Are you struggling to solve the trigonometric ratios and functions in homework and assignments? Then, check out this Big Ideas Math Algebra 2 Answers Chapter 9 Trigonometric Ratios and Functions and solve your issues within seconds. The perfect way to learn the concepts of ch 9 is by practicing the questions given in the BIM Math Book Algebra 2 Chapter 9 Trigonometric Ratios and Functions Solution Key. Look no further and start your preparation by using this quick & ultimate guide. Constant practice can make you succeed in your math learning journey.

Big Ideas Math Book Algebra 2 Answer Key Chapter 9 Trigonometric Ratios and Functions

Enhance your math skills by taking the help of our provided Big Ideas Math Book Algebra 2 Ch 9 Trigonometric Ratios and Functions Solution key. After learning the concepts of the Algebra 2 9th chapters from the BIM Textbook Answers of Algebra 2, students can become math proficient. Access the Topicwise Big Ideas Math Book Algebra 2 Ch 9 Trigonometric Ratios and Functions Answers for free from the respective links presented below and prepare well for the exams.

• Trigonometric Ratios and Functions Maintaining Mathematical Proficiency – Page 459
• Trigonometric Ratios and Functions Mathematical Practices – Page 460
• Lesson 9.1 Right Triangle Trigonometry – Page(461-468)
• Right Triangle Trigonometry 9.1 Exercises – Page(466-468)
• Lesson 9.2 Angles and Radian Measure – Page(469-476)
• Angles and Radian Measure 9.2 Exercises – Page(474-476)
• Lesson 9.3 Trigonometric Functions of Any Angle – Page(477-484)
• Trigonometric Functions of Any Angle 9.3 Exercises – Page(482-484)
• Lesson 9.4 Graphing Sine and Cosine Functions – Page(485-494)
• Graphing Sine and Cosine Functions 9.4 Exercises – Page(491-494)
• Trigonometric Ratios and Functions Study Skills: Form a Final Exam Study Group – Page 495
• Trigonometric Ratios and Functions 9.1–9.4 Quiz – Page 496
• Lesson 9.5 Graphing Other Trigonometric Functions – Page(497-504)
• Graphing Other Trigonometric Functions 9.5 Exercises – Page(502-504)
• Lesson 9.6 Modeling with Trigonometric Functions – Page(505-512)
• Modeling with Trigonometric Functions 9.6 Exercises – Page(510-512)
• Lesson 9.7 Using Trigonometric Identities – Page(513-518)
• Using Trigonometric Identities 9.7 Exercises – Page(517-518)
• Lesson 9.8 Using Sum and Difference Formulas – Page(519-524)
• Using Sum and Difference Formulas 9.8 Exercises – Page(523-524)
• Trigonometric Ratios and Functions Performance Task: Lightening the Load – Page 525
• Trigonometric Ratios and Functions Chapter Review – Page(526-530)
• Trigonometric Ratios and Functions Chapter Test – Page 531
• Trigonometric Ratios and Functions Cumulative Assessment – Page(532 – 533)

Trigonometric Ratios and Functions Maintaining Mathematical Proficiency

Order the expressions by value from least to greatest.
Question 1.
log₂ x
∣4∣, 2 − 9∣, ∣6 + 4∣, − ∣7∣
Answer:

Question 2.
∣9 − 3∣, ∣0∣, ∣−4∣, \(\frac{|-5|}{|2|}\)
Answer:

Question 3.
∣−8³∣,∣−2 • 8 ∣, ∣9 − 1∣, ∣9∣ + ∣−2∣ − ∣1 ∣
Answer:

Question 4.
∣−4 + 20∣, −∣4²∣, ∣5∣−∣3 • 2 ∣, ∣−15∣
Answer:

Find the missing side length of the triangle.
Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.
ABSTRACT REASONING
The line segments connecting the points (x₁, y₁), (x₂, y₁), and (x₂, y₂) form a triangle. Is the triangle a right triangle? Justify your answer.
Answer:

Trigonometric Ratios and Functions Mathematical Practices

Mathematically proficient students reason quantitatively by creating valid representations of problems.

Monitoring Progress

Find the exact coordinates of the point (x, y) on the unit circle.
Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Lesson 9.1 Right Triangle Trigonometry

Essential Question How can you find a trigonometric function of an acute angle θ?
Consider one of the acute angles θ of a right triangle. Ratios of a right triangle’s side lengths are used to define the six trigonometric functions, as shown.

EXPLORATION 1

Trigonometric Functions of Special Angles
Work with a partner. Find the exact values of the sine, cosine, and tangent functions for the angles 30°, 45°, and 60° in the right triangles shown.

EXPLORATION 2

Exploring Trigonometric Identities
Work with a partner.

Use the definitions of the trigonometric functions to explain why each trigonometric identity is true.
a. sin θ = cos(90° − θ)
b. cos θ = sin(90° − θ)
c. sin θ =\(\frac{1}{\csc \theta}\)
d. tan θ = \(\frac{1}{\cot \theta}\)
Use the definitions of the trigonometric functions to complete each trigonometric identity.

Communicate Your Answer

Question 3.
How can you find a trigonometric function of an acute angle θ?
Answer:

Question 4.
Use a calculator to find the lengths x and y of the legs of the right triangle shown.

Answer:

Monitoring Progress

Evaluate the six trigonometric functions of the angle θ.
Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.
In a right triangle, θ is an acute angle and cos θ = \(\frac{7}{10}\). Evaluate the other five trigonometric functions of θ.
Answer:

Question 5.
Find the value of x for the right triangle shown.

Answer:

Solve △ABC using the diagram at the left and the given measurements.

Question 6.
B = 45°, c = 5
Answer:

Question 7.
A = 32°, b = 10
Answer:

Question 8.
A = 71°, c = 20
Answer:

Question 9.
B = 60°, a = 7
Answer:

Question 10.
In Example 5, find the distance between B and C.
Answer:

Question 11.
WHAT IF?
In Example 6, estimate the height of the parasailer above the boat when the angle of elevation is 38°.
Answer:

Right Triangle Trigonometry 9.1 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
In a right triangle, the two trigonometric functions of θ that are defined using the lengths of the hypotenuse and the side adjacent to θ are __________ and __________.
Answer:

Question 2.
VOCABULARY
Compare an angle of elevation to an angle of depression.
Answer:

Question 3.
WRITING
Explain what it means to solve a right triangle.
Answer:

Question 4.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, evaluate the six trigonometric functions of the angle θ.
Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.
REASONING
Let θ be an acute angle of a right triangle. Use the two trigonometric functions tan θ = \(\frac{4}{9}\) and sec θ = \(\frac{\sqrt{97}}{9}\) to sketch and label the right triangle. Then evaluate the other four trigonometric functions of θ.
Answer:

Question 12.
ANALYZING RELATIONSHIPS
Evaluate the six trigonometric functions of the 90° − θ angle in Exercises 5–10. Describe the relationships you notice.
Answer:

In Exercises 13–18, let θ be an acute angle of a right triangle. Evaluate the other five trigonometric functions of θ.
Question 13.
sin θ = \(\frac{7}{11}\)
Answer:

Question 14.
cos θ = \(\frac{5}{12}\)
Answer:

Question 15.
tan θ = \(\frac{7}{6}\)
Answer:

Question 16.
csc θ = \(\frac{15}{8}\)
Answer:

Question 17.
sec θ = \(\frac{14}{9}\)
Answer:

Question 18.
cot θ = \(\frac{16}{11}\)
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in finding sin θ of the triangle below.

Answer:

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding csc θ, given that θ is an acute angle of a right triangle and cos θ = \(\frac{7}{11}\).

Answer:

In Exercises 21–26, find the value of x for the right triangle.
Question 21.

Answer:

Question 22.

Answer:

Question 23.

Answer:

Question 24.

Answer:

Question 25.

Answer:

Question 26.

Answer:

USING TOOLS In Exercises 27–32, evaluate the trigonometric function using a calculator. Round your answer to four decimal places.
Question 27.
cos 14°
Answer:

Question 28.
tan 31°
Answer:

Question 29.
csc 59°
Answer:

Question 30.
sin 23°
Answer:

Question 31.
cot 6°
Answer:

Question 32.
sec 11°
Answer:

In Exercises 33–40, solve △ABC using the diagram and the given measurements.

Question 33.
B = 36°, a = 23
Answer:

Question 34.
A = 27°, b = 9
Answer:

Question 35.
A = 55°, a = 17
Answer:

Question 36.
B = 16°, b = 14
Answer:

Question 37.
A = 43°, b = 31
Answer:

Question 38.
B = 31°, a = 23
Answer:

Question 39.
B = 72°, c = 12.8
Answer:

Question 40.
A = 64°, a = 7.4
Answer:

Question 41.
MODELING WITH MATHEMATICS
To measure the width of a river, you plant a stake on one side of the river, directly across from a boulder. You then walk 100 meters to the right of the stake and measure a 79° angle between the stake and the boulder. What is the width w of the river?

Answer:

Question 42.
MODELING WITH MATHEMATICS
Katoomba Scenic Railway in Australia is the steepest railway in the world. The railway makes an angle of about 52° with the ground. The railway extends horizontally about 458 feet. What is the height of the railway?
Answer:

Question 43.
MODELING WITH MATHEMATICS
A person whose eye level is 1.5 meters above the ground is standing 75 meters from the base of the Jin Mao Building in Shanghai, China. The person estimates the angle of elevation to the top of the building is about 80°. What is the approximate height of the building?
Answer:

Question 44.
MODELING WITH MATHEMATICS
The Duquesne Incline in Pittsburgh, Pennsylvania, has an angle of elevation of 30°. The track has a length of about 800 feet. Find the height of the incline.
Answer:

Question 45.
MODELING WITH MATHEMATICS
You are standing on the Grand View Terrace viewing platform at Mount Rushmore, 1000 feet from the base of the monument.

a. You look up at the top of Mount Rushmore at an angle of 24°. How high is the top of the monument from where you are standing? Assume your eye level is 5.5 feet above the platform.
b. The elevation of the Grand View Terrace is 5280 feet. Use your answer in part (a) to find the elevation of the top of Mount Rushmore.
Answer:

Question 46.
WRITING
Write a real-life problem that can be solved using a right triangle. Then solve your problem.
Answer:

Question 47.
MATHEMATICAL CONNECTIONS
The Tropic of Cancer is the circle of latitude farthest north of the equator where the Sun can appear directly overhead. It lies 23.5° north of the equator, as shown.

a. Find the circumference of the Tropic of Cancer using 3960 miles as the approximate radius of Earth.
b. What is the distance between two points on the Tropic of Cancer that lie directly across from each other?
Answer:

Question 48.
HOW DO YOU SEE IT?
Use the figure to answer each question.

a. Which side is adjacent to θ?
b. Which side is opposite of θ?
c. Does cos θ = sin(90° − θ)? Explain.
Answer:

Question 49.
PROBLEM SOLVING
A passenger in an airplane sees two towns directly to the left of the plane.

a. What is the distance d from the airplane to the first town?
b. What is the horizontal distance x from the airplane to the first town?
c. What is the distance y between the two towns? Explain the process you used to find your answer.
Answer:

Question 50.
PROBLEM SOLVING
You measure the angle of elevation from the ground to the top of a building as 32°. When you move 50 meters closer to the building, the angle of elevation is 53°. What is the height of the building?
Answer:

Question 51.
MAKING AN ARGUMENT
Your friend claims it is possible to draw a right triangle so the values of the cosine function of the acute angles are equal. Is your friend correct? Explain your reasoning.
Answer:

Question 52.
THOUGHT PROVOKING
Consider a semicircle with a radius of 1 unit, as shown below. Write the values of the six trigonometric functions of the angle θ. Explain your reasoning.

Answer:

Question 53.
CRITICAL THINKING
A procedure for approximating π based on the work of Archimedes is to inscribe a regular hexagon in a circle.

a. Use the diagram to solve for x. What is the perimeter of the hexagon?
b. Show that a regular n-sided polygon inscribed in acircle of radius 1 has a perimeter of 2n • sin (\(\frac{180}{n}\))°.
c. Use the result from part (b) to find an expression in terms of n that approximates π. Then evaluate the expression when n= 50.
Answer:

Maintaining Mathematical Proficiency.

Perform the indicated conversion.
Question 54.
5 years to seconds
Answer:

Question 55.
12 pints to gallons
Answer:

Question 56.
5.6 meters to millimeters
Answer:

Find the circumference and area of the circle with the given radius or diameter.
Question 57.
r = 6 centimeters
Answer:

Question 58.
r = 11 inches
Answer:

Question 59.
d = 14 feet
Answer:

Lesson 9.2 Angles and Radian Measure

Essential Question How can you find the measure of an angle in radians?
Let the vertex of an angle be at the origin, with one side of the angle on the positive x-axis. The radian measure of the angle is a measure of the intercepted arc length on a circle of radius 1. To convert between degree and radian measure, use the fact that \(\frac{\pi \text { radians }}{180^{\circ}}\) = 1.

EXPLORATION 1

Writing Radian Measures of Angles
Work with a partner. Write the radian measure of each angle with the given degree measure. Explain your reasoning.

EXPLORATION 2

Writing Degree Measures of Angles
Work with a partner. Write the degree measure of each angle with the given radian measure. Explain your reasoning.

Communicate Your Answer

Question 3.
How can you find the measure of an angle in radians?
Answer:

Question 4.
The figure shows an angle whose measure is 30 radians. What is the measure of the angle in degrees? How many times greater is 30 radians than 30 degrees? Justify your answers.

Answer:

Monitoring Progress

Draw an angle with the given measure in standard position.
Question 1.
65°
Answer:

Question 2.
300°
Answer:

Question 3.
−120°
Answer:

Question 4.
−450°
Answer:

Find one positive angle and one negative angle that are coterminal with the given angle.
Question 5.
80°
Answer:

Question 6.
230°
Answer:

Question 7.
740°
Answer:

Question 8.
−135°
Answer:

Convert the degree measure to radians or the radian measure to degrees.
Question 9.
135°
Answer:

Question 10.
−40°
Answer:

Question 11.
\(\frac{5 \pi}{4}\)
Answer:

Question 12.
−6.28
Answer:

Question 13.
WHAT IF?
In Example 4, the outfield fence is 220 feet from home plate. Estimate the length of the outfield fence and the area of the field.
Answer:

Angles and Radian Measure 9.2 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
An angle is in standard position when its vertex is at the __________ and its __________ lies on the positive x-axis.
Answer:

Question 2.
WRITING
Explain how the sign of an angle measure determines its direction of rotation.
Answer:

Question 3.
VOCABULARY
In your own words, define a radian.
Answer:

Question 4.
WHICH ONE DOESN’T BELONG?
Which angle does not belong with the other three? Explain your reasoning.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, draw an angle with the given measure in standard position.
Question 5.
110°
Answer:

Question 6.
450°
Answer:

Question 7.
−900°
Answer:

Question 8.
−10°
Answer:

In Exercises 9–12, find one positive angle and one negative angle that are coterminal with the given angle.
Question 9.
70°
Answer:

Question 10.
255°
Answer:

Question 11.
−125°
Answer:

Question 12.
−800°
Answer:

In Exercises 13–20, convert the degree measure to radians or the radian measure to degrees.
Question 13.
40°
Answer:

Question 14.
315°
Answer:

Question 15.
−260°
Answer:

Question 16.
−500°
Answer:

Question 17.
\(\frac{\pi}{9}\)
Answer:

Question 18.
\(\frac{3 \pi}{4}\)
Answer:

Question 19.
−5
Answer:

Question 20.
12
Answer:

Question 21.
WRITING
The terminal side of an angle in standard position rotates one-sixth of a revolution counterclockwise from the positive x-axis. Describe how to find the measure of the angle in both degree and radian measures.
Answer:

Question 22.
OPEN-ENDED
Using radian measure, give one positive angle and one negative angle that are coterminal with the angle shown. Justify your answers.

Answer:

ANALYZING RELATIONSHIPS In Exercises 23–26, match the angle measure with the angle.
Question 23.
600°
Answer:

Question 24.
\(-\frac{9 \pi}{4}\)
Answer:

Question 25.
\(\frac{5 \pi}{6}\)
Answer:

Question 26.
−240°
Answer:

Question 27.
MODELING WITH MATHEMATICS
The observation deck of a building forms a sector with the dimensions shown. Find the length of the safety rail and the area of the deck.

Answer:

Question 28.
MODELING WITH MATHEMATICS
In the men’s shot put event at the 2012 Summer Olympic Games, the length of the winning shot was 21.89 meters. A shot put must land within a sector having a central angle of 34.92° to be considered fair.

a. The officials draw an arc across the fair landing area, marking the farthest throw. Find the length of the arc.
b. All fair throws in the 2012 Olympics landed within a sector bounded by the arc in part (a). What is the area of this sector?
Answer:

Question 29.
ERROR ANALYSIS
Describe and correct the error in converting the degree measure to radians.

Answer:

Question 30.
ERROR ANALYSIS
Describe and correct the error in finding the area of a sector with a radius of 6 centimeters and a central angle of 40°.

Answer:

Question 31.
PROBLEM SOLVING
When a CD player reads information from the outer edge of a CD, the CD spins about 200 revolutions per minute. At that speed, through what angle does a point on the CD spin in oneminute? Give your answer in both degree and radian measures.
Answer:

Question 32.
PROBLEM SOLVING
You work every Saturday from 9:00 A.M. to 5:00 P.M. Draw a diagram that shows the rotation completed by the hour hand of a clock during this time. Find the measure of the angle generated by the hour hand in both degrees and radians. Compare this angle with the angle generated by the minute hand from 9:00 A.M. to 5:00 P.M.
Answer:

USING TOOLS In Exercises 33–38, use a calculator to evaluate the trigonometric function.
Question 33.
cos \(\frac{4 \pi}{3}\)
Answer:

Question 34.
sin \(\frac{7 \pi}{8}\)
Answer:

Question 35.
csc \(\frac{10 \pi}{11}\)
Answer:

Question 36.
cot (− \(\frac{6 \pi}{5}\))
Answer:

Question 37.
cot(−14)
Answer:

Question 38.
cos 6
Answer:

Question 39.
MODELING WITH MATHEMATICS
The rear windshield wiper of a car rotates 120°, as shown. Find the area cleared by the wiper.

Answer:

Question 40.
MODELING WITH MATHEMATICS
A scientist performed an experiment to study the effects of gravitational force on humans. In order for humans to experience twice Earth’s gravity, they were placed in a centrifuge 58 feet long and spun at a rate of about 15 revolutions per minute.

a. Through how many radians did the people rotate each second?
b. Find the length of the arc through which the people rotated each second.
Answer:

Question 41.
REASONING
In astronomy, the terminator is the day-night line on a planet that divides the planet into daytime and nighttime regions. The terminator moves across the surface of a planet as the planet rotates. It takes about 4 hours for Earth’s terminator to move across the continental United States. Through what angle has Earth rotated during this time? Give your answer in both degree and radian measures.

Answer:

Question 42.
HOW DO YOU SEE IT?
Use the graph to find the measure of θ. Explain your reasoning.

Answer:

Question 43.
MODELING WITH MATHEMATICS
A dartboard is divided into 20 sectors. Each sector is worth a point value from 1 to 20 and has shaded regions that double or triple this value. A sector is shown below. Find the areas of the entire sector, the double region, and the triple region.

Answer:

Question 44.
THOUGHT PROVOKING
π is an irrational number, which means that it cannot be written as the ratio of two whole numbers. π can, however, be written exactly as a continued fraction, as follows.

Answer:

Question 45.
MAKING AN ARGUMENT
Your friend claims that when the arc length of a sector equals the radius, the area can be given by A = \(\frac{s^{2}}{2}\). Is your friend correct? Explain.
Answer:

Question 46.
PROBLEM SOLVING
A spiral staircase has 15 steps. Each step is a sector with a radius of 42 inches and a central angle of \(\frac{\pi}{8}\).
a. What is the length of the arc formed by the outer edge of a step?
b. Through what angle would you rotate by climbing the stairs?
c. How many square inches of carpeting would you need to cover the 15 steps?
Answer:

Question 47.
MULTIPLE REPRESENTATIONS
There are 60 minutes in 1 degree of arc, and 60 seconds in 1 minute of arc. The notation 50° 30′ 10″ represents an angle with a measure of 50 degrees, 30 minutes, and 10 seconds.
a. Write the angle measure 70.55° using the notation above.
b. Write the angle measure 110° 45′ 30″ to the nearest hundredth of a degree. Justify your answer.
Answer:

Maintaining Mathematical Proficiency

Find the distance between the two points.
Question 48.
(1, 4), (3, 6)
Answer:

Question 49.
(−7, −13), (10, 8)
Answer:

Question 50.
(−3, 9), (−3, 16)
Answer:

Question 51.
(2, 12), (8, −5)
Answer:

Question 52.
(−14, −22), (−20, −32)
Answer:

Question 53.
(4, 16), (−1, 34)
Answer:

Lesson 9.3 Trigonometric Functions of Any Angle

Essential Question How can you use the unit circle to define the trigonometric functions of any angle?
Let θ be an angle in standard position with (x, y) a point on the terminal side of θ and r = \(\sqrt{x^{2}+y^{2}}\) ≠ 0. The six trigonometric functions of θ are defined as shown.

EXPLORATION 1

Writing Trigonometric Functions
Work with a partner. Find the sine, cosine, and tangent of the angle θ in standard position whose terminal side intersects the unit circle at the point (x, y) shown.

Communicate Your Answer

Question 2.
How can you use the unit circle to define the trigonometric functions of any angle?
Answer:

Question 3.
For which angles are each function undefined? Explain your reasoning.

a. tangent
b. cotangent
c. secant
d. cosecant
Answer:

Monitoring Progress

Evaluate the six trigonometric functions of θ.
Question 1.

Answer:

Question 2.

Answer:

Question 3.

Answer:

Question 4.
Use the unit circle to evaluate the six trigonometric functions of θ = 180º.
Answer:

Sketch the angle. Then find its reference angle.
Question 5.
210°
Answer:

Question 6.
−260°
Answer:

Question 7.
\(\frac{-7 \pi}{9}\)
Answer:

Question 8.
\(\frac{15 \pi}{4}\)
Answer:

Evaluate the function without using a calculator.
Question 9.
cos(−210º)
Answer:

Question 10.
sec \(\frac{11 \pi}{4}\)
Answer:

Question 11.
Use the model given in Example 5 to estimate the horizontal distance traveled by a track and field long jumper who jumps at an angle of 20° and with an initial speed of 27 feet per second.
Answer:

Trigonometric Functions of Any Angle 9.3 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A(n) ___________ is an angle in standard position whose terminal side lies on an axis.
Answer:

Question 2.
WRITING
Given an angle θ in standard position with its terminal side in Quadrant III, explain how you can use a reference angle to find cos θ.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, evaluate the six trigonometric functions of θ.
Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

In Exercises 9–14, use the unit circle to evaluate the six trigonometric functions of θ.
Question 9.
θ = 0°
Answer:

Question 10.
θ = 540°
Answer:

Question 11.
θ = \(\frac{\pi}{2}\)
Answer:

Question 12.
θ = \(\frac{7 \pi}{2}\)
Answer:

Question 13.
θ = −270°
Answer:

Question 14.
θ = −2π
Answer:

In Exercises 15–22, sketch the angle. Then find its reference angle.
Question 15.
−100°
Answer:

Question 16.
150°
Answer:

Question 17.
320°
Answer:

Question 18.
−370°
Answer:

Question 19.
\(\frac{15 \pi}{4}\)
Answer:

Question 20.
\(\frac{8 \pi}{3}\)
Answer:

Question 21.
−\(\frac{5 \pi}{6}\)
Answer:

Question 22.
−\(\frac{13 \pi}{6}\)
Answer:

Question 23.
ERROR ANALYSIS
Let (−3, 2) be a point on the terminal side of an angle θ in standard position. Describe and correct the error in finding tan θ.

Answer:

Question 24.
ERROR ANALYSIS
Describe and correct the error in finding a reference angle θ′ for θ = 650°.

Answer:

In Exercises 25–32, evaluate the function without using a calculator.
Question 25.
sec 135°
Answer:

Question 26.
tan 240°
Answer:

Question 27.
sin(−150°)
Answer:

Question 28.
csc(−420°)
Answer:

Question 29.
tan (−\(\frac{3 \pi}{4}\))
Answer:

Question 30.
cot (\(\frac{-8 \pi}{3}\))
Answer:

Question 31.
cos \(\frac{7 \pi}{4}\)
Answer:

Question 32.
sec \(\frac{11 \pi}{6}\)
Answer:

In Exercises 33–36, use the model for horizontal distance given in Example 5.
Question 33.
You kick a football at an angle of 60° with an initial speed of 49 feet per second. Estimate the horizontal distance traveled by the football.
Answer:

Question 34.
The “frogbot” is a robot designed for exploring rough terrain on other planets. It can jump at a 45° angle with an initial speed of 14 feet per second. Estimate the horizontal distance the frogbot can jump on Earth.

Answer:

Question 35.
At what speed must the in-line skater launch himself off the ramp in order to land on the other side of the ramp?

Answer:

Question 36.
To win a javelin throwing competition, your last throw must travel a horizontal distance of at least 100 feet. You release the javelin at a 40° angle with an initial speed of 71 feet per second. Do you win the competition? Justify your answer.
Answer:

Question 37.
MODELING WITH MATHEMATICS
A rock climber is using a rock climbing treadmill that is 10 feet long. The climber begins by lying horizontally on the treadmill, which is then rotated about its midpoint by 110° so that the rock climber is climbing toward the top. If the midpoint of the treadmill is 6 feet above the ground, how high above the ground is the top of the treadmill?

Answer:

Question 38.
REASONING
A Ferris wheel has a radius of 75 feet. You board a car at the bottom of the Ferris wheel, which is 10 feet above the ground, and rotate 255°counterclockwise before the ride temporarily stops. How high above the ground are you when the ride stops? If the radius of the Ferris wheel is doubled, is your height above the ground doubled? Explain your reasoning.
Answer:

Question 39.
DRAWING CONCLUSIONS
A sprinkler at ground level is used to water a garden. The water leaving the sprinkler has an initial speed of 25 feet per second.
a. Use the model for horizontal distance given in Example 5 to complete the table.

b. Which value of θ appears to maximize the horizontal distance traveled by the water? Use the model for horizontal distance and the unit circle to explain why your answer makes sense.
c. Compare the horizontal distance traveled by the water when θ = (45 − k)° with the distance when θ = (45 + k)°, for 0 < k < 45.
Answer:

Question 40.
MODELING WITH MATHEMATICS
Your school’s marching band is performing at halftime during a football game. In the last formation, the band members form a circle 100 feet wide in the center of the field. You start at a point on the circle 100 feet from the goal line, march 300° around the circle, and then walk toward the goal line to exit the field. How far from the goal line are you at the point where you leave the circle?

Answer:

Question 41.
ANALYZING RELATIONSHIPS
Use symmetry and the given information to label the coordinates of the other points corresponding to special angles on the unit circle.

Answer:

Question 42.
THOUGHT PROVOKING
Use the interactive unit circle tool at BigIdeasMath.com to describe all values of θ for each situation.
a. sin θ > 0, cos θ < 0, and tan θ > 0
b. sin θ > 0, cos θ < 0, and tan θ < 0
Answer:

Question 43.
CRITICAL THINKING
Write tan θ as the ratio of two other trigonometric functions. Use this ratio to explain why tan 90° is undefined but cot 90° = 0.
Answer:

Question 44.
HOW DO YOU SEE IT?
Determine whether each of the six trigonometric functions of θ is positive, negative, or zero. Explain your reasoning.

Answer:

Question 45.
USING STRUCTURE
A line with slope m passes through the origin. An angle θ in standard position has a terminal side that coincides with the line. Use a trigonometric function to relate the slope of the line to the angle.
Answer:

Question 46.
MAKING AN ARGUMENT
Your friend claims that the only solution to the trigonometric equation tan θ = \(\sqrt{3}\) is θ= 60°. Is your friend correct? Explain your reasoning.
Answer:

Question 47.
PROBLEM SOLVING
When two atoms in a molecule are bonded to a common atom, chemists are interested in both the bond angle and the lengths of the bonds. An ozone molecule is made up of two oxygen atoms bonded to a third oxygen atom, as shown.

a. In the diagram, coordinates are given in picometers (pm). (Note: 1 pm = 10⁻¹² m) Find the coordinates (x, y) of the center of the oxygen atom in Quadrant II.
b. Find the distance d (in picometers) between the centers of the two unbonded oxygen atoms.
Answer:

Question 48.
MATHEMATICAL CONNECTIONS
The latitude of a point on Earth is the degree measure of the shortest arc from that point to the equator. For example, the latitude of point P in the diagram equals the degree measure of arc PE. At what latitude θ is the circumference of the circle of latitude at P half the distance around the equator?

Answer:

Maintaining Mathematical Proficiency

Find all real zeros of the polynomial function.
Question 49.
f (x) = x⁴ + 2x³ + x² + 8x − 12
Answer:

Question 50.
f(x) = x⁵ + 4x⁴ − 14x³ − 14x² − 15x− 18
Answer:

Graph the function.
Question 51.
f(x) = 2(x+ 3)² (x − 1)
Answer:

Question 52.
f(x) = \(\frac{1}{2}\) (x − 4)(x + 5)(x + 9)
Answer:

Question 53.
f(x) = x²(x + 1)³ (x − 2)
Answer:

Lesson 9.4 Graphing Sine and Cosine Functions

Essential Question What are the characteristics of the graphs of the sine and cosine functions?

EXPLORATION 1

Graphing the Sine FunctionWork with a partner.
a. Complete the table for y= sin x, where x is an angle measure in radians.

b. Plot the points (x, y) from part (a). Draw a smooth curve through the points to sketch the graph of y = sin x.

c. Use the graph to identify the x-intercepts, the x-values where the local maximums and minimums occur, and the intervals for which the function is increasing or decreasing over −2π ≤ x ≤ 2π. Is the sine function even, odd, or neither?

EXPLORATION 2

Graphing the Cosine Function
Work with a partner.
a. Complete a table for y= cos x using the same values of x as those used in Exploration 1.
b. Plot the points (x, y) from part (a) and sketch the graph of y= cos x.
c. Use the graph to identify the x-intercepts, the x-values where the local maximums and minimums occur, and the intervals for which the function is increasing or decreasing over −2π ≤ x ≤ 2π. Is the cosine function even, odd, or neither?

Communicate Your Answer

Question 3.
What are the characteristics of the graphs of the sine and cosine functions?
Answer:

Question 4.
Describe the end behavior of the graph of y = sin x

Answer:

Monitoring Progress

Identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of its parent function.
Question 1.
g(x) = \(\frac{1}{4}\)sin x
Answer:

Question 2.
g(x) = cos 2x
Answer:

Question 3.
g(x) = 2 sin πx
Answer:

Question 4.
g(x) = \(\frac{1}{3}\) cos \(\frac{1}{2}\)x
Answer:

Graph the function.
Question 5.
g(x) = cos x+ 4
Answer:

Question 6.
g(x) = \(\frac{1}{2}\)sin (x − \(\left.\frac{\pi}{2}\right\))
Answer:

Question 7.
g(x) = sin(x + π) − 1
Answer:

Graph the function.
Question 8.
g(x) = −cos (x + \(\left.\frac{\pi}{2}\right\))
Answer:

Question 9.
g(x) = −3 sin \(\frac{1}{2}\)x + 2
Answer:

Question 10.
g(x) = −2 cos 4x − 1
Answer:

Graphing Sine and Cosine Functions 9.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The shortest repeating portion of the graph of a periodic function is called a(n) _________.
Answer:

Question 2.
WRITING
Compare the amplitudes and periods of the functions y = \(\frac{1}{2}\)cos x and y = 3 cos 2x.
Answer:

Question 3.
VOCABULARY
What is a phase shift? Give an example of a sine function that has a phase shift.
Answer:

Question 4.
VOCABULARY
What is the midline of the graph of the function y = 2 sin 3(x + 1) − 2?
Answer:

Monitoring Progress and Modeling with Mathematics

USING STRUCTURE In Exercises 5–8, determine whether the graph represents a periodic function. If so, identify the period.
Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Question 8.

Answer:

In Exercises 9–12, identify the amplitude and period of the graph of the function.
Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Question 12.

Answer:

In Exercises 13–20, identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of its parent function.
Question 13.
g(x) = 3 sin x
Answer:

Question 14.
g(x) = 2 sin x
Answer:

Question 15.
g(x) = cos 3x
Answer:

Question 16.
g(x) = cos 4x
Answer:

Question 17.
g(x) = sin 2πx
Answer:

Question 18.
g(x) = 3 sin 2x
Answer:

Question 19.
g(x) = \(\frac{1}{3}\)cos 4x
Answer:

Question 20.
g(x) = \(\frac{1}{2}\)cos 4πx
Answer:

Question 21.
ANALYZING EQUATIONS
Which functions have an amplitude of 4 and a period of 2?
A. y = 4 cos 2x
B. y = −4 sin πx
C. y = 2 sin 4x
D. y = 4 cos πx
Answer:

Question 22.
WRITING EQUATIONS
Write an equation of the form y = a sin bx, where a > 0 and b > 0, so that the graph has the given amplitude and period.
a. amplitude: 1
period: 5
b. amplitude: 10
period: 4
c. amplitude: 2
period: 2π
d. amplitude: \(\frac{1}{2}\)
period: 3π
Answer:

Question 23.
MODELING WITH MATHEMATICS
The motion of a pendulum can be modeled by the function d = 4 cos 8πt, where d is the horizontal displacement (in inches) of the pendulum relative to its position at rest and t is the time (in seconds). Find and interpret the period and amplitude in the context of this situation. Then graph the function.
Answer:

Question 24.
MODELING WITH MATHEMATICS
A buoy bobs up and down as waves go past. The vertical displacement y (in feet) of the buoy with respect to sea level can be modeled by y = 1.75 cos \(\frac{\pi}{3}\)t, where t is the time (in seconds). Find and interpret the period and amplitude in the context of the problem. Then graph the function.

Answer:

In Exercises 25–34, graph the function.
Question 25.
g(x) = sin x + 2
Answer:

Question 26.
g(x) = cos x − 4
Answer:

Question 27.
g(x) = cos (x − \(\frac{\pi}{2}\))
Answer:

Question 28.
g(x) = sin (x + \(\frac{\pi}{4}\))
Answer:

Question 29.
g(x) = 2 cos x − 1
Answer:

Question 30.
g(x) = 3 sin x + 1
Answer:

Question 31.
g(x) = sin 2(x + π)
Answer:

Question 32.
g(x) = cos 2(x − π)
Answer:

Question 33.
g(x) = sin \(\frac{1}{2}\)(x + 2π) + 3
Answer:

Question 34.
g(x) = cos \(\frac{1}{2}\)(x − 3π) − 5
Answer:

Question 35.
ERROR ANALYSIS
Describe and correct the error in finding the period of the function y = sin \(\frac{2}{3}\)x.

Answer:

Question 36.
ERROR ANALYSIS
Describe and correct the error in determining the point where the maximum value of the function y = 2 sin (x − \(\frac{\pi}{4}\)) occurs.

Answer:

USING STRUCTURE In Exercises 37–40, describe the transformation of the graph of f represented by the function g
Question 37.
f(x) = cos x, g(x) = 2 cos (x − \(\frac{\pi}{2}\)) + 1
Answer:

Question 38.
f(x) = sin x, g(x) = 3 sin (x + \(\frac{\pi}{4}\)) − 2
Answer:

Question 39.
f(x) = sin x, g(x) = sin 3(x + 3π) − 5
Answer:

Question 40.
f(x) = cos x, g(x) = cos 6(x − π) + 9
Answer:

In Exercises 41–48, graph the function.
Question 41.
g(x) = −cos x + 3
Answer:

Question 42.
g(x) = −sin x − 5
Answer:

Question 43.
g(x) = −sin \(\frac{1}{2}\)x − 2
Answer:

Question 44.
g(x) = −cos 2x + 1
Answer:

Question 45.
g(x) = −sin(x − π) + 4
Answer:

Question 46.
g(x) = −cos(x + π) − 2
Answer:

Question 47.
g(x) = −4 cos (x + \(\frac{\pi}{4}\)) − 1
Answer:

Question 48.
g(x) = −5 sin (x − \(\frac{\pi}{2}\)) + 3
Answer:

Question 49.
USING EQUATIONS
Which of the following is a point where the maximum value of the graph of y =−4 cos (x − \(\frac{\pi}{2}\))occurs?
A. (−\(\frac{\pi}{2}\), 4 )
B. (\(\frac{\pi}{2}\), 4 )
C. (0, 4)
D. (π, 4)
Answer:

Question 50.
ANALYZING RELATIONSHIPS
Match each function with its graph. Explain your reasoning.
a. y = 3 + sin x
b. y = −3 + cos x
c. y = sin 2(x − \(\frac{\pi}{2}\))
d. y = cos 2 (x − \(\frac{\pi}{2}\))

Answer:

WRITING EQUATIONS In Exercises 51–54, write a rule for g that represents the indicated transformations of the graph of f.
Question 51.
f(x) = 3 sin x; translation 2 units up and π units right
Answer:

Question 52.
f(x) = cos 2πx; translation 4 units down and 3 units left
Answer:

Question 53.
f(x) = \(\frac{1}{3}\)cos πx; translation 1 unit down, followed by a reflection in the line y =−1
Answer:

Question 54.
f(x) = \(\frac{1}{2}\) sin 6x; translation \(\frac{3}{2}\) units down and 1 unit right, followed by a reflection in the line y = −\(\frac{3}{2}\)
Answer:

Question 55.
MODELING WITH MATHEMATICS
The height h(in feet) of a swing above the ground can be modeled by the function h = −8 cos θ+ 10, where the pivot is 10 feet above the ground, the rope is 8 feet long, and θ is the angle that the rope makes with the vertical. Graph the function. What is the height of the swing when θ is 45°?

Answer:

Question 56.
DRAWING A CONCLUSION
In a particular region, the population L (in thousands) of lynx (the predator) and the population H (in thousands) of hares (the prey) can be modeled by the equations
L = 11.5 + 6.5 sin\(\frac{\pi}{5}\)t
H = 27.5 + 17.5 cos \(\frac{\pi}{5}\)t
where t is the time in years.
a. Determine the ratio of hares to lynx when t = 0, 2.5, 5, and 7.5 years.
b. Use the figure to explain how the changes in the two populations appear to be related.

Answer:

Question 57.
USING TOOLS
The average wind speed s (in miles per hour) in the Boston Harbor can be approximated by s = 3.38 sin\(\frac{\pi}{180}\)(t + 3) + 11.6 where t is the time in days and t = 0 represents January 1. Use a graphing calculator to graph the function. On which days of the year is the average wind speed 10 miles per hour? Explain your reasoning.
Answer:

Question 58.
USING TOOLS
The water depth d (in feet) for the Bay of Fundy can be modeled by d = 35 − 28 cos \(\frac{\pi}{6.2}\)t, where t is the time in hours and t = 0 represents midnight. Use a graphing calculator to graph the function. At what time(s) is the water depth 7 feet? Explain.

Answer:

Question 59.
MULTIPLE REPRESENTATIONS
Find the average rate of change of each function over the interval 0 < x < π.
a. y = 2 cos x

Answer:

Question 60.
REASONING
Consider the functions y = sin(−x) and y = cos(−x).
a. Construct a table of values for each equation using the quadrantal angles in the interval −2π ≤ x ≤ 2π.
b. Graph each function.
c. Describe the transformations of the graphs of the parent functions.
Answer:

Question 61.
MODELING WITH MATHEMATICS
You are riding a Ferris wheel that turns for 180 seconds. Your height h (in feet) above the ground at any time t (in seconds) can be modeled by the equation h = 85 sin\(\frac{\pi}{20}\)(t − 10) + 90.

a. Graph the function.
b. How many cycles does the Ferris wheel make in 180 seconds?
c. What are your maximum and minimum heights?
Answer:

Question 62.
HOW DO YOU SEE IT?
Use the graph to answer each question.

a. Does the graph represent a function of the form f(x) = a sin bx or f(x) =a cos bx? Explain.
b. Identify the maximum value, minimum value, period, and amplitude of the function.
Answer:

Question 63.
FINDING A PATTERN
Write an expression in terms of the integer n that represents all the x-intercepts of the graph of the function y = cos 2x. Justify your answer.
Answer:

Question 64.
MAKING AN ARGUMENT
Your friend states that for functions of the form y = a sin bx and y = a cos bx, the values of a and b affect the x-intercepts of the graph of the function. Is your friend correct? Explain.
Answer:

Question 65.
CRITICAL THINKING
Describe a transformation of the graph of f(x) = sin x that results in the graph of g(x) = cos x.
Answer:

Question 66.
THOUGHT PROVOKING
Use a graphing calculator to find a function of the form y = sin b₁x + cos b₂xwhose graph matches that shown below.

Answer:

Question 67.
PROBLEM SOLVING
For a person at rest, the blood pressure P (in millimeters of mercury) at time t (in seconds) is given by the function
P = 100 − 20 cos \(\frac{8 \pi}{3}\)t.
Graph the function. One cycle is equivalent to one heartbeat. What is the pulse rate (in heartbeats per minute) of the person?

Answer:

Question 68.
PROBLEM SOLVING
The motion of a spring can be modeled by y = A cos kt, where y is the vertical displacement (in feet) of the spring relative to its position at rest, A is the initial displacement (in feet), k is a constant that measures the elasticity of the spring, and t is the time (in seconds).
a. You have a spring whose motion can be modeled by the function y= 0.2 cos 6t. Find the initial displacement and the period of the spring. Then graph the function.
b. When a damping force is applied to the spring, the motion of the spring can be modeled by the function y = 0.2e^−4.5t cos 4t. Graph this function. What effect does damping have on the motion?
Answer:

Maintaining Mathematical Proficiency

Simplify the rational expression, if possible.
Question 69.
\(\frac{x^{2}+x-6}{x+3}\)
Answer:

Question 70.
\(\frac{x^{3}-2 x^{2}-24 x}{x^{2}-2 x-24}\)
Answer:

Question 71.
\(\frac{x^{2}-4 x-5}{x^{2}+4 x-5}\)
Answer:

Question 72.
\(\frac{x^{2}-16}{x^{2}+x-20}\)
Answer:

Find the least common multiple of the expressions.
Question 73.
2x, 2(x − 5)
Answer:

Question 74.
x² − 4, x + 2
Answer:

Question 75.
x² + 8x + 12, x + 6
Answer:

Trigonometric Ratios and Functions Study Skills: Form a Final Exam Study Group

9.1–9.4 What Did You Learn?

Core Vocabulary

Core Concepts

Mathematical Practices
Question 1.
Make a conjecture about the horizontal distances traveled in part (c) of Exercise 39 on page 483.
Answer:

Question 2.
Explain why the quantities in part (a) of Exercise 56 on page 493 make sense in the context of the situation.
Answer:

Study Skills: Form a Final Exam Study Group

Form a study group several weeks before the final exam. The intent of this group is to review what you have already learned while continuing to learn new material.

Trigonometric Ratios and Functions 9.1–9.4 Quiz

Question 1.
In a right triangle, θ is an acute angle and sin θ = \(\frac{1}{2}\). Evaluate the other five trigonometric functions of θ.
Answer:

Find the value of x for the right triangle.
Question 2.

Answer:

Question 3.

Answer:

Question 4.

Answer:

Draw an angle with the given measure in standard position. Then find one positive angle and one negative angle that are coterminal with the given angle.
Question 5.
40°
Answer:

Question 6.
\(\frac{5 \pi}{6}\)
Answer:

Question 7.
−960°
Answer:

Convert the degree measure to radians or the radian measure to degrees.
Question 8.
\(\frac{3 \pi}{10}\)
Answer:

Question 9.
−60°
Answer:

Question 10.
72°
Answer:

Evaluate the six trigonometric functions of θ.
Question 11.

Answer:

Question 12.

Answer:

Question 13.

Answer:

Question 14.
Identify the amplitude and period of g(x) = 3 sin x. Then graph the function and describe the graph of g as a transformation of the graph of f (x) = sin x.
Answer:

Question 15.
Identify the amplitude and period of g(x) = cos 5πx + 3. Then graph the function and describe the graph of g as a transformation of the graph of f(x) = cos x.
Answer:

Question 16.
You are flying a kite at an angle of 70°. You have let out a total of 400 feet of string and are holding the reel steady 4 feet above the ground.

a. How high above the ground is the kite?
b. A friend watching the kite estimates that the angle of elevation to the kite is 85°. How far from your friend are you standing?
Answer:

Question 17.
The top of the Space Needle in Seattle, Washington, is a revolving, circular restaurant. The restaurant has a radius of 47.25 feet and makes one complete revolution in about an hour. You have dinner at a window table from 7:00 P.M. to 8:55 P.M. Compare the distance you revolve with the distance of a person seated 5 feet away from the windows.
Answer:

Lesson 9.5 Graphing Other Trigonometric Functions

Essential Question What are the characteristics of the graph of the tangent function?

EXPLORATION 1

Graphing the Tangent Function
Work with a partner. a. Complete the table for y = tan x, where x is an angle measure in radians.

b. The graph of y = tan x has vertical asymptotes at x-values where tan x is undefined. Plot the points (x, y) from part (a). Then use the asymptotes to sketch the graph of y = tan x.

c. For the graph of y = tan x, identify the asymptotes, the x-intercepts, and the intervals for which the function is increasing or decreasing over −\(\frac{\pi}{2}\) ≤ x ≤ \(\frac{3 \pi}{2}\). Is the tangent function even, odd, or neither?

Communicate Your Answer

Question 2.
What are the characteristics of the graph of the tangent function?
Answer:

Question 3.
Describe the asymptotes of the graph of y = cot x on the interval −\(\frac{\pi}{2}\) < x < \(\frac{3 \pi}{2}\).

Answer:

Monitoring Progress

Graph one period of the function. Describe the graph of g as a transformation of the graph of its parent function.
Question 1.
g(x) = tan 2x
Answer:

Question 2.
g(x) = \(\frac{1}{3}\)cot x
Answer:

Question 3.
g(x) = 2 cot 4x
Answer:

Question 4.
g(x) = 5 tan πx
Answer:

Graph one period of the function. Describe the graph of g as a transformation of the graph of its parent function.
Question 5.
g(x) = csc 3x
Answer:

Question 6.
g(x) = \(\frac{1}{2}\)sec x
Answer:

Question 7.
g(x) = 2 csc 2x
Answer:

Question 8.
g(x) = 2 sec πx
Answer:

Graphing Other Trigonometric Functions 9.5 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Explain why the graphs of the tangent, cotangent, secant, and cosecant functions do not have an amplitude.
Answer:

Question 2.
COMPLETE THE SENTENCE
The _______ and _______ functions are undefined for x-values at which sin x = 0.
Answer:

Question 3.
COMPLETE THE SENTENCE
The period of the function y = sec x is _____, and the period of y = cot x is _____.
Answer:

Question 4.
WRITING
Explain how to graph a function of the form y = a sec bx.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, graph one period of the function. Describe the graph of gas a transformation of the graph of its parent function.
Question 5.
g(x) = 2 tan x
Answer:

Question 6.
g(x) = 3 tan x
Answer:

Question 7.
g(x) = cot 3x
Answer:

Question 8.
g(x) = cot 2x
Answer:

Question 9.
g(x) = 3 cot \(\frac{1}{4}\)x
Answer:

Question 10.
g(x) = 4 cot\(\frac{1}{2}\)x
Answer:

Question 11.
g(x) = \(\frac{1}{2}\)tan πx
Answer:

Question 12.
g(x) = \(\frac{1}{3}\) tan 2πx
Answer:

Question 13.
ERROR ANALYSIS
Describe and correct the error in finding the period of the function y = cot 3x.

Answer:

Question 14.
ERROR ANALYSIS
Describe and correct the error in describing the transformation of f(x) = tan x represented by g(x) = 2 tan 5x.

Answer:

Question 15.
ANALYZING RELATIONSHIPS
Use the given graph to graph each function.

Answer:

Question 16.
USING EQUATIONS
Which of the following are asymptotes of the graph of y = 3 tan 4x?
A. x = \(\frac{\pi}{8}\)
B. x = \(\frac{\pi}{4}\)
C. x = 0
D. x = −\(\frac{5 \pi}{8}\)
Answer:

In Exercises 17–24, graph one period of the function. Describe the graph of gas a transformation of the graph of its parent function.
Question 17.
g(x) = 3 csc x
Answer:

Question 18.
g(x) = 2 csc x
Answer:

Question 19.
g(x) = sec 4x
Answer:

Question 20.
g(x) = sec 3x
Answer:

Question 21.
g(x) = \(\frac{1}{2}\)sec πx
Answer:

Question 22.
g(x) = \(\frac{1}{4}\) sec 2πx
Answer:

Question 23.
g(x) = csc \(\frac{\pi}{2}\)x
Answer:

Question 24.
g(x) = csc \(\frac{\pi}{4}\)x
Answer:

ATTENDING TO PRECISION In Exercises 25–28, use the graph to write a function of the form y = a tan bx.
Question 25.

Answer:

Question 26.

Answer:

Question 27.

Answer:

Question 28.

Answer:

USING STRUCTURE In Exercises 29–34, match the equation with the correct graph. Explain your reasoning.
Question 29.
g(x) = 4 tan x
Answer:

Question 30.
g(x) = 4 cot x
Answer:

Question 31.
g(x) = 4 csc πx
Answer:

Question 32.
g(x) = 4 sec πx
Answer:

Question 33.
g(x) = sec 2x
Answer:

Question 34.
g(x) = csc 2x
Answer:

Question 35.
WRITING
Explain why there is more than one tangent function whose graph passes through the origin and has asymptotes at x = −π and x = π.
Answer:

Question 36.
USING EQUATIONS
Graph one period of each function. Describe the transformation of the graph of its parent function.
a. g(x) = sec x + 3
b. g(x) = csc x − 2
c. g(x) = cot(x − π)
d. g(x) = −tan x
Answer:

WRITING EQUATIONS In Exercises 37–40, write a rule for g that represents the indicated transformation of the graph of f.
Question 37.
f(x) = cot 2x; translation 3 units up and \(\frac{\pi}{2}\) units left
Answer:

Question 38.
f(x) = 2 tan x; translation π units right, followed by a horizontal shrink by a factor of \(\frac{1}{3}\)
Answer:

Question 39.
f(x) = 5 sec (x − π); translation 2 units down, followed by a reflection in the x-axis
Answer:

Question 40.
f(x) = 4 csc x; vertical stretch by a factor of 2 and a reflection in the x-axis
Answer:

Question 41.
MULTIPLE REPRESENTATIONS
Which function has a greater local maximum value? Which has a greater local minimum value? Explain.

Answer:

Question 42.
ANALYZING RELATIONSHIPS
Order the functions from the least average rate of change to the greatest average rate of change over the interval −\(\frac{\pi}{4}\) < x < \(\frac{\pi}{4}\).

Answer:

Question 43.
REASONING
You are standing on a bridge 140 feet above the ground. You look down at a car traveling away from the underpass. The distance d (in feet) the car is from the base of the bridge can be modeled by d= 140 tan θ. Graph the function. Describe what happens to θ as d increases

Answer:

Question 44.
USING TOOLS
You use a video camera to pan up the Statue of Liberty. The height h (in feet) of the part of the Statue of Liberty that can be seen through your video camera after time t (in seconds) can be modeled by h= 100 tan \(\frac{\pi}{36}\)t. Graph the function using a graphing calculator. What viewing window did you use? Explain.
Answer:

Question 45.
MODELING WITH MATHEMATICS
You are standing 120 feet from the base of a 260-foot building. You watch your friend go down the side of the building in a glass elevator.

a. Write an equation that gives the distance d (in feet) your friend is from the top of the building as a function of the angle of elevation θ.
b. Graph the function found in part (a). Explain how the graph relates to this situation.
Answer:

Question 46.
MODELING WITH MATHEMATICS
You are standing 300 feet from the base of a 200-foot cliff. Your friend is rappelling down the cliff.
a. Write an equation that gives the distance d(in feet) your friend is from the top of the cliff as a function of the angle of elevation θ.
b. Graph the function found in part (a).
c. Use a graphing calculator to determine the angle of elevation when your friend has rappelled halfway down the cliff.

Answer:

Question 47.
MAKING AN ARGUMENT
Your friend states that it is not possible to write a cosecant function that has the same graph as y = sec x. Is your friend correct? Explain your reasoning.
Answer:

Question 48.
HOW DO YOU SEE IT?
Use the graph to answer each question.

a. What is the period of the graph?
b. What is the range of the function?
c. Is the function of the form f(x) = a csc bx or f(x) = a sec bx? Explain.
Answer:

Question 49.
ABSTRACT REASONING
Rewrite a sec bx in terms of cos bx. Use your results to explain the relationship between the local maximums and minimums of the cosine and secant functions.
Answer:

Question 50.
THOUGHT PROVOKING
A trigonometric equation that is true for all values of the variable for which both sides of the equation are defined is called a trigonometric identity.Use a graphing calculator to graph the function
y = \(\frac{1}{2}\)(tan \(\frac{x}{2}\) + cot \(\frac{x}{2}\)) .
Use your graph to write a trigonometric identity involving this function. Explain your reasoning.
Answer:

Question 51.
CRITICAL THINKING
Find a tangent function whose graph intersects the graph of y = 2 + 2 sin x only at minimum points of the sine function.
Answer:

Maintaining Mathematical Proficiency

Write a cubic function whose graph passes through the given points.
Question 52.
(−1, 0), (1, 0), (3, 0), (0, 3)
Answer:

Question 53.
(−2, 0), (1, 0), (3, 0), (0, −6)
Answer:

Question 54.
(−1, 0), (2, 0), (3, 0), (1, −2)
Answer:

Question 55.
(−3, 0), (−1, 0), (3, 0), (−2, 1)
Answer:

Find the amplitude and period of the graph of the function.
Question 56.

Answer:

Question 57.

Answer:

Question 58.

Answer:

Lesson 9.6 Modeling with Trigonometric Functions

Essential Question What are the characteristics of the real-life problems that can be modeled by trigonometric functions?

EXPLORATION 1

Modeling Electric Currents
Work with a partner. Find a sine function that models the electric current shown in each oscilloscope screen. State the amplitude and period of the graph.

Communicate Your Answer

Question 2.
What are the characteristics of the real-life problems that can be modeled by trigonometric functions?
Answer:

Question 3.
Use the Internet or some other reference to find examples of real-life situations that can be modeled by trigonometric functions.
Answer:

Monitoring Progress

Question 1.
WHAT IF?
In Example 1, how would the function change when the audiometer produced a pure tone with a frequency of 1000 hertz?
Answer:

Write a function for the sinusoid.
Question 2.

Answer:

Question 3.

Answer:

Question 4.
WHAT IF?
Describe how the model in Example 3 changes when the lowest point of a rope is 5 inches above the ground and the highest point is 70 inches above the ground.
Answer:

Question 5.
The table shows the average daily temperature T (in degrees Fahrenheit) for a city each month, where m = 1 represents January. Write a model that gives T as a function of m and interpret the period of its graph.

Answer:

Modeling with Trigonometric Functions 9.6 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
Graphs of sine and cosine functions are called __________.
Answer:

Question 2.
WRITING
Describe how to find the frequency of the function whose graph is shown.

Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the frequency of the function.
Question 3.
y = sin x
Answer:

Question 4.
y = sin 3x
Answer:

Question 5.
y = cos 4x + 2
Answer:

Question 6.
y =−cos 2x
Answer:

Question 7.
y = sin 3πx
Answer:

Question 8.
y = cos \(\frac{\pi x}{4}\)
Answer:

Question 9.
y = \(\frac{1}{2}\) cos 0.75x − 8
Answer:

Question 10.
y = 3 sin 0.2x + 6
Answer:

Question 11.
MODELING WITH MATHEMATICS
The lowest frequency of sounds that can be heard by humans is 20 hertz. The maximum pressure P produced from a sound with a frequency of 20 hertz is 0.02 millipascal. Write and graph a sine model that gives the pressure P as a function of the time t(in seconds).
Answer:

Question 12.
MODELING WITH MATHEMATICS
A middle-A tuning fork vibrates with a frequency f of 440 hertz (cycles per second). You strike a middle-A tuning fork with a force that produces a maximum pressure of 5 pascals. Write and graph a sine model that gives the pressure Pas a function of the time t (in seconds).

Answer:

In Exercises 13–16, write a function for the sinusoid.
Question 13.

Answer:

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

Question 17.
ERROR ANALYSIS
Describe and correct the error in finding the amplitude of a sinusoid with a maximum point at (2, 10) and a minimum point at (4, −6).

Answer:

Question 18.
ERROR ANALYSIS
Describe and correct the error in finding the vertical shift of a sinusoid with a maximum point at (3, −2) and a minimum point at (7, −8).

Answer:

Question 19.
MODELING WITH MATHEMATICS
One of the largest sewing machines in the world has a flywheel (which turns as the machine sews) that is 5 feet in diameter. The highest point of the handle at the edge of the flywheel is 9 feet above the ground, and the lowest point is 4 feet. The wheel makes a complete turn every 2 seconds. Write a model for the height h(in feet) of the handle as a function of the time t(in seconds) given that the handle is at its lowest point when t = 0.
Answer:

Question 20.
MODELING WITH MATHEMATICS
The Great LaxeyWheel, located on the Isle of Man, is the largest working water wheel in the world. The highest point of a bucket on the wheel is 70.5 feet above the viewing platform, and the lowest point is 2 feet below the viewing platform. The wheel makes a complete turn every 24 seconds. Write a model for the height h(in feet) of the bucket as a function of time t (in seconds) given that the bucket is at its lowest point when t = 0.

Answer:

USING TOOLS In Exercises 21 and 22, the time t is measured in months, where t = 1 represents January. Write a model that gives the average monthly high temperature D as a function of t and interpret the period of the graph.
Question 21.

Answer:

Question 22.

Answer:

Question 23.
MODELING WITH MATHEMATICS
A circuit has an alternating voltage of 100 volts that peaks every 0.5 second. Write a sinusoidal model for the voltage Vas a function of the time t (in seconds).

Answer:

Question 24.
MULTIPLE REPRESENTATIONS
The graph shows the average daily temperature of Lexington, Kentucky. The average daily temperature of Louisville, Kentucky, is modeled by y =−22 cos \(\frac{\pi}{6}\)t + 57, where y is the temperature (in degrees Fahrenheit) and t is the number of months since January 1. Which city has the greater average daily temperature? Explain.

Answer:

Question 25.
USING TOOLS
The table shows the numbers of employees N (in thousands) at a sporting goods company each year for 11 years. The time t is measured in years, with t = 1 representing the first year.

a. Use sinusoidal regression to find a model that gives N as a function of t.
b. Predict the number of employees at the company in the 12th year.
Answer:

Question 26.
THOUGHT PROVOKING
The figure shows a tangent line drawn to the graph of the function y = sin x. At several points on the graph, draw a tangent line to the graph and estimate its slope. Then plot the points (x, m), where m is the slope of the tangent line. What can you conclude?

Answer:

Question 27.
REASONING
Determine whether you would use a sine or cosine function to model each sinusoid with the y-intercept described. Explain your reasoning.
a. The y-intercept occurs at the maximum value of the function.
b. The y-intercept occurs at the minimum value of the function.
c. The y-intercept occurs halfway between the maximum and minimum values of the function.
Answer:

Question 28.
HOW DO YOU SEE IT?
What is the frequency of the function whose graph is shown? Explain.

Answer:

Question 29.
USING STRUCTURE
During one cycle, a sinusoid has a minimum at (\(\frac{\pi}{2}\), 3 ) and a maximum at (\(\frac{\pi}{4}\), 8 ). Write a sine function and a cosine function for the sinusoid. Use a graphing calculator to verify that your answers are correct.
Answer:

Question 30.
MAKING AN ARGUMENT
Your friend claims that a function with a frequency of 2 has a greater period than a function with a frequency of \(\frac{1}{2}\). Is your friend correct? Explain your reasoning.
Answer:

Question 31.
PROBLEM SOLVING
The low tide at a port is 3.5 feet and occurs at midnight. After 6 hours, the port is at high tide, which is 16.5 feet.

a. Write a sinusoidal model that gives the tide depth d(in feet) as a function of the time t(in hours). Let t = 0 represent midnight.
b. Find all the times when low and high tides occur in a 24-hour period.
c. Explain how the graph of the function you wrote in part (a) is related to a graph that shows the tide depth d at the port t hours after 3:00 A.M.
Answer:

Maintaining Mathematical Proficiency

Simplify the expression.
Question 32.
\(\frac{17}{\sqrt{2}}\)
Answer:

Question 33.
\(\frac{3}{\sqrt{6}-2}\)
Answer:

Question 34.
\(\frac{8}{\sqrt{10}+3}\)
Answer:

Question 35.
\(\frac{13}{\sqrt{3}+\sqrt{11}}\)
Answer:

Expand the logarithmic expression.
Question 36.
log₈\(\frac{x}{7}\)
Answer:

Question 37.
ln 2x
Answer:

Question 38.
log₃ 5x³
Answer:

Question 39.
ln \(\frac{4 x^{6}}{y}\)
Answer:

Lesson 9.7 Using Trigonometric Identities

Essential Question How can you verify a trigonometric identity?

EXPLORATION 1

Writing a Trigonometric Identity
Work with a partner. In the figure, the point (x, y) is on a circle of radius c with center at the origin.

a. Write an equation that relates a, b, and c.
b. Write expressions for the sine and cosine ratios of angle θ.
c. Use the results from parts (a) and (b) to find the sum of sin²θ and cos²θ. What do you observe?
d. Complete the table to verify that the identity you wrote in part (c) is valid for angles (of your choice) in each of the four quadrants.

EXPLORATION 2

Writing Other Trigonometric Identities
Work with a partner. The trigonometric identity you derived in Exploration 1 is called a Pythagorean identity. There are two other Pythagorean identities. To derive them, recall the four relationships:

a. Divide each side of the Pythagorean identity you derived in Exploration 1 by cos²θ and simplify. What do you observe?
b. Divide each side of the Pythagorean identity you derived in Exploration 1 by sin²θ and simplify. What do you observe?

Communicate Your Answer

Question 3.
How can you verify a trigonometric identity?
Answer:

Question 4.
Is sin θ = cos θ a trigonometric identity? Explain your reasoning.
Answer:

Question 5.
Give some examples of trigonometric identities that are different than those in Explorations 1 and 2.
Answer:

Monitoring Progress

Question 1.
Given that cos θ = \(\frac{1}{6}\) and 0 < θ < \(\frac{\pi}{2}\), find the values of the other five trigonometric functions of θ.
Answer:

Simplify the expression.
Question 2.
sin x cot x sec x
Answer:

Question 3.
cos θ − cos θ sin²θ
Answer:

Question 4.
\(\frac{\tan x \csc x}{\sec x}\)
Answer:

Verify the identity.
Question 5.
cot(−θ) =−cot θ
Answer:

Question 6.
csc²x(1 − sin2x) = cot²x
Answer:

Question 7.
cos x csc x tan x = 1
Answer:

Question 8.
(tan²x + 1)(cos²x− 1) = −tan²x
Answer:

Using Trigonometric Identities 9.7 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe the difference between a trigonometric identity and a trigonometric equation.
Answer:

Question 2.
WRITING
Explain how to use trigonometric identities to determine whether sec(−θ) = sec θ or sec(−θ) = −sec θ.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the values of the other five trigonometric functions of θ.
Question 3.
sinθ = \(\frac{1}{3}\), 0 < θ < \(\frac{\pi}{2}\)
Answer:

Question 4.
sin θ = −\(\frac{7}{10}\), π < θ < \(\frac{3 \pi}{2}\)
Answer:

Question 5.
tanθ = −\(\frac{3}{7}\), \(\frac{\pi}{2}\) < θ < π
Answer:

Question 6.
cot θ = −\(\frac{2}{5}\), \(\frac{\pi}{2}\) < θ < π
Answer:

Question 7.
cos θ = −\(\frac{5}{6}\), π < θ < \(\frac{3 \pi}{2}\)
Answer:

Question 8.
sec θ = \(\frac{9}{4}\), \(\frac{3 \pi}{2}\) < θ < 2π
Answer:

Question 9.
cot θ = −3, \(\frac{3 \pi}{2}\) < θ < 2π
Answer:

Question 10.
csc θ = −\(\frac{5}{3}\), π < θ < \(\frac{3 \pi}{2}\)
Answer:

In Exercises 11–20, simplify the expression.
Question 11.
sin x cot x
Answer:

Question 12.
cos θ (1 + tan²θ)
Answer:

Question 13.
\(\frac{\sin (-\theta)}{\cos (-\theta)}\)
Answer:

Question 14.
\(\frac{\cos ^{2} x}{\cot ^{2} x}\)
Answer:

Question 15.
\(\frac{\cos \left(\frac{\pi}{2}-x\right)}{\csc x} \)
Answer:

Question 16.
sin(\(\frac{\pi}{2}\) – θ) sec θ
Answer:

Question 17.
\(\frac{\csc ^{2} x-\cot ^{2} x}{\sin (-x) \cot x}\)
Answer:

Question 18.
\(\frac{\cos ^{2} x \tan ^{2}(-x)-1}{\cos ^{2} x}\)
Answer:

Question 19.
\(\frac{\cos \left(\frac{\pi}{2}-\theta\right)}{\csc \theta}\) + cos² θ
Answer:

Question 20.
\(\frac{\sec x \sin x+\cos \left(\frac{\pi}{2}-x\right)}{1+\sec x}\)
Answer:

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in simplifying the expression.
Question 21.

Answer:

Question 22.

Answer:

In Exercises 23–30, verify the identity.
Question 23.
sin x csc x = 1
Answer:

Question 24.
tan θ csc θ cos θ = 1
Answer:

Question 25.
cos (\(\frac{3 \pi}{2}\) − x)cot x = cos x
Answer:

Question 26.
sin (\(\frac{3 \pi}{2}\) − x)tan x = sin x
Answer:

Question 27.
\(\frac{\cos \left(\frac{\pi}{2}-\theta\right)+1}{1-\sin (-\theta)}\) = 1
Answer:

Question 28.
\(\frac{\sin ^{2}(-x)}{\tan ^{2} x}\) = cos² x
Answer:

Question 29.
\(\frac{1+\cos x}{\sin x}+\frac{\sin x}{1+\cos x}\) = 2 csc x
Answer:

Question 30.
\(\frac{\sin x}{1-\cos (-x)}\) = csc x + cot x
Answer:

Question 31.
USING STRUCTURE
A function f is odd when f(−x) = −f(x). A function f is even when (−x) = f (x). Which of the six trigonometric functions are odd? Which are even? Justify your answers using identities and graphs.
Answer:

Question 32.
ANALYZING RELATIONSHIPS
As the value of cos θ increases, what happens to the value of sec θ? Explain your reasoning.
Answer:

Question 33.
MAKING AN ARGUMENT
Your friend simplifies an expression and obtains sec x tan x− sin x. You simplify the same expression and obtain sin x tan2x. Are your answers equivalent? Justify your answer.
Answer:

Question 34.
HOW DO YOU SEE IT?
The figure shows the unit circle and the angle θ.
a. Is sin θ positive or negative? cos θ? tan θ?
b. In what quadrant does the terminal side of −θ lie?
c. Is sin(−θ) positive or negative? cos(−θ)? tan(−θ)?

Answer:

Question 35.
MODELING WITH MATHEMATICS
A vertical gnomon(the part of a sundial that projects a shadow) has height h. The length s of the shadow cast by the gnomon when the angle of the Sun above the horizon is θ can be modeled by the equation below. Show that the equation below is equivalent to s = h cot θ.

Answer:

Question 36.
THOUGHT PROVOKING
Explain how you can use a trigonometric identity to find all the values of x for which sin x = cos x.
Answer:

Question 37.
DRAWING CONCLUSIONS
Static friction is the amount of force necessary to keep a stationary object on a flat surface from moving. Suppose a book weighing W pounds is lying on a ramp inclined at an angle θ. The coefficient of static friction u for the book can be found using the equation uW cos θ = W sin θ.
a. Solve the equation for u and simplify the result.
b. Use the equation from part (a) to determine what happens to the value of u as the angle θ increases from 0° to 90°.
Answer:

Question 38.
PROBLEM SOLVING
When light traveling in a medium (such as air) strikes the surface of a second medium (such as water) at an angle θ₁, the light begins to travel at a different angle θ₂. This change of direction is defined by Snell’s law, n₁ sin θ₁ = n₂ sin θ₂, where n₁ and n₂ are the indices of refraction for the two mediums. Snell’s law can be derived from the equation.

a. Simplify the equation to derive Snell’s law.
b. What is the value of n1 when θ₁ = 55°, θ₂ = 35°, and n₂ = 2?
c. If θ₁ = θ₂, then what must be true about the values of n₁ and n₂? Explain when this situation would occur.
Answer:

Question 39.
WRITING
Explain how transformations of the graph of the parent function f(x) = sin x support the cofunction identity sin (\(\frac{\pi}{2}\) − θ) = cos θ.
Answer:

Question 40.
USING STRUCTURE
Verify each identity.
a. ln ∣sec θ∣= −ln ∣cos θ∣
b. ln ∣tan θ∣= ln ∣sin θ∣− ln ∣cos θ∣
Answer:

Maintaining Mathematical Proficiency

Find the value of x for the right triangle.
Question 41.

Answer:

Question 42.

Answer:

Question 43.

Answer:

Lesson 9.8 Using Sum and Difference Formulas

Essential Question How can you evaluate trigonometric functions of the sum or difference of two angles?

EXPLORATION 1

Deriving a Difference Formula
Work with a partner.
a. Explain why the two triangles shown are congruent.

b. Use the Distance Formula to write an expression for d in the first unit circle.
c. Use the Distance Formula to write an expression for d in the second unit circle.
d. Write an equation that relates the expressions in parts (b) and (c). Then simplify this equation to obtain a formula for cos(a − b).

EXPLORATION 2

Deriving a Sum Formula

Work with a partner. Use the difference formula you derived in Exploration 1 to write a formula for cos(a+b) in terms of sine and cosine of a and b. Hint: Use the fact that cos(a+b) = cos[a − (−b)].

EXPLORATION 3

Deriving Difference and Sum Formulas
Work with a partner. Use the formulas you derived in Explorations 1 and 2 to write formulas for sin(a − b) and sin(a + b) in terms of sine and cosine of a and b. Hint: Use the cofunction identities sin (\(\frac{\pi}{2}\) − a)= cos a and cos (\(\frac{\pi}{2}\) − a)= sin a and the fact that
cos [(\(\frac{\pi}{2}\) − a) + b ]= sin(a − b) and sin(a+b) = sin[a − (−b)].

Communicate Your Answer

Question 4.
How can you evaluate trigonometric functions of the sum or difference of two angles?
Answer:

Question 5.
a. Find the exact values of sin 75° and cos 75° using sum formulas. Explain your reasoning.
b. Find the exact values of sin 75° and cos 75° using difference formulas. Compare your answers to those in part (a).
Answer:

Monitoring Progress

Find the exact value of the expression.
Question 1.
sin 105°
Answer:

Question 2.
cos 15°
Answer:

Question 3.
tan \(\frac{5 \pi}{2}\)
Answer:

Question 4.
cos \(\frac{\pi}{12}\)
Answer:

Question 5.
Find sin(a−b) given that sin a = \(\frac{8}{17}\) with 0 < a < \(\frac{\pi}{2}\) and cos b = −\(\frac{24}{25}\) with π < b < \(\frac{3 \pi}{2}\) .
Answer:

Simplify the expression.
Question 6.
sin(x + π)
Answer:

Question 7.
cos(x − 2π)
Answer:

Question 8.
tan(x − π)
Answer:

Question 9.
Solve sin (\(\frac{\pi}{4}\) − x)− sin (x + \(\frac{\pi}{4}\))= 1 for 0 ≤ x < 2π.
Answer:

Using Sum and Difference Formulas 9.8 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
Write the expression cos 130° cos 40°− sin 130° sin 40° as the cosine of an angle.
Answer:

Question 2.
WRITING
Explain how to evaluate tan 75° using either the sum or difference formula for tangent.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, find the exact value of the expression.
Question 3.
tan(−15°)
Answer:

Question 4.
tan 195°
Answer:

Question 5.
sin \(\frac{23 \pi}{12}\)
Answer:

Question 6.
sin(−165°)
Answer:

Question 7.
cos 105°
Answer:

Question 8.
cos \(\frac{11 \pi}{12}\)
Answer:

Question 9.
tan \(\frac{17 \pi}{12}\)
Answer:

Question 10.
sin (−\(\frac{7 \pi}{12}\))
Answer:

In Exercises 11–16, evaluate the expression given that cos a = \(\frac{4}{5}\) with 0 < a < \(\frac{\pi}{2}\) and sin b = –\(\frac{15}{17}\) with \(\frac{3 \pi}{2}\) < b < 2π.
Question 11.
sin(a + b)
Answer:

Question 12.
sin(a − b)
Answer:

Question 13.
cos(a − b)
Answer:

Question 14.
cos(a + b)
Answer:

Question 15.
tan(a + b)
Answer:

Question 16.
tan(a − b)
Answer:

In Exercises 17–22, simplify the expression.
Question 17.
tan(x + π)
Answer:

Question 18.
cos (x − \(\frac{\pi}{2}\))
Answer:

Question 19.
cos(x + 2π)
Answer:

Question 20.
tan(x − 2π)
Answer:

Question 21.
sin (x − \(\frac{3 \pi}{2}\))
Answer:

Question 22.
tan (x + \(\frac{\pi}{2}\))
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in simplifying the expression.
Question 23.

Answer:

Question 24.

Answer:

Question 25.
What are the solutions of the equation 2 sin x − 1 = 0 for 0 ≤ x < 2π?
A. \(\frac{\pi}{3}\)
B. \(\frac{\pi}{6}\)
C. \(\frac{2 \pi}{3}\)
D. \(\frac{5 \pi}{6}\)
Answer:

Question 26.
What are the solutions of the equation tan x + 1 = 0 for 0 ≤ x < 2π?
A. \(\frac{\pi}{4}\)
B. \(\frac{3 \pi}{4}\)
C. \(\frac{5 \pi}{4}\)
D. \(\frac{7 \pi}{4}\)
Answer:

In Exercises 27–32, solve the equation for 0 ≤ x < 2π.
Question 27.
sin (x + \(\frac{\pi}{2}\)) = \(\frac{1}{2}\)
Answer:

Question 28.
tan (x − \(\frac{\pi}{4}\)) = 0
Answer:

Question 29.
cos (x +\(\frac{\pi}{6}\)) − cos (x −\(\frac{\pi}{6}\)) = 1
Answer:

Question 30.
sin (x + \(\frac{\pi}{4}\)) + sin (x − \(\frac{\pi}{4}\)) = 0
Answer:

Question 31.
tan(x + π) − tan(π − x) = 0
Answer:

Question 32.
sin(x + π) + cos(x + π) = 0
Answer:

Question 33.
USING EQUATIONS
Derive the cofunction identity sin (\(\frac{\pi}{2}\) − θ)= cos θ using the difference formula for sine.
Answer:

Question 34.
MAKING AN ARGUMENT
Your friend claims it is possible to use the difference formula for tangent to derive the cofunction identity tan (\(\frac{\pi}{2}\) − θ) = cot θ. Is your friend correct? Explain your reasoning.
Answer:

Question 35.
MODELING WITH MATHEMATICS
A photographer is at a height h taking aerial photographs with a 35-millimeter camera. The ratio of the image length WQ to the length NA of the actual object is given by the formula

where θ is the angle between the vertical line perpendicular to the ground and the line from the camera to point A and t is the tilt angle of the film. When t = 45°, show that the formula can be rewritten as \(\frac{W Q}{N A}=\frac{70}{h(1+\tan \theta)}\).
Answer:

Question 36.
MODELING WITH MATHEMATICS
When a wave travels through a taut string, the displacement y of each point on the string depends on the time t and the point’s position x. The equation of a standing wave can be obtained by adding the displacements of two waves traveling in opposite directions. Suppose a standing wave can be modeled by the formula
y = A cos (\(\frac{2 \pi t}{3}-\frac{2 \pi x}{5}\)) + A cos (\(\frac{2 \pi t}{3}+\frac{2 \pi x}{5}\)) .When t= 1, show that the formula can be rewritten as y = −A cos \(\frac{2 \pi x}{5}\).
Answer:

Question 37.
MODELING WITH MATHEMATICS
The busy signal on a touch-tone phone is a combination of two tones with frequencies of 480 hertz and 620 hertz. The individual tones can be modeled by the equations:
480 hertz: y₁ = cos 960πt
620 hertz: y₂ = cos 1240πt
The sound of the busy signal can be modeled by y₁ + y₂. Show that y₁ + y₂ = 2 cos 1100πt cos 140πt.
Answer:

Question 38.
HOW DO YOU SEE IT?
Explain how to use the figure to solve the equation sin (x + \(\frac{\pi}{4}\)) − sin (\(\frac{\pi}{4}\) − x) = 0 for 0 ≤ x < 2π.

Answer:

Question 39.
MATHEMATICAL CONNECTIONS
The figure shows the acute angle of intersection, θ₂ − θ₁, of two lines with slopes m₁ and m₂.

a. Use the difference formula for tangent to write an equation for tan (θ₂ − θ₁) in terms of m₁ and m₂.
b. Use the equation from part (a) to find the acute angle of intersection of the lines y = x− 1 and y = \(\left(\frac{1}{\sqrt{3}-2}\right)\)x + \(\frac{4-\sqrt{3}}{2-\sqrt{3}}\).
Answer:

Question 40.
THOUGHT PROVOKING
Rewrite each function. Justify your answers.
a. Write sin 3x as a function of sin x.
b. Write cos 3x as a function of cos x.
c. Write tan 3x as a function of tan x.
Answer:

Maintaining Mathematical Proficiency

Solve the equation. Check your solution(s).
Question 41.
1 − \(\frac{9}{x-2}\) = −\(\frac{7}{2}\)
Answer:

Question 42.
\(\frac{12}{x}\) + \(\frac{3}{4}\) = \(\frac{8}{x}\)
Answer:

Question 43.
\(\frac{2 x-3}{x+1}\) = \(\frac{10}{x^{2}-1}\) + 5
Answer:

Trigonometric Ratios and Functions Performance Task: Lightening the Load

9.5–9.8 What Did You Learn?

Core Vocabulary
frequency, p. 506
sinusoid, p. 507
trigonometric identity, p. 514

Core Concepts
Section 9.5
Characteristics of y = tan x and y = cot x, p. 498
Period and Vertical Asymptotes of y = a tan bx and y = a cot bx, p. 499
Characteristics of y = sec x and y = csc x, p. 500

Section 9.6
Frequency, p. 506
Writing Trigonometric Functions, p. 507
Using Technology to Find Trigonometric Models, p. 509

Section 9.7
Fundamental Trigonometric Identities, p. 514

Section 9.8
Sum and Difference Formulas, p. 520
Trigonometric Equations and Real-Life Formulas, p. 522

Mathematical Practices
Question 1.
Explain why the relationship between θ and d makes sense in the context of the situation in Exercise 43 on page 503.
Answer:

Question 2.
How can you use definitions to relate the slope of a line with the tangent of an angle in Exercise 39 on page 524?
Answer:

Performance Task: Lightening the Load

You need to move a heavy table across the room. What is the easiest way to move it? Should you push it? Should you tie a rope around one leg of the table and pull it? How can trigonometry help you make the right decision?
To explore the answers to these questions and more, go to BigIdeasMath.com.

Trigonometric Ratios and Functions Chapter Review

9.1 Right Triangle Trigonometry (pp. 461−468)

Question 1.
In a right triangle, θ is an acute angle and cos θ = \(\frac{6}{11}\). Evaluate the other five trigonometric functions of θ.
Answer:

Question 2.
The shadow of a tree measures 25 feet from its base. The angle of elevation to the Sun is 31°. How tall is the tree?

Answer:

9.2 Angles and Radian Measure (pp. 469−476)

Question 3.
Find one positive angle and one negative angle that are coterminal with 382°.
Answer:

Convert the degree measure to radians or the radian measure to degrees.
Question 4.
30°
Answer:

Question 5.
225°
Answer:

Question 6.
\(\frac{3 \pi}{4}\)
Answer:

Question 7.
\(\frac{5 \pi}{3}\)
Answer:

Question 8.
A sprinkler system on a farm rotates 140°and sprays water up to 35 meters. Draw a diagram that shows the region that can be irrigated with the sprinkler. Then find the area of the region.
Answer:

9.3 Trigonometric Functions of Any Angle (pp. 477−484)

Evaluate the six trigonometric functions of θ.
Question 9.

Answer:

Question 10.

Answer:

Question 11.

Answer:

Evaluate the function without using a calculator.
Question 12.
tan 330°
Answer:

Question 13.
sec(−405°)
Answer:

Question 14.
sin \(\frac{13 \pi}{6}\)
Answer:

Question 15.
sec \(\frac{11 \pi}{3}\)
Answer:

9.4 Graphing Sine and Cosine Functions (pp. 485−494)

Identify the amplitude and period of the function. Then graph the function and describe the graph of g as a transformation of the graph of the parent function.
Question 16.
g(x) = 8 cos x
Answer:

Question 17.
g(x) = 6 sin πx
Answer:

Question 18.
g(x) = \(\frac{1}{4}\) cos 4x
Answer:

Graph the function.
Question 19.
g(x) = cos(x + π) + 2
Answer:

Question 20.
g(x) = −sin x − 4
Answer:

Question 21.
g(x) = 2 sin (x + \(\frac{\pi}{2}\))
Answer:

9.5 Graphing Other Trigonometric Functions (pp. 497−504)

Graph one period of the function. Describe the graph of g as a transformation of the graph of its parent function.
Question 22.
g(x) = tan \(\frac{1}{2}\)x
Answer:

Question 23.
g(x) = 2 cot x
Answer:

Question 24.
g(x) = 4 tan 3πx
Answer:

Graph the function.
Question 25.
g(x) = 5 csc x
Answer:

Question 26.
g(x) = sec \(\frac{1}{2}\)x
Answer:

Question 27.
g(x) = 5 sec πx
Answer:

Question 28.
g(x) = \(\frac{1}{2}\) csc \(\frac{\pi}{4}\)x
Answer:

9.6 Modeling with Trigonometric Functions (pp. 505−512)

Write a function for the sinusoid.
Question 29.

Answer:

Question 30.

Answer:

Question 31.
You put a reflector on a spoke of your bicycle wheel. The highest point of the reflector is 25 inches above the ground, and the lowest point is 2 inches. The reflector makes 1 revolution per second. Write a model for the height h (in inches) of a reflector as a function of time t (in seconds) given that the reflector is at its lowest point when t = 0.
Answer:

Question 32.
The table shows the monthly precipitation P (in inches) for Bismarck, North Dakota, where t = 1 represents January. Write a model that gives P as a function of t and interpret the period of its graph.

Answer:

9.7 Using Trigonometric Identities (pp. 513−518)

Simplify the expression.
Question 33.
cot²x − cot²x cos²x
Answer:

Question 34.
\(\frac{(\sec x+1)(\sec x-1)}{\tan x}\)
Answer:

Question 35.
sin (\(\frac{\pi}{2}\) − x)tan x
Answer:

Verify the identity.
Question 36.
\(\frac{\cos x \sec x}{1+\tan ^{2} x}\) = cos²x
Answer:

Question 37.
tan (\(\frac{\pi}{2}\) − x)cot x = csc²x − 1
Answer:

9.8 Using Sum and Difference Formulas (pp. 519−524)

Find the exact value of the expression.
Question 38.
sin 75°
Answer:

Question 39.
tan(−15°)
Answer:

Question 40.
cos \(\frac{\pi}{12}\)
Answer:

Question 41.
Find tan(a + b), given that tan a = \(\frac{1}{4}\) with π < a < \(\frac{3 \pi}{2}\) and tan b = \(\frac{3}{7}\) with 0 < b < \(\frac{\pi}{2}\) .
Answer:

Solve the equation for 0 ≤ x < 2π.
Question 42.
cos (x + \(\frac{3 \pi}{4}\)) + cos (x − \(\frac{3 \pi}{4}\)) = 1
Answer:

Question 43.
tan(x + π) + cos (x + \(\frac{\pi}{2}\))= 0
Answer:

Trigonometric Ratios and Functions Chapter Test

Verify the identity.
Question 1.
\(\frac{\cos ^{2} x+\sin ^{2} x}{1+\tan ^{2} x}\) = cos²x
Answer:

Question 2.
\(\frac{1+\sin x}{\cos x}+\frac{\cos x}{1+\sin x}\) = 2 sec x
Answer:

Question 3.
cos (x + \(\frac{3 \pi}{2}\)) = sin x
Answer:

Question 4.
Evaluate sec(−300°) without using a calculator.
Answer:

Write a function for the sinusoid.
Question 5.

Answer:

Question 6.

Answer:

Graph the function. Then describe the graph of g as a transformation of the graph of its parent function.
Question 7.
g(x) = −4 tan 2x
Answer:

Question 8.
g(x) = −2 cos \(\frac{1}{3}\)x + 3
Answer:

Question 9.
g(x) = 3 csc πx
Answer:

Convert the degree measure to radians or the radian measure to degrees. Then find one positive angle and one negative angle that are coterminal with the given angle.
Question 10.
−50°
Answer:

Question 11.
\(\frac{4 \pi}{5}\)
Answer:

Question 12.
\(\frac{8 \pi}{3}\)
Answer:

Question 13.
Find the arc length and area of a sector with radius r = 13 inches and central angle θ = 40°.
Answer:

Evaluate the six trigonometric functions of the angle θ.
Question 14.

Answer:

Question 15.

Answer:

Question 16.
In which quadrant does the terminal side of θ lie when cos θ < 0 and tan θ > 0? Explain.
Answer:

Question 17.
How tall is the building? Justify your answer.

Answer:

Question 18.
The table shows the average daily high temperatures T (in degrees Fahrenheit) in Baltimore, Maryland, where m= 1 represents January. Write a model that gives T as a function of m and interpret the period of its graph.

Answer:

Trigonometric Ratios and Functions Cumulative Assessment

Question 1.
Which expressions are equivalent to 1?

Answer:

Question 2.
Which rational expression represents the ratio of the perimeter to the area of the playground shown in the diagram?

Answer:

Question 3.
The chart shows the average monthly temperatures (in degrees Fahrenheit) and the gas usages (in cubic feet) of a household for 12 months.

a. Use a graphing calculator to find trigonometric models for the average temperature y₁ as a function of time and the gas usage y₂ (in thousands of cubic feet) as a function of time. Let t = 1 represent January.
b. Graph the two regression equations in the same coordinate plane on your graphing calculator. Describe the relationship between the graphs.
Answer:

Question 4.
Evaluate each logarithm using log₂ 5 ≈ 2.322 and log₂ 3 ≈ 1.585, if necessary. Then order the logarithms by value from least to greatest.
a. log 1000
b. log₂ 15
c. ln e
d. log₂ 9
e. log₂\(\frac{5}{3}\)
f. log₂ 1
Answer:

Question 5.
Which function is not represented by the graph?

Answer:

Question 6.
Complete each statement with < or > so that each statement is true.

Answer:

Question 7.
Use the Rational Root Theorem and the graph to find all the real zeros of the function f(x) = 2x³ − x² − 13x− 6. (HSA-APR.B.3)

Answer:

Question 8.
Your friend claims −210° is coterminal with the angle \(\frac{5 \pi}{6}\). Is your friend correct? Explain your reasoning.
Answer:

Question 9.
Company A and Company B offer the same starting annual salary of $20,000. Company A gives a $1000 raise each year. Company B gives a 4% raise each year.
a. Write rules giving the salaries an and bn for your nth year of employment at Company A and Company B, respectively. Tell whether the sequence represented by each rule is arithmetic, geometric, or neither.
b. Graph each sequence in the same coordinate plane.
c. Under what conditions would you choose to work for Company B?
d. After 20 years of employment, compare your total earnings.
Answer:

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Big Ideas Math Book Geometry Answer Key Chapter 8 Similarity

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• Similarity Maintaining Mathematical Proficiency – Page 415
• Similarity Mathematical Practices – Page 416
• 8.1 Similar Polygons – Page 417
• Lesson 8.1 Similar Polygons – Page (418-426)
• Exercise 8.1 Similar Polygons – Page (423-426)
• 8.2 Proving Triangle Similarity by AA – Page 427
• Lesson 8.2 Proving Triangle Similarity by AA – Page (428-432)
• Exercise 8.2 Proving Triangle Similarity by AA – Page (431-432)
• 8.1 & 8.2 Quiz – Page 434
• 8.3 Proving Triangle Similarity by SSS and SAS – Page 435
• Lesson 8.3 Proving Triangle Similarity by SSS and SAS – Page (436-444)
• Exercise 8.3 Proving Triangle Similarity by SSS and SAS – Page (441-444)
• 8.4 Proportionality Theorems – Page 445
• Lesson 8.4 Proportionality Theorems – Page (446-452)
• Exercise 8.4 Proportionality Theorems – Page (450-452)
• Similarity Chapter Review – Page (454-456)
• Similarity Chapter Test – Page 457
• Similarity Cumulative Assessment – Page (458-459)

Similarity Maintaining Mathematical Proficiency

Tell whether the ratios form a proportion.

Question 1.
\(\frac{5}{3}, \frac{35}{21}\)

Answer:
Yes, the ratios \(\frac{5}{3}, \frac{35}{21}\) form a proportion.

Explanation:
A proportion means two ratios are equal.
So, cross product of \(\frac{5}{3}, \frac{35}{21}\) is 21 x 5 = 105 = 35 x 3
Therefore, \(\frac{5}{3}, \frac{35}{21}\) form a proportion.

Question 2.
\(\frac{9}{24}, \frac{24}{64}\)

Answer:
Yes, the ratios \(\frac{9}{24}, \frac{24}{64}\) form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 24 x 24 = 576 = 64 x 9
Therefore, the ratios \(\frac{9}{24}, \frac{24}{64}\) form a proportion.

Question 3.
\(\frac{8}{56}, \frac{6}{28}\)

Answer:
The ratios \(\frac{8}{56}, \frac{6}{28}\) do not form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 56 x 6 = 336, 28 x 8 = 224
Therefore, the ratios \(\frac{8}{56}, \frac{6}{28}\) do not form a proportion.

Question 4.
\(\frac{18}{4}, \frac{27}{9}\)

Answer:
The ratios \(\frac{18}{4}, \frac{27}{9}\) do not form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 9 x 18 = 162, 27 x 4 = 108
Therefore, the ratios \(\frac{18}{4}, \frac{27}{9}\) do not form a proportion.

Question 5.
\(\frac{15}{21}, \frac{55}{77}\)

Answer:
The ratios \(\frac{15}{21}, \frac{55}{77}\) form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 15 x 77 = 1155, 55 x 21 = 1155
Therefore, the ratios \(\frac{15}{21}, \frac{55}{77}\) form a proportion.

Question 6.
\(\frac{26}{8}, \frac{39}{12}\)

Answer:
The ratios \(\frac{26}{8}, \frac{39}{12}\) form a proportion.

Explanation:
If the cross product of two ratios is equal, then it forms a proportion.
So, 26 x 12 = 312, 39 x 8 = 312
Therefore, the ratios \(\frac{26}{8}, \frac{39}{12}\) form a proportion.

Find the scale factor of the dilation.

Question 7.

Answer:
k = \(\frac { 3 }{ 7 } \)

Explanation:
The scale factor k = \(\frac { CP’ }{ CP } \)
= \(\frac { 6 }{ 14 } \)
= \(\frac { 3 }{ 7 } \)

Question 8.

Answer:
k = \(\frac { 3 }{ 8 } \)

Explanation:
The scale factor k = \(\frac { CP }{ CP’ } \)
= \(\frac { 9 }{ 24 } \)
= \(\frac { 3 }{ 8 } \)

Question 9.

Answer:
k = \(\frac { 1 }{ 2 } \)

Explanation:
The scale factor k = \(\frac { MK }{ M’K’ } \)
= \(\frac { 14 }{ 28 } \)
= \(\frac { 1 }{ 2 } \)

Question 10.
ABSTRACT REASONING
If ratio X and ratio Y form a proportion and ratio Y and ratio Z form a proportion, do ratio X and ratio Z form a proportion? Explain our reasoning.

Answer:
Yes, ratio X and ratio Z form a proportion.

Explanation:
If ratios are proportional means they are equal.
So, ratio X and ratio Y form a proportion that means X = Y
ratio Y and ratio Z form a proportion that means Y = Z
From the above two equations, we can say that X = Z
So, ratio X and ratio Z also form a proportion.

Similarity Mathematical Practices

Monitoring Progress

Question 1.
Find the perimeter and area of the image when the trapezoid is dilated by a scale factor of
(a) 2, (b) 3, and (c) 4.

Answer:
(a) Perimeter is 32 cm, area is 48 sq cm.
(b) Perimeter is 48 cm, the area is 108 sq cm.
(c) Perimeter is 64 cm, the area is 192 sq cm.

Explanation:
The perimeter of trapezoid P = 2 + 5 + 6 + 3 = 16 cm
Area of trapezoid A = \(\frac { (2 + 6)3 }{ 2 } \)
= \(\frac { 3(8) }{ 2 } \)
= \(\frac { 24 }{ 2 } \)
= 12 sq cm
(a) If scale factor k = 2, then
Perimeter = kP
= 2 x 16 = 32
Area = k²A
= 2² x 12
= 4 x 12
= 48
(b) If scale factor k = 3, then,
Perimeter = kP
= 3 x 16 = 48 cm
Area = k²A
= 3² x 12 = 9 x 12 = 108
(c) If scale factor k = 4, then
Perimeter = kP
= 4 x 16 = 64
Area = k²A
= 4² x 12 = 16 x 12 = 192

Question 2.
Find the perimeter and area of the image when the parallelogram is dilated by a scale factor of
(a) 2, (b) 3, and (c) \(\frac{1}{2}\)

Answer:
(a) Perimeter is 28 ft, area is 32 sq ft
(b) Perimeter is 42 ft, area is 72 sq ft
(c) Perimeter is 7 ft, area is 2 sq ft

Explanation:
Perimeter of parallelogram P = 2(2 + 5) = 7 x 2 = 14 ft
Area of the parallelogram = 2 x 4 = 8 sq ft
(a) If scale factor k = 2, then
Perimeter = kP
= 2 x 14 = 28
Area = k²A
= 2² x 8 = 4 x 8 = 32
(b) If scale factor k = 3, then
Perimeter = kP
= 3 x 14 = 42
Area = k²A
= 3² x 8 = 72
(c) If scale factor k = \(\frac{1}{2}\), then
Perimeter = kP
= \(\frac{1}{2}\) x 14 = 7
Area = k²A
= \(\frac{1}{2²}\) x 8 = \(\frac{1}{4}\) x 8 = 2

Question 3.
A rectangular prism is 3 inches wide, 4 inches long, and 5 inches tall. Find the surface area and volume of the image of the prism when it is dilated by a scale factor of
(a) 2, (b) 3, and (c) 4.

Answer:
(a) Surface area is 376 sq in, volume is 480 cubic in
(b) Surface area is 846 sq in, volume is 1620 cubic in
(c) Surface area is 1504 sq in, volume is 3840 cubic in

Explanation:
The surface area of the rectangular prism A = 2(3 x 4 + 4 x 5 + 5 x 3)
= 2(12 + 20 + 15) = 2(47) = 94 in
Volume of the rectangular prism V = 3 x 4 x 5 = 60 in³
(a) If the scale factor k = 2, then
Surface Area = k²A
= 2² x 94 = 4 x 94 = 376 sq in
Volume = k³V
= 2³ x 60 = 8 x 60 = 480 cubic in
(b) If the scale factor k = 3, then
Surface Area = k²A
= 3² x 94 = 9 x 94 = 846
Volume = k³V
= 3³ x 60 = 27 x 60 = 1620
(c) If the scale factor k = 4, then
Surface Area = k²A
= 4² x 94 = 16 x 94 = 1504
Volume = k³V
= 4³ x 60 = 64 x 60 = 3840

8.1 Similar Polygons

Exploration 1

Comparing Triangles after a Dilation

Work with a partner: Use dynamic geometry software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using an scale factor k and an center of dilation.

a. Compare the corresponding angles of ∆A’B’C and ∆ABC.
Answer:

b. Find the ratios of the lengths of the sides of ∆A’B’C’ to the lengths of the corresponding sides of ∆ABC. What do you observe?
Answer:

c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results?
Answer:

Exploration 2

Comparing Triangles after a Dilation

Work with a partner: Use dynamic geometry Software to draw any ∆ABC. Dilate ∆ABC to form a similar ∆A’B’C’ using any scale factor k and any center of dilation.

a. Compare the perimeters of ∆A’B’C and ∆ABC. What do you observe?
Answer:

b. Compare the areas of ∆A’B’C’ and ∆ABC. What do you observe?
Answer:

c. Repeat parts (a) and (b) for several other triangles, scale factors, and centers of dilation. Do you obtain similar results?
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Communicate Your Answer

Question 3.
How are similar polygons related?

Answer:
if two polygons are similar means they have the same shape, corresponding angles are congruent and the ratios of lengths of their corresponding sides are equal. The common ratio is called the scale factor.

Question 4.
A ∆RST is dilated by a scale factor of 3 to form ∆R’S’T’. The area of ∆RST is 1 square inch. What is the area of ∆R’S’T’?

Answer:
Area of ∆R’S’T’ = 9 sq in

Explanation:
Given that,
Area of ∆RST = 1 sq inch
Scale factor k = 3
Area of ∆R’S’T’ = k² x Area of ∆RST
= 3² x 1 = 9 x 1 = 9

Lesson 8.1 Similar Polygons

Monitoring Progress

Question 1.
In the diagram, ∆JKL ~ ∆PQR. Find the scale factor from ∆JKL to ∆PQR. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

Answer:
The pairs of congruent angles are ∠K = ∠Q, ∠J = ∠P, ∠ L = ∠R
The scale factor is \(\frac { 3 }{ 2 } \)
The ratios of the corresponding side lengths in a statement of proportionality are \(\frac { PQ }{ JK } = \frac { PR }{ JL } = \frac { QR }{ LK } \)

Explanation:
Given that,
∆JKL ~ ∆PQR
The pairs of congruent angles are ∠K = ∠Q, ∠J = ∠P, ∠ L = ∠R
To find the scale factor,
\(\frac { PQ }{ JK } = \frac { 9 }{ 6 } \) = \(\frac { 3 }{ 2 } \), \(\frac { PR }{ JL } = \frac { 12 }{ 8 } \) = \(\frac { 3 }{ 2 } \), \(\frac { QR }{ LK } = \frac { 6 }{ 4 } \) = \(\frac { 3 }{ 2 } \)
So, the scale factor is \(\frac { 3 }{ 2 } \)

Question 2.
Find the value of x.

ABCD ~ QRST

Answer:
x = 2

Explanation:
The triangles are similar, so corresponding side lengths are proportional.
\(\frac { RS }{ BC } \) = \(\frac { AB }{ QR } \)
\(\frac { 4 }{ x } \) = \(\frac { 12 }{ 6 } \)
\(\frac { 4 }{ x } \) = 2
4 = 2x
x = 2

Question 3.
Find KM

∆JKL ~ ∆EFG

Answer:
KM = 42

Explanation:
Scale factor = \(\frac { JM }{ GH } \)
= \(\frac { 48 }{ 40 } \)
= \(\frac { 6 }{ 5 } \)
Because the ratio of the lengths of the altitudes in similar triangles is equal to the scale factor, you can write the following proportion
\(\frac { KM }{ HF } \) = \(\frac { 6 }{ 5 } \)
\(\frac { KM }{ 35 } \) = \(\frac { 6 }{ 5 } \)
KM = \(\frac { 6 }{ 5 } \) x 35
KM = 42

Question 4.
The two gazebos shown are similar pentagons. Find the perimeter of Gazebo A.

Answer:
Perimeter of Gazebo A = 46 m

Explanation:
Scale factor = \(\frac { AB }{ FG } \)
= \(\frac { 10 }{ 15 } \)
= \(\frac { 2 }{ 3 } \)
So, \(\frac { AE }{ FK } \) = \(\frac { 2 }{ 3 } \)
\(\frac { x }{ 18 } \) = \(\frac { 2 }{ 3 } \)
x = 12
\(\frac { ED }{ KJ } \) = \(\frac { 2 }{ 3 } \)
\(\frac { ED }{ 15 } \) = \(\frac { 2 }{ 3 } \)
ED = 10
\(\frac { DC }{ JH } \) = \(\frac { 2 }{ 3 } \)
\(\frac { DC }{ 12 } \) = \(\frac { 2 }{ 3 } \)
DC = 8
\(\frac { BC }{ GH } \) = \(\frac { 2 }{ 3 } \)
\(\frac { BC }{ 9 } \) = \(\frac { 2 }{ 3 } \)
BC = 6
Therefore, perimeter = 6 + 8 + 10 + 12 + 10 = 46

Question 5.
In the diagram, GHJK ~ LMNP. Find the area of LMNP.

Area of GHJK = 84m²

Answer:
Area of LMNP = 756 m²

Explanation:
As shapes are similar, their corresponding side lengths are proportional.
Scale Factor k = \(\frac { NP }{ JK } \)
= \(\frac { 21 }{ 7 } \)
= 3
Area of LMNP = k² x Area of GHJK
= 3² x 84
= 756 m²

Question 6.
Decide whether the hexagons in Tile Design 1 are similar. Explain.

Answer:

Question 7.
Decide whether the hexagons in Tile Design 2 are similar. Explain.

Answer:

Exercise 8.1 Similar Polygons

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
For two figures to be similar, the corresponding angles must be ____________ . and the corresponding side lengths must be _____________ .

Answer:

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

What is the scale factor?
Answer:
Scale Factor = \(\frac{image-length}{actual-length}\)
= \(\frac{3}{12}\)= \(\frac{4}{16}\)= \(\frac{5}{20}\)
= \(\frac{1}{4}\)

What is the ratio of their areas?

Answer:

What is the ratio of their corresponding side lengths?
Answer:

What is the ratio of their perimeters?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

Question 3.
∆ABC ~ ∆LMN

Answer:

Question 4.
DEFG ~ PQRS

Answer:

In Exercises 5-8, the polygons are similar. Find the value of x.

Question 5.

Answer:

Question 6.

Answer:
x = 20

Explanation:
\(\frac { DF }{ GJ } \) = \(\frac { DE }{ GH } \)
\(\frac { 16 }{ 12 } \) = \(\frac { x }{ 15 } \)
x = \(\frac { 16 x 15 }{ 12 } \) = \(\frac { 240 }{ 12 } \)
x = 20

Question 7.

Answer:

Question 8.

Answer:
x = 12

Explanation:
\(\frac { PN }{ KJ } \) = \(\frac { MN }{ JH } \)
\(\frac { x }{ 8 } \) = \(\frac { 9 }{ 6 } \)
x = \(\frac { 9 x 8 }{ 6 } \) = \(\frac { 72 }{ 6 } \)
x = 12

In Exercises 9 and 10, the black triangles are similar. Identify the type of segment shown in blue and find the value of the variab1e.

Question 9.

Answer:

Question 10.

Answer:

Explanation:
\(\frac { y }{ 18 } \) = \(\frac { y – 1 }{ 16 } \)
18(y – 1) = 16y
18y – 18 = 16y
18y – 16y = 18
2y = 18
y = 9

In Exercises 11 and 12, RSTU ~ ABCD. Find the ratio of their perimeters.

Question 11.

Answer:

Question 12.

Answer:
\(\frac { RS + ST + TU + UR }{ AB + BC + CD + DA } \) = \(\frac { RS }{ AB } \) = \(\frac { 18 }{ 24 } \)
The ratio of perimeter is \(\frac { 3 }{ 4 } \).

In Exercises 13-16, two polygons are similar. The perimeter of one polygon and the ratio of the corresponding side lengths are given. Find the perimeter of the other polygon.

Question 13.
perimeter of smaller polygon: 48 cm: ratio: \(\frac{2}{3}\)
Answer:

Question 14.
perimeter of smaller polygon: 66 ft: ratio: \(\frac{3}{4}\)

Answer:
The perimeter of larger polygon is 88 ft.

Explanation:
\(\frac { smaller }{ larger } \) = \(\frac { 66 }{ x } \) = \(\frac{3}{4}\)
66 x 4 = 3x
3x = 264
x = \(\frac { 264 }{ 3 } \) = 88

Question 15.
perimeter of larger polygon: 120 yd: rttio: \(\frac{1}{6}\)
Answer:

Question 16.
perimeter of larger polygon: 85 m; ratio: \(\frac{2}{5}\)

Answer:
The perimeter of smaller polygon is 34 m.

Explanation:
\(\frac { smaller }{ larger } \) = \(\frac { x }{ 85 } \) = \(\frac{2}{5}\)
85 x 2 = 5x
5x = 170
x = \(\frac { 170 }{ 5 } \) = 34

Question 17.
MODELING WITH MATHEMATICS
A school gymnasium is being remodeled. The basketball court will be similar to an NCAA basketball court, which has a length of 94 feet and a width of 50 feet. The school plans to make the width of the new court 45 feet. Find the perimeters of ail NCAA court and of the new court in the school.
Answer:

Question 18.
MODELING WITH MATHEMATICS
Your family has decided to put a rectangular patio in your backyard. similar to the shape of your backyard. Your backyard has a length of 45 feet and a width of 20 feet. The length of your new patio is 18 feet. Find the perimeters of your backyard and of the patio.

Answer:
The perimeter of the backyard is 130 ft.
Perimeter of patio is 52 ft

Explanation:
Draw a rectangle to represent the patio and a larger rectangle to represent our backyard and its going to similar figures

Perimeter of backyard = 2(45 + 20) = 2(65)
= 130 ft
Scale factor = \(\frac { 18 }{ 45 } \) = \(\frac { 2 }{ 5 } \)
So, \(\frac { perimeter of patio }{ perimeter of backyard } \) = \(\frac { 2 }{ 5 } \)
\(\frac { perimeter of patio }{ 130 } \) = \(\frac { 2 }{ 5 } \)
Perimeter of patio = \(\frac { 260 }{ 5 } \) = 52 ft

In Exercises 19-22, the polygons are similar. The area of one polygon is given. Find the area of the other polygon.

Question 19.

Answer:

Question 20.

Answer:
Area of the larger triangle is 90 cm²

Explanation:
\(\frac { 10 }{ A } \) = (\(\frac { 4 }{ 12 } \))²
\(\frac { 10 }{ A } \) = \(\frac { 1 }{ 9 } \)
A = 10 x 9
A = 90

Question 21.

Answer:

Question 22.

Answer:
Area of smaller triangle = 6 sq cm

Explanation:
\(\frac { A }{ 96 } \) = (\(\frac { 3 }{ 12 } \))²
\(\frac { A }{ 96 } \) = \(\frac { 1 }{ 16 } \)
16A = 96
A = \(\frac { 96 }{ 16 } \)
A = 6

Question 23.
ERROR ANALYSIS
Describe and correct the error in finding the perimeter of triangle B. The triangles are similar.

Answer:

Question 24.
ERROR ANALYSIS
Describe and correct the error in finding the area of triangle B. The triangles are similar.

Answer:
Because the first ratio has a side of A over the side length of B, the square of the second ratio should have the area of B over the area of A.
\(\frac { 24 }{ x } \) = (\(\frac { 6 }{ 18 } \))²
\(\frac { 24 }{ x } \) = \(\frac { 1 }{ 9 } \)
x = 24 x 9
x = 216

In Exercises 25 and 26, decide whether the red and blue polygons are similar.

Question 25.

Answer:

Question 26.

Answer:
Yes
Because both shapes are apparent and their side lengths are proportional and their corresponding angles are congruent.

Question 27.
REASONING
Triangles ABC and DEF are similar. Which statement is correct? Select all that apply.
(A) \(\frac{B C}{E F}=\frac{A C}{D F}\)
(B) \(\frac{A B}{D E}=\frac{C A}{F E}\)
(C) \(\frac{A B}{E F}=\frac{B C}{D E}\)
(D) \(\frac{C A}{F D}=\frac{B C}{E F}\)
Answer:

ANALYZING RELATIONSHIPS
In Exercises 28 – 34, JKLM ~ EFGH.

Question 28.
Find the scale factor of JKLM to EFGH.

Answer:
scale factor = \(\frac { EF }{ JK } \) =\(\frac { 8 }{ 20 } \)
k = \(\frac { 2 }{ 5 } \)

Question 29.
Find the scale factor of EFGH to JKLM.
Answer:

Question 30.
Find the values of x, y, and z.

Answer:
x = \(\frac { 55 }{ 2 } \)
y = 12
z = 65°

Explanation:
\(\frac { KL }{ GF } \) = \(\frac { x }{ 11 } \) = \(\frac { 5 }{ 2 } \)
2x = 55
x = \(\frac { 55 }{ 2 } \)
\(\frac { MJ }{ HE } \) = \(\frac { 30 }{ y } \) = \(\frac { 5 }{ 2 } \)
5y = 60
y = 12

Question 31.
Find the perimeter of each polygon.
Answer:

Question 32.
Find the ratio of the perimeters of JKLM to EFGH.

Answer:
The perimeter of JKLM : Perimeter of EFGH = 85 : 34

Question 33.
Find the area of each polygon.
Answer:

Question 34.
Find the ratio of the areas of JKLM to EFGH.

Answer:
Area of JKLM : Area of EFGH = 378.125 : 60.5 = 25 : 4

Question 35.
USING STRUCTURE
Rectangle A is similar to rectangle B. Rectangle A has side lengths of 6 and 12. Rectangle B has a side length of 18. What are the possible values for the length of the other side of rectangle B? Select all that apply.
(A) 6
(B) 9
(C) 24
(D) 36
Answer:

Question 36.
DRAWING CONCLUSIONS
In table tennis, the table is a rectangle 9 feet long and 5 feet wide. A tennis Court is a rectangle 78 feet long and 36 feet wide. Are the two surfaces similar? Explain. If so, find the scale factor of the tennis court to the table.

Answer:
The tennis table and court are not similar

Explanation:
If two figures are similar then their angles are congruent and sides are proportional.
If the tennis court and table are similar, then
\(\frac { length of table }{ length of court } \) = \(\frac { width of table }{ width of court } \)
\(\frac { 9 }{ 78 } \) = \(\frac { 5 }{ 36 } \)
9 • 36 = 5 • 78
324 = 390
So, Table and court are not similar.

MATHEMATICAL CONNECTIONS
In Exercises 37 and 38, the two polygons are similar. Find the values of x and y.

Question 37.

Answer:

Question 38.

Answer:
x = 7.5

Explanation:
\(\frac { x }{ 5 } \) = \(\frac { 6 }{ 4 } \)
x = \(\frac { 15 }{ 2 } \)

ATTENDING TO PRECISION
In Exercises 39 – 42. the figures are similar. Find the missing corresponding side length.

Question 39.
Figure A has a pen meter of 72 meters and one of the side lengths is 18 meters. Figure B has a perimeter of 120 meters.
Answer:

Question 40.
Figure A has a perimeter of 24 inches. Figure B has a perimeter of 36 inches and one of the side lengths is 12 inches.

Answer:
The corresponding side length of figure A is 8 in

Explanation:
\(\frac { Perimeter of A }{ Perimeter of B } \) = \(\frac { Side length of A }{ Side length of B } \)
\(\frac { 24 }{ 36 } \) = \(\frac { x }{ 12 } \)
\(\frac { 2 }{ 3 } \) = \(\frac { x }{ 12 } \)
12 • 2 = 3x
x = 8

Question 41.
Figure A has an area of 48 square feet and one of the side lengths is 6 feet. Figure B has an area of 75 square feet.
Answer:

Question 42.
Figure A has an area of 18 square feet. Figure B has an area of 98 square feet and one of the side lengths is 14 feet.

Answer:
The corresponding side length of figure A is 6 ft.

Explanation:
\(\frac { Area of A }{ Area of B } \) = (\(\frac { Side length of A }{ Side length of B } \))²
\(\frac { 18 }{ 98 } \) = (\(\frac { x }{ 14 } \))²
\(\frac { 9 }{ 49 } \) = \(\frac { x² }{ 196 } \)
x² = 36
x = 6

CRITICAL THINKING
In Exercises 43-48, tell whether the polygons are always, sometimes, or never similar.

Question 43.
two isosceles triangles
Answer:

Question 44.
two isosceles trapezoids

Answer:
Two isosceles trapezoids are sometimes similar.

Question 45.
two rhombuses
Answer:

Question 46.
two squares

Answer:
Two squares are always similar.

Question 47.
two regular polygons
Answer:

Question 48.
a right triangle and an equilateral triangle

Answer:
A right triangle and an equilateral triangle are never similar.

Question 49.
MAKING AN ARGUMENT
Your sister claims that when the side lengths of two rectangles are proportional, the two rectangles must be similar. Is she correct? Explain your reasoning.
Answer:

Question 50.
HOW DO YOU SEE IT?
You shine a flashlight directly on an object to project its image onto a parallel screen. Will the object and the image be similar? Explain your reasoning.

Answer:
The object and image are similar.

Question 51.
MODELING WITH MATHEMATICS
During a total eclipse of the Sun, the moon is directly in line with the Sun and blocks the Sun’s rays. The distance DA between Earth and the Sun is 93,00,000 miles. the distance DE between Earth and the moon is 2,40,000 miles, and the radius AB of the Sun is 432,5000 miles. Use the diagram and the given measurements to estimate the radius EC of the moon.

Answer:

Question 52.
PROVING A THEOREM
Prove the Perimeters of Similar Polygons Theorem (Theorem 8.1) for similar rectangles. Include a diagram in your proof.

Answer:

\(\frac { PQ + QR + RS + SP }{ KL + LM + MN + NK } \) = \(\frac { PQ }{ KL } \) = \(\frac { QR }{ LM } \) = \(\frac { RS }{ MN } \) = \(\frac { SP }{ NK } \)

Question 53.
PROVING A THEOREM
Prove the Areas of Similar Polygons Theorem (Theorem 8.2) for similar rectangles. Include a diagram in our proof.
Answer:

Question 54.
THOUGHT PROVOKING
The postulates and theorems in this book represent Euclidean geometry. In spherical geometry. all points are points on the surface of a sphere. A line is a circle on the sphere whose diameter is equal to the diameter of the sphere. A plane is the surface of the sphere. In spherical geometry, is it possible that two triangles are similar but not congruent? Explain your reasoning.
Answer:

Question 55.
CRITICAL THINKING
In the diagram, PQRS is a square, and PLMS ~ LMRQ. Find the exact value of x. This value is called the golden ratio. Golden rectangles have their length and width in this ratio. Show that the similar rectangles in the diagram are golden rectangles.

Answer:

Question 56.
MATHEMATICAL CONNECTIONS
The equations of the lines shown are y = \(\frac{4}{3}\)x + 4 and y = \(\frac{4}{3}\)x – 8. Show that ∆AOB ~ ∆COD.

Answer:
The two lines slopes are equal and triangles angles are congruent and side lengths are proportional. So, triangles are similar.

Maintaining Mathematical proficiency

Find the value of x.

Question 57.

Answer:

Question 58.

Answer:
x = 66°

Explanation:
x + 24 + 90 = 180
x + 114 = 180
x = 180 – 114
x = 66

Question 59.

Answer:

Question 60.

Answer:
x = 60°

Explanation:
x + x + x = 180
3x = 180
x = 60

8.2 Proving Triangle Similarity by AA

Exploration 1

Comparing Triangles

Work with a partner. Use dynamic geometry software.

a. Construct ∆ABC and ∆DEF So that m∠A = m∠D = 106°, m∠B = m∠E = 31°, and ∆DEF is not congruent to ∆ABC.

Answer:
m∠C ≠ m∠F

b. Find the third angle measure and the side lengths of each triangle. Copy the table below and record our results in column 1.

Answer:

c. Are the two triangles similar? Explain.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results in constructing arguments.
Answer:

d. Repeat parts (a) – (c) to complete columns 2 and 3 of the table for the given angle measures.
Answer:

e. Complete each remaining column of the table using your own choice of two pairs of equal corresponding angle measures. Can you construct two triangles in this way that are not similar?
Answer:

f. Make a conjecture about any two triangles with two pairs of congruent corresponding angles.
Answer:

Communicate Your Answer

Question 2.
What can you conclude about two triangles when you know that two pairs of corresponding angles are congruent?
Answer:

Question 3.
Find RS in the figure at the left.
Answer:

Lesson 8.2 Proving Triangle Similarity by AA

Monitoring Progress

Show that the triangles are similar. Write a similarity statement.

Question 1.
∆FGH and ∆RQS

Answer:
∆FGH and ∆RQS are similar by the AA similarity theorem.

Question 2.
∆CDF and ∆DEF

Answer:

Question 3.
WHAT IF?
Suppose that \(\overline{S R}\) \(\overline{T U}\) in Example 2 part (b). Could the triangles still be similar? Explain.
Answer:

Question 4.
WHAT IF?
A child who is 58 inches tall is standing next to the woman in Example 3. How long is the child’s shadow’?
Answer:

Question 5.
You are standing outside, and you measure the lengths 0f the shadows cast by both you and a tree. Write a proportion showing how you could find the height of the tree.
Answer:

Exercise 8.2 Proving Triangle Similarity by AA

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
If two angles of one triangle are congruent to two angles of another triangle. then the triangles are __________ .
Answer:

Question 2.
WRITING
Can you assume that corresponding sides and corresponding angles of any two similar triangles are congruent? Explain.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6. determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.

Question 3.

Answer:

Question 4.

Answer:

Question 5.

Answer:

Question 6.

Answer:

In Exercises 7 – 10. show that the two triangles are similar.

Question 7.

Answer:

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

In Exercises 11 – 18, use the diagram to copy and complete the statement.

Question 11.
∆CAG ~ _________
Answer:

Question 12.
∆DCF ~ _________
Answer:

Question 13.
∆ACB ~ _________
Answer:

Question 14.
m∠ECF = _________
Answer:

Question 15.
m∠ECD = _________
Answer:

Question 16.
CF = _________
Answer:

Question 17.
BC = _________
Answer:

Question 18.
DE = _________
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in using the AA Similarity Theorem (Theorem 8.3).

Answer:

Question 20.
ERROR ANALYSIS
Describe and correct the error in finding the value of x.

Answer:

Question 21.
MODELING WITH MATHEMATICS
You can measure the width of the lake using a surveying technique, as shown in the diagram. Find the width of the lake, WX. Justify your answer.

Answer:

Question 22.
MAKING AN ARGUMENT
You and your cousin are trying to determine the height of a telephone pole. Your cousin tells you to stand in the pole’s shadow so that the tip of your shadow coincides with the tip of the pole’s shadow. Your Cousin claims to be able to use the distance between the tips of the shadows and you, the distance between you and the pole, and your height to estimate the height of the telephone pole. Is this possible? Explain. Include a diagram in your answer.
Answer:

REASONING
In Exercises 23 – 26, is it possible for ∆JKL and ∆XYZ to be similar? Explain your reasoning.

Question 23.
m∠J = 71°, m∠K = 52°, m∠X = 71°, and m∠Z = 57°
Answer:

Question 24.
∆JKL is a right triangle and m∠X + m∠Y= 150°.
Answer:

Question 25.
m∠L = 87° and m∠Y = 94°
Answer:

Question 26.
m∠J + m∠K = 85° and m∠Y + m∠Z = 80°
Answer:

Question 27.
MATHEMATICAL CONNECTIONS
Explain how you can use similar triangles to show that any two points on a line can be used to find its slope.

Answer:

Question 28.
HOW DO YOU SEE IT?
In the diagram, which triangles would you use to find the distance x between the shoreline and the buoy? Explain your reasoning.

Answer:

Question 29.
WRITING
Explain why all equilateral triangles are similar.
Answer:

Question 30.
THOUGHT PROVOKING
Decide whether each is a valid method of showing that two quadrilaterals are similar. Justify your answer.
a. AAA
Answer:

b. AAAA
Answer:

Question 31.
PROOF
Without using corresponding lengths in similar polygons. prove that the ratio of two corresponding angle bisectors in similar triangles is equal to the scale factor.
Answer:

Question 32.
PROOF
Prove that if the lengths of two sides of a triangle are a and b, respectively, then the lengths of the corresponding altitudes to those sides are in the ratio \(\frac{b}{a}\).
Answer:

Question 33.
MODELING WITH MATHEMATICS
A portion of an amusement park ride is shown. Find EF. Justify your answer.

Answer:

Maintaining Mathematical Practices

Determine whether there is enough information to prove that the triangles are congruent. Explain your reasoning.

Question 34.

Answer:

Question 35.

Answer:

Question 36.

Answer:

8.1 & 8.2 Quiz

List all pairs of congruent angles. Then write the ratios of the corresponding side lengths in a statement of proportionality.

Question 1.
∆BDG ~ ∆MPQ

Answer:

Question 2.
DEFG ~ HJKL

Answer:

The polygons are similar. Find the value of x.

Question 3.

Answer:

Question 4.

Answer:

Determine whether the polygons are similar. If they are, write a similarity statement. Explain your reasoning. (Section 8.1 and Section 8.2)

Question 5.

Answer:

Question 6.

Answer:

Question 7.

Answer:

Show that the two triangles are similar.

Question 8.

Answer:

Question 9.

Answer:

Question 10.

Answer:

Question 11.
The dimensions of an official hockey rink used by the National Hockey League (NHL) are 200 feet by 85 feet. The dimensions of an air hockey table are 96 inches by 408 inches. Assume corresponding angles are congruent. (Section 8.1)
a. Determine whether the two surfaces are similar.
Answer:

b. If the surfaces are similar, find the ratio of their perimeters and the ratio ol their areas. If not, find the dimensions of an air hockey table that are similar to an NHL hockey rink.
Answer:

Question 12.
you and a friend buy camping tents made by the same company but in different sizes and colors. Use the information given in the diagram to decide whether the triangular faces of the tents are similar. Explain your reasoning. (Section 8.2)

Answer:

8.3 Proving Triangle Similarity by SSS and SAS

Exploration 1

Deciding Whether Triangles Are Similar

Work with a partner: Use dynamic geometry software.

a. Construct ∆ABC and ∆DEF with the side lengths given in column 1 of the table below.

Answer:

b. Copy the table and complete column 1.
Answer:

c. Are the triangles similar? Explain your reasoning.
Answer:

d. Repeat parts (a) – (c) for columns 2 – 6 in the table.
Answer:

e. How are the corresponding side lengths related in each pair of triangles that are similar? Is this true for each pair of triangles that are not similar?
Answer:

f. Make a conjecture about the similarity of two triangles based on their corresponding side lengths.
CONSTRUCTING VIABLE ARGUMENTS
To be proficient in math, you need to analyze situations by breaking them into cases and recognize and use counter examples.
Answer:

g. Use your conjecture to write another set of side lengths of two similar triangles. Use the side lengths to complete column 7 of the table.
Answer:

Exploration 2

Deciding Whether Triangles Are Similar

Work with a partner: Use dynamic geometry software. Construct any ∆ABC.
a. Find AB, AC, and m∠A. Choose any positive rational number k and construct ∆DEF so that DE = k • AB, DF = k • AC, and m∠D = m∠A.
Answer:

b. Is ∆DEF similar to ∆ABC? Explain your reasoning.
Answer:

c. Repeat parts (a) and (b) several times by changing ∆ABC and k. Describe your results.
Answer:

Communicate Your Answer

Question 3.
What are two ways to use corresponding sides of two triangles to determine that the triangles are similar?
Answer:

Lesson 8.3 Proving Triangle Similarity by SSS and SAS

Monitoring progress

Use the diagram.

Question 1.
Which of the three triangles are similar? Write a similarity statement.

Answer:
The ratios are equal. So, △LMN, △XYZ are similar.
The ratios are not equal. So △LMN, △RST are not similar.

Explanation:
Compare △LMN, △XYZ by finding the ratios of corresponding side lengths
Shortest sides: \(\frac { LM }{ YZ } \) = \(\frac { 20 }{ 30 } \) = \(\frac { 2 }{ 3 } \)
Longest sides: \(\frac { LN }{ XY } \) = \(\frac { 26 }{ 39 } \) = \(\frac { 2 }{ 3 } \)
Remaining sides: \(\frac { MN }{ ZX } \) = \(\frac { 24 }{ 36 } \) = \(\frac { 2 }{ 3 } \)
The ratios are equal. So, △LMN, △XYZ are similar.
Compare △LMN, △RST by finding the ratios of corresponding side lengths
Shortest sides: \(\frac { LM }{ RS } \) = \(\frac { 20 }{ 24 } \) = \(\frac { 5 }{ 6 } \)
Longest sides: \(\frac { LN }{ ST } \) = \(\frac { 26 }{ 33 } \)
Remaining sides: \(\frac { MN }{ RT } \) = \(\frac { 24 }{ 30 } \) = \(\frac { 4 }{ 5 } \)
The ratios are not equal. So △LMN, △RST are not similar.

Question 2.
The shortest side of a triangle similar to ∆RST is 12 units long. Find the other side 1enths of the triangle.

Answer:
The other side lengths of the triangle are 15 units, 16.5 units.

Explanation:
The shortest side of a triangle similar to ∆RST is 12 units
Scale factor = \(\frac { 12 }{ 24 } \) = \(\frac { 1 }{ 2 } \)
So, other sides are 33 x \(\frac { 12 }{ 2 } \) = 16.5, 30 x \(\frac { 12 }{ 2 } \) = 15.

Explain how to show that the indicated triangles are similar.

Question 3.
∆SRT ~ ∆PNQ

Answer:
The shorter sides: \(\frac { 18 }{ 24 } \) = \(\frac { 3 }{ 4 } \)
Longer sides: \(\frac { 21 }{ 28 } \) = \(\frac { 3 }{ 4 } \)
The side lengths are proportional. So ∆SRT ~ ∆PNQ

Question 4.
∆XZW ~ ∆YZX

Answer:
∆XZW and ∆YZX are not proportional.

Explanation:
The shorter sides: \(\frac { 9 }{ 16 } \)
Longer sides: \(\frac { 15 }{ 20 } \) = \(\frac { 3 }{ 4 } \)
The side lengths are not proportional. So ∆XZW and ∆YZX are not proportional.

Exercise 8.3 Proving Triangle Similarity by SSS and SAS

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
You plan to show that ∆QRS is similar to ∆XYZ by the SSS Similarity Theorem (Theorem 8.4). Copy and complete the proportion that you will use:

Answer:

Question 2.
WHICH ONE DOESN’T BELONG?
Which triangle does not belong with the other three? Explain your reasoning.

Answer:

Monitoring progress and Modeling with Mathematics

In Exercises 3 and 4, determine whether ∆JKL or ∆RST is similar to ∆ABC.

Question 3.

Answer:

Question 4.

Answer:

In Exercises 5 and 6, find the value of x that makes ∆DEF ~ ∆XYZ.

Question 5.

Answer:

Question 6.

Answer:

In Exercises 7 and 8, verify that ∆ABC ~ ∆DEF Find the scale factor of ∆ABC to ∆DEF

Question 7.
∆ABC: BC = 18, AB = 15, AC = 12
∆DEF: EF = 12, DE = 10, DF = 8
Answer:

Question 8.
∆ABC: AB = 10, BC = 16, CA = 20
∆DEF: DE = 25, EF = 40, FD =50
Answer:

In Exercises 9 and 10. determine whether the two triangles are similar. If they are similar, write a similarity statement and find the scale factor of triangle B to triangle A.

Question 9.

Answer:

Question 10.

Answer:

In Exercises 11 and 12, sketch the triangles using the given description. Then determine whether the two triangles can be similar.

Question 11.
In ∆RST, RS = 20, ST = 32, and m∠S = 16°. In ∆FGH, GH = 30, HF = 48, and m∠H = 24°.
Answer:

Question 12.
The side lengths of ∆ABC are 24, 8x, and 48, and the side lengths of ∆DEF are 15, 25, and 6x.

Answer:
\(\frac { AB }{ DE } \) = \(\frac { AC }{ DF } \) = \(\frac { BC }{ EF } \)
\(\frac { 24 }{ 15 } \) = \(\frac { 8x }{ 25 } \)
x = 5

In Exercises 13 – 16. show that the triangles are similar and write a similarity statement. Explain your reasoning.

Question 13.

Answer:

Question 14.

Answer:

Question 15.

Answer:

Question 16.

Answer:

In Exercises 17 and 18, use ∆XYZ.

Question 17.
The shortest side of a triangle similar to ∆XYZ is 20 units long. Find the other side lengths of the triangle.
Answer:

Question 18.
The longest side of a triangle similar to ∆XYZ is 39 units long. Find the other side lengths of the triangle.
Answer:

Question 19.
ERROR ANALYSIS
Describe and correct the error in writing a similarity statement.

Answer:

Question 20.
MATHEMATICAL CONNECTIONS
Find the value of n that makes ∆DEF ~ ∆XYZ when DE = 4, EF = 5, XY = 4(n + 1), YZ = 7n – 1, and ∠E ≅ ∠Y. Include a sketch.

Answer:
\(\frac { DE }{ XY } \) = \(\frac { EF }{ YZ } \)
\(\frac { 4 }{ 4(n + 1) } \) = \(\frac { 5 }{ 7n – 1 } \)
cross multiply the fractions
4(7n – 1) = 20(n + 1)
28n – 4 = 20n + 20
28n – 20n = 20 + 4
8n = 24
n = \(\frac { 24 }{ 8 } \)
n = 3

ATTENDING TO PRECISION
In Exercises 21 – 26, use the diagram to copy and complete the statement.

Question 21.
m∠LNS = ___________
Answer:

Question 22.
m∠NRQ = ___________
Answer:
m∠NRQ = m∠NRP = 91° by the vertical congruence

Question 23.
m∠NQR = ___________
Answer:

Question 24.
RQ = ___________

Answer:
RQ = 4√3

Explanation:
Using the pythogrean theorem
NQ² = NR² + RQ²
8² = 4² + RQ²
64 = 16 + RQ²
64 – 16 = RQ²
48 = RQ²
RQ = 4√3

Question 25.
m∠NSM = ___________
Answer:

Question 26.
m∠NPR = ___________

Answer:
m∠NPR = 28°

Explanation:
m∠NPR + m∠NRP + m∠RNP = 180°
m∠NPR + 91° + 61° = 180°
m∠NPR + 152° = 180°
m∠NPR = 180° – 152°
m∠NPR = 28°

Question 27.
MAKING AN ARGUMENT
Your friend claims that ∆JKL ~ ∆MNO by the SAS Similarity Theorem (Theorem 8.5) when JK = 18, m∠K = 130° KL = 16, MN = 9, m∠N = 65°, and NO = 8, Do you support your friend’s claim? Explain your reasoning.
Answer:

Question 28.
ANALYZING RELATIONSHIPS
Certain sections of stained glass are sold in triangular, beveled pieces. Which of the three beveled pieces, if any, are similar?

Answer:
Out of three triangles, violet and blue triangles are similar.

Explanation:
Check the similarity of maroon and violet triangles.
longest sides: \(\frac { 5 }{ 7 } \)
shortest sides: \(\frac { 3 }{ 4 } \)
remaining sides: \(\frac { 3 }{ 4 } \)
The ratios are not equal. So those traingles are not similar.
Check the similarity of blue and violet triangles.
longest sides: \(\frac { 5 }{ 5.25 } \) = 1
shortest sides: \(\frac { 3 }{ 3 } \) = 1
remaining sides: \(\frac { 3 }{ 3 } \) = 1
The ratios are equal. So those traingles are similar.

Question 29.
ATTENDING TO PRECISION
In the diagram, \(\frac{M N}{M R}=\frac{M P}{M Q}\) Which of the statements must be true?
Select all that apply. Explain your reasoning.

(A) ∠1 ≅∠2
(B) \(\overline{Q R}\) || \(\overline{N P}\)
(C)∠1 ≅ ∠4
(D) ∆MNP ~ ∆MRQ
Answer:

Question 30.
WRITING
Are any two right triangles similar? Explain.

Answer:
Yes, any two right triangles can be similar. If two right triangles are similar, then the ratio of their longest, smallest and remaining side lengths must be equal and their angles must be congruent.

Question 31.
MODELING WITH MATHEMATICS
In the portion of the shuffleboard court shown, \(\frac{B C}{A C}=\frac{B D}{A E}\)

a. What additional information do you need to show that ∆BED ~ ∆ACE using the SSS Similarity Theorem (Theorem 8.4)?
b. What additional information do, you need to show that ∆BCD ~ ∆ACE using the SAS Similarity Theorem (Theorem 8.5)?
Answer:

Question 32.
PROOF
Given that ∆BAC is a right triangle and D, E, and F are midpoints. prove that m∠DEF = 90°.

Answer:
By observing the triangle ABC, m∠BAC = 90°
Join the midpoints of the sides of the triangle.
m∠DEF = 90°

Question 33.
PROVING A THEOREM
Write a two-column proof of the SAS Similarity Theorem (Theorem 8.5).
Given ∠A ≅ ∠D, \(\frac{A B}{D E}=\frac{A C}{D F}\)
Prove ∆ABC ~ ∆DEF

Answer:

Question 34.
CRITICAL THINKING
You are given two right triangles with one pair of corresponding legs and the pair of hypotenuses having the same length ratios.
a. The lengths of the given pair of corresponding legs are 6 and 18, and the lengths of the hypotenuses are 10 and 30. Use the Pythagorean Theorem to find the lengths of the other pair of corresponding legs. Draw a diagram.
Answer:

b. Write the ratio of the lengths of the second pair of corresponding legs.
Answer:
First find the length of AC using pythagorean theorem
AC² + AB² = BC²
AC² + 36 = 100
AC² = 64
AC = 8
Find the length of DF using pythagorean theorem
DF² + DE² = EF²
DF² + 18² = 30²
DF² = 900 – 324
DF² = 576
DF= 24

c. Are these triangles similar? Does this suggest a Hypotenuse-Leg Similarity Theorem for right triangles? Explain.
Answer:
k = \(\frac { AC }{ DF } \) = \(\frac { 8 }{ 24 } \) = \(\frac { 1 }{ 3 } \)
k = \(\frac { AB }{ DE } \) = \(\frac { 6 }{ 18 } \) = \(\frac { 1 }{ 3 } \)
So, triangles are similar.

Question 35.
WRITING
Can two triangles have all three ratios of corresponding angle measures equal to a value greater than 1 ? less than 1 ? Explain.
Answer:

Question 36.
HOW DO YOU SEE IT?
Which theorem could you use to show that ∆OPQ ~ ∆OMN in the portion of the Ferris wheel shown when PM = QN = 5 feet and MO = NO = 10 feet?

Answer:
The corresponding angle theorem states that ∆OPQ is similar to ∆OMN.

Question 37.
DRAWING CONCLUSIONS
Explain why it is not necessary to have an Angle-Side-Angle Similarity Theorem.
Answer:

Question 38.
THOUGHT PROVOKING
Decide whether each is a valid method of showing that two quadrilaterals are similar. Justify your answer.
a. SASA
Answer:
If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar.

b. SASAS
Answer:
If two sides and the included angle of one triangle are equal to the corresponding sides and angle of another triangle, the triangles are congruent.

c. SSSS
Answer:
If the corresponding side lengths of two triangles are proportional, then the triangles are similar.

d. SASSS
Answer:
If two sides in one triangle are proportional to two sides in another triangle and the included angle in both are congruent, then those two triangles are similar.

Question 39.
MULTIPLE REPRESENTATIONS
Use a diagram to show why there is no Side-Side-Angle Similarity Theorem.
Answer:

Question 40.
MODELING WITH MATHEMATICS
The dimensions of an actual swing set are shown. You want to create a scale model of the swing set for a dollhouse using similar triangles. Sketch a drawing of your swing set and label each side length. Write a similarity statement for each pair of similar triangles. State the scale factor you used to create the scale model.

Answer:
Here we have to check the similarity statement for △ABC, △DEF.
The scale factor k = \(\frac { AB }{ DE } \) = \(\frac { 8 }{ 6 } \) = \(\frac { 4 }{ 3 } \)

Question 41.
PROVING A THEOREM
Copy and complete the paragraph proot of the second part of the Slopes of Parallel Lines Theorem (Theorern 3. 13) from page 439.
Given m_l = m_n, l and n are nonvertical.
Prove l || n
You are given that m_l = m_n. By the definition of slope. m_l = \(\frac{B C}{A C}\) and m_n = \(\frac{E F}{D F}\) By ____________, \(\frac{B C}{A C}=\frac{E F}{D F}\). Rewriting this proportion yields ___________,

By the Right Angles Congruence Theorem (Thin. 2.3), ___________, So. ∆ABC ~ ∆DEF by ___________ . Because corresponding angles of similar triangles are congruent, ∠BAC ≅∠EDF. By ___________, l || n.

Answer:

Question 42.
PROVING A THEOREM
Copy and complete the two-column proof 0f the second part of the Slopes of Perpendicular Lines Theorem (Theorem 3.14)
Given m_l m_n = – 1, l and n are nonvertical.
Prove l ⊥ n

Answer:

Maintaining Mathematical proficiency

Find the coordinates of point P along the directed line segment AB so that AP to PB is the given ratio.

Question 43.
A(- 3, 6), B(2, 1); 3 to 2
Answer:

Question 44.
A(- 3, – 5), B(9, – 1); 1 to 3
Answer:

Question 45.
A(1, – 2), B(8, 12); 4 to 3
Answer:

8.4 Proportionality Theorems

Exploration 1

Discovering a Proportionality Relationship

Work with a partner. Use dynamic geometry software to draw any ∆ABC.
a. Construct \(\overline{D E}\) parallel to \(\overline{B C}\) with endpoints on \(\overline{A B}\) and \(\overline{A C}\), respectively.

Answer:

b. Compare the ratios of AD to BD and AE to CE.
Answer:

c. Move \(\overline{D E}\) to other locations Parallel to \(\overline{B C}\) with endpoints on \(\overline{A B}\) and \(\overline{A C}\), and repeat part (b).
Answer:

d. Change ∆ABC and repeat parts (a) – (c) several times. Write a conjecture that summarizes your results.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Answer:

Exploration 2

Discovering a Proportionality Relationship

Work with a partner. Use dynamic geometry software to draw any AABC.

a. Bisect ∆B and plot point D at the intersection of the angle bisector and \(\overline{A C}\).
Answer:

b. Compare the ratios of AD to DC and BA to BC.
Answer:

c. Change ∆ABC and repeat parts (a) and (b) several times. Write a conjecture that summarizes your results.
Answer:

Communicate Your Answer

Question 3.
What proportionality relationships exist in a triangle intersected by an angle bisector or by a line parallel to one of the sides?

Answer:

Question 4.
Use the figure at the right to write a proportion.

Answer:

Lesson 8.4 Proportionality Theorems

Monitoring Progress

Question 1.
Find the length of \(\overline{Y Z}\).

Answer:
YZ = \(\frac { 315 }{ 11 } \)

Explanation:
Triangle property thorem is \(\frac { XW }{ WV } \) = \(\frac { XY }{ YZ } \)
\(\frac { 44 }{ 35 } \) = \(\frac { 36 }{ YZ } \)
cross multiply the fractions
44 • YZ = 36 • 35
44 • YZ = 1260
YZ = \(\frac { 1260 }{ 44 } \)
YZ = \(\frac { 315 }{ 11 } \)

Question 2.
Determine whether \(\overline{P S}\) || \(\overline{Q R}\)

Answer:
\(\frac { PQ }{ PN } \) = \(\frac { 50 }{ 90 } \) = \(\frac { 5 }{ 9 } \)
\(\frac { SR }{ SN } \) = \(\frac { 40 }{ 72 } \) = \(\frac { 5 }{ 9 } \)
\(\frac { PQ }{ PN } \) = \(\frac { SR }{ SN } \) so PS is parallel to QR

Find the length of the given line segment.

Question 3.
\(\overline{B D}\)

Answer:
\(\overline{B D}\) = 12

Explanation:
All the angles are congruent. So, \(\overline{A B}\), \(\overline{C D}\), \(\overline{E F}\) are parallel.
using the three parallel lines theorem
\(\frac { BD }{ DF } \) = \(\frac { AC }{ CE } \)
\(\frac { [latex]\overline{B D}\) }{ 30 } [/latex] = \(\frac { 16 }{ 40 } \)
\(\overline{B D}\) = \(\frac { 16 }{ 40 } \) • 30
\(\overline{B D}\) = 12

Question 4.
\(\overline{J M}\)

Answer:
\(\overline{J M}\) = \(\frac { 96 }{ 5 } \)

Explanation:
All the angles are congruent. So, \(\overline{G H}\), \(\overline{J K}\), \(\overline{M N}\) are parallel.
using the three parallel lines theorem
\(\frac { HK }{ KN } \) = \(\frac { GJ }{ JM } \)
\(\frac { 15 }{ 18 } \) = \(\frac { 16 }{ [latex]\overline{J M}\) } [/latex]
Cross multiply
15 • \(\overline{J M}\) = 16 • 18 = 288
\(\overline{J M}\) = \(\frac { 288 }{ 15 } \)
\(\overline{J M}\) = \(\frac { 96 }{ 5 } \)

Find the value of the variable.

Question 5.

Answer:
x = 28

Explanation:
\(\overline{T V}\) is the angle bisector
So, \(\frac { ST }{ TU } \) = \(\frac { SV }{ VU } \)
\(\frac { 14 }{ x } \) = \(\frac { 24 }{ 48 } \)
cross multiply
24x = 14 • 48 = 672
x = \(\frac { 672 }{ 24 } \)
x = 28

Question 6.

Answer:
x = 4√2

Explanation:
\(\overline{W Z}\) is the angle bisector
So, \(\frac { YZ }{ ZX } \) = \(\frac { YW }{ WX } \)
\(\frac { 4 }{ 4 } \) = \(\frac { 4√2 }{ x } \)
cross multiply
4x = 4 • 4√2 = 16√2
x = 4√2

Exercise 8.4 Proportionality Theorems

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE STATEMENT
If a line divides two sides of a triangle proportionally, then it is ____________ to the third side. This theorem is knon as the ____________ .
Answer:

Question 2.
VOCABULARY
In ∆ABC, point R lies on \(\overline{B C}\) and \(\vec{A}\)R bisects ∆CAB. Write the proportionality statement for the triangle that is based on the Triangle Angle Bisector Theorem (Theorem 8.9).

Answer:
According to the triangle angle bisector theorem \(\frac { CR }{ BR } \) = \(\frac { AC }{ AB } \)

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, find the length of \(\overline{A B}\) .

Question 3.

Answer:

Question 4.

Answer:
\(\frac { AE }{ ED } \) = \(\frac { AB }{ BC } \)
\(\frac { 14 }{ 12 } \) = \(\frac { AB }{ 18 } \)
AB = \(\frac { 14 }{ 12 } \) • 18
AB = 21 units.

In Exercises 5 – 8, determine whether \(\overline{K M}\) || \(\overline{J N}\).

Question 5.

Answer:

Question 6.

Answer:
If \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \), then KM || JN
\(\frac { JK }{ KL } \) = latex]\frac { 22.5 }{ 25 } [/latex] = latex]\frac { 9 }{ 10 } [/latex]
\(\frac { NM }{ ML } \) = \(\frac { 18 }{ 20 } \) = latex]\frac { 9 }{ 10 } [/latex]
\(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \)
Hence KM || JN

Question 7.

Answer:

Question 8.

Answer:
If \(\frac { JK }{ KL } \) = \(\frac { NM }{ ML } \), then KM || JN
\(\frac { JK }{ KL } \) = latex]\frac {35 }{ 16 } [/latex]
\(\frac { NM }{ ML } \) = \(\frac { 34 }{ 15 } \)
\(\frac { JK }{ KL } \) ≠ \(\frac { NM }{ ML } \)
So, KM is not parallel to JN

CONSTRUCTION
In Exercises 9 – 12, draw a segment with the given length. Construct the point that divides the segment in the given ratio.
Question 9.
3 in.; 1 to 4
Answer:

Question 10.
2 in.; 2 to 3

Answer:
Construct a 2 inch segment and divide the segment into 2 + 3 or 5 congruent pieces. Point P is the point that is \(\frac { 1 }{ 5 } \) of the way from point A to point B.

Question 11.
12 cm; 1 to 3
Answer:

Question 12.
9 cm ; 2 to 5
Answer:
Construct a 9 cm segment and divide the segment into 2 + 5 or 7 congruent pieces. Point p is the point that is \(\frac { 1 }{ 7 } \) of the way from point A to point B.

In Exercises 13 – 16, use the diagram to complete the proportion.

Question 13.

Answer:

Question 14.

Answer:
\(\frac { CG }{ EG } \) = \(\frac { BF }{ DF } \)

Question 15.

Answer:

Question 16.

Answer:
\(\frac { BF }{ BD } \) = \(\frac { CG }{ CE } \)

In Exercises 17 and 18, find the length of the indicated line segment.

Question 17.
\(\overline{V X}\)

Answer:

Question 18.
\(\overline{S U}\)

Answer:
\(\frac { SU }{ NS } \) = \(\frac { RT }{ PR } \)
\(\frac { SU }{ 10 } \) = \(\frac { 12 }{ 8 } \)
SU = \(\frac { 12 }{ 8 } \) • 10
SU = 10

In Exercises 19 – 22, find the value of the variable.

Question 19.

Answer:

Question 20.

Answer:
\(\frac { z }{ 1.5 } \) = \(\frac { 3 }{ 4.5 } \)
z = \(\frac { 3 }{ 4.5 } \) • 1.5
z = 1

Question 21.

Answer:

Question 22.

Answer:
\(\frac { q }{ 16 – q } \) = \(\frac { 36 }{ 28 } \)
28q = 36 (16 – q)
28q = 576 – 36q
28q + 36q = 576
64q = 576
q = 9

Question 23.
ERROR ANALYSIS
Describe and correct the error in solving for x.

Answer:

Question 24.
ERROR ANALYSIS
Describe and correct the error in the students reasoning.

Answer:
\(\frac { BD }{ CD } \) = \(\frac { AB }{ AC } \)
BD = CD
So, 1 = \(\frac { AB }{ AC } \)
AC = AB

MATHEMATICAL CONNECTIONS
In Exercises 25 and 26, find the value of x for which \(\overline{P Q}\) || \(\overline{R S}\).

Question 25.

Answer:

Question 26.

Answer:
\(\frac { PR }{ RT } \) = \(\frac { QS }{ ST } \)
\(\frac { 12 }{ 2x – 2 } \) = \(\frac { 21 }{ 3x – 1 } \)
12(3x – 1) = 21(2x – 2)
36x – 12 = 42x – 42
42x – 36x = 42 – 12
6x = 30
x = 5

Question 27.
PROVING A THEOREM
Prove the Triangle Proportionality Theorem (Theorem 8.6).

Given \(\overline{Q S}\) || \(\overline{T U}\)
Prove \(\frac{Q T}{T R}=\frac{S U}{U R}\)
Answer:

Question 28.
PROVING A THEOREM
Prove the Converse of the Triangle Proportionality Theorem (Theorem 8.7).

Given \(\frac{Z Y}{Y W}=\frac{Z X}{X V}\)
Prove \(\overline{Y X}\) || \(\overline{W V}\)
Answer:

Question 29.
MODELING WITH MATHEMATICS
The real estate term lake frontage refers to the distance along the edge of a piece of property that touches a lake.

a. Find the lake frontage (to the nearest tenth) of each lot shown.
b. In general, the more lake frontage a lot has, the higher its selling price. Which lot(s) should be listed for the highest price?
c. Suppose that low prices are in the same ratio as lake frontages. If the least expensive lot is $250,000, what are the prices of the other lots? Explain your reasoning.
Answer:

Question 30.
USING STRUCTURE
Use the diagram to find the values of x and y.

Answer:
\(\frac { 5 }{ 2 } \) = \(\frac { x }{ 1.5 } \)
x = \(\frac { 5 }{ 2 } \) • 1.5
x = 3.75
\(\frac { 3 }{ 7 } \) = \(\frac { y }{ 5.25 } \)
y = \(\frac { 3 }{ 7 } \) • 5.25
y = 2.25

Question 31.
REASONING
In the construction on page 447, explain why you can apply the Triangle Proportionality Theorem (Theorem 86) in Step 3.
Answer:

Question 32.
PROVING A THEOREM
Use the diagram with the auxiliary line drawn to write a paragraph proof of the Three Parallel Lines Theorem (Theorem 8.8).
Given K₁ || K₂ || K₃
Prove \(\frac{C B}{B A}=\frac{D E}{E F}\)

Answer:
From the diagram, we can see that K₁ || K₂ || K₃
Those three parallel lines interest two traversals t₁, t₂
So, \(\frac{C B}{B A}=\frac{D E}{E F}\)

Question 33.
CRITICAL THINKING
In ∆LMN, the angle bisector of ∠M also bisects \(\overline{L N}\). Classify ∆LMN as specifically as possible. Justify your answer.
Answer:

Question 34.
HOW DO YOU SEE IT?
During a football game, the quarterback throws the ball to the receiver. The receiver is between two defensive players, as shown. If Player 1 is closer to the quarterback when the ball is thrown and both defensive players move at the same speed, which player will reach the receiver first? Explain your reasoning.

Answer:
As per the image, player 1 is closer to the receiver. So, player 1 will reach the receiver first.

Question 35.
PROVING A THEOREM
Use the diagram with the auxiliary lines drawn to write a paragraph proof of the Triangle Angle Bisector Theorem (Theorem 8.9).
Given ∠YXW ≅ ∠WXZ
prove \(\frac{Y W}{W Z}=\frac{X Y}{X Z}\)

Answer:

Question 36.
THOUGHT PROVOKING
Write the converse of the Triangle Angle Bisector Theorem (Theorem 8.9). Is the converse true? Justify your answer.
Answer:

Question 37.
REASONING
How is the Triangle Midsegment Theorem (Theorem 6.8) related to the Triangle Proportionality Theorem (Theorem 8.6)? Explain your reasoning.
Answer:

Question 38.
MAKING AN ARGUMENT
Two people leave points A and B at the same time. They intend to meet at point C at the same time. The person who leaves point A walks at a speed of 3 miles per hour. You and a friend are trying to determine how fast the person who leaves point B must walk. Your friend claims you need to know the length of \(\overline{A C}\). Is your friend correct? Explain your reasoning.

Answer:
My self starts walking from point A with a speed of 3 miles per hour and reaches point C.
My friend starts walking from point B with x speed and reaches point C.
\(\frac { AD }{ DC } \) = \(\frac { BE }{ CE } \)
I have to travel from A to C. So, I need to know distance between AC.
Therefore, my friend is correct.

Question 39.
CONSTRUCTION
Given segments with lengths r, s, and t, construct a segment of length x, such that \(\frac{r}{s}=\frac{t}{x}\)

Answer:

Question 40.
PROOF
Prove Ceva’s Theorem: If P is any point inside ∆ABC, then \(\frac{A Y}{Y C} \cdot \frac{C X}{X B} \cdot \frac{B Z}{Z A}\) = 1

(Hint: Draw segments parallel to \(\overline{B Y}\) through A and C, as shown. Apply the Triangle Proportionality Theorem (Theorem 8.6) to ∆ACM. Show that ∆APN ~ ∆MPC, ∆CXM ~ ∆BXP, and ∆BZP ~ ∆AZN.)
Answer:

Maintaining Mathematical Proficiency

Use the triangle.

Question 41.
Which sides are the legs?
Answer:

Question 42.
Which side is the hypotenuse?
Answer:
The leg c is the hypotenuse.

Solve the equation.

Question 43.
x² = 121
Answer:

Question 44.
x² + 16 = 25
Answer:
x² + 16 = 25
x² = 25 – 16
x² = 9
x = ∓3

Question 45.
36 + x² = 85
Answer:

Similarity Review

8.1 Similar Polygons

Find the scale factor. Then list all pairs of congruent angles and write the ratios of the corresponding side lengths in a statement of proportionality.

Question 1.
ABCD ~ EFGH

Answer:
\(\frac { BC }{ CD } \) = \(\frac { 8 }{ 12 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { EH }{ GH } \) = \(\frac { 6 }{ 9 } \) = \(\frac { 2 }{ 3 } \)
So, scale factor = \(\frac { 2 }{ 3 } \)

Question 2.
∆XYZ ~ ∆RPQ

Answer:
longer sides: \(\frac { 10 }{ 25 } \) = \(\frac { 2 }{ 5 } \)
shorter sides: \(\frac { 6 }{ 15 } \) = \(\frac { 2 }{ 5 } \)
remaining sides: \(\frac { 8 }{ 20 } \) = \(\frac { 2 }{ 5 } \)
So, scale factor = \(\frac { 2 }{ 5 } \)

Question 3.
Two similar triangles have a scale factor of 3 : 5. The altitude of the larger triangle is 24 inches. What is the altitude of the smaller triangle?
Answer:
Scale factor of smaller triangle to larger traingle is \(\frac { 3 }{ 5 } \) and larger traingle altitude is 24 inches
Let x be the smaller triangle altitude
\(\frac { altitude of smaller triangle }{ altitude of larger traingle } \) = scale factor
\(\frac { x }{ 24 } \) = \(\frac { 3 }{ 5 } \)
x = \(\frac { 3 }{ 5 } \) • 24
x = 14.4

Question 4.
Two similar triangles have a pair of corresponding sides of length 12 meters and 8 meters. The larger triangle has a perimeter of 48 meters and an area of 180 square meters. Find the perimeter and area of the smaller triangle.
Answer:
Scale factor = \(\frac { 2 }{ 3 } \)
perimeter of smaller triangle = 32
Area of smaller triangle = 80

Explanation:
The scale factor of smaller to larger traingle = \(\frac { 8 }{ 12 } \) = \(\frac { 2 }{ 3 } \)
\(\frac { perimeter of smaller triangle }{ perimeter of larger triangle } \) = scale factor\(\frac { perimeter of smaller triangle }{ 48 } \) = \(\frac { 2 }{ 3 } \)
perimeter of smaller triangle = \(\frac { 2 }{ 3 } \) • 48
perimeter of smaller triangle = 32
\(\frac { Area of smaller triangle }{ Area of larger triangle } \) = (scale factor)²
\(\frac { Area of smaller triangle }{ 180 } \) = ( \(\frac { 2 }{ 3 } \))²
Area of smaller triangle = \(\frac { 4 }{ 9 } \) • 180
= 80

8.2 Proving Triangle Similarity by AA

Show that the triangles are similar. Write a similarity statement.

Question 5.

Answer:
m∠RQS = m∠UTS = 30°.
△QRS and △STU are similar as per the AA similarity theorem.

Question 6.

Answer:
m∠CAB = 60°, m∠DEF = 30°
△ABC and △DEF are not similar as per the AA similarity theorem.

Question 7.
A cellular telephone tower casts a shadow that is 72 feet long, while a nearby tree that is 27 feet tall casts a shadow that is 6 feet long. How tall is the tower?
Answer:
\(\frac { shadow of tree }{ shadow of tower } \) = \(\frac { height of tree }{ height of tower } \)
\(\frac { 6 }{ 72 } \) = \(\frac { 27 }{ x } \)
6x = 1944
x = 324 ft
The height of the tower is 324 ft.

8.3 Proving Triangle Similarity by SSS and SAS

Use the SSS Similarity Theorem (Theorem 8.4) or the SAS Similarity Theorem (Theorem 8.5) to show that the triangles are similar.

Question 8.

Answer:
\(\frac { DE }{ CD } \) = \(\frac { 7 }{ 3.5 } \) = 2
\(\frac { AB }{ BC } \) = \(\frac { 8 }{ 4 } \) = 2
\(\frac { DE }{ CD } \) = \(\frac { AB }{ BC } \)
So, BD is parallel to AE.

Question 9.

Answer:
\(\frac { QU }{ TU } \) = \(\frac { 9 }{ 4.5 } \) = 2
\(\frac { QR }{ SR } \) = \(\frac { 14 }{ 7 } \) = 2
\(\frac { QU }{ TU } \) = \(\frac { QR }{ SR } \)
So, ST is parallel to RU.

Question 10.
Find the value of x that makes ∆ABC ~ ∆DEF

Answer:
\(\frac { 24 }{ 6 } \) = 4
\(\frac { 32 }{ 2x } \) = 4
32 = 8x
x = \(\frac { 32 }{ 8 } \)
x = 4

8.4 Proportionality Theorems

Determine whether \(\overline{A B}\) || \(\overline{C D}\)

Question 11.

Answer:
\(\frac { DB }{ BE } \) = \(\frac { 10 }{ 16 } \) = \(\frac { 5 }{ 8 } \)
\(\frac { CA }{ AE } \) = \(\frac { 20 }{ 28 } \) = \(\frac { 5 }{ 7 } \)
\(\frac { DB }{ BE } \) ≠ \(\frac { CA }{ AE } \)
So, CD and AB are not parallel.

Question 12.

Answer:
\(\frac { DB }{ BE } \) = \(\frac { 12 }{ 20 } \) = \(\frac { 3 }{ 5 } \)
\(\frac { CA }{ AE } \) = \(\frac { 13.5 }{ 22.5 } \) = \(\frac { 3 }{ 5 } \)
\(\frac { DB }{ BE } \) = \(\frac { CA }{ AE } \)
So, AB and CD are parallel.

Question 13.
Find the length of \(\overline{Y B}\).

Answer:
\(\frac { ZC }{ AZ } \) = \(\frac { 24 }{ 15 } \) = \(\frac { 8 }{ 5 } \)
\(\frac { YB }{ AY } \) = \(\frac { ZC }{ AZ } \)
\(\frac { YB }{ 7 } \) = \(\frac { 8 }{ 5 } \)
YB = \(\frac { 8 }{ 5 } \) • 7
YB = \(\frac { 56 }{ 5 } \)
The length of \(\overline{Y B}\) is \(\frac { 56 }{ 5 } \)

Find the length of \(\overline{A B}\).

Question 14.

Answer:
\(\frac { AB }{ 7 } \) = \(\frac { 6 }{ 4 } \)
\(\overline{A B}\) = \(\frac { 6 }{ 4 } \) • 7
\(\overline{A B}\) = \(\frac { 42 }{ 4 } \)
\(\overline{A B}\) = \(\frac { 21 }{ 2 } \)
The length of \(\overline{A B}\) is \(\frac { 21 }{ 2 } \).

Question 15.

Answer:
\(\frac { DB }{ CD } \) = \(\frac { AB }{ AC } \)
\(\frac { 4 }{ 10 } \) = \(\frac { AB }{ 18 } \)
\(\frac { 2 }{ 5 } \) = \(\frac { AB }{ 18 } \)
AB = \(\frac { 36 }{ 5 } \)

Similarity Test

Determine whether the triangles are similar. If they are, write a similarity statement. Explain your reasoning.

Question 1.

Answer:
Longer sides: \(\frac { 32 }{ 24 } \) = \(\frac { 4 }{ 3 } \)
shorter sides: \(\frac { 18 }{ 14 } \) = \(\frac { 9 }{ 7 } \)
remaining sides: \(\frac { 20 }{ 15 } \) = \(\frac { 4 }{ 3 } \)
Those are not equal.
So, triangles are not similar.

Question 2.

Answer:
\(\frac { AC }{ KJ } \) = \(\frac { 6 }{ 8 } \) = \(\frac { 3 }{ 4 } \)
\(\frac { BC }{ JL } \) = \(\frac { 8 }{ [latex]\frac { 32 }{ 3 } \) } [/latex] = \(\frac { 3 }{ 4 } \)
\(\frac { AC }{ KJ } \) = \(\frac { BC }{ JL } \)
∠C = ∠J
So, △ABC and △JLK are similar.

Question 3.

Answer:
\(\frac { XY }{ XW } \) = \(\frac { PZ }{ PW } \)

Find the value of the variable.

Question 4.

Answer:
\(\frac { 9 }{ w } \) = \(\frac { 15 }{ 5 } \)
\(\frac { 9 }{ w } \) = 3
9 = 3w
w = 3

Question 5.

Answer:
\(\frac { 17.5 }{ 21 } \) = \(\frac { q }{ 33 } \)
q = \(\frac { 17.5 }{ 21 } \) • 33
q = \(\frac { 55 }{ 2 } \)

Question 6.

Answer:
\(\frac { 21 – p }{ p } \) = \(\frac { 12 }{ 24 } \)
\(\frac { 21 – p }{ p } \) = \(\frac { 1 }{ 2 } \)
cross multiply
2(21 – p) = p
42 – 2p = p
42 = p + 2p
42 = 3p
p = 14

Question 7.
Given ∆QRS ~ ∆MNP, list all pairs of congruent angles, Then write the ratios of the corresponding side lengths in a statement of proportionality.

Answer:
The pairs of congruent anglres are m∠QRS, m∠RSQ, m∠SQR, m∠MNP, m∠NPM, m∠PMN.
The ratios of side lengths are \(\frac { RQ }{ MN } \), \(\frac { QS }{ MP } \), \(\frac { RS }{ NP } \)

Use the diagram.

Question 8.
Find the length of \(\overline{E F}\).

Answer:
\(\frac { DE }{ EF } \) = \(\frac { CD }{ BC } \)
\(\frac { 3.2 }{ EF } \) = \(\frac { 2.8 }{ 1.4 } \)
EF = 1.6
The length of \(\overline{E F}\) is 1.6

Question 9.
Find the length of \(\overline{F G}\).

Answer:
\(\frac { EF }{ FG } \) = \(\frac { BC }{ AB } \)
\(\frac { 1.6 }{ FG } \) = \(\frac { 1.4 }{ 4.2 } \)
FG = 4.8
The length of \(\overline{F G}\) is 4.8

Question 10.
Is quadrilateral FECB similar to quadrilateral GFBA? If so, what is the scale factor of the dilation that maps quadrilateral FECB to quadrilateral GFBA?

Answer:
The scale factor of dilation from quadrilateral FECB to quadrilateral GFBA is \(\frac { GF }{ FE } \)

Question 11.
You are visiting the Unisphere at Flushing Meadows Corona Park in New York. To estimate the height of the stainless steel model of Earth. you place a mirror on the ground and stand where you can see the top of the model in the mirror. Use the diagram to estimate the height of the model. Explain why this method works.

Answer:

Question 12.
You are making a scale model of a rectangular park for a school project. Your model has a length of 2 feet and a width of 1.4 feet. The actual park is 800 yards long. What are the perimeter and area of the actual park?

Answer:
we know that 1 yard = ft
As per the similarity theorem
\(\frac { AD }{ EH } \) = \(\frac { AB }{ EF } \)
\(\frac { 2 }{ 800.3 } \) = \(\frac { 1.4 }{ EF } \)
EF = 1200 • 1 • 4
EF = 1680
Perimete of the park P = 2(1680 + 2400)
= 2(4080) = 8160 ft
Area of the actual park = 1680 • 2400 = 4,032,000 sq ft
Therefore, perimeter of the actual park = 8160 ft
area of the actual park is 4,032,000 sq ft.

Question 13.
In a Perspective drawing, lines that are parallel in real life must meet at a vanishing point on the horizon. To make the train cars in the drawing appear equal in length, they are drawn so that the lines connecting the opposite corners of each car are parallel. Use the dimensions given and the yellow parallel lines to find the length of the bottom edge of the drawing of Car 2.

Answer:
\(\frac { 5.4 }{ 10.6 } \) = \(\frac { 5.4 + C }{ 19 } \)
C = \(\frac { 45.36 }{ 10.6 } \)
\(\frac { (19 – x – 8.4) }{ (19 – 8.4) } \) = \(\frac { 5.4 }{ 5.4 + c } \)
\(\frac { 19 – 8.4 }{ 19 } \) = \(\frac { c2 + 5.4 }{ c1 + c2 + 5.4 } \)
\(\frac { 19 – 8.4 }{ 19 } \) = \(\frac { 5.4 }{ 5.4 + c2 } \)
c1 = 7.6, c2 = 4.2, x = 4.6
The length of car 2 is 4.2 cm.

Similarity Cumulative Assessment

Question 1.
Use the graph of quadrilaterals ABCD and QRST.

a. Write a composition of transformations that maps quadrilateral ABCD to quadrilateral QRST.
Answer:
The scale factor = \(\frac { AD }{ QT } \) = \(\frac { 2 }{ 1.5 } \)

b. Are the quadrilaterals similar? Explain your reasoning.
Answer:
No.
\(\frac { AD }{ QT } \) = \(\frac { 2 }{ 1.5 } \)
\(\frac { CD }{ TS } \) = \(\frac { 2.8 }{ 1.4 } \) = 2
So, quadrilaterals are not similar.

Question 2.
In the diagram. ABCD is a parallelogram. Which congruence theorem(s) could you Use to show that ∆AED ≅ ∆CEB? Select all that apply.

SAS Congruence Theorem (Theorem 5.5)
Answer:
It states that if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then these two triangles are congruent.

SSS Congruence Theorem (Theorem 5.8)
Answer:
If all the three sides of one triangle are equivalent to the corresponding three sides of the second triangle, then the two triangles are said to be congruent by SSS rule.

HL Congruence Theorem (Theorem 5.9)
Answer:
A given set of triangles are congruent if the corresponding lengths of their hypotenuse and one leg are equal.

ASA Congruence Theorem (Theorem 5. 10)
Answer:
If any two angles and the side included between the angles of one triangle are equivalent to the corresponding two angles and side included between the angles of the second triangle, then the two triangles are said to be congruent by ASA rule

AAS Congruence Theorem (Theorem 5. 11)
Answer:
AAS stands for Angle-angle-side. When two angles and a non-included side of a triangle are equal to the corresponding angles and sides of another triangle, then the triangles are said to be congruent.

Question 3.
By the Triangle Proportionality Theorem (Theorem 8.6), \(\frac{V W}{W Y}=\frac{V X}{X Z}\) In the diagram, VX > VW and XZ > WY. List three possible values for VX and XZ.

Answer:
\(\frac{V W}{W Y}=\frac{V X}{X Z}\)
\(\frac{ 4 }{6}=\frac{V X}{X Z}\)
The possible values of VX are greater than 4 means 5, 6, 7, . . .
The possible values of XZ are greater than 6 means 7, 8, 9, . .

Question 4.
The slope of line l is – \(\frac{3}{4}\). The slope of line n is \(\frac{4}{3}\) What must be true about lines l and n ?
(A) Lines l and n are parallel.
(B) Lines l and n arc perpendicular.
(C) Lines l and n are skew.
(D) Lines l and n are the same line.
Answer:
The slope of l = – \(\frac{3}{4}\)
Slope of n = \(\frac{4}{3}\)
lines slopes are reciprocal and opposite. So, they are perpendicular.

Question 5.
Enter a statement or reason in each blank to complete the two-column proof.

Given \(\frac{K J}{K L}=\frac{K H}{K M}\)
Prove ∠LMN ≅ ∠JHG

Answer:

Question 6.
The coordinates of the vertices of ∆DEF are D(- 8, 5), E(- 5, 8), and F(- 1, 4), The coordinates of the vertices of ∆JKL are J(16, – 10), K(10, – 16), and L(2, – 8), ∠D ≅ ∠J. Can you show that ∆DEF ∆JKL by using the AA Similarity Theorem (Theorem 8.3)? If so, do so by listing the congruent corresponding angles and writing a similarity transformation that maps ∆DEF to ∆JKL. If not, explain why not.
Answer:
AA similarity theorem states that ∠D = ∠J. So, ∆DEF and ∆JKL are similar.

Question 7.
Classify the quadrilateral using the most specific name.

rectangle     square    parallelogram    rhombus
Answer:

Question 8.
‘Your friend makes the statement “Quadrilateral PQRS is similar to quadrilateral WXYZ.” Describe the relationships between corresponding angles and between corresponding sides that make this statement true.

Answer:
When 2 figures are similar, then their corresponding angles are congruent and their corresponding lengths are proportional. hence if PQRS is similar to wxyZ, then the following statements are true.
∠P = ∠W, ∠Q = ∠X, ∠R = ∠Y and ∠S = ∠Z and
\(\frac { PQ }{ WX } \) = \(\frac { QR }{ XY } \) = \(\frac { RS }{ YZ } \) = \(\frac { PS }{ WZ } \) = k
Here is a constant of proportionality.

Big Ideas Math 7th Grade Chapter 3 Expressions Answer Key: Download free step-by-step problem and solution pdf of Big Ideas Math Grade 7 Chapter 3 Expressions from here. Follow the various concepts and topics involved in this chapter and make your preparation easy and efficient. We are providing the free Big Ideas Math Book 7th Grade Answer Key Chapter 3 Expressions pdf links in the below sections.

Refer to the Big Ideas Math Answers Grade 7 Chapter 3 Expressions and finish your preparation and assignment. The solutions for 7th-grade chapter 3 expressions give you a clear idea of how to solve and understand the problems. With the help of experts, we have given all the problems of a chapter in a single pdf. To avoid time waste, download the pdf and start your preparation.

Big Ideas Math Book 7th Grade Answer Key Chapter 3 Expressions

There are various topics available in the Expressions Chapter. Those are Adding and Subtracting Linear Expressions, The Distributive Property, Algebraic Expressions, Factoring Expressions, and so on. Find out the solutions and problems involved in these chapters. To understand the concept more clearly, compare the problems with real-life situations.

Follow BIM 7th Grade Chapter 3 Expressions Solution Key and know your strengths and weaknesses. First of all, know the syllabus of expressions and then based on the various concepts, prepare the timetable. Involve the number of concepts, time remaining to learn all the topics, etc, and prepare the timetable according to it. Know the example problems and formulae in the below sections.

Performance

• Expressions STEAM VIDEO/Performance
• Expressions Getting Ready for Chapter 3

Lesson: 1 Algebraic Expressions

• Lesson 3.1 Algebraic Expressions
• Algebraic Expressions Homework & Practice 3.1

Lesson: 2 Adding and Subtracting Linear Expressions

• Lesson 3.2 Adding and Subtracting Linear Expressions
• Adding and Subtracting Linear Expressions Homework & Practice 3.2

Lesson: 3 The Distributive Property

• Lesson 3.3 The Distributive Property
• The Distributive Property Homework & Practice 3.3

Lesson: 4 Factoring Expressions

• Lesson 3.4 Factoring Expressions
• Factoring Expressions Homework & Practice 3.4

Chapter: 3 – Expressions 

• Expressions Connecting Concepts
• Expressions Chapter Review
• Expressions Practice Test
• Expressions Cumulative Practice

Expressions STEAM VIDEO/Performance

STEAM Video

Trophic Status

In an ecosystem, energy and nutrients flow between abiotic and components. Biotic components are the living parts of an ecosystem. Abiotic components are the non-living parts of an ecosystem. What is an example of an ecosystem?

Watch the STEAM video “Trophic Status.” Then answer the following questions.

Question 1.
Give examples of both biotic and abiotic components in an ecosystem. Explain.

Answer:
Examples of biotic components are plants, animals, bacteria, fungi. Examples of abiotic components are water, soil, sunlight, gases, etc.

Explanation:
Biotic components are the living organisms in an ecosystem. Examples of biotic components are plants, animals, bacteria, fungi. Abiotic components are the non-living physical and chemical factors in the environment that affect ecosystems. Examples of abiotic components are water, soil, sunlight, gases, etc. Abiotic factors affect the ability of organisms to reproduce, survive, help determine the types and number of organisms able to exist in the environment. Limiting factors restrict growth. Biotic factors are the living things that directly or indirectly affect organisms in the environment.

Question 2.
When an organism is eaten, its energy flows into the organism that consumes it. Explain how to use an expression to represent the total energy that a person gains from eating each of the items shown.

Answer:
bx + lx + fx

Explanation:
Let us take the energy gained by eating a banana is bx, by eating leaves is lx, by eating fish is fx.
So, the expression becomes the sum of all those energies.
bx + lx + fx

Performance Task

Chlorophyll in Plants

After completing this chapter, you will be able to use the STEAM concepts you learned to answer the questions in the Video Performance Task. You will be given the numbers of atoms found in molecules involved in photosynthesis.


You will be asked to determine the total cost for a model of a molecule given the costs of different types of atom models. How can you find the total cost of purchasing several identical objects?

Expressions Getting Ready for Chapter 3

Chapter Exploration

Work with a partner. Rewrite the algebraic expression so that it has fewer symbols x but still has the same value when evaluated for any value of x.

Answer:
1. 3x + 4
2. 3x – 1
3. 2x – 3
4. 2(1 + x)
5. 3x – 1
6. x + 5

Explanation:
1. Original expression is 2x + 4 + x
Simplified expression is 3x + 4
2. Original expression is 3(x + 1) – 4
3x + 3 – 4 = 3x – 1
Simplified expression is 3x – 1
3. Original expression is x – (3 – x)
x – 3 + x
Simplified expression is 2x – 3
4. Original expression is 5 + 2x – 3
Simplified expression is 2 + 2x = 2(1 + x)
5. Original expression is x + 3 + 2x – 4
Simplified expression is 3x – 1
6. Original expression is 2x + 2 – x + 3
Simplified expression is x + 5

Question 7.
WRITING GUIDELINES
Work with a partner. Use your answers in Exercises 1-6 to write guidelines for simplifying an expression.

Answer:
First, you need to take the given original expression as it is.
If there are any braces, then eliminate them by expanding the expression
Then, add or subtract variables and constants together to get the solution

Explanation:
Let us take question 2. 3(x + 1) – 4
Expand 3
3x + 3 – 4
Simplify
3x – 1

APPLYING A DEFINITION
Work with a partner. Two expressions are equivalent if x they have the same value when evaluated for any value of x. Decide which two expressions are equivalent. Explain your reasoning.

Answer:
8. x – (2x + 1) is equivalent to -x – 1
9. 2x + 3 – x + 4 is equivalent to x + 7 in expression B
10. 3 + x -2(x + 1) is equivalent to -x + 1
11. 2 – 2x – (x + 2) is equivalent to -3x

Explanation:
8. x – (2x + 1) = x – 2x – 1
= -x – 1
So, x – (2x + 1) is equivalent to -x – 1
9. 2x + 3 – x + 4 = x + 7
So, 2x + 3 – x + 4 is equivalent to x + 7 in expression B
10. 3 + x -2(x + 1) = 3 + x -2x – 2
= 1 – x
So, 3 + x -2(x + 1) is equivalent to -x + 1
11. 2 – 2x – (x + 2) = 2 – 2x – x – 2
= -3x
So, 2 – 2x – (x + 2) is equivalent to -3x

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.

Lesson 3.1 Algebraic Expressions

EXPLORATION 1
Simplifying Algebraic Expressions
Work with a partner.
a. Choose a value of other than 0 or 1 for the last column in the table. Complete the table by evaluating each algebraic expression for each value of x. What do you notice?

b. How can you use properties of operations to justify your answers in part(a)? Explain your reasoning.
c. To subtract a number, you can add its opposite. Does a similar rule apply to the terms of an algebraic expression? Explain your reasoning.

Answer:
a.
b. Just simplify the expression and put the values of x in simplified expression
c. Yes

Explanation:
a.
A. -1/3 + x + 7/3 = (-1 + 7)/3 + x
= -6/3 + x = -2 + x
If x = 0, then -2 + 0 = -2
If x = 1, then -2 + 1 = -1
If x = 4, then -2 + 4 = 2
B. 0.5x + 3 – 1.5x – 1 = -x + 4
If x = 0, then -0 + 4 = 4
If x = 1, then -1 + 4 = 3
If x = 4, then -4 + 4 = 0
C. 2x + 6
If x = 0, then 2(0) + 6 = 6
If x = 1, then 2(1) + 6 = 8
Ifx = 4, then 2(4) + 6 = 14
D. x + 4
If x = 0, then 0 + 4 = 4
If x = 1, then 1 + 4 = 5
If x = 4, then 4 + 4 = 8
E. -2x + 2
If x = 0, then -2(0) + 2 = 2
If x = 1, then -2(1) + 2 = 0
If x = 4, then -2(4) + 2 = -6
F. 1/2 x – x + 3/2 x + 4 = 4/2 x – x + 4
= 2x – x + 4 = x + 4
If x = 0, then 0 + 4 = 4
If x = 1, then 1 + 4 = 5
If x = 4, then 4 + 4 = 8
G. -4.8x + 2 – x + 3.8x = -2x + 2
If x = 0, then -2(0) + 2 = 2
If x = 1, then -2(1) + 2 = 0
If x = 4, then -2(4) + 2 = -6
H. x + 2
If x = 0, then 0 + 2 = 2
If x = 1, then 1 + 2 = 3
If x = 4, then 4 + 2 = 6
I. -x + 2
If x = 0, then -0 + 2 = 2
If x = 1, then -1 + 2 =1
If x = 4, then -4 + 2 = -2
J. 3x + 2 – x + 4 = 2x + 6
If x = 0, then 2(0) + 6 = 6
If x = 1, then 2(1) + 6 = 8
If x = 4, then 2(4) + 6 = 14

3.1 Lesson

In an algebraic expression, are terms that have the same variables raised to the same exponents. Constant terms are also like terms. To identify terms and like terms in an expression, first write the expression as a sum of its terms.

Try It
Identify the terms and like terms in the expression.

Question 1.

Answer:
Terms are 10, y, -3/2y, like terms are y, -3/2y.

Explanation:
The given expression is y + 10 – 3/2 y
Like terms are y, -3/2y
Because they have the same variable y
Terms are 10, y, -3/2y

Question 2.
2r² + 7r – r² – 9

Answer:
The like terms are 2r², -r², Terms are -9, 7r, 2r², -r²

Explanation:
Given expression is 2r² + 7r -r² – 9
The like terms are 2r², -r²
Terms are -9, 7r, 2r², -r²

Question 3.
7 + 4p – 5 + p + 2q

Answer:
The like terms are 4p, p
The terms are 7, -5, 2q, 4p, p

Explanation:
Given expression is 7 + 4p – 5 + p + 2q
The like terms are 4p, p
The terms are 7, -5, 2q, 4p, p

Try It
Simplify the expression.

Question 4.
-10y + 15y

Answer:
-10y + 15y = 5y

Explanation:
Given expression is -10y + 15y
5y

Question 5.

Answer:
3/8 b – 3/4 b = -3/8 b

Explanation:
Given expression is 3/8 b – 3/4 b
L.C.M of 8, 4 is 8
(3 – 6)/8 b = -3/8 b

Question 6.
2.4g – 2.4g – 9.8g

Answer:
-9.8g

Explanation:
Given expression is 2.4g – 2.4g – 9.8g
-9.8g

Try It

Simplify the expression.

Question 7.
14 – 3z + 8 + z

Answer:
22 – 2z

Explanation:
Given expression is 14 – 3z + 8 + z
Add constants and like terms
22 – 2z

Question 8.
2.5x + 4.3x – 5

Answer:
6.8x – 5

Explanation:
Given expression is 2.5x + 4.3x – 5
6.8x – 5

Question 9.
2s – 9s + 8t – t

Answer:
7(t – s)

Explanation:
Given expression is 2s – 9s + 8t – t
-7s + 7t = 7(t – s)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
WRITING
Explain how to identify the terms and like terms of 3y – 4 – 5y.

Answer:
The like terms are 3y, -5y
Terms are -4, 3y, -5y

Explanation:
Given expression is 3y – 4 – 5y
Like terms are the terms in the expression which have the same variable
The following are the like terms because each term contains a variable y and numeric coefficient
3y, -5y
Other than like terms are called the terms
Terms are -4, 3y, -5y

SIMPLIFYING ALGEBRAIC EXPRESSIONS
Simplify the expression.

Question 11.
7p + 6p

Answer:
13p

Explanation:
Given expression is 7p + 6p
13p

Question 12.

Answer:
3n/2 – 3

Explanation:
Given expression is 4/5 n – 3 + 7/10 n
(8n + 7n)/10 – 3
= 15n/10 – 3
= 3n/2 – 3

Question 13.
2w – g – 7w + 3g

Answer:
-5w + 2g

Explanation:
Given expression is 2w – g – 7w + 3g
Perform the arithmetic operations between the like terms
-5w + 2g

Question 14.
VOCABULARY
Is the expression 3x + 2x – 4 in simplest form? Explain.

Answer:
The expression 3x + 2x – 4 is not in the simplest form.

Explanation:
Given expression is 3x + 2x – 4
= 5x – 4
The simplest form of 3x + 2x – 4 is 5x – 4

Question 15.
WHICH ONE DOESN’T BELONG?
Which expression does not belong with the other three? Explain your reasoning.

Answer:
5x – 10 – 2x does not belong with the other three.

Explanation:
The first expression is -4 + 6 + 3x = 2 + 3x
The second expression is 3x + 9 – 7 = 3x + 2
The third expression is 5x – 10 – 2x = 3x – 10
Fourth expression is 5x – 4 + 6 – 2x = 3x + 2
So, out of all 5x – 10 – 2x does not belong with the other three.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 16.
MODELING REAL LIFE
An exercise mat is 3.3 times as long as it is wide. Write expressions in simplest form that represent the perimeter and the area of the exercise mat.

Answer:
The perimeter of the exercise mat is 8.6l, Area of the exercise mat is 3.3l²

Explanation:
Let us take the l is the length of the exercise mat
Then, the width of the exercise mat = 3.3l
Perimeter of exercise mat = 2(length + width)
= 2(l + 3.3l)
= 2(4.3l) = 8.6 l
Area of the exercise mat = length x width
= l x 3.3l
= 3.3l²
Therefore, the perimeter of the exercise mat is 8.6l, Area of the exercise mat is 3.3l²

Question 17.
DIG DEEPER!
A group of friends visits the movie theater in Example 4. Each person buys a daytime ticket and a small drink. The group shares 2 large popcorns. What is the average cost per person when there are 4 people in the group?

Answer:
The total cost for a group of 4 people is $35.

Explanation:
Number of tickets . cost per ticket + number of small drinks . cost per small drink + Number of large popcorns . Cost per large popcorn
The same number of tickets, small drinks are purchased. So, can represent the number of tickets, the number of small drinks.
5x + 1.75x + $8 = x(5 + 1.75) + $8
= 6.75x + $8
2 popcorns cost is 2 x 4 = $8
The expression 6.75x + $8 indicates that the cost per person is $6.75 + $8 = 14.75
To find the cost for a group of 4 people, evaluate the expression when x = 4
= 6.75 (4 ) + 8 = 27 + 8 = 35
The total cost for a group of 4 people is $35.

Algebraic Expressions Homework & Practice 3.1

Review & Refresh

Find the product or quotient. Write fractions in simplest form.

Question 1.

Answer:
-1/2

Explanation:
Given that,
-2/7 x 7/4 = -2/4
= -1/2

Question 2.

Answer:
3/5

Explanation:
Given that,
-2/3(-9/10) = 2/3 x 9/10
= 6/10 = 3/5

Question 3.

Answer:
-13/6

Explanation:
Given that,
1(4/9) ÷ (-2/3) = 13/9 ÷ (-2/3)
= 13/9 x (-3/2) = -13/6

Order the numbers from least to greatest.

Question 4.

Answer:
3/4, 78%, 0.85, 87%, 7/8

Explanation:
Given that
7/8 = 0.875, 0.85, 87% = 87/100 = 0.87 3/4 = 0.75, 78% = 78/100 = 0.78
So, the numbers are 0.875, 0.85, 0.87, 0.75, 0.78
Ordering from the least to the greatest is 0.75,0.78,0.85,0.87,0.875
i.e 3/4, 78%, 0.85, 87%, 7/8

Question 5.

Answer:
15%, 1450%, 14.8, 15(4/5)

Explanation:
Given that,
15% = 15/100 = 0.15, 14.8, 15(4/5) = 79/5 = 15.8, 1450% = 1450/100 = 14.5
So the numbers are 0.15, 14.8, 15.8, 14.5
The order of number from least to greatest is 0.15, 14.5, 14.8, 15.8
i.e 15%, 1450%, 14.8, 15(4/5)

Question 6.
A bird’s nest is 12 feet above the ground. A mole’s den is 12 inches below the ground. What is the difference in height of these two positions?
A. 24 in.
B. 11 ft
C. 13 ft
D. 24 ft

Answer:
The difference in height of these two positions is 13 feet.

Explanation:
Given that,
The bird nest is 12 feet above the ground and a mole’s den is 12 inches below the ground.
So, the mole’s den have to travel 12 inches +12 feet to reach the bird nest
So, the difference in height of these two positions is 12 feet + 12 inches
= 12 feet + 1 feet = 13 feet

Concepts, Skills, & Problem Solving
REASONING
Determine whether the expressions are equivalent. Explain your reasoning. (See Exploration 1, p. 91.)

Question 7.

Answer:
The expressions are not equivalent

Explanation:
Expression 1 is 3 – 5x
Expression 2 is 4.25 – 5x – 4.25
= -5x
So, expressions are not equivalent

Question 8.

Answer:
Both the expressions are equivalent.

Explanation:
Expression 1 is 1.25x + 4 + 0.75x – 3
= 2x + 4 – 3 = 2x + 1
Expression 2 is 2x + 1
So, both the expressions are equivalent.

IDENTIFYING TERMS AND LIKE TERMS
Identify the terms and like terms in the expression.

Question 9.
t + 8 + 3t

Answer:
Like terms are t, 3t
Terms are 8, t, 3t

Explanation:
Given expression is t + 8 + 3t
Like terms are t, 3t
Terms are 8, t, 3t

Question 10.
3z + 4 + 2 + 4z

Answer:
Like terms are 3z, 4z
terms are 4, 2, 3z, 4z

Explanation:
Given expression is 3z + 4 + 2 + 4z
Like terms are 3z, 4z
terms are 4, 2, 3z, 4z

Question 11.
2n – n – 4 + 7n

Answer:
Like terms are 2n, -n, 7n
Terms are -4, 2n, -n, 7n

Explanation:
Given expression is 2n – n – 4 + 7n
Like terms are 2n, -n, 7n
Terms are -4, 2n, -n, 7n

Question 12.
-x – 9x² + 12x² + 7

Answer:
Like terms are -9x², 12x²
Terms are 7, -x, -9x², 12x²

Explanation:
Given expression is -x – 9x² + 12x² + 7
Like terms are -9x², 12x²
Terms are 7, -x, -9x², 12x²

Question 13.
1.4y + 5 – 4.2 – 5y² + z

Answer:
There are no like terms
Terms are 1.4y, 5, -4.2, -5y², z

Explanation:
Given expression is 1.4y + 5 – 4.2 – 5y² + z
There are no like terms
Terms are 1.4y, 5, -4.2, -5y², z

Question 14.

Answer:
Like terms are 1/2 s, 3/4 s
Terms are -4, 1/8, -s³, 1/2 s, 3/4 s

Explanation:
Given expression is 1/2 s – 4 + 3/4 s + 1/8 – s³
Like terms are 1/2 s, 3/4 s
Terms are -4, 1/8, -s³, 1/2 s, 3/4 s

Question 15.
YOU BE THE TEACHER
Your friend identifies the terms and like terms in the expression 3x – 5 – 2x + 9x. Is your friend correct? Explain your reasoning.

Answer:
Wrong

Explanation:
Given expression is 3x – 5 – 2x + 9x
like terms are 3x, -2x, 9x
Terms are 3x, -2x, 9x, 5

SIMPLIFYING ALGEBRAIC EXPRESSIONS
Simplify the expression.

Question 16.
12g + 9g

Answer:
21g

Explanation:
The given expression is 12g + 9g
= 21g

Question 17.
11x + 9 – 7

Answer:
11x + 2

Explanation:
Given expression is 11x + 9 – 7
= 11x + 2

Question 18.
8s – 11s + 6s

Answer:
3s

Explanation:
Given expression is 8s – 11s + 6s
= 14s – 11s = 3s

Question 19.
4b – 24 + 19

Answer:
4b – 5

Explanation:
Given expression is 4b – 24 + 19
= 4b – 5

Question 20.
4p – 5p – 30p

Answer:
-31p

Explanation:
Given expression is 4p – 5p – 30p
= 4p – 35p = -31p

Question 21.
4.2v – 5 – 6.5v

Answer:
-2.3v – 5

Explanation:
Given expression is 4.2v – 5 – 6.5v
= -2.3v – 5

Question 22.
8 + 4a + 6.2 – 9a

Answer:
14.2 – 5a

Explanation:
Given expression is 8 + 4a + 6.2 – 9a
= 14.2 – 5a

Question 23.

Answer:
3 – 1/2 y

Explanation:
Given expression is 2/5y – 4 + 7 – 9/10y
= (4 – 9)/10 y + 3
= -5/10 y + 3
= -1/2 y + 3

Question 24.

Answer:
40c/3 – 3/2

Explanation:
Given expression is -2/3 c -9/5 + 14c + 3/10
= (-2 + 42)c/3 – (18 – 3)/10
= 40c/3 – 15/10
= 40c/3 – 3/2

Question 25.
MODELING REAL LIFE
On a hike, each hiker carries the items shown. Write and interpret an expression in simplest form that represents the weight carried by x hikers. How much total weight is carried when there are 4 hikers?

Answer:
When there are 4 hikers the total weight is 40.8 lb.

Explanation:
A hiker has thw following items:
A sleeping bag of 3.4 lb weight, camping bag of 4.6 lb weight, and a water bottle of 2.2 lb weight
The expression can be
3.4x + 4.6x + 2.2x = x
Here x is the weight each hiker is carrying
10.2x = x
So, each hiker carries a total weight of 10.2 pounds
When there are 4 hikes, the total weight is 10.2 x 4 = 40.8

Question 26.
STRUCTURE
Evaluate the expression -8x + 5 – 2x – 4 + 5 when x = 2 before and after simplifying. Which method do you prefer? Explain.

Answer:
-14

Explanation:
Given expression is -8x + 5 – 2x – 4 + 5
Put x = 2
-8(2) + 5 -2(2) – 4 + 5
= -16 + 5 – 4 – 4 + 5 = -14
Simplify -8x + 5 – 2x – 4 + 5
= -10x + 10 – 4 = -10x + 6
Put x = 2 in -10x + 6
= -10(2) + 6
= -20 + 6 = -14
I just prefer putting x = 2 after simplifying the expression.

Question 27.
OPEN-ENDED
Write an expression with five different terms that is equivalent 8x² + 3x² + 3y. Justify your answer.

Answer:
3x² + 2y + y + 7x² + x²

Explanation:
Given expression is 8x² + 3x² + 3y.
The expression which is equivalent to 8x² + 3x² + 3y and having 5 different terms can be as follows
3x² + 2y + y + 7x² + x²
by simplifying the above expression, you will get the original expression.

Question 28.
STRUCTURE
Which of the following shows a correct way of simplifying 6 + (3 – 5x)? Explain the errors made in the other choices.

Answer:
C. 6 + (3 – 5x) = (6 + 3) – 5x = 9 – 5x

Explanation:
A. 6 + (3 – 5x) = (6 + 3 – 5)x
Here, x is applicable to only -5x, it is not multiplied with whole expression
B. 6 + (3 – 5x) = 6 + (3 – 5)x
Here, x is the multiple of only -5x, but you have included 3 also which is false
C. 6 + (3 – 5x) = (6 + 3) – 5x = 9 – 5x
This is true
D. 6 + (3 – 5x) = (6 + 3 + 5) – x
5 is the multiple of x, it is not constant.

Question 29.
PRECISION
Two comets orbit the Sun. One comet travels 30,000 miles per hour and the other comet travels 28,500 miles per hour. What is the most efficient way to calculate the difference of the distances traveled by the comets for any given number of minutes? Justify your answer.

Answer:
The difference between those distances is 1500 miles for 60 minutes.

Explanation:
Given that,
One comet travels 30,000 miles per hour and the other comet travels 28,500 miles per hour.
Let us say, the time given for the comets is 60 minutes.
So, in 60 minutes, the first one travels 30,000 miles, and the second one travels 28,500 miles
The difference between those distances = 30,000 – 28,500 = 1500 miles

Question 30.
MODELING REAL LIFE
Find the earnings for washing and waxing 12 cars and 8 trucks. Justify your answer.

Answer:
and waxing 12 cars and 8 trucks is $440

Explanation:
The earnings for washing and waxing 12 cars = 12(washing one car cost) + 12(waxing one car cost)
= 12($8) + 12($12)
= 12(8 + 12) = $240
The earnings for washing and waxing 8 trucks = 8(washing one truck cost) + 8(waxing one truck cost)
= 8($10) + 8($15)
= 8(10 + 15) = 8(25) = $200
So, the earnings for washing and waxing 12 cars and 8 trucks is $ 240 + $200 = $440

Question 31.
CRITICAL THINKING
You apply gold foil to a piece of red poster board to make the design shown.
a. Find the area of the gold foil when x = 3. Justify your answer.
b. The pattern at the right is called “St. George’s Cross.” Find a country that uses this pattern as its flag.

Answer:
a. Area of the gold foil = 87 sq in
b. England

Explanation:
You apply gold foil to a piece of red poster board to make the design.
a. Area of rectangle = length x breadth
= 20 x 12 = 240 sq in.
Length of the gold foil = (12 – x) in
The breadth of gold foil = (20 – x) in
Area of horizontal gold foil = length x breadth
= 20 * x = 20x sq. in
Area of the vertical gold foil = length x breath
= 12 * x = 12x sq. in
Area of the middle gold foil = length x breadth
= x * x = x² sq. in
Area of the gold foil = Area of horizontal gold foil + Area of vertical gold foil – Area of the middle gold foil
= 20x + 12x – x²
The area of gold foil when x = 3 is 20(3) + 12(3) – (3)²
= 60 + 36 – 9
= 87 sq in
b. The pattern at the right is called “St. George’s Cross.” Find a country that uses this pattern as its flag.
England

Question 32.
GEOMETRY
Two rectangles have different dimensions. Each rectangle has a perimeter of (7x + 5) inches. Draw and label diagrams that represent possible dimensions of the rectangles.

Answer:

Rectangle 1 dimensions are 3.5x and 2.5, rectangle 2 dimensions are 1/2(5x + 1), x + 2

Explanation:
Perimeter of rectangle = (7x + 5) inches
2(length + width) = 7x + 5
length + width = (7x + 5)/2
Rectangle 1 dimensions can be 3.5x and 2.5
Rectangle 2 dimensions can be 1/2(5x + 1), x + 2

Lesson 3.2 Adding and Subtracting Linear Expressions

EXPLORATION 1

Using Algebra Tiles
Work with a partner. You can use the algebra tiles shown at the left to find sums and differences of algebraic expressions.

a. How can you use algebra tiles to model a sum of terms that equals 0? Explain your reasoning.
b. Write each sum or difference modeled below. Then use the algebra tiles to simplify the expression.

c. Write two algebraic expressions of the form ax + b, where a and b are rational numbers. Find the sum and difference of the expressions.

EXPLORATION 2
Using Properties of Operations
Work with a partner.
a. Do algebraic expressions, such as 2x, -3y, and 3z + 1 have additive inverses? How do you know?
b. How can you find the sums and differences modeled in Exploration 1 without using algebra tiles? Explain your reasoning.

3.2 Lesson

Try It

Find the sum.

Question 1.
(x + 3) + (2x – 1)

Answer:
(x + 3) + (2x – 1) = 3x + 2

Explanation:
The given expression is (x + 3) + (2x – 1)
Add the like terms and constant
= 3x + 3 – 1
= 3x + 2

Question 2.
(-8z + 4) + (8z – 7)

Answer:
(-8z + 4) + (8z – 7) = -3

Explanation:
The given expression is (-8z + 4) + (8z – 7)
= -8z + 4 + 8z – 7
= -3

Question 3.
(4.5 – n) + (-10n + 6.5)

Answer:
(4.5 – n) + (-10n + 6.5) = 11(1 – n)

Explanation:
The given expression is (4.5 – n) + (-10n + 6.5)
= 4.5 – n – 10n + 6.5
= 11 – 11n
= 11(1 – n)

Question 4.

Answer:
(1/2w – 3) + (1/4w + 3) = 3w/4

Explanation:
The given expression is (1/2w – 3) + (1/4w + 3)
= 1/2w – 3 + 1/4w + 3
= (2w + w)/4 = 3w/4

Try It

Find the difference.

Question 5.
(m – 3) – (-m + 12)

Answer:
(m – 3) – (-m + 12) = 2m – 15

Explanation:
The given expression is (m – 3) – (-m + 12)
= m – 3 + m – 12
= 2m – 15

Question 6.
(-2c + 5) – (6.3c + 20)

Answer:
(-2c + 5) – (6.3c + 20) = -83c – 15

Explanation:
The given expression is (-2c + 5) – (6.3c + 20)
= -2c + 5 – 6.3c – 20
= -83c – 15

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
WRITING
Describe how to distinguish a linear expression from a nonlinear expression. Give an example of each.

Answer:
An example of a linear expression is 2x + 1.
An example of a non-linear expression is 2x² + 1

Explanation:
Linear expressions are the expressions where the sum of constants and products of constants and raised to a power of 0 or 1. Non-linear expressions are the expressions where the sum of constants and products of constants and raised to a power that is not 0 or 1.
So, an example of a linear expression is 2x + 1.
An example of a non-linear expression is 2x² + 1

Question 8.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Answer:
What is x more than 3x – 1 is different.

Explanation:
What is x more than 3x – 1 = x + 3x – 1 = 4x – 1
Find 3x – 1 decreased by x = 3x – 1 – x = 2x – 1
What is the difference of 3x – 1 and x = 3x – 1 – x = 2x – 1
Subtract (x + 1) from 3x = 3x – (x + 1) = 3x – x – 1 = 2x – 1

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
DIG DEEPER!
In a basketball game, the home team scores (2m + 39) points and the away team scores (3m + 40) points, where m is the number of minutes since halftime. Who wins the game? What is the difference in the scores minutes after halftime? Explain.

Answer:
The away team wins the game. And the difference in the scores minutes after halftime is 2m + 2.

Explanation:
Given that,
The score of home team = 2m + 39 points
The score of away team = 3m + 40 points
Here, m is the number of minutes since halftime
Let us consider the teams that score the same points after the half time
The score of home team = 2m + 39 + 2m + 39 = 4m + 78
The score of away team = 3m + 40 + 3m + 40 = 6m + 80
The difference between those scores = 6m + 80 – (4m + 78)
= 6m + 80 – 4m – 78 = 2m + 2
The away team wins the game.

Question 10.
Electric guitar kits originally cost d dollars online. You buy the kits on sale for 50% of the original price, plus a shipping fee of $4.50 per kit. After painting and assembly, you sell each guitar online for (1.5d + 4.5) dollars. Find and interpret your profit on each guitar sold.

Answer:
Your profit on each electric guitar kit is d dollars

Explanation:
Electric guitar kits originally cost d dollars online.
You buy the kits on sale for 50% of the original price. So, the purchase price of kit is = d/2 + 4.50
After painting and assembly, you sell each guitar online for (1.5d + 4.5) dollars
Profit = Selling price – Purchased price
= 1.5d + 4.5 – (d/2+ 4.50)
= 1.5d + 4.5 – 0.5d – 4.50
= d dollars
Your profit on each electric guitar kit is d dollars. You pay (0.5d + 4.50) dollars for each kit. so, you are getting 4.50 dollars less than double your money.

Adding and Subtracting Linear Expressions Homework & Practice 3.2

Review & Refresh

Simplify the expression.

Question 1.
4f + 11f

Answer:
4f + 11f = 15f

Explanation:
Given expression is 4f + 11f
Add the like terms
= 15f

Question 2.
b + 4b – 9b

Answer:
b + 4b – 9b = -4b

Explanation:
Given expression is b + 4b – 9b
Perform required operations between like terms
= 5b – 9b = -4b

Question 3.
-4z – 6 – 7z + 3

Answer:
-4z – 6 – 7z + 3 = -11z- 3

Explanation:
Given expression is -4z – 6 – 7z + 3
Perform required operations between like terms and constants
= -11z- 3

Evaluate the expression when x = \(-\frac{4}{5}\) and y = \(\frac{1}{3}\).

Question 4.
x + y

Answer:
-7/15

Explanation:
Given expression is x + y
Put x = -4/5, y = 1/3
= -4/5 + 1/3
= (-12 + 5)/15 = -7/15

Question 5.
2x + 6y

Answer:
2/5

Explanation:
Given expression is 2x + 6y
Put x = -4/5, y = 1/3
= 2(-4/5) + 6(1/3) = -8/5 + 6/3
= (-24 + 30)/15 = 6/15
= 2/5

Question 6.
-x + 4y

Answer:
32/15

Explanation:
Given expression is -x + 4y
Put x = -4/5, y = 1/3
= -(-4/5) + 4(1/3)
= 4/5 + 4/3
= (12 + 20)/15
= 32/15

Question 7.
What is the surface area of a cube that has a side length of 5 feet?
A. 25 ft²
B. 75 ft²
C. 125 ft²
D. 150 ft²

Answer:
D. 150 ft²

Explanation:
Given that,
the side length of cube = 5 feet
The surface area of a cube = 6side²
= 6 x (5)² = 6 x 25
= 150 ft²

Concepts, Skills, & Problem Solving
USING ALGEBRA TILES
Write the sum or difference modeled by the algebra tiles. Then use the algebra tiles to simplify the expression. (See Exploration 1, p. 97.)

Question 8.

Answer:
3x – 1

Explanation:
Convert the given algebra tiles to the expressions
(x – 1 – 1 – 1 + x – 1 – 1 – 1) + (x + 1 + 1+ 1+ 1+ 1)
= (2x – 6) +(x + 5)
= 2x – 6 + x + 5
= 3x – 1

Question 9.

Answer:
11

Explanation:
Convert the given algebra tiles to the expressions
(x + 1 + 1 + 1+ 1 + 1 + x + 1 + 1) – (x – 1 – 1 – 1 – 1 + x)
= (2x + 7) – (2x – 4)
= 2x + 7 – 2x + 4
= 11

ADDING LINEAR EXPRESSIONS
Find the sum.

Question 10.
(n + 8) + (n – 12)

Answer:
(n + 8) + (n – 12) = 2n – 4

Explanation:
Given linear expression is (n + 8) + (n – 12)
= n + 8 + n – 12
= 2n – 4

Question 11.
(7 – b) + (3b + 2)

Answer:
(7 – b) + (3b + 2) = 9 + 2b

Explanation:
Given linear expression is (7 – b) + (3b + 2)
= 7 – b + 3b + 2
= 9 + 2b

Question 12.
(2w – 9) + (-4w – 5)

Answer:
(2w – 9) + (-4w – 5) = -2w – 14

Explanation:
Given linear expression is (2w – 9) + (-4w – 5)
= 2w – 9 – 4w- 5
= -2w – 14

Question 13.
(2x – 6) + (4x – 12)

Answer:
(2x – 6) + (4x – 12) = 8x – 18

Explanation:
Given linear expression is (2x – 6) + (4x – 12)
= 2x – 6 + 4x – 12
= 8x – 18

Question 14.
(-3.4k – 7) + (3k + 21)

Answer:
(-3.4k – 7) + (3k + 21) = -0.4k + 14

Explanation:
Given linear expression is (-3.4k – 7) + (3k + 21)
= -3.4k – 7 + 3k + 21
= -0.4k + 14

Question 15.

Answer:
(-7/2 z+ 4) + (1/5 z- 15) = -33z/10 – 11

Explanation:
Given linear expression is (-7/2 z+ 4) + (1/5 z- 15)
= -7/2 z + 4 + 1/5 z- 15
= (-35z + 2z)/10 – 11
= -33z/10 – 11

Question 16.
(6 – 2.7h) + (-1.3j – 4)

Answer:
(6 – 2.7h) + (-1.3j – 4) = 2 – 2.7h – 1.3j

Explanation:
Given linear expression is (6 – 2.7h) + (-1.3j – 4)
= 6 – 2.7h – 1.3j – 4
= 2 – 2.7h – 1.3j

Question 17.

Answer:
(7/4 x – 5) + (2y – 3.5) + (-1/4 x + 5) = 3x/2 + 2y – 3.5

Explanation:
Given linear expression is (7/4 x – 5) + (2y – 3.5) + (-1/4 x + 5)
= 7/4 x – 5 + 2y – 3.5 – 1/4x + 5
= (7x – x)/4 + 2y – 3.5
= 6x/4 + 2y – 3.5
= 3x/2 + 2y – 3.5

Question 18.
MODELING REAL LIFE
While catching fireflies, you and a friend decide to have a competition. After m minutes, you have (3m + 13) fireflies and your friend has (4m + 6) fireflies.
a. How many total fireflies do you and your friend catch? Explain your reasoning.
b. The competition lasts 3 minutes. Who has more fireflies? Justify your answer.

Answer:
a. The total number of fireflies collected is 7m + 19
b. You have more fireflies when compared with your friend.

Explanation:
The number of fireflies you collected after m minutes is (3m + 13)
The number of fireflies your friend collected after m minutes is (4m + 6)
a.
The number of fireflies you collected is (3m + 13)
The number of fireflies your friend collected is (4m + 6)
The total number of fireflies collected = (3m + 13) + (4m + 6)
= 3m + 13 + 4m + 6 = 7m + 19
b. The competition lasts 3 minutes
So, The number of fireflies you collected is 3(1) + 13 = 3 + 13 = 16
The number of fireflies your friend collected is 4(1) + 6 = 4 + 6 = 10
So, you have more fireflies when compared with your friend.

SUBTRACTING LINEAR EXPRESSIONS

Find the difference.

Question 19.
(-2g + 7) – (g + 11)

Answer:
(-2g + 7) – (g + 11) = -3g – 4

Explanation:
The given linear expression is (-2g + 7) – (g + 11)
= -2g + 7 – g – 11
= -3g – 4

Question 20.
(6d + 5) – (2 – 3d)

Answer:
(6d + 5) – (2 – 3d) = 9d + 3

Explanation:
The given linear expression is (6d + 5) – (2 – 3d)
= 6d + 5 – 2 + 3d
= 9d + 3

Question 21.
(4 – 5y) – (2y – 16)

Answer:
(4 – 5y) – (2y – 16) = 20 – 7y

Explanation:
The given linear expression is (4 – 5y) – (2y – 16)
= 4 – 5y – 2y + 16
= 20 – 7y

Question 22.
(2n – 9) – (-2.4n + 4)

Answer:
(2n – 9) – (-2.4n + 4) = 4.4n – 13

Explanation:
The given linear expression is (2n – 9) – (-2.4n + 4)
= 2n – 9 + 2.4n – 4
= 4.4n – 13

Question 23.

Answer:
(-1/8 c + 16) – (3/8 + 3c) = (-25c + 125)/8

Explanation:
The given linear expression is (-1/8 c + 16) – (3/8 + 3c)
= -1/8 c + 16 – 3/8 – 3c
= (-c – 24c)/8 + (128 – 3)/8
= -25c/8 + 125/8
= (-25c + 125)/8

Question 24.

Answer:
(9/4 x + 6) – (-5/4 x – 24) = 7x/2 + 30

Explanation:
The given linear expression is (9/4 x + 6) – (-5/4 x – 24)
= 9x/4 + 6 + 5x/4 + 24
= (9x + 5x)/4 + 30
= 14x/4 + 30
= 7x/2 + 30

Question 25.

Answer:
(1/3 – 6m) – (1/4 n – 8) = 25/4 – 6m – n/4

Explanation:
The given linear expression is (1/3 – 6m) – (1/4 n – 8)
= 1/3 – 6m – 1/4 n + 8
= (1 + 24)/3 – 6m – n/4
= 25/4 – 6m – n/4

Question 26.
(1 – 5q) – (2.5s + 8) – (0.5q + 6)

Answer:
(1 – 5q) – (2.5s + 8) – (0.5q + 6) = -5.5q – 2.5s – 13

Explanation:
The given linear expression is (1 – 5q) – (2.5s + 8) – (0.5q + 6)
= 1 – 5q – 2.5s – 8 – 0.5q – 6
= 1 – 5.5q – 2.5s – 14
= -5.5q – 2.5s – 13

Question 27.
YOU BE THE TEACHER
Your friend finds the difference (4m + 9) – (2m – 5). Is your friend correct? Explain your reasoning.

Answer:
Wrong

Explanation:
The given expression is (4m + 9) – (2m – 5)
= 4m + 9 – 2m + 5
= 2m + 14

Question 28.
GEOMETRY
The expression 17n + 11 represents the perimeter of the triangle. What is the length of the third side? Explain your reasoning.

Answer:
The length of the third side is 8n.

Explanation:
Perimeter of the triangle = 17n + 11
5n + 6 + 4n + 5 + third side = 17n + 11
9n + 11 + third side = 17n + 11
third side = 17n + 11 – (9n +11)
= 17n + 11 – 9n – 11
= 8n
The length of the third side is 8n.

Question 29.
LOGIC
Your friend says the sum of two linear expressions is always a linear expression. Is your friend correct? Explain.

Answer:
yes, the sum of two linear expressions is always a linear expression.

Explanation:
The sum of two linear expressions is always a linear expression. But the product of linear expressions may not be the linear expression.

Question 30.
MODELING REAL LIFE
You burn 265 calories running and then 7 calories per minute swimming. Your friend burns 273 calories running and then 11 calories per minute swimming. You each swim for the same number of minutes. Find and interpret the difference in the amounts of calories burned by you and your friend.

Answer:
The difference in the amounts of calories burned by you and your friend is 8 + 4m.

Explanation:
The number of calories your burn by running and swimming is 265 + 7m
The number of calories your friend burns by running and swimming is 273 + 11m
The difference in amount of calories burn = 273 + 11m – (265 + 7m)
= 273 + 11m – 265 – 7m
= 8 + 4m
The difference in the amounts of calories burned by you and your friend is 8 + 4m.

Question 31.
DIG DEEPER!
You start a new job. After w weeks, you have (10w + 120) dollars in your savings account and (45w + 25) dollars in your checking account.
a. What is the total amount of money in the accounts? Explain.
b. How much money did you have before you started your new job? How much money do you save each week?
c. You want to buy a new phone for $150, and still have $500 left in your accounts afterwards. Explain how to determine when you can buy the phone.

Answer:
a. The total amount of money in the accounts = (55w + 145) dollars
b. You started a new job with 120 dollars savings and you earned 10 dollars savings each week.
c. You can buy the phone after 9 weeks of starting new job.

Explanation:
You have (10w + 120) dollars in your savings account and (45w + 25) dollars in your checking account.
a.
The total amount of money in the accounts = 10w + 120 + 45w + 25
= 55w + 145
b.
In the saving’s account the multiple of w is the amount earned during job and other constant is the amount you have before the new job.
So, you started a new job with 120 dollars savings and you earned 10 dollars savings each week.
c.
The total amount in accounts is (55w + 145) dollars
You want to buy a new phone for $150, and still have $500 left in your accounts afterwards
Deduct $150 from total amount so that the remaining amount should be $500.
55w + 145 – 150 = 500
55w – 5 = 500
55w = 500 + 5
55w = 505
w = $505/55
w = 9.1818
So, you can buy the phone after 9 weeks of starting new job.

Question 32.
REASONING
Write an expression in simplest form that represents the vertical distance between the two lines shown. What is the distance when x = 3? when x = -3?

Answer:
The vertical distance between the two lines is 3 units.

Explanation:
Given equations are y = x – 1, y = 2x – 4
Equating those two
x – 1 = 2x – 4
-1 + 4 = 2x – x
3 = x
So, the vertical distance between the two lines is 3 units.

Lesson 3.3 The Distributive Property

EXPLORATION 1

Using Models to Write Expressions
Work with a partner.
a. Write an expression that represents the area of the shaded region in each figure.


b. Compare your expressions in part(a) with other groups in your class. Did other groups write expressions that look different than yours? If so, determine whether the expressions are equivalent.

3.3 Lesson

Try It
Simplify the expression.

Question 1.
-1(x + 9)

Answer:
-1(x + 9) = 9 – x

Explanation:
Given expression is -1(x + 9)
= -x + 9

Question 2.

Answer:
2/3 (-3z – 6) = -2z – 4

Explanation:
Given expression is 2/3 (-3z – 6)
= 2/3(-3z) – 2/3(6)
= -2z – 2(2)
= -2z – 4

Question 3.
-1.5(8m – n)

Answer:
-1.5(8m – n) = -12m + 1.5n

Explanation:
Given expression is -1.5(8m – n)
= -1.5(8m) + 1.5(n)
= -12m + 1.5n

Try It

Simplify the expression.

Question 4.
2(-3s + 1 – 5)

Answer:
2(-3s + 1 – 5) = -6s – 8

Explanation:
Given expression is 2(-3s + 1 – 5)
= 2(-3s) + 2(1) + 2(-5)
= -6s + 2 – 10
= -6s – 8

Question 5.

Answer:
-3/2 (a – 4 – 2a) = 3a/2 + 6

Explanation:
Given expression is -3/2 (a – 4 – 2a)
= -3/2 (a) + 3/2(4) + 3/2(2a)
= -3a/2 + 3(2) + 3(a)
= -3a/2 + 6 + 3a
= (-3a + 6a)/2 + 6
= 3a/2 + 6

Try It

Simplify the expression.

Question 6.
3.5m – 1.5(m – 10)

Answer:
3.5m – 1.5(m – 10) = 2m + 15

Explanation:
Given expression is 3.5m – 1.5(m – 10)
= 3.5m – 1.5(m) + 1.5(10)
= 3.5m – 1.5m + 15
= 2m + 15

Question 7.

Answer:
4/5(10w – 5) – 2(w + 9) = 6w – 22

Explanation:
Given expression is 4/5(10w – 5) – 2(w + 9)
= 4/5(10w) – 4/5(5) – 2w – 18
= 4(2w) – 4 – 2w – 18
= 8w – 2w – 22
= 6w – 22

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
WRITING
Explain how to use the Distributive Property when simplifying an expression.

Answer:
Distributive property says that a(b+ c) = ab + ac and a(b – c) = ab – ac
So, you need to expand the outer number with the sum or difference inside the braces
And perform the required arithmetical operations to get the simplified expression.

USING THE DISTRIBUTIVE PROPERTY
Simplify the expression.

Question 9.

Answer:
5/6 (-2y + 3) = -5y/3 + 5/2

Explanation:
Given expression is 5/6 (-2y + 3)
= 5/6(-2y) + 5/6(3)
= -10y/6 + 15/6
= -5y/3 + 5/2

Question 10.
6(3s – 2.5 – 5s)

Answer:
6(3s – 2.5 – 5s) = -15 – 12s

Explanation:
Given expression is 6(3s – 2.5 – 5s)
= 6(-2.5 – 2s)
= 6(-2.5) – 6(2s)
= -15 – 12s

Question 11.

Answer:
3/10(4m – 8) + 9m = (51m – 12)/5

Explanation:
Given expression is 3/10(4m – 8) + 9m
= 3/10(4m) – 3/10(8) + 9m
= 3/5(2m) – 3/5(4) + 9m
= 6m/5 – 12/5 + 9m
= (6m + 45m)/5 – 12/5
= 51m/5 – 12/5
= (51m – 12)/5

Question 12.
2.25 – 2(7.5 – 4h)

Answer:
2.25 – 2(7.5 – 4h) = 8h – 12.75

Explanation:
Given expression is 2.25 – 2(7.5 – 4h)
= 2.25 – 2(7.5) + 2(4h)
= 2.25 – 15 + 8h
= 8h – 12.75

Question 13.
STRUCTURE

Use the terms at the left to complete the expression below so that it is equivalent to 9x – 12. Justify your answer.

Answer:
3/2 (4x – 8) + 3x

Explanation:
The expression can be
3/2 (4x – 8) + 3x
= 3/2(4(x – 2)) + 3x
= 3(2(x – 2)) + 3x
= 6(x – 2) + 3x
= 6x – 12 + 3x
= 9x – 12

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 14.
A rectangular room is 10 feet longer than it is wide. How many 1-foot square tiles does it take to tile along the inside walls of the room?

Answer:
The number of tiles required is 4w + 20.

Explanation:
Let the width of the rectangular room is w
The length of the rectangular room is w + 10
Perimeter of the rectangular room = 2(length + with)
= 2(w + 10 + w)
= 2(2w +10)
= 4w + 20
So, the number of tiles required is 4w + 20.

Question 15.
How many 2-foot square tiles does it take to tile the border of the pool in Example 4? Explain.

Answer:
(4s + 8) 2 foot square tiles are required.

Explanation:

The diagram shows that the tiled border can be divided into two sections that each requires s + 4 tiles and two sections that each requires s tiles. So, the number of tiles can be represented as 2(s + 4) + 2s
2(s + 4) + 2s = 2(s) + 2(4) + 2s
= 4s + 8
So, (4s + 8) 2 foot square tiles are required.

The Distributive Property Homework & Practice 3.3

Review & Refresh

Find the sum or difference.

Question 1.
(5b – 9) + (b + 8)

Answer:
(5b – 9) + (b + 8) = 6b – 1

Explanation:
Given expression is (5b – 9) + (b + 8)
= 5b – 9 + b + 8
= 6b – 1

Question 2.
(3m + 5) – (6 – 5m)

Answer:
(3m + 5) – (6 – 5m) = 8m – 1

Explanation:
Given expression is (3m + 5) – (6 – 5m)
= 3m + 5 – 6 + 5m
= 8m – 1

Question 3.
(1 – 9z) + 3(z – 2)

Answer:
(1 – 9z) + 3(z – 2) = -5 – 6z

Explanation:
Given expression is (1 – 9z) + 3(z – 2)
= 1 – 9z + 3(z) – 3(2)
= 1 – 9z + 3z – 6
= -5 – 6z

Question 4.
(7g – 6) – (-3n – 4)

Answer:
(7g – 6) – (-3n – 4) = 7g + 3n – 2

Explanation:
Given expression is (7g – 6) – (-3n – 4)
= 7g – 6 -(-3n) -(-4)
= 7g – 6 + 3n + 4
= 7g + 3n – 2

Evaluate the expression.

Question 5.
-6²

Answer:
-6² = -36

Explanation:
Given expression is -6²
-6 x 6 = -36

Question 6.
-9² . 3

Answer:
-9² . 3 = -243

Explanation:
Given expression is -9² . 3
= -9 . 9 . 3
= -9 . 27
= -243

Question 7.
(-7) . (-2) . (-4)

Answer:
(-7) . (-2) . (-4) = -56

Explanation:
Given expression is (-7) . (-2) . (-4)
= -7 . -2 . -4
= 14 . -4
= -56

Copy and complete the statement using <, >, or =.

Question 8.

Answer:
11 = | -11 |

Explanation:
11 = | -11 |
mod of -11 is positive 11
So, they are equal.

Question 9.

Answer:
| 3.5 | <  | -5.8 |

Explanation:
| 3.5 | _  | -5.8 |
3.5 < 5.8

Question 10.

Answer:
| -3.5 | <  | 17/5 |

Explanation:
| -3.5 | _  | 17/5 |
3.5 < 3.4

Concepts, Skills, & Problem Solving

USING MODELS
Write two different expressions that represent the area of the shaded region. Show that the expressions are equivalent. (See Exploration 1, p. 103.)

Question 11.

Answer:
6.5(2x + 2), 13x + 13

Explanation:
The length of the shaded region is 6.5
The width of the shaded region is (3x + 4) – (x + 2) = 3x + 4 – x – 2
= 2x + 2
The area of the shaded region = length . width
= 6.5 . (2x + 2)
= 6.5(2x + 2)
= 6.5(2x) + 6.5(2)
= 13x + 13
The second way is as follows
The length of bigger rectangle is 3x + 4
Width of the bigger rectangle is 6.5
Area of the bigger rectangle = length . width
= 6.5(3x + 4)
The length of unshaded region is x + 2
width of the unshaded region is 6.2
Area of the unshaded region = length . width
= 6.5(x + 2)
Area of the shaded region = Area of the bigger rectangle – area of the shaded region
= 6.5(3x + 4) – 6.5(x + 2)
= 6.5(3x + 4 -(x + 2))
= 6.5(3x + 4 – x – 2)
= 6.5(2x + 2)
= 6.5(2x) + 6.5(2)
= 13x + 13