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UNIT 7 Similarity and Right Triangles CONTENTS COMMON CORE G-SRT.A.1a G-SRT.A.2 G-SRT.A.2 G-SRT.A.3 COMMON CORE G-SRT.B.4 G-GPE.B.6 G-SRT.B.5 G-SRT.B.4 COMMON CORE G-SRT.C.6 G-SRT.C.6 G-SRT.C.8 G-SRT.C.8 F-TF.C.8 823A Unit 7 MODULE 16 Similarity and Transformations Lesson 16.1 Lesson 16.2 Lesson 16.3 Lesson 16.4 Dilations . . . . . . . . . . . . . . . . . . . . . . . . . . . . Proving Figures are Similar Using Transformations . Corresponding Parts of Similar Figures . . . . . . . . AA Similarity of Triangles . . . . . . . . . . . . . . . . . MODULE 17 Using Similar Triangles Lesson 17.1 Lesson 17.2 Lesson 17.3 Lesson 17.4 Triangle Proportionality Theorem . . . . . Subdividing a Segment in a Given Ratio . Using Proportional Relationships . . . . . Similarity in Right Triangles . . . . . . . . . MODULE 18 Trigonometry with Right Triangles Lesson 18.1 Lesson 18.2 Lesson 18.3 Lesson 18.4 Lesson 18.5 Tangent Ratio . . . . . . . . . . . . . . . Sine and Cosine Ratios . . . . . . . . . Special Right Triangles . . . . . . . . . Problem Solving with Trigonometry Using a Pythagorean Identity. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 827 837 851 861 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 881 891 903 913 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 931 941 953 967 981 UNIT 7 Unit Pacing Guide 45-Minute Classes Module 16 DAY 1 DAY 2 DAY 3 DAY 4 DAY 5 Lesson 16.1 Lesson 16.1 Lesson 16.2 Lesson 16.3 Lesson 16.4 DAY 6 Module Review and Assessment Readiness Module 17 DAY 1 DAY 2 DAY 3 DAY 4 DAY 5 Lesson 17.1 Lesson 17.2 Lesson 17.3 Lesson 17.3 Lesson 17.4 DAY 1 DAY 2 DAY 3 DAY 4 DAY 5 Lesson 18.1 Lesson 18.1 Lesson 18.2 Lesson 18.2 Lesson 18.3 DAY 6 DAY 7 DAY 8 DAY 9 DAY 10 Lesson 18.4 Lesson 18.4 Lesson 18.5 Module Review and Assessment Readiness Unit Review and Assessment Readiness DAY 6 Module Review and Assessment Readiness Module 18 90-Minute Classes Module 16 DAY 1 DAY 2 DAY 2 Lesson 16.1 Lesson 16.2 Lesson 16.3 Lesson 16.4 Module Review and Assessment Readiness Module 17 DAY 1 DAY 2 DAY 3 Lesson 17.1 Lesson 17.2 Lesson 17.3 Lesson 17.4 Module Review and Assessment Readiness DAY 1 DAY 2 DAY 3 DAY 4 DAY 5 Lesson 18.1 Lesson 18.2 Lesson 18.3 Lesson 18.4 Module Review and Assessment Readiness Lesson 18.4 Lesson 18.5 Unit Review and Assessment Readiness Module 18 Unit 7 823B Program Resources PLAN ENGAGE AND EXPLORE HMH Teacher App Access a full suite of teacher resources online and offline on a variety of devices. Plan present, and manage classes, assignments, and activities. Real-World Videos Engage students with interesting and relevant applications of the mathematical content of each module. Explore Activities Students interactively explore new concepts using a variety of tools and approaches. ePlanner Easily plan your classes, create and view assignments, and access all program resources with your online, customizable planning tool. Professional Development Videos Authors Juli Dixon and Matt Larson model successful teaching practices and strategies in actual classroom settings. QR Codes Scan with your smart phone to jump directly from your print book to online videos and other resources. DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A Teacher’s Edition Support students with point-of-use Questioning Strategies, teaching tips, resources for differentiated instruction, additional activities, and more. NOT EDIT--Changes must made through "File info" DODO NOT EDIT--Changes must bebe made through "File info" CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A 22.2 Name Name Isosceles and Equilateral Triangles NOT EDIT--Changes must made through "File info" DODO NOT EDIT--Changes must bebe made through "File info" CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A Class Class straightedge to draw segment to draw lineline segment BCBC . . CCUseUsethethestraightedge 22.2 Isosceles Isoscelesand andEquilateral Equilateral 22.2 Triangles Triangles Common Core Math Standards Investigating Isosceles Triangles INTEGRATE TECHNOLOGY Resource Resource Locker Locker G-CO.C.10 Explore Explore Prove theorems about triangles. Vertex angle Vertex angle MP.3 Logic angles have base a side base angles. TheThe angles thatthat have thethe base as aasside areare thethe base angles. ENGAGE work in the space provided. a straightedge to draw angle. in the space provided. UseUse a straightedge to draw an an angle. AADoDoyouryourwork a different each time. is aisdifferent sizesize each time. Label your angle ∠A, as shown in the figure. Label your angle ∠A, as shown in the figure. A A Reflect Reflect © Houghton Mifflin Harcourt Publishing Company Make a Conjecture Looking at your results, what conjecture made about base angles, 2. 2. Make a Conjecture Looking at your results, what conjecture cancan be be made about thethe base angles, ∠C? ∠B∠B andand ∠C? The base angles congruent. The base angles areare congruent. Using a compass, place point vertex draw intersects a compass, place thethe point onon thethe vertex andand draw an an arcarc thatthat intersects thethe BBUsing Explain Explain 1 1 Proving Provingthe theIsosceles Isosceles Triangle Theorem Triangle Theorem sides of the angle. Label points B and sides of the angle. Label thethe points B and C. C. andItsItsConverse Converse and A A In the Explore, made a conjecture base angles of an isosceles triangle congruent. In the Explore, youyou made a conjecture thatthat thethe base angles of an isosceles triangle areare congruent. This conjecture proven it can stated a theorem. This conjecture cancan be be proven so so it can be be stated as aastheorem. C C Isosceles Triangle Theorem Isosceles Triangle Theorem If two sides a triangle congruent, then angles opposite sides If two sides of aoftriangle areare congruent, then thethe twotwo angles opposite thethe sides areare congruent. congruent. This theorem is sometimes called Base Angles Theorem stated as “Base angles This theorem is sometimes called thethe Base Angles Theorem andand cancan alsoalso be be stated as “Base angles of an isosceles triangle congruent. of an isosceles triangle areare congruent. ” ” Module Module 22 22 must be EDIT--Changes A;CA-A DO NOT CorrectionKey=NL- made through Lesson Lesson 2 2 1097 1097 "File info" The angles that have the base as a side are the base angles. In this activity, you will construct isosceles triangles and investigate other potential characteristics/properties of these special triangles. How know triangles constructed isosceles triangles? 1. 1. How do do youyou know thethe triangles youyou constructed areare isosceles triangles? ――. ―― The compass marks equal lengths both sides ∠A; therefore, ABAB ≅≅ ACAC . The compass marks equal lengths onon both sides of of ∠A; therefore, Check students’ construtions. Check students’ construtions. B B The side opposite the vertex angle is the base. How could you draw isosceles triangles without using a compass? Possible answer: Draw ∠A and plot point B on one side of ∠A. Then _ use a ruler to measure AB and plot point C on the other side of ∠A so that AC = AB. Repeat steps A–D at least more times record results in the table. Make sure steps A–D at least twotwo more times andand record thethe results in the table. Make sure ∠A∠A EERepeat Module Module 22 22 1098 1098 Lesson Lesson 2 2 Date EXPLAIN 1 Proving the Isosceles Triangle Theorem and Its Converse Essential COMMON CORE IN1_MNLESE389762_U8M22L2 IN1_MNLESE389762_U8M22L2 10971097 Question: G-CO.C.10 relationships the special What are triangles? and equilateral Prove theorems triangle is The congruent The angle The side a triangle sides are formed by opposite with at least called the the legs is the vertex among in isosceles Resource Locker HARDCOVER PAGES 10971110 HARDCOVER PAGES 10971110 PROFESSIONALDEVELOPMENT DEVELOPMENT PROFESSIONAL about triangles. Investigating Explore An isosceles sides angles and legs of the the vertex angle is the Isosceles two congruent Triangles sides. Legs Vertex angle triangle. Base angle. Base angles base. the as a side are base angles. other potential the base and investigate that have triangles isosceles you will construct special triangles. angle. es of these In this activity, to draw an characteristics/properti Use a straightedge space provided. figure. work in the in the Do your as shown angle ∠A, A Label your The angles Check students’ construtions. 4/19/14 12:10 4/19/14 12:10 PM PM Watch the hardcover Watch forfor the hardcover student edition page student edition page numbers this lesson. numbers forfor this lesson. IN1_MNLESE389762_U8M22L2 IN1_MNLESE389762_U8M22L2 10981098 LearningProgressions Progressions Learning this lesson, students add their prior knowledge isosceles and equilateral InIn this lesson, students add toto their prior knowledge ofof isosceles and equilateral 4/19/14 12:10 4/19/14 12:10 PM PM Do your work in the space provided. Use a straightedge to draw an angle. Label your angle ∠A, as shown in the figure. A CONNECT VOCABULARY Ask a volunteer to define isosceles triangle and have students give real-world examples of them. If possible, show the class a baseball pennant or other flag in the shape of an isosceles triangle. Tell students they will be proving theorems about isosceles triangles and investigating their properties in this lesson. Class al and Equilater 22.2 Isosceles Triangles Name Legs The angle formed by the legs is the vertex angle. What must be true about the triangles you construct in order for them to be isosceles triangles? They must have two congruent sides. m∠B m∠B m∠C m∠C Vertex angle The congruent sides are called the legs of the triangle. QUESTIONING STRATEGIES Possible answer Triangle m∠A 70°; m∠B ∠55°; m∠C 55°. Possible answer forfor Triangle 1: 1: m∠A == 70°; m∠B == ∠55°; m∠C == 55°. In this activity, construct isosceles triangles investigate other potential In this activity, youyou willwill construct isosceles triangles andand investigate other potential characteristics/properties of these special triangles. characteristics/properties of these special triangles. © Houghton Mifflin Harcourt Publishing Company Triangle Triangle 4 4 © Houghton Mifflin Harcourt Publishing Company View the Engage section online. Discuss the photo, explaining that the instrument is a sextant and that long ago it was used to measure the elevation of the sun and stars, allowing one’s position on Earth’s surface to be calculated. Then preview the Lesson Performance Task. Triangle Triangle 3 3 © Houghton Mifflin Harcourt Publishing Company PREVIEW: LESSON PERFORMANCE TASK Triangle Triangle 2 2 m∠mA∠A Base Base Base angles Base angles opposite vertex angle is the base. TheThe sideside opposite thethe vertex angle is the base. Explain to a partner what you can deduce about a triangle if it has two sides with the same length. In an isosceles triangle, the angles opposite the congruent sides are congruent. In an equilateral triangle, all the sides and angles are congruent, and the measure of each angle is 60°. Triangle Triangle 1 1 angle formed is the vertex angle. TheThe angle formed by by thethe legslegs is the vertex angle. Language Objective Essential Question: What are the special relationships among angles and sides in isosceles and equilateral triangles? Legs Legs congruent sides called of the triangle. TheThe congruent sides areare called thethe legslegs of the triangle. ing Company COMMON CORE DD InvestigatingIsosceles Isosceles Triangles Investigating Triangles isosceles triangle a triangle with at least congruent sides. AnAn isosceles triangle is aistriangle with at least twotwo congruent sides. Mathematical Practices Investigating Isosceles Triangles An isosceles triangle is a triangle with at least two congruent sides. Students have the option of completing the isosceles triangle activity either in the book or online. a protractor to measure each angle. Record measures in the table under column UseUse a protractor to measure each angle. Record thethe measures in the table under thethe column Triangle forfor Triangle 1. 1. Resource Locker G-CO.C.10 Prove theorems about triangles. Explore C C The student is expected to: COMMON CORE COMMON CORE EXPLORE A A B B Essential Question: What special relationships among angles and sides in isosceles Essential Question: What areare thethe special relationships among angles and sides in isosceles and equilateral triangles? and equilateral triangles? Date Essential Question: What are the special relationships among angles and sides in isosceles and equilateral triangles? NOT EDIT--Changes must made through "File info" DODO NOT EDIT--Changes must bebe made through "File info" CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A __ Date Date Class 22.2 Isosceles and Equilateral Triangles DONOT NOTEDIT--Changes EDIT--Changesmust mustbebemade madethrough through"File "Fileinfo" info" DO CorrectionKey=NL-A;CA-A CorrectionKey=NL-A;CA-A DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A LESSON Name Base Base angles PROFESSIONAL DEVELOPMENT TEACH ASSESSMENT AND INTERVENTION Math On the Spot video tutorials, featuring program authors Dr. Edward Burger and Martha Sandoval-Martinez, accompany every example in the textbook and give students step-by-step instructions and explanations of key math concepts. Interactive Teacher Edition Customize and present course materials with collaborative activities and integrated formative assessment. C1 Lesson 19.2 Precision and Accuracy Evaluate 1 Lesson XX.X ComparingLesson Linear, Exponential, and Quadratic Models 19.2 Precision and Accuracy teacher Support 1 EXPLAIN Concept 1 Explain The Personal Math Trainer provides online practice, homework, assessments, and intervention. Monitor student progress through reports and alerts. Create and customize assignments aligned to specific lessons or Common Core standards. • Practice – With dynamic items and assignments, students get unlimited practice on key concepts supported by guided examples, step-by-step solutions, and video tutorials. • Assessments – Choose from course assignments or customize your own based on course content, Common Core standards, difficulty levels, and more. • Homework – Students can complete online homework with a wide variety of problem types, including the ability to enter expressions, equations, and graphs. Let the system automatically grade homework, so you can focus where your students need help the most! • Intervention – Let the Personal Math Trainer automatically prescribe a targeted, personalized intervention path for your students. 2 3 4 Question 3 of 17 Concept 2 Determining Precision ComPLEtINg thE SquArE wIth EXPrESSIoNS Avoid Common Errors Some students may not pay attention to whether b is positive or negative, since c is positive regardless of the sign of b. Have student change the sign of b in some problems and compare the factored forms of both expressions. questioning Strategies In a perfect square trinomial, is the last term always positive? Explain. es, a perfect square trinomial can be either (a + b)2 or (a – b)2 which can be factored as (a + b)2 = a 2 + 2ab = b 2 and (a – b)2 = a 2 + 2ab = b 2. In both cases the last term is positive. reflect 3. The sign of b has no effect on the sign of c because c = ( b __ 2 ) 2 and a nonzero number squared is always positive. Thus, c is always positive. c = ( b __ 2 ) 2 and a nonzero number c = ( b __ 2 ) 2 and a nonzero number 5 6 7 View Step by Step 8 9 10 11 - 17 Video Tutor Personal Math Trainer Textbook X2 Animated Math Solve the quadratic equation by factoring. 7x + 44x = 7x − 10 As you have seen, measurements are given to a certain precision. Therefore, x= the value reported does not necessarily represent the actual value of the measurement. For example, a measurement of 5 centimeters, which is , Check given to the nearest whole unit, can actually range from 0.5 units below the reported value, 4.5 centimeters, up to, but not including, 0.5 units above it, 5.5 centimeters. The actual length, l, is within a range of possible values: Save & Close centimeters. Similarly, a length given to the nearest tenth can actually range from 0.05 units below the reported value up to, but not including, 0.05 units above it. So a length reported as 4.5 cm could actually be as low as 4.45 cm or as high as nearly 4.55 cm. ? ! Turn It In Elaborate Look Back Focus on Higher Order Thinking Raise the bar with homework and practice that incorporates higher-order thinking and mathematical practices in every lesson. Differentiated Instruction Resources Support all learners with Differentiated Instruction Resources, including • Leveled Practice and Problem Solving • Reading Strategies • Success for English Learners • Challenge Calculate the minimum and maximum possible areas. Round your answer to Assessment Readiness the nearest square centimeters. The width and length of a rectangle are 8 cm and 19.5 cm, respectively. Prepare students for success on high stakes tests for Integrated Mathematics 2 with practice at every module and unit Find the range of values for the actual length and width of the rectangle. Minimum width = 7.5 cm and maximum width < 8.5 cm My answer Find the range of values for the actual length and width of the rectangle. Minimum length = 19.45 cm and maximum length < 19.55 Name ________________________________________ Date __________________ Class __________________ LESSON 1-1 Assessment Resources cm Name ____________ __________________ __________ Date __________________ LESSON Class ____________ ______ Precision and Significant Digits 6-1 Success for English Learners Linear Functions Reteach The graph of a linear The precision of a measurement is determined bythe therange smallest unit or Find of values for the actual length and width of the rectangle. function is a straig ht line. fraction of a unit used. Ax + By + C = 0 is the standard form for the equat ion of a linear functi • A, B, and C are on. Problem 1 Minimum Area = Minimum width × Minimum length real numbers. A and B are not both zero. • The variables x and y Choose the more precise measurement. = 7.5 cm × 19.45 cm have exponents of 1 are not multiplied together are not in denom 42.3 g is to the 42.27 g is to the inators, exponents or radical signs. nearest tenth. nearest Examples These are NOT hundredth. linear functions: 2+4=6 no variable x2 = 9 exponent on x ≥ 1 xy = 8 x and y multiplied 42.3 g or 42.27 g together 6 =3 Because a hundredth of a gram is smaller than a tenth of a gram, 42.27 g x in denominator x is more precise. 2y = 8 y in exponent Problem 2 In the above exercise, the location of the uncertainty in the linear y = 5 y in a square root measurements results in different amounts of uncertainty in the calculated Choose the more precise measurement: 36 inches or 3 feet. measurement. Explain how to fix this problem. Tell whether each function is linear or not. 1. 14 = 2 x 2. 3xy = 27 3. 14 = 28 4. 6x 2 = 12 x ____________ Reflect ____ ________________ _______________ The graph of y = C is always a horiz ontal line. The graph always a vertical line. of x = C _______________ is Unit 7 Send to Notebook _________________________________________________________________________________________ 2. An object is weighed on three different scales. The results are shown Explore in the table. Which scale is the most precise? Explain your answer. Measurement ____________________________________________________________ • Tier 1, Tier 2, and Tier 3 Resources Examples 1. When deciding which measurement is more precise, what should you Formula consider? Scale Tailor assessments and response to intervention to meet the needs of all your classes and students, including • Leveled Module Quizzes • Leveled Unit Tests • Unit Performance Tasks • Placement, Diagnostic, and Quarterly Benchmark Tests Your Turn y=1 T x=2 y = −3 x=3 823D Math Background Transformations COMMON CORE G-SRT.A.2 LESSONS 16.1 and 16.2 A transformation is a function that changes the position, size, or shape of a figure. In this course, the emphasis is on transformations that are most closely linked to congruence and similarity: reflections, translations, rotations, and dilations. However, it is important to understand that there exist many other transformations. Perhaps the simplest transformation is the transformation that maps every point to itself. This is known as the identity transformation. Another simple transformation is the one that maps every point to the origin. Dilations COMMON CORE G-SRT.A.1 LESSON 16.1 A dilation is a transformation that changes the size of a figure, but not its shape. As such, a dilation is an example of a transformation that is not an isometry (unless the scale factor of the dilation is 1). Every dilation has exactly one fixed point, which is the center of the dilation. Although dilations do not preserve distance, they do preserve other properties of a figure. For example, dilations preserve angle measure. In other words, under a dilation, an angle in the preimage is congruent to the corresponding angle in the image. Dilations also preserve parallel lines. That is, two lines that are parallel in the preimage are mapped to two parallel lines in the image. If two figures are congruent, then there is an isometry that maps one figure onto the other. If two figures are similar, one may be mapped onto the other through a combination of a dilation and an isometry. A dilation with a scale factor greater than 1 is an enlargement. A dilation with a scale factor greater than 0 but less than 1 is a reduction. It is also possible to extend the definition of a dilation to allow a scale factor of 0 (in this case, the entire preimage 823E Unit 7 is collapsed to a single point, the center of dilation) and negative scale factors. A dilation with a scale factor of –k , where k > 0, is equivalent to the dilation with scale factor k followed by a rotation of 180° about the center of dilation. Similarity COMMON CORE G-SRT.A.2 LESSONS 16.2 to 16.4 Recall that two figures can be defined as congruent if there is a sequence of isometries—reflections, translations, and/or rotations—that maps one figure onto the other. Likewise, similarity may be defined in terms of transformations. In particular, two figures are similar if one may be obtained from the other through a combination of a dilation and one or more isometries. Dilations transform figures by enlarging or reducing them. Finally, it is worth noting that similarity is an equivalence relation; that is, similarity is reflexive, symmetric, and transitive. For figures F 1, F 2, and F 3, F 1 ∼ F 1 (reflexivity); if F 1 ∼ F 2, then F 2 ∼ F 1 (symmetry); and if F 1 ∼ F 2 and F 2 ∼ F 3, then F 1 ∼ F 3 (transitivity). Triangle Proportionality COMMON CORE G-SRT.B.4 LESSON 17.1 The Triangle Proportionality Theorem states that if a line is parallel to a side of a triangle and intersects the other two sides, then it divides those sides proportionally. In the ‹ › _ − AE AF figure, EF ∥ BC. Therefore, __ = __ . EB FC A E B F C PROFESSIONAL DEVELOPMENT The proof, which uses similarity and facts about proportions, runs as follows. Because they are corresponding angles, ∠AEF ≅ ∠ABC. Since ∠A ≅ ∠A, ▵AEF ∼ ▵ABC by AA Similarity. Similarity in Right Triangles AE AF Thus, __ = __ , so the reciprocals are also equal, AB AC In 1816, the French mathematician August Leopold Crelle stated, “It is indeed wonderful that so simple a figure as the triangle is so inexhaustible in properties.” It is possible to develop and prove theorems that are increasingly elegant and unexpected, for example, the rather surprising fact that the three altitudes of any triangle are concurrent. AC AB __ = __ . AE AF By the Segment Addition Postulate, AB = AE + EB and AE + EB AF + AC AC = AF + FC. By substitution, ______ = ______ or AE AF FC FC EB EB 1 + __ = 1 + __ . Therefore, __ = __ , which is equivalent to AF AE AF AE the required proportion. Proportional Relationships COMMON CORE G-SRT.B.5 LESSONS 17.2 and 17.3 A proportion is an equation that states that two ratios are equal. Students should realize that the steps for solving a proportion such as __3x = __58 are the same as those for solving any other type of equation; that is, the equation must be transformed into simpler equivalent equations with the goal of isolating the variable on one side of the equation. A possible first step in solving __3x = __58 is to clear the fractions by multiplying both sides of the equation by 3 ⋅ 8 or 24. This results in the equivalent equation 8x = 15, which may be solved by dividing both sides by 8. Students may already be familiar with the shortcut that results from performing the initial multiplication on a general proportion—the Cross Product Property. This property states that if __ba = __dc , where b ≠ 0 and d ≠ 0, then ad = bc. To see why the Cross Product Property is true, begin by multiplying both sides of the proportion __ba = __dc by bd. a _ c _ = , b d a ( )_ c (bd)_ = bd , b d bda _ bdc _ = b d Multiplication Property of Equality Multiply. da = bc Simplify. ad = bc Commutative Property of Mult. COMMON CORE G-SRT.B.4 LESSON 17.4 This lesson contains the following important theorems: • The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle. • The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the two segments of the hypotenuse. • The length of a leg of a right triangle is the geometric mean of the lengths of the hypotenuse and segment of the hypotenuse adjacent to that leg. • The Pythagorean Theorem The Pythagorean Theorem, one of the best-known relationships in mathematics, can be traced back almost as far as recorded history itself. Most of the early appearances of the theorem occur in the form of Pythagorean triples (sets of three nonzero whole numbers a, b, and c such that a 2 + b 2 = c 2). One example is the clay tablet known as Plimpton 322, which was written in Babylonia around 1800 B.C.E.; it contains 15 rows of numbers based on Pythagorean triples. Such archaeological findings show that the relationship was known long before the time of the Greek mathematician Pythagoras (c. 582 B.C.E.–507 B.C.E.); it is because of the work of later Greek and Roman historians that the theorem has come to bear Pythagoras’s name. Regardless of its beginnings, the theorem has continued to inspire both professional and amateur mathematicians because of its elegance and adaptability to diverse methods of proof. In fact, The Pythagorean Proposition by Elisha Scott Loomis presents more than 350 different proofs of the Pythagorean Theorem! Unit 7 823F UNIT 7 UNIT 7 Similarity and Right Triangles MODULE Similarity and Right Triangles MATH IN CAREERS Unit Activity Preview 16 Similarity and Transformations MODULE 17 Using Similar Triangles MODULE 18 Trigonometry with Right Triangles After completing this unit, students will complete a Math in Careers task by using similarity to make a calculation for a special effects engineer. Critical skills include modeling real-world situations and using similar figures to find missing measurements. © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©TriStar Pictures & Touchstone Pictures/Everett Collection, Inc. For more information about careers in mathematics as well as various mathematics appreciation topics, visit The American Mathematical Society at http://www.ams.org. MATH IN CAREERS Special Effects Engineer Special effects engineers make movies come to life. With the use of math and some creative camera angles, special effects engineers can make big things appear small and vice versa. If you’re interested in a career as a special effects engineer, you should study these mathematical subjects: • Algebra • Geometry • Trigonometry Research other careers that require the use of engineering to understand real-world scenarios. See the related Career Activity at the end of this unit. Unit 7 811 TRACKING YOUR LEARNING PROGRESSION IN2_MNLESE389847_U7UO.indd 811 4/12/14 1:44 PM Before In this Unit After Students understand: • parallel lines, transversals, and angle relationships • perpendicular lines and bisectors • slopes and equations of parallel and perpendicular lines • properties of parallelograms, rectangles, rhombuses, and squares • theorems about parallelograms • special segments of triangles Students will learn about: • similarity and dilations • similarity of circles • corresponding parts of similar figures • proving triangles similar • the triangle proportionality theorem • dividing segments in a given ratio • geometric means theorems • Using tangents, sine, and cosine Students will study: • central and inscribed angles • chords, secants, tangent lines, and arcs • segment lengths in circles • angles formed by intersecting lines of a circle • formulas for circumference and area of a circle • area of a sector 823 Unit 7 Reading Start -Up Visualize Vocabulary Use the ✔ words to complete the main idea web. Write the review words in the squares and include definitions. Translation A translation slides all points of a figure the same distance in the same direction. Images formed by rigid transformations are congruent to their preimages. Reflection Rotation A reflection across a line maps each point to its image such that the line forms the perpendicular bisector of the point and its image. If the point is on the line, then the image is as well. A rotation moves each point about a center such that the distance from the center doesn’t change and all angles formed by a point and its image have the same measures. Complete the sentences using the preview words. 2. 3. The image formed by a(n) similarity transformation/dilation has the same shape as its pre-image. The scale factor indicates the ratio of the lengths of corresponding sides of two similar figures. geometric mean . x a = __ In the proportion __ x b , x is called the Active Reading Review Words ✔ betweenness (intermediación) ✔ collinearity (colinealidad) ✔ congruent (congruente) ✔ hypotenuse (hipotenusa) ✔ legs (catetos) ✔ orientation (orientación) ✔ parallel (paralelo) ✔ reflection (reflejo) ✔ rotation (rotación) ✔ transformation (transformación) ✔ translation (traslación) Have students complete the activities on this page by working alone or with others. VISUALIZE VOCABULARY The word web graphic helps students review vocabulary associated with transformations. If time allows, brainstorm other connections among the review words. Preview Words center of dilation (centro de la dilatación) cosine (coseno) dilation (dilatación) geometric mean (media geométrica) indirect measurement (medición indirecta) scale factor (factor de escala) similar (similar) similarity transformation (transformación de semejanza) sine (seno) tangent (tangente) trigonometric ratio (razón trigonométrica) UNDERSTAND VOCABULARY Use the following explanations to help students learn the preview words. A dilation or similarity transformation is a transformation that changes the size of a figure but not its shape. The image of a dilation is a figure similar to the preimage. The center of dilation is the intersection of the lines that connect each point of the image with the corresponding point of the preimage. Indirect measurement is a method of measurement that uses similar figures. The multiplier used on each dimension to change one figure into a similar figure is the scale factor. © Houghton Mifflin Harcourt Publishing Company Understand Vocabulary 1. Reading Start Up Vocabulary ACTIVE READING Students can use these reading and note-taking strategies to help them organize and understand the new concepts and vocabulary. Encourage students to ask questions about any references to new vocabulary and associated concepts that seem unclear. Emphasize the importance of understanding precise mathematical language to accurately work through problems. Double-Door Fold Create a Double-Door Fold prior to starting the unit. Write characteristics of congruency under one flap. Fill out the other flap with corresponding characteristics of similarity so that the two topics can be compared more easily. Unit 7 IN2_MNLESE389847_U7UO.indd 812 812 4/12/14 1:43 PM ADDITIONAL RESOURCES Differentiated Instruction • Reading Strategies Unit 7 824 MODULE 16 Similarity and Transformations Similarity and Transformations Essential Question: How can you use similarity ESSENTIAL QUESTION: and transformations to solve real-world problems? Answer: Similarity and transformations are useful tools in any real-world things that have the same shape but different sizes. 16 MODULE LESSON 16.1 Dilations LESSON 16.2 Proving Figures Are Similar Using Transformations LESSON 16.3 Corresponding Parts of Similar Figures This version is for Algebra 1 and Geometry only PROFESSIONAL DEVELOPMENT VIDEO LESSON 16.4 AA Similarity of Triangles Professional Development Video Professional Development my.hrw.com © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©lzzy Schwartz/Digital Vision/Getty Images Author Juli Dixon models successful teaching practices in an actual high-school classroom. REAL WORLD VIDEO Check out how properties of similarity and transformations can be used to create scale models of large, real-world structures like monuments. MODULE PERFORMANCE TASK PREVIEW Modeling the Washington Monument In this module, you will be challenged to create a plan for a scale model of the Washington Monument. How can you use similarity and dilations to help you produce an accurate model? Let’s find out. Module 16 DIGITAL TEACHER EDITION IN2_MNLESE389847_U7M16MO 825 Access a full suite of teaching resources when and where you need them: • Access content online or offline • Customize lessons to share with your class • Communicate with your students in real-time • View student grades and data instantly to target your instruction where it is needed most 825 Module 16 825 PERSONAL MATH TRAINER Assessment and Intervention Assign automatically graded homework, quizzes, tests, and intervention activities. Prepare your students with updated, Common Core-aligned practice tests. 18/04/14 7:47 PM Are YOU Ready? Are You Ready? Complete these exercises to review skills you will need for this module. ASSESS READINESS Properties of Transformations Example 1 Stretch △ABC with points A(1, 2), B(3, 2), and C(3, –1) horizontally and vertically by a factor of 4. (x, y) → (4x, 4y) Use the assessment on this page to determine if students need strategic or intensive intervention for the module’s prerequisite skills. • Online Homework • Hints and Help • Extra Practice Write the transformation rule. A′(4, 8), B(12, 8), C′(12, –4) Use the transformation to write each transformed point. Describe the transformation. 1. Stretch △DEF with points D(–2, 1), E(–1, –1), and F(–2, –2) horizontally and vertically by a factor of –3. D′(6, –3), E′(3, 3), F′(6, 6) 2. Is the stretch a rigid motion? ASSESSMENT AND INTERVENTION No 3. Is it true that △DEF ≅ △D′E′F′? No 3 2 1 Personal Math Trainer will automatically create a standards-based, personalized intervention assignment for your students, targeting each student’s individual needs! Similar Figures Example 2 Transform △ABC with points A(3, 4), B(–1, 6), and C(0, 1) by shifting it 2 units to the right and 1 unit up. (x, y) → (x + 2, y + 1) Write the transformation rule. A′(5, 5), B′(1, 7), C′(2, 2) Write each transformed point. 5. 4 ) Write the rule used to transform △ABC. y x + 1, _ (x, y) → _ –4 (2 2 Describe in words the transformation shown in the figure. Answers may vary. Sample: Compress △ABC vertically and IN2_MNLESE389847_U7M16MO 826 Tier 1 Lesson Intervention Worksheets Reteach 16.1 Reteach 16.2 Reteach 16.3 Reteach 16.4 B 2 -4 -2 horizontally by a factor of 2, and then shift it to the right 1 unit and down 4 units. Module 16 y A 0 -2 -4 C 2 A′ x 4 B′ C′ © Houghton Mifflin Harcourt Publishing Company Describe the transformation shown in the graph. 4. TIER 1, TIER 2, TIER 3 SKILLS ADDITIONAL RESOURCES See the table below for a full list of intervention resources available for this module. Response to Intervention Resources also includes: • Tier 2 Skill Pre-Tests for each Module • Tier 2 Skill Post-Tests for each skill 826 Response to Intervention Tier 2 Strategic Intervention Skills Intervention Worksheets 35 Properties of Dilations 41 Similar Figures 29 Geometric Drawings Differentiated Instruction 18/04/14 7:47 PM Tier 3 Intensive Intervention Worksheets available online Building Block Skills 36, 50, 80, 86, 95, 99, 103, 104, 112 Challenge worksheets Extend the Math Lesson Activities in TE Module 16 826 LESSON 16.1 Name Dilations Class Date 16.1 Dilations Essential Question: How does a dilation transform a figure? Common Core Math Standards The student is expected to: COMMON CORE G-SRT.A.1a, G-SRT.A.1.b Verify experimentally the properties of dilations given by a center and a scale factor … . Also G-CO.A.2 Use △ABC and its image △A'B'C' after a dilation to answer the following questions. MP.5 Using Tools B Language Objective B' Work with a partner to identify the center of dilation and scale factor in a dilation. A A ENGAGE View the Engage section online. Discuss the photo. Ask students to describe the relationship between the forms created by the hands and the shapes that appear on the wall. Then preview the Lesson Performance Task. © Houghton Mifflin Harcourt Publishing Company PREVIEW: LESSON PERFORMANCE TASK A' C B Use a ruler to measure the following lengths. Measure to the nearest tenth of a centimeter. Essential Question: How does a dilation transform a figure? A dilation changes the size of a figure without changing its shape, so the image is similar to the pre-image but not congruent. Resource Locker A dilation is a transformation that can change the size of a polygon but leaves the shape unchanged. A dilation has a center of dilation and a scale factor which together determine the position and size of the image of a figure after the dilation. Mathematical Practices COMMON CORE Investigating Properties of Dilations Explore 1 C AB = 6.0 cm A'B' = 3.0 cm AC = 4.0 cm A'C' = 2.0 cm BC = 3.0 cm B'C' = 1.5 cm C' Use a protractor to measure the corresponding angles. m∠A = 22° m∠A' = 22° m∠B = 33° m∠B' = 33° m∠C = 125° m∠C' = 125° Complete the following ratios _ 3.0 A'B' = _ = 1 _ AB 2 6.0 _ _ 2.0 A'C' = _ = 1 _ AC 2 4.0 1.5 B'C' = _ = 1 _ BC 2 3.0 Reflect 1. What do you notice about the corresponding sides of the figures? What do you notice about the corresponding angles? The ratios of the lengths of corresponding sides are equal. Corresponding angles are congruent because the measures of the corresponding angles are the same. 2. Discussion What similarities are there between reflections, translations, rotations, and dilations? What is the difference? Reflections, translations, rotations, and dilations all preserve angle measure. Reflections, translations, and rotations all preserve distance but dilations do not preserve distance. Module 16 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction Lesson 1 827 gh "File info" made throu Date Class Name ons 16.1 Dilati IN2_MNLESE389847_U7M16L1 827 a figure? a transform center and a dilation given by a of dilations How does properties Question: entally the Essential Verify experim G-SRT.A.1.b Dilations .2 COMMON G-SRT.A.1a, CORE … . Also G-CO.A perties of scale factor the shape ating Pro n but leaves ine 1 Investig er determ size of a polygo togeth the Explore which can change factor rmation that n and a scale n is a transfo has a center of dilatio after the dilation. A dilatio ns. A dilation of a figure ing questio unchanged. of the image r the follow n and size n to answe the positio after a dilatio △A'B'C' B and its image Use △ABC Resource Locker HARDCOVER PAGES 827836 Watch for the hardcover student edition page numbers for this lesson. B' C A y g Compan Publishin Harcour t © Houghto n Mifflin _ Lesson 16.1 m∠A = 22° m∠B = 33° m∠C = 125° the m∠A' = 22° m∠B' = 33° m∠C' = 125° _ 1 1.5 B'C' = _ = 2 _ BC 3.0 you s? What do the figure ng sides of angles are correspondi sponding about the equal. Corre you notice ponding angles? sides are What do same. sponding the corres s are the notice about the lengths of corre ding angle of correspon ns, The ratios ures of the tions, rotatio the meas because ions, transla ctions, en reflect congruent ure. Refle there betwe meas are rve angle similarities distance. ons all prese ssion What is the difference? preserve and dilati 2. Discu ns? What ons do not s, rotations, and dilatio nce but dilati Lesson 1 s, translation rve dista Reflection ions all prese s, and rotat translation 827 Module 16 827 _ re ctor to measu Use a protra angles. ng correspondi Reflect 1. 6L1 827 47_U7M1 ESE3898 IN2_MNL ing re the follow to measu nearest Use a ruler re to the lengths. Measu eter. centim tenth of a 3.0 cm A'B' = cm AB = 6.0 = 2.0 cm A'C' 4.0 cm AC = 1.5 cm B'C' = cm BC = 3.0 ing ratios the follow 1 2.0 Complete A'C' = _ = 2 _ 1 3.0 AC 4.0 A'B' = _ = 2 _ AB 6.0 C' A' 18/04/14 7:51 PM 18/04/14 7:53 PM Explore 2 Dilating a Line Segment EXPLORE 1 The dilation of a line segment (the pre-image) is a line segment whose length is the product of the scale factor and the length of the pre-image. ‹ › − Use the following steps to apply a dilation by a factor of 3, with center at the point O, to AC . O A Investigating Properties of Dilations C INTEGRATE TECHNOLOGY B Students have the option of doing the dilation activity either in the book or online. QUESTIONING STRATEGIES A To locate the point A', draw a ray from O through A. Place A' on this ray so that the distance from O to A' is three times the distance from O to A. B To locate point B′, draw a ray from O through B. Place B′ on this ray so that the distance from O to B′ is three times the distance from O to B. C To locate point C′, draw a ray from O through C. Place C′ on this ray so that the distance from O to C′ is three times the distance from O to C. D Draw a line through A', B', and C'. E _ _ _ _ _ _ Measure AB, AC, and BC. Measure A'B', A'C', and B'C'. Make a conjecture about the lengths of segments that have been dilated. Do dilations appear to preserve angle measure? yes Do dilations appear to preserve distance? no What happens when you dilate with a scale factor of 1? The image and pre-image coincide. EXPLORE 2 The length of a segment that has been dilated is the original length times the scale factor. 3. Make a conjecture about the length of the image of a 4 cm segment after a dilation with scale factor k. Can the image ever be shorter than the preimage? The image will be 4k cm long. Yes, if k is between 0 and 1, the image will be shorter than the preimage. 4. What can you say about the image of a segment under a dilation? Does your answer depend upon the location of the segment? Explain The image of segment m is parallel to m. The only exception is when line m passes through © Houghton Mifflin Harcourt Publishing Company Reflect Dilating a Line Segment INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Use an overhead projector to project a shape onto the board and then trace the shape. Move the projector closer and trace the shape again to demonstrate a reduction. Move it farther away and trace it again to demonstrate an enlargement. the center of dilation. In that case, the dilation lies along line m. Module 16 828 Lesson 1 QUESTIONING STRATEGIES PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M16L1 828 Math Background Dilations are one of the major transformations that students study. Unlike earlier transformations (reflections, translations, and rotations), dilations are not rigid motions. That is, they do not preserve both the shape and the size of a figure. However, dilations do preserve the shape of a figure. Thus, every dilation is either an enlargement or a reduction. The image that results from a dilation is generally not congruent to the pre-image, so a dilation is not an isometry. An exception is a dilation with a scale factor of –1, which is equivalent to a 180° rotation about the center of dilation. 18/04/14 7:53 PM Is it ever possible for the image of a point to coincide exactly with the pre-image? Explain. yes; when the scale factor is 1 Dilations 828 Explain 1 EXPLAIN 1 Applying Properties of Dilations The center of dilation is the fixed point about which all other points are transformed by a dilation. The ratio of the lengths of corresponding sides in the image and the preimage is called the scale factor. Applying Properties of Dilations Properties of Dilations • Dilations preserve angle measure. • Dilations preserve betweenness. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Have students review the characteristics of • Dilations preserve collinearity. • Dilations preserve orientation. • Dilations map a line segment (the pre-image) to another line segment whose length is the product of the scale factor and the length of the pre-image. • Dilations map a line not passing through the center of dilation to a parallel line and leave a line passing through the center unchanged. similar figures. Help them see that a dilation is a transformation that changes the size of a figure but not the shape. The image and the pre-image of a figure under a dilation are similar. Example 1 Determine if the transformation on the coordinate plane is a dilation. If it is, give the scale factor. Preserves angle measure: yes 6 Preserves betweenness: yes D' A' 4 Preserves collinearity: yes y A D 2 Preserves orientation: no Ratio of corresponding sides: 1 : 1 B B' C' -6 -4 -2 0 2 Cx 4 6 8 Is this transformation a dilation? No, it does not preserve orientation. Preserves angle measure (Y/N) © Houghton Mifflin Harcourt Publishing Company Preserves betweenness (Y/N) Preserves collinearity (Y/N) Preserves orientation (Y/N) Scale Factor Is this transformation a dilation? Module 16 yes y 4 C' yes 2 yes yes x -4 829 -2 A 2 Yes C A' 0 -2 2 B B' -4 Lesson 1 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M16L1 829 Peer-to-Peer Activity Divide students into pairs. Have one student draw a pre-image and an image, and then have the other determine whether the image is a dilation and explain why or why not. If it is a dilation, they should find the scale factor. Then have students switch roles. 829 Lesson 16.1 18/04/14 7:53 PM Your Turn QUESTIONING STRATEGIES Determine if the transformations are dilations. 5. If a transformation preserves angle measure, betweenness, and collinearity, is the transformation a dilation? Explain. Not necessarily; rigid transformations such as translations also preserve these things. y 4 2 -4 0 -2 D The transformation preserves angle measures, betweenness, collinearity, and orientation but not A' C' C D' B' B 2 -2 A 4 x 3 same. Therefore, it is a dilation. The scale factor is __ . 2 E E' -4 6. AVOID COMMON ERRORS C 4 A The transformation preserves betweenness and x B -6 A' -4 B' C' Explain 2 0 -2 -2 Some students may think that any larger or smaller figure is a dilation. Use examples to show that the shape and angles of the pre-image and image must be the same, and the lengths of the corresponding sides proportional. y 2 -8 distance. The ratios of corresponding lengths are all the collinearity, angle measure, or ratio of corresponding sides. It does not preserve orientation. This is not a dilation. EXPLAIN 2 -4 Determining the Center and Scale of a Dilation Determining the Center and Scale of a Dilation When you have a figure and its image after dilation, you can find the center of dilation by drawing lines that connect corresponding vertices. These lines will intersect at the center of dilation. Determine the center of dilation and the scale factor of the dilation of the triangles. ‹ › − − ‹ › ‹ › − Draw AA', BB' , and CC'. The point where the lines cross is the center of dilation. Label the intersection O. Measure to find the scale factor. OA = 25 mm OB = 13 mm OC = 19 mm OA′ = 50 mm OB′ = 26 mm OC′ = 38 mm A' B' C' A The scale factor is 2 to 1. INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Discuss the reason that dilation does not © Houghton Mifflin Harcourt Publishing Company Example 2 affect a line through the center of dilation, but does affect a line not through the center. Remind students that the image of a point on a line through the center of dilation will remain on the line, and move only along it. B C O Module 16 830 Lesson 1 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M16L1 830 Critical Thinking 18/04/14 7:53 PM Students may wonder how to construct the dilation image of a figure when the scale factor is not a whole number. While it is not possible to use a compass and straightedge to construct the image for every scale factor, it is possible for certain scale factors, using a combination of the construction for copying a segment, the construction of a segment bisector, and the construction for dividing a segment into a given ratio. You may want to challenge students to use some combination of these constructions to draw the image of a triangle after a dilation with center of dilation O and scale factor 1.75. Dilations 830 ‹ › − − ‹ › ‹ › − Draw AA', BB' , and CC'. Measure from each point to the intersection O to the nearest millimeter. B QUESTIONING STRATEGIES How can you use the given scale factor to predict whether the image is an enlargement? A scale factor >1 produces an enlargement. OA′ = 30 mm OB′ = 19 mm OC = 52 mm B' The scale factor is O A' A 26 mm OC′ = QUESTIONING STRATEGIES C' B OB = 38 mm ELABORATE 1 to 2 . Reflect How can a dilation produce an image in which the pre-image and the image are two different congruent figures? The image under a dilation with a scale factor of –1 is congruent to the pre-image and is rotated 180°. 7. For the dilation in Your Turn 5, what is the center of dilation? Explain how you can tell without drawing lines. The origin; several of the points and their images lie on axes, which intersect at the origin. For the points that are not on axes, you can use slopes to check that the lines connecting the point and image go through the origin. A Your Turn 8. SUMMARIZE THE LESSON Determine the center of dilation and the scale factor of the dilation. Measurements may vary but scale factor should not. © Houghton Mifflin Harcourt Publishing Company How can you determine whether the image of a figure is the result of a dilation? How might you determine the scale factor? If the lengths of the sides of the image are proportional to the lengths of the sides of the pre-image, and the corresponding angles are congruent, the image is the result of a dilation. If the figures are on a coordinate gird, you could divide the coordinates of the points of the image by the coordinates of the corresponding points of the pre-image to find the scale factor. C OA = 60 mm OA' = 18 mm cm, OA = 54 mm The scale factor of the dilation is 1 to 3 A' Elaborate 9. B C . C' B' How is the length of the image of a line segment O under a dilation related to the length of its preimage? The ratio of the length of the image to the length of the preimage is the scale factor. 10. Discussion What is the result of dilating a figure using a scale factor of 1? For this dilation, does the center of dilation affect the position of the image relative to the preimage? Explain. A dilation by a scale factor of 1 will leave the figure unchanged. It will remain in the same position no matter what point is used as the center of dilation. Module 16 831 Lesson 1 LANGUAGE SUPPORT IN2_MNLESE389847_U7M16L1 831 Connect Vocabulary Relate the word dilation to its meaning of becoming larger or smaller by using the example of the human eye. Pupils dilate in response to changes in light; in darkness, they get larger to let in more light, and in bright light, pupils get smaller to reduce the light entering the eye. Explain to students that when people talk about pupils dilating, they are using the mathematical term correctly. 831 Lesson 16.1 18/04/14 7:53 PM 11. Essential Question Check-In In general how does a dilation transform a figure? A dilation changes the size of a figure without changing the shape. Dilations preserve EVALUATE angle measures and orientation, but not length. Evaluate: Homework and Practice 1. • Online Homework • Hints and Help • Extra Practice Consider the definition of a dilation. A dilation is a transformation that can change the size of a polygon but leaves the shape unchanged. In a dilation, how are the ratios of the measures of the corresponding sides related? ASSIGNMENT GUIDE The ratios of the lengths of the corresponding sides are equal. Tell whether one figure appears to be a dilation of the other figure Explain. 2. 3. No, this is not a dilation, since the ratios of the lengths of corresponding sides are not equal. Some sides are about the same length and others are not. It appears to be a dilation that preserves angle measure, betweenness, collinearity and orientation. Also, the ratios of the lengths of corresponding sides appears to be equal. 4. 1 ? Explain. Is the scale factor of the dilation of △ABC equal to _ 2 6 4 A' A 5. B' B Exercise IN2_MNLESE389847_U7M16L1 832 6 8 B 28 Exercises 1–5 Explore 2 Dilating a Line Segment Exercises 6–11 Example 1 Applying Properties of Dilations Exercises 12–13 Example 2 Finding the Center and Scale of a Dilation Exercises 14–15 similar? Why or why not? Yes; if figures are congruent, then there is a sequence of rigid motions that maps one to the other. Rigid motions are also similarity transformations, so the figures must also be similar. Lesson 1 832 Depth of Knowledge (D.O.K.) x 4 A _ _ Module 16 2 35 Square A is a dilation of square B. What is the scale factor? 1 5 35 a. _ = The answer is (c). The ratio is 7 4 28 4 b. _ 5 5 c. _ 4 d. 7 25 e. _ 16 Explore 1 Investigating Properties of Dilations INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 If two figures are congruent, are they also C 2 0 Practice C' © Houghton Mifflin Harcourt Publishing Company No, the scale factor is 2. The center of dilation is at (0, 0). The measure from the center of dilation to A' is twice the measure from the center of dilation to A. The image is larger than the pre-image so the scale factor must be greater than 1. y Concepts and Skills COMMON CORE Mathematical Practices 1–5 2 Skills/Concepts MP.2 Reasoning 6–15 2 Skills/Concepts MP.5 Using Tools 16–17 2 Skills/Concepts MP.3 Logic 18 2 Skills/Concepts MP.4 Modeling 19 3 Strategic Thinking MP.1 Problem Solving 20 3 Strategic Thinking MP.1 Problem Solving 18/04/14 7:53 PM Dilations 832 6. AVOID COMMON ERRORS _ Apply a dilation to AC with a scale factor of 2 and center at the point O. 7. _ 1 Apply a dilation to AC with a scale factor of _ 3 and center at the point O. O Some students may have trouble determining scale factors. In particular, they may have difficulty distinguishing a dilation with a scale factor of k from a dilation with a scale factor of __1k . Remind students that they should always compare the image to the pre-image. Also, remind students that if the image is an enlargement of the pre-image, then the scale factor must be greater than 1. O C C' B A A' B' C B' B A' 8. C' A What happens when a triangle is dilated using one of the vertices as the center of dilation? 9. Draw an image of WXYZ. The center of the dilation is O, and the scale factor is 2. The sides of the triangles adjacent to the center of dilation will be collinear. The third sides of the preimage and image will be parallel. The vertex used as the center of dilation will be in the same location in both triangles. X' Y' X Y O Z W Z' W' 10. Draw an image of △ABC . The center of dilation is C, and the scale factor is 1.5. 11. Compare dilations to rigid motions. How are they similar? How are they different? Rigid motions preserve angle measure, betweenness, and collinearity. Dilations preserve all of these except distance. The dilation of a line segment (preimage) is another line segment whose length is the product of the scale factor and the length of the preimage. B' B © Houghton Mifflin Harcourt Publishing Company C' C A A' Determine if the transformation of figure A to figure B on the coordinate plane is a dilation. Verify ratios of corresponding side lengths for a dilation. 12. 13. y 6 y 8 4 6 B 4 B 2 A 0 2 x A 2 4 6 x 0 2 4 6 8 10 It is a dilation. The ratio for corresponding It is a dilation. The ratio for corresponding 2 . side lengths is _ 1 side lengths is _ . 2 1 Module 16 IN2_MNLESE389847_U7M16L1 833 833 Lesson 16.1 833 Lesson 1 18/04/14 7:53 PM Determine the center of dilation and the scale factor of the dilation. 14. 15. E A' F D C' E' B' A F' D' B C O O The scale factor is 1 to 2 . The scale factor is 3 to 1 . 16. You work at a photography store. A customer has a picture that is 4.5 inches tall. The customer wants a reduced copy of the picture to fit a space of 1.8 inches tall on a postcard. What scale factor should you use to reduce the picture to the correct size? 1.8 in _ 2 _ = 5 4.5 in 2 A scale factor of should be used. 5 _ y The length of the pre-image is ⎜5 - 2⎟ = 3 units. 14 C '(15, 12) B'(6, 12) 12 The height of the pre-image is ⎜4 - 2⎟ = 2 units. 10 The height of the image is ⎜12 - 6⎟ = 6 units. 8 6 4 2 The length of the image is ⎜15 - 6⎟ = 9 units. B(2, 4) C(5, 4) Module 16 IN2_MNLESE389847_U7M16L1 834 6 3 lengths is _ =_ . The scale factor is 3 to 1. 1 2 D(5, 2) A(2, 2) 2 3 9 The ratio of the heights is _ =_ . The ratio of the 1 3 D '(15, 6) A'(6, 6) x 4 6 8 10 12 14 16 18 834 © Houghton Mifflin Harcourt Publishing Company ∙ Image Credits: ©Digital Vision/Getty Images 17. Computer Graphics An artist uses a computer program to enlarge a design, as shown. What is the scale factor of the dilation? Lesson 1 18/04/14 7:53 PM Dilations 834 18. Explain the Error What mistakes did the student make when trying to determine the center of dilation? Determine the center of dilation. JOURNAL Have students write an explanation of how, given a triangle and its image under a dilation, you could use a ruler to find the scale factor of the dilation. P P' O O R' Q' R The lines were constructed ‹ › − incorrectly. PO should go through the ‹ › − points P and P' to form PP' . The lines must go through the vertex and the corresponding vertex to meet at the center of dilation O. Q F' y H.O.T. Focus on Higher Order Thinking 19. Draw △DEF with vertices D (3, 1) E (3, 5) F (0, 5). 12 a. Determine the perimeter and the area of △DEF. 10 Perimeter is 12 units, Area is 6 square units 8 b. Draw an image of △DEF after a dilation having a scale factor of 3, with the center of dilation at the origin (0, 0). Determine the perimeter and area of the image. 6 E F 4 Perimeter is 36 units, Area is 54 square units c. E' 14 D' 2 perimeter △D'E'F' How is the scale factor related to the ratios _____________ perimeter △DEF area △D'E'F' and __________ ? x D area △DEF 0 2 4 6 perimeter △D'E'F' 36 3 area △D'E'F' 54 9 ____________ = __ =_ = scale factor; ________ = __ =_ = scale factor squared perimeter △DEF 12 area △DEF 1 6 1 © Houghton Mifflin Harcourt Publishing Company 20. Draw △WXY with vertices (4, 0), (4, 8), and (-2, 8). a. Dilate △WXY using a factor of _14 and the origin as the center. Then dilate its image using a scale factor of 2 and the origin as the center. Draw the final image. b. Use the scale factors given in part (a) to determine the scale factor you could use to dilate △WXY with the origin as the center to the final image in one step. 1 1 ×2=_ = the scale factor you Multiply the scale factors: _ 4 2 multiply the pre-image by to draw the final image in one step. c. 8 y Y X 8 6 Y" X" 4 Y'2 X' x -2 0 W' 2W" 4W Do you get the same final image if you switch the order of the dilations in part (a)? Explain your reasoning. 1 1 =_ it will give you the same scale factor Yes. If you multiply 2 × _ 4 2 because multiplication is commutative. Module 16 IN2_MNLESE389847_U7M16L1 835 835 Lesson 16.1 835 Lesson 1 18/04/14 7:53 PM Lesson Performance Task INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 A rectangle measuring 3 inches by 6 inches is You’ve hung a sheet on a wall and lit a candle. Now you move your hands into position between the candle and the sheet and, to the great amusement of your audience, create an image of an animal on the sheet. Compare and contrast what you’re doing with what happens when you draw a dilation of a triangle on a coordinate plane. Point out ways that dilations and hand puppets are alike and ways they are different. Discuss measures that are preserved in hand-puppet projections and those that are not. Some terms you might like to discuss: projected on a wall. The image of the rectangle on the wall has an area of 200 square inches. What are the dimensions of the wall rectangle? Explain. 10 in. x 20 in. ; sample answer: Since the length of the smaller rectangle is twice its width, the length of the larger rectangle must also be twice its width. The problem becomes one of finding two numbers in the ratio 2:1 with a product of 200. The solutions 10 in. and 20 in. can be found using the guess-and-check problem solving strategy. • pre-image • image • center of dilation • scale factor • transformation • input • output Points that students may make: • A coordinate-plane dilation takes place in two dimensions, a shadow puppet in three dimensions. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 A small rectangle is suspended halfway • When a shadow puppet is projected on a surface, angle measures are preserved. Lengths are not preserved. The depth of the hands is not preserved but is transformed into a two-dimensional image. © Houghton Mifflin Harcourt Publishing Company ∙ Image Credits: ©Digital Vision/Getty Images • The pre-image is the hands. The image is the shadow. The center of dilation is the light source. • The scale factor for a shadow puppet projection is the ratio of a given measurement on the wall to the corresponding hand measurement in a plane parallel to the wall. Module 16 836 between a flashlight and a wall. It is parallel to the wall. Viewed from the side, what figure and special segment does this situation resemble? How could you use your knowledge of the segment to find unknown dimensions? Sample answer: The light from the flashlight, a side of the rectangle, and a side of the shadow form a triangle with a midsegment drawn. By the Midsegment Theorem, the side of the rectangle has half the length of the side of the shadow. Lesson 1 EXTENSION ACTIVITY IN2_MNLESE389847_U7M16L1 836 Teams of three or four will need a flashlight, an index card, and a way of supporting the card at a fixed distance from the wall and parallel to it (one team member can hold the card, keeping it as parallel as possible). One student holds the flashlight, another measures the flashlight’s distance from the rectangle, and a third measures the dimensions of the projected rectangle. Teams should take measurements with the flashlight at various distances from the rectangle. They can then study their data and hypothesize about the relationship between an object’s dimensions and the dimensions of its image when it is projected on a wall. 18/04/14 7:53 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Dilations 836 LESSON 16.2 Name Proving Figures are Similar Using Transformations Essential Question: How can similarity transformations be used to show two figures are similar? A similarity transformation is a transformation in which an image has the same shape as its pre-image. Similarity transformations include reflections, translations, rotations, and dilations. Two plane figures are similar if and only if one figure can be mapped to the other through one or more similarity transformations. G-SRT.A.2 Given two figures, use the definition of similarity in terms of similarity transformations to decide if they are similar … . Also G-C.A.1 A grid shows a map of the city park. Use tracing paper to confirm that the park elements are similar. A Mathematical Practices Trace patio EFHG. Turn the paper so that patio EFHG is mapped onto patio LMON. Describe the transformation. What does this confirm about the patios? y MP.6 Precision 4 Work with a partner to list the essential components needed to prove a figure is similar to another. -8 View the Engage section online. Discuss the photo. Ask students to identify the game that’s being played and the game board on which it is being played. Then preview the Lesson Performance Task. -6 -4N -2 H 0 2 x G 4 6 8 -2 L -4 O © Houghton Mifflin Harcourt Publishing Company PREVIEW: LESSON PERFORMANCE TASK E 2 ENGAGE If a sequence of similarity transformations can be shown to map one figure to another, the figures are similar. F 6 Language Objective Essential Question: How can similarity transformations be used to show two figures are similar? Resource Locker Confirming Similarity Explore The student is expected to: COMMON CORE Date 16.2 Proving Figures are Similar Using Transformations Common Core Math Standards COMMON CORE Class -6 M A rotation of 180° around the origin; The fountains are similar. B Trace statues ABCDEF and JKLMNO. Fold the paper so that statue ABCDEF is mapped onto statue JKLMNO. Describe the transformation. 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Descr statue JKLM y B 4 K C A figures are J 2 y-axis; The x D across the M L A reflection 6 4 2 0 -2 similar. -6 -4 E F -2 © Houghto n Mifflin Harcour t Publishin A rotation N Lesson 2 O 837 Module 16 6L2 837 47_U7M1 ESE3898 IN2_MNL 837 Lesson 16.2 18/04/14 7:58 PM 18/04/14 7:58 PM Describe the transformation you can use to map vertices of garden RST to corresponding vertices of garden DEF. What does this confirm about the gardens? EXPLORE y 4 S Confirming Similarity 2 E x -8 -6 -4 -2 D R 0 2 -2 4 F 6 8 T 1 A dilation with scale factor _ ; The 2 INTEGRATE TECHNOLOGY gardens are similar. Students have the option of doing the transformation activity either in the book or online. Reflect 1. Look back at all the steps. Were any of the images congruent to the pre-images? If so, what types of similarity transformations were performed with these figures? What does this tell you about the relationship between similar and congruent figures? The two figures in Steps A and B are congruent to each other. The types of transformations QUESTIONING STRATEGIES What kind of transformations result in similar figures? similarity transformations. that were performed with these figures were rotations and reflections, which are rigid motions. So congruent figures are also similar figures. 2. Of these transformations, what kind give congruent figures? Which transformation could result in figures that are similar but not congruent? rigid motions; dilation If two figures are similar, can you conclude that corresponding angles are congruent? Why or why not? Yes, the corresponding angles are congruent because rigid motions and dilations preserve angle measures. Determining If Figures are Similar Explain 1 You can represent dilations using the coordinate notation (x, y) → (kx, ky), where k is the scale factor and the center of dilation is the origin. If 0 < k < 1, the dilation is a reduction. If k > 1, the dilation is an enlargement. △RST and △XYZ X To map △RST onto △XYZ, there must be some factor k that dilates △RST. y R x -5 0 T Z Pre-image Image R(0, 1) X(0, 3) 5 S Y Module 16 S(1, -1) T(-1, -1) Y(3, -3) Z(-3, -3) 838 Determining If Figures Are Similar © Houghton Mifflin Harcourt Publishing Company Determine whether the two figures are similar using similarity transformations. Explain. Example 1 EXPLAIN 1 QUESTIONING STRATEGIES Are dilations the only transformations that result in an image that is similar to the pre-image? Explain. No, rigid motions like translations result in figures that are congruent, and therefore similar. Lesson 2 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M16L2 838 18/04/14 7:58 PM Math Background Two conditions must be met in order to state that two polygons are similar. Two polygons are similar if and only if their corresponding angles are congruent and their corresponding sides are proportional. Similarity is an equivalence relation; that is, similarity is reflexive, symmetric, and transitive. For figures A, B, and C, A ~ A (reflexivity); if A ~ B, then B ~ A (symmetry); and if A ~ B and B ~ C, then A ~ C (transitivity). Proving Figures are Similar Using Transformations 838 You can see that each coordinate of the pre-image is multiplied by 3 to get the image, so this is a dilation with scale factor 3. Therefore, △RST can be mapped onto △XYZ by a dilation with center at the origin, which is represented by the coordinate notation (x, y) → (3x, 3y). A dilation is a similarity transformation, so △RST is similar to △XYZ. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Have students discuss the various approaches B PQRS and WXYZ to defining similarity, including the intuitive approach (same shape, may be different size); meeting the conditions of congruent angles and proportional sides; and the transformational approach–if one can be obtained from the other by similarity transformations. 10 To map PQRS onto WXYZ, there must be some factor k that enlarges PQRS. y X(5, 9) Y(12, 9) 8 6 W(5, 5) Pre-image Image P(2, 2) W(5, 5) R(6, 4) Y(12, 9) Z(12, 5) 4 Q(2, 4) R(6, 4) 2 P(2, 2) S(6, 2) Q(2, 4) S(6, 2) x 0 2 4 6 8 10 12 X(5, 9) Z(12, 5) 14 Find each distance: PQ = 2, QR = 4 , WX = 4 , and XY = 7 If kPQ = WX, then k = 2. However. 2QR = / ≠ XY. No value of k can be determine that will map PQRS to WXYZ. So, the figures are/are not similar. Your Turn Determine whether the two figures are similar using similarity transformations. Explain. 3. LMNO and GHJK y © Houghton Mifflin Harcourt Publishing Company 14 K(2, 12) 12 J(8, 12) 10 8 6 O(1, 6) 4 2 N(4, 6) G(2, 4) L(1, 2) 0 2 H(8, 4) M(4, 2) 4 x 6 8 10 Yes, you can use a dilation with scale factor 2 and center at the origin to map LMNO onto GHJK. The figures are similar because a dilation is a similarity transformation. Module 16 839 Lesson 2 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M16L2 839 Small Group Activity Have students work in small groups to design a logo for an imaginary company; the logo should repeat a single design as several similar figures. Have them write a description accompanying the logo that describes the transformations they used. 839 Lesson 16.2 18/04/14 7:58 PM 4. △JKL and △MNP CDEF and TUVF 5. J y C K -5 L 0 M T N F Finding a Sequence of Similarity Transformations x x 5 U 0 -5 P D No, the angles are different. △JKL and △MNP are not similar figures. Explain 2 EXPLAIN 2 y V 5 E INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Help students see that scale factors are the Yes, there is a dilation centered at point F with a scale factor of 2. CDEF is similar to TUVF. Finding a Sequence of Similarity Transformations same as similarity ratios. In order for two figures to be similar, there has to be some sequence of similarity transformations that maps one figure to the other. Sometimes there will be a single similarity transformation in the sequence. Sometimes you must identify more than one transformation to describe a mapping. Example 2 Find a sequence of similarity transformations that maps the first figure to the second figure. Write the coordinate notation for each transformation. ABDC to EFHG y A 5 B D C F x E -5 0 5 G H Original Coordinates A(1, 6) B(5, 6) C(-2, 2) D(2, 2) Coordinates after 1 dilation k = __ 2 1 A′ __, 3 (2 ) 5 B′ __, 3 (2 ) C′(-1, 1) D′(1, 1) © Houghton Mifflin Harcourt Publishing Company Since EFHG is_ smaller than ABDC, the scale _ factor k of the dilation must be between 0 and 1. The length of AB is 4 and the length of EF is 2; therefore, the scale factor is __12 . Write the new coordinates after the dilation: A translation right 2 units and down 3 units completes the mapping. ( ) ( ) Coordinates after dilation 1, 3 A′ _ 2 5, 3 B′ _ 2 C′(-1, 1) D′(1, 1) Coordinates after translation (x + 2, y - 3) 5 E __ ,0 2 ( ) 9 F __ ,0 2 ( ) G(1, -2) H(3, -2) The coordinates after translation are the same as the coordinates of EFGH, so you can map ABDC to EFHG by the dilation (x, y) → __12 x, __12 y followed by a translation (x, y) → (x + 2, y - 3). ( Module 16 ) 840 Lesson 2 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M16L2 840 18/04/14 7:57 PM Modeling To help students understand the concepts in this lesson, have them draw figures and their transformed images on a coordinate plane. Then have them cut out the figures and see when they can be made to coincide, and when they cannot. Proving Figures are Similar Using Transformations 840 △JKL to △PQR B QUESTIONING STRATEGIES y How do you use the scale factor of a dilation to find the coordinates of points on the image of a figure? Multiply the coordinates of points on the pre-image by the scale factor. R 8 6 4 -8 AVOID COMMON ERRORS Students may think that all rectangles are similar. Challenge them to draw two rectangles that do not have proportional sides. -6 -4 J 0 -2 Q K 2 x L 2 P4 6 8 You can map △JKL to △PQR with a reflection across the x-axis followed by a dilation followed by a 90 ° counterclockwise rotation about the origin. ° Reflection: (x, y) → (x, -y) Dilation : (x, y) → ( 3x, 3y ) 90 counterclockwise rotation: ( ) -y, x x, y → ( ) Reflect 6. Using the figure in Example 3A, describe a single dilation that maps ABDC to EFHG. By connecting the corresponding vertices, you can identify (4, -6) as the center of dilation for a dilation with scale factor __12 that maps ABDC to EFHG. 7. Using the figure in Example 3B, describe a different sequence of transformations that will map △JKL to △PQR. Reflect △JKL across the y-axis. Then dilate with center at the origin and scale factor 3. Then © Houghton Mifflin Harcourt Publishing Company rotate 90° clockwise around the origin. Your Turn For each pair of similar figures, find a sequence of similarity transformations that maps one figure to the other. Use coordinate notation to describe the transformations. 8. PQRS to TUVW y Q 6 R 4 U V -8 -6 -4 W -2 T 0 Reflection: (x, y) → (-x, y) S P 2 Followed by… x 2 Module 16 4 6 You can map PQRS to TUVW by a reflection followed by a dilation. 8 841 Dilation: (x, y) → (__13 x, __13 y) Lesson 2 LANGUAGE SUPPORT IN2_MNLESE389847_U7M16L2 841 Connect Vocabulary Make sure students are not confused by the meanings of similarity in everyday English and in math. In everyday life, things may be called similar based only on how they appear. But, in math, we must look at the scale factors in corresponding sides and the angle measures in two similar figures before we call the figures similar. 841 Lesson 16.2 18/04/14 7:57 PM 9. △ABC to △DEF EXPLAIN 3 y F 8 You can map △ABC to △DEF by a rotation about the origin 180° followed by a dilation followed by a translation. 6 Rotation: (x, y) → (-x, -y) 4 2 D E Followed by… x -8 -6 B -4 -2 A 0 2 4 6 Proving All Circles Are Similar 8 ( INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Help students realize that every dilation has ) 3 _ Dilation: (x, y) → _ x, 3 y 2 2 Followed by… Translation: (x, y) (x - 3, y + 1.5) -2 -4 an inverse, and the scale factors are reciprocals. C 10. Describe a sequence of similarity transformations that maps JKLMN to VWXYZ. QUESTIONING STRATEGIES L How can you perform the dilation in the proof that two circles are similar without knowing the lengths of their radii? Using the ratio of the lengths of the radii as the scale factor ensures that the dilation will map the circles onto one another. This is true of all circles of any size, so you do not need to know specific lengths. Translate JKLMN right 7 units so that J maps to V. ‹ › − Reflect JKLMN across JN . M K 1 Dilate JKLMN with center J and scale factor _ . 2 J V N Z W X Explain 3 Y Proving All Circles Are Similar You can use the definition of similarity to prove theorems about figures. © Houghton Mifflin Harcourt Publishing Company Circle Similarity Theorem All circles are similar. Example 3 Prove the Circle Similarity Theorem. Given: Circle C with center C and raduis r. Circle D with center D and raduis s. s r D C Prove: Circle C is similar to circle D. To prove similarity, you must show that there is a sequence of similarity transformations that maps circle C to circle D. Module 16 IN2_MNLESE389847_U7M16L2 842 842 Lesson 2 18/04/14 7:57 PM Proving Figures are Similar Using Transformations 842 → ‾ . Start by transforming circle C with a translation along the vector CD A ELABORATE s QUESTIONING STRATEGIES D r C What scale factor is needed to dilate a circle with radius 2 to get a circle with radius 3? __32 r ⇀ CD Circle C' translation , the image of point C is Through this point D . Let the image of circle C be circle Cʹ. The center of circle Cʹ coincides with point SUMMARIZE THE LESSON . Transform circle Cʹ with the dilation with center of dilation D and scale factor _rs . B How do you use transformations to determine whether two polygons are similar? Determine whether one figure can be mapped to the other by a similarity transformation. D Circle Cʹ is made up of all the points at distance r from point D . _s After the dilation, the image of circle Cʹ will consist of all the points at distance r × r = s from point D. translation followed by the dilation These are the same points that form circle D . Therefore, the similarity transformations translations D maps circle C to circle . Because and dilations are , you can conclude that circle C is similar to circle D . Reflect 11. Can you show that circle C and circle D are similar through another sequence of similarity transformations? Explain. _ Yes, can reflect circle C across the perpendicular bisector of CD, mapping point C to point D. © Houghton Mifflin Harcourt Publishing Company Then, follow the same steps for dilation. 12. Discussion Is it possible that circle C and circle D are congruent? If so, does the proof of the similarity of the circles still work? Explain. Yes, if the ratio of each circle is the same positive value. The proof still works, because the ratio of s to r is 1, so the dilation does not affect the size of the circle. Elaborate 13. Translations, reflections, and rotations are rigid motions. What unique characteristic keeps dilations from being considered a rigid motion? Unlike the other transformations, dilations don't preserve distance, meaning the length of the sides will not stay the same between the pre-image and its image. The lengths of corresponding sides will be proportional according to the scale factor used. 14. Essential Question Check-In Two squares in the coordinate plane have horizontal and vertical sides. Explain how they are similar using similarity transformations. Possible answer: translate the bottom left vertex of one square to the other square. Then dilate the first square by the ratio of the side lengths. Module 16 IN2_MNLESE389847_U7M16L2 843 843 Lesson 16.2 843 Lesson 2 18/04/14 7:57 PM Evaluate: Homework and Practice EVALUATE In Exercises 1–4, determine if the two figures are similar using similarity transformations. Explain. 1. Looking at the two figures, you can see that EFGH has to be EFGH and ABCD 5 enlarged by some factor k to be mapped to ABCD. y A E -5 D F H 0 Pre-Image Image E(0, 1) A(0, 4) G(0, -1) C(0, -4) x B F(1, 0) 5 G -5 • Online Homework • Hints and Help • Extra Practice H(-1, 0) C ASSIGNMENT GUIDE B(4, 0) Concepts and Skills Practice D(-4, 0) Explore Confirming Similarity Exercises 11–12 Example 1 Determining If Figures Are Similar Exercises 1–4 Example 2 Finding a Sequence of Similarity Transformations Exercises 5–10 Example 3 Proving All Circles Are Similar Lesson Performance Task From the table, you can determine that k = 4. Therefore, EFGH can be mapped onto ABCD with a dilation of 4 with center at the origin, which is represented by the coordinate notation (x, y) → (4x, 4y). 2. △PQR and △STU. y No value of k will enlarge △PQR onto △STU. Therefore, there is no similarity transformation that maps △PQR onto △STU, and △PQR is not similar to △STU. S x -5 0 5 R Q T U © Houghton Mifflin Harcourt Publishing Company 3. R JKLMN and JPQRS In order to map JKLMN to JPQRS, there must be a scale factor k that reduces JKLMN. In this situation, the center of the dilation is not the origin but point J. 3 The scale factor is _ for the dilation. 5 Since dilations are similarity transformations, JKLMN is similar to JPQRS. y L K -8 -6 -4 Exercise IN2_MNLESE389847_U7M16L2 844 M 4 Q 2 P -2 J Module 16 6 R J 0 x 2 4 -4 S N Lesson 2 COMMON CORE Mathematical Practices 1–4 1 Recall of Information MP.1 Problem Solving 5–10 2 Skills/Concepts MP.4 Modeling 11–13 2 Skills/Concepts MP.3 Logic 14–17 2 Skills/Concepts MP.4 Modeling 18–19 2 Skills/Concepts MP.2 Reasoning 3 Strategic Thinking MP.2 Reasoning 20 8 -2 844 Depth of Knowledge (D.O.K.) 6 18/04/14 7:57 PM Proving Figures are Similar Using Transformations 844 4. AVOID COMMON ERRORS △UVW and △GHI y Some students may confuse congruence and similarity. Have them draw two similar triangles that are not congruent, and two similar triangles that are congruent, and label each set of figures with the appropriate term. Then have them repeat with pairs of circles, to make clear that all circles are similar, but all circles are not congruent. U W V -5 0 No, the angles are different. Similarity transformations preserve angle measures in similar figures, so △UVW and △GHI are not similar figures. x 5 H I G For the pair of similar figures in each of Exercises 5–10, find a sequence of similarity transformations that maps one figure to the other. Provide the coordinate notation for each transformation. 5. Map △ABC to △PQR. A -6 -4 B -2 6. y C y 4 B 2 A J 0 R 2 4 5 G F -5 P Q You can map △ABC to △PQR by a reflection followed by a translation. You can map ABCD to EFGH by a reflection followed by a dilation. Reflection: (x, y)→(-x, y) Reflection: (x, y)→(x,-y) Dilation: (x, y)→(2x, 2y) Map △CED to △CBA. 8. Map ABCDE to JKLMN. y D A 0C A B C D 5 J -2 K -5 You can map △CED to △CBA by a reflection followed by a dilation. Reflection: (x, y)→(-x, y) 1 __ Dilation: (x, y)→ __ , 1 2x 2y ( B 5 x -5 E y E 2 ) N x L 0 M 5 You can map ABCDE to JKLMN by a reflection followed by a dilation centered at the origin followed by a translation. Reflection: (x, y)→(-x, y) ( ) 1 _ Dilation: (x, y)→ _ x, 1 y 2 2 ( Translation: (x, y)→ x + Module 16 Exercise IN2_MNLESE389847_U7M16L2 845 845 Lesson 16.2 x H 6 -2 Translation: (x, y)→(x, y-6 ) © Houghton Mifflin Harcourt Publishing Company C D 0 E -5 x -4 7. Map ABCD to EFGH. 2 Lesson 2 845 Depth of Knowledge (D.O.K.) _1 , y − 2) COMMON CORE Mathematical Practices 21 3 Strategic Thinking MP.3 Logic 22 3 Strategic Thinking MP.3 Logic 23 3 Strategic Thinking MP.3 Logic 18/04/14 7:57 PM 9. 10. Map △JKL to △PQR Map ABCD to JKLM. y K -8 -6 -4 L K 6 4 C D -2 B y 4 2 x A 0 2 2 4 6 8 -2 J -4 L -6 -4 -2 Q J R P 0 x 2 4 6 -2 -4 M You can map ABCD to JKLM by a reflection followed by a dilation centered at the origin followed by a translation. You can map △JKL to △PQR by a reflection followed by a dilation followed by a 90° clockwise rotation about the origin. Reflection: (x, y)→(x, y) Reflection: (x, y)→(x,-y) ( ) 1 _ Dilation: (x, y)→ _ x, 1 y 3 3 Rotation: (x, y)→(y,-x) Dilation: (x, y)→(3x, 3y) Translation: (x, y)→(x + 1, y − 2) Complete the proof. 11. Given: Square ABCD with side length x. Square EFGH with side length y. E A B y C H F x D G Step A: Dilate ABCD with center of dilation A and scale factor _x , producing square A′B′C′D′. Square ABCD has four sides of length x. Square A′B′C′D′ will have four sides of y length _x (x) = y. These are the same side lengths as EFGH. The angles are all 90° in each square, so A′B′C′D′ is congruent to EFGH. → ‾ , producing A″B″C″D″ . Step B: Translate A′B′C′D′ with a translation along the vector A′E Through this translation, A″ is mapped to E. It may be true that B″ is mapped to F, C″ is mapped to G, and D″ is mapped to H. If not, rotate A″B″C″D″ about E so that B‴ is mapped to F. Then, C‴ lands on G and D‴ lands on H. y Module 16 IN2_MNLESE389847_U7M16L2 846 846 © Houghton Mifflin Harcourt Publishing Company Prove: Square ABCD is similar to square EFGH. Lesson 2 18/04/14 7:57 PM Proving Figures are Similar Using Transformations 846 12. Given: Equilateral △JKL with side length j. Equilateral △PQR with side length p Multiple Representations Ask students to rewrite dilation rules that use fractions without using fractions. For example: 3 3 (x, y) → __4 x, __4 y . (x, y) → (0.75x, 0.75y) ( J P Prove: △JKL is similar to △PQR. ) p j R L Q K Step A: Dilate △JKL with center of dilation J and scale factor __j , producing △J′K′L′. p △JKL has three sides of length j. After the dilation △J′K′L′ will have three sides p of length __j (j) = p. These are the same side lengths as △PQR. By SSS Triangle Congruence, △J′K′L′ is congruent to △PQR. → ‾ , producing △J″K″L″. Through this translation, Step B: Translate △J′K′L′ along the vector J′P J″ is mapped to P. It may be true that K″ and L″ are mapped to Q and R. If not, rotate △J″K″L″ about P so that K‴ is mapped to Q. Then, L‴ is mapped to R. 13. Given: △ABC with AB = c, BC = a, AC = b cx bx __ △XYZ with YZ = x, XY = __ a , XZ = a Y B Prove: △ABC is similar to △XYZ. cx a c © Houghton Mifflin Harcourt Publishing Company ∙ Image Credits: ©Billy Hustace/Corbis A x a b C X bx a Z x Step A: Dilate △ABC with center of dilation B and scale factor k _ a , producing △A′B′C′. After x the dilation △A′B′C′ will have sides of length B′C′ = ka = _ a (a) = x, bx cx x ( ) __ _x (c) = __ b = , and A′B′ = kc = . These are the same side lengths of A′C′ = kb = _ a a a a △XYZ. By SSS Triangle Congruence, △A′B′C′ is congruent to △XYZ. → ‾ , producing △A″B″C″. Through this translation, Step B: Translate △A′B′C′ along the vector B′Y B″ is mapped to Y. It may be true that A″ and C″ are mapped to X and Z. If not, rotate △A″B″C″ about Y so that A‴ is mapped to X. Then, L‴ is mapped to R. 14. The dimensions of a standard tennis court are 36 feet. × 78 feet. with a net that is 3 feet. high in the center. The court is modified for players aged 10 and under such that the dimensions are 27 feet × 60 feet., and the same net is used. Use similarity to determine if the modified court is similar to the standard court. 4 Dilation factor for the width is _ and 3 13 dilation factor for the length is __ . 10 Therefore the courts are not similar because the dilation factor is not constant between the width and length. Module 16 IN2_MNLESE389847_U7M16L2 847 847 Lesson 16.2 847 Lesson 2 18/04/14 7:57 PM 15. Represent Real-World Problems A scuba flag is used to indicate there is a diver below. In North America, scuba flags are red with a white stripe from the upper left corner to the lower right corner. Justify the triangles formed on the scuba flag are similar triangles. Since the base and the height of the triangles are equal, no dilation has occurred. The triangles are 180° reflections of each other. The triangles are similar to each other. 16. The most common picture size is 4 inches. ×6 inches. Other common pictures sizes (in inches) are 5 × 7, 8 × 10, 9 × 12, 11 × 14, 14 × 18, and 16 × 20. a. Are any of these picture sizes similar? Explain using similarity transformations. In order for any of the pictures to be similar, there needs to be a scale factor k which maps one picture to another. The ratio of the sides of each picture are 2:3, 5:7, 4:5, 3:4, 11:14, 7:9, and 4:5, respectively. Therefore, the only picture sizes that are similar are the 8” × 10” and 16” × 20”, with a scale factor of k = 2. b. What does your conclusion indicate about resizing pictures? When resizing pictures one of two things will occur: the picture will be distorted or part of the picture will need to be cropped out. © Houghton Mifflin Harcourt Publishing Company ∙ Image Credits: (t) ©Chuck Wagner/Shutterstock; (b) ©Dave Reede/Design Pics Inc./Alamy 17. Nicole wants to know the height of the snow sculpture but it is too tall to measure. Nicole measured the shadow of the snow sculpture's highest point to be 10 feet long. At the same time of day Nicole's shadow was 40 inches long. If Nicole is 5 feet tall, what is the height of the snow sculpture? Since the shadows were measured at the same time of day the angle measures are equal and the triangles are similar. Therefore a proportion can be used to determine the height. 10 ft. = 120 inches 5 ft. = 60 inches Let h represent the height of the snow sculpture. 60 h = ; h = 180 in., or 15 ft 120 40 18. Which of the following is a dilation? A. (x, y) → (x, 3y) _ _ 19. What is not preserved under dilation? Select all that apply. B. (x, y) → (3x, -y) A. Angle measure D. (x, y) → (x, y - 3) C. Collinearity C. (x, y) → (3x, 3y) B. Betweenness E. (x, y) → (x - 3, y - 3) D. Distance Module 16 IN2_MNLESE389847_U7M16L2 848 E. Proportionality 848 Lesson 2 18/04/14 7:57 PM Proving Figures are Similar Using Transformations 848 H.O.T. Focus on Higher Order Thinking JOURNAL 20. Analyze Relationships Consider the transformations below. Have students define similarity in their own words, and then justify the definition using properties of similar figures. I. Translation II. Reflection III. Rotation IV. Dilation a. Which transformations preserve distance? I, II, III b. Which transformations preserve angle measure? I, II, III, IV c. Use your knowledge of rigid transformations to compare and contrast congruency and similarity. Rigid transformations preserve angle measure and distance. Therefore, rigid transformations (translations, reflections, rotations) guarantee congruency. Similarity transformations (dilations) only preserve angle measures. Justify Reasoning For Exercises 21–23, use the figure shown. Determine whether the given assumptions are enough to prove that the two triangles are similar. Write the correct correspondence of the vertices. If the two triangles must be similar, describe a sequence of similarity transformations that maps one triangle to the other. If the triangles are not necessarily similar, explain why. AX DX 21. The lengths AX, BX, CX, and DX satisfy the equation ___ = ___ . BX CX BX AX __ AX __ AX = DX into __ = CX . Let k = __ . Step A: Rearrange __ BX DX DX CX Step B: Rotate △DXC 180° around point X so that ∠DXC coincides with ∠AXB. Step C: Dilate ∠DXC by a factor of k about the center X. This dilation moves the point D to A, since k(DX) = AX, and moves C to B, since k(CX) = BX. Since the dilation is through point X and dilations take line segments to line segments, △DXC is mapped to △AXB. So △DXC is similar to △AXB. A C X D B © Houghton Mifflin Harcourt Publishing Company 22. Lines AB and CD are parallel. ¯ is parallel to Step A: Rotate △DXC so that △DXC coincides with △AXB. Then the image C′D′ ¯ ¯ ¯ the pre-image of CD. So the new side C′D′ is still parallel to side AB. ¯ to a Step B: Dilate △D′X′C′ with center at point X. This moves vertex C′ to point B and C′D′ ‹ › ‹ › − − ‹ › ‹ › − − line through B parallel to CD . Since AB is parallel to C′D′, the dilation moves C′D′ ‹ › ‹ › → − − ‾ and on AB , D′ must move to A. Therefore, onto AB . Since D′ moves to a point on XA the rotation and dilation map △DXC to △AXB. So △DXC is similar to △AXB. 23. ∠XAB is congruent to ∠XCD. Step A: Draw the bisector of ∠AXC. ¯ onto XA ¯. Since reflections Step B: Reflect △CXD across the angle bisector. This maps XC ¯ is ¯ onto XB ¯. Since △XCD ≅ △XAB, the image of CD preserve angles, it also maps XD ¯. parallel to AB ‹ › − Step C: Dilate △XCD about point X. This moves the new point C to A. Since AB is parallel ‹ › ‹ › ‹ › − − − to CD , the new CD moves onto AB . Therefore, the new point D is mapped to B and △XCD is mapped to △XAB. So △XCD is similar to △XAB. Module 16 IN2_MNLESE389847_U7M16L2 849 849 Lesson 16.2 849 Lesson 2 18/04/14 7:57 PM Lesson Performance Task AVOID COMMON ERRORS Students may correctly calculate the areas of circles A, B, C, and D, but may forget that in order to calculate the required probabilities, they need to find the areas of rings. So, for example, Area Ring C = Area Circle C – Area Circle B. Answer the following questions about the dartboard pictured here. 1. Are the circles similar? Explain, using the concept of a dilation in your explanation. 2. You throw a dart and it sticks in a random location on the board. What is the probability that it sticks in Circle A? Circle B? Circle C? Circle D? Explain how you found your answers. INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Study the pattern in the probabilities you 2′′ 2′′ 2′′ 2′′ A B C obtained as answers in the Lesson Performance Task. Then, predict the probabilities of landing darts in each ring if a fifth ring of radius 10 inches were 5 ___ 9 1 ___ , 3 , ___ , 7 , ___ added outside the 8-inch circle. ___ 25 25 25 25 25 D 1. Yes. Each circle can be regarded as a pre-image that can be transformed through a dilation with its center at the center of the dartboard into any of the other circles. 2. Area Circle A = πr 2 = π(2) = 4π 2 Area Circle B = πr = π(4) = 16π 2 2 Area Circle C = πr = π(6) = 36π 2 2 Area Circle D = πr 2 = π(8) = 64π 2 Area Ring B = Area Circle B - Area Circle A = 16π - 4π = 12π Area Ring C = Area Circle C - Area Circle B = 36π - 16π = 20π Area Ring D = Area Circle D - Area Circle C = 64π - 36π = 28π © Houghton Mifflin Harcourt Publishing Company Area of dartboard = Area Circle D = 64π 4π area Circle A 1 P(A) = __ = _ = _ area of dartboard 64π 16 area Ring B 12π 3 P(B) = __ = _ = _ area of dartboard 64π 16 area Ring C 20π 5 P(C) = __ = _ = _ area of dartboard 64π 16 area Ring D 28π 7 P(D) = __ = _ = _ area of dartboard 64π 16 Module 16 850 Lesson 2 EXTENSION ACTIVITY IN2_MNLESE389847_U7M16L2 850 Have students design dartboards in basic shapes different from the circular board (for example, squares or equilateral triangles) but with the same basic properties: each should have four sections, and the sections should differ in dimensions from adjacent sections by the same amount. Have students answer the same questions about their designs that they answered about the circular dartboard. For squares and equilateral triangles, they should obtain the same results: the figures are similar and the probabilities of landing a dart in each are (smallest to largest) 5 7 1 __ __ , 3 , __ , and __ . 16 16 16 16 18/04/14 7:57 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Proving Figures are Similar Using Transformations 850 LESSON 16.3 Name Corresponding Parts of Similar Figures Class Date 16.3 Corresponding Parts of Similar Figures Essential Question: If you know two figures are similar, what can you determine about measures of corresponding angles and lengths? Common Core Math Standards Resource Locker The student is expected to: COMMON CORE G-SRT.A.2 Explore … explain using similarity transformations the meaning of similarity for triangles … . Also G-CO.A.2, G-CO.A.5 Mathematical Practices COMMON CORE A MP.4 Modeling ENGAGE View the Engage section online. Discuss the floor plan and ask students to explain the relationship of the floor plan to the house. Then preview the Lesson Performance Task. N 4 D m∠A = 80.5° m∠K = 80.5° m∠B = 63.4° m∠L = 63.4° m∠C = 116.6° m∠M = 116.6° m∠D = 99.5° m∠N = 99.5° M C x -8 -4 0 4 -4 B A K 8 -8 L Accept reasonable measures; opposite angles should be supplementary B © Houghton Mifflin Harcourt Publishing Company PREVIEW: LESSON PERFORMANCE TASK y 8 Measure the angles. Play “similar or not” with a partner given pairs of figures on coordinate planes. Similar figures have corresponding angles that are congruent and have corresponding sides that are proportional. Consider the graph of ABCD and KLMN. Are corresponding angles congruent? Yes/No Language Objective Essential Question: If two figures are similar, what can you determine about measures of corresponding angles and lengths? Connecting Angles and Sides of Figures You know that if figures are similar, the side lengths are proportional and the angle measures are equal. If you have two figures with proportional side lengths and congruent angles, can you conclude that they are similar? Are the ratios of corresponding side lengths equal? Yes/No 1 AB = _ _ KL 2 C 1 BC = _ _ LM 2 1 AD = _ _ KN 2 1 CD = _ _ MN 2 Are the figures similar? Describe how you know using similarity transformations. y 8 Yes, there is a dilation centered at the origin that you can N 4 D find by connecting the corresponding vertices. x -8 -4 0 -4 A K Module 16 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction -8 4 8 B L Lesson 3 851 gh "File info" made throu M C Date Class Parts sponding 16.3 Corre ilar Figures of Sim Name HARDCOVER PAGES 851860 Resource Locker about determine can you similar, what lengths? figures are angles and know two ponding ion: If you of corres ity for measures ng of similar ns the meani transformatio similarity Figures explain using G-CO.A.5 COMMON .2, G-SRT.A.2 … Sides of CORE . 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Yes/N side length 1 CD = _ _ MN 2 1 AD = _ _ KN 2 y similarity know using x -4 63.4° opposite 1 BC = _ _ LM 2 M 4 D -4 -8 80.5° m∠L = 63.4° 116.6° m∠C = 99.5° m∠D = Watch for the hardcover student edition page numbers for this lesson. y 8 N how you Describe s similar? you can Are the figure ions. origin that transformat red at the on cente es. is a dilati ding vertic Yes, there correspon ecting the find by conn 8 N -8 M 4 C D 0 -4 -4 A K -8 x 8 4 B L Lesson 3 851 Module 16 6L3 851 47_U7M1 ESE3898 IN2_MNL 851 Lesson 16.3 18/04/14 9:04 PM 18/04/14 9:05 PM D y Consider the graph of ABCD and EFGH. E The figures are both rectangles, so the measure of each -8 B 0 -4 D H 4 Connecting Angles and Sides of Figures -4 8 C G -8 INTEGRATE TECHNOLOGY Are the ratios of corresponding side lengths equal? Yes/No 4 AB = _ _ EF 7 F F 4 A x angle is 90°. E EXPLORE 8 Are corresponding angles congruent? Yes/No Explain. 3 BC = _ _ FG 5 4 CD = _ _ GH 7 Students have the option of doing the angle measure activity either in the book or online. 3 AD = _ _ EH 5 QUESTIONING STRATEGIES Are the figures similar? Describe how you know using similarity transformations. EF , D′ will No. If you translate ABCD so that A maps to E, and dilate A′B′C′D′ by the ratio __ Given a figure that appears to be a dilation of another, how could you check whether a center of dilation exists? Draw lines through corresponding vertices of the figures. If they intersect at a point, it is the center of dilation. If they do not intersect at the point, no center of dilation exists. AD map to H but B′ and C′ will not map to F and G. G Consider the graph of PQRS and WXYZ. y Are corresponding angles congruent? Yes/No 8 Measure the angles. Z m∠W = 76° m∠Q = 153.4° m∠X = 104° m∠R = 26.6° m∠Y = 76° m∠S = 153.4° m∠Z = 104° 4 Y R S -8 -4 P W 0 -4 Q 4 x 8 X -8 © Houghton Mifflin Harcourt Publishing Company m∠P = 26.6° Accept reasonable measures; opposite angles should be supplementary Are the ratios of corresponding side lengths equal? Yes/No 1 PQ _ =_ WX 2 1 QR _ _ = XY 2 1 RS = _ _ YZ 2 1 PS = _ _ WZ 2 Are the figures similar? Describe how you know using similarity transformations. No. Dilations map line segments to parallel line segments, so there is no dilation that will ¯ to ZW ¯. The other similarity transformations are all rigid motions, which preserve map SP length and angle, so no sequence of similarity transformations will map PQRS to WXYZ. Module 16 852 Lesson 3 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M16L3 852 Math Background 18/04/14 9:05 PM Students use similarity to solve a variety of real-world problems. The general process is this: First, show that two figures are similar; then, use the fact that corresponding sides are proportional to find an unknown side length. Corresponding Parts of Similar Figures 852 Reflect EXPLAIN 1 If two figures have the same number of sides and the corresponding angles are congruent, does this mean that a pair of corresponding sides are either congruent or proportional? No, congruence of corresponding angles does not mean there is any relationship between 1. Justifying Properties of Similar Figures Using Transformations a pair of corresponding sides. For example, consider a square and a rectangle. If two figures have a center of dilation, is a corresponding pair of sides necessarily proportional? Yes. If two figures have a center of dilation, the figures are similar and any pair of 2. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Use drawings to discuss how similar triangles corresponding sides is proportional. If two figures have a correspondence of proportional sides, do the figures necessarily have a center of dilation? No. It’s possible to establish a correspondence of proportional sides between two figures 3. that are not similar (for example, if they don’t have a center of dilation). can be used to find the heights of buildings or trees that are difficult to measure directly. The objects’ shadows, measured at the same time of day, are proportional to the heights of the objects. Explain 1 Justifying Properties of Similar Figures Using Transformations Two figures that can be mapped to each other by similarity transformations (dilations and rigid motions) are similar. Similar figures have certain properties. Properties of Similar Figures QUESTIONING STRATEGIES Corresponding angles of similar figures are congruent. How do you know when two rectangles are similar? They are similar if corresponding sides are proportional. Y B Corresponding sides of similar figures are proportional. If △ABC ∼ △XYZ, then ∠A ≅ ∠X ∠B ≅ ∠Y ∠C ≅ ∠Z © Houghton Mifflin Harcourt Publishing Company BC = _ AC AB = _ _ XY YZ XZ A C X Z To show that two figures with all pairs of corresponding sides having equal ratio k and all pairs of corresponding angles congruent are similar, you can use similarity transformations. Dilate one figure using k. The dilated figure is congruent to the second figure by the definition of congruence. So, there is a sequence of rigid motions (which are also similarity transformations) that maps one to the other. Example 1 Identify properties of similar figures. Figure EFGH maps to figure RSTU by a similarity transformation. Write a proportion that contains EF and RU. List any angles that must be congruent to ∠G or congruent to ∠U. EF = _ EH _ RU RS ∠T is congruent to ∠G, and ∠H is congruent to ∠U. Figure JKLMN maps to figure TUVWX by a similarity transformation. Write a proportion that contains TX and LM. List any angles that must be congruent to ∠V or congruent to ∠K. JN LM _=_ TX VW ∠ L is congruent to ∠V, and ∠ U is congruent to ∠K. Module 16 853 Lesson 3 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M16L3 853 Whole Class Activity Have students bring in magazines and newspapers, and challenge them to find sets of similar figures. Discuss the use of similar figures in art and design, and the visual effects that such designs create. 853 Lesson 16.3 18/04/14 9:05 PM Reflect 4. EXPLAIN 2 If you know two figures are similar, what angle or side measurements must you know to find the dilation used in the transformations mapping one figure to another? You must know measurements of one pair of corresponding sides to find the scale factor. Applying Properties of Similar Figures Your Turn 5. Triangles △PQR and △LMN are similar. If QR = 6 and MN = 9, what similarity transformation (in coordinate notation) maps △PQR to △LMN? QUESTIONING STRATEGIES MN The ratio ___ = 1.5, therefore the similarity transformation is (x, y) → (1.5x, 1.5y). QR 6. How can you use similar figures to solve problems? Use the fact that corresponding sides are proportional to find an unknown side length. VW Is the statement true? DE = _ Error Analysis Triangles △DEF and △UVW are similar. _ UV EF DE EF = ___ . No. The proportion must compare corresponding sides. One possibility is ___ UV VW Explain 2 Applying Properties of Similar Figures The properties of similar figures can be used to find the measures of corresponding parts. Example 2 AVOID COMMON ERRORS Given that the figures are similar, find the values of x and y. Find the value of x. Find the value of y. ∠C ≅ ∠R, so m∠C = m∠R AB = _ AD _ PS PQ 3y - 5 4y _ _ = 5 10 3y 5 4y _ ⋅ 10 = _ ⋅ 10 5 10 4y = 6y - 10 4x + 27 = 95 4x = 68 x = 17 Students may sometimes write proportions that fail to use corresponding sides. While there are multiple ways to write a correct proportion, encourage students to use a pattern that they can remember. For example, so that the numerators always give side lengths from one triangle and the denominators always give side lengths from the other triangle. A 4y ft (3y - 5) ft B D (4x + 27)° P 5 ft Q 10 ft 95° S C R y=5 (2y - 8) cm Find the value of x. m∠LMN = m∠XYZ 5(x - 5) = 5x - 25 = x = 4x 4x 25 Z L 1 cm Y Find the value of y. JK MN _ =_ VW YZ 5(x - 5)° J 1.5 2x - 8 = _ _ M V 4x° X 4 cm 1.5 cm N W 1 4 2x - 8 = 1.5(4) 2x - 8 = 6 © Houghton Mifflin Harcourt Publishing Company K COMMUNICATE MATH Give student pairs different pictures of two figures drawn on coordinate planes. The first student decides whether a pair of figures is similar or not, and states three reasons why. The other student must agree or provide an alternative explanation. Students switch roles and repeat the process with another pair of figures. 2x = 14 x= 7 Module 16 854 Lesson 3 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M16L3 854 Modeling 18/04/14 9:05 PM Have students work with pattern blocks to explore the concept of similarity. Students can work in pairs, with one student creating a block design and the other creating a figure that is similar, but not congruent, to the first one. For more sets of pattern blocks, search for websites that feature “virtual pattern blocks.” These sites allow students to select, combine, and rearrange blocks on a computer screen. Corresponding Parts of Similar Figures 854 Reflect ELABORATE 7. QUESTIONING STRATEGIES Discussion What are some things you need to be careful about when solving problems involving finding the values of variables in similar figures? Possible Answer: When finding side lengths, you need to be sure to set up the proportion correctly, and you need to remember that the side lengths are proportional rather than Is an image always similar to its pre-image? Explain. No; for example, stretching and shrinking a figure disproportionally can produce an image that is not similar. equal. Once you have the algebraic equation set up, you need to be careful not to make errors in the calculations. Your Turn Use the diagram, in which △ABE ∼ △ACD. Can two congruent angles have sides that are of different lengths? Explain. Yes, the measure of the angle refers to the amount of opening between the two sides; the length of the sides of the angle does not affect its size. 5.6 cm Find the value of x. 9. 5 cm 50° C Find the value of y. CD AD _ _ = AE 50 = 3x + 14 BE 5.6 + y _ 5 _ = 4 5.6 5 5.6 + y = _ ⋅ 5.6 36 = 3x When two figures are given as similar, what can you conclude about their sides and angles? In similar figures, the corresponding sides are proportional and the corresponding angles are congruent. 4 cm B m∠C = m∠ABE SUMMARIZE THE LESSON E (3x + 14)° A 8. D y cm 12 = x 4 y = 1.4 © Houghton Mifflin Harcourt Publishing Company Elaborate 10. Consider two similar triangles △ABC and △A'B'C'. If both m∠A' = m∠C and m∠B' = m∠A, what can you conclude about triangle △ABC ? Explain your reasoning. Since the two triangles are similar, we know m∠A' = m∠A and m∠B' = m∠B . Along with the given information, this tells us m∠A = m∠C and m∠B = m∠A. Therefore, ΔABC is equilateral. 11. Rectangle JKLM maps to rectangle RSTU by the transformation (x, y) → (4x, 4y). If the perimeter of RSTU is x, what is the perimeter of JKLM in terms of x? The ratio of corresponding sides is 1:4, and therefore the ratio of the perimeters is 1:4. x The perimeter of JKLM is _ . 4 12. Essential Question Check-In If two figures are similar, what can we conclude about their corresponding parts? Similar figures have corresponding angles that are congruent and have corresponding sides that are proportional. Module 16 855 Lesson 3 LANGUAGE SUPPORT IN2_MNLESE389847_U7M16L3 855 Connect Vocabulary Students may be very familiar with the connotation of the word translation as it refers to interpretation between languages. In fact, this use of the word comes from its mathematically rigorous sense: translating between languages “moves” the meaning from one to another, just as translating a figure moves it from one place to another. Make the connection that mathematical translation does not change the essence of the original figure, much as a linguistic translation does not significantly alter the original meaning of the message as it enters another language. 855 Lesson 16.3 18/04/14 9:05 PM Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice In the figures, are corresponding angles congruent? Are corresponding sides proportional? Are the figures similar? Describe how you know using similarity transformations. 1. 2. ASSIGNMENT GUIDE Yes; no; no. Rotate the smaller square 90° around its upper right corner, then translate it down and left until the lower left vertices coincide. If you dilate it with a scale factor of 3 to 2.5, the short edges will be congruent but the long edges will not be. Yes; yes; yes. Translate the smaller square down and right so its upper left vertex coincides with that of the larger square. Then dilate with a scale factor of 2 about this vertex. 3. 4. 5. Figure ABCD is similar to figure MNKL. Write a proportion that contains BC and KL. CD BC = NK KL 6. △XYZ is similar to △XVW. Write the congruence statements that must be true. 8. _ _ _ _ 7. ∠X ≅ ∠X, ∠Y ≅ ∠V, and ∠Z ≅ ∠W. 9. CDEF maps to JKLM with the transformations (x, y) → (5x, 5y) → (x - 4, y - 4). What is the 1 EF ? __ value of ___ LM 5 Module 16 Exercise IN2_MNLESE389847_U7M16L3 856 △DEF is similar to △STU. Write a proportion that contains ST and SU. ST SU = DE DF △MNP is similar to △HJK, and both triangles are isosceles. If m∠P > 90°, name all angles that are congruent to ∠H. If m∠P > 90°, then ∠M ≅ ∠N. So, ∠M, ∠J and ∠N are all ≅ to ∠H. Practice Explore Connecting Angles and Sides of Figures Exercises 1–4 Example 1 Justifying Properties of Similar Figures Using Transformations Exercises 9–12 Example 2 Applying Properties of Similar Figures Exercises 5–8 INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Compare the symbols for equality, similarity, © Houghton Mifflin Harcourt Publishing Company Yes; yes; yes. Rotate the smaller figure 90°, then translate it so that two corresponding vertices coincide. Then dilate the smaller figure the centers by a factor of 2 about that vertex. No; no; no. You can rotate and translate the triangle on the right so the right angles coincide, but no similarity transformation will make the acute angles congruent. Concepts and Skills and congruence. Point out that combining the equality and similarity symbols gives the symbol for congruence. Connect the composition of the symbol to its meaning: congruence combines equality and similarity, because congruent figures have the same size and shape. 10. △PQR maps to △VWX with the transformation (x, y) →(x + 3, y - 1) → (2x, 2y). If WX = 12, what does QR equal? QR = 1 (12) = 6 2 __ Lesson 3 856 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–4 2 Skills/Concepts MP.3 Logic 5–17 2 Skills/Concepts MP.5 Using Tools 18–20 2 Skills/Concepts MP.2 Reasoning 21–22 2 Skills/Concepts MP.4 Modeling 23–24 2 Skills/Concepts MP.4 Modeling 25 3 Strategic Thinking MP.3 Logic 26 3 Strategic Thinking MP.3 Logic 27 3 Strategic Thinking MP.3 Logic 18/04/14 9:05 PM Corresponding Parts of Similar Figures 856 11. △QRS maps to △XYZ with the transformation (x, y) → (6x, 6y). If QS = 7, what is the length of XZ? AVOID COMMON ERRORS When solving a proportion about a scale model, some students may make mistakes because they failed to read the units carefully. Remind them that the lengths of corresponding sides do not necessarily appear in the same units, and they may need to convert one of the measures. 12. Algebra Two similar figures are similar based on the transformation (x, y) → (12x, 3a 2y). What is/ are the value(s) of a? XZ = 42 If the figures are similar, then 12 = 3a 2, and a = 2 or a = -2. 13. Algebra △PQR is similar to △XYZ. If PQ = n + 2, QR = n - 2, and XY = n 2 - 4, what is the value of YZ, in terms of n? Since n 2 - 4 = (n + 2)(n - 2), the ratio of a pair corresponding sides is (n - 2) :1. Then the value of YZ is (n - 2)(n - 2) = n 2 - 4n + 4. 14. Which transformations will not produce similar figures? Select all that apply and explain your choices. Choices B and E will not produce similar A. (x, y) → (x - 4,y) → (-x, -y) → (8x, 8y) B. (x, y) → (x + 1, y + 1) → (3x, 2y) → (-x, -y) C. (x, y) → (5x, 5y) → (x, -y) → (x + 3, y - 3) D. (x, y) → (x, 2y) → (x + 6, y - 2) → (2x, y) E. (x, y) → (x, 3y) → (2x, y) → (x - 3, y - 2) figures, because each sequence contains transformations that are not dilations or rigid motions, and are not balanced elsewhere in the sequence. Choice D contains two transformations that are not dilations or rigid motions, but together create a dilation. 15. The figures in the picture are similar to each other. Find the value of x. The longer bases of the trapezoids are 6 units and 3 units, so the scale factor is 2:1. Set up the equation x + 1 = 2(x - 3) and solve to get x = 7. x-3 6 3 x+1 16. In the diagram, △NPQ ∼ △NLM and PL = 5. M a. Find the value of x. © Houghton Mifflin Harcourt Publishing Company 3x + 18 = 60 3x = 42 4 cm x = 14 P b. Find the lengths NP and NL. NP NL = NM NQ y 5-y = 4 3.2 3.2(5 - y) = 4y 3.2 cm (3x + 18)° L 16 = 7.2y Module 16 IN2_MNLESE389847_U7M16L3 857 Lesson 16.3 y N 20 _ =y 9 25 20 So NP = _ and NL = _. 9 857 60° _ _ _ _ Q 9 857 Lesson 3 18/04/14 9:05 PM 17. △CDE maps to △STU with the transformations (x, y) → (x - 2, y -2) → (3x, 3y) → (x, -y). If CD = a + 1, DE = 2a - 1, ST = 2b + 3, and TU = b + 6, find the values of a and b. Lengths of sides of ∆STU are 3 times the lengths of corresponding sides of △CDE. This produces the equations 3(a + 1) = 2b + 3 and 3(2a - 1) = b + 6. Solve this system of equations to get a = 2 and b = 3. 18. If a sequence of transformations contains the transformation (ax, by), with a ≠ b, could the pre-image and image represent congruent figures? Could they represent similar, non-congruent figures? Justify your answers with examples. Yes; if ab = ±1, then the pre-image and image are congruent (and therefore also similar). The pre-image and image could represent similar, non-congruent figures if the sequence of transformations also contains (bx, ay). The transformation (ax, by) followed by (bx, ay) is equivalent to the dilation (abx, aby). 19. Is any pair of equilateral triangles similar to each other? Why or why not? Yes, the triangles will be similar. All the angles of each triangle are equal to 60°, and therefore corresponding angles are congruent. The sides of each triangle are congruent, and therefore the ratio of corresponding sides is constant. 20. Figure CDEF is similar to figure KLMN. Which statements are false? Select all that apply and explain why. CD = _ EF A. _ KL MN CF = _ EF B. _ KN MN CF DE = _ C. _ LM KN LM = _ KL D. _ DE CD KN LM = _ E. _ DE CD Choice E is false. The proportion doesn’t match corresponding sides with each other. © Houghton Mifflin Harcourt Publishing Company ∙ Image Credits: ©Jeff Dalton/Alamy Consider this model of a train locomotive when answering the next two questions. 21. If the model is 18 inches long and the actual locomotive is 72 feet long, what is the similarity transformation to map from the model to the actual locomotive? Express the answer using the notation x → ax, where x is a measurement on the model and ax is the corresponding measurement on the actual locomotive. The length of the model is 1.5 feet and so the similarity transformation is x → 48x. 22. If the diameter of the front wheels on the locomotive is 4 feet, what is the diameter of the front wheels on the model? Express the answer in inches. The diameter of the front wheels on the model is 1 inch. Module 16 IN2_MNLESE389847_U7M16L3 858 858 Lesson 3 18/04/14 9:05 PM Corresponding Parts of Similar Figures 858 Use the following graph to answer the next two problems. JOURNAL y Have students write a journal entry in which they make up their own problems using an unknown length that must be found using similar triangles. Remind students to include the solutions to their problems. 8 B C K 4 A -8 D -4 L x 0 4 J -4 -8 8 M 23. Specify a sequence of two transformations that will map ABCD onto JKLM. Possible answer: (x, y) → (2x, 2y) → (x + 14, y - 8) or (x, y) → (x + 7, y - 4) → (2x, 2y) AC + BD 24. Find the value of _______ . JL + KM AC The two figures are similar, so all corresponding sides/diagonals have the same ratio. ___ JL BD 1 1 1 = __ and ___ = __ , and so _______ = __ . 2 2 2 KM JL + KM AC + BD H.O.T. Focus on Higher Order Thinking 25. Counterexamples Consider the statement “All rectangles are similar.” Is this statement true or false? If true, explain why. If false, provide a counterexample. The statement is false. For example, a rectangle measuring 5 units by 2 units is not similar to a rectangle measuring 4 units by 3 units. © Houghton Mifflin Harcourt Publishing Company 26. Justify Reasoning If ABCD is similar to KLMN and MNKL, what special type of quadrilateral is KLMN? Justify your reasoning. Looking at the corresponding angles, ∠A ≅ ∠K and ∠A ≅ ∠M, which means ∠K ≅ ∠M by the Transitive Property of Congruence. Also, ∠B ≅ ∠L and ∠B ≅ ∠N, which means ∠L ≅ ∠N by the Transitive Property of Congruence. If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram. Therefore, KLMN is a parallelogram. It could also be a rhombus, rectangle, or square, but more information would be needed to justify that conclusion. 27. Critique Reasoning Consider the statement “If △PQR is similar to △QPR, then △PQR is similar to △RPQ.” Explain whether or not this statement is true. The statement is false. If △PQR is similar to △QPR, then ∠P ≅ ∠Q and the triangle is isosceles. This does not prove any correspondence between ∠P and ∠R, or between ∠Q and ∠R. For △PQR to be similar to △RPQ, the triangle would have to be equilateral. Module 16 IN2_MNLESE389847_U7M16L3 859 859 Lesson 16.3 859 Lesson 3 18/04/14 9:05 PM Lesson Performance Task You’ve hired an architect to design your dream house and now the house has been built. Before moving in, you’ve decided to wander through the house with a tape measure to see how well the builders have followed the architect’s floor plan. Describe in as much detail as you can how you could accomplish your goal. Then discuss how you can decide whether the room shapes and other features of the house are similar to the corresponding shapes on the floor plan. INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Rectangle 1 is a units in length and b units in BEDROOM KITCHEN BEDROOM LIVING RM 17.5’ x 18.7’ width. Rectangle 2 is obtained by multiplying the sides of Rectangle 1 by 10. FOYER • Is Rectangle 2 similar to Rectangle 1? Explain. Yes; multiplying the sides by the same scale factor changes the size of the figure but not the shape, so the rectangles are similar. SCALE 1“ 2 = 10 ft Among things students should discuss: •the scale of the floor plan and how they could use it to check the dimensions of the rooms; • a method they could use to check whether the shapes of the rooms and other features on the floor plan could be mapped to the corresponding shapes in the house by a series of transformations, including dilations; • How does the area of Rectangle 2 compare to that of Rectangle 1? Explain. The area of Rectangle 1 is ab, and the area of Rectangle 2 is 100ab, so the area of Rectangle 2 is 100 times as great as the area of Rectangle 1. • the measurements that should be preserved in the transformation from floor plan to house (e.g. angles) and those that would not be (e.g. lengths). INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Three streets meet to form an equilateral © Houghton Mifflin Harcourt Publishing Company triangle 100 yards on each side. In a photo of the triangle taken directly overhead, each side measures 4 inches. How are the actual triangle of streets and the triangle in the photo alike? How are they different? Explain. Their sizes are different but their shapes are the same; the sides of the triangle in the photograph are shorter than the actual triangle, but because they are both equilateral triangles, their sides are proportional and each has three 60° angles. Module 16 860 Lesson 3 EXTENSION ACTIVITY IN2_MNLESE389847_U7M16L3 860 Supply yardsticks or meter sticks to students and have them make a floor plan of the classroom. Students should concentrate on features of the room that are permanent, such as dimensions and shape, omitting those that are not—desks and tables, for example. You may wish to specify the approximate size of the floor plan (for example, an 8.5 in. × 11 in. sheet of paper), thereby forcing students to calculate an appropriate scale. Students may exchange floor plans with a partner and check the partner’s scale, shapes, and calculations. 18/04/14 9:05 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Corresponding Parts of Similar Figures 860 LESSON 16.4 Name AA Similarity of Triangles Class 16.4 AA Similarity of Triangles Essential Question: How can you show that two triangles are similar? Common Core Math Standards Explore The student is expected to: COMMON CORE G-SRT.A.3 Exploring Angle-Angle Similarity for Triangles Resource Locker Two triangles are similar when their corresponding sides are proportional and their corresponding angles are congruent. There are several shortcuts for proving triangles are similar. Use the properties of similarity transformations to establish the AA criterion for two triangles to be similar. Also G-SRT.B.5 A Mathematical Practices COMMON CORE Date Draw a triangle _ and label it △ABC. Elsewhere on your page, draw a segment longer than AB and label the endpoints D and E. MP.5 Using Tools C Language Objective F Explain to a partner how to use the Angle-Angle criterion to show similarity in triangles. B A D D B Copy ∠CAB and ∠ABC to points D and E, respectively. Extend the rays of your copied angles, if necessary, and label their intersection point F. You have constructed △DEF. Essential Question: How can you show that two triangles are similar? C You constructed angles D and E to be congruent to angles A and B, respectively. Therefore, We can use the AA, SSS, or SAS similarity criteria to prove that triangles are similar. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the illustration and ask students to speculate on what it depicts. Then preview the Lesson Performance Task. © Houghton Mifflin Harcourt Publishing Company ENGAGE angles C and F must also be congruent because of the Third Angle Theorem. D Check the proportionality of the corresponding sides. Possible answer (ratios should be equal): 4 AB = _ = 0.4 _ DE 10 3 AC = _ = _ 0.4 DF 7.5 2 BC = _ _ = 0.4 EF 5 Since the ratios are equal, the sides of the triangles are proportional . Reflect 1. Discussion Compare your results with your classmates. What conjecture can you make about two triangles that have two corresponding congruent angles? 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Lesson 4 861 Module 16 6L4.indd 47_U7M1 ESE3898 IN2_MNL 861 Lesson 16.4 861 18/04/14 9:00 PM 18/04/14 9:01 PM Explain 1 Proving Angle-Angle Triangle Similarity EXPLORE The Explore suggests the following theorem for determining whether two triangles are similar. Angle-Angle (AA) Triangle Similarity Theorem Exploring Angle-Angle Similarity for Triangles If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar. Example 1 Prove the Angle-Angle Triangle Similarity Theorem. Given: ∠A ≅ ∠X and ∠B ≅ ∠Y B INTEGRATE TECHNOLOGY Y Prove: △ABC ∼ △XYZ Students have the option of doing the similar triangles activity either in the book or online. C A Z X QUESTIONING STRATEGIES XY . Let the image of △ABC be △A´B´C. Apply a dilation to △ABC with scale factor k = _ AB B’ What does Angle-Angle similarity claim about triangles? According to the Angle-Angle similarity criterion, triangles with two pairs of congruent angles are similar. Y B dilation A C C’ A’ Z X ∠A and ∠B´ ≅ ∠B corresponding angles of similar triangles are congruent . because XY ∙ AB = XY AB Also, A´B´ = k ∙ AB = . EXPLAIN 1 △A´B´C is similar to △ABC, and ∠A´ ≅ _ It is given that ∠A ≅ ∠X and ∠B ≅ ∠Y By the Transitive Property of Congruence, ∠A´ ≅ ∠X and ∠B´ ≅ ∠Y . the ASA Triangle Congruence Theorem . So, △A´B´C´ ≅ △XYZ by This means there is a sequence of rigid motions that maps △A´B´C´ to △YYZ. The dilation followed by this sequence of rigid motions shows that there is a sequence of similarity transformations that maps △ABC to △XYZ. Therefore, △ABC ∼ △XYZ. Reflect 2. Discussion Compare and contrast the AA Similarity Postulate with the ASA Congruence Postulate. Both postulates require that two pairs of angles be congruent, but for congruence you also © Houghton Mifflin Harcourt Publishing Company Proving Angle-Angle Triangle Similarity INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Compare proving two triangles similar with proving two triangles congruent. QUESTIONING STRATEGIES need to know that the included sides are congruent, so that the figures are the same size. How do you use AA similarity to show two triangles are similar? Show that two angles of one triangle are congruent to two angles of the other triangle. This lets you conclude that the two triangles are similar. The AA Similarity Postulate shows only that the two triangles are the same shape. Module 16 862 Lesson 4 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M16L4.indd 862 Learning Progression 18/04/14 9:01 PM Students have already proved triangles congruent using SAS, SSS, and ASA. The same kind of reasoning is used here to explore AA similarity, and to use this similarity to solve problems. AA Similarity of Triangles 862 3. EXPLAIN 2 In △JKL, m∠J = 40° and m∠K = 55°. In △MNP, m∠M = 40° and m∠P = 85°. A student concludes that the triangles are not similar. Do you agree or disagree? Why? Disagree; by the Triangle Sum Theorem, m ∠ N = 55°, so the triangles are similar by the AA Similarity Criterion. Applying Angle-Angle Similarity Explain 2 Applying Angle-Angle Similarity Architects and contractors use the properties of similar figures to find any unknown dimensions, like the proper height of a triangular roof. They can use a bevel angle tool to check that the angles of construction are congruent to the angles in their plans. INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Remind students that, in triangle similarity, they should identify sides that are proportional, rather than congruent. Example 2 Find the indicated length, if possible. BE First, determine whether △ABC ∼ △DBE. By the Alternate Interior Angles Theorem, ∠A ≅ ∠D and ∠C ≅ ∠E, so △ABC ∼ △DBE by the AA Triangle Similarity Theorem. E Find BE by solving a proportion. © Houghton Mifflin Harcourt Publishing Company ∙ Image Credits: ©John Lund/Drew Kelly/Blend Images/Corbis BD = _ BE _ BA BC A 36 54 = _ BE _ 54 36 54 54 54 ∙ 54 = _ BE ∙ 54 _ 54 36 BE = 81 B D C RT Check whether △RSV ∼ △RTU: It is given in the diagram that ∠ RSV ≅ ∠ T . ∠R is shared by both triangles, so ∠R ≅ ∠R by the Reflexive Property of Congruence. AA Triangle Similarity Theorem , △RST ∼ △RTU . So, by the R S 10 T 8 12 Find RT by solving a proportion. V TU RT = _ _ RS SV U 12 RT = _ ∠RT = 15 _ 10 8 Module 16 863 Lesson 4 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M16L4.indd 863 Small Group Activity Have students work in small groups and draw diagrams to illustrate each of these statements: all squares are similar; all rectangles are not similar; if two polygons are congruent, they are also similar; all right triangles are not similar. 863 Lesson 16.4 18/04/14 9:01 PM Reflect 4. In Example 2A, is there another way you can set up the proportion to solve for BE? BD BA Yes; = would also give the correct result for BE. BE BC 5. Discussion When asked to solve for y, a student sets up the proportion as shown. Explain why the proportion is wrong. How should you adjust the proportion so that it will give the correct result? QUESTIONING STRATEGIES _ _ A y y _ 14 _ = 8 10 The variable y does not refer to the side of a triangle but just a 14 B 10 C How can you use the AA similarity postulate to find unknown dimensions? You can use AA similarity to establish that two triangles with two congruent pairs of angles are similar, and then write a proportion to find unknown lengths of sides. D segment of a side. y____ +8 14 = __ would be a correct way to solve for y. 8 10 8 AVOID COMMON ERRORS E Some students may use an incorrect sequence of points when writing a similarity statement. Compare the process to writing a congruence statement and remind them to list corresponding vertices in the same order. Your Turn 6. A builder was given a design plan for a triangular roof as shown. Explain how he knows that △AED ∼ △ACB. Then find AB. A 9 ft. E D 15 ft. C 7. 6 ft. B By the Corresponding Angles Theorem, ∠AED ≅ ∠C and ∠ADE ≅ ∠B (or ∠A ≅ ∠A by the Reflexive Property of Congruence), so △AED ∼ △ACB by the AA Triangle Similarity Theorem. AB _6 = _ → AB = 10 feet 9 Find PQ, if possible. _ _ ∠TSQ is a right angle because PS and TR are perpendicular. So △PRS ∼ △TQS by the AA Triangle Similarity Theorem. Let x = PQ. P Q 9 15 Explain 3 S 12 R x+9 _ 12 _ → x = 11 = 15 9 © Houghton Mifflin Harcourt Publishing Company T 15 So, PQ = 11. Applying SSS and SAS Triangle Similarity In addition to Angle-Angle Triangle Similarity, there are two additional shortcuts for proving two triangles are similar. Side-Side-Side (SSS) Triangle Similarity Theorem If the three sides of one triangle are proportional to the corresponding sides of another triangle, then the triangles are similar. Side-Angle-Side (SAS) Triangle Similarity Theorem If two sides of one triangle are proportional to the corresponding sides of another triangle and their included angles are congruent, then the triangles are similar. Module 16 864 Lesson 4 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M16L4.indd 864 Cognitive Strategies 18/04/14 9:01 PM Have students write their own AAA Similarity Postulates. Then ask them to explain why this postulate is not necessary. An AAA Similarity Postulate is not required because of this theorem: Given two triangles, if two pairs of corresponding angles are congruent, then the remaining pair of corresponding angles must also be congruent. AA Similarity of Triangles 864 Example 3 EXPLAIN 3 Determine whether the given triangles are similar. Justify your answer. N 4 Applying SSS and SAS for Triangle Similarity 4 M 8 P Q 6 R You are given two pairs of corresponding side lengths and one pair of congruent corresponding angles, so try using SAS. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Remind students that similarity statements Check that the ratios of corresponding sides are equal. 8 =_ MN = _ 8 =_ 4=_ MP = _ 2 2 _ _ 8+ 4 3 12 3 6 MR MQ Check that the included angles are congruent: ∠NMP ≅ ∠QMR is given in the diagram. Therefore △NMP ∼ △RMQ by the SAS Triangle Similarity Theorem. indicate corresponding parts in the same way congruence statements do. M G 4 8 H 15 12 10 J N L 6 You are given three pairs of corresponding side lengths and zero congruent corresponding angles, so try using the SSS Triangle Similarity Theorem . © Houghton Mifflin Harcourt Publishing Company Check that the ratios of corresponding sides are equal. 3 12 LM = _ _ =_ GH 8 2 Therefore △ GHJ 3 15 MN = _ _ =_ HJ 10 2 ∼△ LMN 3 6 GJ _ =_=_ LN 2 4 by the SSS Triangle Similarity Theorem . Since you are given all three pairs of sides, you don’t need to check for congruent angles. Reflect 8. Are all isosceles right triangles are similar? Explain why or why not. Let the legs of one isoc. triangle be x and the legs of another be y. Then the ratios of the sides would be _xy and _xy . The included angle in each triangle is the right angle, so two isosceles right triangles are similar by SAS. 9. Why isn't Angle-Side-Angle (ASA) used to prove two triangles similar? ASA implies two pairs of angles are congruent, which is sufficient to show the triangles similar, so checking the included sides is unnecessary. Module 16 865 Lesson 4 LANGUAGE SUPPORT IN2_MNLESE389847_U7M16L4.indd 865 Connect Vocabulary Relate the idea of proof to justifying ideas in math. You use established rules and conventions to draw some conclusions. In real life, proof means showing something by gathering evidence by established rules and conventions. 865 Lesson 16.4 18/04/14 9:01 PM Your Turn QUESTIONING STRATEGIES If possible, determine whether the given triangles are similar. Justify your answer. 10. C 5 10 6 F B 11. H The two triangles cannot be proven similar. Although 3 the two given sides are in proportion, there is not a G pair of included congruent angles. A AVOID COMMON ERRORS By the Pythagorean Theorem, NO = 6 and GH = 4, M 8 Why are ASA and AAS not similarity theorems? Both of these contain two pairs of corresponding congruent angles, so ASA and AAS triangles are already similar by the AA similarity theorem. Some students may have difficulty identifying corresponding sides in similar triangles because of the orientation of the figures. Show these students how they can copy one of the triangles onto a piece of paper, then cut it out and rotate it, so that the two triangles have the same orientation. HJ GJ GH 1 so ___ = ___ = ___ =_ . △MNO ∼ △GHI by the MN 2 NO MO SSS Triangle Similarity Theorem. 10 J 5 N O G 3 H ELABORATE Elaborate 12. Is triangle similarity transitive? If you know △ABC ∼ △DEF and △DEF ∼ △GHJ, is △ABC ∼ △GHJ? Explain. Yes. If the first two triangles have three pairs of congruent angles and the second two QUESTIONING STRATEGIES Two isosceles triangles have congruent vertex angles. Explain why the two triangles must be similar. Let the measure of the vertex angles be x°. Then, by the Isosceles Triangle Theorem, the base angle in each of the triangles must measure half of (180 - x)°. So, the triangles are similar by AA similarity. triangles do as well, then the first and third triangles will also have those three pairs of 13. The AA Similarity Postulate applies to triangles. Is there an AAA Similarity Postulate for quadrilaterals? Use your geometry software to test your conjecture or create a counterexample. No; a square and a rectangle each have three pairs of right angles, but they won’t be similar because the sides aren’t proportional. 14. Essential Question Check-In How can you prove triangles are similar? Triangles are similar when their corresponding angles are congruent and their © Houghton Mifflin Harcourt Publishing Company congruent angles. SUMMARIZE THE LESSON corresponding sides are in proportion, but it is sufficient to use AA Similarity (show two pairs of congruent angles), SSS Similarity (show three pairs of sides in proportion), or SAS Similarity (show two pairs of sides are in proportion and their included angles are congruent). Module 16 IN2_MNLESE389847_U7M16L4.indd 866 866 Lesson 4 18/04/14 9:01 PM Which postulates allow you to conclude that triangles are similar without using transformations to map one to the other? What do you need to know before you can apply them? The AA similarity postulate, the SSS similarity postulate, and the SAS similarity postulate; you need to know than the triangles have two pairs of congruent angles, or that all three pairs of sides are proportional, or that two pairs of sides are proportional and the included angles are congruent. AA Similarity of Triangles 866 Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice Show that the triangles are similar by measuring the lengths of their sides and comparing the ratios of the corresponding sides. 1. 2. F A B D E ASSIGNMENT GUIDE Concepts and Skills D B C A C E F Practice 4.5 DE = _ _ = AB 3 By the Pythagorean Theorem, NO = 6 and GH = 4, Explore Exercises 1–2 HJ GJ GH 1 so ___ = ___ = ___ =_ . △MNO ∼ △GHI by the MN 2 NO MO Exploring Angle-Angle Similarity in SSSTriangles Triangle Similarity Theorem. Example 1 Proving Angle-Angle Triangle Similarity Exercises 3–6 Example 2 Applying Angle-Angle Similarity Exercises 7–10 Example 3 Applying SSS and SAS for Triangle Similarity Exercises 11–14 3 __ or 1.5 2 2.1 DF = _ _ = AC 1.4 _3 or 1.5 3.9 EF = _ _ = BC 2.6 _3 or 1.5 2 2 3 AB = _ _ = DE 6 _1 2 AC = _ _ = DF 4 _1 2.5 BC = _ _ = EF 5 _1 2 2 2 Determine whether the two triangles are similar. If they are similar, write the similarity statement. 3. 4. D B A 65° 16° 48° © Houghton Mifflin Harcourt Publishing Company E A 65° C 67° F Exercise IN2_MNLESE389847_U7M16L4.indd 867 Lesson 16.4 B By the AA Triangle Similarity Postulate, △ABC ∼ △DEF. Module 16 867 78° D By the Triangle Angle Sum Theorem, m∠C = 67°. So ∠A ≅ ∠D and ∠C ≅ ∠F. C △ABC and △ACD are isosceles triangles, so ∠B ≅ ∠ACB and ∠D ≅ ∠ACD. 16 + 2 ∙ m ∠B = 180, so m ∠B = 82. However, m ∠D = m ∠ACD = 78. None of the angles in △ABC is congruent to ∠D, so the triangles are not similar. Lesson 4 867 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–2 2 Skills/Concepts MP.5 Using Tools 3–17 2 Skills/Concepts MP.2 Reasoning 18 2 Skills/Concepts MP.4 Modeling 19 3 Strategic Thinking MP.3 Logic 18/04/14 9:01 PM Determine whether the two triangles are similar. If they are similar, write the similarity statement. 5. 6. D A D C A AVOID COMMON ERRORS B Because they need to know only that two angles of two triangles are congruent in order to prove similarity, students might think they need to know only three angles of two quadrilaterals to do the same, and so on for any n-gon. Use counter examples to show that this is incorrect. Stress that triangles are a special case because they are rigid structures. C ∠D ≅ ∠C, and ∠BAC ≅ ∠DCA by the Alternate Interior Angles Theorem. Therefore △ADC ∼ △BCA by the AA Triangle Similarity. E B ∠A ≅ ∠E, and ∠ACB ≅ ∠ECD by the Vertical Angles Theorem. Therefore △ABC ∼ △EDC by the AA Triangle Similarity Theorem. Explain how you know whether the triangles are similar. If possible, find the indicated length. 7. AC 8. AD A F C 5.0 10.2 9.0 B A B D 7.5 15.0 E D The triangles are similar by the AA Triangle Similarity Theorem. It is not possible to find the indicated length because, the length of the corresponding side, DF, is not known. 9. QR 12.0 C E The triangles are similar by AA Similarity. 5 AD __ =_ → AD = 6.7 12 9 10. Find BD. A P A 2.0 B 1.6 B C R Not possible. Only one congruent angle is identified between △ABC and △DBC, so similarity cannot be established. The triangles are similar by AA Similarity. QR 1.2 ___ = ___ → QR = 0.96 1.6 2 Module 16 Exercise IN2_MNLESE389847_U7M16L4.indd 868 C D Q 1.2 17.0 Lesson 4 868 Depth of Knowledge (D.O.K.) © Houghton Mifflin Harcourt Publishing Company 8.3 COMMON CORE Mathematical Practices 20 3 Strategic Thinking MP.2 Reasoning 21–22 3 Strategic Thinking MP.3 Logic 18/04/14 9:01 PM AA Similarity of Triangles 868 Show whether or not each pair of triangles are similar, if possible. Justify your answer, and write a similarity statement when the triangles are similar. INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 When using the SSS and SAS Similarity 11. 12. A D B 5.6 3.5 C 4.0 16 Theorems, some students have difficulty matching up the corresponding sides. Tell these students to match up the smallest side to the smallest side, the longest side to the longest side, and match the sides that are neither longest nor shortest. 16 A E 9 D 6.4 E 3 9 F ∠ACB ≅ ∠ECD by the Vertical Angle equal, so the two triangles are not similar. 3.5 BC Theorem. __ = ___ = 0.625; 5.6 CD AC ( = 4.0/6.4 ) = 0.625. Therefore CE △ABC ∼ △DEC by SAS Similarity. 13. 14. B C 5 _ 16 __ 5 AC AB __ = __ = __ ; BC = _ . The ratios are not DE 5 EF DF 3 B H 6 15 D 8 A C 20 3.2 6 BD G © Houghton Mifflin Harcourt Publishing Company Therefore △ABC ∼ △BDC by SSS Similarity. R The triangles cannot be proven similar using the given information, because the congruent angle is not an included angle. BC 5.5 AB ___ = ___ = 1.15, and ___ = __6 = 1 B 5.5 6.0 EF F 4.8 A 6.0 E 4.8 DF 6 Because the two ratios are not equal, the two triangles are not similar. 4.4 7.5 C D ¯ is the The student did not compare corresponding sides of the two triangles. AB _ shortest side of △ABC, so its corresponding side is DE the shortest side of △DEF. AC AB ___ The ratios __ , BC and __ are equal, so the triangles are similar by SSS. DE EF DF IN2_MNLESE389847_U7M16L4.indd 869 Lesson 16.4 60° J 15. Explain the Error A student analyzes the two triangles shown below. Explain the error that the student makes. Module 16 869 9.6 60° 12 8 BC P 10 8 15 20 BC AC AB __ = ___ = 2.5; __ = __ = 2.5; __ = __ = 2.5. DC Q 8 3.2 869 Lesson 4 18/04/14 9:01 PM 16. Algebra Find all possible values of x for which these two triangles are similar. (x + 10)° (2x - 50)° x° 70° The possible values of x are the solutions of x = 70, x = 2x - 50, x + 10 = 70, x + 10 = 2x - 50, x + x + 10 + 70 = 180, and x + x + 10 + 2x - 50 = 180. These result in 50, 55, 60, or 70, of which only 50 results in similar triangles. So x = 50° is the only possible value. 17. Multi-Step Identify two similar triangles in the figure, and explain why they are similar. Then find AB. B ∠A ≅ ∠A, and ∠ABD ≅ ∠C , so △ABD ≅ △ACB by the AA A AB AD AB 4 = __ → __ = __ → AB = 8 Triangle Similarity Theorem. __ 16 AB AB AC 4 D C 12 18. The picture shows a person taking a pinhole photograph of himself. Light entering the opening reflects his image on the wall, forming similar triangles. What is the height of the image to the nearest inch? 15 in. 4 ft 6 in. 5 ft 5 in. 5'5" 65 h _ = _ = _ → h = 18 inches 15 4'6" 54 H.O.T. Focus on Higher Order Thinking Y B XZ and ∠A ≅ ∠X XY = _ Given: _ AB AC Prove: △ABC ∼ △XYZ A C ge07sec07l03004a AB Z X XY Apply a dilation to △ABC with scale factor k = __ and let the image of △ABC be △A'B'C'. AB Then ∠A' ≅ ∠A. It is given that ∠A ≅ ∠X, so by transitivity ∠A' ≅ ∠X. XY XY XZ Also A'B' = k ∙ AB = __ ∙ AB = XY and A'C' = k ∙ AC = __ ∙ AC = __ ∙ AC = XZ. Therefore, AB AB AC © Houghton Mifflin Harcourt Publishing Company 19. Analyze Relationships Prove the SAS Triangle Similarity Theorem. △A'B'C ≅ △XYZ by SAS Congruence. So a sequence of rigid motions maps △A'B'C to △XYZ. The dilation followed by this sequence of rigid motions shows that there is a sequence of similarity transformations that maps △ABC to △XYZ. So △ABC ∼ △XYZ. Module 16 IN2_MNLESE389847_U7M16L4.indd 870 870 Lesson 4 18/04/14 9:01 PM AA Similarity of Triangles 870 20. Analyze Relationships Prove the SSS Triangle Similarity Theorem. JOURNAL Have students write a journal entry to explain what a scale on a map means, how it is used, and how it is related to the concept of similarity. XZ = _ XY = _ YZ Given: _ AB AC BC Y B Prove: △ABC ∼ △XYZ (Hint: The main steps of the proof are similar to those of the proof of the AA Triangle Similarity Theorem.) A C Z X XY Apply a dilation to △ABC with scale factor k = __ and let the image of AB △ABC be △A'B'C'. Then: XY A'B' = k ⋅ AB = __ ⋅ AB = XY AB XY XZ A'C' = k ∙ AC = __ ∙ AC = __ ∙ AC = XZ AB AC XY YZ B'C' = k ∙ BC = __ ∙ BC = __ ∙ BC = YZ. Therefore, △A'B'C ≅ △XYZ by the AB BC SSS Congruence Theorem. This means there is a sequence of similarity transformations that maps △ABC to △XYZ. So △ABC ∼ △XYZ. ‹ › − 21. Communicate Mathematical Ideas A student is asked to find point X on BC such that △ABC ∼ ‹ › − △XBA and XB is as small as possible. The student does so by constructing a perpendicular line to AC at ‹ › − point A, and then labeling X as the intersection of the perpendicular line with BC . Explain why this procedure generates the similar triangle that the student was requested to construct. A 10 6 © Houghton Mifflin Harcourt Publishing Company X B 8 C Possible Answer: For XB to be as small as possible, it should correspond to ― the shortest side of △ABC, which is AB. Thus, X corresponds to A. 22. Make a Conjecture Builders and architects use scale models to help them design and build new buildings. An architecture student builds a model of an office building in which the height of the model 1 1 is ___ of the height of the actual building, while the width and length of the model are each ___ of the 400 200 corresponding dimensions of the actual building. The model includes several triangles. Describe how a triangle in this model could be similar to the corresponding triangle in the actual building, then describe how a triangle in this model might not be similar to the corresponding triangle in the actual building. Use a similarity theorem to support each answer. A triangle that lies entirely in a plane parallel to the ground will have side lengths that are 1 of the side lengths of the corresponding triangle in the actual building. So the two each ___ 200 triangles are similar by the SSS Triangle Similarity Theorem. A triangle that has vertices at several heights will not have side lengths that form a constant ratio with the side lengths of the corresponding triangle in the actual building, and so would not be similar. Module 16 IN2_MNLESE389847_U7M16L4.indd 871 871 Lesson 16.4 871 Lesson 4 18/04/14 9:01 PM Lesson Performance Task CONNECT VOCABULARY The figure shows a camera obscura and the object being “photographed.” Answer the following questions about the figure: 1. Explain how the image of the object would be affected if the camera were moved closer to the object. How would that limit the height of objects that could be photographed? 2. How do you know that △ADC is similar to △GDE? The name camera obscura comes from the Latin words camera, meaning room, and obscura, meaning dark. The plural of Latin words ending in the letter a is formed by changing a to ae. So, more than one of these instruments would be referred to as camerae obscurae. A B D E F G C 3. Write a proportion you could use to find the height of the pine tree. INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 A camera obscura is square. This means that 4. DF = 12 in., EG = 8 in., BD = 96 ft. How tall is the pine tree? 1. As the object gets closer the vertical angle at D will get larger, and since △ADC remains similar to △GDE, the height of the image on the back surface will increase. So the camera must be placed far enough from the object that the image is no taller than the height of the back surface, or else the object will not be able to be photographed. in the figure in the Lesson Performance Task, DF = EG. Suppose you want to photograph an object that is n feet from the front of the camera. What is the maximum height of such an object if you want to photograph its entire height? Explain your reasoning. The maximum height is n feet. AC EG Sample answer: Because DF = EG, and , ___ = ___ , BD DF AC = BD = 1. 2. Possible answer: ∠A ≅ ∠G because the angles are alternate interior angles ― ― for parallel lines AC and EG . ∠ADC ≅ ∠GDE because vertical angles are congruent. So, △ADC ∼ △GDE by the AA Similarity Theorem. AC EG = __ 3. __ BD DF AC __ = 8 ⇒ AC = 64 ft 4. __ 96 12 © Houghton Mifflin Harcourt Publishing Company Module 16 872 Lesson 4 EXTENSION ACTIVITY IN2_MNLESE389847_U7M16L4.indd 872 Have students research methods of making simple models of a camera obscura. For some models, no more than a cereal box or shoe box, tape, scissors, and a pin are required. A more elaborate model can be made by blocking a window of the classroom and projecting an outside scene onto a sheet in the classroom. Either method will give reasonable results when students compare the dimensions and angles of the external object with those of the projected image. 18/04/14 9:01 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. AA Similarity of Triangles 872 MODULE 16 MODULE STUDY GUIDE REVIEW 16 Similarity and Transformations Study Guide Review Essential Question: How can you use similarity and transformations to solve real-world problems? ASSESSMENT AND INTERVENTION KEY EXAMPLE (Lesson 16.1) Determine the center of dilation and the scale factor of the dilation. 8 B′ y B 6 Assign or customize module reviews. 4 O A′ A 2 0 C 2 4 C′ x 6 Key Vocabulary center of dilation (centro de dilatación) dilation (dilatación) scale factor (factor de escala) Side-Side-Side Similarity (Similitud Lado -Lado -Lado) Side-Angle-Side Similarity (Similitud Lado-ÁnguloLado) similar (semejantes) similarity transformation (transformación de semejanza) 8 Draw a line through A and A′. Draw a line through B and B′. Draw a line through C and C′. The three lines intersect at point O(2, 3). Find the distance from point O to points A and A'. ___ _ d A = √(2 - 3) + (3 - 4) = √ 2 2 2 ___ _ d A′ = √ (2 - 4) + (3 - 5) = 2√ 2 2 2 Find the distance to point A. Find the distance to point A'. © Houghton Mifflin Harcourt Publishing Company The distance from point O to point A′ is twice the distance from point O to point A. The scale factor of dilation is 2 to 1. KEY EXAMPLE Corresponding sides of similar _figures _are proportional. _ _ _ _ BC AD _ = ____ _. BC = FG and EH = AD, so ___ EH FG Corresponding angles of similar figures are congruent. ∠D ≌ ∠H and ∠E ≌ ∠A. Module 16 IN2_MNLESE389847_U7M16MC 873 873 Module 16 (Lesson 16.3) △ABCD maps to △EFGH by a similarity transformation. Write a proportion that BC and ¯ EH. Then list any angles that are congruent to ∠D or ∠E. contains ¯ 873 Study Guide Review 18/04/14 9:35 PM KEY EXAMPLE (Lesson 16.4) Determine whether △ABC and △DEF are similar. If so, justify by SSS or SAS. A 3 E B 6 7.5 8 4 C D 10 F Check that the ratios of corresponding sides are equal. 3 __ 4 6 = __ 3 __ 8 4 7.5 = __ 3 _ 4 10 Since the ratios of all corresponding sides are equal, the triangles are similar by SSS. EXERCISES Determine the following for the dilation. (Lesson 16.1) 8 6 y A B 4 D 2 A′ D′ 0 center Module 16 IN2_MNLESE389847_U7M16MC 874 (3, 0) 2 O 4 C B′ x 6 2. 874 8 scale factor © Houghton Mifflin Harcourt Publishing Company 1. C′ 1 to 3 Study Guide Review 18/04/14 9:35 PM Study Guide Review 874 Determine whether the two figures are similar using similarity transformations. (Lesson 16.2) MODULE PERFORMANCE TASK 3. similar △ABC to △DEF 8 B COMMON CORE Mathematical Practices: MP.1, MP.2, MP.4, MP.6 G-MG.A.1, G-MG.A.3 -12 -8 D A -4 E SUPPORTING STUDENT REASONING Students should begin this problem by focusing on what information they will need. Here are some issues they might bring up. B y 0 A x C C D F 0 -4 2 x G -2 4 F E -8 H △ABC maps to △DEF by a similarity transformation. (Lesson 16.3) 5. • What should be the maximum height of the model: Students can either work backwards from the desired height or find a percent of the actual height and use it to determine the scale. 6. _ _ Write a proportion that contains BC and DF. 7. List any angles that are congruent to ∠A or ∠E. • Which dimensions are not needed for a scale model: Students can decide which dimensions might not be needed, such as the height at which the stone color changes. 8. ¯ ___ ¯ AB __ = BC ¯ DE ¯ EF ¯ ___ ¯ BC __ = AC ¯ EF ¯ DF ∠A ≅ ∠D; ∠E ≅ ∠B Determine whether △ABC and △DEF are similar. If so, justify by SSS or SAS. (Lesson 16.4) not similar 6 A 9 13 SAS 9. B C D 2 E 4 E 3 B F 9 6 A 76° C D 8 76° 12 F © Houghton Mifflin Harcourt Publishing Company • What materials should be used: Students can research the appropriate materials to use or you can make suggestions. not similar y 2 4 _ _ Write a proportion that contains AB and EF. • How to determine the dimensions of the model: Students should use proportions to convert the actual dimensions to the scaled dimensions. Students may first want to convert the actual dimensions to decimals. △ABCD to △EFGH 4. Module 16 875 Study Guide Review SCAFFOLDING SUPPORT IN2_MNLESE389847_U7M16MC 875 • Students should recognize that they will need to convert units of the given measurements so that all units are the same. Possible units are feet, inches, meters, or centimeters. • If students choose to solve the problem by first deciding on a scale factor, they will need to make sure the scale factor they use results in a model which will fit inside a classroom. 875 Module 16 18/04/14 9:35 PM MODULE MODULE PERFORMANCE TASK SAMPLE SOLUTION Designing a Model of the Washington Monument Use the scale 1 in. = 10 ft. Your challenge is to design a scale model of the Washington Monument that would be small enough to fit inside your classroom. Here are some key dimensions of the Washington Monument for you to consider in determining the scale factor for your model. (Note that the color of the stone changes part way up the monument because of a halt in construction between 1854 and 1877.) Total height Height to top of trapezoidal side First, convert units so that all measurements are in 1 ft feet, by multiplying inches by the ratio and 12 in. adding that to the number of feet. The results are: _ Total height: 555.42 ft 555 ft. 5 in. Width at base: 55.08 ft 500 ft Width at base 55 ft. 1 in. Width at top of trapezoidal side 34 ft. 5 in. Height at which stone color changes 16 Width at top of trapezoidal side: 34.42 ft To find the model’s dimensions, solve the proportion: 151 ft What scale factor will you use for your model? What are the key dimensions of your model? model height 1 in. = _____________ _____ 10 ft Begin by making some notes in the space below about your strategy for designing the model. Then use your own paper to complete the task. Present your plan using diagrams, words, and/or numbers. actual height Using the proportion, check the total height of the model to see if it will fit inside the classroom. 1 in. = _________ x _____ → x ≈ 55.5 in. 10 ft © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©lzzy Schwartz/Digital Vision/Getty Images Module 16 876 555.42 ft 1 ft ≈ 4.6 ft tall, so the scale factor is (____ 12 in. ) This is 55.54 reasonable. The remaining dimensions are as follows. Height to top of trapezoidal side: 50 in. Width at base: 5.5 in. Width at top of trapezoidal side: 3.4 in. Height at which stone color changes: 15.1 in. Study Guide Review DISCUSSION OPPORTUNITIES IN2_MNLESE389847_U7M16MC 876 18/04/14 9:35 PM • If students choose to work backwards from a desired height of exactly six feet, how will this change the calculations and scale factor used? • What are some advantages to choosing a scale that makes the model no more than a foot tall? Sample answer: The amount of material needed to construct the model will be much less, which in turn will lower both the cost and the weight and make the model more portable. Assessment Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain. 0 points: Student does not demonstrate understanding of the problem. Study Guide Review 876 Ready to Go On? Ready to Go On? 16.1–16.4 Similarity and Transformations ASSESS MASTERY Use the assessment on this page to determine if students have mastered the concepts and standards covered in this module. • Online Homework • Hints and Help • Extra Practice Answer each problem about the image. (Lesson 16.1) 1. ASSESSMENT AND INTERVENTION Yes Are the two shapes similar? 16 3 y B′ 2. Find the scale factor k. 3. Find the center of dilation. (4, 2) 12 4. area △A′B′C′D′ . Compare k to the ratio __ area △ABCD 8 A′ 4 A C The ratio equals 9, which is the square of the scale factor. D 0 Determine which of the following transformations are dilations. (Lesson 16.1) 5. Access Ready to Go On? assessment online, and receive instant scoring, feedback, and customized intervention or enrichment. (x, y) → (4x, 4y) 6. (x, y) → (x - 2, y - 2) 8. Response to Intervention Resources 9. • Success for English Learners • Challenge Worksheets Assessment Resources © Houghton Mifflin Harcourt Publishing Company Find the missing length. (Lesson 16.3) • Reading Strategies D′ x 16 ) △XYZ maps to △MNO with the transformation (x, y) → (7x, 7y). If XY = 3, what is the length of MN? Find the appropriate statements about the triangles. (Lesson 16.4) 10. △ABC is similar to △RTS. Write a proportion that contains AC and RT. Also write the congruence statements that must be true. AC _ __ = AB , ∠A ≅ ∠R, ∠B ≅ ∠T, ∠C ≅ ∠S RS RT ESSENTIAL QUESTION 11. How can you determine whether a shape is similar to another shape? Answers may vary. Sample: Two shapes are only similar if one shape can be mapped to the other through similarity transformations. These transformations are the rigid motions, meaning reflections, translations, and rotations, as well as dilations. Module 16 COMMON CORE IN2_MNLESE389847_U7M16MC 877 Module 16 12 21 • Leveled Module Quizzes 877 ( 1 x, _ 1y (x, y) → _ 3 3 Dilation ADDITIONAL RESOURCES Differentiated Instruction Resources 8 (x, y) → (-x, 3y) Not a dilation • Reteach Worksheets 4 Not a dilation Dilation 7. C′ B Study Guide Review 877 Common Core Standards Lesson Items 16.2 1 16.1 18/04/14 9:35 PM Content Standards Mathematical Practices G-SRT.A.2, G-SRT.A.1, G-CO.A.2 MP.7 2–4 G-SRT.A.1 MP.7 16.2 5–8 G-SRT.A.1, G-CO.A.2 MP.2 16.2 9 G-SRT.A.1, G-CO.A.2 MP.4 16.3 10 G-SRT.A.2 MP.4 MODULE MODULE 16 MIXED REVIEW MIXED REVIEW Assessment Readiness Assessment Readiness 1. Consider each transformation. Does the transformation preserve distance? Select Yes or No for A–C. A. Dilations Yes No B. Reflections C. Rotations Yes Yes ASSESSMENT AND INTERVENTION No No ( ) 1 _ x, 1 y . Choose True or False 2. △MNO maps to △RST with the transformation (x, y) → _ 3 3 for each statement. A. If RT = 3, MO = 9. True False B. If RT = 12, MO = 4. True False C. If RT = 9, MO = 27. True False 3. Determine if the following pair of triangles _ _are similar. If so, explain how. Note that AC || BD. A 70° Possible Answer: ∠ACB and ∠CBD are congruent because they are alternate interior angles. ∠ABC and ∠CDB are congruent because it is given they have the same measure. So the two triangles are similar by Angle–Angle Similarity. Assign ready-made or customized practice tests to prepare students for high-stakes tests. B ADDITIONAL RESOURCES Assessment Resources 70° C • Leveled Module Quizzes: Modified, B D 4. If △ABC is similar to △XYZ and △YZX, what special type of triangle is △ABC? Justify your reasoning. COMMON CORE AVOID COMMON ERRORS Item 2 Some students have trouble following a dilation rule backward. Remind students that if moving from the image to the original figure, multiply by the reciprocal of the original dilation. © Houghton Mifflin Harcourt Publishing Company Answers may vary. Sample: Examine the corresponding angles. Since △ABC is similar to △XYZ, this means ∠A ≅ ∠X. Likewise, since △ABC is similar to △YZX, this means ∠A ≅ ∠Y. Thus, by the Transitive Property of Congruence, it can be said that ∠X ≅ ∠Y. Also, ∠B ≅ ∠Y and ∠B ≅ ∠Z, which means ∠Y ≅ ∠Z by the Transitive Property of Congruence. Putting these facts together, it can be shown that ∠X ≅ ∠Y ≅ ∠Z. Therefore, △XYZ, △YZX, and by similarity, △ABC are all equilateral triangles. Module 16 16 Study Guide Review 878 Common Core Standards IN2_MNLESE389847_U7M16MC 878 18/04/14 9:35 PM Content Standards Mathematical Practices Lesson Items IM1 15.3, IM2 16.1 1* G-SRT.A.2 MP.7 16.3 2 G-SRT.A.1 MP.7 16.4 3 G-SRT.A.3 MP.7 16.3, 15.2 4* G-SRT.A.2 MP.3 * Item integrates mixed review concepts from previous modules or a previous course. Study Guide Review 878 MODULE 17 Using Similar Triangles Using Similar Triangles ESSENTIAL QUESTION: Answer: For one example, similar triangles can be used to find the height of a tall object by measuring its shadow. 17 MODULE Essential Question: How can you use similar triangles to solve real-world problems? LESSON 17.1 Triangle Proportionality Theorem LESSON 17.2 Subdividing a Segment in a Given Ratio This version is for Algebra 1 and Geometry only PROFESSIONAL DEVELOPMENT VIDEO LESSON 17.3 Using Proportional Relationships Professional Development Video Author Juli Dixon models successful teaching practices in an actual high-school classroom. LESSON 17.4 Similarity in Right Triangles Professional Development © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Lightguard/iStockPhoto.com my.hrw.com REAL WORLD VIDEO Check out how properties of similar triangles can be used to determine real-world areas of geographic regions like the Bermuda Triangle. MODULE PERFORMANCE TASK PREVIEW How Large Is the Bermuda Triangle? In this module, you will be asked to determine the area of the Bermuda Triangle from a map. How can indirect measurement and the properties of similar triangles help you find the answer? Let’s get started on solving this “mystery” of the Bermuda Triangle! Module 17 DIGITAL TEACHER EDITION IN2_MNLESE389847_U7M17MO 879 Access a full suite of teaching resources when and where you need them: • Access content online or offline • Customize lessons to share with your class • Communicate with your students in real-time • View student grades and data instantly to target your instruction where it is needed most 879 Module 17 879 PERSONAL MATH TRAINER Assessment and Intervention Assign automatically graded homework, quizzes, tests, and intervention activities. Prepare your students with updated, Common Core-aligned practice tests. 18/04/14 10:00 PM Are YOU Ready? Are You Ready? Complete these exercises to review skills you will need for this module. Scale Factor and Scale Drawings Example 1 ASSESS READINESS Determine the length of the segment A′B′ given AB = 4, BC = 3, and B′C′ = 6. A′ A The image of △ABC is created as a result of a scale drawing, so the transformation is a dilation. C Use the assessment on this page to determine if students need strategic or intensive intervention for the module’s prerequisite skills. • Online Homework • Hints and Help • Extra Practice B 2BC = B′C′, so 2AB = A′B′. ASSESSMENT AND INTERVENTION B′ C′ A′B′ = 2(4) = 8 Give the side length. 1. 6 BC, given AC = 5, A′C′ = 15, and B′C′ = 18 Similar Figures Example 2 The figures PQRS and KLMN are similar. Determine the angle in figure KLMN that is congruent to ∠Q and find its measure if m∠Q = 45°. ∠Q ≅ ∠L Corresponding angles of similar figures are congruent. m∠Q = m∠L = 45° Definition of congruency of angle. 3 2 1 Personal Math Trainer will automatically create a standards-based, personalized intervention assignment for your students, targeting each student’s individual needs! Give each angle measure. 2. 3. 67° m∠A, given △ABC ≅ △DEF and m∠D = 67° m∠E, given △PQR ≅ △DEF, m∠R = 13°, and m∠D = 67° `` 100° Example 3 A right triangle △ABC has side lengths AB = 3 and BC = 4. Find the length of the hypotenuse AC. △ABC is a right triangle, so the Pythagorean Theorem can be used. __ ac = √ ab 2 + bc 2 __ _ AC = √ 3 2 + 4 2 = √ 9 + 16 AC = 5 Write the Pythagorean Theorem. Substitute and simplify. Simplify. Find the side length for each right triangle. 4. 5. DE, given AB = 15, AC = 17, and AC is the hypotenuse IN2_MNLESE389847_U7M17MO 880 ADDITIONAL RESOURCES See the table below for a full list of intervention resources available for this module. Response to Intervention Resources also includes: • Tier 2 Skill Pre-Tests for each Module • Tier 2 Skill Post-Tests for each skill 13 DE, given DF = 5, EF = 12, and DE is the hypotenuse Module 17 © Houghton Mifflin Harcourt Publishing Company The Pythagorean Theorem TIER 1, TIER 2, TIER 3 SKILLS 8 880 Response to Intervention Tier 1 Lesson Intervention Worksheets Tier 2 Strategic Intervention Skills Intervention Worksheets Reteach 17.1 Reteach 17.2 Reteach 17.3 Reteach 17.4 40 Scale Factor and Scale Drawings 41 Similar Figures 43 The Pythagorean Theorem 46 Proportional Relationships Differentiated Instruction 18/04/14 9:59 PM Tier 3 Intensive Intervention Worksheets available online Building Block Skills 36, 38, 46, 48, 50, 63, 80, 82, 86, 90, 95, 100 Challenge worksheets Extend the Math Lesson Activities in TE Module 17 880 LESSON 17.1 Name Triangle Proportionality Theorem Essential Question: When a line parallel to one side of a triangle intersects the other two sides, how does it divide those sides? Resource Locker Constructing Similar Triangles Explore The student is expected to: In the following activity you will see one way to construct a triangle similar to a given triangle. G-SRT.B.4 A Prove theorems about triangles. Also G-CO.C.10, G-CO.D.12, G-SRT.B.5 Mathematical Practices COMMON CORE Date 17.1 Triangle Proportionality Theorem Common Core Math Standards COMMON CORE Class Do your work for Steps A–C in the space provided. Draw a triangle. Label it ABC as shown. A MP.5 Using Tools Check students' constructions. Language Objective B Work with a partner to describe the triangle proportionality theorem and its converse. B C _ Select a point on AB. Label it E. A E ENGAGE The line cuts two sides of the triangle into segments, and the ratios between the two segments of each side are equal. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photograph and ask students to identify the geometrical shapes seen in the sail. Then preview the Lesson Performance Task. C © Houghton Mifflin Harcourt Publishing Company Essential Question: When a line parallel to one side of a triangle intersects the other two sides, how does it divide those sides? B C Construct an angle with vertex E that is congruent to ∠B. Label the point where the _ side of the angle you constructed intersects AC as F. A E B D E F C _ ‹ › − Why are EF and BC parallel? ‹ › ‹ › ‹ › − − − Coplanar lines EF and BC are cut by transversal AC so that ∠AEF ≅ ∠B. By the Converse _ ‹ › − of the Corresponding Angles Theorem, EF ∥ BC. _ _ _ _ Graphics\02192014\From Ramesh\Econ 2016\W Use a ruler to measure AE, EB\\192.168.9.251\07Macdata\07Vol1Data\From , AF, and FB. Then compare the AE AF ratios ___ and ___ . EB FB Measurements will vary, but, if the constructions are accurate, the ratios should be approximately equal. 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E that is congr AC as with vertex an angle intersects Construct constructed angle you side of the the B y g Compan A E F Publishin C Converse ∠B. By the ‹ › so that ∠AEF ≅ − _ ‹ › and BC parallel? − versal AC EF ‹ › are cut by trans _ ‹ › and − orksheet − BC ‹ › ∥ BC. − h\Econ 2016\W lines EF rem, EF \From Rames Coplanar Angles Theo ics\02192014 sponding From Graph re the1Data\ _ of the Corre compa\07Vol _ Thenacdata FB.1\07M _ _\\192.1 68.9.25 ld be , AF, and re AE, EB ratios shou the measu to ate, Use a ruler ___ AF . ns are accur AE and ___ FB constructio ratios EB but, if the vary, ents will Lesson 1 Measurem l. ely equa approximat Why are © Houghto n Mifflin Harcour t B 881 Module 17 ESE3898 7L1 881 47_U7M1 IN2_MNL 881 Lesson 17.1 18/04/14 10:03 PM 18/04/14 10:04 PM Reflect 1. EXPLORE Discussion How can you show that △AEF ∼ △ABC? Explain. You can show that there are three pairs of congruent angles. ∠A ≅ ∠A (Reflexive Property ‹ › _ − of Equality) and ∠AEF ≅ ∠B (by construction). Also, because EF || BC, ∠AFE ≅ ∠C. c. Then Constructing Similar Triangles you can use the AA Similarity Criterion because you can show that there are two pairs of congruent angles. For instance, 2. INTEGRATE TECHNOLOGY AE AF What do you know about the ratios ___ and ___ ? Explain. EB FB Because △ABC ∼ △AEF and corresponding sides of similar triangles are Students have the option of doing the Explore activity either in the book or online. AE AF proportional, __ and ___ ? Explain. AB AB 3. Make a Conjecture Use your answer to Step E to make a conjecture about the line segments produced when a line parallel to one side of a triangle intersects the other two sides. The parallel line divides the other two sides so the lengths of the segments are QUESTIONING STRATEGIES How do you use the AA Similarity Criterion to show two triangles are similar? Show that two angles of one triangle are congruent to two angles of the other triangle. This lets you conclude that the two triangles are similar. proportional. Explain 1 Proving the Triangle Proportionality Theorem As you saw in the Explore, when a line parallel to one side of a triangle intersects the other two sides of the triangle, the lengths of the segments are proportional. Triangle Proportionality Theorem Theorem Hypothesis A E B EXPLAIN 1 Conclusion AE = _ AF _ EB FC F C EF ∥ BC Example 1 Prove the Triangle Proportionality Theorem ‹ › _ − Given: EF ∥ BC A AE AF = ___ Prove: ___ EB FB Step 1 Show that △AEF ∼ △ABC. ‹ › _ − Because EF ∥ BC, you can conclude that ∠1 ≅ ∠2 and by ∠3 ≅ ∠4 the Corresponding Angles Theorem. E 1 Proving the Triangle Proportionality Theorem © Houghton Mifflin Harcourt Publishing Company If a line is parallel to a side of a triangle intersects the other two sides, then it divides those sides proportionally. 3 F 4 C INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Review with students which geometric statements can be used to provide justification for each step of a proof. 2 B QUESTIONING STRATEGIES So, △AEF ∼ △ABC by the AA Similarity Criterion . Module 17 882 x+y Is it always true that the fraction _____ z can be y x + __? Yes, provided that z ≠ 0. written as __ z z Lesson 1 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M17L1 882 Math Background The Triangle Proportionality Theorem and its converse may be stated as a single biconditional: a line that intersects two sides of a triangle is parallel to the third side of the triangle if and only if it divides the intersected sides proportionally. 18/04/14 10:04 PM How do you take the reciprocal of both sides of an equation such as x = y? The reciprocal of an expression is 1 divided by the expression. One 1 way is to multiply both sides by ___ xy (xy ≠ 0). Triangle Proportionality Theorem 882 Step 2 Use the fact that corresponding sides of similar triangles are proportional to AE AF prove that ___ = ___ . EB FC EXPLAIN 2 AC ___ AB = _ AE Applying the Triangle Proportionality Theorem AF+ FC ______ AE + EB _ = AE Segment Addition Postulate AF EB = 1+_ AB INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Some students may misunderstand the Corresponding sides are proportional. AF 1+ EB = _ AE FC _ a+b a b __ __ Use the property that ____ c = c + c. AF FC ___ Subtract 1 from both sides. AF ___ Take the reciprocal of both sides. AF AE = _ EB FC Reflect Triangle Proportionality Theorem to mean that a line parallel to one side of a triangle divides the other two sides into congruent segments rather than proportional ones. Show these students that congruent segments result only in the special case of the line intersecting the two sides at their midpoints. 4. Explain how you conclude that ▵AEF ∼▵ABC without using ∠3 and ∠4. ∠A ≅ ∠A by the Reflexive Property of Congruence, and ∠1 ≅ ∠2 since they are corresponding angles; △AEF ∼ △ABC by the AA Similarity Criterion. Explain 2 Example 2 Applying the Triangle Proportionality Theorem Find the length of each segment. _ CY _ _ AX ___ AY by the Triangle Proportionality It is given that XY ∥ BC so ___ = YC BY Theorem. B 4 X 9 Substitute 9 for AX, 4 for XB, and 10 for AY. © Houghton Mifflin Harcourt Publishing Company Then solve for CY. C 9=_ 10 _ 4 CY Take the reciprocal of both sides. 4=_ CY _ 9 10 Next, multiply both sides by 10. () ( ) CY 10 4 = _ 10 _ 9 10 NP ___ PL by the L Triangle Proportionality Theorem. for N Q, 2 Multiply both sides by 3 A () _ _ NQ It is given that PQ ∥ LM, so ___ = QM 5=_ NP _ 2 3 10 40 = CY, or 4 _ 4 = CY _ 9 9 → Find PN. Substitute 5 Y 3 5 for QM, and 3 for PL . : 3 Module 17 () 5 = 3 _ 2 NP → (_ 3 ) 883 M Q 2 P N 15 2 _ or 7_ 1 2 = NP Lesson 1 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M17L1 883 Small Group Activity Give small groups of students each a sheet with three parallel lines cut by a transversal. Have students measure the segments of the transversal and verify the ratio formed. Then have them draw another transversal across the parallel lines and measure the resulting segments. Have students try to form a conjecture based on their results. 883 Lesson 17.1 18/04/14 10:04 PM Your Turn Find the length of each segment. _ 5. DG E 32 24 F 6. QUESTIONING STRATEGIES _ RN 40 C 8 D M G How can you use the Triangle Proportionality Theorem to find unknown dimensions? If a line cutting two sides of a triangle is parallel to the third, you know that it cuts the two sides proportionally. Knowing this allows you to set up and solve a proportion for the missing side length. 5 P Q 10 N R EC DG ED 32 24 40 24 10 _ 8 RN 5 5 _ _ = MQ _; _ = _; _ =_; _ = _; 40(_) = DG MR = ; _ = _; RN = (_)10 DG 24 CF DG 32 40 32 RN 8 8 4 4 8 Explain 3 5 10 QP RN 1 50 25 RN = _ = _ or 6_ Proving the Converse of the Triangle Proportionality Theorem AVOID COMMON ERRORS Some students may write proportions that do not compare corresponding parts of the figure. Have them use different colors to show the proportional parts. The converse of the Triangle Proportionality Theorem is also true. Converse of the Triangle Proportionality Theorem Theorem Hypothesis If a line divides two sides of a triangle proportionally, then it is parallel to the third side. Example 3 A Conclusion AE AF = EB FC F E B ‹ › _ − EF || BC EXPLAIN 3 C Proving the Converse of the Triangle Proportionality Theorem Prove the Converse of the Triangle Proportionality Theorem AE = _ AF Given: _ EB FC _ ‹ › − Prove: EF ∥ BC A F E AE = _ AF , and taking the reciprocal It is given that _ EB FC EB = FC AF . Now add 1 to of both sides shows that AE ___ ___ C B AF to the right side. AE to the left side and _ both sides by adding _ AE AF AF AE FC EB + AE = AF + AF This gives AE . AB __ __ = AC Adding and using the Segment Addition Postulate gives AE AF . ___ ___ ___ ___ Since ∠A ≅ ∠A, △AEF ∼ △ABC by the SAS Similarity Criterion. ‹ › _ − Step 2 Use corresponding angles of similar triangles to show that EF ∥ BC. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Remind students that the converse of a © Houghton Mifflin Harcourt Publishing Company Step 1 Show that △AEF ∼ △ABC. theorem switches the condition and the conclusion. Review converses from previous lessons, such as converses of triangle congruence theorems, and discuss why they are useful. ∠AEF ≅ ∠ ABC and are corresponding angles, and ∠AEF ≅ ABC . ‹ › _ − So, EF ∥ BC by the Converse of the Corresponding Angles Theorem. Module 17 884 QUESTIONING STRATEGIES What are the main steps of the proof? First, show that the two triangles are similar by SAS. Then, use congruent corresponding angles to show that the lines are parallel. Lesson 1 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M17L1 884 Modeling Use transparencies to explain the Triangle Proportionality Theorem. Put the parallel line on a separate transparency from the diagram of the triangle. Slide the line up and down relative to the triangle to show that regardless of the line’s position, the sides will be divided proportionally. 18/04/14 10:04 PM How is this proof similar to that of the Triangle Proportionality Theorem? Both proofs use properties of fractions and the Segment Addition Postulate to work with proportions. Both proofs also use the fact that you can take the reciprocal of both sides of a proportion. Triangle Proportionality Theorem 884 R Reflect EXPLAIN 4 _ Critique Reasoning A student states that UV must _ be parallel to ST. Do you agree? Why or why not? 7. RU RV Yes; because RU = US and RV = VS, ___ = ___ = 1. US VT Applying the Converse of the Triangle Proportionality Theorem S T ― ― So UV ∥ ST by the Converse of the Triangle Proportionality Theorem. Applying the Converse of the Triangle Proportionality Theorem Explain 4 INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Have students discuss the special case in You can use the Converse of the Triangle Proportionality Theorem to verify that a line is parallel to a side of a triangle. Example 4 Verify that the line segments are parallel. _ _ MN and KL JN JM 30 = 2 42 = 2 _ _ =_ =_ 21 15 MK NL JM JN _ _ Since _ = _, MN || KL by the Converse of the MK NL which the parallel line intersects the two sides at their midpoints. 42 21 M K L 15 J 30 N Triangle Proportionality Theorem. QUESTIONING STRATEGIES _ _ DE and AB (Given that AC = 36 cm, and BC = 27 cm) A D 20 cm AD = AC - DC = 36 - 20 = 16 B BE = BC - EC = 27 - 15 = 12 15 cm E C 20 5 CD = _ = _ _ DA 4 16 15 5 CE = _ = _ _ 4 EB 12 CE , DE || AB by the Converse of the Triangle Proportionality Theorem. CD = _ Since _ DA EB © Houghton Mifflin Harcourt Publishing Company How is using the Converse of the Triangle Proportionality Theorem different from using the theorem itself? You use the TPT when you are given that a line cutting two sides of a triangle is parallel to the third side, and want to prove that it cuts the sides proportionally. You use the converse when you are given that the line cuts two sides proportionally, and want to prove that it is parallel to the third side. V U Reflect 8. Communicate Mathematical Ideas In △ABC, in the example, what is the value AB of ___ ? Explain how you know. DE _ _ 5 CD AB ___ = __ =_ ; Possible answer: because DE || AB, corresponding angles A and CDE are 4 DA DE CD CE AB congruent, as are corresponding angles B and CDE. So, △ABC ̴ △DEC and __ = __ = __ . DE DA EB Your Turn 9. _ _ Verify that TU and RS are parallel. R 72 TR U 4 54 108 4 _ _ 90 67.5 72 US _ _ VU VT = , so RS || TU. TR US T V 90 _ 5 VU 67.5 135 5 VT _ _ = = , _=_=_=_ 54 S Module 17 885 Lesson 1 LANGUAGE SUPPORT IN2_MNLESE389847_U7M17L1 885 Connect Vocabulary Some students may have difficulty identifying a transversal. Remind them that the prefix trans- means across, as in transportation, transfer, and transatlantic. Transverse means situated or lying across. Therefore, the transversal is the line that lies across other lines. 885 Lesson 17.1 18/04/14 10:04 PM Elaborate _ _ 10. In △ABC, XY || BC. Use what you know about similarity and proportionality to identify as many different proportions as possible. AC AX YC B AY AB AY XB AX = ; = ; = ; = Possible answers: AY XB AY AB AC AX YC AX _ __ __ __ _ ELABORATE A X Y C QUESTIONING STRATEGIES What is the difference between triangles having congruent sides and triangles having proportional sides? Congruent sides have equal measures, but proportional sides have equal ratios to one another. 11. Discussion What theorems, properties, or strategies are common to the proof of the Triangle Proportionality Theorem and the proof of Converse of the Triangle Proportionality Theorem? Possible answers: Segment Addition Postulate; properties of fractions; taking reciprocals of both sides of a proportion, and use of Similarity Criteria (AA for the Triangle Proportionality Theorem and SAS for its converse) ― 12. Essential Question Check-In Suppose a line parallel to side BC of ― ― AX ▵ABC intersects sides AB and AC at points X and Y, respectively, and ___ = 1. What XB do you know about X and Y? Explain. _ _ __ AY X and Y are the midpoints of sides AB and AC. If AX = 1, then __ = 1. XB YC _ Then AX = XB, so X is the midpoint of AB. Similarly, Y is the midpoint _ of AC. SUMMARIZE THE LESSON When does a line cut two sides of a triangle proportionally? What does it mean for the sides to be cut proportionally? When it is parallel to the third side; the ratios of the corresponding parts are equal. Evaluate: Homework and Practice 1. • Online Homework • Hints and Help • Extra Practice Check students’ constructions. _ − ‹ › XN XM ZY || MN . Write a paragraph proof to show that ___ = ___ . NY MZ Y X N _ − ‹ › Since ZY|| MN , ∠XNM ≅ ∠XYZ and ∠X MN ≅ ∠XZY by the Corresponding Angles Theorem. Z M So △XYZ ~ △XNM by the AA Similarity Criterion. Since XZ XY corresponding sides of similar triangles are proportional, ___ = __ . Use the Segment XM XN Addition Postulate to rewrite XZ as XM + MZ, and rewrite XY as XN + NY. So ______ XM XM + MZ © Houghton Mifflin Harcourt Publishing Company 2. Copy the triangle ABC that you drew for the Explore activity. Construct a line _ ‹ › − FG parallel to AB using the same method you used in the Explore activity. XN + NY XM MZ XN NY = ______ . This can be rewritten as ___ + ___ = __ + __ . Since any number divided by itself XN XM XM XN XN MZ NY MZ ___ __ is 1, you can rewrite the expression as 1 + XM = 1 + XN . Subtracting 1 from both sides, ___ XM NY XM XN = __ . Taking the reciprocals of both sides, ___ = __ . XN MZ NY Module 17 IN2_MNLESE389847_U7M17L1 886 886 Lesson 1 18/04/14 10:04 PM Triangle Proportionality Theorem 886 Find the length of each segment. _ 3. KL EVALUATE H G 6 4. X m M n K Concepts and Skills 18 Explore Constructing Similar Triangles Exercise 1 Example 1 Proving the Triangle Proportionality Theorem Exercise 2 Example 2 Applying the Triangle Proportionality Theorem Exercises 3–5 Example 3 Proving the Converse of the Triangle Proportionality Theorem Exercise 13 Example 4 Applying the Converse of the Triangle Proportionality Theorem Exercises 6–9 6 8 _ _ AB and CD 6. 7. D 14 B © Houghton Mifflin Harcourt Publishing Company 9. 18 30 _ _ MN and QR 8. 9 10 N 3 R 2.7 M Q PM = 9 - 2.7 = 6.3 PN = 10 - 3 = 7 6.3 PM 7 PN = = = 2.7 3 N MQ _ _ So, MN || QR. _ _ _ _ CA 3.5 2.5 E 1.5 FW ___ = ___ = 0.6; 2.5 WD 2.1 FX __ = ___ = 0.6 XE 3.5 Since 16 20 16 5 AM CN _ 12 _ 4 CL 4 AL _ = = ; _ = _ = _; _ = _ = _; _ = _ = 2 15 5 LA 5 LC 4 MB 16 20 8 NB _ _ CL CN _ _ = and LN||AB. So, LA NB _ _ AM AL So, _ ≠ _ and LM is not parallel to BC. FX _ _ FW _ _ = , WX || DE. XE WD M 16 A 8 B 15 20 MB LC N L 16 12 C 10. On the map, 1st Street and 2nd Street are parallel. What is the distance from City Hall to 2nd Street along Cedar Road? 2.8 2.1 mi mi 2.4 mi 1st St. As p en Rd. Library Cedar Rd. 2nd St. Let x be the distance in miles from City Hall to 1st Street. Exercise IN2_MNLESE389847_U7M17L1 887 ( ) 2.1 x ___ = ___ 2.4 2.1 x = ___ 2.4 = 1.8 2.8 2.8 1.8 + 2.4 = 4.2 miles The distance is 4.2 miles. Module 17 Lesson 17.1 14 _ _ WX and DE F 1.5 2.1 W X 18 Use the Converse of the Triangle Proportionality Theorem to identify parallel lines in the figure. City Hall 887 49 V T N 14 D 3 ED _ 14 _ = 14 ∙ _ = 3 = 14 2 DB 4_ 3 EC _ 12 _ = =3 4 CA _ _ EC ED So, _ = _. So, AB || CD. DB U V P 2 43 8 35 _ XY XZ VN _ VM _ VM _ XY = 12; _ = _; = ; = ; 8 YU ZV NT MU 14 35 XZ 12 _ 12 _ = ; XZ = (_)30 = 20 VM = (_)8 = 20 3 A C 4 12 E 30 U KL 4 1 _4 = _ ; KL = (_)8 = 5_ Practice Z Y 30 L 6 _ VM 5. 8 J 4 ASSIGNMENT GUIDE _ XZ Lesson 1 887 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–9 2 Skills/Concepts MP.4 Modeling 10–12 2 Skills/Concepts MP.4 ge07se_c07l04007a Modeling 13–15 2 Skills/Concepts MP.2 Reasoning 16 3 Strategic Thinking MP.2 Reasoning 17 3 Strategic Thinking MP.3 Logic 18 3 Strategic Thinking MP.1 Problem Solving 18/04/14 10:04 PM 11. On the map, 5th Avenue, 6th Avenue, and 7th Avenue are parallel. What is the length of Main Street between 5th Avenue and 6th Avenue? 5th Ave. 6th Ave. Main St. t. gS S 0.3 5th Avenue and 6th Ave. 0.4 0.5 ___ = ___ m 0.4 km in pr 0.3 0.3 m ___ = ___ 0.4 0.5 In exercises that give a length with an algebraic expression, a common error is to give the value of the variable as the answer. Remind students they need to use the value of the variable to evaluate the expressions and find the segment lengths. ( ) 0.3 m = ___ 0.4 = 0.24 0.5 The length is 0.24 kilometer. 0.5 AVOID COMMON ERRORS Let m be the length in kilometers of Main Street between 7th Ave. km km INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 For problems with two triangles, encourage G 12. Multi-Step The storage unit has horizontal siding that is parallel to the base. a. Find LM. 2.6 GL GH ___ 10.4 ___ 2.6 __ = __ ; 11.3 = ___ ; LM = ___ ; LM = 11.3(___ ) = 2.825 ft LM HJ LM 2.6 11.3 11.3 ft 10.4 10.4 b. Find GM. L GM = GL + LM = 11.3 + 2.825 = 14.125 M N H D 2.6 ft E F J 2.2 ft K Find MN to the nearest tenth of a foot. c. 10.4 ft students to draw (or color) the two triangles separately to identify corresponding parts. GJ _____ 2.2 GM 13 _____ 2.2 ___ = __ ; 14.125 = ___ ; MN = ___ ; MN = 14.125(___ ) ≈ 2.4ft MN JK MN 2.2 14.125 13 13 HJ HJ LM ___ ≈ 1.18 and __ = 1.18. LM and __ as d. Make a Conjecture Write the ratios ___ MN JK decimals to the nearest hundredth and compare them. Make a conjecture about the relationship between parallel ‹ › ‹ › ‹ › − ‹ › ‹ › − − − − lines LD , ME , and NF and transversals GN and GK. MN JK The parallel lines divide the transversals proportionally. E Statements ‹ › − ‹ › − ‹ › − ‹ › − 1. AB ∥ CD , CD ∥ AF ‹ › ‹ › − − 2. Draw EB intersecting CD at X. F Reasons 1. Given 2. Two points determine a line. AC BX 3. ___ = __ XE CE 3. Triangle Proportionality Theorem BD BX 4. __ = ___ XE DF 4. Triangle Proportionality Theorem AC BD 5. ___ = ___ DF CE 5. Transitive Property of Equality Module 17 IN2_MNLESE389847_U7M17L1 888 © Houghton Mifflin Harcourt Publishing Company 13. A corollary to the Converse of the Triangle Proportionality Theorem states that if three or more parallel lines intersect two transversals, then they divide the transversals proportionally. Complete the proof of the corollary. ‹ › − ‹ ›− ‹ › − ‹ › − Given: Parallel lines AB ∥ CD, CD ∥ EF B A AC AC BX BX BD BD = ___ , ___ = ___ , ___ = ___ Prove: ___ XE XE DF CE DF CE X D C 888 Lesson 1 18/04/14 10:03 PM Triangle Proportionality Theorem 888 14. Suppose that LM = 18.75. Use the Triangle Proportionality Theorem to find PM. JOURNAL L 15 ______ 3 PM ___ = __ ; PM = _ 10 24 - PM 2 LP 3( PM 3 16 16 ) PM = _ 24 PM ; PM = 36 - _ PM ___ = __ PM = (__ 22 = 14.08 2 22 2 25 25 ) 3 _ PM = 14 or 14.4 K Have students write a journal entry in which they explain in their own words the Triangle Proportionality Theorem and its converse. Ask students to include figures with their explanations. P 10 2 15. ― ― Which of the given measures allow you to conclude that UV ∥ ST? Select all that apply. A. SR = 12, TR = 9 B. SR = 16, TR = 20 C. SR = 35, TR = 28 D. SR = 50, TR = 48 E. SR = 25, TR = 20 20 5 __ 16 __ =_ ; 20; ≠ __ 12 3 12 9 20 5 __ 16 4 __ __ =_ ; 16 = _ ; 20 ≠ __ 4 20 5 16 16 20 20 16 20 16 4 __ 4 __ __ _ _ __ = ; = ; = 7 28 7 35 35 28 20 16 20 16 __ _2; __ _1; __ __ = = ≠ 5 48 50 3 50 48 20 16 4 __ 4 __ __ =_ ; 16 = _ ; 20 = __ 5 20 25 5 25 15 N M U 20 S T V 16 R 20 ― ― Answer: Only C and E allow you to conclude that UV ∥ ST. H.O.T. Focus on Higher Order Thinking 16. Algebra For what value of x is GF ∥ HJ? G 30 F 45 ― ― H EJ +1 EH ____ GF ∥ HJ if __ = __ , that is, if _____ = 5x ; 360 45 40 JF HG 4x + 4 J 5x + 1 4x + 4 36x + 36 = 40x; x = 7 ― ―. E Then, for x = 7, GF ∥ HJ 17. Communicate Mathematical Ideas John used △ABC to write a proof ― ― of the Centroid Theorem. He began by drawing medians AK and CL, ― ― intersecting at Z. Next he drew midsegments LM and NP, both ― parallel to median AK. © Houghton Mifflin Harcourt Publishing Company +4 5x + 1 = 360( _____ ; (4x_____ 45 ) 40 ) ― ― ― Given: △ABC with medians AK and CL, and midsegments LM ― and NP A L B Prove: Z is located _23 of the distance from each vertex of △ABC to the midpoint of the opposite site. M L a. Complete each statement to justify the first part of John’s proof. By the definition of midsegment , MK = _12BK. By the definition M KC = 2. of median , BK = _12KC. So, by substitution , MK = _12KC, or ___ MC ― ― ZC = ___ KC by the Triangle Proportionality Consider △LMC. LM∥AK, so ___ LZ Z K N P C Z K P C MK ZC 2LZ = ___ = _23 , and Z is located _23 Theorem, and ZC = 2LZ. Because LC = 3LZ, ___ LC 3LZ of the distance from vertex C of △ABC to the midpoint of the opposite side. b. Explain how John can complete his proof. 2 of the distance He can repeat the same process twice to show that Z is located _ 3 from vertices A and B of △ABC to the midpoints of the opposite side. Module 17 IN2_MNLESE389847_U7M17L1 889 889 Lesson 17.1 889 Lesson 1 18/04/14 10:03 PM 18. Persevere in Problem Solving Given △ABC with FC = 5, you want to find BF. First, find the value that y must have for the Triangle Proportionality to apply. Then describe more than one way to find BF, and find BF. AVOID COMMON ERRORS When students attempt to find the area of △AEF, they may mistake the triangle for a right triangle with 1 bh to find the area. right angle ∠AEF, then use A = __ 2 This gives an incorrect area of 31.6875 ft 2, slightly greater than the actual area, which can be found by adding the areas of two different right triangles, △AGE and △EGF. For the triangle Proportionality Theorem to apply, y y ― 9 B (5.5, y) 8 7 F (7, 6) 6 5 4 3 2 C (10, 2) 1 A (1, 2) E (4, 2) x 0 1 2 3 4 5 6 7 8 9 ― must be such that AB and EF are parallel. ― y-2 y-2 ― -2 4 Slope of EF = 6____ =_ , and slope of AB = _____ = ____. 7-7 y-2 5.5 - 1 4 () 4.5 4 4 Let ____ =_ and solve for y: y - 2 = 4.5 _ = 6; y = 8. 4.5 3 3 To find BF, you can use the Pythagorean Theorem, the Distance Formula, or the Triangle Proportionality Theorem: () 3 3 BF AE __ AE = 3, EC = 6, FC = 5; __ = __ ; BF = _ ; BF = 5 _ = 2.5 6 6 FC EC 5 INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 A theorem for rectangles similar to the Lesson Performance Task Shown here is a triangular striped sail, together with some of its dimensions. Find each of the following: A 1. HG 2.5 ft 2. BC B 3. AJ C 1.5 ft 4. GF 6 ft 7. the number of sails you could make for $10,000 if the sail material costs $30 per square yard HG 4. 1.5(HG) = 42 1.8 BC ___ = ___ 1.5 1.2 1.2(BC) = 2.7 BC = 2.25 2.5 AJ ___ = ___ 3. 2.25 2 2 6. 2 (GF) = 6.25 1 ∙ base ∙ height Area = _ 2 = 30.9 2 5. F 6.5 ft 1( =_ 10.3)(6) 2 36 + (GF) = 42.25 HG = 2.8 2. 6 2 + (GF) = (6.5) E GF = 2.5 Perimeter = 2.5 + 2.25 + 1.5 + 3.5 + 6.5 + 2.5 + 2.8 + 1.2 + 1.8 + 2 = 26.55 7. 30.9 ft _____ = 3.43 yd 2 9 ft $30 3.43 yd 2 ∙ ___ = $103 per sail 2 2 yd 2 $10, 000 total _________ = 97 sails $103 per sail © Houghton Mifflin Harcourt Publishing Company 3.5 G 3.5 ft 6. the area of △AEF 1.5 1.2 ___ = ___ Triangle Proportionality Theorem might go like this: If a line segment parallel to one side of a rectangle intersects the two sides that are perpendicular to the segment, then the segment divides the perpendicular sides proportionally. Is the theorem true? Give an example to support your answer. Yes; possible ¯ is drawn answer: In rectangle ABCD, segment EF ¯. AF = 3 and ¯, intersecting AD ¯ and BC parallel to AB FD = 5. Since AB = FE, ABEF is a rectangle, so BE = 3. Since FE = DC, FECD is a rectangle, so EC = 5. 3 and the segment divides the BE = __ AF = ___ So, ___ FD EC 5 perpendicular sides proportionally. 1.8 ft I 1.2 ft H D 5. the perimeter of △AEF 1. J 1.8 4.5 = 2.25(AJ) 2 = AJ Module 17 890 Lesson 1 EXTENSION ACTIVITY IN2_MNLESE389847_U7M17L1 890 Refer students to the sail in the Lesson Performance Task. Then give these directions: _ _ _ 1. Use your knowledge of similar triangles to find the lengths of BJ, CI, and DH. Round to the nearest hundredth. 1.54 ft; 2.92 ft; 3.85 ft 2. Find the perimeters of △ABJ, △ACI, △ADH, and △AEG. 6.04 ft; 11.47 ft; 15.1 ft; 23.55 ft 3. Are the perimeters of the triangles in the same ratio as the side lengths? Explain. The perimeters are in the same ratio as the side lengths. Students perimeter of ▵ACI AC = 1.9 = _________________ . should give sample results such as ___ AB perimeter of ▵ABJ 18/04/14 10:03 PM Scoring Rubric 2 Points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Triangle Proportionality Theorem 890 LESSON 17.2 Name Subdividing a Segment in a Given Ratio Class Date 17.2 Subdividing a Segment in a Given Ratio Essential Question: How do you find the point on a directed line segment that partitions the given segment in a given ratio? Common Core Math Standards Resource Locker The student is expected to: COMMON CORE G-GPE.B.6 It takes just one number to specify an exact location on a number line. For this reason, a number line is sometimes called a one-dimensional coordinate system. The mile markers on a straight stretch of a highway turn that part of the highway into a one-dimensional coordinate system. Mathematical Practices COMMON CORE MP.5 Using Tools Language Objective On a straight highway, the exit for Arthur Avenue is at mile marker 14. The exit for Collingwood Road is at mile marker 44. The state highway administration plans to put an exit for Briar Street at a point that is _23 of the distance from Arthur Avenue to Collingwood Road. Follow these steps to determine where the new exit should be placed. Work in groups to find ratios of subdivided segments. ENGAGE If the segment lies on a number line, subtract the coordinates to find the distance between the endpoints. Then multiply the length by the ratio to find the coordinate of the point that divides the segment in that ratio. If there are no coordinates, use a compass and straightedge to divide the given segment into equal parts, and identify the point that divides the segment in the given ratio. A Mark Arthur Avenue (point A) and Collingwood Road (point C) on the number line. A © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Condor 36/Shutterstock Essential Question: How do you find the point on a directed line segment that partitions the given segment in a given ratio? Partitioning a Segment in a One-Dimensional Coordinate System Explore Find the point on a directed line segment between two given points that partitions the segment in a given ratio. Also G-CO.D.12 10 B D B 20 C 30 40 50 What is the distance from Arthur Avenue to Collingwood Road? Explain. 30 miles; find the absolute value of the difference of the coordinates: ⎜44 - 14⎟ = 30. C How far will the Briar Street exit be from Arthur Avenue? Explain. 2 20 miles; · 30 miles = 20 miles 3 D What is the mile marker number for the Briar Street exit? Why? _ mile marker 34; 14 + 20 = 34 E Plot and label the Briar Street exit (point B) on the number line. Check students’ number lines. 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The exit for markers on coordinate system The mile marker 14. plans to mensional e is at mile istration into a one-di Arthur Avenu ay admin r Avenue the exit for 44. The state highw ce from Arthu should r t highway, On a straigh Road is at mile marke that is _23 of the distan new exit the where Collingwood Briar Street at a pointsteps to determine for Follow these put an exit od Road. to Collingwo C) on the be placed. r Avenue Mark Arthu number line. A (point A) gwood Road and Collin (point C B D 50 40 30 Credits: 20 Explain. 14⎟ = 30. gwood Road? s: ⎜44 e to Collin coordinate Arthur Avenu ence of the distance from the differ What is the value of absolute ; find the Explain. 30 miles r Avenue? be from Arthu Street exit Briar the How far will = 20 miles 2 ; · 30 miles 20 miles 3 Why? Street exit? for the Briar r number mile marke What is the + 20 = 34 er 34; 14 er line. mile mark on the numb (point B) Street exit Briar label the Plot and er lines. nts’ numb stude Check 10 y • Image g Compan IN2_MNLESE389847_U7M17L2 891 Publishin View the Engage section online. Discuss the photograph and point out that the Golden Ratio is found throughout nature—even, some say, in the ratio of the length of a person’s forearm to the length of the person’s hand. Then preview the Lesson Performance Task. Harcour t n Mifflin © Houghto erstock 36/Shutt ©Condor _ Lesson 2 891 Module 17 7L2 891 47_U7M1 ESE3898 IN2_MNL 891 Lesson 17.2 18/04/14 10:08 PM 18/04/14 10:08 PM The highway administration also plans to put an exit for Dakota Lane at a point that divides the highway from Arthur Avenue to Collingwood Road in a ratio of 2 to 3. What is the mile marker number for Dakota Lane? Why? (Hint: Let the distance from Arthur Avenue to Dakota Lane be 2x and let the distance from Dakota Lane to Collingwood Road be 3x.) EXPLORE Partitioning a Segment in a OneDimensional Coordinate System mile marker 26; since the distance from Arthur Avenue to Collingwood Road is 30 miles, 2x + 3x = 30, 5x = 30, and x = 6. Therefore, the distance from Arthur Avenue to Dakota Lane is 2x = 12 miles. INTEGRATE TECHNOLOGY Plot and label the Dakota Lane exit (point D) on the number line. Show how to use geometry software to partition a segment. Check students’ number lines. Reflect 1. How can you tell that the location at which you plotted point B is reasonable? 2 of the way (or about 67% of the way) from point A to Point B appears to be about _ 3 QUESTIONING STRATEGIES point C. 2. What does it mean to say that point P _ divides AB in the ratio 3 to 1? Point P divides the segment into two segments, and the length of one is three times the length of the other. Would your answer in Step F be different if the exit for Dakota Avenue divided the highway from Arthur Avenue to Collingwood Road in a ratio of 3 to 2? Explain. Yes; in this case, the exit would be closer to Collingwood Avenue. It would be at mile marker 32. Explain 1 Partitioning a Segment in a Two-Dimensional Coordinate System EXPLAIN 1 A directed line segment is a segment between two points A and B with a specified direction, from A to B or from B to A. To partition a directed line segment is to divide it into two segments with a given ratio. Example 1 A (-8, -7), B(8, 5); 3 to 1 Step 1 Write a ratio that expresses the distance of point P along the segment from A to B. 3 =_ 3 of the distance from A to B. Point P is _ 4 3+1 Step 2 Find the run and the rise of the directed line segment. 8 y B(8, 5) 4 x -8 -4 0 -4 A(-8, -7) -8 run = 8 - (-8) = 16 4 Partitioning a Segment in a TwoDimensional Coordinate System © Houghton Mifflin Harcourt Publishing Company Find the coordinates of the point P that divides the directed line segment from A to B in the given ratio. INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Have students draw a segment and find the midpoint to divide it into two segments with a ratio of 1 to 1. Then have them find the midpoint of one of those segments to divide the segment with a ratio of 3 to 1. Ask students what other ratios they could divide the segments into by using midpoints. Rise Run (8, -7) rise = 5 - (-7) = 12 Module 17 892 Lesson 2 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M17L2 892 Learning Progressions 18/04/14 10:08 PM Students have previously studied slope and learned to find the slope of a line. They also used slope to write the equation of a line, either in slope-intercept form (y = mx + b) or in point-slope form (y – y 1 = m (x – x 1)). Here students use the definition of slope (the ratio of the rise to the run) to help them find a point that divides a given line segment in a given ratio. Subdividing a Segment in a Given Ratio 892 3 of the distance from point A to point B, so find _ 3 of both the rise and the run. Step 3 Point P is _ 4 4 3 of run = _ 3 (16) = 12 3 of rise = _ 3 (12) = 9 _ _ 4 4 4 4 y Step 4 To find the coordinates of point P, add the 8 values from Step 3 to the coordinates B(8, 5) of point A. 4 P x x-coordinate of point P = -8 + 12 = 4 -8 -4 0 4 y-coordinate of point P = -7 + 9 = 2 Rise -4 QUESTIONING STRATEGIES How can you predict whether the point to divide a line in a given ratio will be closer to one end or the other? If the ratio is less than onehalf, the point will be closer to the first endpoint, and if it is greater than one-half, the point will be closer to the second endpoint. FPO The coordinates of point P are (4, 2). Can a proportion compare measurements that have different units? Explain. Yes; if the numerators are both in one unit and the denominators in another, or if one fraction is in one unit and the other fraction is in another A(-8, -7) -8 Run (8, -7) A(-4, 4), B(2, 1); 1 to 2 B Step 1 Write a ratio that expresses the distance of point P along the segment from A to B. 1 1 Point P is ___________ = _ of the distance from A to B. 1 + 2 3 A P 4 Step 2 Graph the directed line segment. Find the rise and the run y 2 B of the directed line segment. -4 run = 2 - (-4) = 6 rise = 1 - 4 = -3 -2 0 2 x 4 -2 -4 1 Step 3 Point P is _____ of the distance from point A to point B. 3 ( ) © Houghton Mifflin Harcourt Publishing Company 1 1 _ of run = _ (6) = 2 3 3 1 1 _ of run = _ -3 = -1 3 3 Step 4 To find the coordinates of point P, add the values from Step 3 to the coordinates of point A. x-coordinate of point P = -4 + 2 = -2 ( The coordinates of point P are -2 , 3 ) y-coordinate of point P = 4 + -1 = 3 . Plot point P on the above graph. Reflect 3. In Part A, show how you can use the Distance Formula to check that point P partitions the directed line segment in the correct ratio. 2 2 2 2 AP = (4 - (-8)) + (2 - (-7)) = √225 = 15; PB = √(8 - 4) + (5 - 2) = √25 = 5. ――――――――― √ ―― ――――――― ― The ratio of AP to BP is 15 to 5 or 3 to 1, which is the correct ratio. 4. Discussion What can you conclude about a point that partitions a segment in the ratio 1 to 1? How can you find the coordinates of such a point? The point is the midpoint of the segment. Use the Midpoint Formula. Module 17 893 Lesson 2 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M17L2 893 Peer-to-Peer Activity Have students work in pairs to draw and label a line, a line segment, and a directed line segment. Have them discuss their similarities and differences. 893 Lesson 17.2 18/04/14 10:08 PM Your Turn EXPLAIN 2 Find the coordinates of the point P that divides the directed line segment from A to B in the given ratio. 5. A(-6, 5), B(2, -3); 5 to 3 5 5 Point P is = of the distance 5+3 8 from A to B. _ _ 6. Constructing a Partition of a Segment _ _ run = 2 - (-6) = 8; rise = -3 - 5 = -8 5 5 of run = 5; of rise = -5 8 8 x-coordinate of point P = -6 + 5 = −1; run = -6 - 4 = −10; rise = −13 - 2 = −15 QUESTIONING STRATEGIES 5 5 x-coordinate of point P = 4 - 6 = −2; The coordinates of point P are (−1, 0). The coordinates of point P are (−2, -7). Can you partition a line (as opposed to a line segment)? Explain. No; because a line continues infinitely in both directions, it cannot be divided into segments with a definite ratio. _ _ _3 of run = -6; _3 of rise = -9 y-coordinate of point P = 2 + (-9) = -7 y-coordinate of point P = 5 + (-5) = 0 Explain 2 Example 2 A(4, 2), B(-6, -13); 3 to 2 3 3 = of the distance Point P is 5 3+2 from A to B. Constructing a Partition of a Segment Given the directed line segment from A to B, construct the point P that divides the segment in the given ratio from A to B. 2 to 1 A B → ‾ . The exact measure of the angle Step 1 Use a straightedge to draw AC is not important, but the construction is easiest for angles from about 30° to 60°. C A C F E D A Step 3 Use the straightedge to connect points B and F. Construct an angle congruent to ∠AFB with D as its vertex. Construct an angle congruent to ∠AFB with E as its vertex. _ Step 4 The construction partitions AB into 3 equal parts. Label point P at the point that divides the segment in the ratio 2 to 1 from A to B. A Module 17 894 B B C F E D P B © Houghton Mifflin Harcourt Publishing Company → ‾ . Label Step 2 Place the compass point on A and draw an arc through AC the intersection D. Using the same compass setting, draw an arc centered on D and label the intersection E. Using the same compass setting, draw an arc centered on E and label the intersection F. Lesson 2 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M17L2 894 Multiple Representations 18/04/14 10:08 PM ¯ Discuss whether it matters if the segment with endpoints C and D is named as DC _ or as CD. Both are correct. The order does not matter as long as C and D are endpoints. If the segment is a directed line segment, the order does matter. Subdividing a Segment in a Given Ratio 894 B AVOID COMMON ERRORS 1 to 3 C G Remind students to pay attention to the order of the letters in the problem in order to ensure they find the correct point for the given direction. F E D A B P _ Step 1 Use a straightedge to draw AC. _ Step 2 Place the compass point on A and draw an arc through AC. Label the intersection D. Using the same compass setting, draw an arc centered on D and label the intersection E. Using the same compass setting, draw an arc centered on E and label the intersection F. Using the same compass setting, draw an arc centered on F and label the intersection G. Step 3 Use the straightedge to connect points B and G. Construct angles congruent to ∠AGB with D, E, and F as the vertices. _ Step 4 The construction partitions AB into 4 equal parts. Label point P at the point that divides the segment in the ratio 1 to 3 from A to B. Reflect 7. © Houghton Mifflin Harcourt Publishing Company 8. _ _ In Part A, why is EP is parallel to FB? ∠AEP was constructed to be congruent to ∠AFB. These are congruent corresponding _ _ angles, so EP∥FB. How can you use the Triangle Proportionality Theorem to explain why this construction method works? The construction ensures that the segments of the ray you constructed are congruent. Also, the segments that are drawn in Step 3 are all parallel. By the Triangle Proportionality _ Theorem, you can conclude that the segments along AB are congruent to each other. Your Turn Given the directed line segment from A to B, construct the point P that divides the segment in the given ratio from A to B. 9. 1 to 2 10. 3 to 2 C F A E D E F G H C D A P B P B Module 17 895 Lesson 2 LANGUAGE SUPPORT IN2_MNLESE389847_U7M17L2 895 Connect Vocabulary Help students develop an understanding of the meaning of dimension by considering one-dimensional figures, such as a line, whose only dimension is length, and two-dimensional figures, such as a square. The two dimensions of a square are length and width. 895 Lesson 17.2 18/04/14 10:08 PM Elaborate ELABORATE 11. How is a one-dimensional coordinate system similar to a two-dimensional coordinate system? How is it different? In both types of coordinate systems, numbers are used to specify the locations of points. QUESTIONING STRATEGIES In a one-dimensional coordinate system, a single number is used to specify the location of Can a proportion compare measurements that have different units? Explain. Yes, but each ratio must compare measurements with the same unit; for example, you can write a proportion with a ratio of feet to feet equal to a ratio of meters to meters. a point on a line. In a two-dimensional coordinate system, two numbers are used to specify the location of a point on a plane. 4 _ 12. Is finding a point that _ is 5 of the distance from point A to point B the same as finding a point that divides AB in the ratio 4 to 5? Explain. 4 No; the point that is _ of the distance from point A to point B is 80% of the distance from 5 _ 4 point A to point B; the point that divides AB in the ratio 4 to 5 is _ or approximately 44% of 9 the distance from point A to point B. SUMMARIZE THE LESSON 13. Essential Question Check-In What are some different ways to divide a segment in the ratio 2 to 1? If the segment lies on a number line, subtract the coordinates to find the distance between How can you divide a directed line segment in a given ratio? Directed line segments are always read in the order in which the points are given. A segment partition divides the segment from a given point A to another point B in a ratio of a : b. In a coordinate plane, the rise and run can be used to find the point dividing the segment in the ratio a : b. Segment partitions can also be constructed using a compass and straightedge. 2 the endpoints. Then find the point that is _ of the distance from one endpoint to the other. 3 If the segment is on a coordinate plane, use the run and the rise to find the point that is _2 of the distance from one endpoint to the other. If there are no coordinates, use a 3 compass and straightedge to divide the given segment into 3 equal parts. Then identify the point that divides the segment in the ratio 2 to 1. A choreographer uses a number line to position dancers for a ballet. Dancers A and B have coordinates 5 and 23, respectively. In Exercises 1–4, find the coordinate for each of the following dancers based on the given locations. 1. Dancer C stands at a point that is _56 of the distance from Dancer A to Dancer B. AB = ⎜23 - 5⎟ = 18 2. 3 6 Dancer C is 15 units from Dancer A, so the coordinate for Dancer C is 5 + 15 = 20. IN2_MNLESE389847_U7M17L2 896 Dancer D stands at a point that is _13 of the distance from Dancer A to Dancer B. AB = ⎜23 - 5⎟ = 18 _1 ⋅ 18 = 6 _5 ⋅ 18 = 15 Module 17 • Online Homework • Hints and Help • Extra Practice © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©mcpix/iStockPhoto.com Evaluate: Homework and Practice Dancer D is 6 units from Dancer A, so the coordinate for Dancer D is 5 + 6 = 11. 896 Lesson 2 18/04/14 10:08 PM Subdividing a Segment in a Given Ratio 896 3. EVALUATE Dancer E stands at a point that divides the line segment from Dancer A to Dancer B in a ratio of 2 to 1. AB = ⎜23 - 5⎟ = 18 Let the distance from Dancer A to Dancer E be 2x and let the distance from Dancer E to Dancer B be x. Then 2x + x = 18, 3x = 18, and x = 6. So the distance from Dancer A to Dancer E is 2x = 2(6) = 12. The coordinate for Dancer E is 5 + 12 = 17. 4. Dancer F stands at a point that divides the line segment from Dancer A to Dancer B in a ratio of 1 to 5. AB = ⎜23 - 5⎟ = 18 Let the distance from Dancer A to Dancer F be x and let the distance from Dancer F to Dancer B be 5x. Then x + 5x = 18, 6x = 18, and x = 3. So the distance from Dancer A to Dancer F is x = 3. ASSIGNMENT GUIDE Concepts and Skills Practice Explore Partitioning a Segment in a OneDimensional Coordinate System Exercises 1–4 Example 1 Partitioning a Segment in a TwoDimensional Coordinate System Exercises 5–8 Example 2 Constructing a Partition of a Segment Exercises 9–12 The coordinate for Dancer F is 5 + 3 = 8. Find the coordinates of the point P that divides the directed line segment from A to B in the given ratio. 5. A( −3, −2 ), B(12, 3); 3 to 2 3 3 P is ____ =_ of the distance from A to B. 5 3+2 A(−1, 5), B(7, −3); 7 to 1 6. 7 7 P is ____ =_ of the distance from A to B. 7+1 8 run = 12 - (−3) = 15; rise = 3 - (−2) = 5 The coordinates of point P are (6, 1). The coordinates of point P are (6, −2). 5 © Houghton Mifflin Harcourt Publishing Company constructions and labels in the portioning of segments. _7 of run = _7(8) = 7; _7 of rise = _7(-8) = -7 5 5 5 8 x-coordinate of point P = −3 + 9 = 6; y-coordinate of point P = −2 + 3 = 1 7. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Discuss the importance of careful run = 7 - (−1) = 8; rise = −3 - 5 = -8 _3 of run = _3(15) = 9; _3 of rise = _3(5) = 3 A(−1, 4), (B-9, 0); 1 to 3 1 1 P is ____ =_ of the distance from A to B. 4 1+3 8 run = -9 - (−1) = -8; rise = 0 - 4 = -4 _1 of run = _1(-8) = -2; _1 of rise = _1(-4) = -1 4 4 4 4 x-coordinate of point P = −1 + (−2) = −3; y-coordinate of point P = 4 + (−1) = 3 The coordinates of point P are (−3, 3). 3 3 =_ of the distance from A to B. P is ____ 7 3+4 run = -7 - 7 = −14; rise = 4 - (−3) = 7 _3 of run = _3(-14) = -6; _3 of rise = _3 (7) = 3 7 7 7 7 x-coordinate of point P = 7 + (-6) = 1; y-coordinate of point P = −3 + 3 = 0 The coordinates of point P are (1, 0). Given the directed line segment from A to B, construct the point P that divides the segment in the given ratio from A to B. 9. 3 to 1 10. 2 to 3 C H C G G F F E D A D P B Module 17 Exercise IN2_MNLESE389847_U7M17L2 897 Lesson 17.2 8 A(7, −3), B(-7, 4); 3 to 4 8. E 897 8 x-coordinate of point P = −1 + 7 = 6; y-coordinate of point P = 5 + (-7) = −2 B P A Lesson 2 897 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–4 2 Skills/Concepts MP.4 Modeling 5–8 2 Skills/Concepts MP.2 Reasoning 9–12 2 Skills/Concepts MP.5 Using Tools 13–16 2 Skills/Concepts MP.4 Modeling 17–20 2 Skills/Concepts MP.4 Modeling 21 2 Skills/Concepts MP.2 Reasoning 22 3 Strategic Thinking MP.3 Logic 23 3 Strategic Thinking MP.3 Logic 18/04/14 10:08 PM Given the directed line segment from A to B, construct the point P that divides the segment in the given ratio from A to B. 11. 1 to 4 12. 4 to 1 D A F E G H C AVOID COMMON ERRORS Some students may confuse various ratios. Remind them that the similarity ratio refers only to the ratio of the lengths of the sides. It is equal to the perimeter ratio and the square root of the area ratio. A D E P F G H B C P B Find the coordinate of the point P that divides each directed line segment in the given ratio. J K -10 -20 13. from J to M; 1 to 9 JM = ⎜10 - (-15)⎟ = 25 L M 0 N 10 20 14. from K to L; 1 to 1 KL = ⎜5 - (-6)⎟ = 11 Let JP = x and let PM = 9x. Let KP = x and let PL = x. The coordinate of point P is −15 + 2.5 = 12.5 The coordinate of point P is -6 + 5.5 = −0.5 Then x + 9x = 25, 10x = 25, and x = 2.5. 15. from N to K; 3 to 5 Then x + x = 11, 2x = 11, and x = 5.5. 16. from K to J; 7 to 11 KJ = ⎜-15 - (-6)⎟ = 9 NK = ⎜-6 - 18⎟ = 24 Let KP = 7x and let PJ = 11x. The coordinate of point P is 18 - 9 = 9. The coordinate of point P is -6 - 3.5 = -9.5. Then 7x + 11x = 9, 18x = 9, x = 0.5. So KP = 7(0.5) = 3.5. Then 3x + 5x = 24, 8x = 24, and x = 3. So NP = 3(3) = 9. 17. Communicate Mathematical Ideas Leon constructed a point P that divides the directed segment from A to B in the ratio 2 to 1. Chelsea constructed a point Q that divides the directed segment from B to A in the ratio 1 to 2. How are points P and Q related? Explain. 2 Points P and Q are the same point. Sample explanation: Point P is _ of the 3 1 of the distance from B to A. This means the distance from A to B. Point Q is _ 3 points lie at the same location along the line segment. Module 17 Exercise IN2_MNLESE389847_U7M17L2 898 Lesson 2 898 Depth of Knowledge (D.O.K.) © Houghton Mifflin Harcourt Publishing Company Let NP = 3x and let PK = 5x. COMMON CORE Mathematical Practices 24 3 Strategic Thinking MP.2 Reasoning 25 3 Strategic Thinking MP.2 Reasoning 18/04/14 10:08 PM Subdividing a Segment in a Given Ratio 898 18. City planners use a number line to place landmarks along a new street. Each unit of the number line represents 100 feet. A fountain F is located at coordinate −3 and a plaza P is located at coordinate 21. The city planners place two benches along the street at points that divide the segment from F to P in the ratios 1 to 2 and 3 to 1. What is the distance between the benches? FP = ⎜21 - (-3)⎟ = 24; Then x + 2x = 24, 3x = 24, and x = 8. Let the distance from F to the first bench be x and let the distance from the first bench to P be 2x. The coordinate for the first bench is −3 + 8 = 5. Let the distance from F to the second bench be 3x and let the distance from the second bench to P be x. Then 3x + x = 24, 4x = 24, x = 6. So the distance from F to the second bench is 3x = 3(6) = 18 units. The coordinate for the second bench is −3 + 18 = 15. The distance between the benches is 15 - 5 = 10 units or 1000 feet. 19. The course for a marathon includes a straight segment from city hall to the main library. The planning committee wants to put water stations along this part of the course so that the stations divide the segment into three equal parts. Find the coordinates of the points at the which the water stations should be placed. 4 2 y Main Library M x © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©sportgraphic/Fotolia -4 0 -2 2 4 -2 C City Hall -4 From C to M, run = 3 - (−3) = 6; rise = 3 - (−2) = 5 1 Let the water stations be at points P and Q. Point P is _ of the distance from C to M. 3 _1 of run = _1(6) = 2; _1 of rise = _1(5) = 1_2 3 3 3 3 3 2 1 x-coordinate of P = −3 + 2 = −1; y-coordinate of P = -2 + 1_ = -_ 3 3 ( ) 1 The coordinates of point P are -1, -_ . 3 2 Point Q is _ of the distance from C to M. 3 _2 of run = _2(6) = 4; _2 of rise = _2(5) = 3_1 3 3 3 3 3 1 1 x-coordinate of Q = −3 + 4 = 1; y-coordinate of Q = -2 + 3_ = 1_ 3 3 ( ) 1 The coordinates of point Q are 1, 1_ . 3 1 1 The water stations should be placed at -1, -_ and 1, 1_ . 3 3 Module 17 IN2_MNLESE389847_U7M17L2 899 899 Lesson 17.2 ( ) 899 ( ) Lesson 2 18/04/14 10:08 PM 20. Multi-Step Carlos is driving on a straight section of highway from Ashford to Lincoln. Ashford is at mile marker 433 and Lincoln is at mile marker 553. A rest stop is located along the highway _23 of the distance from Ashford to Lincoln. Assuming Carlos drives at a constant rate of 60 miles per hour, how long will it take him to drive from Ashford to the rest stop? The distance from Ashford to Lincoln is ⎜553 - 433⎟ = 120 miles. AVOID COMMON ERRORS Students may forget to use the correct direction to partition line segments. Remind students to doublecheck direction in order to choose the correct partition point. _2 ∙ 120 = 80, so Carlos must drive 80 miles from Ashford to the rest stop. 3 1 d = rt, where d is the distance, r is the rate, and t is the time, so 80 = 60t, and t = 1_ . 3 1 So it will take 1_ hours (1 hour and 20 minutes) to drive from Ashford to the rest stop. 3 21. The directed segment from J to K is shown in the figure. y 4 J Points divide the segment from J to K in the each of the following ratios. Which points have integer coordinates? Select all that apply 2 x A. 1 to 1 -4 0 -2 2 4 B. 2 to 1 -4 C. 2 to 3 K D. 1 to 3 E. 1 to 2 From J to K, run = 0 - (−3) = 3; rise = −3 - 3 = -6 1 1 1 1( ) 1 _ A. The point is ____ =_ of the distance from J to K. _ of run = _ 3 = 1_ ;1 1+1 2 2 2 2 2 1( 1 1 of rise = _ -6) = -3. x-coordinate of the point is -3 + 1_ = -1_ ; y-coordinate of the 2 2 2 point is 3 + (−3) = 0. The point does not have integer coordinates. 2 2 2 2( ) 2 B. The point is ____ =_ of the distance from J to K. _ of run = _ 3 = 2; _ 2+1 3 3 3 3 2 2 2 2( ) 1 _ C. The point is ____ =_ of the distance from J to K. _ of run = _ 3 = 1_ ;2 5 5 5 5 5 2+3 2( 2 1 4 of rise = _ -6) = -2_ . x-coordinate of the point is -3 + 1_ = -1_ ; y-coordinate of the 5 5 5 5 ( ) 3 2 point is 3 + -2_ =_ . The point does not have integer coordinates. 5 5 1 1 1 1( ) _ 1 ____ _ D. The point is 1 + 3 = 4 of the distance from J to K. _ of run = _ 3 = 34 ; _ 4 4 4 3 1( 1 1 of rise = _ -6) = -1_ . x-coordinate of the point is -3 + _ = -2_ ; y-coordinate of the 4 4 4 2 1 _3. The point does not have integer coordinates. point is 3 + -1_ = 1 4 4 ( ) 1 1 1 1 1( ) E. The point is ____ =_ of the distance from J to K. _ of run = _ 3 = 1; _ 1+2 3 3 3 3 1( of rise = _ -6) = -2. x-coordinate of the point is -3 + 1 = -2; y-coordinate of the 3 © Houghton Mifflin Harcourt Publishing Company 2( of rise = _ -6) = -4. x-coordinate of the point is -3 + 2 = -1; y-coordinate of the 3 point is 3 + (-4) = −1. The point has integer coordinates. point is 3 + (−2) = 1. The point has integer coordinates. Module 17 IN2_MNLESE389847_U7M17L2 900 900 Lesson 2 18/04/14 10:08 PM Subdividing a Segment in a Given Ratio 900 H.O.T. Focus on Higher Order Thinking JOURNAL 22. Critique Reasoning Jeffrey was given a directed line segment and was asked to use a compass and straightedge to construct the point that divides the segment in the ratio 4 to 2. He said he would have to draw a ray and then construct 6 congruent segments along the ray. Tamara said it is not necessary to construct 6 congruent segments along the ray. Do you agree? If so, explain Tamara’s shortcut. If not, explain why not. Yes; the ratio 4 to 2 is equivalent to the ratio 2 to 1. To construct a point that divides a segment in the ratio 2 to 1, it is only necessary to construct 3 congruent segments along the ray. Have students summarize the process for finding the point on a directed line segment that partitions the segment in a given ratio. 23. Explain the Error Point A has coordinate -9 and point B has coordinate 9. A student was asked to find the coordinate of the point P that is _23 of the distance from A to B. The student said the coordinate of point P is −3. a. Without doing any calculations, how can you tell that the student made an error? Point P must be closer to point B than to point A, so the coordinate of point P should be positive. b. What error do you think the student made? 2 of the distance from B to A. Sample answer: The student found the coordinate of the point that is _ 3 24. Analyze Relationships Point P divides the directed segment from A to B in the ratio 3 to 2. The coordinates of point A are (-4, -2) and the coordinates of point P are (2, 1). Find the coordinates of point B. 3 3 =_ of the distance from A to B. Point P is ____ 5 3+2 3 The run from A to P is 2 - (-4) = 6. Let the run from A to B be x. Then 6 = _ x and x = 10. 5 3 The rise from A to P is 1 - (-2) = 3. Let the rise from A to B be y. Then 3 = _ y and y = 5. 5 © Houghton Mifflin Harcourt Publishing Company x-coordinate of point B = -4 + 10 = 6; y-coordinate of point B = -2 + 5 = 3 The coordinates of point B are (6, 3) _ _ 25. Critical Thinking RS passes through R(−3, 1) and S(4, 3). Find a point P on RS such that the ratio of RP to SP is 5 to 4. Is there more than one possibility? Explain. _ 5 5 If point P is on RS, then point P is ____ =_ of the distance from R to S. 5+4 9 run = 4 - (-3) = 7; rise = 3 - 1 = 2 _5 of run = _5(7) = 3_8; _5 of rise = _5(2) = 1_1 9 9 9 9 9 9 8 _8; y-coordinate of point P = 1 + 1_1 = 2_1 x-coordinate of point P = -3 + 3_ = 9 9 9 9 8 _ In this case, the coordinates of point P are (_ , 2 1). 9 9 ¯, that lies beyond point S. Let P have coordinates (x, y). There is also a point P, not on RS ¯ x - (-3) of RP _______ _5 ______ _5 ) ( Then rise ¯ = 4 so x - 4 = 4 , 4(x + 3) = 5 x - 4 , 4x + 12 = 5x - 20, 12 = x - 20, rise of SP and x =32. ¯ y-1 of RP _______ _5 ____ _5 Also, rise ¯ = 4 , so y - 3 = 4 , 4(y - 1) = 5(y - 3), 4y - 4 = 5y - 15, -4 = y - 15, rise of SP and y = 11. In this case, the coordinates of point P are (32, 11). Module 17 IN2_MNLESE389847_U7M17L2 901 901 Lesson 17.2 901 Lesson 2 18/04/14 10:08 PM Lesson Performance Task In this lesson you will subdivide line segments in given ratios. The diagram shows a line segment divided into two parts in such a way that the longer part divided by the shorter part equals the entire length divided by the longer part: a+b a=_ _ a b Each of these ratios is called the Golden Ratio. To find the point on a line segment that divides the segment this way, study this figure: a L S M R Q INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 The opening of the Lesson Performance Task N states that for the divided line segment a + b, “the longer part divided by the shorter part equals the entire length divided by the longer part.” a+b a = _____ So, __ a . Use that same relationship to b complete this equation _ LM LN = ?. __ relating to line segment LN∶ ____ LM MN P b a+ b INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 The Lesson Performance Task shows a line LN In the figure, LMQS is a square. ___ equals the Golden Ratio (the entire segment length divided LM by the longer part). _ 1. Describe how, starting with line segment LM, you can find the location of point N. LN LN 2. Letting LM equal 1, find ___ = ___ = LN, the Golden Ratio. Describe your 1 LM method. _ _ 1. Using LM as one side, construct square _ LMQS. Construct R, the midpoint of SQ. Place the compass point on R _ and use RM as a radius to construct an arc intersecting line SQ. Mark point P where RM intersects SQ. Construct a perpendicular through P intersecting line RM. Mark point N where the perpendicular intersects LM. Module 17 902 © Houghton Mifflin Harcourt Publishing Company 2. RM equals the hypotenuse of a right triangle with sides 1 and 0.5. By the Pythagorean Theorem, RM ≈ 1.118. So, RP ≈ 1.118. SR = 0.5, so SP ≈ 1.118 + 0.5 ≈ 1.618. LN = SP, so LN, the Golden Ratio, equals approximately 1.618. segment divided into two parts a and b such a+b a+1 a = _____ a _____ __ that __ a . If b = 1, then 1 = a . Solve the b equation for a. Compare your results with your results in the Lesson Performance Task and make a conjecture about the ― 1 + √5 value of a. a = ________ ≈ 1.618; a appears to 2 equal the Golden Ratio. Lesson 2 EXTENSION ACTIVITY IN2_MNLESE389847_U7M17L2 902 The first two numbers in the Fibonacci sequence of natural numbers are 1 and 1. To find each new term in the sequence, add the two previous terms. So, the first few terms are 1, 1, 2, 3, 5, and 8. Have students use calculators to write the ratios of each of the first ten numbers in the sequence to the previous number, rounding to the nearest ten-thousandth. Here are the first four 3 = 1.5; __ 5 ≈ 1.333. After students complete their 2 = 2; __ 1 = 1; __ ratios: __ 1 1 2 3 calculations, have them compare their results with their results in the Lesson Performance Task, and then make a conjecture about the ratios they calculated. The ratios get closer and closer to the Golden Ratio the farther the sequence of ratios continues. 18/04/14 10:08 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points:The student’s answer contains no attributes of an appropriate response. Subdividing a Segment in a Given Ratio 902 LESSON 17.3 Name Using Proportional Relationships Class Date 17.3 Using Proportional Relationships Essential Question: How can you use similar triangles to solve problems? Common Core Math Standards The student is expected to: COMMON CORE Explore G-SRT.B.5 Exploring Indirect Measurement In this Explore, you will consider how to find heights, lengths, or distances that are too great to be measured directly, that is, with measuring tools like rulers. Indirect measurement involves using the properties of similar triangles to measure such heights or distances. Use congruence and similarity criteria for triangles to solve problems and to prove relationships in geometric figures. Mathematical Practices COMMON CORE Resource Locker A MP.5 Using Tools During the day sunlight creates shadows, as shown in the figure below. The dashed segment represents the ray of sunlight. What kind of triangle is formed by the flagpole, its shadow, and the ray of sunlight? Language Objective Explain the difference between direct and indirect measurement to a partner. A right triangle ENGAGE You can use similar triangles to measure things indirectly, using the fact that similar triangles have side lengths that are proportional. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photograph and ask if students can cite evidence that suggests the Earth is round. Then preview the Lesson Performance Task. B © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©naphtalina/iStockPhoto.com Essential Question: How can you use similar triangles to solve problems? Suppose the sun is shining, and you are standing near a flagpole, but out of its shadow. You will cast a shadow as well. You can assume that the rays of the sun are parallel. What do you know about the two triangles formed? Explain your reasoning. The triangles are similar. The rays of the sun are parallel and the line representing the ground is a transversal. So the acute angles formed by the bases and hypotenuses of the right triangles are congruent, as are the right angles. So the triangles are similar by the AA Similarity Criterion. C In the diagram, what heights or lengths do you already know? You probably know your own height. D What heights or lengths can be measured directly? Your height (if necessary) and the lengths of your shadow and the shadow of the flagpole. 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Have someone measure the length of your shadow and the shadow of the flagpole. Write Exploring Indirect Measurement and solve a proportion in which the unknown is the height of the flagpole. Use the fact that corresponding sides of similar triangles are proportional. Explain 1 Example 1 INTEGRATE TECHNOLOGY Finding an Unknown Height After you have the known measurements, use geometry software to calculate the missing measurements. Find the indicated dimension using the measurements shown in the figure and the properties of similar triangles. A In order to find the height of a palm tree, you measure the tree’s shadow and, at the same time of day, you measure the shadow cast by a meter stick that you hold at X a right angle to the ground. Find the height h of the tree. _ _ Because ZX ǁ CA, ∠Z ≅ ∠C. All right angles are congruent, so ∠Y ≅ ∠B. So △XYZ ≅ △ABC. Set up proportion. Substitute. Multiply each by 7.2. Simplify. QUESTIONING STRATEGIES 1m Z 1.6 m Y C What assumptions are made, in this kind of indirect measurement, about the rays of the Sun and the time of day when the shadows are measured? The Sun’s rays are parallel and the measurements of the shadows are made at the same time of day. B 7.2 m BC AB = _ _ XY YZ h =_ 1 _ 7.2 1.6 1 h = 7.2 _ 1.6 h = 4.5 ( ) The tree is 4.5 meters high. EXPLAIN 1 h 72 in. The triangles are similar by the AA Similarity Criterion. Set up proportion. 48 in. 128 in. flagpole’s height __ flagpole’s shadow __ = person’s height person’s shadow 128 h =_ _ 72 48 Substitute. ( ) Multiply both sides by 6. 128 h = 72 _ 128 Simplify. x = 192 Finding an Unknown Height © Houghton Mifflin Harcourt Publishing Company Sid is 72 inches tall. To measure a flagpole, Sid stands near the flag. Sid’s friend Miranda measures the lengths of Sid’s shadow and the flagpole’s shadow. Find the height h of the flagpole. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Have students compare the process of using similar triangles to the process of using congruent triangles. The flagpole is 192 inches tall. Module 17 904 Lesson 3 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M17L3.indd 904 Learning Progressions 18/04/14 10:13 PM Students have already used congruence to solve real-world problems. For example, to find the distance across a pond, students showed that two triangles were congruent and then used CPCTC to find an unknown side length that corresponded to the distance across the pond. This lesson presents similar problems (problems in which unknown lengths must be determined), but now students will use similarity and the proportionality of corresponding sides to find the unknown length. Using Proportional Relationships 904 Reflect QUESTIONING STRATEGIES 2. Which line in the figure is the transversal that is cutting two parallel lines? the ground In the tree example, how can you check that your answer is reasonable? The length of the meter stick’s shadow is a little more than 1.5 times the length of the meter stick. So the length of the tree’s shadow, should be a little more than 1.5 times the height of the tree. Since 1.5(4.5) = 6.75, an answer of 4.5 is reasonable EXPLAIN 2 Your Turn Finding an Unknown Distance Liam is 6 feet tall. To find the height of a tree, he measures his shadow and the tree’s shadow. The measurements of the two shadows are shown. Find the height h of the tree. INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Have students name the corresponding parts The triangles are similar by the AA Similarity Criterion. tree’s height tree’s shadow = Liam’s height Liam’s shadow 28 28 h h=6 = 6 8 8 168 x = 21 h= 8 The tree is 28 feet tall. 3. __ __ _ _ (_) _ before they begin to solve for any lengths. Explain 2 h Liam 6 ft 28 ft 8 ft Finding an Unknown Distance In real-world situations, you may not be able to measure an object directly because there is a physical barrier separating you from the object. You can use similar triangles in these situations as well. QUESTIONING STRATEGIES Example 2 © Houghton Mifflin Harcourt Publishing Company How could you predict if an unknown length is longer or shorter than the corresponding given length? If the triangle with the unknown length looks smaller than the triangle with the given length, you can predict that the unknown length will be smaller. If the triangle with the unknown length looks greater than the triangle with the given length, you can predict that the unknown length will be greater. Tree Explain how to use the information in the figure to find the indicated distance. A hiker wants to find the distance d across a canyon. She locates points as described. 1. She identifies a landmark at X. She places a marker (Y) directly across the canyon from X. 2. At Y, she turns 90° away from X and walks 400 feet in a straight line. She places a marker (Z) at this location. 3. She continues walking another 600 feet, and places a marker (W) at this location. 4. She turns 90° away from the canyon and walks until the marker Z aligns with X._ She places a marker (V) at this location and measures WV. ∠VWZ ≅ ∠XYZ (All right angles are congruent) and ∠VZW ≅ ∠XZY (Vertical angles are congruent). So, △VWZ ≅ △XYZ by the AA Similarity Criterion. d =_ 400 , or _ d =_ XY = _ YZ , So _ 2 _ 327 327 3 600 VW WZ 2 = 218. Then d = 327 _ 3 X W d Z 600 ft 400 ft Y 327 ft V () The distance across the canyon is 218 feet. Module 17 905 Lesson 3 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M17L3.indd 905 Small Group Activity Have students work in a small group to estimate the length of something in the school building that would be difficult to measure. Examples include the height of a stairwell or a flagpole. Then have the group write an explanation of how to measure the structure, including diagrams. 905 Lesson 17.3 18/04/14 10:13 PM B To find the distance d across the gorge, a student identifies points as shown in the figure. Find d. J AVOID COMMON ERRORS d 42 m ∠JKL ≅ ∠NML by the AA Similarity Criterion. K 24 m Remind students to pay attention to the units in their final answers, and to make sure that the lengths they find for real-world problems are reasonable ones. M L JK = _ KL _ NM ML 35 m d =_ 24 _ 35 24 N 24 4 d = 35 ⋅ _ = 35 ⋅ _ 7 42 ELABORATE 140 d=_ 7 QUESTIONING STRATEGIES d = 20 What is the difference between direct and indirect measurement? If you can use a measurement tool like a ruler to measure a dimension, the measurement is direct. If you use the measurement tool to measure a corresponding dimension and relate it to the desired dimension using a proportion, the measurement is indirect. The distance across the gorge is 20 meters . Reflect 4. In the example, why is ∠EGF ≅ ∠IGH? ∠EGF ≅ ∠IGH are vertical angles and vertical angles are congruent. HYour Turn 5. To find the distance d across a stream, Levi located points as shown in the figure. Use the given information to find d. A SUMMARIZE THE LESSON △JKL ≈ △NML by the AA Similarity Criterion. d BC _ AB _ 12 12 _ = ; = _; d = 12(_) = 24 DE EC 12 6 d = 24 meters 6m 12 m C E Elaborate D Discussion Suppose you want to help a friend prepare for solving indirect measurement problems. What topics would you suggest that your friend review? Possible answers: The triangle similarity criteria, theorems about congruent angles (such as the Vertical Angles Theorem), writing and solving proportions, and properties of similar triangles, including the relationships between corresponding angles and between © Houghton Mifflin Harcourt Publishing Company B 12 m 6. How can you use indirect measurement and similar triangles to solve problems? You can show that two triangles are similar using the AA or SAS Similarity Criterion. Then, you can use the fact that corresponding sides are proportional to find an unknown side length. 6 corresponding sides. 7. Essential Question Check-In You are given a figure including triangles that represent a real-world situation. What is the first step you should take to find an unknown measurement? You must first be sure that the triangles are similar. Module 17 906 Lesson 3 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M17L3.indd 906 Modeling 18/04/14 10:13 PM If possible, encourage students to go outdoors to construct and solve indirect measurement problems with shadows. Find a distance that may be difficult to measure directly with a measuring tape and use similar triangles to find the distance. Direct students to draw a diagram of the problem, make measurements, write distance measurements on the diagram, and use these measurements and the appropriate steps to find the distance. Using Proportional Relationships 906 Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice Finding distances using similar triangles is called Indirect measurement . 1. Use similar triangles △ABC and △XYZ to find the missing height h. 2. B 3. h h ASSIGNMENT GUIDE A Concepts and Skills Practice Explore Exploring Indirect Measurement Exercise 1 Example 1 Finding an Unknown Height Exercises 2–5 Example 2 Finding an Unknown Distance Exercises 6–9 C 60 ft 6 ft 60 60 AC _ AB _ h _ = ; = _; h = 6(_) = 24 ft XZ 6 XY 15 Y X 15 ft Z 15 4. A C 156 ft XZ 156 16.5 5. X 14 ft Z 4 ft A 16.5 14 C 208 ft X 15.2 ft Y 3.8 ft Z XY XZ 208 h = 52 feet 14 15.2 15.2 Use similar triangles △EFG and △IHG to find the missing distance d. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Discuss how you could use a yardstick or a 6. 7. E 8. E E d © Houghton Mifflin Harcourt Publishing Company d 60 m F d H 48 m G F 78 m G 80 m I d 48 FG _ EF _ _ = ; = _; 60 IH HG 80 48 d = 80(_) = 64 60 d = 34 meters 9. d E F 45 m H 18 m H 180 m 140.4 m I I 78 FG h EF _ = _; _ = _; 45 IH HG 180 78 EF = 180(_) = 312 45 d = 312 meters 388.8 m 211.2 m F 27 m G d EF _ 27 FG _ _ ; = = _; 18 IH HG 140.4 27 EF = 140.4(_) = 210.6 18 d = 210.6 meters H 64.8 m I G d 64.8 64.8 FG _ EF = _ _ = _; d = 211.2(_) = 35.2 ; d = 35.2 meters ; IH HG 211.2 388.8 388.8 Module 17 Exercise IN2_MNLESE389847_U7M17L3.indd 907 907 Lesson 17.3 Z 3.8 3.8 AC _ AB _ h _ = ; = _; h = 208(_) = 52 AC _ AB _ h 4 4 _ = ; = _; h = 108.5(_) = 31 XY XZ 108.5 h = 31 feet 16.5 ft h C 108.5 ft Y X B h A 5.5 ft 5.5 5.5 AC _ AB _ h _ = ; = _; m = (_) 156 = 52 ft XY B Y ruler to take an indirect measurement of objects in the classroom that are difficult to measure directly, such as a chalkboard or a door. B 907 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–5 1 Recall of Information MP.4 Modeling 6–9 2 Skills/Concepts MP.2 Reasoning 10–16 2 Skills/Concepts MP.4 Modeling 17–19 2 Skills/Concepts MP.2 Reasoning 20 3 Strategic Thinking MP.3 Logic 21 3 Strategic Thinking MP.2 Reasoning Lesson 3 18/04/14 10:13 PM 10. To find the height h of a dinosaur in a museum, Amir placed a mirror on the ground 40 feet from its base. Then he stepped back 4 feet so that he could see the top of the dinosaur in the mirror. Amir’s eyes were approximately 5 feet 6 inches above the ground. What is the height of the dinosaur? The two triangles are similar by the AA Similarity Criterion. All the dimensions have to be in the same units, so write 5 feet 6 inches as 5.5 feet. 40 40 h = ; h = 5.5 = 55; The dinosaur is 55 feet tall. 5.5 4 4 5 ft 6 in. 4 ft (_) _ _ 11. Jenny is 5 feet 2 inches tall. To find the height h of a light pole, she measured her shadow and the pole’s shadow. What is the height of the pole? 40 ft ge07sec07l05005aa AB The two triangles are similar by the AA 5 ft 2 in. Similarity Criterion. All the dimensions 15.5 ft 2 1 have to be in the same units, so write 5 feet 2 as 5___ = 5__ feet. Write 7 feet 9 inches as 6 12 3 9 1 ___ __ 7 = 7 feet. To avoid mixing decimals and fractions, write 15.5 feet as 15__ feet. 12 4 _1 3 7_ 2 56 31 31 31 31 __ 31 _ 62 h 2 ___ = __ = __(__ ÷ __ = 20 _ )= __ (31)(__4 ) = __ (4) = __ 1 15_ 2 4 2 6 4 2 3 31 2 3 3 2 feet, or 20 feet 8 inches tall. The light pole is 20 __ 7 ft 9 in. 3 3 D 12. A student wanted to find the height h of a statue of a pineapple in Nambour, Australia. She measured the pineapple’s shadow and her own shadow. The student’s height is 5 feet 4 inches. What is the height of the pineapple? △ABC ≈ △DEP by the AA Similarity Criterion. AC = 5 ft 4 in. = 64 in. BC = 2 ft = 24 in. A 105 64 24 24 EF h DF The height h of the pineapple is 280 inches or 23 feet 4 inches. B 2 1 2 ft C E 8 ft 9 in. 13. To find the height h of a flagpole, Casey measured her own shadow and the flagpole’s shadow. Given that Casey’s height is 5 feet 4 inches, what is the height of the flagpole? ge07se_ c07105002aa The triangles are similar by the AA Similarity Criterion. AB Casey's height: 5 ft 4 in. = 64 in.; Casey's shadow: 3 ft = 36 in. flagpole's shadow: 14 ft 3 in. = 171 in. Casey's shadow 64 36 x 5 ft 4 in. 3 ft flagpole's height __ flagpole's shadow _ h 171 171 __ = ; = _; h = 64(_) = 304 Casey's height F © Houghton Mifflin Harcourt Publishing Company EF = 8 ft 9 in. = 105 in. 64 _ 105 105 AC _ BC _ 24 _ h _ = ; = ; = _; h = 64(_) = 280 14 ft 3 in. 36 The height h of the flagpole is 304 inches or 25 feet 4 inches. Module 17 IN2_MNLESE389847_U7M17L3.indd 908 908 Lesson 3 18/04/14 10:13 PM Using Proportional Relationships 908 A city is planning an outdoor concert for an Independence Day celebration. To hold speakers and lights, a crew of technicians sets up a scaffold with two platforms by the stage. The first platform is 8 feet 2 inches off the ground. The second platform is 7 feet 6 inches above the first platform. The shadow of the first platform stretches 6 feet 3 inches across the ground. AVOID COMMON ERRORS Some students may not think they can find the similarity ratio of two figures without knowing the lengths of the sides. Remind them that the perimeters have the same similarity ratio as the corresponding sides, and the areas have the same ratio as the squares of the corresponding sides. E 7 ft 6 in. C 8 ft 2 in. A 6 ft 3 in. B D 14. Explain why △ABC is similar to △ADE. (Hint: rays of light are parallel.) Communicate Mathematical Ideas Possible answer: Because rays of light are parallel ‹ › − and AD is a transversal, ∠ABC and ∠ADE are corresponding angles, so they are congruent. ∠A is common to both triangles. So △ABC ~ △ADE by the AA Similarity Postulate. 15. Find the length of the shadow of the second platform in feet and inches to the nearest inch. First, convert all lengths to inches: AC = 8 ft 2 in. = 96 in. CE = 7 ft 6 in. = 90 in. © Houghton Mifflin Harcourt Publishing Company AB = 6 ft 3 in. = 75 in. AE = AC + CE = 98 + 90 = 188 188 AD 188 AD AE = ; = ; AD = 75 = 144 75 98 AB 98 AC _ The shadow of the second platform is represented by BD. _ __ _ BD = AD - AB = 144 - 75 = 69 So the shadow of the second platform is 69 inches or 5 feet 9 inches. 16. A technician is 5 feet 8 inches tall. The technician is standing on top of the second platform. Find the length s of the shadow that is cast by the scaffold and the technician to the nearest inch. Height of technician: 5 ft 8 in. = 68 in. Height of scaffold with technician: 188 + 68 = 256 in. 256 256 s = ; s = 75 = 196 75 98 98 The shadow cast by the scaffold and the technician is 196 inches or 16 feet 4 inches. _ _ Module 17 IN2_MNLESE389847_U7M17L3.indd 909 909 Lesson 17.3 (_) (_) 909 Lesson 3 18/04/14 10:13 PM 17. To find the distance XY across a lake, you locate points as shown in the figure. Explain how to use this information to find XY. 300 ft △XYZ ~ △VUZ by the SAS Similarity Criterion, XY XY XY 800 = . Then = , so XY = 1,000 ft. so 500 VU VZ 400 _ _ 500 ft U V 400 ft Z _ _ 600 ft 800 ft Y X 18. In order to find the height of a cliff, you stand at the bottom of the cliff, walk 60 feet from the base, and place a mirror on the ground. Then you face the cliff and step back 5 feet so that can see the top of the cliff in the mirror. Assuming your eyes are 6 feet above the ground, explain how to use this information to find the height of the cliff. (The angles marked congruent are congruent because of the nature of the reflection of light in a mirror.) J P M 6 ft Q 5 ft 60 ft K Mirror △JKM ~ △PQM by the AA Similarity Criterion, so so JK = 72, and the height of the cliff is 72 feet. JK JK 60 MK _ = _. Then _ = _, PQ 5 6 MQ 19. To find the height of a tree, Adrian measures the tree’s shadow and then his shadow. Which proportion could Adrian use to find the height of the tree? Select all that apply. D © Houghton Mifflin Harcourt Publishing Company BC AC = _ A. _ DF EF DF = _ EF B. _ AC BC BC AB = _ C. _ DF EF DF = _ EF D. _ BC AC AC BC = _ E. _ EF DF Tree Adrian A 5.6 ft B 4.2 ft C E 42.3 ft F ― ― ― ― ― ― ― ― DF and AC are corresponding sides, as are EF and BC. ― ― AB and DF are not corresponding sides. ― ― ― ― DF and BC are not corresponding sides, nor are EF and AC. ― ― ― ― BC and EF are corresponding sides, as are AC and DF. A. AC and DF are corresponding sides, as are BC and EF. B. C. D. E. Module 17 IN2_MNLESE389847_U7M17L3.indd 910 910 Lesson 3 18/04/14 10:13 PM Using Proportional Relationships 910 JOURNAL H.O.T. Focus on Higher Order Thinking Have students make up their own problem in which an unknown length must be found using similar triangles. Remind students to include the solutions to their problems. 20. Critique Reasoning Jesse and Kyle are hiking. Jesse is carrying a walking stick. They spot a tall tree and use the walking stick as a vertical marker to create similar triangles and measure the tree indirectly. Later in the day they come upon a rock formation. They measure the rock formation’s shadow and again want to use similar triangles to measure its height indirectly. Kyle wants to use the shadow length they measured earlier for the stick. Jesse says they should measure it again. Who do you think is right? Jesse is right. The length of a shadow is dependent not only on the height of an object, but on the position of the sun in the sky. To create similar triangles, the shadows of the two objects must be measured at the same time of day. 21. Error Analysis Andy wants to find the distance d across a river. He located points as shown in the figure, then use similar triangles to find that d = 220.5 feet. How can you tell without calculating that he must be wrong? Tell what you think he did wrong and correct his error. C A 300 ft E 200 ft D 147 ft B ― ― AB is the shortest side of right △ABE, so corresponding side DC of △DCE must be shorter ― than DE, that is, DE < 200. the triangles are similar, Andy must have used the wrong ( ) d 200 200 proportion. The correct proportion is ____ = ____ , so d = 147 ____ = 98. The distance across 147 300 300 © Houghton Mifflin Harcourt Publishing Company the river is 98 ft. Module 17 IN2_MNLESE389847_U7M17L3.indd 911 911 Lesson 17.3 911 Lesson 3 18/04/14 10:13 PM Lesson Performance Task AVOID COMMON ERRORS Around 240 B.C., the Greek astronomer Eratosthenes was residing in Alexandria, Egypt. He believed that the Earth was spherical and conceived of an experiment to measure its circumference. At noon in the town of Syene, the sun was directly overhead. A stick stuck vertically in the ground cast no shadow. At the same moment in Alexandria, 490 miles from Syene, a vertical stick cast a shadow that veered 7.2° from the vertical. Percent error compares the error in a measurement to the correct measurement. So, to find the percent error in Eratosthenes’ calculations, students should compare the 401-mile error to the correct circumference, not to the incorrect circumference. 401 401 , not _____ . The correct ratio is _____ 24,901 24,500 shadow Alexandria 7.2° 490 mi 7.2° Syene no shadow 1. Refer to the diagram. Explain why Eratosthenes reasoned that the angle at the center of the Earth that intercepted a 490-mile arc measured 7.2 degrees. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 The ancient Greeks used several different 2. Calculate the circumference of the Earth using Eratosthenes’s figures. Explain how you got your answer. 3. Calculate the radius of the Earth using Eratosthenes’s figures. units of length. One was the stadion (plural, stadia), which, by one modern estimate, equaled 185.4 meters. How far was Alexandria from Syene, in stadia? (1 mi ≈ 1609 meters) about 4252 stadia 4. The accepted circumference of the Earth today is 24,901 miles. Calculate the percent error in Eratosthenes’s calculations. 1. The line from the center of the Earth to Syene and the line from the top of the Alexandria stick to the north end of the Alexandria shadow are parallel. By the Alternate Interior Angles Theorem, the angle at the center of the Earth is congruent to the 7.2° angle at Alexandria. 2. In a complete rotation of 360° at the Earth’s center, each 7.2° angle will intercept a 360 = 50 such 490-mile arcs in the entire circumference. So, by 490-mile arc. There are ____ 7.2 Eratosthenes’ figures, the circumference of the Earth measures 50 × 490 = 24, 500 miles. C = 2πr 24, 500 ≈ 2(3.14)r 24, 500 _ ≈r © Houghton Mifflin Harcourt Publishing Company 3. 2(3.14) 3901.3 ≈ r C = 2πr 24, 500 ≈ 2(3.14)r _ 24, 500 ≈r 2(3.14) 3901.3 ≈ r The radius of the Earth is about 3901.3 miles 401 4. 24,901 - 24,500 = 401 miles; ______ ≈ 1.6% 24, 901 Module 17 912 Lesson 3 EXTENSION ACTIVITY IN2_MNLESE389847_U7M17L3.indd 912 Give these directions to students: 1. 2. 3. Choose one of the planets of the Solar System other than Earth and conduct research to find its radius or diameter. Calculate the circumference of the planet (use 3.14 for π). Suppose that you placed two sticks in the ground 490 miles apart on your planet. At the moment the Sun was directly over one of the sticks, what angle from the vertical would the shadow of the second stick cast? Explain. Example: Mars: diameter 4200 mi; circumference: 13,888 mi; 490 = _____ x ; x = 12.7° _______ 13,888 360° 18/04/14 10:13 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Using Proportional Relationships 912 LESSON 17.4 Name Similarity in Right Triangles Date 17.4 Similarity in Right Triangles Essential Question: How does the altitude to the hypotenuse of a right triangle help you use similar right triangles to solve problems? Resource Locker Common Core Math Standards Explore The student is expected to: COMMON CORE Class G-SRT.B.4 Identifying Similar Triangles A Make two copies of the right triangle on a piece of paper and cut them out. B Choose one of the triangles. Fold the paper to find the altitude to the hypotenuse. C Cut the second triangle along the altitude. Label the triangles as shown. Prove theorems about triangles. Also G-SRT.B.5 Mathematical Practices COMMON CORE MP.8 Patterns Language Objective Explain to a partner how to use the Angle/Angle criterion to show similarity in triangles. Essential Question: How does the altitude to the hypotenuse of a right triangle help you use similar right triangles to solve problems? You can use the geometric means theorems to find the missing measurement using indirect measurement. PREVIEW: LESSON PERFORMANCE TASK © Houghton Mifflin Harcourt Publishing Company ENGAGE 2 1 3 View the Engage section online. Discuss the photograph, asking students if they can describe what is happening and where. Then preview the Lesson Performance Task. Module 17 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction Lesson 4 913 gh "File info" made throu Date Class Name ilarity 17.4 Sim angles in Right Tri le help you a right triang enuse of to the hypot altitude problems? does the les to solve ion: How r right triang B.5 use simila Also G-SRT. triangles. ms about Resource Locker Quest Essential Explore IN2_MNLESE389847_U7M17L4 913 s theore G-SRT.B.4 Prove COMMON CORE Identifying copies Make two y g Compan Choose one triangle on of the triang Cut the second a piece les. Fold the triangle along cut them of paper and paper to find HARDCOVER PAGES 913924 out. Watch for the hardcover student edition page numbers for this lesson. enuse. e to the hypot the altitud e. Label the the altitud triangles Publishin of the right ngle Similar Tria . as shown n Mifflin Harcour t 2 1 © Houghto 3 Lesson 4 913 Module 17 7L4 913 47_U7M1 ESE3898 IN2_MNL 913 Lesson 17.4 18/04/14 10:18 PM 18/04/14 10:19 PM D Place triangle 2 on top of triangle 1. What do you notice about the angles? EXPLORE The corresponding angles are congruent. E Identifying Similar Triangles What is true of triangles 1 and 2? How do you know? They are congruent by the AA Similarity Criterion. F INTEGRATE TECHNOLOGY Use geometry software to explore drawing a right angle and an altitude to the hypotenuse. Repeat Steps 1 and 2 for triangles 1 and 3. Does the same relationship hold true for triangles 1 and 3? Yes; the corresponding angles are congruent, and the triangles are similar by the AA Similarity Criterion. QUESTIONING STRATEGIES Reflect 1. How are the hypotenuses of the triangles 2 and 3 related to triangle 1? The hypotenuses of the smaller triangles are the legs of the original triangle. 2. What is the relationship between triangles 2 and 3? Explain. They are similar because triangle similarity is transitive. When you draw the altitude to the hypotenuse of a right triangle, what kinds of figures are produced? Two triangles that are similar to the original triangle and to each other. 4. Suppose you draw △ABC such that _∠B is a right angle and the altitude to the hypotenuse intersects hypotenuse AC at point P. Match each triangle to a similar triangle. Explain your reasoning. B A. △ABC △PAB B. △PBC C △CAB C. △BAP A △BPC What does it mean when you say that two triangles are similar? The triangles have the same shape; that is, corresponding angles are congruent and corresponding sides are proportional. © Houghton Mifflin Harcourt Publishing Company 3. What kind of angle is formed by the altitude to the hypotenuse? The altitude to the hypotenuse forms a right angle with the hypotenuse. A. Angles B and P are corresponding right angles. Angle C corresponds to itself. B. ∠BPC corresponds to ∠APB and ∠BCP corresponds to ∠ABP. C. Angles B and P are corresponding right angles. Angle A corresponds to itself. Module 17 914 Lesson 4 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M17L4 914 Math Background 18/04/14 10:19 PM In mathematics, the word mean describes a relationship between numbers. The arithmetic mean of two numbers is the average of the numbers. This is different from the geometric mean, which, for positive numbers, is the positive square root of their product. The geometric mean of a and b is the positive number x such ― that x = √ab . In later math courses, students may also study harmonic means. Similarity in Right Triangles 914 Explain 1 EXPLAIN 1 Finding Geometric Means of Pairs of Numbers Consider the proportion __ax = __bx where two of the numbers in the proportion are the same. The number x is the geometric mean of a and b. The geometric mean of two positive numbers is the positive square root of their product. So the geometric mean of a and b is the positive number x such that x = √ab or x 2 = ab. ― Finding Geometric Means of Pairs of Numbers Example 1 Find the geometric mean x of the numbers. 4 and 25 x 4 =_ _ x 25 x 4 = 25x · _ 25x · _ x 25 2 100x = _ 25x _ x 25 100 = x 2 Write proportion. INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Point out that only a positive number can Multiply both sides by the product of the denominators. Multiply. Simplify. _ √ 100 Take the square root of both sides. represent the length of a segment in a triangle. 10 = x Simplify. 9 and 20 QUESTIONING STRATEGIES 9 x _=_ x 20 Write proportion. In a proportion involving a and b, where is the geometric mean between the two numbers? The geometric mean is the denominator of one fraction and the numerator of the other. _ = √x 2 9 x _ 20x · _ x = 20x · 20 Multiply both sides by the product of the denominators. 2 180 x _ _ = 20x x 20 Multiply. 180 = x 2 Simplify. _ √ 180 = √_x 5 =x 6 √― 2 Take the square root of both sides. © Houghton Mifflin Harcourt Publishing Company Simplify. Reflect 5. ― How can you show that if positive numbers a and b are such that __ax = __bx , then x = √ab ? Multiply both sides of the proportion by xb, then take the square root of both sides: () () ― ― ― a _x 2 2 xb _ x = xb b ; ab = x ; √ab = √x ; √ab = x. Your Turn Find the geometric mean of the numbers. If necessary, give the answer in simplest radical form. 6. 6 and 24 √―― 6 · 24 = √―― 144 = 12 Module 17 7. 5 and 12 ― ―― ― √60 = √4 · 15 = 2√15 915 Lesson 4 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M17L4 915 Small Group Activity Have small groups of students work together to research and present to the class other proofs of the Pythagorean Theorem. 915 Lesson 17.4 18/04/14 10:19 PM Explain 2 Proving the Geometric Means Theorems EXPLAIN 2 In the Explore activity, you discovered a theorem about right triangles and similarity. The altitude to the hypotenuse of a right triangle forms two triangles that are similar to each other and to the original triangle. Proving the Geometric Means Theorems That theorem lead to two additional theorems about right triangles. Both of the theorems involve geometric means. Geometric Means Theorems Theorem Example The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the segments of the hypotenuse. h = xy or h = √xy The length of a leg of a right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse adjacent to that leg. a 2 = xc or a = √xc b 2 = yc or b = √yc 2 Diagram ― x ― ― Prove the first Geometric Means Theorem. _ Given: Right triangle ABC with altitude BD A CD = _ BD Prove: _ BD AD B Example 2 Statements a c h QUESTIONING STRATEGIES Reasons 2. The altitude to the hypotenuse of a right triangle forms two triangles that are similar to the original triangle and to each other. 3. Corresponding sides of similar triangles are proportional. Reflect 8. Discussion How can you prove the second Geometric Means Theorem? Use the triangle shown in the Example. Show that △ABC ∼ △ADB because the altitude to the hypotenuse of a right triangle forms two triangles that are similar to the original AD AB triangle and to each other. Then show that __ = __ because corresponding sides of similar AB AC How could you express this relationship as a proportion? The ratio of one segment to the altitude is equal to the ratio of the altitude to the other segment. © Houghton Mifflin Harcourt Publishing Company AD What is the relationship between the altitude of a hypotenuse and the segments on the hypotenuse? It is the geometric mean of the lengths of the segments. C 2. △CBD ∼ △BAD BD angle and the hypotenuse, as well as the altitude to the hypotenuse. D 1. Given CD _ BD _ = y b _ 1. △ABC with altitude BD 3. INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 Make sure that students can identify the right DC BC triangles are proportional. Then prove that __ = __ in the same way. BC AC Module 17 916 Lesson 4 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M17L4 916 Modeling 18/04/14 10:19 PM Have students cut a 15–20–25-unit right triangle out of a sheet of graph paper. Ask them to draw the altitude to the hypotenuse, forming a 12–16–20-unit right triangle and a 9–12–15-unit right triangle. Then have students use the side lengths in these triangles to verify the relationships in this lesson. Similarity in Right Triangles 916 Using the Geometric Means Theorems Explain 3 EXPLAIN 3 You can use the Geometric Means Theorems to find unknown segment lengths in a right triangle. Example 3 Using the Geometric Means Theorems Find the indicated value. y 2 10 QUESTIONING STRATEGIES x If you know the lengths of the segments that the altitude divides the hypotenuse into, how can you find the length of the altitude? You can write a proportion with the lengths of the segments and length of the hypotenuse and solve for the unknown. x 2 =_ _ x 10 x 2 = 10x · _ 10x · _ x 10 2 20x = _ 10x _ x 10 20 = x 2 Write proportion. Multiply both sides by the product of the denominators. Multiply. Simplify. ― ― 2√5― = x √20 = √x 2 Take the square root of both sides. Simplify. AVOID COMMON ERRORS y y 10 = _ _ y 12 Write proportion. Some students may find it puzzling that a leg in one triangle can become the hypotenuse in a related triangle. Have them use colored pencils to distinguish the various triangles, marking the hypotenuse in each with a double or darker line. y 10 = 12y _ 12y _ y 12 Multiply both sides by the product of the denominators. 12y 2 120y _=_ y 12 Multiply. 120 = y 2 Simplify. © Houghton Mifflin Harcourt Publishing Company z x ―― ――― √ 120 = √ y Take the square root of both sides. ― =y 2√30 Simplify. 2 Reflect 9. Discussion How can you check your answers? Possible answer: Use the Pythagorean Theorem. Your Turn 10. Find x. y 5 x z _5 = _x ; 7x(_5 ) = 7x(_x ); 35 = x ; √― 35 = x 16 cm x 7 x 7 2 7 Module 17 917 Lesson 4 LANGUAGE SUPPORT IN2_MNLESE389847_U7M17L4 917 Connect Vocabulary Differentiate the geometric mean from previous uses of mean such as the arithmetic mean, in which we add a set of numbers and divide by the number of values. The geometric mean is the nth root of the product of n numbers. That means, we multiply the set of numbers, then take the nth root, where n is the number of values that were multiplied. 917 Lesson 17.4 18/04/14 10:19 PM Explain 4 Proving the Pythagorean Theorem using Similarity EXPLAIN 4 You have used the Pythagorean Theorem in earlier courses as well as in this one. There are many, many proofs of the Pythagorean Theorem. You will prove it now using similar right triangles. Proving the Pythagorean Theorem using Similarity The Pythagorean Theorem In a right triangle, the square of the sum of the lengths of the legs is equal to the square of the length of the hypotenuse. Complete the proof of the Pythagorean Theorem. _ Given: Right △ABC with altitude CD A Example 4 c b Prove: a 2 + b 2 = c 2 Draw the altitude to the hypotenuse. Label the point of intersection X. A . ∠B ≅ ∠B by the Reflexive Property of Congruence . B a C Part 1 ∠BXC ≅ ∠BCA because all right angles are congruent INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 The Pythagorean Theorem was known in many ancient civilizations, including those of Egypt, India, and China. The oldest known axiomatic proof of the theorem, however, dates to Euclid’s Elements (circa 300 bce). Since then, hundreds of proofs have been given, including proofs that rely on area, proofs that use calculus, and the proof that is given in this lesson, which is based on similar triangles. x c b B a C So, △BXC ∼ △BCA by the AA Similarity Criterion . ∠AXC ≅ ∠ACB because all right angles are congruent . ∠A ≅ ∠A by the Reflexive Property of Congruence . So, △AXC ∼ △ACB by the AA Similarity Criterion . Part 2 Let the lengths of the segments of the hypotenuse be d and e, as shown in the figure. Use the fact that corresponding sides of similar triangles are proportional to write two proportions. e a = _. Proportion 1: △BXC ∼ △BCA,so _ a c d x c b C e a B d b = _____ Proportion 2: △AXC ∼ △ACB, so _ . c b Module 17 IN2_MNLESE389847_U7M17L4 918 918 © Houghton Mifflin Harcourt Publishing Company A Lesson 4 18/04/14 10:19 PM Similarity in Right Triangles 918 Part 3 QUESTIONING STRATEGIES Now perform some algebra to complete the proof as follows. To which triangles does the Pythagorean Theorem apply? Why is it useful? To right triangles only; it allows you to relate the lengths of the sides of right triangles. Multiply both sides of Proportion1 by ac. Write the resulting equation. Multiply both sides of Proportion12 by bc. Write the resulting equation. a2 = ce b2 = cd 2 2 Adding the two resulting equations give this: a + b = ce + cd 2 2 Factor the right side of the equation: a + b = c (e + d) Finally, use the fact that e + d = c by the Segment Addition Postulate to rewrite the equation as a + b = c . 2 2 2 Reflect 11. Error Analysis A student used the figure in Part 2 of the example, and wrote the following incorrect proof of the Pythagorean Theorem. Critique the student’s proof. △BXC ∼ △BCA and △BCA ∼ △CXA, so △BXC ∼ △CXA by transitivity of similarity. Since corresponding f sides of similar triangles are proportional, _e = _ and f 2 = ed. Because △BXC ∼ and △CXA f d are right triangles, a2 = e2 + f 2 and b2 = f 2 + d 2. Add the equations. a2 + b2 = e2 + 2f 2 + d 2 Substitute. = e2 + 2ed + d 2 Factor. = (e + d) Segment Addition Postulate = c2 2 The proof is incorrect because the student assumes the result that is to be proved. The student assumes that the Pythagorean Theorem is true in order to write that © Houghton Mifflin Harcourt Publishing Company a 2 = e 2 + f 2 and b 2 = f 2 + d 2. Module 17 IN2_MNLESE389847_U7M17L4 919 919 Lesson 17.4 919 Lesson 4 18/04/14 10:19 PM Elaborate ELABORATE 12. How would you explain to a friend how to find the geometric mean of two numbers? Possible answers: Write a proportion and multiply both sides by the product of the denominators, then simplify and take the square root of both sides; multiply the two QUESTIONING STRATEGIES numbers and then take the square root. What is the relationship between the original triangle and the two smaller triangles formed by the altitude to the hypotenuse? All three triangles are similar to one another. _ 13. △XYZ is an _isosceles right triangle and the right angle is ∠Y. Suppose the altitude to hypotenuse XZ intersects XZ at point P. Describe the relationships among triangles △XYZ, △YPZ and △XPY. Possible answer: All three triangles are similar. In addition, △YPZ ≅ △XPY. How can you identify the hypotenuse in a right triangle? It is opposite the right angle, and it is the longest side. 14. Can two different pairs of numbers have the same geometric mean? If so, give an example. If not, explain why not. Yes; possible answer: the geometric mean of 4 and 16 is 8, and the geometric mean of 2 and 32 is also 8. 15. Essential Question Check-In How is the altitude to the hypotenuse of a right triangle related to the segments of the hypotenuse it creates? The length of the altitude is the geometric mean of the lengths of the segments of the SUMMARIZE THE LESSON How are geometric means related to the Pythagorean Theorem? Geometric means can be used to prove the Pythagorean Theorem. hypotenuse. © Houghton Mifflin Harcourt Publishing Company Module 17 IN2_MNLESE389847_U7M17L4 920 920 Lesson 4 18/04/14 10:19 PM Similarity in Right Triangles 920 Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice Write a similarity statement comparing the three triangles to each diagram. 1. 2. P B 3. C X W R ASSIGNMENT GUIDE E Q S △PQR ∼ △SPR ∼ △SQP D Y △BDE ∼ △DEC ∼ △BEC Z △XYZ ∼ △XWY ∼ △YWZ Concepts and Skills Practice Explore Identifying Similar Right Triangles Exercises 1–3 Find the geometric mean x of each pair of numbers. If necessary, give the answer in simplest radical form. Example 1 Finding Geometric Means of Pairs of Numbers Exercises 4–9 4. Example 2 Proving the Geometric Means Theorems Exercises 13–18 Example 3 Using the Geometric Means Theorems Exercises 10–12 Example 4 Proving the Pythagorean Theorem using Similarity Exercise 22 5 and 20 x ;x _5 = _ x 6. 2 x 7. ― _ _ 1.5 and 84 1.5 x ;x _ =_ 84 2 9. ― 2 = 36; x = 6 3.5 and 20 3.5 _ _ = x ;x 20 2 27 2 and _ _ 40 3 2 3 √5 x 3 9 2 x = 27 ; x = 20 ; x = 10 40 _ _ _ _ = 126; x = 3 √14 ― = 70; x = √70 _― _ Find x, y, and z. 10. 11. 25 65 x y x y z z ――― ―― 30 y = √652 - 252 = √3600 = 60 2 2 2 Exercise 40 30 _ x _ √― x = 40 ; x = 1200; x = 20 3 y 70 _ _ √― y = 30 ; y = 2100; y = 10 21 70 _ z _ = 7 ; z = 2800; z = 20√― 25 25 _ 60 _ _y _ y = x ; 60 = x ; 25x = 3600; x = 144 25 + x _ z z 169 _ _ = x; _ z z = 144 ; z = 24,366; ――― z = √24,336 = 156 IN2_MNLESE389847_U7M17L4 921 Lesson 17.4 12 x Module 17 921 3 and 12 x ;x _3 = _ = 100; x = 10 8 and 13 x © Houghton Mifflin Harcourt Publishing Company with the altitude drawn to the hypotenuse. Have them identify three similar triangles. Ask them to write three different proportions involving a geometric mean. 20 8 x 2 √ x = 13 ; x = 104; x = 2 26 8. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Have students draw and label a right triangle 5. z 40 2 Lesson 4 921 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–9 2 Skills/Concepts MP.2 Reasoning 10–12 2 Skills/Concepts MP.4 Modeling 13–18 2 Skills/Concepts MP.1 Problem Solving 19–22 2 Skills/Concepts MP.5 Using Tools 23 3 Strategic Thinking MP.1 Problem Solving 24 3 Strategic Thinking MP.3 Logic 18/04/14 10:19 PM ―― 12. Some students may have difficulty visualizing the corresponding angles in the overlapping triangles. Suggest that they redraw the diagram to separate the three triangles. 12.8 9.6 _ = _; 12.8z = 92.16; z = 7.2 y 9.6 AVOID COMMON ERRORS 2 2 y2 = (12.8) + (9.6) = 256; y = √256 = 16 12.8 z 9.6 z 12.8 + 7.2 _ x _ = ;x x x 7.2 = 144; x = 12 2 Use the diagram to complete each equation. c a d e b e 13. _ec = _ d c _ a 14. _ a= c + d c+d b 15. _ = _ b d 17. c(c + d) = a d = _e 16. _ c e 2 2 18. e = ab © Houghton Mifflin Harcourt Publishing Company Find the length of the altitude to the hypotenuse under the given conditions. B D A 19. BC = 5 AC = 4 _ _ AB = 3; AD = 4 ; 5 3 AD = 2.4 Module 17 IN2_MNLESE389847_U7M17L4 922 C 20. BC = 17 AC = 15 _ _ _ 15 ; AB = 8; AD = 17 8 120 ≈ 7.06 AD ≈ 17 922 21. BC = 13 AC = 12 _ _ _ 12 AB = 5; AD = ; 5 13 60 ≈ 4.62 AD = 13 Lesson 4 18/04/14 10:19 PM Similarity in Right Triangles 922 22. Communicate Mathematical Ideas The area of a rectangle with a length of ℓ and a width of w has the same area as a square. Show that the side length of the square is the geometric mean of the length and width of the rectangle. ― Area of the square = s2 = Area of rectangle = ℓw; s2 = ℓw; s = √ℓw JOURNAL Ask students to demonstrate, with examples and reasons, that the geometric mean of two positive numbers is always smaller than their arithmetic mean. H.O.T. Focus on Higher Order Thinking 23. Algebra An 8-inch-long altitude of a right triangle divides the hypotenuse into two segments. One segment is 4 times as long as the other. What are the lengths of the segments of the hypotenuse? x in. = shorter segment; 4x in. = longer segment; The length of the hypotenuse is the 8 x geometric mean of se the lengths, so = . Then 4x2 = 64, x2 = 16, and x = 4. 4x 8 The segments of the hypotenuse measure 4 inches and 16 inches. _ _ 24. Error Analysis Cecile and Amelia both found a value for EF in △DEF. Both students work are shown. Which student’s solution is correct? What mistake did the other student make? 12 EF = ___ Cecile: ___ 8 EF E So EF 2 = 12(8) = 96. _ 8 _ © Houghton Mifflin Harcourt Publishing Company Then EF = √96 = 4√6 . _ F _ Then EF = √32 = 4√2 . Cecile’s solution is correct. Amelia wrote the wrong proportion. IN2_MNLESE389847_U7M17L4 923 Lesson 17.4 D So EF 2 = 8(4) = 32. Module 17 923 G 4 8 EF Amelia: ___ = ___ 4 EF 923 Lesson 4 18/04/14 10:19 PM Lesson Performance Task INTEGRATE MATHEMATICAL PRACTICES MP.8 Focus on Patterns In the example at the beginning of the lesson, a $100 investment grew for one year at the rate of 50%, to $150, then fell for one year at the rate of 50%, to $75. The arithmetic mean of +50% and -50%, which is 0%, was not a good predictor of the change, for it predicted the investment would still be worth $100 after two years, not $75. • Find the geometric mean of 1, 1, 2, 4, and 4. 2 1. Find the geometric mean of 1 + 50% and 1 - 50%. (Each 1 represents the fact that at the beginning of each year, an investment is worth 100% of itself.) Round to the nearest thousandth. • Write an expression representing the geometric mean of the numbers a, b, c, d, 5 ―― 5 ― a · b · c · d · e or abcde and e. 2. It is the geometric mean, not the arithmetic mean, that tells you what the interest rate would have had to have been over an entire investment period to achieve the end result. You can use your answer to Exercise 1 to check this claim. Find the value of a $100 investment after it increased or decreased at the rate you found in Exercise 1 for two years. Show your work. AVOID COMMON ERRORS In Question 4 of the Lesson Performance Task, some students may calculate the geometric mean and then assume that it represents the interest rate they are looking for. Remind students that a geometric mean of 1 represents 0% change, and they must subtract the mean they found from 1 to find the average percent change per year. 3. Copy the right triangle shown here. Write the terms “Year 1 Rate”, “Year 2 Rate”, and “Average Rate” to show geometrically how the three investment rates relate to each other. Average Rate. Year 1 Rate Year 2 Rate 4. The geometric mean of n numbers is the nth root of the product of the numbers. Find what the interest rate would have had to have been over 4 years to achieve the result of a $100 investment that grew 20% in Year 1 and 30% in Year 2, then lost 20% in Year 3 and 30% in Year 4. Show your work. Round your answer to the nearest tenth of a percent. ―――――― = √―― .75 ――― √(1 + .5)(1 - .5) = √(1.5)(.5) © Houghton Mifflin Harcourt Publishing Company 1. = .866 2. After 1 year: 100 × 0.866 = 86.6 After 2 years: 86.6 × 0.866 = 74.9956, or $75 3. See diagram. 4 ――――――― 4 ――― 4. √(1.2)(1.3)(0.8)(0.7) = √0.8736 ≈ 0.967 Since 1 represents 0% change, 0.967 represents (1 - 0.967) = 0.033, or a 3.3% average loss per year. Module 17 924 Lesson 4 EXTENSION ACTIVITY IN2_MNLESE389847_U7M17L4 924 The Pythagorean Theorem is only one of many contributions the Greek mathematician and philosopher Pythagoras made to mathematics. He was the first to identify three means for weighting numbers: the arithmetic mean (or average), the geometric mean, and the harmonic mean. Have students research and explain the harmonic mean. Then they should show how to find all three Pythagorean means for the numbers 4 and 9. arithmetic: 6.5; geometric: 6; 72 ≈ 5.54 harmonic: ___ 13 18/04/14 10:19 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Similarity in Right Triangles 924 MODULE 17 MODULE STUDY GUIDE REVIEW 17 Using Similar Triangles Study Guide Review Essential Question: How can you use similar triangles to solve real-world problems? ASSESSMENT AND INTERVENTION KEY EXAMPLE Key Vocabulary indirect measurement (medición indirecta) geometric mean (media geométrica) (Lesson 17.1) Find the missing length x. ZU = _ ZV _ UX VY 16 x =_ ___ 10 9 16 (10) x = ___ 9 ( ) Assign or customize module reviews. x ≈ 17.8 KEY EXAMPLE MODULE PERFORMANCE TASK Mathematical Practices: MP.1, MP.2, MP.4, MP.6 G-SRT.B.5, G-MG.A.1 • The best way to work with the GPS coordinates: Students may want to express the coordinates as (latitude, longitude) so they can use the distance formula to find the lengths of the sides of the triangle (see Module 1 Performance Task). This gives the following coordinates: Miami: (–80.226529, 25.789106); San Juan: (–66.1057427, 18.4663188); Hamilton: (–64.781380, 32.294887) • How to find the area from the lengths of the sides: Students can experiment with ways to find the area or you can make a suggestion. 925 Module 17 © Houghton Mifflin Harcourt Publishing Company 10 U x Z 16 Multiply both sides by 10. Y Simplify. 9 V (Lesson 17.2) Use a straightedge to connect points B and H. SUPPORTING STUDENT REASONING • The GPS coordinates of the vertices of the Bermuda Triangle: Students can research these. The GPS coordinates for Miami are (25.789106, –80.226529), for San Juan are (18.4663188, –66.1057427), and for Hamilton are (32.294887, –64.781380). X Substitute. Given the directed line segment from A to B, construct the point P that divides the segment in the ratio 2 to 3 from A to B. → ‾ . Use a straightedge to draw AC → ‾ . Place a compass on point A and draw an arc through AC Label the intersection D. Continue this for intersections D through H. COMMON CORE Students should begin this problem by focusing on what information they will need. Here are some issues they might bring up. Write a proportion. C H G F E D A B P Construct angles congruent to ∠AHB with points D through G. _ AB is partitioned into five equal parts. Label point P at the point that divides the segment in the ratio 2 to 3 from A to B. KEY EXAMPLE (Lesson 17.3) A 5.8-foot-tall man is standing next to a basketball hoop that casts an 11.2-foot shadow. The man’s shadow is 6.5 feet long. How tall is the basketball hoop? Let x be the height of the basketball hoop. x =_ 5.8 _ Write a proportion. 11.2 6.5 5.8 (11.2) x = ___ Multiply both sides by 11.2. 6.5 x ≈ 10 ( ) Module 17 925 Study Guide Review SCAFFOLDING SUPPORT IN2_MNLESE389847_U7M17MC 925 • For this region of the globe, assume that each unit of latitude or longitude is equal to 100 km. • One way to find the area is to enclose the triangle shown in the figure in a rectangle then find the area of the rectangle, and subtract the areas in the rectangle that are outside of the triangle. • Students can also use the area formula for a triangle with side lengths a, b a+b+c and c given by A = √s(s - a)(s - b)(s - c) , where s = _________. 2 ―――――――― 18/04/14 10:23 PM EXERCISES Find the missing lengths. (Lesson 17.1) 4.5 A 1. _ BG = SAMPLE SOLUTION B 3.5 E 8 F G 5D 15 7 Find the unknown length. (Lesson 17.3) A 5.9-foot-tall-man stands near a 12-foot statue. The man places a mirror on the ground a certain distance from the base of the statue, and then stands another 7 feet from the mirror to see the top of the statue in it. How far is the mirror from the base of the statue? 14.2 feet 4. C 3. G F E D A Find the lengths. (Lesson 17.4) 9 x 6. x = 16.2 7. ――――――― A 45-foot flagpole casts a 22-foot shadow. At the same time of day, a woman casts a 2.7-foot shadow. How tall is the woman? 5.5 feet 5. B P Find the GPS coordinates of the three locations and write them as (x, y) = (latitude, longitude). Then use the distance formula, ――― d = (x 2 - x 1) 2 + (y 2 - y 1) 2 , to find the lengths of the three sides, a, b, and c, of the triangle. To find the distances, multiply the results by 100 km. Finally, use the area formula for a triangle, a+b+c A = √s(s - a)(s - b)(s - c) where s = _________, 2 to determine the area. C _1 _ CE = 10 2 2. Given the directed line segment from A to B, construct the point P that divides the segment in the ratio 3 to 1 from A to B. (Lesson 17.2) Methodology: 7 Coordinates: Miami: (-80.226529, 25.789106); San Juan: (-66.1057427, 18.4663188); Hamilton: (-64.781380, 32.294887) 20 y z y = 13.4 8. Using the distance formula and multiplying the results by 100 km gives the following distances. z = 24.1 Miami to San Juan: 1590 km San Juan to Hamilton: 1389 km Hamilton to Miami: 1676 km MODULE PERFORMANCE TASK The boundaries of the Bermuda Triangle are not well defined, but the region is often represented as a triangle with vertices at Miami, Florida; San Juan, Puerto Rico; and Hamilton, Bermuda. The distance between Miami and San Juan is about 1,034 miles. What is the approximate area of this region? One tool that you may find helpful in solving this problem is the similar triangle shown here with angle measures labeled. Use your own paper to complete the task. Be sure to record all your data and assumptions. Then use graphs, diagrams, words, or numbers to explain how you reached your conclusion. H 61° M 50° 69° S © Houghton Mifflin Harcourt Publishing Company How Large Is the Bermuda Triangle? Calculate s, s - a, s - b, and s - c. s= 1590 + 1389 + 1676 __ = 2327.5 2 s - a = 2327.5 - 1590 = 737.5 s - b = 2327.5 - 1389 = 938.5 s - c = 2327.5 - 1676 = 651.5 Find the area: ――――――― ―――――――――― = √2327.5(737.5)(938.5)(651.5) A= Module 17 926 ≈ 1, 024, 472 km 2 Study Guide Review DISCUSSION OPPORTUNITIES IN2_MNLESE389847_U7M17MC 926 √s(s - a)(s - b)(s - c) 18/04/14 10:23 PM • If students choose to add a height of the triangle to the drawing they are provided, and then use the familiar area of a triangle formula, how does this affect the accuracy of their calculations? • How does the fact that the Bermuda Triangle is not actually a triangle in the plane, but a triangle on a sphere, affect the accuracy of the calculations? Is the actual area more or less than the one calculated? Assessment Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain. 0 points: Student does not demonstrate understanding of the problem. Study Guide Review 926 Ready to Go On? Ready to Go On? 17.1–17.4 Using Similar Triangles ASSESS MASTERY Use the assessment on this page to determine if students have mastered the concepts and standards covered in this module. • Online Homework • Hints and Help • Extra Practice Find the missing lengths. (Lesson 17.1) B 1. 5 ASSESSMENT AND INTERVENTION A 3 2. 7 D E 15 y x 20 12 z G x = 4.2 z = 16 ; y= 9 Given the directed line segment from A to B, construct the point P that divides the segment in the given ratio from A to B. (Lesson 17.2) C 3. 3 to 4 G Access Ready to Go On? assessment online, and receive instant scoring, feedback, and customized intervention or enrichment. D A • Reading Strategies • Success for English Learners • Challenge Worksheets Assessment Resources B P 6.25 feet © Houghton Mifflin Harcourt Publishing Company Differentiated Instruction Resources F The height of a street light is 25 feet. It casts a 12-foot shadow. At the same time, a man standing next to the street light casts a 3-foot shadow. How tall is the man? ADDITIONAL RESOURCES • Reteach Worksheets I Find the missing height. (Lesson 17.3) 4. Response to Intervention Resources E H ESSENTIAL QUESTION How can you use similar triangles to find the missing parts of a triangle? You can use proportions or corresponding parts of a triangle to find the missing parts of a triangle that is similar to another. 5. • Leveled Module Quizzes Module 17 COMMON CORE IN2_MNLESE389847_U7M17MC 927 927 Module 17 Study Guide Review 927 Common Core Standards 18/04/14 10:23 PM Content Standards Mathematical Practices Lesson Items 17.1, 17.3 1 G-SRT.B.4, G-SRT.B.5, G-CO.C.10 MP.7 17.1, 17.3, 17.4 2 G-SRT.B.4, G-SRT.B.5, G-CO.C.10 MP.7 17.2 3 G-CO.D.12, G-GPE.B.6 MP.5 17.1, 17.3, 17.4 4 G-SRT.B.4, G-SRT.B.5, G-CO.C.10 MP.4 MODULE MODULE 17 MIXED REVIEW MIXED REVIEW Assessment Readiness Assessment Readiness SELECTED RESPONSE 1. △XYZ is given by the points X(-1, −1), Y(3, 5), and Z(5, 1). Consider each of the points below. Is each point a vertex of the image under the transformation (x, y) → (x + 3, y - 2) → Select Yes or No for A–C. ( ASSESSMENT AND INTERVENTION ) 1 _ x, y → (y, -x)? 2 A. X‴ (-3, -1) B. Y‴ (3, -3) C. Z‴ (-1, -4) Yes Yes Yes No No No 2. Which of the following statements are true about the triangle at the right? Choose True or False for each statement. A. The value of x is 15. True False B. The value of y is 12. True False C. The value of y is 16. True False Assign ready-made or customized practice tests to prepare students for high-stakes tests. 28 30 Y 7 ADDITIONAL RESOURCES X 3. △ABC is given by the points A(-1, 2), B(2, 5), and C(4, -1). What is (2 2) Assessment Resources 1 _ the point _ , 5 ? Explain what this means. • Leveled Module Quizzes: Modified, B Possible Answer: The orthocenter; When the altitude is drawn from each of the three vertices of the triangle, the point they 1 _ ,5 . intersect at is _ 2 2 ( ) COMMON CORE AVOID COMMON ERRORS Item 1 Some students will stop working on this problem too early, not completing all three transformations. Encourage students to double-check the problem to make sure they have completely finished before making their answer choices. © Houghton Mifflin Harcourt Publishing Company 4. Given the directed segment from A to B, construct I C H the point P that divides the segment in the ratio 1 G F to 5 from A to B. Explain your process and how it E D relates to similar triangles. → A B ‾ . Use a compass to draw 6 arcs Draw AC P → ‾ and label (equidistant spacing) through AC each intersection D through I. Connect the points B and I. Construct angles ¯ is partitioned into 6 congruent to ∠AIB for the remaining intersections. AB equal parts. Label the point P that divides the segment in the ratio 1 to 5 from A to B. Each triangle created is a similar triangle to △AIB. Module 17 17 Study Guide Review 928 Common Core Standards IN2_MNLESE389847_U7M17MC 928 18/04/14 10:23 PM Content Standards Mathematical Practices Lesson Items IM1 17.1 1* G-SRT.B.5, G-CO.A.5 MP.6 17.4 2 G-SRT.C.8 MP.6 15.3 3* G-CO.D.12, G-CO.C.10, G-GPE.B.5 MP.6 17.2 4 G-GPE.B.6, G-CO.D.12 MP.5 * Item integrates mixed review concepts from previous modules or a previous course. Study Guide Review 928 MODULE 18 18 MODULE Trigonometry with Right Triangles Trigonometry with Right Triangles Essential Question: How can you use ESSENTIAL QUESTION: trigonometry with right triangles to solve real-world problems? Answer: Trigonometric ratios allow you to find the side length of a right triangle given an angle measure, or vice versa. This can be useful whenever a triangular shape appears in the real world, such as in a metal bracket or a sculpture. LESSON 18.1 Tangent Ratio LESSON 18.2 Sine and Cosine Ratios LESSON 18.3 Special Right Triangles LESSON 18.4 Problem Solving with Trigonometry This version is for Algebra 1 and PROFESSIONAL DEVELOPMENT Geometry only VIDEO LESSON 18.5 Using a Pythagorean Identity Author Juli Dixon models successful teaching practices in an actual high-school classroom. Professional Development my.hrw.com © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©artzenter/ Shutterstock Professional Development Video REAL WORLD VIDEO Check out how right triangle trigonometry is used in real-world warehouses to minimize the space needed for items being shipped or stored. MODULE PERFORMANCE TASK PREVIEW How Much Shorter Are Staggered Pipe Stacks? In this module, you will investigate how much space can be saved by stacking pipes in a staggered pattern rather than directly on top of each other. How can trigonometry help you find the answer to this problem? Get prepared to discover the “staggering” results! Module 18 DIGITAL TEACHER EDITION IN2_MNLESE389847_U7M18MO 929 Access a full suite of teaching resources when and where you need them: • Access content online or offline • Customize lessons to share with your class • Communicate with your students in real-time • View student grades and data instantly to target your instruction where it is needed most 929 Module 18 929 PERSONAL MATH TRAINER Assessment and Intervention Assign automatically graded homework, quizzes, tests, and intervention activities. Prepare your students with updated, Common Core-aligned practice tests. 18/04/14 11:08 PM Are YOU Ready? Are You Ready? Complete these exercises to review skills you will need for this module. ASSESS READINESS Angle Relationships Example 1 Find the angle complementary to the given angle. 75° x + 75° = 90° Use the assessment on this page to determine if students need strategic or intensive intervention for the module’s prerequisite skills. • Online Homework • Hints and Help • Extra Practice Write as an equation. x = 90° − 75° Solve for x. x = 15° Find the complementary angle. ASSESSMENT AND INTERVENTION 70° 1. 20° 2. 35° 55° 3. 67° 23° Find the supplementary angle. 4. 80° 5. 65° 6. 34° 100° 115° 3 2 146° 1 Find the remaining angle or angles for △ABC. m∠A = 50°, m∠B = 40° m∠C = 90° 8. m∠A = 60°, m∠C = 20° m∠B = 100° 9. m∠B = 70° and ∠A ≅ ∠C Personal Math Trainer will automatically create a standards-based, personalized intervention assignment for your students, targeting each student’s individual needs! m∠A = 55°, m∠C = 55° m∠A = 60°, m∠B = 60°, m∠C = 60° 10. ∠A ≅ ∠B ≅ ∠C 1 m∠C 11. m∠B = 30° and m∠A = _ 2 m∠A = 50°, m∠C = 100° 12. △ABC is similar to △DEF and m∠D = 70° and m∠F = 50° m∠A = 70°, m∠B = 60°, m∠C = 50° 13. △ABC is similar to △PQR and m∠R = 50° and ∠P ≅ ∠Q 14. m∠A = 45° and m∠B = m∠C 15. m∠B = 105° and m∠A = 2 · m∠C 16. m∠A = 5° and m∠B = 9 · m∠C Module 18 IN2_MNLESE389847_U7M18MO 930 Tier 1 Lesson Intervention Worksheets Reteach 18.1 Reteach 18.2 Reteach 18.3 Reteach 18.4 Reteach 18.5 m∠A = 65°, m∠B = 65°, m∠C = 50° m∠B = 67.5°, m∠C = 67.5° m∠A = 50°, m∠C = 25° © Houghton Mifflin Harcourt Publishing Company 7. TIER 1, TIER 2, TIER 3 SKILLS ADDITIONAL RESOURCES See the table below for a full list of intervention resources available for this module. Response to Intervention Resources also includes: • Tier 2 Skill Pre-Tests for each Module • Tier 2 Skill Post-Tests for each skill m∠B = 157.5°, m∠C = 17.5° 930 Response to Intervention Tier 2 Strategic Intervention Skills Intervention Worksheets Tier 3 Intensive Intervention Worksheets available online 38 Angle Relationships 43 The Pythagorean Theorem 46 Proportional Relationships Building Block Skills 7, 15, 16, 38, 46, 53, 56, 63, 66, 82, 90, 95, 98, 100, 102 Differentiated Instruction 18/04/14 11:08 PM Challenge worksheets Extend the Math Lesson Activities in TE Module 18 930 LESSON 18.1 Name Tangent Ratio Class Date 18.1 Tangent Ratio Essential Question: How do you find the tangent ratio for an acute angle? Common Core Math Standards The student is expected to: COMMON CORE Resource Locker G-SRT.C.6 Explore Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. Also G-SRT.C.8 Mathematical Practices COMMON CORE Investigating a Ratio in a Right Triangle In a given a right triangle, △ABC, with a right angle at vertex C, there are three sides. The side adjacent to ∠A is the leg that forms one side of ∠A. The side opposite ∠A is the leg that does not form a side of ∠A. The side that connects the adjacent and opposite legs is the hypotenuse. B Hypotenuse MP.4 Modeling Language Objective Explain to a partner how to find the tangent of an angle given a diagram of a right triangle with given angle measure and leg lengths. A Measurements may vary slightly. opposite 2.5 cm 40° 3.1 cm adjacent F © Houghton Mifflin Harcourt Publishing Company View the Engage section online. Discuss the photograph, asking students to describe the forces that might create the shape of a sand dune. Then preview the Lesson Performance Task. C E Essential Question: How do you find the tangent ratio for an acute angle? PREVIEW: LESSON PERFORMANCE TASK Leg adjacent to ∠A In △DEF, label the legs opposite and adjacent to ∠D. Then measure the lengths of the legs in centimeters and record their values in the rectangles provided. ENGAGE For an acute angle in a right triangle, the tangent ratio is the ratio of the length of the opposite leg to the length of the adjacent leg. Leg opposite to ∠A D What is the ratio of the opposite leg length to the adjacent leg length, rounded to the nearest hundredth? EF ≈ 0.81 Ratios may vary slightly depending on measurements. _ DF Using a protractor and ruler, draw right triangle △JKL with a right angle at vertex L and ∠J = 40° so that △JKL ~ △DEF. Label the opposite and adjacent legs to ∠J and include their measurements. Drawings will vary but should be similar to the triangle in Step A. For all triangles, it is true that opposite length < adjacent length. What is the ratio of the opposite leg length to the adjacent leg length, rounded to the nearest hundredth? KL ≈ _ 0.81 JL Module 18 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction Lesson 1 931 gh “File info” made throu Date Class ent Ratio 18.1 Tang Name angle? an acute nt ratio for the tange angles in ties of the do you find ion: How es are proper G-SRT.C.8 in right triangl . Also ity, side ratios ratios for acute angles by similar metric tand that of trigono G-SRT.C.6 Unders ngle to definitions e, leading a Right Tria the triangl a Ratio in are three sides. The side g atin does C, there Investig the leg that angle at vertex ite ∠A is nuse. Explore , with a right ∠A. 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Label the Using a protra that △JKL ~ △DEF A. For all 40° so le in Step ts. and ∠J = the triang measuremen similar to include their h. should be lengt but ent will vary h < adjac Drawings site lengt that oppo , it is true triangles, nt leg length to the adjace leg length opposite ratio of the What is the t hundredth? to the neares rounded KL ≈ 0.81 _ JL Module 18 8L1 931 47_U7M1 ESE3898 IN2_MNL 931 Lesson 18.1 Lesson 1 931 18/04/14 11:19 PM 18/04/14 11:20 PM Reflect 1. EXPLORE Discussion Compare your work with that of other students. Do all the triangles have the same angles? Do they all have the same side lengths? Do they all have the same leg ratios? Summarize your findings. The triangles are all similar, so they have angles that have the same Investigating a Ratio in a Right Triangle measure but not the same side lengths. Yet, for a given angle, the ratio of the length of the opposite leg to the length of the adjacent leg is constant. 2. INTEGRATE TECHNOLOGY If you repeated Steps A–D with a right triangle having a 30° angle, how would your results be similar? How would they be different? The triangles would all be similar, with the ratio of the length of the leg Students have the option of doing the Explore activity either in the book or online. opposite the 30° angle to the length of the leg adjacent to the 30° angle constant, but the actual value of the ratio would be different from 0.81. CONNECT VOCABULARY Finding the Tangent of an Angle Explain 1 Draw and label a variety of right triangles with different orientations. For each of the triangles, have students name the hypotenuse, the leg adjacent to a given angle, and the leg opposite the angle. The ratio you calculated in the Explore section is called the tangent of an angle. The tangent of acute angle A, written tan ∠A, is defined as follows: length of leg opposite ∠A tan A = ___ length of leg adjacent to ∠A You can use what you know about similarity to show why the tangent of an angle is constant. By the AA Similarity Postulate, given ∠ D ≅ ∠J and also ∠ F ≅ ∠L, then △DEF ~ △JKL. This means the lengths of the sides of △JKL are each the same multiple, k, of the lengths of the corresponding sides of △DEF. Substituting into the tangent equation shows that the ratio of the length of the opposite leg to the length of the adjacent leg is always the same value for a given acute angle. E 40° F tangent defined for specified angle △DEF D J Find the tangent of each specified angle. Write each ratio as a C fraction and as a decimal rounded to the nearest hundredth. ∠A 18 ∠B B length of leg opposite ∠ A 18 = _ 3 = 0.75 tan A = ___ = _ 4 24 length of leg adjacent to ∠ A 4 24 length of leg opposite ∠ B tan B = ___ = _ = _ ≈ 1.33 3 length of leg adjacent to ∠ B 18 Module 18 932 24 A © Houghton Mifflin Harcourt Publishing Company ⎧ ⎪ Example 1 40° ⎪ k∙DF ⎨ L The ratio of the length of the opposite leg to the adjacent leg of an angle of a right triangle is constant for any right triangle with the same angle measures. What does this imply about right triangles with that given angle measure? Explain. The triangles must be similar because the angle measures are constant and the ratios of the sides are proportional. △JKL leg opposite ∠ 40° k·EF = _ KL = _ EF EF = _ tan 40° = __ = _ JL k·DF DF DF leg adjacent to ∠ 40° k∙DE k∙EF ⎪ ⎪ ⎩ K QUESTIONING STRATEGIES AVOID COMMON ERRORS If students use calculators to compute the tangent ratio of an angle, remind them to check that their calculators are in degree mode. Otherwise, they will not find the correct value. 30 Lesson 1 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M18L1 932 Learning Progressions Students work with three trigonometric ratios in this course: tangent, sine, and cosine. The tangent is treated separately for two reasons. First, working with a single ratio gives students a chance to focus on the conceptual underpinnings of trigonometry. That is, right triangles with a given acute angle, ∠A, are all similar, so the ratio of the length of the leg opposite ∠A to the length of the leg adjacent to ∠A is constant for all such triangles. Second, students sometimes have difficulty deciding which trigonometric ratio to use when solving a problem. By working with tangents first, students can instead focus on the general process for solving a problem using trigonometry. 18/04/14 11:20 PM EXPLAIN 1 Finding the Tangent of an Angle CONNECT VOCABULARY The abbreviation tan is read tangent. For example, tan 30° is read as tangent of 30 degrees. Tangent Ratio 932 Reflect QUESTIONING STRATEGIES 3. If you know the tangent ratio for an acute angle of a right triangle, and the length of one of the legs, can you reconstruct the triangle? Explain. Yes; you can use the ratio to find the length of the other leg, and then use the Pythagorean Theorem to find the length of the hypotenuse. What is the relationship between the ratios for tan A and tan B? Do you believe this relationship will be true for acute angles in other right triangles? Explain. The ratios are reciprocals of each other. This will always be true because the opposite side of one acute angle is the adjacent side of the other acute angle, and vice versa. 4. Why does it not make sense to ask for the value of tan L? The tangent ratio is defined only for the acute angles of a right triangle, not for the right angle. You can use a table of values to find the tangent of a given acute angle. How do you think the tables were compiled? by constructing right triangles with the given angle measure and its complement, measuring the opposite and adjacent side lengths, and then computing the tangent ratio Your Turn Find the tangent of each specified angle. Write each ratio as a fraction and as a decimal rounded to the nearest hundredth. 5. ∠Q 6. 15 5 tan ∠ Q = _ = _ ≈ 0.42 36 12 39 S 15 R 36 12 tan ∠ R = _ = _ = 2.4 5 15 When you know the length of a leg of a right triangle and the measure of one of the acute angles, you can use the tangent to find the length of the other leg. This is especially useful in real-world problems. Apply the tangent ratio to find unknown lengths. Example 2 Finding a Side Length using Tangent In order to meet safety guidelines, a roof contractor determines that she must place the base of her ladder 6 feet away from the house, making an angle of 76° with the ground. To the nearest tenth of a foot, how far above the ground is the eave of the roof? C © Houghton Mifflin Harcourt Publishing Company QUESTIONING STRATEGIES What if you want to find the tangent of the other acute angle in the triangle? How would you find the angle? When might this be necessary? The measure of the angle is the complement of the given angle. You might want to do this if the missing side is in the denominator of the tangent ratio. ∠R 36 Finding a Side Length using Tangent Explain 2 EXPLAIN 2 Why is it important to draw and label a diagram when solving a real-world problem about relationships in a right triangle? The diagram makes it possible to identify values and relationships that you can use to solve the problem. Q A 76° 6 ft B Step 1 Write a tangent ratio that involves the unknown length. length of leg opposite ∠A BC tan A = ___ = _ BA length of leg adjacent to ∠A Step 2 Identify the given values and substitute into the tangent equation. Given: BA = 6 ft and m∠A = 76° BC Substitute: tan 76° = _ 6 Module 18 933 Lesson 1 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M18L1 933 Peer-to-Peer Activity Provide pairs of students with graph paper and ask each student to draw two different right triangles, with unit side lengths for the legs. Have them use a protractor to estimate one of the acute angles in one triangle, then have pairs work together to find the tangent. Then, have them choose an acute angle in the remaining triangle and find the inverse tangent of the angle. For additional work, have students exchange triangles with other pairs. 933 Lesson 18.1 18/04/14 11:20 PM Step 3 Solve for the unknown leg length. Be sure the calculator is in degree mode and do not round until the final step of the solution. Multiply each side by 6. BC 6 ·_ 6 · tan 76° = _ 1 6 Use a calculator to find tan 76°. 6 · tan 76° = BC Substitute this value in for tan 76°. 6(4.010780934) = BC Multiply. Round to the nearest tenth. 24.1 ≈ BC AVOID COMMON ERRORS x, When solving an equation of the form tan 35° = ___ 12 students may forget to multiply by 12 to solve for x after they find tan 35°. These students may benefit from rewriting the equation as 12 ∙ tan 35° = x before evaluating tan 35°. So, the eave of the roof is about 24.1 feet above the ground. B For right triangle △STU, what is the length of the leg adjacent to ∠S? S 54° Step 1 Write a tangent ratio that involves the unknown length. EXPLAIN 3 TU length of leg opposite ∠ S tan S = ___ = _ length of leg adjacent to ∠ S SU T 87 U Finding an Angle Measure using Tangent Step 2 Identify the given values and substitute into the tangent equation. Given: TU = 87 and m∠S = 54 ° 87 Substitute: tan 54 ° = _ SU Step 3 Solve for the unknown leg length. QUESTIONING STRATEGIES In the equation y = tan -1x, explain what x and y represent. In the equation, x represents the tangent of the angle with measure y°. 87 Multiply both sides by SU, then divide both sides by 54°. SU = __ tan 54 ° 87 Use a calculator to find 54° and substitute. SU ≈ __ Divide. Round to the nearest tenth. SU ≈ 63.2 © Houghton Mifflin Harcourt Publishing Company Your Turn 7. A ladder needs to reach the second story window, which is 10 feet above the ground, and make an angle with the ground of 70°. How far out from the building does the base of the ladder need to be positioned? 10 10 10 _____ __ 3.6 ft; tan 70° = __ x , so x = tan 70° ≈ 2.747477419 ≈ 3.6 Explain 3 Finding an Angle Measure using Tangent In the previous section you used a given angle measure and leg measure with the tangent ratio to solve for an unknown leg. What if you are given the leg measures and want to find the measures of the acute angles? If you know the tan A, read as “tangent of ∠A,” then you can use the tan -1 A, read as “inverse tangent of ∠A,” to find m∠A. So, given an acute angle ∠A, if tan A = x, then tan -1 x = m∠A. Module 18 If tan x = 1, what is the measure of angle x? Explain. The measure of angle x must be 45° because the legs of the triangle must be the same length to make the tangent ratio 1. 1.37638192 934 How could you evaluate the inverse tangent of an angle without using a calculator’s inverse tangent key or a table of values? Use the measurements of the tangent ratio as the legs to draw the corresponding right triangle. Then use a protractor to measure the angle formed by the hypotenuse and the adjacent side to estimate the angle measure. Lesson 1 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M18L1 934 Kinesthetic Experience 18/04/14 11:20 PM To help students identify the opposite and adjacent legs, model a right triangle on the floor with masking tape. Have a student stand at one acute angle and walk to the leg next to, or touching, that angle. That is the adjacent leg. Then have the student walk from the acute angle to the leg opposite, or across from, the angle. This is the opposite leg. Tangent Ratio 934 Example 3 INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Discuss why students will get more accurate 19 in. What is m∠A? A () 1 results if they evaluate an expression like tan -1 __ 3 rather than finding a decimal approximation for the 1 rounding, and evaluating tan -1(0.33). fraction __ 3, ELABORATE SUMMARIZE THE LESSON How do you find the tangent ratio for an acute angle of a right triangle? Find the ratio of the length of the leg opposite the angle to the length of the leg adjacent to the angle. 935 Lesson 18.1 Step 2 Write the inverse tangent equation. Step 3 Evaluate using a calculator and round as indicated. 19 tan A = _ 36 19 tan -1 _ = m∠ A 36 m∠ A ≈ 27.82409638 ≈ 28° Step 2 Write the inverse tangent equation. 36 tan -1 tan B = _____ 36 _____ = m∠ B 19 19 Step 3 Evaluate using a calculator and round as indicated. ° ° m∠ B ≈ 62.17590362 ≈ 62 Your Turn AVOID COMMON ERRORS 8. J Find m∠J. 46 26°, m∠J = tan -1 __ ≈ 26.31808814 93 K 93 46 © Houghton Mifflin Harcourt Publishing Company How are tangent and inverse tangent related? The tangent of an acute angle of a right triangle is the ratio of the lengths of the opposite side to the adjacent side of the right triangle. The inverse tangent is the angle with that tangent ratio. C Step 1 Write the tangent ratio for ∠ A using the known values. Step 1 Write the tangent ratio for ∠ B using the known values. Have students graph y = tan x on their graphing calculators on the interval [0, 90°]. Discuss how to interpret the graph, especially as x nears 90°. QUESTIONING STRATEGIES 36 in. What is m∠B? INTEGRATE TECHNOLOGY Some students may have the misconception that the opposite leg is always a vertical side of the triangle and the adjacent leg is always a horizontal side of the triangle. Remind students that the opposite and adjacent sides are determined by the location of the associated angle, not by the orientation of the triangle. B Find the measure of the indicated angle. Round to the nearest degree. L Elaborate 9. Explain how to identify the opposite and adjacent legs of a given acute angle. The opposite leg does not form a side of the given angle. The adjacent leg forms a side of the given angle and is not the hypotenuse. 10. Discussion How does tan A change as m∠A increases? Explain the basis for the identified relationship. Tan A increases as m∠A increases. As m∠A increases, the length of the leg adjacent to ∠A remains the same and the length of the leg opposite ∠A increases. Since the length of the opposite leg is the numerator of the tangent ratio, and the denominator of that ratio does not change, the ratio increases. Module 18 935 Lesson 1 LANGUAGE SUPPORT IN2_MNLESE389847_U7M18L1 935 Connect Vocabulary 18/04/14 11:20 PM Have students look up the word inverse in the dictionary. A definition may include reversed in position. Discuss how this corresponds to the relationship between the inverse tangent and the tangent of an angle. 11. Essential Question Check-In Compare and contrast the use of the tangent and inverse tangent ratios for solving problems. The tangent and inverse tangent both apply only to acute angles in right triangles. When the leg lengths are known, the tangent can be used to solve for the constant ratio. Whereas, the inverse tangent can be EVALUATE • Online Homework • Hints and Help • Extra Practice used to solve for the associated acute angle. When the measures of one leg and one angle are known, the tangent can be used to solve for the unknown leg whereas the inverse tangent is not applicable. ASSIGNMENT GUIDE Evaluate: Homework and Practice In each pair of right triangles, measure the lengths of the adjacent side and the opposite side of the angle specified in the figure. Then calculate and compare the ratios. 1. In each triangle, measure the length of the adjacent side and the opposite side of the 22° angle. Then calculate and compare the ratios. 22° 22° Opposite Opposite ______ = 0.40; ______ = 0.40; the ratios are the same. Adjacent Adjacent Concepts and Skills Practice Explore Investigating a Ratio in a Right Triangle Exercise 1 Example 1 Finding the Tangent of an Angle Exercises 2–8 Example 2 Finding a Side Length using Tangent Exercises 9–14, 18–19 Example 3 Finding an Angle Measure using Tangent Exercises 15–17, 20–21 In each right triangle, find the tangent of each angle that is not the right angle. 2. 3. A E 5.0 COMMUNICATING MATH F 10 9.0 C 8 B 10.3 D 8 tan ∠A = __ 6 6 tan ∠B = _ 8 5 tan ∠D = __ 9 tan ∠A = 1.33 tan ∠B = 0.75 tan ∠D = 0.56 tan ∠F = 1.8 4. B 32 81.5 A 75 5. C 9 tan ∠F = _ 5 19 Q 3 P R 19.2 32 tan ∠A = ___ 75 75 tan ∠B = __ 32 19 tan ∠P = ___ 3 3 tan ∠R = __ 19 tan ∠A = 0.43 tan ∠B = 2.34 tan ∠P = 6.33 tan ∠R = 0.16 Module 18 Exercise IN2_MNLESE389847_U7M18L1 936 INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Discuss rounding errors that occur when finding the tangent and inverse tangent of an angle. Have students find the tangent of an angle on their calculators. Then have them use the value to find the inverse tangent for that tangent. Compare results. Repeat with several exercises. Have students also start with the inverse tangent. Lesson 1 936 Depth of Knowledge (D.O.K.) Point out that the hypotenuse of a right triangle will never be the opposite or the adjacent side for the tangent ratio. Discuss why this must be so. © Houghton Mifflin Harcourt Publishing Company 6 COMMON CORE Mathematical Practices 1–4 1 Recall of Information MP.6 Precision 5–17 2 Skills/Concepts MP.5 Using Tools 18–22 2 Skills/Concepts MP.2 Reasoning 23–26 2 Skills/Concepts MP.4 Modeling 18/04/14 11:20 PM Tangent Ratio 936 Let △ABC be a right triangle, with m∠C = 90°. Given the tangent of one of the complementary angles of the triangle, find the tangent of the other angle. AVOID COMMON ERRORS 6. Students may confuse the opposite and the adjacent sides of a given angle in a right triangle. Suggest that students use one color to highlight the opposite side and another color to highlight the adjacent side of a given acute angle. Emphasize that the adjacent side forms the given angle’s ray that is not the hypotenuse of the triangle. tan ∠A = 1.25 1 tan ∠B = ___ 1.25 7. tan ∠B = 0.80 tan ∠B = 0.50 1 tan ∠A = ___ 0.50 8. tan ∠A = 2.0 tan ∠B = 1.0 1 tan ∠A = ___ 1.0 tan ∠A = 1.0 Use the tangent to find the unknown side length. 9. Find QR. 10. Find AC. Q A R C 11. Find PQ. P 7.0 7.0 tan 60° = ___ QR 7.0 QR = _____ E 13 tan 85° 14. Find PR. P 45° 8.4 C R 54° 4.6 Q AB tan 45° = ___ 8.4 DE AB = tan 45°(8.4) PR tan 54° = ___ 4.6 tan 21° AB = 8.4 PR = 6.3 21° 13 tan 21° = ___ DE = 9 PQ = _____ PQ = 0.79 A B D 9 tan 85° = __ PQ AC = 4.1 13. Find AB. F R 85° B AC = tan 27°(7.0) QR = 4.0 12. Find DE. 27° 8.0 AC tan 27° = ___ 8.0 tan 60° 9 Q P 60° 13 _____ © Houghton Mifflin Harcourt Publishing Company DE = 33.9 PR = tan 54°(4.6) Module 18 IN2_MNLESE389847_U7M18L1 937 937 Lesson 18.1 937 Lesson 1 18/04/14 11:20 PM Find the measure of the angle specified for each triangle. Use the inverse tangent (tan -1) function of your calculator. Round your answer to the nearest degree. 15. Find ∠A. 16. Find ∠R. A C B R 9 B 6.8 For right triangles with unit side lengths for the legs, when they are finding the inverse tangent, students may benefit from drawing the right triangle on graph paper and then using a protractor to estimate the acute angle measure. They can also do this to check their work. 17. Find ∠B. 24 Q 3.0 MANIPULATIVES 17 P 6.8 tan -1 ___ = m∠A 3.0 9 tan -1 __ = m∠R 24 m∠A = 66° m∠R = 21° A C 16 16 tan -1 B = __ 17 AVOID COMMON ERRORS m∠B = 43° Call attention to the notation used to represent the inverse tangent: tan -1x. The expression tan -1x is a short way to indicate “the angle whose tangent ratio is x.” Be sure students understand that the raised –1 is not an exponent and that tan -1x does not 1 mean ____ tanx . Emphasize that when students find the inverse tangent, the result will be an angle with a degree measure. -1 Write an equation using either tan or tan to express the measure of the angle or side. Then solve the equation. 18. Find BC. B 40° 10 38 Q P A 20. Find ∠A and ∠C. 19. Find PQ. R 14.5 A B 75° 14.5 38 PQ = _____ = 10.2 tan 75° C C ( ) BC = 10 tan 40° = 8.4 14.5 = 45° m∠A = m∠C = tan -1 A ___ 14.5 21. Multi-Step Find the measure of angle D. Show your work. B BC tan 30° = __ ; BC = 12 tan 30° 12 ( ) ( ) A 30° Module 18 IN2_MNLESE389847_U7M18L1 938 12 cm C 4 cm © Houghton Mifflin Harcourt Publishing Company 12 tan30° BC m∠D = tan -1 __ = tan -1 _______ = 60° 4 CD D 938 Lesson 1 18/04/14 11:20 PM Tangent Ratio 938 22. Engineering A client wants to build a ramp that carries 1.4 m people to a height of 1.4 meters, as shown in the diagram. What additional information is necessary to identify the measure of angle a, the angle the ramp forms with the horizontal? After the additional measurement is made, describe how to find the measure of the angle. Possible answer: Measure the distance x of the ramp across JOURNAL Have students summarize what they know about the tangent ratio. Remind them to include at least one figure and at least one example in their descriptions. a ( ) 1.4 the horizontal. Then find tan -1 ___ x , which equals the measure of angle a. 23. Explain the Error A student uses the triangle shown to 6.5 2.5 calculate a. Find and explain the student’s error. a 6.5 a = tan -1 _ = tan -1(2.6) The student’s calculations are correct only if the triangle 2.5 is a right triangle. a = 69.0° ( ) 24. When m∠A + m∠B = 90°, what relationship is formed by tan ∠A and tan ∠B? Select all that apply. 1 C. (tan∠A)(tan∠B) = 1 A. tan∠A = _ tanB B. tan∠A + tan∠B = 1 D. (tan∠A)(tan∠B) = -1 A, C; Tan A and tan B are reciprocals of each other so they have a product of 1. H.O.T. Focus on Higher Order Thinking 25. Analyze Relationships To travel from Pottstown to Cogsville, a man drives his car 83 miles due east on one road, and then 15 miles due north on another road. Describe the path that a bird could fly in a straight line from Pottstown to Cogsville. What angle does the line make with the two roads that the man used? © Houghton Mifflin Harcourt Publishing Company The bird’s path is the hypotenuse of the right triangle formed by the path of the man. The bird’s path forms an angle of 10.2° with the first road and an angle of 79.8° with the second road. 26. Critical Thinking A right triangle has only one 90° angle. Both of its other angles have measures greater than 0° and less than 90°. Why is it useful to define the tangent of 90° to equal 1, and the tangent of 0° to equal 0? Answers will vary. Possible answer: Consider a right triangle in which the complementary angles are very close to 0° and 90°. Let the length of the 0 hypotenuse be 1. The tangent of the small angle is very close to _ . The 1 1 tangent of the larger angle will be very close to _ . 1 Module 18 IN2_MNLESE389847_U7M18L1 939 939 Lesson 18.1 939 Lesson 1 18/04/14 11:20 PM Lesson Performance Task AVOID COMMON ERRORS When they form conical piles, granular materials such as salt, gravel, and sand settle at different “angles of repose,” depending on the shapes of the grains. One particular 13-foot tall cone of dry sand has a base diameter of 38.6 feet. When using the tangent function to find the angle of repose of a conical pile, students may mistakenly use the diameter of the base of the pile as the side adjacent to the angle of repose. The triangle they should use has the radius of the base as the side adjacent to the angle of repose and the height of the pile as the side opposite. angle of repose 1. To the nearest tenth of a degree, what is the angle of repose of this type of dry sand? 2. A different conical pile of the same type of sand is 10 feet tall. What is the diameter of the cone’s base? 3. Henley Landscaping Supply sells a type of sand with a 30° angle of repose for $32 per cubic yard. Find the cost of an 11-foot-tall cone of this type of sand. Show your work. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Are values of the tangent function 13 tan (x) = ____ 38.6 (___ 2 ) 13 1. tan (x) = ___ 19.3 ( ) 13 x = tan -1 ___ 19.3 x = 34° proportional? If they are, the tangent of 40°, for example, should be double the tangent of 20°. Use tangent tables or a calculator to investigate the question. Give examples to support your conclusion. Not proportional; example: tan 20° ≈ 0.36, but tan 40° ≈ 0.84 ≠ 2tan 20° 10 tan (34) = ___ (_2x ) 20 2. tan (34) = ___ x x = 29.7 3. Let r = radius of cone. Then: 11 tan 30° = __ r © Houghton Mifflin Harcourt Publishing Company 11 0.5774 = __ r r ≈ 19.1 ft Volume of cone: 1 V=_ Bh 3 1 2 =_ πr h 3 2 1( = __ 3.14)(19.1) (11) 3 = 4200.2 ft 3 1 yd 3 = 27 ft 3 so 4200.2 ft 3 = 4200.2 ÷ 27 ≈ 155.6 yd 3 At $32 per yd 3 55.6 yd 3 will cost 155.6 × 32 = $4979.20. Module 18 940 Lesson 1 EXTENSION ACTIVITY IN2_MNLESE389847_U7M18L1 940 Have students research the angle of repose of at least four granular substances. For each one they should draw an equilateral triangle showing a cross-section of a pile of the substance labeled with a base diameter of 20 feet, base angles measuring the angle of repose, and the correct height to the nearest tenth. 18/04/14 11:20 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Tangent Ratio 940 LESSON 18.2 Name Sine and Cosine Ratios Class Date 18.2 Sine and Cosine Ratios Essential Question: How can you use the sine and cosine ratios, and their inverses, in calculations involving right triangles? Common Core Math Standards The student is expected to: COMMON CORE Resource Locker G-SRT.C.6 Understand that by similarity, side ratios in right triangles are properties of the angles in the triangle, leading to definitions of trigonometric ratios for acute angles. Also G-SRT.C.7, G-SRT.C.8 Explore You can use geometry software or an online tool to explore ratios of side lengths in right triangles. Mathematical Practices COMMON CORE Investigating Ratios in a Right Triangle Construct three points A, B, and C. → → ‾ and AC ‾ . Move C so that ∠A is acute. Construct raysAB MP.4 Modeling _ Construct point D_ on AC. Construct a line through D perpendicular to AB. Construct _ point E as the intersection of the perpendicular line and AB. Language Objective Explain to a partner how to find the sine and cosine of an angle given a diagram of a right triangle with given angle measure and opposite or adjacent leg and hypotenuse lengths. Measure ∠A. Measure the side lengths DE, AE, and AD of △ADE. DE and _ AE . Calculate the ratios _ AD AD Essential Question: How can you use the sine and cosine ratios, and their inverses, in calculations involving right triangles? Given the measure of one acute angle and the length of the hypotenuse, you can use the sine and cosine ratios to find the lengths of each leg. Given the length of the hypotenuse and the length of a leg, you can use their inverses to find the measure of each acute angle. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photograph, asking students to speculate on what the person in the photo might be doing. Then preview the Lesson Performance Task. © Houghton Mifflin Harcourt Publishing Company ENGAGE Reflect 1. 2. 3. → → ‾ . What happens to m∠A as D moves along AC ‾ ? What postulate or Drag D along AC theorem guarantees that the different triangles formed are similar to each other? m∠A does not change; AA Similarity Postulate, given also that ∠AED remains a right angle. → DE and _ AE ? ‾ , what happens to the values of the ratios _ As you move D along AC AD AD Use the properties of similar triangles to explain this result. The ratios do not change; Given similar triangles △ADE, and △AD’E’, AD = _ DE →_ DE and _ AD = _ AE → _ AE’ = _ AE . D’E’ = _ _ AD AD’ D’E’ AD’ AE’ AD’ AD AD’ Move C. What happens to m∠A? With a new value of m∠A, note the values of → ‾ ? the two ratios. 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As you properties of simila Given similar triang AE . e; AE’ = _ Use the AE → _ _ do not chang AD AD’ AD The ratios DE and _ = AE’ of D’E’ = _ AD’ the values DE →_ note AD AD = _ _ → AD’ value of m∠A, AC ‾ ? dragged AD’ D’E’ With a new D along e as D is ns to m∠A? ratios if you drag do not chang C. What happe ns to the the ratios 3. Move ratios. What happe values of ; the new the two Lesson 2 es in value m∠A chang → ‾ . 941 along AC Publishin Reflect © Houghto n Mifflin Harcour t 1. Module 18 8L2 941 47_U7M1 ESE3898 IN2_MNL 941 Lesson 18.2 18/04/14 11:31 PM 18/04/14 11:33 PM Explain 1 Finding the Sine and Cosine of an Angle EXPLORE Trigonometric Ratios A trigonometric ratio is a ratio of two sides of a right triangle. You have already seen one trigonometric ratio, the tangent. There are two additional trigonometric ratios, the sine and the cosine, that involve the hypotenuse of a right triangle. The sine of ∠A, written sin A, is defined as follows: lenngth of leg opposite ∠A BC sin A = ___ = _ AB length of hypotenuse The cosine of ∠A, written cos A, is defined as follows: length of leg adjacent to ∠A AC cos A = ___ = _ AB length of hypotenuse B INTEGRATE TECHNOLOGY A C You can use these definitions to calculate trigonometric ratios. Example 1 Investigating Ratios in a Right Triangle Students have the option of doing the Explore activity either in the book or online. D Write sine and cosine of each angle as a fraction and as a decimal rounded to the nearest thousandth. ∠D 15 E length of leg opposite ∠D EF 8 sin D = ___ = _ = _ ≈ 0.471 DF 17 length of hypotenuse INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Have students examine what happens to the 17 8 F ratio of the opposite side length to the hypotenuse length as the acute angle gets closer to 90°. Repeat for the ratio of the adjacent side length to the hypotenuse length as the acute angle gets closer to 0°. length of leg adjacent to ∠D 15 ≈ 0.882 DE = _ cos D = ___ = _ 17 DF length of hypotenuse ∠F 15 length of leg opposite to ∠F DE = _ ≈ cos F = ___ = _ DF length of hypotenuse 0.882 17 8 QUESTIONING STRATEGIES 0.471 Reflect 4. What do you notice about the sines and cosines you found? Do you think this relationship will be true for any pair of acute angles in a right triangle? Explain. sin D = cos F and cos D = sin F; this relationship always holds because the leg opposite one acute angle is adjacent to the other one. 5. In a right triangle △PQR with hypotenuse 5, m∠Q = 90°, and PQ > QR, what are the values of sin P and cos P? sin P = cos P = _3 If the measure of the acute angle of the right triangle does not change but the side lengths of the triangle change, how do the ratios change? Explain. The values of the numerator and the denominators will change but the ratios are equal to the original ratios of opposite length to hypotenuse and adjacent length to hypotenuse because the triangles are similar. © Houghton Mifflin Harcourt Publishing Company length of leg adjacent to ∠F cos F = ___ = _ ≈ length of hypotenuse 17 5 _4 EXPLAIN 1 5 Module 18 942 Finding the Sine and Cosine of an Angle Lesson 2 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M18L2 942 Math Background Trigonometry is the branch of mathematics concerned with angle relationships in triangles. The ancient Egyptians used trigonometry to reset land boundaries after the Nile River flooded each year. The Babylonians used trigonometry to measure distances to nearby stars. Trigonometry is used in modern engineering, cartography, medical imaging, and many other fields. 18/04/14 11:33 PM AVOID COMMON ERRORS Students often use the wrong ratio for sine or cosine. Help students review these relationships by using flashcards, mnemonics, or other memory aids. Encourage students to research mnemonics or produce their own. Sine and Cosine Ratios 942 Explain 2 QUESTIONING STRATEGIES Using Complementary Angles The acute angles of a right triangle are complementary. Their trigonometric ratios are related to each other as shown in the following relationship. If you know only the sine for an acute angle of a right triangle, how could you find the cosine? The sine gives the ratio of the opposite side to the hypotenuse, so you can construct a right triangle with a hypotenuse and a leg that match the ratio. Then you can use the Pythagorean Theorem to find the length of the other leg, and use it to write the cosine ratio. Trigonometric Ratios of Complementary Angles If ∠A and ∠B are the acute angles in a right triangle, then sin A = cos B and cos A = sin B. Therefore, if θ (“theta”) is the measure of an acute angle, then sinθ° = cos (90 - θ)° and cos θ° = sin (90 - θ)°. (90 - θ°) A Do you think it is possible for the value of a sine or cosine to be greater than 1? Why or why not? It is not possible; because the hypotenuse is the longest side of a right triangle, any ratio that has the length of the hypotenuse as the denominator will be less than 1. θ° B C You can use these relationships to write equivalent expressions. Example 2 Write each trigonometric expression. Given that sin 38° ≈ 0.616, write the cosine of a complementary angle in terms of the sine of 38°. Then find the cosine of the complementary angle. Use an expression relating trigonometric ratios of complementary angles. sin θ° = cos(90 - θ)° EXPLAIN 2 Substitute 38 into both sides. sin 38° = cos(90 - 38)° Simplify. sin 38° = cos 52° Using Complementary Angles Substitute for sin 38°. 0.616 ≈ cos 52° QUESTIONING STRATEGIES Why do the sine and cosine have a complementary angle relationship? The ratios are for a right triangle. Since one angle must be a right angle, the other two angles must be complementary because the sum of the measures of the angles of a triangle is 180°. How can you write equivalent expressions for sin x° using cosine and cos y° using sine? Explain. Use the complement to write sin x° = cos (90° - x°) and cos y° = sin (90° - y°). How can the relationship between the sine and cosine of complementary angles help you solve equations involving sines and cosines? Apply the fact that the total measure of the angles is 90° to set up an equation to solve for unknown values. 943 Lesson 18.2 © Houghton Mifflin Harcourt Publishing Company So, the cosine of the complementary angle is about 0.616. Given that cos 60° = 0.5, write the sine of a complementary angle in terms of the cosine of 60°. Then find the sine of the complementary angle. Use an expression relating trigonometric ratios of complementary angles. cos θ° = sin(90 - θ)° ( Substitute 60 into both sides. cos 60 ° = sin 90 - 60 Simplify the right side. cos 60 ° = sin 30 ° Substitute for 60 °. 0.5 = sin 60 ° )° So, the sine of the complementary angle is 0.5. Module 18 943 Lesson 2 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M18L2 943 Small Group Activity 18/04/14 11:33 PM Have students work in small groups to investigate the relationship between the size of an angle and its sine, using geometry software or graph paper, rulers, and protractors. They should draw several triangles with unit side lengths and an angle that increases. Those using graph paper will need to measure with protractor and ruler. Have them determine how the sine changes as the angle size increases; then repeat for cosine. Have students share their results and how they drew their conclusions. Reflect COLLABORATIVE LEARNING 6. What can you conclude about the sine and cosine of 45°? Explain. sin 45° = cos 45°; 45° is complementary to itself. 7. Discussion Is it possible for the sine or cosine of an acute angle to equal 1? Explain. No; the hypotenuse of a right triangle is always longer than its legs, so the side ratios Have students construct right triangles with unit side lengths on graph paper, then use a protractor to measure each of the acute angles to the nearest degree. Have students verify the complementary relationship between the sine and cosine for their triangles for each of the acute angles, and then share their results with the class. defining sine and cosine must always be less than 1. Your Turn Write each trigonometric expression. 8. Given that cos 73° ≈ 0.454, write the sine of a complementary angle. EXPLAIN 3 θ = 73°, so 90 - θ = 27°. sin 27° ≈ 0.454 9. Finding Side Lengths using Sine and Cosine Given that sin 45° ≈ 0.707, write the cosine of a complementary angle. θ = 45°, so 90 - θ = 45°. cos 45° ≈ 0.707 Explain 3 INTEGRATE TECHNOLOGY Finding Side Lengths using Sine and Cosine It may be helpful to review with students how to evaluate expressions of the form asin (x°) and bcos (y°), with given values for a, b, x, and y, using their calculators. You can use sine and cosine to solve real-world problems. Example 3 A 12-ft ramp is installed alongside some steps to provide wheelchair access to a library. The ramp makes an angle of 11° with the ground. Find each dimension, to the nearest tenth of a foot. C 12 ft B y Find the height x of the wall. Multiply both sides by 12. A length of leg opposite ∠A AB sin A = ___ = _ AC length of hypotenuse Use the definition of sine. Substitute 11° for A, x for BC, and 12 for AC. 11° © Houghton Mifflin Harcourt Publishing Company wall x x sin 11° = _ 12 12sin 11° = x QUESTIONING STRATEGIES Why is a trigonometric ratio useful in solving a real-world problem involving right triangles? Sample answer: It makes it possible to use known lengths and angle measures to find unknown lengths that might be difficult to measure. If you use a trigonometric ratio, such as sine or cosine, to find the length of one of the legs of a right triangle, do you have to use a trigonometric ratio to find the length of the other leg as well? Explain. No, you could also use the Pythagorean Theorem to find the other leg since you will know the lengths of the hypotenuse and one leg. x ≈ 2.3 Use a calculator to evaluate the expression. So, the height of the wall is about 2.3 feet. Module 18 944 Lesson 2 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M18L2 944 Critical Thinking 18/04/14 11:32 PM sinx Discuss how to use the trigonometric ratios to prove that tan x = ____ cosx . Ask students if they find this surprising and why or why not. Sine and Cosine Ratios 944 B EXPLAIN 4 Find the distance y that the ramp extends in front of the wall. Substitute 11 ° for A, y for AB, and 12 for AC. Finding Angle Measures using Sine and Cosine Multiply both sides by 12 . y cos 11 ° = _ 12 12 cos 11 ° = y y ≈ 11.8 Use a calculator to evaluate the expression. QUESTIONING STRATEGIES _ length of leg adjacent to ∠A cos A = ___ = AB AC length of hypotenuse Use the definition of cosine. So, the ramp extends in front of the wall about 11.8 feet. In the equation y = sin -1 x, explain what x and y represent. In the equation, x represents the sine of the angle with measure y°. Reflect _ 10. Could you find the height of the wall using the cosine? Explain. x and Since ∠A and ∠B are complementary, m∠C = 79°. Then cos 79° = 12 x = 12cos 79° ≈ 2.3 ft. Your Turn 11. Suppose a new regulation states that the maximum angle of a ramp for wheelchairs is 8°. At least how long must the new ramp be? Round to the nearest tenth of a foot. C 2.3 ft z wall 8° B A The ramp must be at least long enough to create an 8° angle at ∠A. BC 2.3 ⇒ sin 8° ≈ _ sin A = z AC z ≈ 16.5 © Houghton Mifflin Harcourt Publishing Company _ The ramp must be at least 16.5 ft long. Explain 4 Finding Angle Measures using Sine and Cosine 5 =_ 1 . However, you already know that 30° = _ 1 . So you can In the triangle, sinA = _ 10 2 2 conclude that m∠A = 30°, 1 = 30°. and write sin -1 _ 2 Extending this idea, the inverse trigonometric ratios for sine and cosine are A defined as follows: ( ) 10 C 5 B Given an acute angle, ∠A, • if sin A = x, then sin -1 x = m∠A, read as “inverse sine of x” • if cos A = x, then cos -1 x = m∠A, read as “inverse cosine of x” You can use a calculator to evaluate inverse trigonometric expressions. Module 18 945 Lesson 2 LANGUAGE SUPPORT IN2_MNLESE389847_U7M18L2 945 Connect Vocabulary Distinguishing between sine and cosine may be challenging for some students. Explain that the prefix co- can mean together, as it does in the word cooperate. Point out that the cosine ratio for an acute angle of a triangle involves the adjacent leg. Tell students to remember this by thinking of the adjacent leg as “coming together” with the hypotenuse from the angle. 945 Lesson 18.2 18/04/14 11:32 PM Example 4 Find the acute angle measures in △PQR, to the nearest degree. P QUESTIONING STRATEGIES Write a trigonometric ratio for ∠R. Since the lengths of the hypotenuse and the opposite leg are given, PQ use the sine ratio. sinR = _ PR 7 Substitute 7 for PQ and 13 for PR. sinR = _ 13 7 Q How could you evaluate the inverse sine or cosine of an angle without using the calculator’s inverse trigonometric keys or a table of values? Use the measurements of the sine or cosine ratio in the Pythagorean Theorem to find the length of the missing leg. Then draw a right triangle with the side lengths and use a protractor to measure the angle formed by the hypotenuse and the opposite side or the adjacent side to estimate the angle measure. 13 R Write and evaluate an inverse trigonometric ratio to find m∠R and m∠P. 7 __ Start with the trigonometric ratio for ∠R. sinR = 13 7 __ Use the definition of the inverse sine ratio. m∠R = sin -1 13 Use a calculator to evaluate the inverse sine ratio. m∠R = 33 ° Write a cosine ratio for ∠P. cosP = Substitute 7 for PQ and 13 for PR. cosP = Use the definition of the inverse cosine ratio. m∠P = cos - 1 Use a calculator to evaluate the inverse cosine ratio. m∠P = 57 ° PQ _ PR 7 __ 13 7 __ AVOID COMMON ERRORS 13 Remind students about the notation used to represent the inverse sine and cosine: sin -1 x and cos -1 x. As with the inverse tangent, the –1 is not an exponent. Rather, it denotes the inverse trigonometric ratio whose value is an angle with a degree measure. Reflect 12. How else could you have determined m∠P? ∠P and ∠R are complementary. Therefore, m∠P = 90° - m∠R ≈ 90° - 33° = 57°. Your Turn Find the acute angle measures in △XYZ, to the nearest degree. XY = _ 14 Z X cos Y = _ 13. m∠Y YZ 23 14 m∠Y = cos -1 23 14 23 14. m∠Z m∠Z = 90° - m∠Y ≈ 90° - 37° = 53° Y Elaborate B hypotenuse 15. How are the sine and cosine ratios for an acute angle of a right triangle defined? sin A = adjacent opposite BC AC __ = _ and cos A = __ = _ hypotenuse Module 18 IN2_MNLESE389847_U7M18L2 946 AB hypotenuse 946 AB A adjacent ELABORATE © Houghton Mifflin Harcourt Publishing Company (_ ) ≈ 37° INTEGRATE TECHNOLOGY Have students graph y = sin x and y = cos x on their graphing calculators on the interval [0°, 90°]. Discuss similarities and differences. opposite C Lesson 2 18/04/14 11:32 PM Sine and Cosine Ratios 946 16. How are the inverse sine and cosine ratios for an acute angle of a right triangle defined? Because the sine ratio for a given acute angle is always the same, the measure of that angle QUESTIONING STRATEGIES can be defined as an inverse sine ratio: BC BC → m∠A = sin -1 sin A = or sin A = x → m∠A = sin -1 x AB AB Similarly, AC AC → m∠A = cos -1 cos A = or cos A = y → m∠A = cos -1 y AB AB (_ ) _ How are cosine and the inverse cosine related? The cosine of an acute angle of a right triangle is the ratio of the lengths of the adjacent side to the hypotenuse of the right triangle. The inverse cosine is the angle with that cosine ratio. (_ ) _ 17. Essential Question Check-In How do you find an unknown angle measure in a right triangle? First, use two known side lengths to form a ratio. Then use the appropriate inverse trigonometric ratio to find the angle measure. SUMMARIZE THE LESSON Evaluate: Homework and Practice How do you decide which trigonometric ratio to use to find a missing side length or angle measure in a right triangle? Use sine or inverse sine for opposite side and hypotenuse relationships and cosine or inverse cosine for adjacent side and hypotenuse relationships. • Online Homework • Hints and Help • Extra Practice Write each trigonometric expression. Round trigonometric ratios to the nearest thousandth. 1. Given that sin 60° ≈ 0.866, write the cosine of a complementary angle. Given that cos 26° ≈ 0.899, write the sine of a complementary angle. 2. sin 60° = cos(90° - 60°) cos 26° = sin(90° - 26°) 0.866 ≈ cos 30° 0.899 ≈ sin 64° Write each trigonometric ratio as a fraction and as a decimal, rounded (if necessary) to the nearest thousandth. B © Houghton Mifflin Harcourt Publishing Company 13 A Module 18 IN2_MNLESE389847_U7M18L2 947 947 Lesson 18.2 4. cos A adjacent 5 AC cos A = = = ≈ 0.385 13 AB hypotenuse 5. cos B cos B = AB hypotenuse 13 __ _ _ adjacent __ 12 _ ≈ 0.9231 13 hypotenuse C 25 D sin A = 12 5 opposite BC 12 __ = _ = _ ≈ 0.923 3. sin A 24 F 6. sin D cos A = opposite EF 7 __ = _ = _ = 0.28 7. cos F cos F = adjacent EF 7 __ = _ = _ = 0.28 8. sin F sin F = opposite 24 __ = _ = 0.96 7 E 947 hypotenuse hypotenuse hypotenuse DF DF 25 25 25 Lesson 2 18/04/14 11:32 PM Find the unknown length x in each right triangle, to the nearest tenth. 9. 10. C EVALUATE D 37° 15 27 B F A x EF cos E = _ DE x x cos 53° = _ ⇒ 16.2 ≈ x AB _ AC x sin 37° = _ ⇒ 9.0 ≈ x sin C = ASSIGNMENT GUIDE 53° 27 E 15 11. 12. J K 14 L R 19 x 75° JL _ KL 14 cos 75° = _ ⇒ ≈ 54.1 24° P cos L = Q x PR _ PQ 19 sin 24° = _ ⇒ ≈ 46.7 sin P = x x Find each acute angle measure, to the nearest degree. 11 V P W 18 8 15 Q U 13. m∠P 8 PR cos P = _ = _ 18 PQ 8 m∠P = cos (_) ≈ 64° -1 VW 11 sin U = _ = _ 15 UW 11 _ m∠U = sin ( ) ≈ 47° Exercise 16. m∠W m∠W = 90° - m∠U ≈ 90° - 47° = 43° Explore Investigating Ratios in a Right Triangle Exercise 17 Example 1 Finding the Sine and Cosine of an Angle Exercises 3–8 Example 2 Using Complementary Angles Exercises 1–2 Example 3 Finding Side Lengths using Sine and Cosine Exercises 9–12 Example 4 Finding Angle Measures using Sine and Cosine Exercises 13–16 Remind students that their calculators must be set on degrees, not radians, to get the correct values for the trigonometric ratios. Communicating Math Discuss the importance of being able to draw and interpret right triangle diagrams to use trigonometric ratios to solve real-world problems. Lesson 2 948 Depth of Knowledge (D.O.K.) Practice AVOID COMMON ERRORS 15 Module 18 IN2_MNLESE389847_U7M18L2 948 m∠Q = 90° - m∠P ≈ 90° - 64° = 26° 18 15. m∠U -1 14. m∠Q © Houghton Mifflin Harcourt Publishing Company R Concepts and Skills COMMON CORE Mathematical Practices 1–16 1 Recall of Information MP.4 Modeling 17–24 2 Skills/Concepts MP.4 Modeling 25 3 Strategic Thinking MP.3 Logic 26 3 Strategic Thinking MP.2 Reasoning 27 3 Strategic Thinking MP.3 Logic 18/04/14 11:32 PM Sine and Cosine Ratios 948 AVOID COMMON ERRORS Students may have difficulty solving equations such a when the variable is in the denominator. as sin x = __ c Review with students how to use inverse operations to write an equivalent equation with the variable in a the numerator, such as c = _____ sin x. 17. Use the property that corresponding sides of similar triangles are proportional to explain why the trigonometric ratio sin A is the same when calculated in △ADE as in △ABC. C By AA ~, △ABC~△ADE, so corresponding sides are proportional: (BC)(AE) AC BC AC AE = ⇒ AC = ⇒ = A AE DE DE DE BC B E ∠A ≅ ∠A and ∠ABC ≅ ∠ADE, since both are right angles. _ _ _ _ _ D These ratios determine sin A, and since they are equal, sin A is the same when calculated in either right triangle. 18. Technology The specifications for a laptop computer describe its screen as measuring 15.6 in. However, this is actually the length of a diagonal of the rectangular screen, as represented in the figure. How wide is the screen horizontally, to the nearest tenth of an inch? INTEGRATE TECHNOLOGY In addition to calculators, students can use geometry software and spreadsheets to evaluate trigonometric ratios. cos P = R 15.6 in PQ PQ _ ⇒ cos29° = _ ⇒ 13.6 in. = PQ P 15.6 PR 19. Building Sharla’s bedroom is directly under the roof of her house. Given the dimensions shown, how high is the ceiling at its highest point, to the nearest tenth of a foot? sin(m∠AEB) = AB AB _ ⇒ sin 22° = _ ⇒ AB ≈ 4.0 10.6 AE Maximum height: AC = AB + BC ≈ 4.0 + 5.5 = 9.5 ft 29° Q A 10.6 ft roof B E 22° 5.5 ft © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©teekaygee/Shutterstock C 20. Zoology You can sometimes see an eagle gliding with its wings flexed in a characteristic double-vee shape. Each wing can be modeled as two right triangles as shown in the figure. Find the measure of the angle in the middle of the wing, ∠DHG to the nearest degree. cos(m∠DHE) = 32 EH _ =_ 49 32 m∠DHE = cos -1 ≈ 49.2° 49 38 FG sin(m∠FHG) = = 43 GH 38 ≈ 62.1° m∠FHG = sin -1 43 m∠DHG = m∠DHE + m∠FHG ≈ DH (_) _ _ (_) H 43 cm 32 cm 49.2 + 62.1 = 111.3° Module 18 IN2_MNLESE389847_U7M18L2 949 949 Lesson 18.2 D D 949 49 cm F 38 cm G E Lesson 2 18/04/14 11:32 PM 21. Algebra Find a pair of acute angles that satisfy the equation sin(3x + 9) = cos(x + 5). Check that your answers make sense. COGNITIVE STRATEGIES The expressions must be the measures of two complementary angles: Discuss how students can use the facts that sin 0° = 0 and the sine increases as the angle increases (between 0° and 90°), and cos 0° = 1 and the cosine decreases as the angle increases (between 0° and 90°), as a quick check when they evaluate sine and cosine trigonometric ratios. (3x + 9) + (x + 5) = 90 ⇒ x = 19 (3x + 9)° = (3(19) + 9)° = 66° and (x + 5)° = ((19) + 5)° = 24° Check: 66° + 24° = 90° 22. Multi-Step Reginald is planning to fence his back yard. Every side of the yard except for the side along the house is to be fenced, and fencing costs $3.50/yd. How much will the fencing cost? (23 - 13) LM sin(m∠LKM) = ⇒ sin 32° = ⇒ LK ≈ 18.9 LK LK distance to fence: JK + KL + KN ≈ 13 + 18.9 + 23 = 54.9 yd _ _ J K 32° yard house cost of fencing: about 54.9(3.50) = $192.15 13 yd N M 23 yd L 23. Architecture The sides of One World Trade Center in New York City form eight isosceles triangles, four of which are 200 ft long at their base BC. The length AC of each sloping side is approximately 1185 ft. A B 200 ft © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Michael Utech/Alamy 1185 ft C Find the measure of the apex angle BAC of each isosceles _triangle, to the nearest tenth of a degree. (Hint: Use the midpoint D of BC to create two right triangles.) 1 In △ABD, AB = 1185 ft and BD = (200 ft) = 100 ft. Therefore, 2 -1 100 100 BD = ⇒ m∠BAD = sin sin(m∠BAD) = ≈ 4.8° 1185 1185 AD _ _ Module 18 IN2_MNLESE389847_U7M18L2 950 _ (_) 950 Lesson 2 18/04/14 11:32 PM Sine and Cosine Ratios 950 PEERTOPEER DISCUSSION H.O.T. Focus on Higher Order Thinking Have students investigate the following identities by evaluating them for different angles: sin -1(sin x) = x and cos -1(cos x) = x. _ 24. Explain the Error Melissa has calculated the length of XZ in △XYZ. Explain why Melissa’s answer must be incorrect, and identify and correct her error. XZ cos X = _ XY XY XZ = _ cos X XZ = 27 cos 42° ≈ 20.1 JOURNAL Have students describe how they remember the difference between the sine and cosine ratios, and the relationship between the ratios. Remind them to include at least one figure and at least one example in their descriptions. 20.1 < 27, but the hypotenuse should be the longest side. Her definition of cosine was inverted. 27 XY ⇒ XZ = ≈ 36.3 cos X = XZ cos 42° Z Melissa’s solution: _ 42° X _ Y 27 25. Communicate Mathematical Ideas Explain why the sine and cosine of an acute angle are always between 0 and 1. 0 BC AB < < ⇒ 0 < sinA < 1 0 < BC < AB ⇒ AB AB AB The same argument shows that 0 < cos A < 1. B _ _ _ A _ 26. Look for a Pattern In △ABC, the hypotenuse AB has a length of 1. Use the Pythagorean Theorem to explore the relationship between the squares of the sine and cosine of ∠A, written sin 2 A and cos 2A. Could you derive this relationship using a right triangle without any lengths specified? Explain. BC BC AC AC sin A = ___ = ___ = BC and cos A = ___ = __ = AC 1 AB AB 1 C B 1 A C By the Pythagorean Theorem, AC 2 + BC 2 = AB 2 © Houghton Mifflin Harcourt Publishing Company (sin A) 2 + (cos A) 2 = 1 2 sin 2 A + cos 2 A = 1 BC Yes: sin 2 A + cos 2 A = AB AC (_) + (_ AB ) 2 2 = BC + AC AB _ =_=1 2 2 AB 2 2 AB 2 27. Justify Reasoning Use the Triangle Proportionality Theorem to explain why the trigonometric ratio cos A is the same when calculated in △ADE as in △ABC. E C Two segments ⊥ to the same line are ∥ to each other, ¯ ∥ DE ¯ ¯. By the Triangle Proportionality Theorem, BC so BC ¯ and AE ¯ of △ADE proportionally: divides sides AD A B D AB + BD AC + CE CE CE BD BD AD AD AB AE __ = __ ⇒ 1 + __ = 1 + __ ⇒ ______ = ______ ⇒ __ = __ ⇒ __ = __ AB AC AB AC AB AC AB AC AE AC These ratios determine cos A, and since they are equal, cos A is the same when calculated in either ⇒ triangle. Module 18 IN2_MNLESE389847_U7M18L2 951 951 Lesson 18.2 951 Lesson 2 18/04/14 11:32 PM Lesson Performance Task As light passes from a vacuum into another medium, it is refracted—that is, its direction changes. The ratio of the sine of the angle of the incoming incident ray, I, to the sine of the angle of the outgoing refracted ray, r, is called the index of refraction: AVOID COMMON ERRORS incident ray Question 4 in the Lesson Performance Task identifies an error commonly made by students in working with the trigonometric functions. Here is a similar error: gem refracted ray sin I n = ___ . where n is the index of refraction. sin r This relationship is important in many fields, including gemology, the study of precious stones. A gemologist can place an unidentified gem into an instrument called a refractometer, direct an incident ray of light at a particular angle into the stone, measure the angle of the refracted ray, and calculate the index of refraction. Because the indices of refraction of thousands of gems are known, the gemologist can then identify the gem. Gem 1. Identify the gem, given these angles obtained from a refractometer: a. I = 71°, r = 29° b. I = 51°, r = 34° I = 45°, r = 17° c. 2. 3. 4. An incident ray of light struck a slice of serpentine. The resulting angle of refraction measured 21°. Find the angle of incidence to the nearest degree. 2.94 Diamond 2.42 Zircon 1.95 Azurite 1.85 Sapphire 1.77 Tourmaline 1.62 Serpentine 1.56 Coral 1.49 Describe the error(s) in a student’s solution and explain why they Opal 1.39 were error(s): sin I sin(56°) n=_ 2. 1.77 = ______ sin r sin(x°) sin 51° =_ sin(56°) ______ sin 34° sin(x°) = 1.77 51° =_ 34° sin(56°) x = sin -1 ______ = 28° 1.77 = 1.5 → coral a. sin(71°) ______ = 1.95 sin(29°) 3. zircon c. sin(45°) ______ = 2.42 sin(17°) sin(x°) 1.56 = ______ sin(21°) ) x = sin -1(1.56 ⋅ sin(21°)) =34° sin(51°) b. ______ = 1.39 opal sin(34°) 4. Index of Refraction Hematite ( sin 10° 10° Explain to students that the function and the angle cannot be separated. The correct solution is: diamond Possible answer: The student divided out the word “sin” from the sin 50° ≈ _______ 0. 7660 ≈ 4.41 ____ sin 10° 0.1736 INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 Using a refractometer, a gemologist found that the index of refraction of a substance was 2.0. Using the same angle of incidence, the gemologist found that the angle of the refracted ray for a second substance was greater than the angle of the refracted ray for the first substance. Was the index of refraction of the second substance greater or less than 2.0? Explain. Sample answer: Let i = angle of incidence, r 1 = refraction angle 1, and r 2 = refraction angle 2. sin i . Since the sine function increases as Then, 2 = _____ sin r 1 an angle increases from 0° to 90°, sin r 2 > sin r 1 sin i , the index of sin i < _____ sin i = 2. So, _____ and _____ sin r 2 sin r 2 sin r 1 refraction of the second substance, was less than 2. © Houghton Mifflin Harcourt Publishing Company 1. A thin slice of sapphire is placed in a refractometer. The angle of the incident ray is 56°. Find the angle of the refracted ray to the nearest degree. sin 50° = sin ____ 50° = sin 5° ≈ 0.0872 ______ numerator and the denominator in Step 2. This was incorrect because “sin 51°” and “sin 34°” are indivisible units, with each representing a single real number. Module 18 952 Lesson 2 EXTENSION ACTIVITY IN2_MNLESE389847_U7M18L2 952 Physicists define the index of refraction of a material in terms of the speed of light. Research the formula used by physicists that relates the index of refraction of a substance, n, the speed of light in a vacuum, c, and the speed of light through the c substance, v. n = __ v The speed of light in azurite is approximately 100,693 miles per second. Use that fact and the table in the Lesson Performance Task to find the speed of light in a vacuum. Express your answer in miles per second and miles per hour. 186,282 mi/sec; 670,615,200 mi/hr The Sun is about 93 million miles from Earth. How long does it take light from the Sun to reach Earth? about 8.3 min 18/04/14 11:32 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Sine and Cosine Ratios 952 LESSON 18.3 Name Special Right Triangles Class Date 18.3 Special Right Triangles Essential Question: What do you know about the side lengths and the trigonometric ratios in special right triangles? Common Core Math Standards The student is expected to: COMMON CORE Resource Locker G-SRT.C.8 Explore 1 Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. Discover relationships that always apply in an isosceles right triangle. A Mathematical Practices COMMON CORE The figure shows an isosceles right triangle. Identify the base angles, and use the fact that they are complementary to write an equation relating their x ∠A, ∠B; m∠A + m∠B = 90° Language Objective Explain to a partner how to find the sine, cosine, and tangent of a 30° -60° -90° triangle or a 45° -45° -90° triangle. B Use the Isosceles Triangle Theorem to write a different equation relating the base angle measures. C Use the Pythagorean Theorem to find the length of the hypotenuse in terms of the length of each leg, x. AB = x 2 + x 2 ― √3 tan 30° = , 3 _ PREVIEW: LESSON PERFORMANCE TASK © Houghton Mifflin Harcourt Publishing Company 2 _ 2 What must the measures of the base angles be? Why? D The side lengths of a 45° -45° -90° triangle are ― always in the ratio 1: 1: √2 . The trigonometric ― √2 ratios associated with this triangle are sin 45° = 2 and tan 45° = 1. The side lengths of a 30° -60° ― -90° triangle are always in the ratio 1: √3 : 2. The trigonometric ratios associated with this triangle are _1 2 ― √3 sin 60° = cos 30° = _, Reflect 1. Is it true that if you know one side length of an isosceles right triangle, then you know all the side lengths? Explain. Yes; If you know either leg length x, then the other leg length is also equal to x, and the ― length, then each leg length is this value divided by √― 2. length of the hypotenuse is this value multiplied by √2 . If you know the hypotenuse 2. _ What if? Suppose_ you draw the perpendicular from C to AB. Explain how to find the length of CD. Since △ABC is isosceles, m∠A = 45°, and since △ADC is a right triangle, m∠ACD = 90° - m∠A = 45°. Therefore △ADC is an _ √― 2 1 isosceles right triangle, so CD = ___ . ― or ___ 2 √2 Module 18 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction D A 1 C 1 Lesson 3 953 gh "File info" made throu B Date Class s ht Triangle Name al Rig 18.3 Speci in ic ratios trigonometr s and the side length about the es in do you know right triangl ion: What m to solve triangles? orean Theore special right the Pythag ratios and Quest Essential 1 Explore apply es right triang in an isoscel le. that always HARDCOVER PAGES 953966 B x A Watch for the hardcover student edition page numbers for this lesson. C x be? Why? base angles m∠A = 45°. res of the = 90°, so the measu n, 2m∠A What must substitutio length of = 45°; using terms of the enuse in m∠A = m∠B of the hypot the length em to find gorean Theor √2 Use the Pytha AB = x 2 x. leg, 2 each AB = 2x 2 2 2 +x s? 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Identify relatin right triang equation isosceles to write an shows an ementary The figure are compl that they use the fact 90° + m∠B = measures. ∠A, ∠B; m∠A relating the equation nt differe to write a le Theorem les Triang Use the Isosce res. measu m∠A = m∠B base angle relationships Discover Resource Locker ngle trigonometric G-SRT.C.8 Use ms. applied proble © Houghto View the Engage section online. Discuss the photograph, asking students to speculate on whether the Lesson Performance Task will involve the dog or the flying disc, and why. Then preview the Lesson Performance Task. ― AB = x √2 AB 2 = 2x 2 COMMON CORE B 1 C Lesson 3 953 Module 18 8L3 953 47_U7M1 ESE3898 IN2_MNL Lesson 18.3 C x m∠A = m∠B = 45°; using substitution, 2m∠A = 90°, so m∠A = 45°. Essential Question: What do you know about the side lengths and the trigonometric ratios in special right triangles? sin 45° = cos 60° = , A m∠A = m∠B ENGAGE 953 B measures. MP.2 Reasoning ― and tan 60° = √3 . Investigating an Isosceles Right Triangle 18/04/14 11:49 PM 18/04/14 11:50 PM Explore 2 Investigating Another Special Right Triangle EXPLORE 1 Discover relationships that always apply in a right triangle formed as half of an equilateral triangle. A _ _ △ABD is an equilateral triangle and BC is a perpendicular from B to AD. Determine all three angle measures in △ABC. m∠C = 90°; each angle in an equilateral triangle measures 60°, B Investigating an Isosceles Right Triangle so m∠A = 60°, and therefore m∠ABC = 90° - m∠A = 30°. B A Explain why △ABC ≅ △DBC. INTEGRATE TECHNOLOGY D C ∠A ≅ ∠D (Equilateral Triangle Theorem), ∠ACB ≅ ∠DCB (all right angles are congruent), _ _ and BC ≅ BC (Reflexive Property of Congruence), so △ABC ≅ △DBC by AAS Congruence. _ _ _ _ Or, AB ≅ DB (△ABD equilateral), BC ≅ BC (Reflexive Property of Congruence), and since Students have the option of doing the Explore activity either in the book or online. △ABC and △DBC are right triangles, △ABC ≅ △DBC by HL Congruence. C _ _ Let the length of AC be x. What is the length of AB, and why? _ _ From Step B, △ABC ≅ △DBC, so AC ≅ DC (CPCTC) and therefore CD = AC = x. In △ABD, AD = AC + DC = x + x = 2x (Seg. Add. Post.); _ _ since AB ≅ AD (△ABD equilateral), AB = AD = 2x. D _ Using the Pythagorean Theorem, find the length of BC. (2x) 2 = x 2 + BC 2 QUESTIONING STRATEGIES B A x How can you use the relationship between the angle measure and the length of the sides to help you reconstruct the simplest form of an isosceles right triangle? The legs are opposite the 45° angles and measure 1 unit while the hypotenuse is ― opposite the right angle and measures √2 units. C ― 3x 2 = BC 2 x3 = BC Reflect 3. ― ― x : x3 : 2x, or 1 : 3 : 2. 4. Explain the Error A student has drawn a right triangle with a 60° angle and a hypotenuse of 6. He has labeled the other side lengths as shown. Explain how you can tell at a glance that he has made an error and how to correct it. ― Since 63 > 6, leg JK is longer than the hypotenuse, which is ― impossible. The side lengths must be in the ratio 1 : ― For this to be true, the length of JK should be 3 Module 18 ―3 . ―3 : 2. 954 J 6 L 60° 3 6√3 K EXPLORE 2 © Houghton Mifflin Harcourt Publishing Company What is the numerical ratio of the side lengths in a right triangle with acute angles that measure 30° and 60°? Explain. The right triangle is similar to △ABC (AA Similarity), so its side lengths are in the ratio Investigating Another Special Right Triangle QUESTIONING STRATEGIES How can you use the relationship between the angle measure and the length of the side opposite the angle to help you reconstruct the simplest form of a 30° -60° -90° triangle? The shortest side is opposite the 30° angle and the longest side is the hypotenuse. So label the side opposite the 30° angle 1 unit, the side opposite the ― 60° angle √3 units, and the hypotenuse 2 units. Lesson 3 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M18L3 954 18/04/14 11:50 PM Integrate Mathematical Practices This lesson provides an opportunity to address Mathematical Practice MP.2, which calls for students to “reason abstractly and quantitatively.” Students investigate the relationships among the side lengths and angles of the special right triangles, and use them to find the trigonometric ratios and angle measures associated with these relationships. This recognition can often provide a quicker solution to a problem involving a special triangle. Special Right Triangles 954 Applying Relationships in Special Right Triangles Explain 1 AVOID COMMON ERRORS The right triangles you explored are sometimes called 45°-45°-90° and 30°-60°-90° triangles. In a 45°-45°-90° triangle, the hypotenuse is 2 times as long as each leg. In a 30°-60°-90° triangle, the hypotenuse is twice as long as the shorter leg and the longer leg is √3 times as long as the shorter leg. You can use these relationships to find side lengths in these special types of right triangles. ― Students may confuse the side lengths when labeling the sides of a standard 30° -60° -90° triangle. ― Review with students why √3 > 1 and why this side length or its multiple must be opposite the 60° angle. 30° 45° x√2 EXPLAIN 1 45° Applying Relationships in Special Right Triangles Example 1 QUESTIONING STRATEGIES √2 the length without the radical in the denominator ( ) ― ― √2 √2 1 _ · _ = _ . Remind students this is called ― ― 2 √2 √2 x Find the unknown side lengths in △ABC. A ― ― ―2 = BC ―2 The hypotenuse is √2 times as long as each leg. AB = AC2 = BC2 Substitute 10 for AB. 10 = AC ― 45° 10 ― ―2 = AC = BC 102 = 2AC = 2BC Divide by 2. C 45° B 5 ― In right △DEF, m∠D = 30°and m∠E = 60°. The shorter leg measures 53 . Find the remaining side lengths. E © Houghton Mifflin Harcourt Publishing Company 1 unit. The lengths are _ ― . Review how to write x√3 Find the unknown side lengths in each right triangle. Multiply by 2 . legs of an isosceles right triangle with a hypotenuse of one 60° x _ If you know the exact value of the length of one side of a special right triangle, can you always find the exact value of the remaining sides? Explain. Yes, because of the special side-length relationships that exist in the special right triangles. INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Work with students to find the lengths of the 2x x 60° 5√3 D 30° F The hypotenuse is twice as long as the shorter leg. DE = 2 EF Substitute 5 √3 for DE = 2 5 √3 ― EF . The longer leg is ― Substitute 5 √3 for rationalizing the denominator. ― ―3 = EF = 5 √― 3 3― DE = Simplify. _ √3 ― times as long as the shorter leg. EF . DF DF = 15 Simplify. Module 18 DF 10 √3 955 Lesson 3 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M18L3 955 Small Group Activity 955 18/04/14 11:50 PM Have students work in small groups to draw and label right triangles with the following properties: • isosceles right triangles with legs whose lengths are whole numbers greater than 1 • isosceles right triangles with a hypotenuse whose length is a whole number greater than 1 • 30° -60° -90° triangles with a shortest leg whose length is a whole number greater than 1 • 30° -60° -90° triangles with a hypotenuse whose length is a whole number greater than 2 Lesson 18.3 Your Turn EXPLAIN 2 Find the unknown side lengths in each right triangle. L 5. 6. 30° Q Trigonometric Ratios of Special Right Triangles 45° 4√3 J KL = 2JK ― ―3 = JK 43 = 2JK 60° JL = JK K P ―3 ―3 ) ―3 2√6 ― If a trigonometric ratio includes √2 in the ratio, which special right triangle does it likely refer to? Explain. an isosceles right triangle because that is the length of the hypotenuse ― If a trigonometric ratio includes √3 in the ratio, which special right triangle does it likely refer to? Explain. a 30° -60° -90° triangle because that is the length of the side opposite the 60° angle R ― _ _ PR ≅ QR , so QR = PR = 2 √6 JL = (2 QUESTIONING STRATEGIES 45° PQ = PR ―2 ―6 ) √―2 = 4 ―3 2 JL = 6 Explain 2 Trigonometric Ratios of Special Right Triangles = (2 You can use the relationships you found in special right triangles to find trigonometric ratios for the angles 45°, 30°, and 60°. Example 2 For each triangle, find the unknown side lengths and trigonometric ratios for the angles. If you know the exact value of a trigonometric ratio for a special right triangle, can you find the measure of the angle that corresponds to the ratio? Explain. Yes. Draw the corresponding special right triangle and then find the angle that corresponds to the trigonometric ratio. A 45°-45°-90° triangle with a leg length of 1 Step 1 ― ― Since the lengths of the sides _ opposite the 45° angles are congruent, they are both 1. The length of the hypotenuse is √2 times as long as each leg, so it is 1(2 ), or 2 . √2 © Houghton Mifflin Harcourt Publishing Company 45° 1 45° 1 Step 2 Use the triangle to find the trigonometric ratios for 45°. Write each ratio as a simplified fraction. Angle 45° Sine = opp _ hyp ― √2 _ 2 Module 18 Cosine = ―2 _ adj _ hyp 2 956 Tangent = INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Point out to students that the side lengths and the trigonometric ratios for the special right triangles are usually left in exact answer form. This means they are often expressed as ratios in simplest form, with rationalized denominators, when applicable. opp _ adj 1 Lesson 3 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M18L3 956 Kinesthetic Experience 18/04/14 11:50 PM Kinesthetic learners may benefit from verifying the relationships of an isosceles right triangle through a simple paper-folding activity. Have students take a square piece of paper and fold it in half diagonally. This creates an isosceles right triangle. Then ask students to measure one leg of the triangle to the nearest millimeter, and apply their understanding of the relationships among the legs and the hypotenuse to predict the length of the hypotenuse. Finally, have students measure the length of the hypotenuse to check their predictions. Special Right Triangles 956 B A 30°-60°-90° triangle with a shorter leg of 1 Step 1 30° The hypotenuse is twice √― 3 as long as the shorter leg, 2 so the length of the hypotenuse is The longer leg is . 2 √3 times as long as the shorter leg, ―3 so the length of the longer leg is . 60° 1 Step 2 Use the triangle to complete the table. Write each ratio as a simplified fraction. Angle Sine = opp _ hyp Cosine = √― 3 _ _1 2 √― 3 _ 30° 60° adj _ hyp Tangent = √― 3 _ 2 adj 3 ― _1 2 opp _ √3 2 Reflect 7. Write any patterns or relationships you see in the tables in Part A and Part B as equations. Why do these patterns or relationships make sense? ― ― 3 2 1 sin 30° = cos 60° = _ ; sin 45° = cos 45° = ___ ; sin 60° = cos 30° = ___ ; the sine of an angle 2 2 2 equals the cosine of its complement. © Houghton Mifflin Harcourt Publishing Company 8. For which acute angle measures θ, is tanθ less than 1? equal to 1? greater than 1? tan θ < 1 for θ < 45; tan θ = 1 for θ = 45°; tan θ > 1 for θ > 45°. Your Turn Find the unknown side lengths and trigonometric ratios for the 45 angles. 9. 10 CB _ _ ― , AB = ―2 = 5 ―2 ; 2 AC = AB, AC = 5 ― 2 10 sin 45° = cos 45° = _ ―2 = 5 ―2 ; C AB = 45° 45° 10 B tan 45° = 1 A Module 18 957 Lesson 3 LANGUAGE SUPPORT IN2_MNLESE389847_U7M18L3 957 Connect Vocabulary Point out that what makes special right triangles special are the unique relationships among the sides. It may be easier for students to focus on the special triangles by referring to them by their angle measures. Instead of discussing an isosceles right triangle, refer to the triangle as a 45° -45° -90° triangle. 957 Lesson 18.3 18/04/14 11:50 PM Explain 3 Investigating Pythagorean Triples EXPLAIN 3 Pythagorean Triples A Pythagorean triple is a set of positive integers a, b, and c that satisfy the equation a 2 + b 2 = c 2. This means that a, b, and c are the legs and hypotenuse of a right triangle. Right triangles that have non-integer sides will not form Pythagorean triples. B c A b Examples of Pythagorean triples include 3, 4, and 5; 5, 12, and 13; 7, 24, and 25; and 8, 15, and 17. Example 3 Investigating Pythagorean Triples a C QUESTIONING STRATEGIES How can understanding the relationship between a Pythagorean triple and its multiples help you find the missing side in a right triangle? If you can identify the triple and the multiple, you can multiply the missing triple length by the multiple to find the missing side length without having to use the Pythagorean Theorem. Use Pythagorean triples to find side lengths in right triangles. Verify that the side lengths 3, 4, and 5; 5, 12, and 13; 7, 24, and 25; and 8, 15, and 17 are Pythagorean triples. 3 2 + 4 2 = 9 + 16 = 25 = 5 √ 5 2 + 12 2 = 25 + 144 = 169 = 13 2 √ 7 2 + 24 = 49 + 576 = 625 = 25 2 √ 8 2 + 15 2 = 64 + 225 = 289 = 17 2 √ 2 2 The numbers in Step A are not the only Pythagorean triples. In the following steps you will discover that multiples of known Pythagorean triples are also Pythagorean triples. INTEGRATE MATHEMATICAL PRACTICES Focus on Patterns MP.8 Have students create a list of Pythagorean In right triangles DEF and JKL, a, b, and c form a Pythagorean triple, and k is a positive integer greater than 1. Explain how the two triangles are related. J D a F c b kc ka E L kb © Houghton Mifflin Harcourt Publishing Company △DEF is similar to △JKL by the Side-Side-Side (SSS) Triangle Similarity Theorem because the corresponding sides are proportional. Complete the ratios to verify Side-Side-Side (SSS) Triangle Similarity. a : b : c = ka : kb : kc You can use the Pythagorean Theorem to compare the lengths of the sides of ▵JKL. What must be true of the set of numbers ka, kb, and kc? 2 2 (ka 2) + (kb 2) = k a + k 2b 2 ( 2 2) = k 2 a +b = k 2c 2 = triples. Students do not need to memorize every triple on the list, but they can use it as a reference to help them recognize the triples and to look for multiples. Students may want to include side columns with multiples to help them recognize the patterns between the original triple and its multiples. K (kc) 2 The set of numbers ka, kb, and kc form a Pythagorean triple . Module 18 IN2_MNLESE389847_U7M18L3 958 958 Lesson 3 18/04/14 11:50 PM Special Right Triangles 958 Reflect ELABORATE 10. Suppose you are given a right triangle with two side lengths. What would have to be true for you to use a Pythagorean triple to find the remaining side length? The given side lengths would have to be two numbers in a Pythagorean triple. Also, if the AVOID COMMON ERRORS legs are given, the side lengths would have to be the smaller two numbers in the triple, Students may confuse Pythagorean triples. Emphasize that students can always use the Pythagorean Theorem to verify that a triple or a multiple of a triple is correct. whereas if one leg and the hypotenuse are given, the side lengths would have to include the largest number in the triple. Your Turn Use Pythagorean triples to find the unknown side length. 11. SUMMARIZE THE LESSON What can you say about the side lengths and the trigonometric ratios associated with special right triangles? The sides of an isosceles ― right triangle are in the ratio 1 - 1 - √2 . The sides of a 30° -60° -90° triangle are in the ratio ― 1 - √3 - 2. You can find the related trigonometric ratios by drawing the triangles and finding the sine, cosine, and tangent of the corresponding 30°, 45°, and 60° angles from the triangles. R _ 12. In △XYZ, the hypotenuse XY has length 68, and _ the shorter leg XZ has length 32. 32 : 68 = 8 : 17, so the side lengths are multiples of 8, 15, and 17. 24 P 18 8 : 15 : 17 = 4(8) : 4(15) : 4(17) = 32 : 60 : 68, so YZ = 60. Q 18 : 24 = 3 : 4, so the side lengths are multiples of the Pythagorean triple 3, 4, 5. 3 : 4 : 5 = 6(3) : 6(4) : 6(5) = 18 : 24 : 30, so PR = 30. Elaborate 13. Describe the type of problems involving special right triangles you can solve. Once you identify the side (longer leg, shorter leg, hypotenuse) that is given, you can find the lengths of the other two sides by applying relationships such as the length of the © Houghton Mifflin Harcourt Publishing Company hypotenuse in a 30°-60°-90° triangle being twice as long as the length of the shorter leg. 14. How can you use Pythagorean triples to solve right triangles? Suppose two side lengths of a right triangle are given and correspond to two numbers in a Pythagorean triple. If one of the given sides is the hypotenuse and one of the lengths is the largest number of the triple, then the unknown length is the remaining number in the triple. Likewise, if neither given side is the hypotenuse and the lengths are the smaller two numbers of the triple, the unknown side length must be the largest number of the triple. 15. Discussion How many Pythagorean triples are there? Infinitely many; for example, since 3, 4, and 5 is a Pythagorean triple, so are 2(3) = 6, 2(4) = 8, and 2(5) = 10; 9, 12, and 15; 12, 16, and 20; and so on. 16. Essential Question Check-In What is the ratio of the length of the hypotenuse to the length of the shorter leg in any 30°-60°-90° triangle? The ratio is 2 to 1. Module 18 IN2_MNLESE389847_U7M18L3 959 959 Lesson 18.3 959 Lesson 3 18/04/14 11:50 PM Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice For each triangle, state whether the side lengths shown are possible. Explain why or why not. 1. 2. 3√3 45° 6√3 6 3√3 60° 3√3 6 √― 3 6 :_ = No; 3 √― 3 : 6 : 6 √― 3=1:_ ― ― √ √ ― 3 3 3 3 2 √3 1 : _ : 2 ≠ 1 : √― 3:2 ― ― _― ― ― ASSIGNMENT GUIDE Yes; 3 √3 : 3 √3 : 3 √6 = 3 √6 = 1 : 1 : √2 1:1: 3 √3 3 3. 3√6 ― 4. 8√3 12 6 30° 45° 4√3 ― ― Yes; 4 √3 : 12 : 8 √3 = 1 : ― 1 : √3 : 2 2√3 8 √― 3 12 _ :_ = ― 4 √3 4 √― 3 ― 1 : 1 : √― 3 ― 2√3 No; 2 √3 : 2 √3 : 6 = 1 : 1 : 6 _ ≠ 2 √― 3 Find the unknown side lengths in each right triangle. 5. 6. A L 18 B 9 C ― ― ― ― 18 = AB √2 = BC √2 18 √― 2 = 2AB = 2BC 9 √― 2 = AB = BC ― ― AC = AB √2 = BC √2 Module 18 Exercise IN2_MNLESE389847_U7M18L3 960 30° © Houghton Mifflin Harcourt Publishing Company 45° JL = JK √3 9 = JK √3 ― √ 3 ― 3 = JK 9 √3 = 3JK J KL = 2JK K ― KL = 2(3 √3 ) KL = 6 √3 ― Practice Explore 1 Investigating an Isosceles Right Triangle Exercises 1–4 Explore 2 Investigating Another Special Right Triangle Exercises 5–8 Example 1 Applying Relationships in Special Right Triangles Exercises 9–12 Example 2 Trigonometric Ratios of Special Right Triangles Exercises 13–16 Example 3 Investigating Pythagorean Triples Exercises 17–20 INTEGRATE TECHNOLOGY Although students should write answers in exact form for special right triangles, they can use a calculator to check that their answers make sense. Lesson 3 960 Depth of Knowledge (D.O.K.) 60° Concepts and Skills COMMON CORE Mathematical Practices 1–20 2 Skills/Concepts MP.4 Modeling 21 2 Skills/Concepts MP.4 Modeling 22–24 2 Strategic Thinking MP.4 Modeling 25 3 Strategic Thinking MP.3 Logic 26 3 Strategic Thinking MP.3 Logic 18/04/14 11:50 PM Special Right Triangles 960 AVOID COMMON ERRORS 7. Students working with special right triangles can mislabel diagrams when the side lengths differ from the standard reference triangles. Remind students they can refer to those triangles, but that they must identify the appropriate relationship for each triangle based on its corresponding problem statement. Right triangle UVW has acute angles U measuring 30° and W measuring 60°. _ Hypotenuse UW measures 12. (You may want to draw the triangle in your answer.) 30° Q U 45° P 12 = 2VW 6 = VW Have students construct several triangles with legs whose measures are the legs of Pythagorean triples using centimeter graph paper. Then have them use a ruler to verify that the length of the hypotenuse corresponds to the length of the hypotenuse in the corresponding Pythagorean triple. 45° ― ― ― = 5 √― 20 = 10 √― 5 R 5√10 PQ = PR √2 =(5 √10 ) √2 UV = VW √― 3 UV = 6 √― 3 UW = 2VW MANIPULATIVES V Use trigonometric ratios to solve each right triangle. 9. 10. D A 14 45° C AC _ AB ― √2 AC _ =_ 14 2 7 √― 2 = AC © Houghton Mifflin Harcourt Publishing Company 60° E B 10√3 _ ― _ _ ― BC _ AB ― √2 BC _ =_ 14 2 7 √― 2 = BC F _ _ _ ― ― cos 60° = DE sin 60° = DF EF EF √3 DF DE 1 = = 2 2 10 √3 10 √3 15 = DF 5 √3 = DE cos 45° = sin 45° = ― 11. Right △KLM with m∠J = 45°, leg JK = 4 √3 12. Right △PQR with m∠Q = 30°, leg QR = 15 Q L 30° 45° 15 J 45° 4√3 K P KL _ tan 45° = _ JK ― ― 4 √― 3 _ _ 1=_ JK JL = (4 √― 3 ) √― 2 = 4 √― 6 JK= 4 √― 3 sin 45° = JK JL √2 4 √3 = 2 JL Module 18 IN2_MNLESE389847_U7M18L3 961 Lesson 18.3 Right triangle PQR has P and Q _ acute angles _ measuring 45°. Leg PR measures 5√10 . (You may want to draw the triangle in your answer.) 12 W 60° 961 8. 961 PR _ QR √― 3 _ PR _ = 15 3 5 √― 3 = PR tan 30° = R PR _ PQ 5 √― 3 _1 = _ PQ 2 PQ =2(5 √― 3 ) = 10 √― 3 sin 30° = Lesson 3 18/04/14 11:49 PM COGNITIVE STRATEGIES For each right triangle, find the unknown side length using a Pythagorean triple. If it is not possible, state why. 13. 14. K Discuss the importance of being able to draw and interpret right triangle diagrams to use trigonometric ratios to solve real-world problems. B 25 A 60 8 C 25:60 = 5:12, so the side lengths are multiples of 5, 12, and 13. J 5 : 12 : 13 = 5(5) : 5(12) : 5(13) = 25 : 60 : 65 , so AB = 65. L 4 Not possible: 8 - 4 = 64 - 16 = 48; 48 is not a perfect square. 2 2 15. In right △PQR, the legs have lengths PQ = 9 and QR = 21. Not possible: 9 2 + 21 2 = 81 + 441 = 522 ; 522 is not a perfect square. _ _ 16. In right △XYZ, the hypotenuse XY has length 35, and the shorter leg YZ has length 21. 21 : 35 = 3 : 5, so the side lengths are multiples of 3, 4, and 5. 3 : 4 : 5 = 7(3) : 7(4) : 7(5) = 21 : 28 : 35 , so XZ = 28. 17. E 12 12 D x F H © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Exactostock/Superstock G DEFG is a rhombus, so its diagonals bisect each other, and EH = HG. Since EH + HG = EG =12 (Seg. Add. Post.), 2EH = 12, and therefore EH = 6. By the Pythagorean Thm., x 2 + 6 2 = 12 2 x 2 + 36 = 144 x 2 = 108 x = √108 = 6 √3 . ―― ― 18. Represent Real-World Problems A baseball “diamond” actually forms a square, each side measuring 30 yards. How far, to the nearest yard, must the third baseman throw the ball to reach first base? The line from first base to third base forms the diagonal of the square, so with home plate they form a 45°-45°-90° triangle with legs of length 30 yd. Therefore, the third baseman must throw the ball 30 √2 yd ≈ 42 yd. ― Module 18 IN2_MNLESE389847_U7M18L3 962 962 Lesson 3 18/04/14 11:49 PM Special Right Triangles 962 _ 19. In a right triangle, the longer leg is exactly √3 times the length of the shorter leg. Use the inverse tangent trigonometric ratio to prove that the acute angles of the triangle measure 30° and 60°. BC , use an inverse tangent ratio: Given: BC = AC √3 . Since tan A = B AC √ AC 3 BC √ tanA = = = 3 AC AC m∠A = tan -1 √3 = 60° ― _ _― ― A ― _ ∠A and ∠B are complementary, so m∠B = 90° - m∠A = 90° - 60° = 30°. C Algebra Find the value of x in each right triangle. 20. 21. C 5x - 6 B x√3 L x x J A ― Since the longer leg is √3 times the length of the shorter leg, △ABC is a 30°-60°-90° triangle. Therefore, (3x - 25)√2 x√2 2 K ― Since the hypotenuse is √2 times the length of one of the legs, △JKL is a 45°-45°-90° triangle. Therefore, ― _ BC = 2AC JK = KL x √2 = (3x - 25) √2 2 x = 2(3x - 25) 5x - 6 = 2x 3x = 6 x=2 ― x = 6 x - 50 50 = 5 x © Houghton Mifflin Harcourt Publishing Company 10 = x sides of a right triangle 22. Explain the Error Charlene is trying to find the unknown _ with a 30° acute angle, whose hypotenuse measures 12√2 . Identify, explain, and correct Charlene’s error. 12√2 P Module 18 IN2_MNLESE389847_U7M18L3 963 Lesson 18.3 12 R 6√2 ― ― ― Charlene appears to have used the ratio 1 : √2 : 2, whereas the correct ratio for a 30°-60°-90° triangle is 1 : √3 : 2. PR = 6 √2 is correct, but QR = PR √3 = (6 √2 ) √3 = 6 √6 . ― 963 Q 30° ― ― ― 963 Lesson 3 18/04/14 11:49 PM 23. Represent Real-World Problems Honeycomb blinds form a string of almostregular hexagons when viewed end-on. Approximately how much material, to the nearest ten square centimeters, is needed for each 3.2-cm deep cell of a honeycomb blind that is 125 cm wide? (Hint: Draw a picture. A regular hexagon can be divided into 6 equilateral triangles.) ... 3.2 cm ... 125 cm B 3.2 cm A C _ 1 △ABC is a 30°-60°-90° triangle, and BC = (3.2 cm) = 1.6 cm. Therefore, 2 BC cos 30° = AB √3 1.6 cm = 2 AB AB = 3.2 √3 ≈ 5.542 cm _ ― _ _ ― The amount of material needed is the perimeter of the hexagon × the width of the blind: 24. Which of these pairs of numbers are two out of three integer-valued side lengths of a right triangle? (Hint: for positive integers a, b, c, and k, ka, kb, and kc are side lengths of a right triangle if and only if a, b, and c are side lengths of a right triangle.) A. 15, 18 True False B. 15, 30 True False C. 15, 51 True False D. 16, 20 True False E. 16, 24 True False A. False; 15 = 3(5) and 18 = 3(6); 5 2 + 6 2 = 25 + 36 = 61 and 6 2 - 5 2 = 36 - 25 = 11, and neither are perfect squares. B. False; 15 = 15(1) and 30 = 15(2); 1 2 + 2 2 = 5 and 2 2 - 1 2 = 3, and neither are perfect squares. © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Exactostock/Superstock 6(AB) × width = 6(5.542) × 125 ≈ 4,160 cm 2 C. True: 15 = 3(5) and 39 = 3(13); 5, 12, 13 is a Pythagorean triple. D. True; 16 = 4(4) and 20 = 4(5); 3, 4, 5 is a Pythagorean triple. E. False; 16 = 8(2) and 24 = 8(3); 2 2 + 3 2 = 13 and 3 2 - 2 2 = 5, and neither are perfect squares. Module 18 IN2_MNLESE389847_U7M18L3 964 964 Lesson 3 18/04/14 11:49 PM Special Right Triangles 964 PEERTOPEER DISCUSSION H.O.T. Focus on Higher Order Thinking 25. Communicate Mathematical Ideas Is it possible for the three side lengths of a right triangle to be odd integers? Explain. Have students use the special right triangles to make up several inverse trigonometric problems to 1 . Have them exchange and evaluate, such as sin -1 _ 2 solve problems with another pair of students. Have students use a calculator to check their work. No; if the two shorter side lengths are odd, then their squares are odd, because the square of an odd number is always odd. But the sum of their squares is even, because the sum of two odd numbers is always even. Therefore the sum of the squares of the two shorter side lengths cannot itself be the square of an odd number. 26. Make a Conjecture Use spreadsheet software to investigate this question: are there sets of positive integers a, b, and c such that a 3 + b 3 = c 3? You may choose to begin with these formulas: JOURNAL Have students summarize what they know about special right triangles. Remind them to include a chart that summarizes trigonometric ratios for relevant angles and diagrams that support the ratios. © Houghton Mifflin Harcourt Publishing Company Actually there are no sets of positive integers a, b, and c such that a 3 + b 3 = c 3, but this is very hard indeed to prove. Students should be able to extend the spreadsheet example as follows, testing triples with different smallest numbers by changing the value in cell A1; for example: Module 18 IN2_MNLESE389847_U7M18L3 965 965 Lesson 18.3 965 Lesson 3 18/04/14 11:49 PM Lesson Performance Task AVOID COMMON ERRORS Kate and her dog are longtime flying disc players. Kate has decided to start a small business making circles of soft material that dogs can catch without injuring their teeth. Since she also likes math, she’s decided to see whether she can apply Pythagorean principles to her designs. She used the Pythagorean triple 3-4-5 for the dimensions of her first three designs. When students check three measurements to see if they could represent the sides of a right triangle, they may add the two smaller measurements before squaring them to see if the result is the square of the third measurement. Caution students to square each smaller measurement first, before finding their sum. r = 3 in. small r = 4 in. medium r = 5 in. INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 If you look at a list of Pythagorean Triples, large 1. Is it true that the (small area) + (medium area) = (large area)? Explain. 2. If the circles had radii based on the Pythagorean triple 5-12-13, would the above equation be true? Explain. you’ll notice that at least one of the numbers forming the triple is even. Must this be true for all Pythagorean Triples? Explain. Yes; sample answer: The square of an odd number is always odd. The sum of two odd numbers is always even. So, if a and b are both odd, a 2 + b 2 must be even, which means that c must also be even. 3. Three of Kate’s circles have radii of a, b, and c, where a, b, and c form a Pythagorean triple (a 2 + b 2 = c 2). Show that the sum of the areas of the small and medium circles equals the area of the large circle. 5. Explain the discrepancy between your results in Exercises 3 and 4. 1. yes; small area + medium area = 9 in. 2 + 16 in. 2 = 25 in. 2 = large area 2. yes; small area + medium area = 25 in. 2 + 144 in. 2 = 169 in. 2 = large area 3. small area + medium area = a 2 + b 2 =( a 2 + b 2) = c 2 = large area 1 1 2 4. no; small volume + medium volume = 36 in. 3 + 85 in. 3 = 121 in. 3 ≠ 166 in. 3 3 3 3 (large area) _ _ _ 5. Possible answer: Numbers are squared in both the Pythagorean Theorem and the formula for the area of a circle. In the formula for the volume of a sphere, numbers are cubed. There is no theorem analogous to the Pythagorean Theorem for cubes of numbers. Module 18 © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©dlewis33/iStockPhoto.com 4. Kate has decided to go into the beach ball business. Sticking to her Pythagorean principles, she starts with three spherical beach balls--a small ball with radius 3 in., a medium ball with radius 4 in., and a large ball with radius 5 in. Is it true that (small volume) + (medium volume) = (large volume)? Show your work. Lesson 3 966 EXTENSION ACTIVITY IN2_MNLESE389847_U7M18L3 966 Choose two whole numbers m and n, such that m < n. Make a table like the following for at least 10 pairs of values of m and n: M n n2 - m2 2mn n2 + m2 Pythagorean Triple? 1 2 3 4 5 yes Describe your results, and use algebra to explain them. m and n always generate a Pythagorean Triple. (n 2 - m 2) 2 + (2mn) 2 = n 4 - 2n 2m 2 + m 4 + 4m 2n 2 = n 4 + 2n 2 m 2 + m 4 2 = (n 2 + m 2) 18/04/14 11:49 PM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Special Right Triangles 966 LESSON 18.4 Name Problem Solving with Trigonometry Class Date 18.4 Problem Solving with Trigonometry Essential Question: How can you solve a right triangle? Resource Locker Common Core Math Standards The student is expected to: COMMON CORE Explore G-SRT.C.8 Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. Also G-SRT.D.9(+), G-GPE.B.7 _ _ Suppose you draw an altitude AD to side BC of △ABC. Then write an equation using a trigonometric ratio in terms of ∠C, the height h of △ABC, and the length of one of its sides. A Mathematical Practices COMMON CORE Deriving an Area Formula You can use trigonometry to find the area of a triangle without knowing its height. b Language Objective C Explain to a partner how to solve a right triangle, and how to solve a right triangle in the coordinate plane. sin C = ENGAGE B Essential Question: How can you solve a right triangle? C b c C B a h a D c B AD _ h _ = AC b Solve your equation from Step A for h. b sin C = h Complete this formula for the area of △ABC in terms of h and 1 ah another of its side lengths: Area = _ 2 © Houghton Mifflin Harcourt Publishing Company You can use trigonometric ratios to find side lengths, or their inverses to find angle measures; you can use the Pythagorean Theorem to find the third side length; you can use the fact that the acute angles are complementary; if the triangle is in the coordinate plane, you can use the distance formula to find side lengths. A A MP.2 Reasoning your expression for h from Step B into your formula from Step C. D Substitute _1 ab sin C 2 Reflect 1. Does the area formula you found work if ∠C is a right angle? Explain. 1 Yes; in this case, sin C = sin 90° = 1, so the formula becomes Area = __ ab, and this is correct 2 since a and b are now the base and height of △ABC. PREVIEW: LESSON PERFORMANCE TASK View the Engage section online. Discuss the photograph. Ask students if they know why water is sometimes referred to as H 2O. Then preview the Lesson Performance Task. Module 18 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction Lesson 4 967 gh “File info” made throu Date Class ng blem Solvi 18.4 Pro Trigonometry with Name IN2_MNLESE389847_U7M18L4 967 HARDCOVER PAGES 967980 Resource Locker le? es in a right triang right triangl you solve m to solve How can orean Theore Question: the Pythag Essential ratios and trigonometric D.9(+), G-GPE.B.7 COMMON G-SRT.C.8 Use CORE ms. Also G-SRT. Formula applied proble . an Area its height Deriving t knowing le withou triang Explore a a on using the area of an equati y to find its sides. _ △ABC. Then write trigonometr of one of You can use _ length BC of , and the e AD to side h of △ABC draw an altitud of ∠C, the height A Suppose you terms ic ratio in trigonometr A h b B C Watch for the hardcover student edition page numbers for this lesson. c B c b a C D a _ _ AD = h b sin C = AC from Step equation Solve your h b sin C = A for h. of h and in terms area of △ABC la for the this formu 1 ah _ Complete s: Area = 2 C. its side length la from Step another of your formu Step B into for h from expression tute your y g Compan Substi _1 ab sin C © Houghto n Mifflin Harcour t Publishin 2 __ n. this angle? Explai Area = 12 ab, and is a right becomes work if ∠C formula la you found = 1, so the area formu Does the . C = sin 90° case, sin t of △ABC and heigh Yes; in this the base b are now since a and Reflect 1. is correct Lesson 4 967 Module 18 8L4 967 47_U7M1 ESE3898 IN2_MNL 967 Lesson 18.4 19/04/14 12:02 AM 19/04/14 12:03 AM Suppose you used a trigonometric ratio in terms of ∠B, h, and a different side length. How would this change your findings? What does this tell you about the choice of sides and included angle? 2. EXPLORE Step C would be the same, but the other steps would change: AD h = c sin B = AB c sin B = h 1 Area = ac sin B 2 You can choose different sides and included angle and derive a slightly different _ _ Deriving an Area Formula _ INTEGRATE TECHNOLOGY formula for the area, but in the same form. Explain 1 Students have the option of doing the Explore activity either in the book or online. Using the Area Formula INTEGRATE TECHNOLOGY Area Formula for a Triangle in Terms of its Side Lengths The area of △ABC with sides a, b, and c can be found using the lengths of two of its sides and the sine of the included angle: Area = __12 bc sin A, Area = __12 ac sin B, or Area = __12 ab sin C. Students are familiar with the trigonometric ratios from 0° to 90°. Ask them to graph y = sin x from 0° to 180° using their graphing calculators to see that the ratios are defined for angles greater (and less) than 90°. C a B b c A You can use any form of the area formula to find the area of a triangle, given two side lengths and the measure of the included angle. Example 1 QUESTIONING STRATEGIES Find the area of each triangle to the nearest tenth. Trigonometric ratios are defined using right triangles. Why does the area formula work for all types of triangles when it uses the sine ratio? The formula works because when the altitude, or height, is drawn to apply the usual formula for the area of a triangle, a right triangle is created. 3.2 m 142° 4.7 m Let the known angle be ∠C. 1 ab sin C. Substitute in the formula Area = _ 2 Evaluate, rounding to the nearest tenth. © Houghton Mifflin Harcourt Publishing Company a = 3.2 m and b = 4.7 m Let the known side lengths be a and b. m ∠C = 142° 1 (3.2)(4.7)sin 142° Area = _ 2 Area ≈ 4.6 m 2 If you know the area of a triangle and the lengths of two sides, how can you find the measure of the included angle? Find the inverse sine of twice the area divided by the product of the two sides. EXPLAIN 1 Module 18 968 Lesson 4 Using the Area Formula PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M18L4 968 Integrate Mathematical Practices This lesson provides an opportunity to address Mathematical Practice MP.2, which calls for students to “reason abstractly and quantitatively.” Students derive the formula for the area of a triangle by recognizing the relationships that occur within the triangle when the altitude is constructed. They apply this formula to a variety of triangles. As students solve a right triangle, they must identify relationships that can be used to find missing measures, and they can often choose which of the three inverse trigonometric ratios to apply. 19/04/14 12:03 AM AVOID COMMON ERRORS Students may have difficulty substituting values into the sine area formula from a diagram. Emphasize that the two sides are both adjacent to the angle. Or, the angle is the included angle formed by the sides of the triangle. Problem Solving with Trigonometry 968 In △DEF, DE = 9 in., DF = 13 in., and m∠D = 57°. B QUESTIONING STRATEGIES Sketch △DEF and check that ∠D is the included angle. Why is the sine area formula useful when there is already a formula for the area of a triangle? It provides another method to find the area of a triangle without having to construct the height. E 9 in. D 57° Why does the angle have to be the included angle to use the sine area formula? The formula is derived by constructing a right triangle so that the height is opposite the angle and the hypotenuse is adjacent to the angle to make it possible to apply the sine ratio. F 13 in. ( ) ( )( ) 1 (DE) DF sin D Area = _ 2 1 9 13 sin 57 ° Area = _ 2 Write the area formula in terms of △DEF. Substitute in the area formula. Area ≈ 49.1 in. 2 Evaluate, rounding to the nearest tenth. Your Turn Find the area of each triangle to the nearest tenth. 3. CONNECT VOCABULARY 12 mm Remind students that another word for the altitude of a triangle is its height. one version of the sine area formula to apply the formula to find the area of any triangle. 15 mm Let the known side lengths be a and b. a = 12 mm and b = 15 mm Let the known angle be ∠C. m ∠C = 34° 1 Area = (12)(15)sin 34° 2 Area ≈ 50.3 mm 2 1 Substitute in the formula Area = _ ab sin C. 2 Evaluate, rounding to the nearest tenth. © Houghton Mifflin Harcourt Publishing Company INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Discuss why students need remember only 34° 4. _ In △PQR, PQ = 3 cm, QR = 6 cm, and m∠Q = 108°. R 6 cm P 108° Q 3 cm _1 (PQ)(QR)sin Q 2 1 = _(3)(6)sin 108° Area = 2 ≈ 8.6 cm 2 Module 18 969 Lesson 4 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M18L4 969 Small Group Activity Have students verify that they can use different trigonometric ratios and different inverse trigonometric ratios to solve a right triangle. Have students work together to solve a right triangle. Then have each student verify the measures using different trigonometric and inverse trigonometric ratios. 969 Lesson 18.4 19/04/14 12:03 AM Explain 2 Solving a Right Triangle EXPLAIN 2 Solving a right triangle means finding the lengths of all its sides and the measures of all its angles. To solve a right triangle you need to know two side lengths or one side length and an acute angle measure. Based on the given information, choose among trigonometric ratios, inverse trigonometric ratios, and the Pythagorean Theorem to help you solve the right triangle. Solving a Right Triangle A shelf extends perpendicularly 7 in. from a wall. You want to place a 9-in. brace under the shelf, as shown. To the nearest tenth of an inch, how far below the shelf will the brace be attached to the wall? To the nearest degree, what angle will the brace make with the shelf and with the wall? QUESTIONING STRATEGIES What information do you need to solve a right triangle? Explain. You need the lengths of two sides or the length of one side and one acute angle measure. With this information you can find the remaining angle measures and side lengths. Shelf C 7 in. A 9 in. Wall A B Brace Find BC. Substitute 7 for AC and 9 for AB. 7 2 + BC 2 = 9 2 Find the squares. 49 + BC 2 = 81 Be sure students remember how to interpret the notation that represents an inverse function. Remind a them that the –1 in an expression such as tan -1 __ b represents the angle whose tangent is __ab . It does not 1 . It should generally represent the reciprocal _______ a tan __ b be clear from the context whether the –1 is an inverse function or a negative exponent. () BC 2 = 32 Subtract 49 from both sides. BC ≈ 5.7 Find the square root and root. B AVOID COMMON ERRORS AC 2 + BC 2 = AB 2 Use the Pythagorean Theorem to find the length of the third side. () Find m∠A and m∠B. Use an inverse trigonometric ratio to find m∠A. You know the lengths of the adjacent side and the hypotenuse, so use the cosine ratio. cos A = _ 9 Write an inverse cosine ratio. m∠A = cos -1 Evaluate the inverse cosine ratio and round. m∠A ≈ 39 ° © Houghton Mifflin Harcourt Publishing Company Write a cosine ratio for ∠A. 7 ∠ A and ∠B are complementary. m∠ A Substitute 39 ° for m∠ A . (_) 7 9 + m∠B = 90° 39 ° + m∠B ≈ 90° Subtract 39 ° from both sides. m∠B ≈ 51 ° Reflect 5. (_) Is it possible to find m∠B before you find m∠A? Explain. 7 7 ≈ 51°. Yes; use an inverse sine ratio; sin B = , so m∠B = sin -1 9 9 Module 18 _ 970 Lesson 4 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M18L4 970 Modeling 19/04/14 12:03 AM Have students draw an acute triangle with two adjacent sides that have whole number measures. Ask them to use a protractor to measure the included angle. Then have them find the area of the triangle using the sine area formula. Repeat with an obtuse triangle. Auditory Cues Students can use the mnemonic soh-cah-toa to remember that opposite leg adjacent leg opposite leg sin A = __________, cos A = __________, and tan A = __________. hypotenuse hypotenuse adjacent leg Problem Solving with Trigonometry 970 Your Turn EXPLAIN 3 Solving a Right Triangle in the Coordinate Plane 6. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Discuss how to identify the legs and the 7. hypotenuse of a right triangle in the coordinate plane if the legs of the right triangle are not formed by horizontal and vertical line segments. Use the distance formula to find the length of each side. The longest side is the hypotenuse. 8. Use a trigonometric ratio to find the distance EF. DE tan F = EF 33 tan 27° = EF 33 ≈ 65 m EF = tan 27° Use another trigonometric ratio to find the distance DF. DE sin F = DF 33 sin 27° = DF 33 ≈ 73 m DF = sin 27° Use the fact that acute angles of a right triangle are complementary to find m∠D. _ _ _ F Building D 33 m E Shadow _ _ _ m∠D + m∠F = 90° m∠D + 27° = 90° m∠D = 63° QUESTIONING STRATEGIES Explain 3 Solving a Right Triangle in the Coordinate Plane You can use the distance formula as well as trigonometric tools to solve right triangles in the coordinate plane. © Houghton Mifflin Harcourt Publishing Company How is solving a right triangle in the coordinate plane different from solving a labeled right triangle? Neither side lengths nor an angle measure are given in the coordinate plane. You must use the distance formula to find side lengths and then use the inverses of trigonometric ratios to find the angles. 27° A building casts a 33-m shadow when the Sun is at an angle of 27° to the vertical. How tall is the building, to the nearest meter? How far is it from the top of the building to the tip of the shadow? What angle does a ray from the Sun along the edge of the shadow make with the ground? Example 3 Solve each triangle. Triangle ABC has vertices A(-3, 3), B(-3, -1), and C(4, -1). Find the side lengths to the nearest hundredth and the angle measures to the nearest degree. Plot points A, B, and C, and draw △ABC. 5 Find the side lengths: AB = 4, BC = 7 A _ Use the distance formula to find the length of AC. AC = ――――――――― + (-1 - 3) = √― 65 ≈ 8.06 √(4 -(-3)) 2 y x 2 _ _ Find the angle measures: AB ⊥ BC , so m∠B = 90°. -5 B 0 C 5 Use an inverse tangent ratio to find ( ) () AB = tan -1 _ 4 ≈ 30°. m∠C = tan -1 _ 7 BC ∠A and ∠C are complementary, so m∠C ≈ 90° - 30° = 60°. Module 18 971 -5 Lesson 4 LANGUAGE SUPPORT IN2_MNLESE389847_U7M18L4 971 Connect Vocabulary Discuss the meaning of solve in the phrase solve a right triangle. Compare solving a triangle to solving an equation. To use the non-mathematical meaning of solve, compare solving a triangle to solving a mystery. A detective uses clues to find missing information. Similarly, to solve a right triangle means to use given information and tools such as the trigonometric ratios to find all the missing side lengths and angle measures. 971 Lesson 18.4 19/04/14 12:03 AM B Triangle DEF has vertices D(-4, 3), E(3, 4), and F(0, 0). Find the side lengths to the nearest hundredth and the angle measures to the nearest degree. Plot points D, E, and F, and draw △DEF. 5 ∠F appears to be a right angle. To check, find the slope x -5 _ -4 4 0 - -4 __ = _ = _; EF : 0 -3 3 -3 Find the side lengths using the distance formula: ――――――――― ―― ―― √(3 - -4 ) + ( 4 - 3) = √ 50 = 5 √ 2 ――――――――――― ―― DF = √( 0 - -4 ) + ( 0 - 3) = √ 25 = 5 , ――――――――――― ―― EF = √( 0 - 3 ) + ( 0 - 4) = √ 25 = 5 2 2 2 0 F 5 -5 so m∠F = 90 °. DE = E D -3 _ 0 -3 3 of DF: _ = _ = -_ ; 4 0 - -4 4 slope of y ≈ 7.07 , 2 2 2 ( ) Use an inverse sine ratio to find m∠D. ( ) 5 EF = sin -1 _ = m∠D = sin -1 _ 45 ° DE 5 √2 ― ∠D and ∠ E are complementary, so m∠ E = 90° - 45 ° = 45 °. 9. How does the given information determine which inverse trigonometric ratio you should use to determine an acute angle measure? It doesn’t – you can use any inverse trigonometric ratio, as long as you use the correct two side lengths. However, if two sides are vertical and horizontal, it makes sense to use inverse tangent. Module 18 IN2_MNLESE389847_U7M18L4 972 972 © Houghton Mifflin Harcourt Publishing Company Reflect Lesson 4 19/04/14 12:03 AM Problem Solving with Trigonometry 972 Your Turn ELABORATE 10. Triangle JKL has vertices J(3, 5), K(-3, 2), and L(5, 1). Find the side lengths to the nearest hundredth and the angle measures to the nearest degree. Plot points J, K, and L, and draw △JKL. INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 Some students may find it difficult to choose ――――――― ― ― ――――――― JL = √(5 - 3) + (1 - 5) = √― 20 = 2 √― 5 ≈ 4.47, ――――――― ― √ KL = √(-3 - 5) + (1 - 2) = 65 ≈ 8.06 JK = √(-3 - 3) + (2 - 5) = √45 = 3 √5 ≈ 6.71, 2 2 5 -5 2 _ _ _ _ _ (_) ― (_― ) Elaborate 11. Would you use the area formula you determined in this lesson for a right triangle? Explain. Possible answer: No. You could use this formula, but since in a right triangle, the lengths of 1 the legs are the base and height, it is simpler to use the formula A = bh. 2 _ 12. Discussion How does the process of solving a right triangle change when its vertices are located in the coordinate plane? Generally, you need to find all the side lengths and all the angles, checking that one is © Houghton Mifflin Harcourt Publishing Company a right angle. The distance formula is generally used to find side lengths rather than trigonometric ratios or the Pythagorean Theorem (although the distance formula depends on the Pythagorean Theorem). The acute angles are found in the same way in both cases. 13. Essential Question Check-In How do you find the unknown angle measures in a right triangle? You first use two known side lengths to form a ratio and then use the appropriate inverse trigonometric ratio to find one angle measure. Then subtract that measure from 90° to find the measure of the other acute angle. Module 18 IN2_MNLESE389847_U7M18L4 973 Lesson 18.4 2 Lx 0 -5 ¯. ∠J appears to be a right angle. To check, find the slope of JK ― ― -3 2-5 1-5 1 -4 = ; slope of JL: = = = -2; so m∠J = 90°. Slope of JK: -6 -3 - 3 2 5-3 2 -1 JK -1 √45 Use an inverse cosine ratio to find m∠K = cos = cos ≈ 34°. KL √65 ∠J and ∠L are complementary, so m∠L ≈ 90° - 34° = 66°. Students may miss parts when solving a right triangle. Remind students that three side lengths and three angle measures are needed. Encourage students to always draw a diagram and to label it whenever they determine a measure. 973 J 2 2 AVOID COMMON ERRORS How do you solve a right triangle? Use the Pythagorean Theorem, trigonometric ratios, inverse trigonometric ratios, and complementary angle relationships to find missing side lengths and angle measures in a right triangle. y K Find the side lengths using the distance formula: a trigonometric or inverse trigonometric ratio to solve a right triangle. Discuss with the class how students choose a particular relationship. For example, some students may prefer to always use the inverse tangent ratio if the lengths of both legs are known to find an acute angle measure. SUMMARIZE THE LESSON 5 973 Lesson 4 19/04/14 12:03 AM Evaluate: Homework and Practice EVALUATE • Online Homework • Hints and Help • Extra Practice Find the area of each triangle to the nearest tenth. 1. In △PQR, PR = 23 mm, QR = 39 mm, and m∠R = 163°. 2. 62° 4.1 cm 3.2 cm Q 39 mm _ R 163° P 23 mm ASSIGNMENT GUIDE _ 1 1 Area = (PR)(QR)sin R = (23)(39) sin 163° ≈ 131.1 mm 2. 2 2 a = 4.1 cm, b = 3.2 cm, and m∠C = 62°; 1 1 Area = ab sin C = (4.1)(3.2)sin 62° ≈ 5.8 cm 2 2 2 _ _ Solve each right triangle. Round lengths to the nearest tenth and angles to the nearest degree. 3. AC 2 + BC 2 = AB 2 A AC 2 + (2.7) = (3.1) 2 3.1 cm 4. F 5. E ―― (_ ) AC = √2.32 ≈ 1.5 cm 2.7 BC m∠A = sin -1 = sin -1 ≈ 61° 3.1 AB ∠A and ∠B are complementary, so m∠B ≈ 90° - 61° = 29°. (_) ∠D and ∠F are complementary, so m∠F = 90° - 56° = 34°. DE EF cos D = tan D = DE DF 26 EF cos 56° = tan 56° = 26 DF 26 26tan 56° = EF DF = ≈ 46.5 m cos56° 38.5 m ≈ EF _ _ _ _ Right △PQR with PQ ⊥ PR , QR = 47 mm, and m∠Q = 52° ∠Q and ∠R are complementary, so m∠R = 90° - 52° = 38°. PQ PR sin P = cos P = QR QR PQ PR cos 52° = sin 52° = 47 47 R _ _ 47 mm 52° P Q 47 cos 52° = PQ 37.0 mm ≈ PR 28.9 mm ≈ PQ Module 18 Exercise IN2_MNLESE389847_U7M18L4 974 _ _ 47 sin 52° = PR Explore Deriving an Area Formula Exercise 16 Example 1 Using the Area Formula Exercises 1–2 Example 2 Solving a Right Triangle Exercises 3–5 Example 3 Solving a Right Triangle in the Coordinate Plane Exercises 6–8 Students may forget to include units when solving area problems. Remind students that the units are an important part of the answer. COMMUNICATING MATH Encourage students solving right triangles to communicate their understanding of the relationships between the trigonometric ratios and their inverses. Lesson 4 974 Depth of Knowledge (D.O.K.) Practice AVOID COMMON ERRORS _ _ _ © Houghton Mifflin Harcourt Publishing Company 56° 26 m D AC 2 = 2.32 B 2.7 cm C 2 Concepts and Skills COMMON CORE Mathematical Practices 1–8 1 Recall of Information MP.4 Modeling 9–15 2 Skills/Concepts MP.4 Modeling 16 3 Strategic Thinking MP.3 Logic 17 3 Strategic Thinking MP.2 Reasoning 18 3 Strategic Thinking MP.4 Modeling 19 3 Strategic Thinking MP.3 Logic 19/04/14 12:03 AM Problem Solving with Trigonometry 974 Solve each triangle. Find the side lengths to the nearest hundredth and the angle measures to the nearest degree. AVOID COMMON ERRORS Students may have difficulty solving right triangles. They may find it easier when they first identify whether they are looking for a ratio or an angle measure. If they are looking for a ratio, they should use sin, cos, or tan. If they are looking for an angle measure, they should use sin -1, cos -1, or tan -1. 6. Triangle ABC with vertices A(-4, 4), B(3, 4), and C(3, -2) 5 A Plot points A, B, and C, and draw △ABC. y ¯, so m∠B = 90° ¯ ⊥ BC AB B 5 C INTEGRATE MATHEMATICAL PRACTICES Focus on Modeling MP.4 For greater accuracy, remind students to use 2 ( ) x 0 -5 ――――――― ― AB = 7, BC = 6, AC = √(3 - 4) + (-2 - 4) = √85 ≈ 9.2, 6 BC m∠A = tan -1 _ = tan -1 _ ≈ 41° 7 AB ∠A and ∠C are complementary, so m∠C ≈ 90° - 41° = 49°. 2 () -5 7. Triangle JKL with vertices J(-3, 1), K(-1, 4), and L(6, 5) K given measurements instead of calculated values in the later steps of solving a right triangle. 5 Plot points J, K, and L, and draw △JKL. y J x 0 -5 ∠J appears to be a right angle. To check, 3 4-1 ¯: = ; slope of JK 2 -1 - (-3) ¯: -5 - 1 = -6 = - 2 ; so m∠J = 90°. slope of JL 9 3 6 - (-3) 5 __ _ _ _ _ ―――――――― 13 ≈ 3.6, JK = √(-1 - (-3)) + (4 - 1) = √― ―――――――― 117 ≈ 10.8, JL = √(6 - (-3)) + (-5 - 1) = √―― ――――――――― 130 ≈ 11.4 KL = √(6 - (-1)) + (5 - (-4)) = √―― ― √ 13 JK ≈ 18° m∠L = sin (_) = sin _ KL √―― 130 2 2 2 -5 L 2 2 2 ( © Houghton Mifflin Harcourt Publishing Company -1 ) -1 ∠K and ∠L are complementary, so m∠K ≈ 90° - 18° = 72°. 8. Triangle PQR with vertices P(5, 5), Q(-5, 3), and R(-4, -2) 5 y P Q x 0 -5 5 Plot points J, K, and L, and draw △JKL. ∠Q appears to be a right angle. To check, ― 3-5 -2 1 = = ; slope of PQ: -5 - 5 5 -10 ― -2 - 3 = -5 = -5 ; so m∠Q = 90°. slope of PR: 1 -4 - (-5) _ _ _ __ _ ――――――― 104 = 2 √― 26 ≈ 10.2, PQ = √(-5 - 5) + (3 - 5) = √―― ――――――――― 26 ≈ 5.1, QR = √(-4 - (-5)) + (-2 - 3) = √― ―――――――― 130 ≈ 11.4 PR = √(-4 - 5) + (-2 - 5) = √―― 2 2 R 2 -5 2 ( ) 2 2 () QR 1 ≈ 27° m∠P = tan -1 _ = tan -1 _ 2 PR ∠P and ∠R are complementary, so m∠R ≈ 90° - 27° = 63°. Module 18 IN2_MNLESE389847_U7M18L4 975 975 Lesson 18.4 975 Lesson 4 19/04/14 12:03 AM 9. Surveying A plot of land is in the shape of a triangle, as shown. Find the area of the plot, to the nearest hundred square yards. _1 ab sin C 2 1 = _(142)(227) sin 128° ≈ 12,700 yd 142 yd COLLABORATIVE LEARNING Have students construct a right triangle and measure the lengths of two sides. Ask them to find the length of the third side and the acute angle measures without using the Pythagorean Theorem or a protractor. Have students construct a second right triangle and measure one acute angle and a side. Have them find the other acute angle measure and the lengths of the other two sides. Finally, have students exchange triangles and check each other’s work. Area = 128° 2 2 227 yd 10. History A drawbridge at the entrance to an ancient castle is raised and lowered by a pair of chains. The figure represents the drawbridge when flat. Find the height of the suspension point of the chain, to the nearest tenth of a meter and the acute angles the chain makes with the wall and the drawbridge, to the nearest degree. height of wall: AB 2 + BC 2 = AC 2 C (3.2)2 + BC 2 = (5.0)2 10.24 + BC 2 = 25 5.0 m Chain A 3.2 m drawbridge B COMMUNICATE MATH BC 2 = 14.76 Wall Have students explain to each other when they would use the Pythagorean Theorem, the sin, cos, or tan ratios, and the sin -1, cos -1, or tan -1 ratios to solve a right triangle. BC ≈ 3.8 m 3.2 BC -1 _ m∠A = cos = cos -1 _ ≈ 50° 5.0 AC ∠A and ∠C are complementary, so m∠C ≈ 90° - 50° = 40°. ( ) ( ) 11. Building For safety, the angle a wheelchair ramp makes with the horizontal should be no more than 3.5°. What is the maximum height of a ramp of length 30 ft? What distance along the ground would this ramp cover? Round to the nearest tenth of a foot. AB _ AC AB sin 3.5° = _ sin C = 30 ft BC _ AC BC cos 3.5° = _ 30 30 cos 3.5° = BC 1.8 ft ≈ AB 29.9 ft ≈ BC IN2_MNLESE389847_U7M18L4 976 C cos C = 30 30 sin 3.5° = AB Module 18 3.5° © Houghton Mifflin Harcourt Publishing Company A B 976 Lesson 4 19/04/14 12:03 AM Problem Solving with Trigonometry 976 12. Multi-Step The figure shows an origami crane as well as a stage of its construction. The area of each wing is shown by the shaded part of the figure, which is symmetric about its vertical center line. Use the information in the figure to find the total wing area of the crane, to the nearest tenth of a square inch. 2.2 in. A 45° F B C E 4.0 in. 3.7 in. D 22.5° Area of each wing = Area of △ ABF + Area of △ DBF - Area of △ DCE 1 Area of △ ABF = (2.2)(2.2) sin 45°; ≈ 1.711 in. 2 2 1 Area of △ DBF = (4.0)(4.0) sin 22.5°; ≈ 3.061 in. 2 2 1 Area of △ DCE = (3.7 in.)(3.7 in.) sin 22.5°; ≈ 2.619 in. 2 2 Area of each wing: 1.711 in. 2 + 3.061 in. 2 - 2.619 in. 2 = 2.153 in. 2 _ _ _ © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Jules Kitano/iStockPhoto.com Area of both wings: 22.153 in. 2 ≈ 4.3 in. 2 13. Right triangle △XYZ has vertices X(1, 4) and Y(2, -3). The vertex Z has positive integer coordinates, and XZ = 5. Find the coordinates of Z and solve △XYZ, giving exact answers. y 5 X Z x 0 -5 5 Y -5 Based on 3–4–5 right triangles as well as horizontal and vertical displacements of 5, possible coordinates for Z are (1, 9), (4, 8), (5, 7), (6, 4), and (5, 1) . Z(5, 1) is the only possibility that appears to create a right triangle. ― 1 - 4 = -3 = - 3 ; slope of YZ ―: 1 -(-3) = 4 ; so m∠Z = 90°. To check, slope of XZ: 5-1 4 5 -2 3 4 _ _ _ _ _ ――――――― 50 = 5 √―2 , XZ = √――――――― (5 - 1) + (1 - 4) = √― 25 = 5, XY = √(2 - 1) + (-3 - 4) = √― ―――――――― 25 = 5 YZ = √(5 - 2) + (1 - (-3)) = √― 2 2 ― 2 2 2 2 ― Since XZ ≅ YZ, △ XYZ is a 45°-45°-90° triangle. Therefore, m∠X = m∠Y = 45°. Module 18 IN2_MNLESE389847_U7M18L4 977 977 Lesson 18.4 977 Lesson 4 19/04/14 12:03 AM 14. Critique Reasoning Shania and Pedro are discussing whether it is always possible to solve a right triangle, given enough information, without using the Pythagorean Theorem. Pedro says that it is always possible, but Shania thinks that when two side lengths and no angle measures are given, the Pythagorean Theorem is needed. Who is correct, and why? Pedro; Possible answer: When one side and one acute angle are given, trigonometric ratios can be used to find the other sides, and complementary angles give the other acute angle. When two sides are given, an inverse trigonometric ratio can be used to find one acute angle, and this angle together with a given side can be used with a different trigonometric ratio to find the other side. The other acute angle is complementary as before. 15. Design The logo shown is symmetrical about one of its diagonals. Find the angle measures in △CAE, to the nearest degree. (Hint: First find an angle in △ABC, △CDE or △AEF) Then, find the area of △CAE, without first finding the areas of the other triangles. 3 BC 6 mm = tan -1 _ ≈ 26.565° m∠BAC = tan -1 __ A 6 AB ( ) () B ∠BAC ≅ ∠EAF, so m∠EAF = m∠BAC ≈ 26.565° 3 mm m∠BAC + m∠CAE + m∠EAF = 90° C 26.565° + m∠CAE + 26.565° ≈ 90° 3 mm m∠CAE ≈ 90° - 2(26.565°) F ≈ 36.869° ≈ 37° AC 2 = AB 2 + BC 2 ∠AEC ≅ ∠ACE, so m∠AEC = m∠ACE AC 2 = 6 2 + 3 2 = 45 m∠AEC + m∠ACE + m∠CAE = 90° 2(m∠AEC) + 36.869° ≈ 90° m∠AEC ≈ 27° m∠ACE ≈ 27° D E ― ― ― ― AC = √45 = 3 √5 1 Area = (AC)(AE) sin(∠CAE) 2 1 ( √ )( √ ) ≈_ 3 5 3 5 sin(36.869°) ≈ 13.5 mm 2 2 _ A 1 ac sin B Area formula: Area = _ 2 b But △ABC has base a and height h = c sin (∠ABD), 1 (a)(c sin (∠ABD)). so Area = _ 2 C a c h B D Equating these two expressions for the area, sin B = sin ∠ABD But since, for any obtuse ∠B, ∠B and ∠ABD are supplementary in this © Houghton Mifflin Harcourt Publishing Company 16. Use the area formula for obtuse ∠B in the diagram to show that if an acute angle and an obtuse angle are supplementary, then their sines are equal. construction, this proves the result. Module 18 IN2_MNLESE389847_U7M18L4 978 978 Lesson 4 19/04/14 12:02 AM Problem Solving with Trigonometry 978 JOURNAL H.O.T. Focus on Higher Order Thinking Have students create an example that shows how to solve a right triangle. Students should specify the known side or angle measures and explain how to find the unknown side lengths and angle measures. 17. Communicate Mathematical Ideas The HL Congruence Theorem states that for right triangles ABC and DEF such that ― _ ― ― ∠A and ∠D are right angles, BC ≅ EF, and AB ≅ DE, △ABC ≅ △DEF. E C A B Explain, without formal proof, how solving a right triangle F with given leg lengths, or with a given side length and acute angle measure, shows that right triangles with both legs congruent, or with corresponding sides and angles congruent, must be congruent. ¯ ≅ DF ¯ ≅ DE ¯ and AC ¯. Solving either of these right triangles determines the Suppose AB D length of the hypotenuse in the same way, e.g., using the Pythagorean Theorem, so BC = EF and therefore, by SSS ≅, △ABC ≅ △DEF. Suppose ∠B ≅ ∠E. The given corresponding side lengths allow the unknown sides to be calculated in the same way using trigonometric ratios, so that all corresponding side lengths are equal and therefore all corresponding sides are congruent. Again, by SSS ≅, △ABC ≅ △DEF. 18. Persevere in Problem Solving Find the perimeter and area of △ABC, as exact numbers. Then, find the measures of all the angles to the nearest degree. ―――――――― 26 , √(3 -(-2)) + (3 - 2) = √― ――――――――― ― 2 AC = √(3 -(-2)) + (-3 -(-2)) = √50 = 5 √― 26 + 5 √― 2 perimeter: BC + AB + AC = 6 + √― 2 BC = 6, AB = 2 _― _ _ A x 2 _ ― y B 2 0 -5 With base b = BC = 6, height h = 5, 1 1 so area = bh = (6)(5) = 15. 2 2 1 1 Area = ab sin C 15 = (5 √2 )( √26 ) sin A 2 2 3 √13 15 = 1 (6)(5 √2 )sin C = sin A 2 13 √2 = sin C √13 -1 3 2 sin = m∠A 13 45° = m∠C 56° ≈ m∠A _ © Houghton Mifflin Harcourt Publishing Company 5 ― _ ― ) (_ _ ― 5 C -5 ― Angle sum of a triangle: 56° + m∠B + 45° ≈ 180°; m∠B ≈ 79° 19. Analyze Relationships Find the area of the triangle using two different formulas, and deduce an expression for sin2θ. 1 1 Area = (x)(x)sin 2θ Area = bh 2 2 1 1 = x 2 sin 2θ = (2x sin θ)(x cos θ) 2 2 = x 2 sin θ cos θ 1 2 x sin 2θ = x 2 sin θ cos θ → sin 2θ = 2sin θ cos θ 2 _ _ _ _ x θθ x _ Module 18 IN2_MNLESE389847_U7M18L4 979 979 Lesson 18.4 979 Lesson 4 19/04/14 12:02 AM Lesson Performance Task 95 .84 p 1. Draw and label a triangle with the dimensions shown. 2. Find the area of the triangle in square centimeters. Show your work. H O 104.45° Students are probably familiar with the metric prefixes deci-, centi-, and milli-, but may not know the next three smaller prefixes. They are: pm .84 95 Every molecule of water contains two atoms of hydrogen and one atom of oxygen. The drawing shows how the atoms are arranged in a molecule of water, along with the incredibly precise dimensions of the molecule that physicists have been able to determine. (1 pm = 1 picometer = 10 -12m) m CONNECT VOCABULARY micro- = 10 -6 = 0.000 001 H nano- = 10 -9 = 0.000 000 001 pico- = 10 -12 = 0.000 000 000 001 3. Find the distance between the hydrogen atoms in centimeters. Explain your method. 1. A .84 pm 95 104.45° AVOID COMMON ERRORS 95 .84 pm B 2. The conversion from 95.84 picometers to centimeters may trip students up. Here are the steps and reasons: C 95.84 pm = 95.84 × 10 -12 m AB = AC = 95.84 pm = 95.84 × 10 -12 m = 95.84 × 10 -10 cm = 9.584 × 10 -9 cm 1 1 area = AB(AC) sin A = (9.584 × 10 -9 cm)(9.584 × 10 -9 cm) sin 104.45° 2 2 1 = (9.584 × 10 -9 cm)(9.584 × 10 -9 cm)(0.9684) 2 ≈ 44.48 × 10 -18 cm 2 _ _ _ (Definition of pm) = 95.84 × 10 -10 cm (Multiply by 10 2 cm = 1 m.) ≈ 4.45 × 10 -17 cm 2 3. ― ―_ Possible answer: Label the midpoint of BC point D. Draw AD. AD bisects ∠A, forming = 9.584 × 10 -9 cm two 52.225° angles. (Divide 95.84 by 10 and multiply 10 -10 by 10) _ _ -8 BC = 2 × 75.71 = 151.42 pm = 151.42 × 10 -12 m = 151.42 × 10 -10 cm = 1.51 × 10 cm Module 18 980 © Houghton Mifflin Harcourt Publishing Company In right triangle ABD, BD sin 52.225 = AB BD 0.79 ≈ 95.84 75.71 ≈ BD Lesson 4 EXTENSION ACTIVITY IN2_MNLESE389847_U7M18L4 980 This Lesson Performance Task will give students a glimpse into the complex field of molecular geometry. While most of the subject lies beyond the range of high-school geometry students, there are many questions students can research and report on at a basic level. Among them: • What is molecular geometry? • How was the molecular structure of water discovered? • What determines the angles between the atoms in a molecule? • What holds a molecule together? 19/04/14 12:02 AM Scoring Rubric 2 points: The student’s answer is an accurate and complete execution of the task or tasks. 1 point: The student’s answer contains attributes of an appropriate response but is flawed. 0 points: The student’s answer contains no attributes of an appropriate response. Problem Solving with Trigonometry 980 LESSON 18.5 Name Using a Pythagorean Identity Essential Question: How can you use a given value of one of the trigonometric functions to calculate the values of the other functions? Explore The student is expected to: Proving a Pythagorean Identity In the previous lesson, you learned that the coordinates of any point (x, y) that lies on the unit circle where the y terminal ray of an angle θ intersects the circle are x = cos θ and y = sin θ, and y = sin θ that tan θ = _ x . Combining F-TF.C.8 Prove the Pythagorean identity sin 2(θ) + cos 2(θ) = 1 and use it to find sin(θ), cos(θ), or tan(θ) given sin(θ), cos(θ), or tan(θ) and the quadrant of the angle. sin θ these facts gives the identity tan θ = ____ , which is true for all values of θ where cos θ ≠ 0. In the following Explore, cos θ you will derive another identity based on the Pythagorean theorem, which is why the identity is known as a Pythagorean identity. Mathematical Practices COMMON CORE Date 18.5 Using a Pythagorean Identity Common Core Math Standards COMMON CORE Class MP.1 Problem Solving The terminal side of an angle θ intersects the unit circle at the point (a, b) as shown. Write a and b in terms of trigonometric functions involving θ. a= Language Objective b= Explain to a partner what the Pythagorean Theorem and the Pythagorean identity are and how they are related. cos θ sin θ y (a, b) 1 b θ a Essential Question: How can you use a given value of one of the trigonometric functions to calculate the values of the other functions? The values of the sine, cosine, and tangent functions are connected by the identities sinθ sin 2(θ) + cos 2(θ) = 1 and tanθ = . Once you cosθ know the value of one of the functions, you can use the identities to solve for the others, using the quadrant of the terminal side of θ to assign the correct sign. _ PREVIEW: LESSON PERFORMANCE TASK Apply the Pythagorean theorem to the right triangle in the diagram. Note that when a trigonometric function is squared, the exponent is typically written immediately after the name of the function. For 2 instance, (sinθ) = sin 2θ. Write the Pythagorean Theorem. a + b2 = c 2 Substitute for a, b, and c. Square each expression. Module 18 2 ( cos θ ) + ( sin θ ) = 2 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction 2 + 2 cos θ = 2 sin θ 2 1 1 Lesson 5 981 gh "File info" made throu Date Class n Identity thagorea Name a Py 18.5 Using to ic functions trigonometr one of the value of Resource use a given other functions? (θ), cos(θ), or Locker can you it to find sin of the ion: How 2 1 and use the values 2 cos (θ) = calculate y sin (θ) + of the angle. orean identit quadrant the Pythag (θ) and the F-TF.C.8 Prove (θ), cos(θ), or tan tity Iden sin where the tan(θ) given agorean y the unit circle _ ving a Pyth . Combining that lies on y = ) x, Pro θ x point ( e, θ that tan Explore nates of any ing Explor and y = sin the coordi the follow y = sin θ, θ ≠ 0. In learned that cos θ and cos = you , x a as are of θ where us lesson known cts the circle identity is In the previo for all values angle θ interse is why the sin θ , which is true ____ ray of an m, which θ = cos θ terminal orean theore identity tan the Pythag gives the and b y based on these facts . Write a another identit b) as shown you will derive point (a, circle at the identity. cts the unit Pythagorean interse θ angle θ. Quest Essential COMMON CORE IN2_MNLESE389847_U7M18L5.indd 981 an involving al side of ic functions The termin of trigonometr in terms cos θ a= sin θ b= HARDCOVER PAGES 981992 Watch for the hardcover student edition page numbers for this lesson. y (a, b) 1 θ y g Compan View the online Engage. Discuss the photo and how light and shadows can help determine where to place a cell tower. Then preview the Lesson Performance Task. © Houghton Mifflin Harcourt Publishing Company ENGAGE x n Mifflin Harcour t Publishin x ometric when a trigon m. Note that of the function. For the diagra triangle in after the name to the right immediately theorem lly written Pythagorean ent is typica Apply the d, the expon 2 is square 2 . 2 2 2 function c θ a +b = (sinθ) = sin 2 instance, 2 em. Theor 2 1 Pythagorean Write the sin θ = cos θ + for a, Substitute © Houghto b a b, and c. expression. Square each ( 2 cos θ ) ( + 2 sin θ ) = 1 Lesson 5 981 Module 18 8L5.indd 47_U7M1 ESE3898 IN2_MNL 981 Lesson 18.5 981 19/04/14 12:24 AM 19/04/14 12:26 AM Reflect 1. EXPLORE The identity is typically written with the sine function first. Write the identity this way, and explain why it is equivalent to the one in Step B. 2 2 The rewritten identity is sin θ + cos θ = 1 . It is equivalent to the original identity because Proving a Pythagorean Identity of the Commutative Property of Addition. 2. π. Confirm the Pythagorean identity for θ = _ 3 √― 3 π π 1 ; therefore, sin _ = ___ and cos _ = _ 2 3 3 2 2 √― 3 2 2 2 _ 2 _ π π 1 sin θ + cos θ = sin + cos = ____ + _ 2 3 3 2 ( ) 3. INTEGRATE TECHNOLOGY ( ) ( ) ( ) ( ) ( ) = _34 + _14 = 1. Students have the option of completing the Explore activity either in the book or online. 2 3π . Confirm the Pythagorean identity for θ = _ 4 √― √― 2 2 3π 3π and cos _ = -___ . Therefore, sin _ = ___ 2 2 4 4 ( ) ( ) sin θ + cos θ = sin 2 2 Explain 1 2 ( ) ( ) 2 3π 3π _ + cos _ = 4 4 √2 ) (_ _ 2 2 CONNECT VOCABULARY ( √2 _ ) _ + - 2 2 Have the class list expressions that contain the word identity, such as identity theft, or student Identity card. Ask them to discuss the meaning of identity in each expression. Remind them that 0 is the identity element for addition and 1 is the identity element for multiplication. Then have them discuss the use of the word identity in the context presented in the lesson. 1 +_ 1 = 1. =_ 2 2 Finding the Value of the Other Trigonometric Functions Given the Value of sin θ or cos θ You can rewrite the identity sin θ + cos θ = 1 to express one trigonometric function in terms of the other. As shown, each alternate version of the identity involves both positive and negative square roots. You can determine which sign to use based on knowing the quadrant in which the terminal side of θ lies. 2 2 Solve for sin θ Solve for cos θ sin2 θ + cos2 θ = 1 sin2 θ = 1 - cos2θ __ cos θ = ± √1 - sin2θ © Houghton Mifflin Harcourt Publishing Company sin θ = ± √1 - cos2θ How do you know the hypotenuse is equal to 1? The radius of a unit circle is 1. cos2 θ = 1 - cos2θ __ Example 1 QUESTIONING STRATEGIES sin2 θ + cos2 θ = 1 Find the approximate value of each trigonometric function. π , find cos θ. Given that sinθ = 0.766 where 0 < θ < _ 2 __ Use the identity to solve for cosθ. cos θ = ±√1 - sin2 θ __ Substitute for sinθ. = ±√1 -(0.766) Use a calculator, then round. ≈ ±0.643 2 Is (cosθ) the same as cos 2θ? Is (sinθ) the same as sin 2θ? Yes, they are different ways of writing the same thing, because the angle measure is not actually squared. 2 2 If cos 2θ has a value of 0.36, what do you know about sin 2θ and sinθ? 2 sin θ = 1 - 0.36 = 0.64, so sinθ = √0.64 = 0.8. ―― The terminal side of θ lies in Quadrant I, where cos θ > 0. So, cos θ ≈ 0.643. EXPLAIN 1 Module 18 982 Lesson 5 PROFESSIONAL DEVELOPMENT IN2_MNLESE389847_U7M18L5.indd 982 Integrate Mathematical Practices This lesson provides an opportunity to address Mathematical Practice MP.1, which calls for students to “make sense of problems and persevere in solving them.” Students analyze given information and constraints and plan a pathway to the solution. The solution process may involve transforming established identities in order to make use of the given information. In the course of solving the problem, students may revisit the given information to make any additional decisions regarding the solution. 19/04/14 12:26 AM Finding the Value of the Other Trigonometric Functions Given the Value of sin θ or cos θ QUESTIONING STRATEGIES Is 1 - sin 2θ ever negative? Why or why not? No; -1 ≤ sinθ ≤ 1, so 0 ≤ sin 2θ ≤ 1. Therefore, 1 - sin 2θ can never be less than 0. Using a Pythagorean Identity 982 B QUESTIONING STRATEGIES 3π Given that cos θ = -0.906 where π < θ < ___ , find sinθ. 2 Do you need to know the precise value of θ to solve these problems? Why or why not? No; the Pythagorean identity allows you to find the cosine of θ given the sine of θ without knowing the value of θ itself. __ sin θ = ±√1 - cos2θ Use the identity to solve for sinθ. ___ √ ( Substitute for cos θ. = ± 1 - -0.906 Use a calculator, then round. ≈ ± 0.423 ) 2 The terminal side of θ lies in Quadrant III , where sin θ < 0. So, sin θ ≈ -0.423 . Reflect Do you need to know the quadrant where θ terminates to solve these problems? Why or why not? Yes; the sign of the answer varies depending on the quadrant in which θ terminates. 4. π π Suppose that __ < θ < π instead of 0 < θ < __ in part A of this Example. How does 2 2 this affect the value of sin θ? π < θ < π, then the terminal side of θ lies in Quadrant II. Since cosine is negative in If __ 2 Quadrant II, the value of cos θ would be negative. 5. AVOID COMMON ERRORS 3π 3π Suppose that ___ < θ < 2π instead of π < θ < ___ in part B of this Example. How 2 2 does this affect the value of sin θ ? < θ < 2π, then the terminal side of θ lies in Quadrant IV. Sine is also negative in If 3π 2 ___ ― Students may erroneously reason that since sin 2θ + cos 2θ = 1, sinθ + cosθ = √1 = 1. Use a special angle to demonstrate that this is not true, and correct the erroneous reasoning by reminding students that √x 2 + y 2 ≠ x + y. Quadrant IV, so the value of sin θ would not change. 6. ――― Explain how you would use the results of part A of this Example to determine the approximate value for tan θ. Then find it. sin θ To determine the approximate value for tanθ, use the identity tan θ = . cos θ 0.766 ____ tan θ = ≈ 1.191 _ 0.643 Your Turn 7. 3π Given that sinθ = -0.644 where π < θ < ___ , find cosθ. 2 _ © Houghton Mifflin Harcourt Publishing Company cos θ = ±√1 - sin2θ ___ = ±√1 - (-0.644) 2 ≈ ±0.765 Since θ lies in Quadrant III, where cos θ < 0, cos θ ≈ -0.765. 8. π Given that cos θ = -0.994 where __ < θ < π, find sin θ. Then find tanθ. 2 __ sin θ = ±√1 - cos2 θ ___ 2 = ±√1 - (-0.994) = ±0.109 Since θ lies in Quadrant II, where sin θ > 0, sin θ ≈ 0.109. 0.109 tan θ = sinθ ≈ _____ ≈ -0.110 cosθ -0.994 _ Module 18 983 Lesson 5 COLLABORATIVE LEARNING IN2_MNLESE389847_U7M18L5.indd 983 Peer-to-Peer Activity Have students work in pairs. Instruct pairs to write three different identities that show the relationships among the sine, cosine, and tangent functions. Then have them verify each of the three identities for θ = 30°, θ = 45°, and θ = 60°. Have each pair compare their work with that of another pair. 983 Lesson 18.5 19/04/14 12:26 AM Explain 2 Finding the Value of Other Trigonometric Functions Given the Value of tanθ EXPLAIN 2 sinθ If you multiply both sides of the identity tanθ = ____ by cosθ, you get the identity cosθtanθ = sinθ, or cosθ sinθ sinθ = cosθ tanθ. Also, if you divide both sides of sinθ = cosθ tanθ by tanθ, you get the identity cosθ = ____ . tanθ You can use the first of these identities to find the sine and cosine of an angle when you know the tangent. Example 2 Finding the Value of the Other Trigonometric Functions Given the Value of tan θ Find the approximate value of each trigonometric function. π Given that tanθ ≈ -2.327 where __ < θ < π, find the values of sinθ and cosθ. 2 First, write sinθ in terms of cosθ. Use the identity sinθ = cosθ tanθ. QUESTIONING STRATEGIES sinθ = cosθ tanθ ≈ -2.327cosθ Substitute the value of tanθ. Is there another way to begin the problem besides writing sinθ in terms of cosθ? Yes, you can write cosθ in terms of sinθ. Now use the Pythagorean Identity to find cosθ. Then find sinθ. sin2θ = cos θ = 1 2 Use the Pythagorean Identity. Substitute for sinθ. (-2327 cosθ)2 + cos2θ ≈ 1 How can you rewrite the Pythagorean identity with cosθ written in terms of sinθ and tanθ? 5.415 cos2θ + cos2θ ≈ 1 Square. (_) = 1 6.415 cos2θ ≈ 1 Combine like terms. Solve for cos2θ. cos2 ≈ 0.156 Solve for cosθ. cosθ ≈ ±0.395 sin 2θ + sinθ tanθ INTEGRATE TECHNOLOGY The terminal side of θ lies in Quadrant II, where cosθ < 0. Therefore, cosθ ≈ −0.395 and sinθ ≈ −2.327cosθ ≈ 0.919. Students can use a graphing calculator to check that the sum of the squares of the values that they find for sinθ and cosθ is 1. Note that, because the values are approximations, the calculator may return a sum that is not exactly 1. 3π Given that tanθ ≈ -4.366 where ___ < θ < 2π, find the values of sinθ and cosθ. 2 First, write sinθ in terms of cosθ. Use the identity sinθ = cosθ tanθ. sinθ = cosθ tanθ © Houghton Mifflin Harcourt Publishing Company ≈ -4.366 cosθ Substitute the value of tanθ. Now use the Pythagorean Identity to find cosθ . Then find sinθ . Use the Pythagorean Identity. sin2θ + cos2θ = 1 ( -4.366 cosθ) + cos θ ≈ 1 2 Substitute for sinθ. 2 19.062 cos2θ + cos2θ ≈ 1 Square. Combine like terms. 2 20.062 cos2θ ≈ 1 Solve for cos2θ. cos2θ ≈ 0.050 Solve for cosθ. cosθ ≈ 0.223 The terminal side of θ lies in Quadrant IV , where cosθ > 0. Therefore, cosθ ≈ 0.223 and sinθ ≈ -4.366 cosθ ≈ -0.974 . Module 18 984 Lesson 5 DIFFERENTIATE INSTRUCTION IN2_MNLESE389847_U7M18L5.indd 984 19/04/14 12:26 AM Multiple Representations Students may benefit from relating the concepts in this lesson to right triangle trigonometry. For example, if students are given that sinθ = 0.766 and 0 < θ < __π2 , they can draw a right triangle in the first quadrant, with angle θ at the origin and the side opposite to label adjacent leg on the x-axis. They can then use the definition sinθ = _________ hypotenuse the vertical leg 766 units and the hypotenuse 1000 units, and use the Pythagorean side adjacent Theorem to find the length of the missing leg. Finally, the definitions cosθ = _________ hypotenuse side oppositte can be used to find these values. Students can check their and tanθ = _________ side adjacent results using the methods from the lesson. In doing so, students will likely see the connection between the two methods of solution. Using a Pythagorean Identity 984 Reflect ELABORATE 9. QUESTIONING STRATEGIES In part A of this Example, when you multiplied the given value of tanθ by the calculated value of cosθ in order to find the value of sinθ, was the product positive or negative? Explain why this is the result you would expect. The product was positive. This is expected, since sinθ should be positive in Quadrant II. Suppose you know the sine and tangent of an angle. Would you also need to be given information about the quadrant in which the angle terminates in order to find the cosine of the angle? Explain. No; the signs of the sine and tangent of the angle would be enough information to determine the quadrant in which the angle terminates and, thus, the sign of the cosine. π , show that you can solve for sinθ and cosθ exactly 10. If tanθ = 1 where 0 < θ < _ 2 using the Pythagorean identity. Why is this so? If tanθ = 1, then sinθ = cosθ , so sin2θ + cos2 θ = 1 becomes 2 sin2 θ = 1, which gives the _ √2 π result sinθ = cosθ = __ . This occurs because θ is the special angle __ . 4 2 Your Turn 3π , find the values of sinθ and cosθ. 11. Given that tanθ ≈ 3.454 where π < θ < _ 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ 3.454 cosθ Solve for cosθ and sinθ. INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Challenge students to use the identities in the sin2θ + cos2θ = 1 (3.454 cosθ )2+ cos2θ ≈ 1 11.930 cos2θ + cos2θ ≈ 1 12.930 cos2θ ≈ 1 lesson to write a quotient identity that defines in cos2θ ≈ 0.0773 1 - cos θ 1 or tan 2θ = ____ -1 terms of cosθ. tan 2θ = _______ cos 2θ cos 2θ 2 cosθ ≈ ±0.278 Since cosine is negative and sine is negative in Quadrant III, cosθ = 0.278 and If you know the sine of an angle, how can you find the cosine and tangent of the angle? To find the cosine, you can use the identity sin θ + cos θ = 1, substitute the given value for 2 2 sin θ, and solve for cos θ. You can then substitute the values of sin θ and cos θ in the identity sinθ tan θ = ____ to find the tangent. cosθ sinθ ≈ -3.454 cosθ ≈ -0.960. © Houghton Mifflin Harcourt Publishing Company SUMMARIZE THE LESSON Elaborate 12. What conclusions can you draw if you are given only the information that tanθ = −1? Possible Answer: Since the tangent is negative, the terminal angle must be either in Quadrant II, where sine is positive and cosine is negative, or in Quadrant IV, where sine is negative and cosine is positive. More specifically, sinθ = cosθ , so sin 2θ + cos 2θ = 1 ― ― ― ― becomes 2 cos 2 = 1, so cosθ = ± ____ . Then sinθ = ____ = and cosθ = -____ or sinθ = -____ 2 2 2 2 ― √ 2 and cosθ = ____ . 2 √2 Module 18 √2 985 √2 √2 Lesson 5 LANGUAGE SUPPORT IN2_MNLESE389847_U7M18L5.indd 985 Communicate Math Have students work in pairs. The first student explains what the Pythagorean Theorem is while the second student takes notes. They switch roles and repeat the procedure with the Pythagorean identity. Have them collaborate to describe the relationship between the two, as well as similarities and differences. Encourage students to use pictures and symbols as well as verbal descriptions when explaining to their partners. 985 Lesson 18.5 19/04/14 12:26 AM 13. Discussion Explain in what way the process of finding the sine and cosine of an angle from the tangent ratio is similar to the process of solving a linear equation in two variables by substitution. When solving a linear equation in two variables, you use one equation to find an EVALUATE expression for one variable in terms of the other, then substitute that expression in the second equation to obtain an equation in one variable that you can solve for that variable and use to find the value of the other variable. When finding the sine and cosine of an angle from the tangent ratio, you use the known tangent value and the identity sinθ = cosθ tanθ to write an expression for sinθ in terms of cosθ, then substitute that ASSIGNMENT GUIDE expression in the second identity sin 2θ + cos 2θ = 1 to obtain an equation that you can solve for cosθ, and then use the value of cosθ to find the value of sinθ. 14. Essential Question Check-In If you know only the sine or cosine of an angle and the quadrant in which the angle terminates, how can you find the other trigonometric ratios? If you know only sinθ or cosθ, you can find the other by substituting the known value in the Pythagorean identity sin θ + cos θ = 1 and solving for the unknown value. Then 2 2 sinθ you can find the tangent of the angle using the identity tanθ = ____ . In all cases, use the cosθ quadrant in which the angle terminates to choose the correct sign for the ratio. Evaluate: Homework and Practice • Online Homework • Hints and Help • Extra Practice Find the approximate value of each trigonometric function. 1. π Given that sinθ = 0.515 where 0 < θ < __ , find cosθ. ――― 2 ――――― cosθ = ± √1 - sin 2θ = ± √1 - (0.515) ≈ ±0.857 Since θ lies in Quadrant I, where cosθ > 0, cosθ ≈ 0.857. 2. 3π Given that cosθ = 0.198 where ___ < θ < 2π, find sinθ. 2 ―――― ――――― sinθ = ± √1 - cos 2θ = ± √1 - (0.198) ≈ ±0.980 2 Since θ lies in Quadrant IV, where sinθ < 0, sinθ ≈ −0.980. 3. 3π Given that sinθ = −0.447 where ___ < θ < 2π, find cosθ. 2 ――――― ――― cosθ = ± √1 - sin θ = ± √1 - (-0.447) ≈ ±0.895 2 2 ――――― 2 Since θ lies in Quadrant II, where sinθ > 0, sinθ ≈ 0.839. Exercise IN2_MNLESE389847_U7M18L5.indd 986 Lesson 5 986 Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–16 1 Recall of Information MP.7 Using Structure 17 1 Recall of Information MP.6 Precision 18–19 1 Recall of Information MP.7 Using Structure 20 2 Skills/Concepts MP.4 Modeling 21 3 Strategic Thinking MP.4 Modeling 22–23 2 Skills/Concepts MP.3 Logic 24–25 3 Strategic Thinking MP.3 Logic Example 1 Finding the Value of the Other Trigonometric Functions Given the Value of sin θ or cos θ Exercises 1–8, 20 Example 2 Finding the Value of the Other Trigonometric Functions Given the Value of tan θ Exercises 9–17, 21 Suggest that students draw a sketch of the angle described in the problem, placing the terminal side of the angle in the correct quadrant. Students can then write the signs of each of the functions for that quadrant so that they will remember to affix the proper signs to the calculated values. sinθ = ± √1 - cos 2θ = ± √1 - (-0.544) ≈ ±0.839 Module 18 Exercises 18–19 VISUAL CUES π Given that cosθ = −0.544 where __ < θ < 2π, find sinθ. 2 ――― Explore Proving a Pythagorean Identity If you are given the cosine of an angle, why is it necessary to know in which quadrant the terminal side of the angle lies in order to determine the sine of the angle? Knowing the quadrant enables you to determine the sign of the sine. Since θ lies in Quadrant IV, where cosθ > 0, cosθ ≈ 0.895. 4. Practice QUESTIONING STRATEGIES © Houghton Mifflin Harcourt Publishing Company 2 Concepts and Skills 19/04/14 12:26 AM Using a Pythagorean Identity 986 5. AVOID COMMON ERRORS 3π Given that sinθ = −0.908 where π < θ < ___ , find cosθ. 2 ――― ――――― cosθ = ± √1 - sin 2θ = ± √1 - (-0.908) ≈ ±0.419 2 Since θ lies in Quadrant III, where cosθ > 0, cosθ ≈ −0.419. When making a substitution from the transformation sinθ into the Pythagorean of the identity tanθ = ___ cosθ 6. identity, some students may forget to square the coefficient of the substituted expression. Review that π < θ < π, find cosθ. Given that sinθ = 0.313 where __ 2 ――― ――――― cosθ = ± √1 - sin 2θ = ± √1 - (-0.313) ≈ ±0.950 2 Since θ lies in Quadrant II, where cosθ < 0, cosθ ≈ −0.950. sin 2θ means (sinθ) , and the same for cosine, and encourage students to use parentheses when making the substitution. 2 7. π , find sinθ. Given that cosθ = 0.678 where 0 < θ < __ 2 ―――― ――――― sinθ = ± √1 - cos 2θ = ± √1 - (-0.678) ≈ ±0.735 2 Since θ lies in Quadrant I, where sinθ > 0, sinθ ≈ 0.735. INTEGRATE TECHNOLOGY 8. 3π , find sinθ. Given that cosθ = -0.489 where π < θ < ___ 2 ―――― ――――― sinθ = ± √1 - cos 2θ = ± √1 - (-0.489) ≈ ±0.872 Encourage students to use their calculators to check their results, checking that for the given 2 Since θ lies in Quadrant III, where sinθ < 0, sinθ ≈ -0.872. sinθ . angle θ, tanθ = ___ Find the approximate value of each trigonometric function. cosθ 9. __ < θ < π, find the values of sinθ and cosθ. Given that tanθ ≈ −3.966 where π 2 sinθ = cosθ tanθ ≈ -3.966 cosθ sin 2θ + cos 2θ = 1 (-3.966 cosθ)2 + cos 2θ ≈ 1 © Houghton Mifflin Harcourt Publishing Company cos 2θ ≈ 0.060 → cosθ ≈ ±0.245 In Quad. II, cosine is neg. and sine is pos., so cosine is negative and sine is positive. cosθ ≈ −0.245 and sinθ ≈ −3.966 cosθ ≈ 0.972. 3π < θ < 2π, find the values of sinθ and cosθ. 10. Given that tanθ ≈ −4.580 where ___ 2 sinθ = cosθ tanθ ≈ -4.580 cosθ sin 2θ + cos 2θ = 1 (-4.580 cosθ)2 + cos 2θ ≈ 1 cos 2θ ≈ 0.046 → cosθ ≈ ±0.214 Since cosine is positive in Quadrant IV, cosθ ≈ 0.214 and sinθ ≈ −4.580 cosθ ≈ 0.980. Module 18 IN2_MNLESE389847_U7M18L5.indd 987 987 Lesson 18.5 987 Lesson 5 19/04/14 12:26 AM π 11. Given that tanθ ≈ 7.549 where 0 < θ < __ , find the values of sinθ and cosθ. 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ 7.549 cosθ Solve for sinθ and cosθ. sin 2θ + cos 2θ = 1 (7.549 cosθ)2 + cos 2θ ≈ 1 56.987 cos 2θ + cos 2θ ≈ 1 57.987 cos 2θ ≈ 1 cos 2θ ≈ 0.017 cosθ ≈ ±0.130 Since cosine is positive in Quadrant I, cosθ ≈ −0.130 and sinθ ≈ 7.549 cosθ ≈ 0.981. 3π , find the values of sinθ and cosθ. 12. Given that tanθ ≈ 4.575 where π < θ < ___ 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ 4.575 cosθ Solve for sinθ and cosθ. sin 2θ + cos 2θ = 1 (4.575 cosθ)2 + cos 2θ ≈ 1 20.931 cos 2θ + cos 2θ ≈ 1 21.931 cos 2θ ≈ 1 cos 2θ ≈ 0.046 cosθ ≈ ±0.214 Since cosine is negative in Quadrant III, cosθ ≈ −0.214 and sinθ ≈ 4.575 cosθ ≈ -0.979. © Houghton Mifflin Harcourt Publishing Company 3π , < θ < 2π find the values of sinθ and cosθ. 13. Given that tanθ ≈ -1.237 where ___ 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ −1.237 cosθ Solve for sinθ and cosθ. sin 2θ + cos 2θ = 1 (-1.237 cosθ)2 + cos 2θ ≈ 1 1.530 cos 2θ + cos 2θ ≈ 1 2.530 cos 2θ ≈ 1 cos 2θ ≈ 0.395 cosθ ≈ ±0.628 Since cosine is negative in Quadrant IV, cosθ ≈ 0.628 and sinθ ≈ -1.237 cosθ ≈ -0.777. Module 18 IN2_MNLESE389847_U7M18L5.indd 988 988 Lesson 5 19/04/14 12:26 AM Using a Pythagorean Identity 988 3π 14. Given that tanθ ≈ 5.632 where π < θ < ___ , find the values of sinθ and cosθ. 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ 5.632 cosθ Solve for sinθ and cosθ. sin 2θ + cos 2θ = 1 (5.632 cosθ)2 + cos 2θ ≈ 1 31.719 cos 2θ + cos 2θ ≈ 1 32.719 cos 2θ ≈ 1 cos 2θ ≈ 0.031 cosθ ≈ ±0.176 Since cosine is negative in Quadrant III, cosθ ≈ -0.176 and sinθ ≈ 5.632 cosθ ≈ -0.991. π , find the values of sinθ and cosθ. 15. Given that tanθ ≈ 6.653 where 0 < θ < __ 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ 6.653 cosθ Solve for sinθ and cosθ. sin 2θ + cos 2θ = 1 (6.653 cosθ)2 + cos 2θ ≈ 1 44.262 cos 2θ + cos 2θ ≈ 1 45.262 cos 2θ ≈ 1 cos 2θ ≈ 0.022 cosθ ≈ ±0.148 © Houghton Mifflin Harcourt Publishing Company Since cosine is positive in Quadrant I, cosθ ≈ -0.148 and sinθ ≈ 6.653 cosθ ≈ 0.985. π , < θ < π find the values of sinθ and cosθ. 16. Given that tanθ ≈ -9.366 where __ 2 Write sinθ in terms of cosθ. sinθ = cosθ tanθ ≈ -9.366 cosθ Solve for sinθ and cosθ. sin 2θ + cos 2θ = 1 (-9.366 cosθ)2 + cos 2θ ≈ 1 87.722 cos 2θ + cos 2θ ≈ 1 88.722 cos 2θ ≈ 1 cos 2θ ≈ 0.011 cosθ ≈ ±0.105 Since cosine is negative in Quadrant II, cosθ ≈ -0.105 and sinθ ≈ -9.366 cosθ ≈ 0.983. Module 18 IN2_MNLESE389847_U7M18L5.indd 989 989 Lesson 18.5 989 Lesson 5 19/04/14 12:26 AM 17. Given the trigonometric function and the location of the terminal angle, state whether the function value will be positive or negative. PEERTOPEER DISCUSSION A. Positive A. cosθ , Quadrant I B. sinθ , Quadrant IV B. Negative C. tanθ , Quadrant II C. Negative D. sinθ , Quadrant III D. Negative E. tanθ , Quadrant III E. Positive Ask students to discuss with a partner how they could use what they know about the trigonometry of sinθ . a right triangle to prove the identity tanθ = ___ cosθ Have them discuss their proofs with the class. Students’ work should show that since side opposite sinθ = _____________ and hypotenuse side adjacent cosθ = ____________ , then hypotenuse 7π 18. Confirm the Pythagorean identity sin 2θ + cos 2θ = 1 for θ = ___ . 4 √― √― 2 2 7π 7π = -____ and cos __ = ____ . Therefore, sin __ 4 4 2 2 ( ) ( ) 7π 7π sin θ + cos θ = sin (__ ) + cos2(__ )= 2 2 2 4 4 (-____― ) + (____― ) = __ + __ = 1. √2 2 2 √2 2 2 1 2 1 2 side opposite _________ hypotenuse sinθ = __________ _____ side adjacent cosθ _________ hypotenuse hypotenuse side opposite ____________ ____________ = · 19. Recall that the equation of a circle with radius r centered at the origin is x 2 + y 2 = r 2. Use this fact and the fact that the coordinates of a point on this circle are (x, y) = (rcosθ, r sinθ) for a central angle θ to show that the Pythagorean identity derived above is true. 2 2 x 2 + y 2 = r 2 ⇒ (r cosθ) + (rsinθ) = r 2 r 2 cos 2θ + r 2 sin 2θ = r 2 ⇒ cos 2θ + sin 2θ = 1 hypotenuse side opposite = side adjacent ____________ 20. Sports A ski supply company is testing the friction of a new ski wax by placing a waxed block on an inclined plane covered with wet snow. The incline plane is slowly raised until the block begins to slide. At the instant the block starts to slide, the component of the weight of the block parallel to the incline, mgsinθ, and the resistive force of friction, μmgcosθ, are equal, where μ is the coefficient of friction. Find the value of μ to the nearest hundredth if sinθ = 0.139 at the instant the block begins to slide. Find cosθ. Because cosθ will be between 0 and 90 degrees, cosθ will be positive. = tanθ © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Aurora Photos/Alamy ――― cosθ = √1 - sin 2θ = ――――― √1 - (0.139) 2 ≈ 0.928 mg sinθ = μ mg cosθ ⇒ sinθ = μ cosθ 0.139 = μ cosθ 0.139 0.139 μ= ≈ ≈ 0.15 0.928 cosθ 21. Driving Tires and roads are designed so that the coefficient of friction between the rubber of the tires and the asphalt of the roads is very high, which gives plenty of traction for starting, stopping, and turning. For a particular road surface and tire, the steepest angle for which a car could rest on the slope without starting to slide has a sine of 0.643. This value satisfies the equation mgsinθ = μmgcosθ where μ is the coefficient of friction. Find the value of μ to the nearest hundredth. Because cosθ will be between 0 and 90 degrees, cosθ will be positive. _ _ ――― side adjacent ――――― cosθ = √1 - sin 2θ = √1 - (0.643) ≈ 0.766 mg sinθ = μ mg cosθ ⇒ sinθ = μ cosθ 0.643 = μ cosθ ⇒ 0.643 0.643 ≈ ≈ 0.84 μ= 0.766 cosθ 2 _ _ Module 18 IN2_MNLESE389847_U7M18L5.indd 990 990 Lesson 5 19/04/14 12:26 AM Using a Pythagorean Identity 990 H.O.T. Focus on Higher Order Thinking JOURNAL 3π 22. Explain the Error Julian was given that sinθ = -0.555 where ___ , < θ < 2π and 2 told to find cosθ . He produced the following work: Have students explain how knowing that the sine of an angle is approximately 0.2588 enables them to find two possible values for the cosine of the angle. Have them find these values, and tell what information would be necessary to decide which of the two possible values is the correct cosine of the angle. __ cosθ = ±√1 - sin 2θ __ = ±√1 - (-0.555 2) ≈ ±1.144 3π < θ < 2π, cosθ ≈ 1.144. Since cosθ > 0 when _ 2 Explain his error and state the correct answer. Julian added the two quantities inside of the square root together instead of subtracting. He should have noticed, since the cosine of a number is never more than 1. He should have done the following: ―――― ――――― = ± √1 - (-0.555) cosθ = ± √1 - sin 2θ 2 ≈ ±0.832 Since cosθ > 0 when 3π _ < θ < 2π, cosθ ≈ 0.832. 2 Critical Thinking Rewrite each trigonometric expression in terms of cosθ and simplify. sin 2θ 23. _ 1 - cosθ sin 2θ 1 - cosθ 1 - cos 2θ = 1 - cosθ (1- cosθ)(1+ cosθ) = 1- cosθ _ _ __ © Houghton Mifflin Harcourt Publishing Company = 1 + cosθ 24. cosθ + sinθ cosθ - tanθ + tanθ sin 2θ (Hint: Begin by factoring tanθ from the last two terms.) cosθ + sinθ cosθ - tanθ + tanθ sin 2θ = cosθ + sinθ cosθ - tanθ(1 - sin 2θ) = cosθ + sinθ cosθ - tanθ cos 2θ = cosθ + sinθ cosθ - sinθ _ cos θ 2 cosθ = cosθ + sinθ cosθ - sinθ cosθ __ = cosθ √ 1 - cos 2θ __ 25. Critical Thinking To what trigonometric function does the expression _________ √ 1 - sin 2θ simplify? Explain your answer. ―――― ―――― The expression __________ simplifies to tanθ. If you rewrite __________ ――― ――― 2 2 √1 - cos 2θ √1 - sin θ ―― √1 - cos 2θ √1 - sin θ using the Pythagorean Identity, you get the result _______ . Taking the ―― 2 √sin 2θ √cos θ sinθ square root results in the expression ____ , which is equivalent to tanθ. cosθ Module 18 IN2_MNLESE389847_U7M18L5.indd 991 991 Lesson 18.5 991 Lesson 5 19/04/14 12:25 AM Lesson Performance Task INTEGRATE MATHEMATICAL PRACTICES Focus on Math Connections MP.1 Students likely have analyzed the relationships A tower casts a shadow that is 160 feet long at a particular time one morning. With the base of the tower as the origin, east as the positive x-axis, and north as the positive y-axis, the shadow at this time is in the northwest quadrant formed by the axes. Also at this time, the tangent of the angle of rotation measured so that the shadow lies on the terminal ray is tanθ = −2.545. What are the coordinates of the tip of the shadow to the nearest foot, and what do they indicate? between shadow length and other measures, such as object height, in two dimensions. Remind them that they have used indirect measurement and proportion as well as the Pythagorean Theorem to approach these problems. Note that the Lesson Performance Task expands these analyses to allow for positions in all four quadrants using positive and negative coordinate positioning. The coordinates of a point on a circle centered at the origin with radius r are r cosθ and r sinθ. Use identities relating the sine, cosine, and tangent of an angle to find cosθ and sinθ, then use those values to find the coordinates. First use the value of the tangent and the identity sinθ = cosθ tanθ to find an expression for sinθ in terms of cosθ. CURRICULUM INTEGRATION sinθ = cosθ tanθ = -2.545 cosθ Note that a counterclockwise movement of the sun’s shadows occurs only in the Southern Hemisphere. Most sundials, which were first introduced in the Northern Hemisphere, assume that shadows move clockwise, and this is also the reason for the development of clockwise-turning clock hands! Point out to students that the sun appears to move in the opposite direction from its shadows—the sun moves counterclockwise in the Northern Hemisphere, so its shadows move in the opposite direction. Substitute this expression for sinθ into the Pythagorean identity which means that the value of cosθ will be negative. sin 2θ + cos 2θ = 1 (-2.545 cosθ)2 + cos 2θ ≈ 1 6.477 cos 2θ + cos 2θ ≈ 1 7.477 cos 2θ ≈ 1 cos 2θ ≈ 0.1337 cosθ ≈ -0.366 Now substitute to find sinθ. sinθ = -2.545 cosθ ≈ -2.545(-0.366) ≈ 0.931 Use the fact that r = 160 to find the coordinates of the tip of the shadow. r cosθ = 160(-0.366) = -58.56 r sinθ = 160(0.931) = 148.96 © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Peteri/ Shutterstock sin 2θ + cos 2θ = 1 to solve for cosθ. Since the shadow is in the northwest, it is in Quadrant II, The coordinates are (-59,149). This indicates that the tip of the shadow is about 58 feet west and 149 feet north of the base of the tower. Module 18 992 Lesson 5 EXTENSION ACTIVITY IN2_MNLESE389847_U7M18L5.indd 992 Your location with respect to two cell towers can be determined by a process called triangulation. Have students research how triangulation, a use of angles to find distances, is also employed in other activities, such as mapmaking. Students will encounter and may present the Law of Cosines, which states that c 2 = a 2 + b 2 - 2ab cos C. 19/04/14 12:25 AM Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem. Using a Pythagorean Identity 992 MODULE 18 MODULE STUDY GUIDE REVIEW Study Guide Review Essential Question: How can you use trigonometry with right triangles to solve real-world problems? ASSESSMENT AND INTERVENTION KEY EXAMPLE (Lesson 18.1) C 60 Assign or customize module reviews. 11 MODULE PERFORMANCE TASK A 61 length of leg opposite ∠A tanA = ___ length of leg adjacent to ∠A 60 ≈ 5.45 tanA = _ 11 KEY EXAMPLE Mathematical Practices: MP.1, MP.2, MP.3, MP.4, MP.7 G-SRT.C.8, G-MG.A.1 • How to find the total height of layers of staggered pipes: Students need to realize that the centers of three staggered pipes form an equilateral triangle. The height of the triangle is the height of three rows of staggered pipes. The height can be found in multiple ways, including the Pythagorean Theorem or trigonometry, or you can provide an algorithm to find the height. • How to find the height of n layers of staggered pipes: Students need to generalize the relationship from the height of 10 layers of staggered pipes to n layers. • If the thickness of the pipes matters: Students can brainstorm possibilities for both a stack and a staggered stack of pipes. The thickness adds to the height of the stack, but will add less to the height of the staggered stack. 993 Module 18 © Houghton Mifflin Harcourt Publishing Company SUPPORTING STUDENT REASONING Definition of tangent Substitute and simplify (Lesson 18.2) Find the sine and cosine of angle A. length of leg opposite ∠A sinA = ___ length of hypotenuse 60 _ sinA = ≈ .98 61 length of leg adjacent to ∠A cosA = ___ length of hypotenuse 11 _ cosA = ≈ .18 61 KEY EXAMPLE adjacent (adyacente) opposite (opuesto) tangent (tangente) cosine (coseno) inverse trigonometric ratios (proporción inversa trigonométrica) Pythagorean triple (terna pitagórica) Find the tangent of angle A. COMMON CORE Key Vocabulary tan -1 A (tan -1 A ) inverse tangent of ∠A (tangente inversa de ∠A) trigonometric ratio (proporción trigonométrica) sine (seno) B Students should begin this problem by focusing on what information they will need. Here are some issues they might bring up. 18 Trigonometry with Right Triangles Definition of sine 11 C 60 A 61 Simplify. Definition of cosine B Simplify. (Lesson 18.3) _ Given an isosceles right triangle ∠DEF with ∠F = 90° and DE = 7, find the length of the other two sides. _ DE = DF√ 2 Apply the relationship of 45°-45°-90° triangles. _ 7 = DF√ 2 7_ = DF _ √2 Substitute. Simplify. _ 7√ 2 _ 7_ = DE = EF = _ 2 √2 Module 18 Apply properties of isosceles triangles. 993 Study Guide Review SCAFFOLDING SUPPORT IN2_MNLESE389847_U7M18MC 993 • Suggest that students use the strategy of solving a simpler problem to help them find the height of stacked pipes. Have them first find the height of two layers, then find the height as another layer is added until they recognize a pattern. • One way to determine the height for two layers is to look at three pipes that form an equilateral triangle, and determine the total height using the students’ knowledge of triangles. They can continue this process, adding more layers to form larger triangles. 19/04/14 12:34 AM EXERCISES Given a right triangle △XYZ where ∠Z is a right angle, XY = 53, YZ = 28, and XZ = 45, find the following rounded to the nearest hundredth. (Lessons 18.1, 18.2) 1. sinX 0.53 2. cosX 0.85 tanX 3. SAMPLE SOLUTION Methodology: 0.62 The centers of three staggered pipes form an equilateral triangle. Find an expression for the height of this triangle, which is two layers. Next, add another layer to form an equilateral triangle using 6 pipes. Find an expression for the height of this triangle, which is three layers. Continue the process until a clear pattern emerges, then generalize to n layers. Find the area of the following triangle, rounding to the nearest tenth. 4. triangle △ABC, where C = 127°, AC = 5, and BC = 9 18.0 Find the other two sides of the following triangle. Find exact answers in order of least to greatest. (Lesson 18.3) _ 5. 7, 7√ 3 30°-60°-90° triangle with a hypotenuse of 14 The height of an equilateral triangle formed by the centers of three pipes can be found using the Pythagorean Theorem, a 2 + b 2 = c 2, where a = 1 in., c = 2 in. MODULE PERFORMANCE TASK How Much Shorter Are Staggered Pipe Stacks? How much space can be saved by stacking pipe in a staggered pattern? The illustration shows you the difference between layers of pipe stacked directly on top of each other (left) and in a staggered pattern (right). Suppose you have pipes that are 2 inches in diameter. How much shorter will a staggered stack of 10 layers be than a non-staggered stack with the same number of layers? In general, how much shorter are n layers of staggered pipe? 2 in. b Start by listing in the space below how you plan to tackle the problem. Then use your own paper to complete the task. Be sure to write down all your data and assumptions. Then use numbers, graphs, diagrams, or algebraic equations to explain how you reached your conclusion. © Houghton Mifflin Harcourt Publishing Company 1 in. The total height of the three pipes is the height of the triangle plus two times the radius of the pipes. ― h 2 layers = 2 + √3 Continuing the calculations, we get: ― = 2 + 3 √― 3 h 3 layers = 2 + 2 √3 h 4 layers Or, generalizing, Module 13 994 Study Guide Review DISCUSSION OPPORTUNITIES IN2_MNLESE389847_U7M18MC 994 • What assumptions are made when finding the height of a stack of pipes? • What could be sources of error in finding the height of a staggered stack of pipes? ― h n layers = 2 + (n - 1) √3 19/04/14 12:34 AM So the difference in the heights of 10 layers staggered versus non-staggered is ― 10(2) - ⎡⎣2 + (10 - 1) √3 ⎤⎦≈ 2.41 in. Assessment Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain. 0 points: Student does not demonstrate understanding of the problem. Study Guide Review 994 Ready to Go On? Ready to Go On? 18.1–18.5 Trigonometry with Right Triangles ASSESS MASTERY Use the assessment on this page to determine if students have mastered the concepts and standards covered in this module. • Online Homework • Hints and Help • Extra Practice Solve the problem. (Lesson 18.1) 1. ASSESSMENT AND INTERVENTION A painter is placing a ladder to reach the third story window, which is 20 feet above the ground and makes an angle with the ground of 70°. How far out from the building does the base of the ladder need to be positioned? 20 20 20 _________ _ tan70° = __ x , so x = tan70° ≈ 2.747477419 ≈ 7.3 feet 2. Access Ready to Go On? assessment online, and receive instant scoring, feedback, and customized intervention or enrichment. Given the value of cos 30° = 1, write the sine of a complementary angle. Use an expression relating trigonometric ratios of complementary angles. (Lesson 18.2) cos θ° = sin(90 - θ)° Substitute 30 into both sides. cos30° = sin(90 - 30)° Simplify the left side. cos30° = sin60° Substitute for cos30°. 1 = sin60° Find the area of the regular polygon. (Lesson 18.3) 3. What is the area of a regular hexagon with a distance from its center to a vertex of 1 cm? (Hint: A regular hexagon can be divided into six equilateral triangles.) Each of the six equilateral triangles has sides of length 1. 1 ADDITIONAL RESOURCES You can split each of these triangles into two congruent Response to Intervention Resources 30-60-90 triangles. These make twelve 30-60-90 triangles. Differentiated Instruction Resources • Reading Strategies • Success for English Learners • Challenge Worksheets Assessment Resources © Houghton Mifflin Harcourt Publishing Company • Reteach Worksheets ― 3 1 Each of these triangles has a base of _ and a height of ___ . 2 2 √ The area of each of these equilateral triangles is √― √― 3 3 _1bh = _1 · _1 · ___ = ___ . The area of the hexagon is twelve 2 2 2 2 8 ― 12 √3 3 √― 3 _____ times that, or = ____ . 8 2 ESSENTIAL QUESTION 4. How would you go about finding the area of a regular pentagon given the distance from its center to the vertices? Answers may vary. Sample: You can determine the area of a regular pentagon by splitting it into 5 congruent triangles (each including a 360/5 = 72 degree angle). Find the angles of the triangles and use trigonometric ratios to find their height and base. With this information we can find the area of one triangle and multiply by 5 to get the area of the polygon. • Leveled Module Quizzes Module 18 COMMON CORE IN2_MNLESE389847_U7M18MC 995 995 Module 18 Study Guide Review 995 Common Core Standards 19/04/14 12:34 AM Content Standards Mathematical Practices Lesson Items 18.1 1 G-SRT.C.6, G-SRT.C.8 MP.2 18.2 2 G-SRT.C.7 MP.6 18.4 3 G-SRT.C.9 MP.6 MODULE MODULE 18 MIXED REVIEW MIXED REVIEW Assessment Readiness Assessment Readiness 1. Julia is standing 2 feet away from a lamppost. She casts a shadow of 5 feet and the light makes a 20° angle relative to the ground from the top of her shadow. Consider each expression. Does the expression give you the height of the lamppost? Select Yes or No for A–C. A. 7 sin20° Yes No B. 7 tan20° Yes No C. 7 tan70° Yes No ASSESSMENT AND INTERVENTION 2. A right triangle has two sides with lengths 10 and 10. Choose True or False for each statement. A. The triangle has two angles that measure True False 45° each. B. The triangle is equilateral. True False C. _ The length of the third side is 10√2 . True Assign ready-made or customized practice tests to prepare students for high-stakes tests. False 3. The measure of angle 1 is 125° and the measure of angle 2 is 55°. State two different relationships that can be used to prove m∠3 = 125°. 4 Possible Answer: ∠2 and ∠3 are supplementary 1 3 ADDITIONAL RESOURCES 2 Assessment Resources angles, ∠1 and ∠3 are vertical angles. • Leveled Module Quizzes: Modified, B 4. For the rhombus, specify how to find its area using the four congruent right triangles with variable angle θ. θ AVOID COMMON ERRORS Answers may vary. Sample: Find the base and height of each right triangle using trigonometric ratios of θ. Multiply by 4 to 2 1 A r = 4 · __ bh = 2h cos θ sin θ 2 COMMON CORE Item 1 Some students will have a hard time visualizing the problem. Encourage students to draw and label a quick sketch to help them see which trigonometric function best fits the situation. © Houghton Mifflin Harcourt Publishing Company get the area of the rhombus, as all the triangles are congruent. Module 18 18 Study Guide Review 996 Common Core Standards IN2_MNLESE389847_U7M18MC 996 19/04/14 12:34 AM Content Standards Mathematical Practices Lesson Items 18.1 1 G-SRT.C.8 MP.4 18.3, 15.2 2* G-SRT.C.8 MP.2 14.1 3* G-CO.C.9 MP.6 18.2 4 G-SRT.C.8 MP.5 * Item integrates mixed review concepts from previous modules or a previous course. Study Guide Review 996 UNIT 7 UNIT 7 MIXED REVIEW Assessment Readiness MIXED REVIEW Assessment Readiness 1. Determine whether the statements are true. Select True or False for each statement. A. Dilations preserve angle measure. ASSESSMENT AND INTERVENTION B. Dilations preserve distance. C. Dilations preserve collinearity. D. Dilations preserve orientation. B. Reflection across the x-axis C. Dilation D. Translation ADDITIONAL RESOURCES • Leveled Unit Tests: Modified, A, B, C • Performance Assessment © Houghton Mifflin Harcourt Publishing Company AVOID COMMON ERRORS Yes Yes Yes No No No 8 S -4 T Yes No B. Quadrilateral JKLM is a rhombus. Yes No C. Quadrilateral JKLM is a rectangle. Yes No Yes Yes No No Yes No 4. Will the transformation produce similar figures? Select Yes or No for each statement. A. (x, y) → (x - 5, y + 5) → (-x, -y) → (3x, 3y) B. (x, y) → (3x, y + 5) → (x, 3y) → (x - 1, y - 1) C. (x, y) → (x, y + 5) → (2x, y) → (x + 5, y) 4 D y A 0 x B 4 8 R C Q A. Quadrilateral JKLM is a parallelogram. -8 5. Is ABC similar to DEF? Select Yes or No for each statement. A. A (-1, -3), B (1, 3), C (3, -5) D (2, -6), E (3, 0), F (6, -8) B. A (-1, -3), B (1, 3), C (3, -5) D (-5, -1), E (-4, 2), F (-3, -2) C. A (-1, -3), B (1, 3), C (3, -5) D (-2, -2), E (2, 4), F (2, -4) Unit 7 COMMON CORE IN2_MNLESE389847_U7UC 997 Yes No Yes No Yes No 997 Common Core Standards Items Unit 7 False False False 3. The vertices of quadrilateral JKLM are J(-2, 0), K(-1, 2), L(1, 3), and M(0, 1). Can you use slopes and/or the distance formula to prove each statement? Select Yes or No for A–C. Assessment Resources 997 False True True True 2. Was the given transformation used to map ABCD to QRST? Select Yes or No for each statement. Yes No A. Reflection across the y-axis Assign ready-made or customized practice tests to prepare students for high-stakes tests. Item 6 Some students will have trouble deciding which sides correspond on the triangles, especially when the triangles have different orientations. Remind students that the smallest side on one shape will always correspond to the smallest side on the other shape. The same goes for the largest side. True • Online Homework • Hints and Help • Extra Practice Content Standards Mathematical Practices 1 G-SRT.A.1 MP.6 2 G-SRT.A.1 MP.7 3* G-CO.C.11 MP.2 4 G-SRT.A.2 MP.1 5* G-SRT.A.1, S-ID.B.6 MP.5 19/04/14 12:37 AM 6. Are the triangles similar? Select Yes or No for each statement. A. Yes PERFORMANCE TASKS No N There are three different levels of performance tasks: R 14 12 *Novice: These are short word problems that require students to apply the math they have learned in straightforward, real-world situations. 10.5 O P M Q 9 B. Yes T 12.8 S **Apprentice: These are more involved problems that guide students step-by-step through more complex tasks. These exercises include more complicated reasoning, writing, and open ended elements. No 16 U V 20 C. Yes ***Expert: These are open-ended, nonroutine problems that, instead of stepping the students through, ask them to choose their own methods for solving and justify their answers and reasoning. No X 6 B 3 W 1 6 3 6 1 Y 7. Find the missing heights. (ST) + (TU) = (SU) 2 2 P T ― 45 2 Q (45) 2 + (30) 2 = (SU) 2 ―― 50 R S 30 U 5 √117 = (SU) Unit 7 © Houghton Mifflin Harcourt Publishing Company Using AA Similarity, the triangles are similar. 45 x = ___ ___ (PQ) 2 + (QR) 2 = (PR) 2 50 30 45 . 50 (75) 2 + (50) 2 = (PR) 2 x = ___ 30 x = 75 25 √13 = PR 998 COMMON CORE IN2_MNLESE389847_U7UC 998 Common Core Standards Items 19/04/14 12:37 AM Content Standards Mathematical Practices 6 G-CO.C.10, G-SRT.A.2 MP.5 7 G-CO.C.10 MP.2 * Item integrates mixed review concepts from previous modules or a previous course. Unit 7 998 Performance Tasks 2.4 3.2 x = 2.1 Item 9 (6 points) i m B Street Distance: 2.1 + 2.4 = 4.5; 4.5 miles 1 point for finding each coordinate point, 3 in all 5th Ave 9. A city has a walkway between the middle school and the library that can be represented in the image given. The city decides it wants to place three trash cans, equally spaced along the walkway, to help reduce any littering. Find the coordinates of the points at which the trash cans should be placed, and then plot them on the graph. Subdivide the line segment SL into four equal parts. The horizontal distance from point S to point L is 16, and one-fourth of this is 4. So each trash can should be placed at horizontal intervals of 4 units. The vertical distance from point S to point L is 8, and one-fourth of this is 2, so each trash can should be placed at vertical intervals of 2 units. The coordinates are (-4, 4), (0, 2), and (4, 0). Item 10 (6 points) 3 points for correct cost 3 points for explanation © Houghton Mifflin Harcourt Publishing Company The distance from K to L is 35 - (15 + 10) = 10 yd, so 8 School S y 4 x -8 -4 0 -4 L 4 Library -8 (-4, 4), (0, 2), and (4, 0) 10. Sam is planning to fence his backyard. Every side of the yard except for the side along the house is to be fenced, and fencing costs $3/yd and can only be bought in whole yards. (Note that m∠NPM = 28°, and the side of his yard opposite the house measures 35 yd.) How much will the fencing cost? Explain how you found your answer. House J P Yard K triangles JKL and MNP are congruent. The distance JK (and PN) L 15 yd M 10 yd N 10 is _____ = 21.3 yd. So the total length of fencing needed is sin28° (2)(21.3) + 35 = 77.6 yd. Because the fencing can only be bought in whole yards, Sam will need to buy 78 yd. The cost is (78 yd)($3/yd) = $234. Unit 7 IN2_MNLESE389847_U7UC 999 Unit 7 6th Ave The Triangle Proportionality Theorem 1 point for correctly graphing each coordinate point, 3 in all 999 3.2 A Street mi 2.8 x = ____ ____ 2.4 mi 8. The map shows that A Street and B Street are parallel. Find the distance on 6th Ave between A Street and the library. Explain any theorems that come into play here. Item 8 (2 points) Award the student 1 point for the correct distance of 4.5 miles, and 1 point for correctly naming the Triangle Proportionality Theorem. 2.8 SCORING GUIDES 999 19/04/14 12:37 AM math in careers MATH IN CAREERS special effects engineers A special effects engineer is helping create a movie and needs to add a shadow to a tall totem pole that is next to a 6-foot-tall man. The totem pole is 48 feet tall and is next to the man, who has a shadow that is 2.5 feet long. Create an image with the given information and then use the image to find the length of the shadow that the engineer needs to create for the totem pole. Special Effects Engineer In this Unit Performance Task, students can see how a special effects engineer uses mathematics on the job. For more information about careers in mathematics as well as various mathematics appreciation topics, visit the American Mathematical Society http://www.ams.org Totem Pole Man 48 ft 6 ft SCORING GUIDES X 2.5 ft Task (6 points) 2.5 x = ____ ___ 6 48 x = 20 3 points for creating a diagram that represents the situation, with distances labeled 3 points for correctly setting up the proportion and solving © Houghton Mifflin Harcourt Publishing Company Unit 7 IN2_MNLESE389847_U7UC 1000 1000 19/04/14 12:37 AM Unit 7 1000

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