Prasanna

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 10.3 Answer Key Lateral and Total Surface Area.

Texas Go Math Grade 7 Lesson 10.3 Answer Key Lateral and Total Surface Area

Example 1
A net of a rectangular prism is shown. Use the net to find the lateral area and the total surface area of the prism. Each square represents one square inch. The blue regions are the bases, and the green regions are the lateral faces.

Step 1
Find the lateral area of the rectangular prism. There are two 2 in. by 5 in. rectangles, and two 2 in. by 6 in. rectangles.

2 . (2 . 5) = 20 in²
2 . (2 . 6) = 24 in²
The lateral area is 20 + 24 = 44 in².

Step 2
Find the total surface area of the rectangular prism.
Each base is 5 inches by 6 inches. 2 . (5 . 6) = 60 in².
The total surface area is 44 + 60 = 104 in².

Math Talk
Mathematical processes
Explain how the total surface area of a prism differs from the lateral area.

Your Turn

7th Grade Go Math Answer Key Lesson 10.3 Question 1.
Use the net to find the lateral area and the total surface area of the triangular prism described by the net.

Answer:
Find the Lateral area of the triangular prism. We have:
One 4 cm. by 4 cm. rectangle;
One 4 cm. by 5 cm. rectangle;
One 3 cm. by 4 cm. rectangle.
1 . (4 . 4) = 16
1 . (4 . 5) = 20
1 . (3 . 4) = 12
The lateral area is 16 + 20 + 12 = 48 cm².
Each base is 4 cm. by 3 cm.\
2 . (4 . 3) = 2 . 12 = 24 cm²
The total surfance area is 48 – 24 = 72 cm².

Example 2
Use the net of this rectangular pyramid to find the lateral area and the total surface area of this pyramid.

Step 1
Find the lateral area of the rectangular pyramid. There are four triangles with base 16 in. and height 17 in. The lateral area is 4 × \(\frac{1}{2}\)(16)(17) = 544 in².

Step 2
Find the total surface area of the rectangular pyramid.
The area of the base is 16 × 16 = 256 in².
The total surface area is 544 + 256 = 800 in².

Reflect

Question 2.
How many surfaces does a triangular pyramid have? What shape are they?
Answer:
A triangular pyramid has four surfaces and they have the shape of triangle.

Your Turn

Question 3.
The base and all three faces of a triangular pyramid are equilateral triangles with side lengths of 3 ft. The height of each triangle is 2.6 ft. Use the net to find the lateral area and the total surface area of the triangular pyramid.

Answer:
Find the lateral area of the triangular pyramid.
We have three equilateral triangles with base 3 ft. and height 17 ft.
b = 3 ft Length of base
h_B = 2.6 ft Height of base
Lateral area:
3 . (\(\frac{1}{2}\) . b . h_b) = 3 . \(\frac{1}{2}\) . 3 . (2.6) = 11.7 ft²
The base is the same triangle as triangles in lateral faces of a triangular pyramid
Therefore, area of the base is: \(\frac{1}{2}\) . b . h_b = 3.9 ft
The total surface area is 11.7 + 3.9 = 15.6 ft².
The lateral area is 11.1 ft².
The total surface area is 15.6 ft².

Go Math Grade 7 Lesson 10.3 Answer Key Question 4.
Use a net to find the lateral area and the total surface area of the pyramid.

Answer:
Find the Lateral area of the rectangular pyramid.
We have four triangles with base 16 in. and height 20 in.
b = 16 ft Length of base
h_B = 20 ft Height of base
Lateral area:
4. (\(\frac{1}{2}\) . b. h_b) = 4 . \(\frac{1}{2}\) . 16 . 20 = 640 in²
The base is in the shape of rectangle.
Therefore, area of the base is: b . b = 256 in
The total surface area is 640 + 256 = 896 in²
The lateral area is 640 in².
The total surface area is 896 in².

Example 3.
Shoshanna’s team plans to build stands to display sculptures. Each stand will be in the shape of a rectangular prism. The prism will have a square base with side lengths of 2\(\frac{1}{2}\) feet, and it will be 3\(\frac{1}{2}\) feet high. The team plans to cover the stands with metallic foil that costs $0.22 per square foot. How much money will the team save on each stand if they cover only the lateral area instead of the total surface area?

Step 1
Make a net of the rectangular prism.

Step 2
Find the lateral and total surface areas of the prism.
The four lateral faces are 3\(\frac{1}{2}\) feet by 2\(\frac{1}{2}\) feet rectangles. The two bases are 2\(\frac{1}{2}\) feet by 2\(\frac{1}{2}\) feet squares.
Lateral area: 4\(\frac{1}{2}\) . (3\(\frac{1}{2}\) . 2\(\frac{1}{2}\)) = 35 square feet
Total surface area: 2 (2\(\frac{1}{2}\) . 2\(\frac{1}{2}\)) + 35 = 47\(\frac{1}{2}\) square feet

Add the area of the bases to the lateral area.

Step 3
Find and compare the prices.
Cost for total surface area: $0.22(47\(\frac{1}{2}\)) = $10.45
Cost for lateral area: $0.22(35) = $7.70
The team will save $10.45 – $7.70, or $2.75, for each stand by covering only the lateral area.

Your Turn

Question 5.
Kwame’s team will make two triangular pyramids to decorate the entrance to the exhibit. They will be wrapped in the same metallic foil. Each base is an equilateral triangle. If the base has an area of about 3.9 square feet, how much will the team save altogether by covering only the lateral area of the two pyramids?
The foil costs $0.22 per square foot. _______

Answer:
Find the lateral area of the triangular pyramid
We have three triangles with base of 3 ft and height of 6.1 ft.
b = 3 ft Length of base
h_B = 6.1 ft Height of base
Lateral area:
3. (\(\frac{1}{2}\) . b. h_b) = 3 . \(\frac{1}{2}\) . 3. (6.1) = 27.45 ft²
Base area:
B = 3.9 ft²
If they cover both whole pyramids they will spend:
2 (27.45 + 3.9) = 62.7 ft^2<
If we cover the lateral area of two pyramids we will spend:
2 . (27.45) = 54.9 ft²
The team will save:
0.22. (62.7 – 54.9) = 0.22. 7.8 = $1.7
The team will save $1.7

Texas Go Math Grade 7 Lesson 10.3 Guided Practice Answer Key 

A three-dimensional figure is shown sitting on a base. (Example 1)

Question 1.
The figure has a total of _____ rectangular faces.
Answer:
The figure has a total of six rectangular faces.

Question 2.
Of the total number of faces, ___ are lateral faces.
Answer:
Of the total number of faces, four are lateral faces

Question 3.
The figure is a ____.
Answer:
The figure is a rectangular prism

Go Math 7th Grade Practice and Homework Lesson 10.3 Answer Key Question 4.
Sketch a net of the figure.
Answer:

Each box in the figure represent a single unit.

Question 5.
The lateral area of the prism is ___.
Answer:
For lateral area:
Two 6 by 3 rectangtes
Two 7 by 3 rectangles.
2 . (6 . 3) = 36
2 . (7 . 3) = 42
Lateral area: 36 + 42 = 78

The lateral area of the prism is
36 + 42 = 78

Question 6.
The total surface area is ___.
Answer:
The base is in the shape of 6 by 7 rectangle.
Base area: 2 (6 . 7) = 84
The total surface area is 78 + 84 = 162

A triangular prism is shown. (Example 2)

Question 7.
Identify the number and type of faces of the prism.
Answer:
A triangular prism has five face:
Two 6 cm. by 4.3 cm. rectangles;
One 6 cm. by 3 cm. rectangle;
Two triangles with base of 3 cm. and height of 4 cm.

Question 8.
Find the lateral area of the prism. ____
Answer:
For lateral area:
Two 6 cm. by 4.3 cm. rectangles;
One 6 cm. by 3 cm. rectangle;
2 . (6 . (4.3)) = 51.6
1 . (6 . 3) = 18
Lateral area: 51.6 + 18 = 69.6 cm².

The lateral area of the prism is 69.6 cm².

Question 9.
Find the total surface area of the prism.
Answer:
The base is in the shape of triangle with length of 3 cm. and height of 4 cm.
Base area 2 . (\(\frac{1}{2}\) . 3 . 4) = 12 ft²
The total surface area is 69.6 + 12 = 81.6 cm².

Go Math Lesson 10.3 7th Grade Areas of Similar Shapes Answer Key Question 10.
Use a net to find the total surface area of the pyramid. Then find the cost of wrapping the pyramid completely in gold foil which costs $0.05 per square centimeter. (Examples 2 and 3)
Answer:
Find the lateral area of the rectangular pyramid.
we have four triangles with base of 10 cm. ain height of 12 cm.
b = 10 cm Length of base
h_B = 12 cm Height of base
Lateral area:
4 . (\(\frac{1}{2}\) . b . h_b) = 4 . \(\frac{1}{2}\) . 10 . 12 = 240 cm²
Base area:
10 . 10 = 1oo cm²
The total surface area is 240 + 100 = 340 cm².
Wrapping the pyramid completely in gold foil will cost: 340 . (0.05) = $17

Essential Question Check-In

Question 11.
How do you find the lateral and total surface area of a triangular pyramid?
Answer:
Lateral surface area of any three-dimensional figure means the sum of area of all sides except the area of base. And the total surface area means the area of all the sides of figure including the area of base.

Now in case of triangular pyramid it will have total four faces including the base triangle. So the lateral surface area will. be the sum of areas of all the three triangle which are faces of triangular pyramid except the base triangle. And the total surface area will be the sum of areas of all the four triangle including the base triangle

Texas Go Math Grade 7 Lesson 10.3 Independent Practice Answer Key 

Question 12.
Use a net to find the lateral area and the total surface area of the cereal box.
Lateral area: __________________________
Total surface area: _____________________

Answer:
The cereal box is in the shape of a rectangular prism.
Find the lateral area of the rectangular prism.
We have:
Two 2 in. by 12 in. rectangles;
Two 8 in. by 12 in. rectangles;
2 . (2 . 12) = 2 . 24 = 48
2 . (8 . 12) = 2 . 96 = 192
Lateral area: 48 – 192 = 240 in².
Each base is 8 in. by 2 in.\
2 . (8 . 2) = 2 . 16 = 32 in²
The total surface area is 240 + 32 = 272 in².
The lateral area is 240 in².
The total surface area is 272 in².

Question 13.
Describe a net for the shipping carton shown.
Answer:
Net of the given rectangular prism or cuboid will consist of six rectangular surface with front and behind rectangle having dimension of 8 inch by 12 inch. And the dimension of top and bottom rectangle will be 8 inch by 2 inch.

Net of the given figure will consist of 6 rectangular surface.

Question 14.
A shipping carton is in the shape of a triangular prism.
Use a net to find the lateral area and the total surface area of the carton.
Lateral area: __________________________
Total surface area: _____________________

Answer:
Find the lateral area of the triangular prism.
We have:
Two 3.6 in. by 15 in. rectangles;
One 4 in. by 15 in. rectangles.
2. ((3.6) . 15) = 2 . 54 = 108
1 . (4 . 15) = 1 . 60 = 60
Lateral area: 108 + 60 = 168 in²
Each base has the shape of triangle with length of 8 in. and height of 12 in.

Base Area
2 . 48 = 96 in²
The total surface area is 168 + 96 = 264 in².

The lateral area of the carton is 168 in².
The total surface area of the carton is 264 in².

Question 15.
Victor wrapped this gift box with adhesive paper (with no overlaps). How much paper did he use?
Answer:
The gift box is in the shape of a rectangular prism.
Find the lateral area of the rectangular prism.
We have:
Two 6 in. by 5 in. rectangle;
Two 8 in. by 5 in. rectangle.
2 . (6 . 5) = 60
2 . (8 . 5) = 80
The lateral area is 60 + 80 = 140 in².
Each base is 8 in. by 6 in.\
2 . (8 . 6) = 2 . 48 = 96 in²
The total surfance area is 140 + 96 = 236 in².

He used 236 in² of paper.

Question 16.
Vocabulary Name a three-dimensional shape that has four triangular faces and one rectangular face.
Answer:
Three-dimensional The shape that has four triangular faces and one rectangular face is a rectangular pyramid

Go Math 7th Grade Lesson 10.3 Answer Key Question 17.
Cindi wants to cover the top and sides of this box with glass tiles that are 1 cm square. How many tiles will she need?

Answer:
The box is in the shape of a rectangular prism.
Find the lateral area.
Two 15 cm. by 9 cm. rectangLe,
Two 20 cm. by 9 cm. rectangle;
2 . (15 . 9) = 270
2 . (20 . 9) = 360
Lateral area: 270 + 360 = 630 cm².
The base is 20 cm. by 15 cm.
To cover a top of box we need a surface of one base.
1 . (20 . 15) = 1 . 300 = 300 cm²
To cover a top and sides of this box Cindi needs 630 + 300 = 930 cm².
Therefore, she needs 930 tiles.
Cindi needs 930 tiles

Question 18.
A glass paperweight has the shape of a triangular prism. The bases are equilateral triangles with side lengths of 4 inches and heights of 3.5 inches. The height of the prism is 5 inches. Find the lateral area and the total surface area of the paperweight.
Answer:
For lateral area:
There are three 4 in. by 5 in. rectangles.
3 . (4 . 5) = 3 . 20 = 60 in²
Lateral area: 60 in²
Base area:
The bases are equilateral triangles with length of 4 in. and height of 3.5 in.
2 . (\(\frac{1}{2}\) . 4 . (3.5)) = 14 in²
The total surface area is 60 + 14 = 74 in².

Lateral area: 60 in².
The total surface area is 74 in².

Question 19.
The doghouse shown has a floor, but no windows. Find the total surface area of the doghouse (including the door).

Answer:
Base of doghouse is in the shape of a rectangular prism.
For lateral area:
Two 4 ft. by 2 ft rectangles;
Two 3 ft by 2 ft. rectangles;
2 . (4 . 2) = 16
2 . (3 . 2) = 12
Lateral area: 16 + 12 = 28 ft².
The base has the shape of rectangle 3 ft by 4 ft
Base area: 1 . (3 . 4) = 24 ft²
The total surface area is 28 + 24 = 52 ft².
The roof of a doghouse is in the shape of triangular pyramid.
For lateral area:
Two 2.5 ft. by 4 ft. rectangles;
One 3 ft by 4 ft. rectangle.
2 . (2.5 . 4) = 20
1. (3 . 4) = 12
Lateral area: 20 + 12 = 32 ft².
The base has the shape of triangle with length of 3 ft and height of 2 ft.
Base area:
2 . (\(\frac{1}{2}\) . 3 . 2) = 6 ft²
The total surface area is 32 + 6 = 38 ft².
The total surface area of the doghouse is:
52 + 38 = 90 ft²

The total surface area of the doghouse is 90 ft².

Question 20.
Describe the simplest way to find the total surface area of a cube.
Answer:
The surface area of a cube is the area of the six squares that cover it. The area of one of them is b. b. Since these are alt the same, multiply one of them by six, so the surface area of a cube is 6 times one of the sides squared.

Lesson 10.3 Answer Key 7th Grade Go Math Question 21.
Communicate Mathematical Ideas Describe how you approach a problem involving lateral area and total surface area. What do you do first? In what ways can you use the figure that is given with a problem? What are some shortcuts that you might use when you are calculating these areas?
Answer:
When solving a problem involving lateral area and total surface area, determine first the lateral area of the given
figure. This is because the lateral area is needed to determine the total surface area of the given figure. It is easier to determine the lateral area of the given figure if you know what figure are you dealing with, is it a pyramid or a prism, and what is the base of the given figure. The total surface area is the sum of the lateral area and the area of the base. The lateral area should be determined first before determining the total surface area of the given figure.

Solve for the lateral area first.

Texas Go Math Grade 7 Lesson 10.3 H.O.T. Focus on Higher Order Thinking Answer Key 

Question 22.
Persevere in Problem-Solving A pedestal in a craft store is in the shape of a triangular prism. The bases are right triangles with side lengths of 12 cm, 16 cm, and 20 cm. The store owner used 192 cm² of burlap cloth to cover the lateral area of the pedestal. Find the height of the pedestal.
Answer:
Lateral area:
One 20 cm. by h cm. rectangle;
One 12 cm. by h cm. rectangle;
One 16 cm. by h cm. rectangle.
1 . (20 . h) = 20 . h.
1 . (12 . h) = 12 . h.
1 . (16 . h.) = 16 . h.
Lateral area: 20 . h. + 12 . h + 16 . h = 48 . h cm².
The lateral area is 192 cm².
192 = 48 . h.
Divide both sides by 48.

The height of the pedestal is h = 4 cm.

Question 23.
Communicate Mathematical Ideas The base of Prism A has an area of 80 ft², and the base of Prism B has an area of 80 ft². The height of Prism A is the same as the height of Prism B. Is the base of Prism A congruent to the base of Prism B? Explain.
Answer:
The base of Prism A has the same area as the base of Prism B. They could be congruent however, it depends on the shape of the base. If the bases of Prism A and Prism B are both squares, then they are congruent. But if the bases of Prism A and Prism B are triangles or rectangles, there could be a difference in the dimensions of the triangle base and the rectangle base, which makes them not congruent It is because congruent figures have the same shape and size.

It depends on the shape of the base.

Question 24.
Critique Reasoning A triangular pyramid is made of 4 equilateral triangles. The sides of the triangles measure 5 m, and the height of each triangle is 4.3 m. A rectangular prism has a height of 4.3 m and a square base that is 5 m on each side. Susan says that the total surface area of the prism is more than twice the total surface area of the pyramid. Is she correct? Explain.
Answer:
Triangular pyramid:
For lateral area:
We have three triangles with a base of 5 m. and height of 4.3 m.
b = 5m Length of base
h_B = 4.3 m Height of base
Lateral area:
3. (\(\frac{1}{2}\). b . h_b) = 3 . \(\frac{1}{2}\) . 5 . (4.3) = 32.25 m²
Lateral area: 32.25 m².
The base is in the shape of a triangle with a base of 5 m. and height of 4.3 m
Base area: 1 . (\(\frac{1}{2}\) . 5. (4.3)) = 10.75 m²
The total surface area is 32.25 + 10.75 = 43 m²
Rectangular prism:
For lateral area:
We have four 5 m. by 4.3 m. rectangles.
4 . (5 . (4.3)) = 86
Lateral area: 86 m²
The base is in the shape of 5 m. by 5 m. rectangle.
Base area:
2 (5 . 5) = 50 m²
The total surface area is 86 + 50 = 136 m².

Susan is correct. The total surface area of the prism is three times of total surface area of the pyramid.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 11.2 Answer Key Comparing Data Displayed in Dot Plots.

Texas Go Math Grade 7 Lesson 11.2 Answer Key Comparing Data Displayed in Dot Plots

Essential Question
How do you compare two sets of data displayed in dot plots?

Texas Go Math Grade 7 Lesson 11.2 Explore Activity Answer Key  

Analyzing Dot Plots
You can use dot plots to analyze a data set, especially with respect to its center and spread.

People once used body parts for measurements. For example, an inch was the width of a man’s thumb. In the 12th century, King Henry I of England stated that a yard was the distance from his nose to his outstretched arm’s thumb. The dot plot shows the different lengths, in inches, of the “yards” for students in a 7th grade class.

A. Describe the shape of the dot plot. Are the dots evenly distributed or grouped on one side?

B. What value best describes the center of the data? Explain how you chose this value.

C. Describe the spread of the dot plot. Are there any outliers?

Reflect

Lesson 11.2 Comparing Data Displayed in Dot Plots Answer Key Question 1.
Calculate the mean, median, and range of the data in the dot plot.
Answer:
To find the mean, add the numbers and divide the sum by the number of addends.
28 + 29 + 29 + 30 + 30.5 + 30.5 + 31 + 31 + 31 + 31.5 + 31.5 + 32 + 32 + 32.5 + 32.5 + 33 + 33 + 33.5 + 34 + 34 + 35 = 664.5
\(\frac{664.5}{21}\) = 31.64
Mean is 31.64
The median is the middle number of the data in the dot plot that is ordered from least to greatest

Range is the difference between the greatest and least number of the data in the dot plot
range = 35 – 28 = 7

Mean : 31.64
Median : 31.7
Range : 7

Comparing Dot Plots Visually
You can compare dot plots visually using various characteristics, such as center, spread, and shape.

Example 1.
The dot plots show the heights of 15 high school basketball players and the heights of 15 high school softball players.

A. Visually compare the shapes of the dot plots.
Softball: All the data is 5’6” or less.
Basketball: Most of the data is 5’8” or greater.
As a group, the softball players are shorter than the basketball players.

B. Visually compare the centers of the dot plots.
Softball: The data is centered around 5’4’
Basketball: The data is centered around 5’8’
This means that the most common height for the softball players is 5 feet 4 inches, and for the basketball players 5 feet 8 inches.

C. Visually compare the spreads of the dot plots.
Softball: The spread is from 4’11 “to 5’6”.
Basketball: The spread is from 5’2” to 6’0’
There is a greater spread in heights for the basketball players.

Math Talk
How do the heights of field hockey players compare with the heights of softball and basketball players?

Your Turn

Question 2.
Visually compare the dot plot of heights of field hockey players to the dot plots for softball and basketball players.
Shape: _______________
Center: _______________
Spread: _______________
Answer:
Shapes:
Hockey: All the data is 5’6” or less.
The hockey players are shorter than the basketball players, but taller than the softball players.
Center:
Hockey: The data is centered around of 5’2”.
Spread:
Hockey: The spreads is from 4’9” to 5’6”.
They have same spread as spread for softball players, but there is a greater spread in heights for the basketball players.

All the data is 5’6″ or Less.
The data is centered around of 5’2”.
The spreads is from 4’9” to 5’6″.

Comparing Dot Plots Numerically
You can also compare the shape, center, and spread of two dot plots numerically by calculating values related to the center and spread. Remember that outliers can affect your calculations.

Example 2
Numerically compare the dot plots of the number of hours a class of students exercises each week to the number of hours they play video games each week.

A. Compare the shapes of the dot plots.
Exercise: Most of the data is less than 4 hours.
Video games: Most of the data is 6 hours or greater.

B. Compare the centers of the dot plots by finding the medians.
Median for exercise: 2.5 hours. Even though there are outliers at
12 hours, most of the data is close to the median.
Median for video games: 9 hours. Even though there is an outlier at 0 hours, these values do not seem to affect the median.

C. Compare the spreads of the dot plots by calculating the range.
Exercise range with outlier: 12 – 0 = 12 hours
Exercise range without outlier: 7 – 0 = 7 hours
Video games range with outlier: 14 – 0 = 14 hours
Video games range without outlier: 14 – 6 = 8 hours

Math Talk
Mathematical Processes
How do outliers affect the results of this data?

Your Turn

Comparing Data Displayed in Box Plots Lesson 11.2 Answer Key Question 3.
Calculate the median and range of the data in the dot plot. Then compare the results to the dot plot for Exercise in Example 2.

Answer:
Median for Internet Usage: 6 hours
Most of the data is close to the median
Median for Exercise: 2.5 hours
Internet Usage range with outliers: 11 – 1 = 10 hours
Internet usage range without outliers: 8 – 4 = 4 hours
Exercise range with outlier: 12 hours
Exercise range without outlier: 7 hours

Median for Internet Usage : 6 hours
Internet Usage range:

1. With outliers: 10 hours
2. Without outliers: 4 hours

Texas Go Math Grade 7 Lesson 11.2 Guided Practice Answer Key

The dot plots show the number of miles run per week for two different classes. For 1-5, use the dot plots shown.

Question 1.
Compare the shapes of the dot plots.
Answer:
Class A: Most of the the data is 6 mi. or less.
Class B: All the data is 3 mi. or greater
As a class, Students from Class A run more than students from Class B

Class A: Most of the the data is 6 mi. or less.
Class B: All the data is 3 mi. or greater
As a class, Students from Class A run more than students from Class B.

Question 2.
Compare the centers of the dot plots.
Answer:
Class A: The data is centred around 4 mi.
Class B: The data is centered around 7 mi.

Question 3.
Compare the spreads of the dot plots.
Answer:
Class A: The spread is from 4 mi. to 14 mi
Class B: The spread is from 3 mi. to 9 mi.

Go Math 7th Grade Practice and Homework Lesson 11.2 Answer Key Question 4.
Calculate the medians of the dot plots.
Answer:

Median for Class A: 7.5 mi.
Median for Class B: 5.75 mi.

Question 5.
Calculate the ranges of the dot plots.
Answer:
Class A range with outliers: 14 – 4 = 10 mi
Class A range without outliers: 6 – 4 = 2 mi
Class B range: 9 – 3 = 6 mi
For Class B the outlier doesn’t exist.

Class A range with outliers: 10 mi
Class A range without outliers: 2 mi
Class B range: 6 mi
For Class B the outlier doesn’t exist.

Essential Question Check-In

Question 6.
What do the medians and ranges of two dot plots tell you about the data?
Answer:
The medians of any data represents the centre or almost centre point of the given data. So in above two dot plots
the medians tells that which of the given dots have higher value. And the range of any data set represent the value around which all the data are spread. So in second dot plot the dots are very close as compared to first dot plot which means in second dot plots the value of data are closer.

Median of any data represent centre value.

Texas Go Math Grade 7 Lesson 11.2 Independent Practice Answer Key

The dot plot shows the number of letters in the spellings of the 12 months. Use the dot plot for 7-10.

Question 7.
Describe the shape of the dot plot.
Answer:
All of the data are 3 letters or more.

Go Math 7th Grade Lesson 11.2 Independent Practice Answer Key Question 8.
Describe the center of the dot plot.
Answer:
The data are centered around 8 letters.

Question 9.
Describe the spread of the dot plot.
Answer:
The spread is from 3 to 9 letters.

Question 10.
Calculate the mean, median, and range of the data in the dot plot.
Answer:

Mean: 6.16 letters
Median: 6 Letters
Range: 6 Letters

The dot plots show the mean number of days with rain per month for two cities.

Question 11.
Compare the shapes of the dot plots.
Answer:
Number of Days of rain for Montogomery, AL: All of the data is 12 days or less.
Number of Days of rain for Lynchburg, VA: All of the data is 8 days or more.
There are more days of rain in Lynchburg than in Montogomery.

Question 12.
Compare the centers of the dot plots.
Answer:
Montgomery: The data is centered around 8 days
Lynchburg: The data is centered around 10 days.

Question 13.
Compare the spreads of the dot plots.
Answer:
Montgomery: The spread is from 1 day to 12 days.
Lynchburg: The spread is from 8 days to 12 days.

Go Math Lesson 11.2 Answer Key Compare Dot Plots Question 14.
What do the dot plots tell you about the two cities with respect to their average monthly rainfall?
Answer:
The dot plot for Montgomery, AL showed that the average monthly rainfall in their city is inconsistent. The dot plot showed that there are days when the rainfall is heavy then it changes. Also, there is an outlier which made it more inconsistent For Lynchburg, VA, the average monthly rainfall in their city is somewhat consistent. Also, the days were relatively closer.

The average monthly rainfall in Lynch burg, VA is more consistent compared to Montgomery, AL.

The dot plots show the shoe sizes of two different groups of people.

Question 15.
Compare the shapes of the dot plots.
Answer:
Group A: All of the data is 65 or greater
Group B: All of the data is 8.5 or greater.
Group B has greater sizes of shoe than Group A.

Question 16.
Compare the medians of the dot plots.
Answer:
Median for Group A: 8 size of shoe
Median for Group B:

Median for Group A: 8 size of shoe
Median for Group B: 9.71 size of shoe

Question 17.
Compare the ranges of the dot plots (with and without the outliers).
Answer:
Group A range with outliers: 13 – 6.5 = 6.5 size of shoe
Group A range without outliers: 9 – 6.5 = 2.5 size of shoe
Group B range with outliers: 11.5 – 8.5 = 3 size of shoe
Group B range without outliers: 11.5 – 8.5 = 3 size of shoe

Group A range with outliers is 65 size of shoe
Group A range without outliers is 2.5 size of shoe
Group B range with outliers is 3 size of shoe
Group B range without outliers is 3 size of shoe

Question 18.
Make A Conjecture Provide a possible explanation for the results of the dot plots.
Answer:
Group A probably consists mostly of younger people and/or women (who tends to have smaller feet). Group B, on
the other hand, consists mostly of adults (who tends to have bigger feet).

Group A consists of younger people while Group B consists of adults.

Texas Go Math Grade 7 Lesson 11.2 H.O.T. Focus On Higher Order Thinking Answer Key

Question 19.
Analyze Relationships Can two dot plots have the same median and range but have completely different shapes? Justify your answer using examples.
Answer:
Yes, two dot plots can have the same median and range but completely different shapes.
In both the given dot plot the median is 6th term which is 7 and also the range of both dot plot is 7.

Dot plot of type A data is shown below:

Dot plot of type B data is shown below:

Yes, two dot plots can have the same median and range.

Comparing Dot Plots Answer Key Go Math Lesson 11.2 Question 20.
Draw Conclusions What value is most affected by an outlier, the median or the range? Explain. Can you see these effects in a dot plot?
Answer:
The range is the most affected by an outlier.
This can be understood by the example given below:
For example the data of the sample are 3, 5, 5, 5, 7, 8, 8, 9, 10, 12, 15.
Median = 6th term of data
= 8
Range = 15 – 3
= 12
New sample after excluding outlier: 3, 5, 5, 5, 7, 8, 8, 9, 10, 12.

Range = 12 – 3
= 9
Here we can see that the median decreases by only 0.5 but the range decreases by 3. Hence we can see that the range is most affected by outliers.

The range is most affected by outlier

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 10.1 Answer Key Volume of Rectangular Prisms and Pyramids.

Texas Go Math Grade 7 Lesson 10.1 Answer Key Volume of Rectangular Prisms and Pyramids

Texas Go Math Grade 7 Lesson 10.1 Example Answer Key  

Example 1.
Find the volume of the rectangular prism.

Step 1
Find the area of the base.
B = lw Use the formula.
B = 12 × 15 Substitute for l and w.
B = 180 mm²

Step 2
Find the volume.
V= Bh Use the formula.
V = 180 × 7 Substitute for sand h.
V = 1,1260 mm³
The volume of the rectangular prism is 1,260 cubic millimeters.

Reflect

Lesson 10.1 Volume of Prisms and Cylinders Answer Key Question 1.
What If? If you know the volume V and the height h of a prism, how would you find the area of the base B?
Answer:
The volume V of a prism is the area of its base B times its height h.
V = B.h…(1)
We know the volume V and the height of a prism.
Divide (1) by h.

We find the area of the base B as \(\frac{V}{h}\) = B

Your Turn

Question 2.
Use the formula V = Bh to find the volume of a gift box that is 3.5 inches high, 7 inches long, and 6 inches wide.
Answer:
The volume V of a prism is the area of its base B times its height h.
V = B . h The volume of a prism
h = 3.5 Height of a prism
w = 6 Width of base
l = 7 Lenght of base

Find the area of the base.
B = w . l = 6 . 7 = 42 in²

Find the volume.

V = B . h = 42 . (3.5) = 147 m³
The volume V of a prism is V = 147 in³.

Texas Go Math Grade 7 Lesson 10.1 Explore Activity Answer Key  

Nets
A net is a two-dimensional pattern of shapes that can be folded into a three-dimensional figure. The shapes in the net become the faces of the three-dimensional figure.

Step 1
Copy Net A and Ñet B on graph paper, and cut them out along the blue lines.

One of these nets can be folded along the black lines to make a cube. Which net will not make a cube? _________________

Step 2
See if you can find another net that can be folded into a cube.
Draw a net that you think will make a cube on your graph paper, and then cut it out. Can you fold it into a cube? Sketch your net below.

Step 3
Compare your results with several of your classmates. How many different nets for a cube did you and your classmates find?

Reflect

Question 3.
What shapes will appear in a net for a rectangular prism that is not a cube? How many of these shapes will there be?
Answer:
A rectangular prism is a three-dimensional shape with six rectangular-shaped sides, so will appear as six rectangles.

How do you know that each net cannot be folded into a cube without actually cutting and folding it?

Question 4.

Answer:
This net cannot be folded in the shape of a cube because a cube is a three-dimensional figure with six equal square faces. The net is missing one square

7th Grade Volume Rectangular Prism Answer Key  Question 5.

Answer:
This net cannot be folded in the shape of a cube because a cube is a three-dimensional figure with six equal square faces. In this net the squares overlap.

Question 6.
Make a Conjecture If you draw a net for a cylinder, such as a soup can, how many two-dimensional geometric shapes would this net have? Name the shapes in the net for a cylinder.
Answer:
It will be three two-dimensional shapes; one rectangle and two circles

Exploring the Volume of a Rectangular Pyramid

A pyramid is a three-dimensional shape whose base is a polygon and whose other faces are all triangles. Like a prism, a pyramid is named by the shape of its base.

The faces of a pyramid that are not the base have a common vertex, called the vertex of the pyramid. The perpendicular distance from the vertex to the base is the height of the pyramid.

Explore Activity 2
In this activity, you will compare the volumes of a pyramid and a prism with congruent bases and equal heights. Remember that congruent figures have the same shape and size.

Step 1
Make three-dimensional models. Make larger versions of the nets shown. Make sure the bases and heights in each net are the same size. Fold each net, and tape it together to form a prism or a pyramid.

Step 2
Fill the pyramid with beans. Make sure that the beans are level with the opening of the pyramid. Then pour the beans into the prism. Repeat until the prism is full. How many times did you fill the prism from the pyramid? __________________________

Step 3
Write a fraction that compares the volume of the pyramid to the volume of the prism.

Math Talk
Mathematical Processes
Describe ways in which a prism and a pyramid are different.

Reflect

Question 7.
Draw Conclusions A rectangular pyramid has a base of B and a height of h. What is a formula for the volume of the pyramid? Justify your reasoning.
Answer:
Given base area of the rectangular Pyramid = B
Given height of the rectangular Pyramid = h
Volume of rectangular pyramid is the one-third of the base area times the height of pyramid.
Volume of rectangular pyramid = \(\frac{1}{3}\)Bh
Hence, the volume of the rectangular pyramid is \(\frac{1}{3}\)Bh

volume of the rectangular pyramid is \(\frac{1}{3}\)Bh.

Texas Go Math Volume of Rectangular Prism and Pyramid Answer Key Question 8.
Communicate Mathematical Ideas The prism and the pyramid in this activity have congruent bases and equal heights. Are they congruent three-dimensional shapes? Explain.
Answer:
The prism and pyramid cannot be considered as congruent three-dimensional shapes even if they have the same bases and equal heights. This is because their volumes are different. When a prism and pyramid have congruent bases and equal heights, the prism will have a greater volume than the pyramid. This is basically, the pyramid can fit inside the prism. The volume of the pyramid is only \(\frac{1}{3}\) of the volume of the prism.

No

Your Turn

Question 9.
Thé volume of a rectangular prism is 4\(\frac{1}{2}\) in³. What is the volume of a rectangular pyramid with a congruent base and the same height? Explain your reasoning.
Answer:
ragged right The volume of a rectangular prism is:
V = B . h (1)
The volume of a rectangular pyramid is:

They have congruent base and same height
Divide (1) by 3

The volume of a rectangular pyramid is three time less than the volume of a rectangular prism
The volume of a rectangular pyramid is:

The volume of a rectangular pyramid is three time less than the volume of a rectangular prism.
V_P = \(\frac{3}{2}\)in³

Solving Volume Problems

You can use the formulas for the volume of a rectangular prism and the volume of a rectangular pyramid to solve problems.

Volume of a Rectangular Pyramid
The volume V of a pyramid is one-third the area of its base B times its height h.
V = \(\frac{1}{3}\)Bh

Example 2
‘A Kyle needs to build a crate in the shape of a rectangular prism. The crate must have a volume of 38\(\frac{1}{2}\) cubic feet, and a base area of 15\(\frac{2}{3}\) square feet. Find the height of the crate.
V= Bh Use the formula.
38\(\frac{1}{2}\) = 15\(\frac{2}{5}\) . h Substitute for V and B.
\(\frac{77}{2}\) = \(\frac{77}{5}\)h Change the mixed numbers to fractions.
\(\frac{5}{77}\) . \(\frac{77}{2}\) = \(\frac{77}{5}\)h . \(\frac{5}{77}\) To divide both sides by \(\frac{77}{5}\), multiply both sides by the reciprocal.
\(\frac{5}{2}\) = h
The height of the crate must be \(\frac{5}{2}\) or 2\(\frac{1}{2}\), feet.

B. A glass paperweight in the shape of a rectangular pyramid has a base that is 4 inches by 3 inches and a height of 5 inches. Find the volume of the paperweight.
V= \(\frac{1}{3}\)Bh Use the formula.
V = \(\frac{1}{3}\) . 12 . 5 Think B = lw = 4 . 3 = 12
V = 20
The paperweight has a volume of 20 cubic inches.

Your Turn

Question 10.
A rectangular prism has a volume of 160 cubic centimeters and a height of 4 centimeters. What is the area of its base? _______
Answer:
V = B . h … (1)
V = 160 cm³ The volume of a prism
h = 4 cm Height of a prism
Divide (1) by h = 4

The area of the base is B = 40 cm².

Go Math Grade 7 Lesson 10.1 Answer Key Question 11.
A square pyramid has a base edge of 5.5 yards and a height of 3.25 yards. Find the volume of the pyramid to the nearest tenth. _______
Answer:
V = \(\frac{1}{3} B \cdot h\)
l = 5.5 yd Edge of base
h = 3.2 yd Height of pyramid
Find the base of a pyramid
B = (5.5). (5.5) = 30.25 yd²
Find the volume of a pyramid.
V = B . h = (30.25) . (3.25) = 98.3215 ≈ 98.3 yd³

The volume of a pyramid is V = 98.3 yd³.

Texas Go Math Grade 7 Lesson 10.1 Guided Practice Answer Key 

Question 1.
Find the volume of the rectangular prism. (Example 1)
V = Bh

Answer:
l = 6ft Length of base
w = 4\(\frac{1}{2}\) ft Width of a base
h = 9 ft Height of a prism
The volume of a prism:

V = 243 ft³

Identify the three-dimensional shape that can be formed from each net.
(Explore Activity 1 and Explore Activity 2)

Question 2.

Answer:
The three-dimensional shape that can be formed from the given net is Rectangular Pyramid. Because the in given
net all the four side have triangular shape and the base is of rectangular shape.

Question 3.

Answer:
The three-dimensional shape that can be formed from the given net is Pentagonal Prism. Because the in given net
all the four side and base have rectangular shape and front and behind of the prism is of pentagonal shape.
The three-dimensional shape that will be formed for the net ¡s also called as Heptahedron.

Question 4.

Answer:
The three-dimensional shape that can be formed from the given net is the Rectangular Prism. Because the in given net all the four sides have rectangular shapes and the base and top of the prism are square shapes.
The three-dimensional shape that will be formed for the net is also called a cuboid.

Go Math Lesson 10.1 Answer Key Volume of Mixed Pyramids Question 5.
The volume of a rectangular prism is 161.2 m³. The prism has a base that is 5.2 m by 3.1 m. Find the height of the prism. (Example 2)

Answer:
B = l . w Base of prism
l = 5.2 m Length of base
w = 3.1 m Width of base
h = ? Height of a prism
V = 161.2 m³ The volume of a prism
Use the formula B = l . w to find a base.
B = l . w = (5.2). (3.1) = 16.12 m²
Use the formula for the volume of a prism to find the height of the prism.
V = l . w . h

The height of the prism is h = 10 m.

Essential Question Check-In

Question 6.
Explain how to use models to show the relationship between the volume of a rectangular prism and a rectangular pyramid with congruent bases and heights.
Answer:
When you have a rectangular prism and rectangular pyramid with congruent bases and heights, you would notice that the rectangular pyramid can fit in the rectangular prism and there is still available space. This is because the prism will have a greater volume compared to the pyramid. Let’s say, that the rectangular prism will be filled with stones. The stones from the rectangular prism will be transferred to the rectangular pyramid notice that after the rectangular pyramid will be filled with stones and there are still remaining stones in the rectangular prism. This is because the volume of the rectangular pyramid is only \(\frac{1}{3}\) of the volume of the rectangular prism.

The relationship between a rectangular prism and rectangular pyramid with congruent bases and heights.

Texas Go Math Grade 7 Lesson 10.1 Independent Practice Answer Key  

Question 8.
Explain the Error A student found the volume of a rectangular pyramid with a base area of 92 square meters and a height of 54 meters to be 4,968 cubic meters. Explain and correct the error.
Answer:
Error in this conclusion is:
A student used formula for the volume of a prism (V = B . h), not for the volume of a pyramid (V_P = \(\frac{1}{3}\) . B. h).
That mistake can be corrected when the result divided with 3.
V = \(\frac{1}{3}\)B . h = \(\frac{1}{3}\) . 92 . 54 = \(\frac{1}{3}\) . 4968 = 1656

Error in this conclusion is:
A student used formula for the volume of a prism, not for the volume of a pyramid.
That mistake can be corrected when the result divided with 3.

Question 9.
A block of marble is in the shape of a rectangular prism. The block is 3 feet long, 2 feet wide, and 18 inches high. What is the volume of the block? _________________
Answer:
l = 3 ft Length of base
w = 2 ft Width of base
h = 1.5 ft Height of a prism
Use the formula B = l . w to find the base.
B = l . w = 3 . 2 = 6 ft²
Use the formula for the volume of a prism.
V = B . h
V = 6. (1.5)
V = 9 ft³

The volume of the block is V = 9 ft³.

Question 10.
Multistep Curtis builds a doghouse with base shaped like a cube and a roof shaped like a pyramid. The cube has an edge length of 3\(\frac{1}{2}\) feet.The height of the pyramid is 5 feet. Find the volume of the doghouse rounded to the nearest tenth.

Answer:
Base of doghouse is in the shape of a rectangular prism.
b = 3\(\frac{1}{2}\) = \(\frac{7}{2}\) ft Length of the base
h = \(\frac{7}{2}\)ft Height of a prism
Use the formula for the volume of a rectangular prism.
V₁ = b . b . h
= \(\frac{7}{2}\) . \(\frac{7}{2}\) . \(\frac{7}{2}\)
= \(\frac{343}{8}\) ft³
The roof of a doghouse is in the shape of rectangular pyramid.
b = \(\frac{7}{2}\) ft Length of the base
h = 5 ft Height of a pyramid
Use the formula for the volume of a rectangular pyramid.
V₂ = \(\frac{1}{3}\) . b . b . h
= \(\frac{1}{3}\) . \(\frac{7}{2}\) . \(\frac{7}{2}\) . 5
= \(\frac{245}{12}\) ft³
The volume of the doghouse is:
V = V₁ + V₂ = \(\frac{343}{8}\) . \(\frac{245}{12}\) = (42.875) + (20.416) ≈ 63.3 ft³
The Volume of the doghouse is V = 63.3 ft³

Question 11.
Miguel has an aquarium in the shape of a rectangular prism. The base is 30.25 inches long and 12.5 inches wide. The aquarium is 12.75 inches high. What is the volume of the aquarium to the nearest cubic inch?
Answer:
l = 30.25 in length of base
w = 12.5 in Width of base
h = 12.75 in Height of a prism
Use the formula B = l . w to find a base.
B = l . w = (30.25). (12.5) = 378.125 in²
Use the formula for the volume of a prism.
V = B . h.
= (378.125) . (12.75)
= 4, 821.09375 ≈ 4,821.09 in³

The volume of the aquarium is V = 4,821.09 in³.

Volume of Pyramid Answer Key Go Math Lesson 10.1 Question 12.
After a snowfall, Sheree built a snow pyramid. The pyramid had a square base with side lengths of 32 inches and a height of 28 inches. What was the volume of the pyramid to the nearest cubic inch?
Answer:
b = 32 in length of base
h = 28 in Height of a pyramid
Use the formula
B = b. b to find the base.
B = b . b = 32 . 32 = 1024 in²
Use the formula for the volume of a pyramid.
V = \(\frac{1}{3}\) . B . h
= \(\frac{1}{3}\) . 1024. 28
= \(\frac{1024.28}{3}\)
= 9, 557.333 ≈ 9, 557.3 in³

The volume of the pyramid is V = 9, 557.3 in³.

Question 13.
A storage chest has the shape of a rectangular prism with the dimensions shown. The volume of the storage chest is
18,432 cubic inches. What is its height?

Answer:
V = 18, 432 in³ The volume of the storage chest
l = 48 in Length of a base
w = 16m Width of a base
B = l . w = 48 . 16 = 768 m² Base area of a prism
The storage chest has the shape of a prism. Use the formula for the volume of a prism and substitute values for B and h.
V = B . h
18432 = 768 . h
Divide by 768 on both sides.
\(\frac{18432}{768}\) = h
24 = h

The height of the storage chest is 24 in.

Question 14.
Draw Conclusions A shipping company ships certain boxes at a special rate. The boxes must not have a volume greater than 2,500 cm³. Can the box shown be shipped at the special rate? Explain.

Answer:
Length of the given box = 10 cm
Width of the given box = 12 cm
H eight of the given box = 20 cm
We know that the volume of the box = Length × width × height
Volume of box = 10 × 12 × 20 cm³
= 2400 cm³
It is given in the problem that a shipping company ships certain boxes at a special rate only if the volume of the box is not greater than 2, 500 cm3 And the volume of the given box is 2, 400 cm³. So the box will be shipped at special rate by shipping company.
Yes, the box will be shipped at a special rate.

Yes, the box will be shipped at a special rate.

Question 15.
Communicate Mathematical Ideas Is the figure shown a prism or a pyramid? Justify your answer.

Answer:
Given figure is triangular Prism
Because prism is a figure which has parallel top and bottom bases. And pyramid has a base and triangular faces
which meet at common vertex point. In the given figure we can see that both triangle faces are parallel to each
other which has base length of 8 inch and height of 4 inch.

Question 16.
A cereal box can hold 144 cubic inches of cereal. Suppose the box is 8 inches long and 1.5 inches wide. How tall is the box?
Answer:
The cereal box is in the shape of rectangular prism.
l = 8 in Length of base
w = 1.5 in Width of base
h = ? Height of a prism
V = 144 in³ The volume of a prism
Use the formula B = l . w to find the base.
B = l . w = 8 . (1.5) = 12 in²
Use the formula for the volume of a prism to find the height of the prism.
V = B . h
144 = 12 . h
Divide both sides by 12.

The box is 12 in. tall.

Question 17.
A Shed in the shape of a rectangular prism has a volume of 1,080 cubic feet. The height of the shed is 8 feet, and the width of its base is 9 feet. What is the length of the shed?
Answer:
V = 1080 ft³ The volume of a rectangular prism
h = 8 ft Length of prism
l = ? Length of base
w = 9 ft Width of base
Use the formula for the volume of a prism to find a length of the base.
V = l . w . h.
1080 = l . 9 . 8
1080 = l . 72
Divide both sides by 72.

The length of shed is 15 ft.

H.O.T. Focus on Higher Order Thinking

Question 18.
Draw Conclusions Sue has a plastic paperweight shaped like a rectangular pyramid. The volume is 120 cubic inches, the height is 6 inches, and the length is 10 inches. She has a gift box that is a rectangular prism with a base that is 6 inches by 10 inches. How tall must the box be for it to hold the pyramid?
Answer:
Given volume of pyramid = 120 inch³
Height of pyramid = 6 inch
Length of pyramid = 10 inch
Volume of pyramid = \(\frac{\text { length } \times \text { width } \times \text { height }}{3}\)
120 = \(\frac{10 \times \text { width } \times 6}{3}\)
width = \(\frac{120 \times 3}{10 \times 6}\)
width = 6 inch
So, base of the pyramid is 6 inch by 10 inch and also the base of prism (box) is 6 inch by 10 inch. This means the height of pyramid to fit in completely but the volume of prism or box will be 3 times the volume of pyramid.
Hence, the box must be 6 inch tall to hold the pyramid.
This can be best understood by the given below figure in which pyramid is inside the prism with both having same base area.

Required height is 6 inch.

Question 19.
Represent Real-World Problems A public swimming pool is in the shape of a rectangular prism. The pool is 20 meters long and 16 meters wide. The pool is filled to a depth of 1.75 meters.
a. Find the volume of water in the pool.
h = 1.75 m Height of prism
l = 20 m Length of base
w = 16 m Width of base
Use the formula for the volume of a rectangular prism.
V = l . w . h
= 20 . 16 . (1.75)
= 320 . (1.75)
= 560 m³

The volume of water in the pool is 560 m³.

b. A cubic meter of water has a mass of 1,000 kilograms. Find the mass of the water in the pool.
Answer:
Mass of the water in the pool is:
560 . 1000 = 560,000 kg
The mass of the water in the pool is 560, 000 kg.

Question 20.
Analyze Relationships There are two glass pyramids at the Louvre Museum in Paris, France. The outdoor pyramid has a square base with side lengths of 35.4 meters and a height of 21.6 meters. The indoor pyramid has a square base with side lengths of 15.5 meters and a height of 7 meters. How many times as great is the volume of the outdoor pyramid than that of the indoor pyramid?
Answer:
The outdoor pyramid is in the shape of a rectangular pyramid.
b = 35.4 m Length of the base
h = 21.6 m Height of a pyramid
Use the formula for the volume of a rectangular pyramid.
V₁ = \(\frac{1}{3}\) . b . b . h
= \(\frac{1}{3}\)(35.4). (35.4). (21.6)
= 9,022.752 m³
The indoor pyramid is in the shape of a rectangular pyramid.
b = 15.5 m Length of the base
h = 7 m Height of a pyramid
Use the formula for the volume of a rectangular pyramid.
V₂ = \(\frac{1}{3}\) . b . b . h
= \(\frac{1}{3}\) . (15.5) . (15.5) . 7
= 560.583 m³
\(\frac{V_{1}}{V_{2}}\) = \(\frac{9022.752}{560.583}\) = 16.095 ≈ 16
The volume of the outdoor pyramid is 16 times greater then the volume of the indoor pyramid.

Question 21.
Persevere in Problem-Solving A small solid pyramid was installed on top of the Washington Monument in 1884. The square base of the pyramid is 13.9 centimeters on a side, and the height of the pyramid is 22.6 centimeters. The pyramid has a mass of 2.85 kilograms.

a. Find the volume of the pyramid. Round to the nearest hundredth.
Answer:
b = 13.9 cm length of the base
h = 22.6 cm Height of a pyramid
Use the formula for the volume of a rectangular pyramid
V₁ = \(\frac{1}{3}\) . b . b . h
= \(\frac{1}{3}\) . (13.9) . (13.9). (22.6)
= 1455.51 cm³
a. The volume of the pyramid is:
V = 1455.51 ≈ 1456 cm³

b. Find the mass of the pyramid in grams.
Answer:
The pyramid has mass of 285 kg. There are 1000 grams in 1 kilogram. Mass of the pyramid in grams is: 2.85 .
1000 = 2850 gr.

c. Science The density of a substance is the ratio of its mass to its volume. Find the density of the pyramid in grams per cubic centimeter. Round to the nearest hundredth.
Answer:
The density of the pyramid is \(\frac{\text { Mass }}{\text { Volume }}\) = \(\frac{2850}{1456}\) ≈ 2g/cm³.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 9.4 Answer Key Area of Composite Figures.

Texas Go Math Grade 7 Lesson 9.4 Answer Key Area of Composite Figures

Essential Question
How do you find the area of composite figures?

Texas Go Math Grade 7 Lesson 9.4 Explore Activity Answer Key

Exploring Areas of Composite Figures
Aaron was plotting the shape of his garden on grid paper. While it was an irregular shape, it was perfect for his yard. Each square on the grid represents 1 square meter.

A. Describe one way you can find the area of this garden.

Answer:

B. The area of the garden is ___ square meters.
Answer:

C. Compare your results with other students. What other methods were used to find the area?
Answer:

D. How does the area you found compare with the area found using different methods?
Answer:

Reflect

Area of Composite Figures Worksheet 7th Grade Answer Key Question 1.
Use dotted lines to show two different ways Aaron’s garden could be divided up into simple geometric figures.

Answer:

Separate the figure into a triangle and two rectangles.

Separate figure into a triangle and three rectangles.

One way is to separate figure into a triangle and two rectangles, and other way is to separate figure into a triangle and three rectangles

Finding the Area of a Composite Figure

A composite figure is made up of simple geometric shapes. To find the area of a composite figure or other irregular-shaped figure, divide it into simple, nonoverlapping figures. Find the area of each simpler figure, and then add the areas together to find the total area of the composite figure.
Use the chart below to review some common area formulas.

Find the area of the figure.
Step 1
Separate the figure into smaller, familiar figures: a parallelogram and a trapezoid.

Step 2
Find the area of each shape.

Step 3
Add the areas to find the total area.
A = 15 + 12.75 = 27.75 cm²
The area of the figure is 27.75 cm².

Your Turn

Find the area of each figure. Use 3.14 for π.

Question 2.

Answer:
Separate the figure into 2 triangles and one rectangle.
Area of the first triangle
base = 2 ft
height = 3 ft
Use the formula for the area of the triangle.
A₁ = \(\frac{1}{2} \cdot b \cdot h\)
A₁ = \(\frac{1}{2} \cdot 2 \cdot 3\)
A₁ = \(\frac{6}{2}\)
A₁ = 3
The area of the first triangle is 3 ft².
Area of the second triangle
base = 3 ft
height = 3 ft
Use the forumula for the area of the triangle.
A₂ = \(\frac{1}{2} \cdot b \cdot h\)
A₂ = \(\frac{1}{2} \cdot 3 \cdot 3\)
A₁ = \(\frac{9}{2}\)
A₁ = 4.5
The area of the second triangle is 4.5 ft².

Area of the rectangle
length = 8 ft
width = 4 ft
Use the forum ula for the area of the rectangle.
À₃ = l ∙ w
A₃ = 8 ∙ 4
A₃ = 24
The area of the rectangle is 24 ft².
Add the areas to find the total area.
A = A₁ + A₂ + A₃ = 3 + 4.5 + 24 = 31.5
The area of the figure is 31.5 ft².

7th Grade Area of Composite Figures Lesson 9.4 Question 3.

Answer:
Separate the figure into a square and a semicircle.

Area of the square
side = 10 m
Use the formula for the area of the square.
A₁ = s²
A₁ = 10²
A₁ = 100
The area of the sqaure is 100 m².
Area of the circle
diameter = 10 m
Use the formula for the area of the circle when given diameter
A_c = \(\pi\left(\frac{d}{2}\right)^{2}\) Subsitute 10 ford and 3.14 for π.
A_c = \(3.14\left(\frac{10}{2}\right)^{2}\)
A_c = 3.14 ∙ 5²
A_c = 3.14 ∙ 25
A_c = 78.5
Area of the semicircle is half the area of the circle.
A₂ = \(\frac{A_{c}}{2}\) = \(\frac{78.5}{2}\) = 39.25
The area of the semicircle is 39.25 m₂.
Add the areas to find the total area.
A = A₁ + A₂ = 100 + 39.25 = 139.25
The area of the figure is 139.25 m²

Using Area to Solve Problems

Example 2
A banquet room is being carpeted. A floor plan of the room is shown at right. Each unit represents 1 yard. The carpet costs $23.50 per square yard. How much will it cost to carpet the room?

Step 1
Separate the composite figure into simpler shapes as shown by the dashed lines: a parallelogram, a rectangle, and a triangle.

Step 2
Find the area of the simpler figures. Count units to find the dimensions.

Step 3
Find the area of the composite figure.
A = 8 + 24 + 1 = 33 square yards

Step 4
Calculate the cost to carpet the room.
Area ∙ Cost per yard = Total cost
33 ∙ $23.50 = $775.50
The cost to carpet the banquet room is $775.50.

Your Turn

Area of Composite Figures Answer Key Go Math 7th Grade Lesson 9.4 Question 4.
A window is being replaced with tinted glass. The plan at the right shows the design of the window. Each unit length represents 1 foot. The glass costs $28 per square foot. How much will it cost to replace the glass? Use 3.14 for π.

Answer:
Seperate the figure into a rectangle and two semicircles.
Since each unit length represents 1 foot, we see from the picture that the diameter of the semicircles is 4 ft,
and length and width of the rectangle are 5 ft and 4 ft.
We have two semicircles with same diameter, which make up the whole circle. Hence, find the area of the circle.
Area of the circle
diameter = 4
Use the formula for the area of the circle when given diameter.
A₁ = \(\pi\left(\frac{d}{2}\right)^{2}\) Substitute 4 for d and 3.14 for π.
A₁ = \(3.14\left(\frac{4}{2}\right)^{2}\)
A₁ = 3.14 ∙ 2²
A₁ = 3.14 ∙ 4
A₁ = 12.56
Area of the circle is 12.56 ft².
Area of the rectangle
length = 5 ft
width = 4 ft
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 5 ∙ 4
A₂ = 20
The area of the rectangle is 20 ft².

Add the areas to find the total area.
A = A₁ + A₂ = 12.56 + 20 = 32.56
The area of the figure is 32.56 ft²..
To find how will it cost to replace the glass multiply the area of the figure by the cost of the glass per square foot.
The total cost = 32.56 ∙ 28 = 911.68
$911.68 will cost the replacement of the glass.

Texas Go Math Grade 7 Lesson 9.4 Guided Practice Answer Key

Question 1.
A tile installer plots an irregular shape on grid paper. Each square on the grid represents 1 square centimeter. What is the area of the irregular shape? (Explore Activity, Example 2)

Step 1
Separate the figure into a triangle, a ____ and a parallelogram.
Step 2
Find the area of each figure.
triangle: ____ cm²; rectangle: ___ cm²; parallelogram: ___ cm²
The area of the irregular shape is ____ cm².
Answer:
A rectangle
Area of the triangle
base = 4 cm
height = 2 cm
Use the formula for the area of the triangle.
A₁ = \(\frac{1}{2} b \cdot h\)
A₁ = \(\frac{1}{2} 4 \cdot 2\)
A₁ = \(\frac{8}{2}\)
A₁ = 4
The area of the triangle is 4 cm².
Area of the rectangle
length = 5 cm
width = 3 cm
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 5 ∙ 3
A₂ = 15
The area of the rectangle is 15 cm²
Area of the parallelogram
length = 5 cm
width = 3 cm
Use the formula for the area of the parallelogram.
A₃ = l ∙ w
A₃ = 5 ∙ 3
A₃ = 15
The area of the parallelogram is 15 cm².

Add the areas to find the total area.
A = A₁ + A₂ + A₃ = 4 + 15 + 15 = 34
The area of the irregular shape is 34 cm².

Area of Composite Figures 7th Grade Go Math Question 2.
Show two different ways to divide the composite figure. Find the area both ways. Show your work below. (Example
Answer:
First way
Separate the figure into two rectangles.
Area of the first rectangle
length = 20 cm
width = 9 cm
Use the formula for the area of the rectangle.
A₁ = l ∙ w
A₁ = 20 ∙ 9
A₁ = 180
The area of the first rectangle is 180 cm².
Area of the second rectangle
length = 12 cm
width = 9 cm
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 20 ∙ 9
A₂ = 180
The area of the second rectangle is 108 cm².
Add the areas to find the total area.
A = A₁ + A₂ = 180 + 108
The area of the figure is 288 cm².

Second way
Separate the figure into three rectangles.
Area of the first rectangle
length = 9 cm
width = 8 cm
Use the formula for the area of the rectangle.
A₁ = l ∙ w
A₁ = 9 ∙ 8
A₁ = 78

The area of the first rectangle is 72 cm².
Area of the second rectangle
length = 12 cm
width = 9 cm
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 12 ∙ 9
A₂ = 108
The area of the second rectangle is 108 cm².

Area of the third rectangle
length = 12 cm
width = 9 cm
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 12 ∙ 9
A₂ = 108
The area of the third rectangle is 108 cm².

Add the areas to find the total area.
A = A₁ + A₂ + A₃ = 72 + 108 + 108 = 288
The area of the figure is 288 cm².

Go Math Grade 7 Answer Key Pdf Area of Composite Figures Question 3.
Sal is tiling his entryway. The floor plan is drawn on a unit grid. Each unit length represents 1 foot. Tile costs $2.25 per square foot. How much will Sal pay to tile his entryway? (Example 2)

Answer:
Separate the figure into a parallelogram and a trapezoid.
Area of the parallelogram
length = 4 ft
width = 4 ft
Use the formula for the area of the parallelogram.
A₁ = l ∙ w
A₁ = 4 ∙ 4
A₁ = 16
The area of the parallelogram is 16 ft².
Area of the trapezoid
b₁ = 7 ft
b₂ = 4 ft
h = 5 ft
Use the formula for the area of the trapezoid.

The area of the trapezoid is 27.5 ft².

Add the areas to find the total area.
A = A₁ + A₂ = 16 + 27.5 = 43.5
The area of the figure is 43.5 ft².
The total area of the floor is 43.3 ft².
Total cost of tilling = 43.5 ∙ 2.25 = 97.87
Sal will pay $97.87 to tile his entryway.

Essential Question Check-In

Question 4.
What is the first step in finding the area of a composite figure?
Answer:
First step is to separate the composite figure into smaller familiar figures.

Texas Go Math Grade 7 Lesson 9.4 Independent Practice Answer Key

Question 5.
A banner is made of a square and a semicircle. The square has side lengths of 26 inches. One side of the square is also the diameter of the semicircle. What is the total area of the banner? Use 3.14 for π.
Answer:
Area of the square
side = 26 in.
Use the formula for the area of the square.
A₁ = s²
A₁ = 26²
A₁ = 676
The area of the sqaure is 676 in.².
Area of the circle
diameter = 26 in.
Use the formula for the area of the circle when given diameter

Area of the semicircle is half the area of the circle.
A₂ = \(\frac{A_{c}}{2}\) = \(\frac{530.66}{2}\) = 265.33
The area of the semicircle is 265.33 in.².

Add the areas to find the total area.
A = A₁ + A₂ = 676 + 265.33 = 941.33
The area of the banner is 941.33 in.²

Lesson 9.3 Area of Composite Figures Answer Key Pdf Question 6.
Multistep Erin wants to carpet the floor of her closet. A floor plan of the closet is shown.

Answer:
Separate the floor into a rectangle and a triangle.
Area of the triangle
base = 6 ft
height = 7 ft
Use the formula for the area of the triangle.
A₁ = \(\frac{1}{2} b \cdot h\)
A₁ = \(\frac{1}{2} 6 \cdot 7\)
A₁ = \(\frac{42}{2}\)
A₁ = 21
The area of the triangle is 21 ft².
Area of the rectangle
length = 10 ft
width = 4 ft
Use the formula for the area of the rectangle
A₃ = l ∙ w
A₃ = 10 ∙ 4
A₃ =40
The area of the rectangle is 40 ft².

Add the areas to find the total area.
A = A₁ + A₂ = 21 + 40 = 61
The area of the floor is 61 ft².

a. How much carpet does Erin need?
Answer:
The area of the floor is 61 ft².

b. The carpet Erin has chosen costs $2.50 per square foot. How much will it cost her to carpet the floor?
Answer:
To find the total cost of the carpet multiply the total area of the carpet by the cost of the carpet per square foot.
Total cost of the carpet = 61 ∙ 2.5 = 152.5
Total cost of the carpet = $152.5

Question 7.
Multiple Representations Hexagon ABCDEF has vertices A(-2, 4), B(0, 4), C(2, 1), D(5, 1), E(5, -2), and F(-2, -2). Sketch the figure on a coordinate plane. What is the area of the hexagon?

Answer:

The area of the hexagon consists of a rectangle and a trapezoid

Area of Composite Figures Worksheet Lesson 9.3 Answer Key Question 8.
A field is shaped like the figure shown. What is the area of the field? Use 3.14 for π.

Answer:
Separate the field into a triangle, a square, and a quarter of a circle.
Area of the triangle
base = 8 m
height = 8 m
Use the formula for the area of the triangle.
A₁ = \(\frac{1}{2} b \cdot h\)
A₁ = \(\frac{1}{2} 8 \cdot 8\)
A₁ = \(\frac{64}{2}\)
A₁ = 32
The area of the triangle is 32 m².
Area of the square
side = 8 m
Use the formula for the area of the square.
A₂ = s²
A₂ = 8²
A₂ = 64
The area of the sqaure is 64 m².

To find the area of the quarter of a circle, find the area of the circle and divide it by 4.
Area of the circle
radius = 8 m
Use the formula for the area of the circle when given radius.
A_c = π(r)² Substitute 10 for r and 3.14 for π.
A_c = 3.14(8)²
A_c = 3.14 ∙ 64
A_c = 200.96
A₃ = \(\frac{A_{c}}{4}\) = \(\frac{200.96}{4}\) = 50.24
The area of the quarter of the circle is 50.24 m².
Add the areas to find the total area.
A = A₁ + A₂ + A₃ = 32 + 64 + 50.24 = 146.24
The area of the field is 146.24 m².

Question 9.
A bookmark is shaped like a rectangle with a semicircle attached at both ends. The rectangle is 12 cm long and 4 cm wide. The diameter of each semicircle is the width of the rectangle. What is the area of the bookmark? Use 3.14 for π.
Answer:

Separate the figure into a rectangle and two semicircles.
Two semicircles with the same diameter make a circle with the same diameter
Area of the circle
diameter = 4 cm
Use the formula for the area of the circle when given diameter

Area of the circle is 12.56 cm².
Area of the rectangle
length = 12 cm
width = 4 cm
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 12 ∙ 4
A₂ = 48
The area of the rectangle is 48 cm2.
Add the areas to find the total area.
A = À₁ + À₂ = 12.56 + 48 = 60.56
The area of the bookmark is 60.56 cm²

Question 10.
Multistep Alex is making 12 pennants for the school fair. The pattern he is using to make the pennants is shown in the figure. The fabric for the pennants costs $1.25 per square foot. How much will it cost Alex to make 12 pennants?

Answer:
Separate the figure into a triangles and one rectangle.
Area of the triangle
base = 1 ft
height = 1 ft
Use the formula for the area of the triangle.

The area of the triangle is 1.5 ft².
Area of the rectangle
length = 3 ft
width = 1 ft
Use the formula for the area of the rectangle.
A₂ = l ∙ w
A₂ = 3 ∙ 1
A₂ = 3
The area of the rectangLe is 3 ft².

Add the area to find the total area.
A = A₁ + A₂ = 1.5 + 3 = 4.5
The area of the pennants is 4.5 ft².
The cost of one pennants = 4 ∙ 1 .25 = 5
The cost of one pennants is 85.
Alex is making 12 of them so the total cost of 12 is
The total cost = total cost of one pennants∙12 = 5 ∙ 12 = 60
Alex will pay $60 for 12 pennants.

Question 11.
Reasoning A composite figure is formed by combining a square and a triangle. Its total area is 32.5 ft². The area of the triangle is 7.5 ft². What is the length of each side of the square? Explain.
Answer:

A = the area of the composite figure
A₁ = the area of the triangle
A₂ = the area of the square
A₁ = 7.5ft
A = 32.5 ft
Composite figure consists a triangle and a square.
A = A₁ + A₂ Substitute 7.5 for A_1 and 32.5 for A.
32.5 = 7.5 + A₂ Subtract 7.5 from both sides.
32.5 – 7.5 = A₂
A₂ = 25
The area of the square is 25 ft².
A₂ = s². Substititute 25 for A_2
25 = s₂ Root both sides.
\(\sqrt{25}\) = \(\sqrt{s^{2}}\)
5 = s
s = 5
The Length of the side of the square is 5 ft

Texas Go Math Grade 7 Lesson 9.4 H.O.T. Focus On Higher Order Thinking Answer Key

Question 12.
Represent Real-World Problems Christina plotted the shape of her garden on graph paper. She estimates that she will get about 15 carrots from each square unit. She plans to use the entire garden for carrots. About how many carrots can she expect to grow? Explain.

Answer:
Separate the figure into a rectangle and a trapezoid.
Area of the parallelogram
length = 4
width = 2
Use the formula for the area of the rectangle.
A₁ = l ∙ w
A₁ = 4 ∙ 2
A₁ = 8
The area of the parallelogram is 8 square units.
Area of the trapezoid
b₁ = 8
b₂ = 6
h = 2
Use the formula for the area of the trapezoid.

Add the areas to find the total area.
A = A₁ + A₂ = 8 + 14 = 22
The area of the garden is 22 square units.
Christina estimates that she’ll get 15 carrots from each square unit hence, the total number of carrots she can expect
is 22 ∙ 15 = 330.

Total number of carrots she can expect is 330.

Go Math Lesson 9.4 7th Grade Area of Composite Figures Worksheet Question 13.
Analyze Relationships The figure shown is made up of a triangle and a square. The perimeter of the figure is 56 inches. What is the area of the figure? Explain.

Answer:
The perimeter of the figure consists 3 sides of a square and 2 sides of a triangle.
Hence
P = 3 s₁ – 2 s₂
where s₁ represents side of a square, and s₂ represents side of a triangle.
P = 56 in.
s₂ = 10 in.
P = 3 ∙ s₁ + 2 ∙ s₂ Substitute 56 for P, and 10 for s_2.
56 = 3s₁ + 2 ∙ 10
56 = 3s₁ + 20 Subtract 20 from both sides.
56 – 20 = 3s₁
36 = 3s₁ Divide both sides by 3.

The side of the square is 12 in.
The side of the square is also base of the triangle.

Area of the triangle
base = 12 in.
height = 8 in.
Use the formula for the area of the triangle.

The area of the triangle is 48 in.².
Area of the square
side = 12 in.
Use the formula for the area of the square.
A₂ = s²
A₂ = 12²
A₂ = 144
The area of the square is 144 in.².
Add the areas to find the total area.
A = A₁ + A₂ = 48 + 144 = 192
The area of the figure is 192 in.².

Question 14.
Critical Thinking The pattern for a scarf is 28 in. shown at right. What is the area of the scarf? Use 3.14 for π.

Answer:
The pattern for a scarf we get when we subtract two semicircles with a diameter 15 in. from the rectangle with 28 in. length and 15 in.width.
Hence, the area of the scarf is
A = A₁ – A₂
A₁ = the area of the rectangle
A₂ = area of the two semicircles with same diameter, which make a circle with the same diameter.
Area of the rectangle
Length = 28 in.
width = 15 in.
Use the formula for the area of the rectangle.
A₁ = l ∙ w
A₁ = 28 ∙ 15
A₁ = 420
The area of the rectangle is 420 in ²

Area of the circle
diameter = 15 in.
Use the formula for the area of the circle when given diameter.

Area of the circle is 176.62 in.².
A = A₁ – A₂ = 420 – 176.62 = 243.38
The area of the scarf is 243.38 in.².

Texas Go Math Grade 7 Lesson 9.4 Composite Figure Answer Key Question 15.
Persevere in Problem-Solving The design for the palladium window shown includes a semicircular shape at the top. The bottom is formed by squares of equal size. A shade for the window will extend 4 inches beyond the perimeter of the window, as shown by the dashed line around the window. Each square in the window has an area of 100 in².

a. What is the area of the window? Use 3.14 for π.
Answer:
The length of the side of small, squares is 10 inches since their area is loo square inches. Since there are four small
squares for the square part of the palladium window, the length will be 40 inches. Determine the area of the
square part of the window.
A = s² Write the formula for the area of square
A = 40² Substitute the value
A = 1, 600 Evaluate the exponent
The diameter of the semicircle is 40 inches (Length of the square of the window). The radius is half of the diameter,
therefore, the radius is 20 inches. Determine the area of the semicircle.

Determine the area of the window.
A_window = A_square + A_semicircle
A_window = 1, 600 in² + 628 in²
A_window = 2, 228 in²

b. What is the area of the shade? Round your answer to the nearest whole number.
Answer:
From the length of the window which is 40 inches, adding the shadow will give the dimensions of the square part
of the window as 48 inches by 44 inches. Determine the area of the rectangle part of the window
A = lw Write the formula for the area of a rectangle
A = 48(44) Substitute the value
A = 2, 112 Multiply the values
The diameter of the semicircle with the shadow will be 48 inches. The radius is half of the diameter, therefore, the
radius is 24 inches. Determine the area of the semicircle.

Determine the area of the shade.
A_shade = A_rectangle + A_semicircle
A_shade = 2, 112 in² + 904.32 in²
A_shade = 3, 016.32 in²
A_shade = 3, 016 in²

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 9 Quiz Answer Key.

Texas Go Math Grade 7 Module 9 Quiz Answer Key

Texas Go Math Grade 7 Module 9 Ready to Go On? Answer Key

9.1 Angle Relationships

Use the diagram to name a pair of each type of angle.

Question 1.
Supplementary angles _____
Answer:
∠DFC and ∠BFC

Go Math Module 9 Answer Key Module 9 Test Question 2.
Complementary angles. _____
Answer:
∠AFB and ∠BFC

Question 3.
Vertical angles. _____
Answer:
∠EFD and ∠BFC

9.2, 9.3 Finding Circumference and Area of Circles

Find the circumference and area of each circle. Use 3.14 for π.

Question 4.
Circumference ≈ _____
Answer:
d = 36 cm
Use the formula for the circumference of the circle when given diameter
C = π(d) Substitute 36 for d, and 3.14 for π.
C ≈ 3.14(36)
C ≈ 113.04
The circumference of the circle is 113.04 cm
C ≈ 113.04 cm

Question 5.
area ≈ ___
Answer:
Use the formula for the area of the circle.
Since the diameter is twice a radius, the formula for area of a circle, when given the diameter, is
A = π\(\left(\frac{d}{2}\right)^{2}\) Substitute 36 for d, and 3.14 for π.
A ≈ 3.14 . \(\left(\frac{36}{4}\right)^{2}\)
A ≈ 3.14 18²
A ≈ 3.14 . 324
A ≈ 1017.36
The area of the circle is about 1017.36 cm².

A ≈ 1017.36 cm²

Question 6.
Circumference ≈ _____
Answer:
The radius of the circle is 7 m.
Use the formula for the circumference of the circle.
C = 2π(r) Substitute 7 for r, and \(\frac{22}{7}\) for π.

The circumference is about 44 m.

C ≈ 44

Grade 7 Math Module Answer Key Pdf Circles Question 7.
area ≈ ___
Answer:
Use the formula for the area of the circle
A = Substitute 7 for r, and 314 for w.
A ≈ 3.14(7)²
A ≈ 3.14 49
A ≈ 153.86
The area of the circle is about 153.86 m²

A ≈ 153.86 m².

9.4 Area of Composite Figures

Find the area of each figure. Use 3.14 for π.

Question 8.

Answer:
Separate the figure into a triangle and one semicircle.
Area of the triangle
base = 14 m
height = 10 m
Use the formula for the area of the triangle.
A₁ = \(\frac{1}{2} b \cdot h\)
A₁ = \(\frac{1}{2} 14 \cdot 10\)
A₁ = \(\frac{70}{2}\)
A₁ = 35
The area of the triangle is 35 m₂.

Area of the circle
diameter = 14 m
Use the formula for the area of the circle when given the diameter
A_c = \(\pi\left(\frac{d}{2}\right)^{2}\) Subsitute 14 for d and 3.14 for π.
A_c = 3.14\(\left(\frac{14}{2}\right)^{2}\)
A_c = 3.14 . 7²
A_c = 3.14 . 49
A_c = 153.86
Area of the semicircle is half the area of the circle.
A₂ = \(\frac{A_{c}}{2}\) = \(\frac{153.86}{2}\) = 76.93
The area of the semicircle is 76.93 m².
Add the areas to find the total area.
A = A₁ + A₂ = 35 + 76.93 = 111.93
The area of the figure is 111.93 m².

Quiz for Grade 7 Math Module 9 Question 9.

Answer:
Separate the figure into a rectangle and a parallelogram.
Area of the rectangle
length = 20 cm
width = 5.5 cm
Use the formula for the area of the rectangle.
A₁ = l . w
A₁ = 20 . 5.5
A₁ = 110
The area of the rectangle is 110 cm².
Area of the parallelogram
b = 20 cm
h = 4.5 cm
Use the formula for the area of the parallelogram.
A₂ = b . h
A₂ = 20 . 4.5
A₂ = 90
The area of the parallelogram is 90 cm²

Add the areas to find the total area.
A = A¹ + A² = 110 + 90 = 200
The area of the figure is 200 cm².

Essential Question

Question 10.
How can you use geometric formulas in real-world situations?
Answer:
There are times that we need to determine the area perimeter or volume of a certain object or even a place and
the only given is the dimensions, the geometric formulas are very helpful in such a way that we can calculate it by ourselves.

The geometric formulas are very helpful.

Texas Go Math Grade 7 Module 9 Mixed Review Texas Test Prep Answer Key 

Selected Response

Use the diagram for Exercises 1-3.

Question 1.
What is the measure of ∠BFC?
(A) 18°
(B) 72°
(C) 108°
(D) 144°
Answer:
(C) 108°

Explanation:
∠AFB and ∠BFC are supplement angles, hence
∠AFB + ∠BFC = 180° Substitute 72° for ∠AFB
72° + ∠BFC = 180° Subtract 72° from both sides.
∠BFC = 180° – 72°
∠BFC = 108°

Math Quiz for Grade 7 Module 9 Test Question 2.
Which describes the relationship between ∠BFA and ∠CFD?
(A) adjacent angles
(B) complementary angles
(C) supplementary angles
(D) vertical angles
Answer:
(D) vertical angles

Question 3.
Which information would allow you to identify ∠BFA and ∠AFE as complementary angles?
(A) m∠AFE = 108°
(B) ∠DFE is a right angle.
(C) ∠BFA and ∠BFC are supplementary angles.
(D) ∠BFA and ∠BFC are adjacent angles.
Answer:
(B) ∠DFE is a right angle.

Explanation:
∠DFE is a right angle, as we see in the diagram.

Question 4.
David pays $7 per day to park his car. He uses a debit card each time. By what amount does his bank account change due to parking charges over a 40-day period?
(A) -$280
(B) -$47
(C) $47
(D) $280
Answer:
(A) -$280

Explanation:
David pays $7 each day for 40 days, so amount on his bank account decreases for $7 . 40 = $280.

Question 5.
What is the circumference of the circle? Use 3.14 for π.

(A) 34.54 m
(B) 69.08 m
(C) 379.94 m
(D) 1,519.76 m
Answer:
(B) 69.08 m

Explanation:
r = 11 m
Use the formula for circumference.
C = 2πr(r) Substitute 11 for r and 314 for π
C ≈ 2 3.14 . 11
C ≈ 69.08
The circumference of the circle is 69.08 m.

7th Grade Module 9 Quiz Answers Question 6.
What is the area of the circle? Use 3.14 for π.

(A) 23.55 m²
(B) 176.625 m²
(C) 47.1 m²
(D) 706.5 m²
Answer:
(B) 176.625 m²

Explanation:
d = 15m
Use the formula for the area of the circle.
A = πr²
Since the diameter is twice a radius, the formula for area of a circle, when given the diameter, is
A = π(r)² Substitute \(\frac{d}{2}\) for r.
A = \(\pi\left(\frac{d}{2}\right)^{2}\) Substitute 15 for d, and 3.14 for π.
A ≈ 3.14 . \(\left(\frac{15}{2}\right)^{2}\)
A ≈ 3.14 . (7.5)²
A ≈ 3.14 . 56.25
A ≈ 176.62
The area of the circle is about 176.62 m².

Gridded Response

Grade 7 Math Module 9 Answer Key Question 7.
Find the area in square meters of the figure below. Use 3.14 for π.

Answer:
Separate the figure into a square and a quarter of circle.
Area of the square
side = 6 m
Use the formula for the area of the square
A₁ = 8²
A₁ = 6²
A₁ = 36
The area of the square is 36 m².
To find the area of the quarter of a circle, find the area of the circle and divide it by 4.
Area of the circle
radius = 6 m
Use the formula for the area of the circle when given a radius.
A_c = πr² Substitute 10 for r and 314 for w.
A_c = 3.14(6)²
A_c = 3.14 . 36
A_c = 113.04
A₂ = \(\frac{A_{c}}{4}\) = \(\frac{113}{404}\) = 28.26
The area of the quarter of the circle is 28.26 m².
Add the areas to find the total area.
A = A₁ + A₂ = 36 + 28.26 = 64.26
The area of the figure is 64.26 m².

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 13 Answer Key Taxes, Interest, and Incentives.

Texas Go Math Grade 7 Module 13 Answer Key Taxes, Interest, and Incentives

Texas Go Math Grade 7 Module 13 Are You Ready? Answer Key

Write each percent as a decimal.

Question 1.
45% ______
Answer:
Write the percent as the sum of 1 whole and a percent remainder.
100% – 55% = \(\frac{100}{100}\) – \(\frac{55}{100}\) Write the percents as fractions.
= 1 – 0.55 Write the fractions as decimals.
= 0.45

Question 2.
91% ______
Answer:
Write the percent as the sum of 1 whole and a percent remainder.
100% – 9% = \(\frac{100}{100}\) – \(\frac{9}{100}\) Write the percents as fractions.
= 1 – 0.09 Write the fractions as decimals.
= 0.91

Question 3.
8%
Answer:
Write the percent as the sum of 1 whole and a percent remainder.
100% – 92% = \(\frac{100}{100}\) – \(\frac{92}{100}\) Write the percents as fractions
= 1 – 0.92 Write the fractions as decimals
= 0.08

Question 4.
111%
Answer:
Write the percent as the sum of 1. whole and a percent remainder.
100% + 11% = \(\frac{100}{100}\) – \(\frac{11}{100}\) Write the percents as fractions.
= 1 + 0.11 Write the fractions as decimals.
= 1.11

Write each decimal as a percent.

Question 5.
0.79 _______
Answer:
Write the decimal as the sum of 1 whole and a decimal remainder.
1 – 0.21 = \(\frac{100}{100}\) – \(\frac{21}{100}\) Write the decimals as fractions.
= 100% – 21% Write the fractions as percents.
= 79%

Question 6.
0.8 ________
Answer:
Write the decimal as the sum of 1 whole and a decimal remainder.
1 – 0.21 = \(\frac{100}{100}\) – \(\frac{20}{100}\) Write the decimals as fractions.
= 100% – 20% Write the fractions as percents.
= 80%

Question 7.
0.05 ____
Answer:
Write the decimal as the sum of 1 whole and a decimal remainder.
1 – 0.95 = \(\frac{100}{100}\) – \(\frac{95}{100}\) Write the decimals as fractions.
= 100% – 95% Write the fractions as percents.
= 5%

Question 8.
1.98
Answer:
Write the decimal as the sum of 1 whole and a decimal remainder.
1 + 0.98 = \(\frac{100}{100}\) + \(\frac{98}{100}\) Write the decimals as fractions.
= 100% + 98% Write the fractions as percents.
= 198%

Find each sum or difference.

Question 9.
11.9 – 7.6 _________
Answer:

4.3

Question 10.
24.1 – 9.25 ______
Answer:

14.85

Question 11.
45 – 10.6 ________
Answer:

34.4

Question 12.
6.04 – 3.5 _________
Answer:

2.54

Question 13.
5.17 – 5.09 ______
Answer:

0.08

Question 14.
100 – 3.77 _______
Answer:

96.23

Multiply.

Question 15.

Answer:
Multiply as whole numbers.
Count the total number of decimal places in the both factors, and put the same number of decimals in the product.

66.48

Question 16.

Answer:
Multiply as whole numbers.
Count the total number of decimal places in the both factors, and put the same number of decimals in the product.

6.965

Question 17.

Answer:
Multiply as whole numbers.
Count the total number of decimal places in the both factors, and put the same number of decimals in the product.

38.528

Question 18.

Answer:
Multiply as whole numbers.
Count the total number of decimal places in the both factors, and put the same number of decimals in the product.

27.648

Texas Go Math Grade 7 Module 13 Reading Start-Up Answer Key

Visualize Vocabulary

Use the ✓ words to complete the graphic. You will put a different word in each box.

Understand Vocabulary

Complete each sentence using the preview words.

Question 1.
The amount of money earned by bank customers based on the amount of principal in their savings account is _____
Answer:
The amount of money earned by bank customers based on the amount of principal in their savings account is Interest

Question 2.
_____________________ is interest paid only on the principal, according to an agreed upon interest rate.
Answer:
Simple Interest is interest paid only on the principal, according to an agreed upon interest rate.

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 9.3 Answer Key Area of Circles.

Texas Go Math Grade 7 Lesson 9.3 Answer Key Area of Circles

Essential Question
How do you find the area of a circle?

Texas Go Math Grade 7 Lesson 9.3 Explore Activity 1 Answer Key
You can use what you know about circles and π to help find the formula for the area of a circle.
Step 1
Use a compass to draw a circle and cut it out.

Step 2
Fold the circle three times as shown to get equal wedges.
Step 3
Unfold and shade one-half of the circle.

Step 4
Cut out the wedges, and fit the pieces together to form a figure that looks like a parallelogram.

The base and height of the parallelogram relate to the parts of the circle.
base b = the circumference of the circle, or ______

height h = the ___________ of the circle, or ______
To find the area of a parallelogram, the equation is A = ___
To find the area of the circle, substitute for b and h in the area formula.

Reflect

Go Math Grade 7 Pdf Practice and Homework Lesson 9.3 Question 1.
How can you make the wedges look more like a parallelogram?
Answer:
In order to make a parallelogram for the wedges of the circle, cut one wedge into half. Then re-arrange the
wedges alternating each wedge to form a parallelogram. The two halves of one wedge will go on each end side.
This will form a rectangle, a kind of parallelogram.

Making a circle into a parallelogram.

Finding the Area of a Circle

Area of circle
The area of a circle is equal to π times the radius squared.
A = πr²

Remember that area is given in square units.

Example 1
A biscuit recipe calls for the dough to be rolled out and circles to be cut from the dough. The biscuit cutter has a radius of 4 cm. Find the area of the biscuit once it is cut. Use 3.14 for π.

Reflect

Question 2.
Compare finding the area of a circle when given the radius with finding the area when given the diameter.
Answer:
The area of the circle when given the radius is
A = πr²
Since the diameter is twice the radius, the formula for the area of a circle, when given the diameter, is
A = πr² Substitute \(\frac{d}{2}\) for r.

Question 3.
Why do you evaluate the power in the equation before multiplying?
Answer:
Because the power refers to the radius only, not the number π.

Your Turn

Practice and Homework Lesson 9.3 Answer Key 7th Grade Question 4.
A circular pool has a radius of 10 feet. What is the area of the pool? Use 3.14 for π. ______
Answer:
r = 10 feet
Use the formula for the area of the circle, cause the pool is circular.
A = πr(r)² Substitute 10 for r, and 3.14 for π.
A ≈ 3.14(2)²
A ≈ 3.14 . 100
A ≈ 314
The area of the pool is about 314 feet

Explore Activity 2
Finding the Relationship between Circumference and Area
You can use what you know about the circumference and area of circles to find a relationship between them.

Find the relationship between the circumference and area of a circle.

The circumference of the circle squared is equal to
___________
Answer:

Reflect

Question 5.
Does this formula work for a circle with a radius of 3 inches? Show your work.
Answer:
The radius of the circle is 3 in.
The circumference of the circle squared is equal to
C² = 4π . A,
where A represents the area of the circle.
First find the area of the circle with the radius of 3 in.
Use the formula for the area of the circle.
A = π(r)² Substitute 3 for r, and 3.14 for π.
A ≈ 3.14(3)²
A ≈ 3.14 . 9
A ≈ 28.26
The area of the circle is about 28.26 in.
Use the formula for the circumference when given the area
C² = 4π . A Substitute 28.26 for A and 3.14 for π.
C² ≈ 4 . 3.14 ∙ 28.26
C² ≈ 354.94 Root both sides.
\(\sqrt{C^{2}}\) ≈ \(\sqrt{354.94}\)
The circumference of the circle is 18.84 in.

Let’s see if we get the same result if we use the original formula for circumference
Use the formula for the circumference of the circle.
C = 2πr(r) Substitute 3 for r and 3.14 for π.
C ≈ 23.14 ∙ 3
C ≈ 18.84
The circumference of the circle is 18.84 in.
Hence, we can use both formulas for the circumference of the circle.

The formula works, so we can use both formulas for the circumference of the circle.

Texas Go Math Grade 7 Lesson 9.3 Guided Practice Answer Key

Find the area of each circle. Round to the nearest tenth if necessary. Use 3.14 for π. (Explore Activity 1)

Question 1.

Answer:
The diameter of the circle is 14,m.
Use the formula for the area of the circle
A = π(r)²
Since the diameter is twice the radius, the formula for the area of a circle, when given the diameter, is

The area of the circle is about 153.86 m².

Go Math Grade 7 Lesson 9.3 Answer Key Question 2.

Answer:
The radius of the circle is 12 mm.
Use the formula for the area of the circle.
A = π(r)² Substitute 12 mm for r, and 3.14 for π.
A ≈ 3.14 ∙ (12 mm)²
A ≈ 3.14 ∙ 144 mm²
A ≈ 452.16 mm²
The area of the circle is about 452.16 mm².

Question 3.

Answer:
The diameter of the circle is 20 yd
Use the formula for the area of the circle
A = π(r)²
Since the diameter is twice the radius, the formula for the area of a circle, when given the diameter, is

The area of the circle is about 314 yd².

Solve. Use 3.14 for π. (Example 1)

Question 4.
A clock face has a radius of 8 inches. What is the area of the clock face? Round your answer to the nearest hundredth.
Answer:
The radius of the circle is 8 in.
Use the formula for the area of the circle
A = π(r)² Substitute 8 in. for r, and 3.14 for π.
A ≈ 3.14. (8 in)²
A ≈ 3.14 ∙ 64 in²
A ≈ 200.96 in²
The area of the circle is about 200.96 in²

Question 5.
A DVD has a diameter of 12 centimeters. What is the area of the DVD? Round your answer to the nearest hundredth.
Answer:
The diameter of the circle is 12 cm.
Use the formula for the area of the Circle.
A = π(r)²
Since the diameter is twice a radius, the formula for area of a circle, when given the diameter, is

The area of the circle is about 226.08 cm².

Go Math Book 7th Grade Area of a Circle Answer Key Question 6.
A company makes steel lids that have a diameter of 13 inches. What is the area of each lid? Round your answer to the nearest hundredth.
Answer:
The diameter of the circle is 13 in.
Use the formula for the area of the circle.
A = π(r)²
Since the diameter is twice a radius, the formula for area of a circle, when given the diameter, is

The area of the circle is about 132.66 in².

Find the area of each circle. Give your answers in terms of π. (Explore Activity 2)

Question 7.
C = 4π
A = ____
Answer:
Use the formula for the circumference of the circle.

Use the formula for area of the circle

Question 8.
C = 12π
A = ___
Answer:
Use the formula for the circumference of the circle.

Use the formula for area of the circle.

Area in Terms of Circumference Answer Key Pdf Grade 7 Question 9.
C = \(\frac{\pi}{2}\)
A = ____________
Answer:
Use the formula for the circumference of the circle.

Use the formula for area of the circle.

Question 10.
A circular pen has an area of 64π square yards. What is the circumference of the pen? Give your answer in terms of π. (Explore Activity 2)
___________
Answer:
Use the formula for the circumference of the circle.

Use the formula for area of the circle.

Essential Question Check-In

Question 11.
What is the formula for the area A of a circle in terms of the radius r?
Answer:
A = r²π

Independent Practice

Question 12.
The most popular pizza at Pavone’s Pizza is the 10-inch personal pizza with one topping. What is the area of a pizza with a diameter of 10 inches? Round your answer to the nearest hundredth.
Answer:
The diameter of the pizza is 10 in.
Use the formula for area of the circle.
Since the radius is half the diameter, formula for the area of the pizza will be
A = π(r)² Substitute \(\frac{d}{2}\) for r.

The area of the pizza is about 78.5 in.².

Lesson 9.3 Area of Circles Answer Key Go Math Grade 7 Question 13.
A hubcap has a radius of 16 centimeters, What is the area of the hubcap? Round your answer to the nearest hundredth.

Answer:
The radius of the hubcap is 16 cm.
Use the formula for area of the circle to find area of the hubcap.
A = π(r)² Substitute 16 cm for r, and 3.14 for π.
A ≈ 3.14(16 cm)²
A ≈ 3.14 256 cm²
A ≈ 803.84 cm²
The area of the hubcap is about 803.84 cm².

Question 14.
A stained glass window is shaped like a semicircle. The bottom edge of the window is 36 inches long. What is the area of the stained glass window? Round your answer to the nearest hundredth.
Answer:
The bottom edge of the window represents the diameter of the circle whose half is a window, hence, the area of the window is half of the area of the circle.
Use the formula for area of the circle, where instead of the radius, we place the diameter which is twice the radius

The area of the circle is about 1017.36 in.².
Hence, the area of the window is half of the area of the circle
A ÷ 2 ≈ 1017.36 ÷ 2 ≈ 508.68 in.²
The area of the window is about 508.68 in.².

Question 15.
Analyze Relationships The point (3, 0) lies on a circle with the center at the origin. What is the area of the circle to the nearest hundredth?
Answer:

As the center of the circle is origin, the distance from it to the point(3, 0) is the radius.
From the diagram above, we see that the radius of the circle is 3.
Use the formula for the area of the circle.
A = π(r)² substitute 3 for r, and 3.14 for π.
A ≈ 3.14(3)²
A ≈ 3.14 ∙ 9
A ≈ 28.26
The area of the circle is about 28.26

Question 16.
Multistep A radio station broadcasts a signal over an area with a radius of 50 miles. The station can relay the signal and broadcast over an area with a radius of 75 miles. How much greater is the area of the broadcast region when the signal is relayed? Round your answer to the nearest square mile.
Answer:
We have to find the area of the signal that the station can relay, and the area of the signal that station broadcasts.
The station can relay the signal over an area with a radius of 75 miles, so the area of it will be
A₁ ≈ π(r)² Substitute 75 miles for r, and 3.14 for π.
A₁ ≈ 3.14(75 miles)²
A₁ ≈ 3.14 ∙ 5625 miles²
A₁ ≈17, 662.5 miles²
The area of the signal that the station can relay is about 17, 662.5 miles².
On the other side, the station broadcasts a signal over an area with a radius of 50 miles, so the area of it will be
A₂ = π(r)² Substitute 50 miles for r, and 3.14 for π.
A₂ ≈ 3.14(50 miles)²
A₂ ≈ 3.14 ∙ 2500 miles²
A₂ ≈ 7 miles
A₂ ≈ 7,850 miles²
The area of the signal that the station broadcasts is about 7,850 miles².
The area of the broadcast region when the signal is relayed, is greater for the difference between the area of the signal that the station can relay, and the area of the signal that station broadcasts.
A₁ – A₂ ≈ 17,662.5 – 7,850 ≈ 9,812.5 miles²

Question 17.
Multistep The sides of a square field are 12 meters. A sprinkler in the center of the field sprays a circular area with a diameter that corresponds to a side of the field. How much of the field is not reached by the sprinkler? Round your answer to the nearest hundredth.
Answer:

The area marked with red color is the area sprinkler does not reach.
To find it, first we have to find the area of the square, and a circular area that a sprinkler sprays.
Use the formula for the area of a square.
A₁ = a² where a represents side of a square. Substitute 12 m for a.
A₁ = 12 m²
A₁ = 144 m²
The area of the square is 144 m².
Use the formula for the area of the circle with a diameter same as a square.
Since the radius is half of the diameter, it follows

The area of the circular area that a sprinkler sprays is about 113.04 m²
From the area of the square subtract the area of the circular area to get the area the sprinkler does not reach.
À₁ – A₂ ≈ 144 m² – 113.04 m² ≈ 30.96 m²
The area the sprinkler does not reach is about 30.96 m².

Lesson 9.3 Area of Circles Answer Key Go Math Pdf Grade 7 Question 18.
Justify Reasoning A small silver dollar pancake served at a restaurant has a circumference of 2π inches. A regular pancake has a circumference of 4π inches. Is the area of the regular pancake twice the area of the silver dollar pancake? Explain.
Answer:
Find the area of the small silver dollar pancake and the area of the regular pancake.
First, find the formula for the area of the circle when given circumstance.
Use the formula for the circumference of the circle.

Use the formula for area of the circle.

The circumference of the small silver dollar pancake is 2π in.
Use the formula above for the area.

The area of the small silver dollar pancake is π in.².

The circumference of the regular pancake is 4π in.
Use the formula above for the area.

The area of the regular pancake is 4π in.².
A₂ – A₁ =4π – π = 3π
The area of the regular pancake is more than twice the area of the small silver dollar pancake.

Question 19.
Analyze Relationships A bakery offers a small circular cake with a diameter of 8 inches. It also offers a large circular Cake with a diameter of 24 inches. Does the top of the large cake have three times the area of that of the small cake? If not, how much greater is its area? Explain.
Answer:
Find the area of the small and large circular cakes.
Use the formula for the area of a circle.
Since the diameter is twice the radius, the formula for the area of a circle when given the diameter is
A = π(r)² Substitute \(\frac{d}{2}\) for r.

Let d₁ be the diameter of the larger cake and d₂ the diameter of the smaller cake.
d₁ = 24 inches
d₂ = 8 inches
Use the formula above to find the area of the large cake.

The area of the large cake is about 452.16 in.².
Use the formula above to find the area of the small cake.

The area of the small cake is about 50.24 in.²
The top of the large cake have area greater than the area of the small cake for about
A₁ – A₂ ≈ 452.16 – 50.24 ≈ 401.92 in.²
Hence, the area of the large cake is more than 3 times the area of the small cake.

Question 20.
Communicate Mathematical Ideas You can use the formula A = \(\frac{C^{2}}{4 \pi}\) to find the area of a circle given the circumference. Describe another way to find the area of a circle when given the circumference.
Answer:
The other way is to find the radius first Since C = 2πr, we have r = \(\frac{C}{2 \pi}\). After that, we can find the area by substituting for r in A = πr².
Determine the radius first with the formula r = \(\frac{C}{2 \pi}\)

Question 21.
Draw Conclusions Mark wants to order a pizza. Which is the better deal? Explain.

Answer:


Hence, a pizza with a diameter of 18 inches for $20 will better deal because in this case will get a larger pizza as compared to a tile pizza with a diameter 12 inch for $10.

Question 22.
Multistep A bear was seen near a campground. Searchers were dispatched to the region to find the bear.
a. Assume the bear can walk in any direction at a rate of 2 miles per hour. Suppose the bear was last seen 4 hours ago. How large an area must the searchers cover? Use 3.14 for π. Round your answer to the nearest square mile. _______
Answer:
Determine the area the searchers must cover The radius of the area the bear walks is 8 miles since the bear was
last seen 4 hours ago with the rate of 2 miles per hour
A = πr² Write the formula for the area of a circle
A = (3.14)(8)² Substitute the values
A = (3.14)(64) Evaluate the exponent
A = 200.96 Multiply the values
A = 201 Round off

b. What If? How much additional area would the searchers have to cover if the bear were last seen 5 hours ago?
Answer:
Determine the area the searchers must cover. The radius of the area the bear walks is 10 miles since the bear was
last seen 5 hours ago with the rate of 2 miles per hour.
A = πr² Write the formula for the area of a circle
A = (3.14)(10)² Substitute the values
A = (3.14)(100) Evaluate the exponent
A = 314 Multiply the values
Determine the additional area the searchers need to cover
314 – 201 = 113
The searchers need to cover an additional 113 square miles.

Texas Go Math Grade 7 Lesson 9.3 H.O.T. Focus on Higher Order Thinking Answer Key

Question 23.
Analyze Relationships Two circles have the same radius. Is the combined area of the two circles the same as the area of a circle with twice the radius? Explain.
Answer:
The combined area of the two circles, with the same radius, is twice the area of that circle.
Use the formula for the area of the circle.
A = the combined area
A₁ the area of the circle
A = 2 ∙ A₁
A = 2πr²
The area of the circle with twice the radius is
A = π(2r)²
A = π2²r²
A = 4πr²
Hence, the combined area of the two circles with the same radius isn’t the same as the area of the circle with twice
the radius.

Question 24.
Look for a Pattern How does the area of a circle change if the radius is multiplied by a factor of n, where n is a whole number?
Answer:
If we multiply the radius by a whole number n, the area of the circle will be
A = π(nr)²
A = πn²r²
Hence, the area of the circle will be n² times larger.

Question 25.
Represent Real World Problems The bull’s eye on a target has a diameter of 3 inches. The whole target has a diameter of 15 inches. What part of the whole target is the bull’s-eye? Explain.
Answer:
Use the formula for the circle to find the area of the whole target with a diameter of 15 in.

The area of the whole target is 176.62 in.²
Use the formula for the circle to find the area of the bull’s – eye with the diameter of 3 in

The area of the bull’s – eye is 7.06 in.²

To find what part of the whole target is the bull’s – eye, we have to divide the area of the whole target by the area
of the bull’s – eye.
\(\frac{A_{1}}{A_{2}}\) = \(\frac{176.62}{7.06}\) = 25.01
Hence, the bull’s – eye is the 25th part of the whole target

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Unit 6 Study Guide Review Answer Key.

Texas Go Math Grade 7 Unit 6 Study Guide Review Answer Key

Analyzing and Comparing Data

Texas Go Math Grade 7 Unit 6 Exercises Answer Key

Question 1.
Five candidates are running for the position of School Superintendent. Find the percent of votes that each candidate received. (Lesson 11.1)

Answer:
Number votes A gets = 70
Number votes B gets = 84
Number votes C gets = 80
Number votes D gets = 82
Number votes E gets = 98
Total number of votes = 70 + 84 + 80 + 82 + 98 = 414

Percentage of votes of A = \(\frac{70}{414}\) × 100 = 16.90%
Percentage of votes of B = \(\frac{84}{414}\) × 100 = 20.29%
Percentage of votes of C = \(\frac{80}{414}\) × 10o = 19.32%
Percentage of votes of D = \(\frac{82}{414}\) × 100 = 19.80%
Percentage of votes of E = \(\frac{98}{414}\) × 100 = 23.67%
Percentage of vote A gets is 16.90%

The dot plots show the number of hours a group of students spend online each week, and how many hours they spend reading. Compare the dot plots visually. (Lesson 11.2)

Grade 7 Unit 6 Practice Problems Answer Key Question 2.
Compare the shapes, centers, and spreads of the dot plots.
Shape: ___________________
Center: _________________________
Spread: ______________
Answer:
Comparing the shapes, centers, and spreads of the dot plots:
Shape: In case of Online time most of the data are greater than 4 but in case of Reading time data is almost symmetric about the point 3.
Centre: Centre of the both given dot plot will be the median of both dot plot. Median or centre of Online time is 6
and the median or centre of Reading time is 5.
Spread: Spread of dot plot is the range of dot pot. Range of Online time is 7 – 0 = 7 and range of Reading time
is 6 – 0 = 6.

Median of Online time is 6 and of Reading time is 5.

Question 3.
Calculate the medians of the dot plots. _______
Answer:
Arranging Online time data: 0, 4, 4, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7
Arranging Reading time data: 0, 0, 0, 0, 1, 1, 2, 5, 5, 5, 6, 6, 6, 6, 6
We know that when total number of given data in a sample is “n” and it is an odd number then the median is the \(\frac{(n+1)}{2} t h\) term after arranging the data in increasing order. In above given dot plot total number of data is 15. So, \(\frac{15+1}{2}\) = 8th term will be the median of both the dot plots.
Median of Online time = 6
Median of Reading time = 5

Medians of dot plots are 6 and 5

Question 4.
Calculate the ranges of the dot plots. _______
Answer:
For Online Time:
Maximum value = 7
Minimum value = 0
Range = Maximum value – Minimum value = 7 – 0 = 7
For Reading Time:
Maximum value = 6
Minimum value = 0
Range = Maximum value – Minimum value = 6 – 0 = 6
Hence, range of online time is 7 hrs and range of reading time is 0 hrs.

Range of online time is 7 hrs and range of reading time is 6 hrs.

The box plots show the math and reading scores on a standardized test for a group of students. Use the box
plots shown to answer the following questions.

Texas Go Math Grade 7 Unit 6 Assessment Math Answer Key Question 5.
Compare the maximum and minimum values of the box plots.
Answer:
For math score:
Maximum value = 20
Minimum vaLue = 4
Range = Maximum value – Minimum value = 20 – 4 = 16
For Reading score:
Maximum value = 20
Minimum vaLue = 4
Range = Maximum value – Minimum value = 20 – 4 = 16
On comparing both the box plot we can see that Minimum and maximum value of both are same so their range are also same. Median of math score is 8 and median of reading score is 14.

Maximum and minimum value of both box plot are same

Question 6.
Compare the interquartile range of the box plots.
Answer:
For math score:
Lower interquartile (Q₁) = 14
Upper interquartile (Q₂) = 6
interquartie Range = Q₂ – Q₁ = 14 – 6 = 8
For Reading score:
Lower interquartile (Q₁) = 18
Upper interquartile (Q₂) = 10
interquartile Range = Q₂ – Q₁ = 18 – 10 = 8
On comparing both the box plot we can see that their interquartite range are also same.

Random Samples and Populations

Question 1.
Molly uses the school directory to select 25 students at random from her school for a survey on which sports people like to watch on television. She calls the students and asks them, “Do you think basketball is the best sport to watch on television?” (Leson 12.1)

a. Did Molly survey a random sample or a biased sample of the students at her school?
Answer:
It is given in problem that Molly uses the school, directory to select 25 students for her survey. She select students
randomly from the school directory. So every student of school have an equal chance of being selected and we
know that the sample in which every person or object has an equal chance of being selected is called as random
sample. Hence, Molly survey a random sample of students at her school.

b. Was the question she asked an unbiased question? Explain your answer.
Answer:
The question asked by Molly is “Do you think basketball is the best sport to watch on television?” Here she has already mentioned basketball sports so every other sport will not have an equal chance of being selected and we know that a sample in which every person, object, or event does not have an equal chance of being selected is called a biased sample. Hence, the question asked by Molly is not unbiased.

Yes Molly surveyed a random sample

Go Math Grade 7 Book Pdf Review and Preview Answer Key Question 2.
There are 2,300 licensed dogs in Clarkson. A ràndom sample of 50 of the dogs in Clarkson shows that 8 have ID microchips implanted. How many dogs in Clarkson are likely to have ID microchips implanted? (Lesson 1 2.2)
Answer:
Total number of licensed dogs = 2,300
Number of dogs in random sample = 50
Number of dogs with ID microchips = 8

Hence, 368 dogs in Clarkson are likely to have ID microchips implanted.

Question 3.
Mr. Puccia teaches Algebra 1 and Geometry. He randomly selected 10 students from each class. He asked the students how many hours they spend on math homework in a week. He recorded each set of data in a list. (Lesson 12.3)
Algebra 1: 4, 0, 5, 3, 6, 3, 2, 1, 1, 4
Geometry: 7, 3, 5, 6, 5, 3, 5, 3, 6, 5

a. Make a dot plot for Algebra 1. Then find the mean and the range for Algebra 1.

Answer:

Minimum value of Algebra 1 = 0
Maximum valueof Algebra 1 = 6
Range of Algebra 1 = Maximum value – Minimum value
= 6 – 0
= 6

b. Make a dot plot for Geometry. Then find the mean and the range for Geometry.

Answer:

Minimum value of Geometry 1 = 3
Maximum valueof Geometry 1 = 7
Range of Geometry 1 = Maximum value – Minimum value
= 7 – 3
= 4

c. What can you infer about the students in the Algebra 1 class compared to the students in the Geometry class?
Answer:
Mean of geometry 1 is greater than algebra 1 which means that the students spend more hours on Geometry 1 as compared to algebra 1 for their math homework. And the range or spread of data is more in case of aLgebra 1 as compared to geometry 1.

Mean of algebra 1 is 2.9 hrs and of Geometry 1 is 4 hrs.

Texas Go Math Grade 7 Unit 6 Performance Tasks Answer Key

Go Math Grade 7 Answer Key Pdf Unit 6 Study Guide Question 1.
CAREERS IN MATH Entomologist An entomologist is studying how two different types of flowers appeal to butterflies. The box-and-whisker plots show the number of butterflies that visited one of two different types of flowers in a field. The data were collected over a two-week period, for one hour each day.

a. Find the median, range, and interquartile range for each data set.
Answer:
In box plot representation the middle point of box is median point and left and right point of box is lower and
upper quartile respectively.
For type A :
Median = Middle point of box
= 11
Maximum value = 13
Minimum value = 9
Range = Maximum vaLue – Minimum value
= 13 – 9
= 4
Lower quartile = 9
Upper quartile = 12
Interquartile Range = Upper quartile – lower quartile
= 12 – 9
= 3

Fortype B:
Median = Middle point of box
= 11
Maximum value = 17
Minimum vaLue = 7
Range = Maximum value – Minimum vaWe
= 17 – 7
= 10
Lower quartile = 10
Upper quartile = 12
Interquartite Range = Upper quartile – Lower quartile
= 12 – 10
= 2

b. Which measure makes it appear that flower type A had a more consistent number of butterfly visits? Which measure makes it appear that flower type B did? If you had to choose one flower as having the more consistent visits, which would you choose? Explain your reasoning.
Answer:
We can see that range of type A flower is 4 and of type B flower is 10 Range of type A flower is less as compared
to type B flower so we can say that type A flower had more consistent number of butterflies visit
We can see that interquartiLe range of type B flower is 2 and of type A flower is 3 Interquartile range of type B
flower is less as compared to type A flower so we can say that type B flower had more consistent number of
butterflies visit.

Hence, median of both type A and type B fLower is same which is 11 but the range of type A flower is very Less as compared to type B flower. In case of type A flower spread of data is less as most of the data are near its centre as median point. So i will choose type A flower as having more consistent visit.
I will choose type A flower as having more consistent visit.

Texas Go Math Grade 7 Unit 6 Mixed Review Texas Test Prep Answer Key

Selected Response

Question 1.
Which is a true statement based on the dot plots below?

(A) Set B has the greater range.
(B) Set B has the greater median.
(C) Set B has the greater mean.
(D) Set A is less symmetric than Set B.
Answer:
(A) Set B has the greater range.

We know that Range = Maximum value – Minimum value
For set A:
Maximum value = 60
Minimum value = 20
Range of set A = 60 – 20 = 40
For set B:
Maximum value = 60
Minimum value = 10
Range of set B = 60 – 10 = 50
So, set B has greater range than set A
Hence, option A is correct answer.

Question 2.
Which is a solution to the equation 7g – 2 = 47?
(A) g = 5
(B) g = 6
(C) g = 7
(D) g = 8
Answer:
(C) g = 7

Given equation in problem: 7g – 2 = 47
7g – 2 = 49 (Given)
7g – 2 + 2 = 47 + 2 (Adding 2 on both side)
7g = 49 (Simplifying)
g = \(\frac{49}{7}\) (Dividing both side by 7)
g = 7 (Solution)
Hence, option C is the correct answer.

Unit 6 Math Test 7th Grade Answer Key Question 3.
Which is a true statement based on the box plots below?

(A) The data for Team B have the greater range.
(B) The data for Team A are more symmetric.
(C) The data for Team B have the greater interquartile range.
(D) The data for Team A have the greater median.
Answer:
(C) The data for Team B have the greater interquartile range.

In boxplot representation of data the box represent the interquartile range. First or left end of the box is called the
lower quartile (Q₁) and the second or right end of box is called the upper quartile (Q₂). So the quartile range is the difference of upper quartile and lower quartile (Q₂ – Q₁). So if box is larger in length means interquartile range is greater as compared to the box having smaller Length.
Now on comparing the boxplot of team A and team B we can see that the box length of team B is larger as compared to the team A. So the interquartile range of team B will be more than team A.
Hence, option C is correct answer.

Question 4.
Which is a random sample?
(A) 10 students in the Spanish Club are asked how many languages they speak.
(B) 20 customers at an Italian restaurant are surveyed on what their favorite food is.
(C) 15 students were asked what their favorite color is.
(D) 10 customers at a pet store were asked whether or not they had pets.
Answer:
(C) 15 students were asked what their favorite color is.

Random Sample : A sample in which every person, object or event has an equal chance of being selected is called as the Random Sample.
So incase of option C there are 15 students who are being asked that “what is there favourite colour”. Here in asked question name of any colour is not mentioned. Hence each colour has equal chance of being selected in the sample. Thus this sample will be random sample.
Hence, option C is correct answer

Question 5.
Find the percent change from 84 to 63.
(A) 30% decrease
(B) 30% increase
(C) 25% decrease
(D) 25% increase
Answer:
Origina[ value = 84
New value = 63
Changed value = 63 – 84 = – 21
Negative sign means the new value has decreased from original value.

Hence, option C is correct answer.

Question 6.
A survey asked 100 students in a school to name the temperature at which they feel most comfortable. The box plot below shows the results for temperatures in degrees Fahrenheit. Which could you infer based on the box plot below?

(A) Most students prefer a temperature less than 65 degrees.
(B) Most students prefer a temperature of at least 70 degrees.
(C) Almost no students prefer a temperature of less than 75 degrees.
(D) Almost no students prefer a temperature of more than 65 degrees.
Answer:
(B) Most students prefer a temperature of at least 70 degrees.

In boxplot representation of data the box has 3 points first and third points are the lower quartile and upper
quartile respectively and the middle point of box is the median And we know that median is the almost centre
point of the data around which the values of other data lies.

So, when we observe the given boxplot we can see that centre point of box or median is 73 which means that the values of most of the data are around 73. Hence we can say that most of the students prefer a temperature of at least 70 degrees.
hence, option B is correct answer.

Unit 6 Study Guide Answer Key Geometry Question 7.
The box plots below show data from a survey of students under 14 years old. They were asked on how many days in a month they read and draw. Based on the box plots, which is a true statement about students?

(A) Most students a month.
(B) Most students a month.
(C) Most students they draw.
(D) Most students they read.
Answer:
(C) Most students they draw.

In boxplot representation of data the box has 3 points first and third points are the lower quartile and upper quartile respectively and the middle point of box is the median And we know that median is the almost centre
point of the data around which the values of other data lies.
Now on comparing the boxplot of Read and Draw we can see that the median or centre point of Read is greater than 18 and median or centre point of Draw is less than 12. And also box length of Read is larger as compared to the draw. So now we can say that most of the students read more often than they draw.
Hence, option C ¡s correct answer.

Hot tip!

Use logic to eliminate answer choices that are incorrect. This will help you to make an educated guess if you are having trouble with the question.

Question 8.
Which describes the relationship between ∠NOM and ∠JOK ¡n the diagram?

(A) adjacent angles
(B) complementary angles
(C) supplementary angles
(D) vertical angles
Answer:
In given figure when we observe line JM and NK we can see that ∠NOM and ∠JOK are just opposite of each
other and vertex point of both angle is point O. So ∠NOM and ∠JOK are vertically opposite angle and both
vertically opposite angLe will be equal in magnitude.
∠NOM = ∠JOK
Hence, option D will be correct answer.

Gridded Response

Question 9.
Katie is reading a 200-page book that is divided into five chapters. The bar graph shows the number of pages in each chapter.

What percent of the pages are in Chapter 2?

Answer:
Total number of pages in book that Katie is reading = 200
Number of pages in chapter 2 of book = 30

Hence, 15% of pages are in the chapter 2 of book.

Steps for marking the given box:
1st column : mark + sign
2nd column: mark 0
3rd column : mark 0
4th column: mark 1
5th column : mark ‘5
6th column : mark t,
7th column : mark ‘0’

Texas Go Math Book 7th Grade Unit 6 Answer Key Question 10.
Lee Middle School has 420 students. Irene surveys a random sample of 45 students and finds that 18 of them have pet cats. How many students are likely to have pet cats?

Answer
Total number of students in Lee middle school = 420
Number of students in random sample = 45
Number of students who have pet cats = 18

So, 196 students in Lee Middle School have pet cats
Step 2
Steps for marking the given box:
1st column : mark ‘+’ sign
2nd column : mark ‘0’
3rd column : mark ‘1’
4th column: mark ‘9’
5th column : mark ‘6’
6th column: mark ‘0’
7th column : mark ‘0’

Texas Go Math Grade 7 Unit 6 Vocabulary Preview Answer Key

Use the puzzle to preview key vocabulary from this unit. Unscramble the circled letters to answer the riddle at the bottom of the page.

Across
3. A sample in which every person, object, or event has an equal chance of being selected (2 words). (Lesson 12.1)
5. A display that shows how the values in a data set are distributed (2 words). (Lesson 11.3)
6. A display in which each piece of data is represented by a dot above a number line (2 words). (Lesson 11.2)

Down

1. A round chart divided into pieces that represent a portion of a set of data (2 words). (Lesson 11.1)
2. The entire group of objects, individuals, or events in a set of data. (Lesson 12.1)
4. Part of a population chosen to represent the entire group. (Lesson 12.1)

Q: Where do cowboys who love statistics live?
A: onthe ___ ___ ___ ___ !

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Module 11 Answer Key Analyzing and Comparing Data.

Texas Go Math Grade 7 Module 11 Answer Key Analyzing and Comparing Data

Texas Go Math Grade 7 Module 11 Are You Ready? Answer Key

Write each fraction as a decimal and a percent.

Question 1.
\(\frac{7}{8}\) _____
Answer:
Divide 7 by 8. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.875
Move the decimal point two places to the right and add the “%“ sign.
0.875 = 87.5%
\(\frac{7}{8}\) = 87.5%

Go Math Grade 7 Answer Key Pdf Module 11 Question 2.
\(\frac{4}{5}\) _____
Answer:
Divide 4 by 5. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.8 = 0.800
Move the decimal point two places to the right and add the “%” sign.
0.8 = 80%
\(\frac{4}{5}\) = 80%

Question 3.
\(\frac{4}{5}\) _____
Answer:
Divide 1 by 4. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.25
Move the decimal point two places to the right mid add the”%” sign.
0.25 = 25%
\(\frac{1}{4}\) = 25%

Question 4.
\(\frac{3}{10}\) _____
Answer:
Divide 3 by 10. Write a decimal point, and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.3 = 0.300
Move the decimal point two places to the right and add the “%” sign.
0.3 = 0.300 = 30%
\(\frac{3}{10}\) = 30%

Grade 7 Math Module 11 Answer Key Analyzing and Comparing Data Question 5.
\(\frac{19}{20}\) _____
Answer:
Divide 19 by 20. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.95
Move the decimal point two places to the right and add the “ %“ sign.
0.95 = 95%
\(\frac{19}{20}\) = 95%

Question 6.
\(\frac{7}{25}\) _____
Answer:
Divide 7 by 25. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.28
Move the decimal point two places to the right and add the ”%” sign.
0.28 = 28%
\(\frac{7}{25}\) = 28%

Question 7.
\(\frac{37}{50}\) _____
Answer:
Divide 37 by 50. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.74
Move the decimal point two places to the tile right awl add the %“ sign.
0.74 = 74%
\(\frac{37}{50}\) = 74%

Grade 7 Mathematics Answer Key Module 11 Question 8.
\(\frac{29}{100}\) _____
Answer:
Divide 29 by 100. Write a decimal point and zeros in the dividend and place a decimal point in the quotient.

The result is: 0.29
Move the decimal point two places to the right and add the “%“ sign.
0.29 = 29%
\(\frac{29}{100}\) = 29%

Find the median and the mode of the data.

Question 9.
11, 17, 7, 6, 7, 4, 15, 9 _________
Answer:
Put number in order centering 4, 6, 7, 7, 9, 11, 15, 17
Median is the middle number; found by ordering alt data points and picking out the one in the middle. There are three middle numbers, so take the mean of those three numbers).
centering Median = \(\frac{6+7+11}{3}\) = \(\frac{24}{3}\) = 8
centering Mode is the number that occurs the highest number of times.
Mode = 7
centering Median= 8
centering Mode= 7

Question 10.
43, 37, 49, 51, 56, 40, 44, 50, 36 _____
Answer:
Put number in order
centering 36, 37, 40, 43, 44, 49, 50, 51, 56
Median is the middle number; found by ordering all, data points and picking out the one in the middle. There are
four middle numbers,so take the mean of those four numbers.
Median = \(\frac{37+43+49+51}{4}\) = \(\frac{180}{4}\) = 45
centering Mode is the number that occurs the highest number of times.
Mode: All numbers only appear.

Median= 45
Mode: All numbers only appear once.

Find the mean of the data.

Question 11.
9, 16, 13, 14, 10, 16, 17,9 ________
Answer:
Mean: found by adding all data points and dividing by the number of data points.

mean = 13

Analyzing and Comparing Data Module 11 Grade 7 Answer Key Question 12.
108, 95, 104, 96, 97, 106, 94 ________
Answer:
Mean: found by adding all data points and dividing by the number of data points.

mean = 100

Texas Go Math Grade 7 Module 11 Reading Start-Up Answer Key

Visualize Vocabulary
Use the ✓ words to complete the right column of the chart.

Complete each sentence using the preview words.

Question 1.
A display that uses values from a data set to show how the values are spread out is a ____.
Answer:
A display that uses values from a data set to show how the values are spread out is a box plot

Question 2.
A ____ uses vertical or horizontal bars to display data.
Answer:
A bar graph uses vertical or horizontal bars to display data

Refer to our Texas Go Math Grade 7 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 7 Lesson 9.1 Answer Key Angle Relationships.

Texas Go Math Grade 7 Lesson 9.1 Answer Key Angle Relationships

Essential Question
How can you use angle relationships to solve problems?

Texas Go Math Grade 7 Lesson 9.1 Explore Activity Answer Key  

Measuring Angles
It is useful to work with pairs of angles and to understand how pairs of angles relate to each other. Congruent angles are angles that have the same measure.

Step 1
Using a ruler, draw a pair of intersecting lines. Label each angle from 1 to 4.

Step 2
Use a protractor to help you complete the chart.

Reflect

Question 1.
Conjecture Share your results with other students. Make a conjecture about pairs of angles that are opposite each other.
Answer:
The measures of the vertically opposite angles are equal, so we have two pairs of angles with the same measure.

Go Math Answer Key Grade 7 Practice and Homework Lesson 9.1 Question 2.
Conjecture When two lines intersect to form two angles, what conjecture can you make about the pairs of angles that are next to each other?
Answer:
Every pair of angles, obtained by the intersection of two lines, that are next to each other form a straight line, or an angle of 180°. Hence, they are supplementary angles.

Example 1
Use the diagram.

A. Name a pair of vertical angles.
∠AFB and ∠DFE
B. Name a pair of adjacent angles.
∠AFB and ∠BFD
C. Name a pair of supplementary angles.
∠AFB and ∠BFD
D. Find the measure of ∠AFB.
Use the fact that ∠AFB and ∠BFD in the diagram are supplementary angles to find m∠AFB.

The measure of ∠AFB is 40°

Reflect

Question 3.
Analyze Relationships What is the relationship between ∠AFB and ∠BFC? Explain.
Answer:
The measure of ∠DFC is 90°
m∠AFB + m∠BFC + m∠DFC = 180°
m∠AFB + m∠BFC + 90° = 180°
Subtract 90° from both sides
m∠AFB + m∠BFC = 90°
∠AFB and ∠BFC are complementary angles.

∠AFB and ∠BFC are complementary angles.

Lesson 9.1 Practice A Geometry Answers Go Math Grade 7 Question 4.
Draw Conclusions Are ∠AFC and ∠BFC adjacent angles? Why or why not?
Answer:
Yes, they are, because they have a common side, a common vertex(corner point), and they don’t overlap.

Your Turn

Use the diagram.

Question 5.
Name a pair of supplementary angles.
Answer:
∠EGF and ∠FGB.

Question 6.
Name a pair of vertical angles.
Answer:
∠FGA and ∠DGC

Question 7.
Find the measure of ∠CGD. ______
Answer:
We see from the diagram that
m∠CGD + m∠DGE + m∠EGF = 180°.
∠DGE and ∠AGB are vertically opposite angles, so their measures are equal.
m∠DGE + m∠AGB = 90°.
m∠CGD + m∠DGE + m∠EGF = 180° Substitute 90° for m∠DGE, and 35° for m∠EGF.
m∠CGD + 90° + 35° = 180°
m∠CGD + 125° = 180° Subtract 125° from both sides.
m∠CGD = 180° – 125°
m∠CGD = 55°

m∠CGD = 55°

Example 2
A. Find the measure of ∠EHF.

Since m∠EHF = 2x, then m∠EHE = 132°

B. Find the measure of ∠ZXY.

Your Turn

Go Math Grade 7 Answer Key Pdf Lesson 9.1 Question 8.
Find the measure of ∠JML.

Answer:
From the diagram we see
m∠JML + m∠LMN = 180° Substitute 3x for m∠JML, and 54° for m∠LMN.
3x + 54° = 180° Subtract 54° from both sides.
3x = 180° – 54°
3x = 126° Divide both sides by 3

m∠JML = 3x Substitute 41° for x.
m∠JML = 3. 41°
m∠JML = 126°

Question 9.
Critique Reasoning Cory says that to find m∠JML above you can stop when you get to the solution step 3x = 126°. Explain why this works.
Answer:
Both angle ∠JML and ∠LMN are on same straight line and the angle of straight line is 180° So sum of the ∠JML and ∠LMN will be 180° this means both angle are supplementary to each other.
∠JML + ∠LMN = 180°
3x + 54° = 180°
3x + 54 – 54 = 180 – 54
3x = 126°
x = 42°
Hence, 3x = 126° works both angle 3x and 54° are supplementary to each other. So the sum of both the angle will be equal to the 180°
∠JML and ∠LMN are supplementary to each other.

Example 3
The front of the top story of a house is shaped like an isosceles triangle. The measure of the angle at the top of the triangle is 70. Find the measure of each of the base angles.

Your Turn

Use the diagram.

Lesson 9.1 Answer Key 7th Grade Angle Relationships Question 10.
Find the value of x. ____________
Answer:
The sum of the measures of angles in a triangle is
∠C + ∠A + ∠B = 180° Substitute 800° for ∠C, x for ∠A, and 3x for ∠B.
80 + x + 3x = 180°
4x + 80° = 180° Subtract 80° from both sides.
4x = 180° – 80°
4x = 100° Divide both sides by 4.

x = 25°

Question 11.
Find the measures of
∠A and ∠B. _____________
Answer:
x = 25°
m∠A = x Substitute 25° for x
m∠A = 25°
m∠B = 3x Substitute 25° for x
m∠B = 3 . 25°
m∠B = 75°
m∠A = 25°, m∠B = 75°

Texas Go Math Grade 7 Lesson 9.1 Guided Practice Answer Key  

For Exercises 1-2, use the figure. (Example 1)

Question 1.
Vocabulary The sum of the measures of ∠UWV and ∠UWZ is 90°, so ∠UWV and ∠UWZ are ____ angles.

Answer:
Complementary angles

Question 2.
Vocabulary ∠UWV and ∠VWX share a vertex and one side. They do not overlap, so ∠UWV and ∠VWX are
____ angles.
Answer:
Adjacent angles

For Exercises 3-4, use the figure.

Angle Relationship Texas Go Math Grade 7 Answer Key Question 3.
∠AGB and ∠DGE are ________________ angles, so m∠DGE= . (Example 1)
Answer:
– Vertical angles.
-m∠DGE = 30°.

Question 4.
Find the measure of ∠EGF. (Example 2)
m∠CGD + m∠DGE + m∠EGF = 180°
____ + ____ + ____ = 180°
____ + 2x = 180°
2x = _____
m∠EGF = 2x = ___
Answer:
50° + 30° + 2x = 180°
80° + 2x = 180°
2x = 100°
m∠EGF = 2x = 100°
m∠EGF = 100°

Question 5.
Find the measures of ∠A and ∠B. (Example 3)
m∠A + m∠B + m∠C = 180
____ + _______ + ____ = 180
2x + __ = 180

2x = _____
x = __, so, m∠A = ___.
x + 10 = _____, so m∠B = ___.
Answer:
x + x + 10° + 40° = 180°
2x + 50° = 180°
2x = 130°
x = 65°, so m∠A = 65°
x + 10° = 75°, so m∠A = 75°
m∠A = 65° m∠B = 75°

Essential Question Check-In

Go Math Lesson 9.1 Angle Relationships Answers Question 6.
Suppose that you know that ∠T and ∠S are supplementary and that m∠T = 3 . (m∠S). How can you find m∠T?
Answer:
∠T and ∠S are supplementary, so they form an angle of 180°
m∠T = 180°
m∠T + m∠S = 180°
3(m∠S) + m∠S = 180°
4 . m∠S = 180° Divide both sides by 4.

m∠T = 3 . 45° = 125°

Texas Go Math Grade 7 Lesson 9.1 Independent Practice Answer Key  

For Exercises 7-11, use the figure.

Question 7.
Name a pair of adjacent angles. Explain why they are adjacent.
Answer:
∠RUS and ∠SUT because they have a common side and a common vertex.

Question 8.
Name a pair of acute vertical angles.
Answer:
∠RUS and ∠QUP

Question 9.
Name a pair of supplementary angles.
Answer:
∠TUP and ∠PUN

Lesson 9.1 Understand Angle Relationships Answer Key Question 10.
Justify Reasoning Find m∠QUR. Justify your answer.
Answer:
∠QUR and ∠RUS are supplementary angles, so
m∠QUR + m∠RUS = 180° Substitute 41° for both sides
m∠QUR + 41° = 180° Subtract 41° from both sides.
m∠QUR = 180° – 41°
m∠QUR = 139°

Question 11.
Draw Conclusions Which is greater, m∠TUR or m∠RUQ? Explain.
Answer:
From the diagram we see
m∠TUR = m∠TUS + m∠RUS
m∠RUQ = m∠NUQ + m∠RUN
∠TUS and ∠NUQ are vertical opposite angles, so their measures are equal.
m∠TUS = m∠NUQ = 90°
∠RUN and ∠RUS are complementary angles, so they form an angle of 90°
Since m∠RUS = 41°, we have
m∠RUS + m∠RUN = 90° Substitute 41° for ∠RUS.
41° + m∠RUN = 90° Subtract 41° from both sides
m∠RUN = 90° – 41°
m∠RUN = 49°
Put these reasults into the following:
m∠TUR = m∠TUS + m∠RUS
m∠TUR = 90° + 41°
m∠TUR = 131°
m∠RUQ = m∠NUQ + m∠RUN
m∠RUQ = 90° + 49°
m∠RUQ = 139°

m∠TUR < m∠RUQ

m∠RUQ is greater than m∠TUR

Solve for each indicated angle measure or variable in the figure.

Question 12.
x ____________
Answer:
∠KMI and ∠HMG are vertical opposite angles, so they have the same measure.
4x = 84° Divide both sides by 4.

x = 21°

Go Math Answer Key Angle Pair Relationships Worksheet Pdf Question 13.
m∠KMH _______
Answer:
m∠KMH and m∠KMI are supplementary angles
m∠KMH – m∠KMI = 180° Substitute 84° for ∠KMI.
m∠KMH + 84° = 180° Subtract 84° from both sides
m∠KMH = 180° – 84°
m∠KMH = 96°

Solve for each indicated angle measure or variable in the figure.

Question 14.
m∠CBE ______
Answer:
m∠CBE and m∠EBF are supplementary angles.
m∠CBE + m∠EBF = 180° Substitute 62° for ∠EBF.
m∠CBE + 62° = 180° Subtract 62° from both sides.
m∠CBE = 180° – 62°
m∠CBE = 118°

Question 15.
m∠ABF _____
Answer:
m∠ABF and m∠FBE are complementary angles.
m∠ABF + m∠FBE = 90° Substitute 62° for ∠FBE.
m∠ABF + 62° = 90° Subtract 62° from both sides.
m∠ABF = 90° – 62°
m∠ABF = 28°

Question 16.
m∠CBA _____
Answer:
m∠CBA and m∠ABF are supplementary angles.
m∠CBA + m∠ABF = 180° Substitute 28° for ∠ABF.
m∠CBA + 28° = 180° Subtract 28° from both sides.
m∠CBA = 180° – 28°
m∠CBA = 152°

Solve for each indicated angle measure or variable in the figure.

Question 17.
x ___________
Answer:
The sum of the measures of angles in a triangle is 180°, so we have
∠P + ∠Q + ∠R = 180° Substitute 25° for ∠P, 3x for ∠Q and 20° for ∠R.
25° + 3 + 20° = 180°
3x + 45° = 180° Subtract 45° from both sides.
3x = 180° – 45°
3x = 135° Divide both sides by 3.

Texas Go Math Grade 7 Solutions Lesson 9.1 Answer Key Question 18.
m∠Q _______
Answer:
x = 45°
m∠Q = 3x Supstitute 45° for z.
= 3 . 45°
= 135°

Texas Go Math Grade 7 Lesson 9.1 H.O.T. Focus on Higher Order Thinking Answer Key  

Let ∆ABC be a right triangle with m∠C = 90°.

Question 19.
Critical Thinking An equilateral triangle has three congruent sides and three congruent angles. Can ∆ABC be an equilateral triangle? Explain your reasoning.
Answer:
In given triangle ABC one angle is m∠C = 90° so the triangle ABC cannot be a equilateral triangle. Because in
equilateral triangle all three sides and angle are congruent to each other. So each angle of equilateral triangle is
60° because sum of all the interior angle of triangle is 180°. Hence each angle of equilateral triangle is \(\frac{180}{3}\) = 60°.
Thus the given triangle ABC cannot be a equilateral triangle.

Given triangle ABC cannot be a equilateral triangle.

Question 20.
Counterexample An isosceles triangle has two congruent sides, and the angles opposite those sides are congruent. River says that right triangle ABC cannot be an isosceles triangle. Give a counterexample to show that his statement is incorrect.
Answer:
There is a special right triangle that can be considered as an isosceles triangle. This is the 45° right triangle wherein, the two sides have 45° angles. The measure of the sides are congruent and the angles are also congruent.

The 45° right triangle is an isosceles triangle.

Angle Relationships Practice Answer Key Lesson 9.1 Go Math 7th Grade Question 21.
Make a Conjecture In a scalene triangle, no two sides have the same length, and no two angles have the same measure. Do you think a right triangle can be a scalene triangle? Explain your reasoning.
Answer:
Any right triangle can be a scalene triangle except for the 45°-45° right triangle. This is because the two angles of
a right triangle can be of any measure for as long as the total of the two angles is 90°. The measurement of the
sides of the right triangle will also vary.

Yes

Question 22.
Represent Real-World Problems The railroad tracks meet the road as shown. The town will allow a parking lot at angle J if the measure of angle J is greater than 38°. Can a parking lot be built at an angle? Why or why not?

Answer:
All the upper angles are on a straight line so they must be supplementary to each other, thus their sum will be equal to 180°
50° + 90° + J = 180°
140° + J = 180°
J = 180 – 140
J = 40°
It is given in the problem that the minimum required angle J for the parking lot should be greater than 38° and we have found that angle J is 40°. So a parking lot can be built there.
Hence, the parking lot can be built there because angle J is 40°

Question 23.
Analyze Relationships In triangle XYZ, m∠X = 30°, and all the angles have measures that are whole numbers. Angle Y is an obtuse angle. What is the greatest possible measure that angle Z can have? Explain your answer.
Answer:
In triangle XYZ, m∠X = 30° so the sum of remaining angles ∠Y and ∠Z will be 180 – 30 = 150°. Since ∠Y is
an obtuse angle and all angle measures the whole number, so minimum value of ∠Y will be 91°. And in equation
∠Y + ∠Z = 180° if ∠Y is minimum then ∠Z must be maximum. Thus the maximum value of ∠Y = 150 – 91 = 59°.
Hence, the greatest possible measure of angle Z is 59°.