\). The triangles are identical by the two sides and included angle condition. The correspondence △ABM ↔ △DCM matches two pairs of equal sides and one pair of included equal angles. Since the triangles are identical, we can use the correspondence to conclude that BM = CM, which makes M the midpoint, by definition.

Eureka Math Grade 7 Module 6 Lesson 15 Exit Ticket Answer Key

Question 1.
Alice is cutting wrapping paper to size to fit a package. How should she cut the rectangular paper into two triangles to ensure that each piece of wrapping paper is the same? Use your knowledge of conditions that determine unique triangles to justify that the pieces resulting from the cut are the same.

Answer:
Alice should cut along the diagonal of rectangle ABCD. Since ABCD is a rectangle, the opposite sides will be equal in length, or AB = DC and AD = BC. A rectangle also has four right angles, which means a cut along the diagonal will result in each triangle with one 90° angle. The correspondence △ABD ↔ △CDB matches two equal pairs of sides and an equal, included pair of angles; the triangles are identical by the two sides and included angle condition.

Engage NY Eureka Math 7th Grade Module 6 Lesson 14 Answer Key

Eureka Math Grade 7 Module 6 Lesson 14 Example Answer Key

Example 1.
What is the relationship between the two triangles below?

Answer:
What is the relationship between the two triangles below?
The triangles share a common side.
Imagine that △WXY and △WZY were pulled apart and separated. Sketch the triangles separately. Based on how they were joined, what kind of tick mark should be added to each triangle?
\(\overline{W Y}\) is a common side. Since it belongs to each triangle, we should put a triple tick mark on \(\overline{W Y}\) to indicate that it is a part of equal measure in both triangles.

Are the triangles identical? How do you know?
The triangles are identical by the three sides condition. The correspondence that matches the three equal pairs of sides is △WXY↔△YZW.

Example 2.
Are the triangles identical? Justify your reasoning.

Answer:
The triangles are identical by the two sides and the included angle condition. The correspondence △AOC ↔ △BOD matches two equal pairs of sides and a pair of equal angles, ∠AOC = ∠BOD, which we know to be equal in measurement because they are vertical angles.

Eureka Math Grade 7 Module 6 Lesson 14 Exercise Answer Key

Exercise 1.
Are the triangles identical? Justify your reasoning.

Answer:
The triangles are not necessarily identical. The correspondence △QRT ↔ △SRT matches a pair of equal angles and a pair of equal sides. The correspondence also matches a common side, \(\overline{R T}\), to both triangles. Two sides and a non-included acute angle do not necessarily determine a unique triangle.

Exercise 2.
Are the triangles identical? Justify your reasoning.

Answer:
These triangles are identical by the two angles and side opposite a given angle condition. The correspondence △EFH ↔ △GFH matches two pairs of equal angles and a pair of equal sides. ∠HFE must be a right angle since ∠HFG is a right angle, and they are both on a line. Also, \(\overline{H F}\) is a common side to both triangles. Since both triangles have parts under the condition of the same measurement, the triangles must be identical.

Exercise 3.
Are the triangles identical? Justify your reasoning.

Answer:
These triangles are identical by the two angles and side opposite a given angle condition. The correspondence △AOB ↔ △COD matches the two pairs of equal angles and one pair of equal sides. There is a marked pair of equal sides and one pair of marked, equal angles. The second pair of equal angles, ∠AOB = ∠COD, are vertical angles.

Exercise 4.
Are the triangles identical? Justify your reasoning.

Answer:
The triangles are not necessarily identical. The correspondence △AEB ↔ △CED matches three pairs of equal angles, including the unmarked angles, ∠AEB and ∠CED, which are equal in measurement because they are vertical angles. The triangles could have different side lengths; therefore, they are not necessarily identical.

Exercise 5.
Are the triangles identical? Justify your reasoning.

Answer:
The triangles are identical by the two sides and non-included 90° (or greater) angle condition. The correspondence △MPO ↔ △NPO matches two pairs of equal sides and one pair of equal angles. One of the two pairs of equal sides is side \(\overline{O P}\), which is common to both triangles.

Exercise 6.
Are the triangles identical? Justify your reasoning.

Answer:
These triangles are not necessarily identical. The triangles have a pair of marked equal sides and equal angles; side \(\overline{C E}\) is also common to both triangles. The triangles satisfy the two sides and non-included acute angle condition, which does not determine a unique triangle.

Exercise 7.
Are the triangles identical? Justify your reasoning.

Answer:
These triangles are identical by the two angles and a side opposite a given angle condition. The triangle correspondence △RWS ↔ △RWT matches two pairs of equal angles and one pair of equal sides. The equal pair of sides, \(\overline{R W}\), is common to both triangles.

Exercise 8.
Create your own labeled diagram and set of criteria for a pair of triangles. Ask a neighbor to determine whether the triangles are identical based on the provided information.
Answer:
Answers will vary.

Eureka Math Grade 7 Module 6 Lesson 14 Problem Set Answer Key

In the following problems, determine whether the triangles are identical, not identical, or not necessarily identical; justify your reasoning. If the relationship between the two triangles yields information that establishes a condition, describe the information. If the triangles are identical, write a triangle correspondence that matches the sides and angles.
Question 1.

Answer:
These triangles are identical by the two sides and the included angle condition. The triangle correspondence △ABC ↔ △ADC matches two pairs of equal sides and one pair of equal angles. One of the equal pairs of sides is shared side \(\overline{A C}\).

Question 2.

Answer:
These triangles are not necessarily identical. The triangles have a pair of marked equal sides, and a pair of marked, acute, equal angles; side \(\overline{A C}\) is also common to both triangles. The triangles satisfy the two sides and non-included acute angle condition, which does not determine a unique triangle.

Question 3.

Answer:
These triangles are identical by the two angles and included side condition. The triangle correspondence △QPR ↔ △TSR matches the two pairs of equal angles and one pair of equal sides. One pair of equal angles is the vertical angle pair ∠PRQ = ∠SRT.

Question 4.

Answer:
These triangles are not necessarily identical. A correspondence that matches up the equal pair of sides and the equal pair of vertical angles does not match the marked equal pair of angles.

Question 5.

Answer:
The triangles are identical by the two sides and non-included 90° (or greater) angle condition. The correspondence △IFG ↔ △GHI matches two pairs of equal sides and one pair of equal angles. One of the two pairs of equal sides is side \(\overline{I G}\), which is common to both triangles.

Question 6.

Answer:
The triangles are not identical since a correspondence that matches the two marked equal pairs of sides also matches sides \(\overline{V X}\) and \(\overline{X Y}\), which are not equal in length.

Question 7.

Answer:
These triangles are identical by the two angles and side opposite a given angle condition. The correspondence △ABD ↔ △CBD matches the two pairs of equal angles and one pair of equal sides. The pair of equal sides is the common side, \(\overline{B D}\). We know ∠ADB must be a right angle since ∠CDB is a right angle; they are both on a line, and of course angles ∠A and ∠C are equal in measurement.

Question 8.
Are there any identical triangles in this diagram?

Answer:
△TUX and △VWX are not necessarily identical; we only know about a single pair of equal angles in the triangles, which are vertical angles. △TUV and △VWT are also not necessarily identical. We only know about a single pair of equal angles and a side common to both triangles, which is not enough information to determine the triangles as identical or non-identical.

Question 9.

Answer:
The triangles are not necessarily identical since there is no correspondence that matches the two marked equal pairs of sides as well as the two pairs of equal angles. One of the pairs of equal angles is the pair of vertical angles.

Question 10.

Answer:
△ABE and △CBD are not necessarily identical. The triangles satisfy the two sides and non-included acute angle condition, which does not determine a unique triangle.

Eureka Math Grade 7 Module 6 Lesson 14 Exit Ticket Answer Key

Question 1.
Are △DEF and △DGF identical, not identical, or not necessarily identical? Justify your reasoning. If the relationship between the two triangles yields information that establishes a condition, describe the information. If the triangles are identical, write a triangle correspondence that matches the sides and angles.

Answer:
These triangles are identical by the two angles and side opposite a given angle condition. The triangle correspondence △DEF ↔ △DGF matches the two pairs of equal angles and one pair of equal sides condition. The pair of equal sides, \(\overline{D F}\), is common to both triangles.