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Interactive Notebook Foldables, Organizers and Templates

Interactive Notebook Foldables, Organizers and Templates

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  • 1. Contributing AuthorDinah ZikeConsultantDouglas Fisher, Ph.D.Director of Professional DevelopmentSan Diego, CA
  • 2. Copyright © by The McGraw-Hill Companies, Inc. All rights reserved. Printed in theUnited States of America. Except as permitted under the United States CopyrightAct, no part of this book may be reproduced in any form, electronic or mechanical,including photocopy, recording, or any information storage or retrieval system,without prior written permission of the publisher.Send all inquiries to:The McGraw-Hill Companies8787 Orion PlaceColumbus, OH 43240-4027ISBN: 978-0-07-877340-2 Geometry Noteables™: InteractiveMHID: 0-07-877340-7 Study Notebook with Foldables™1 2 3 4 5 6 7 8 9 10 005 16 15 14 13 12 11 10 09 08 07
  • 3. Glencoe Geometry iiiContentsFoldables. . . . . . . . . . . . . . . . . . . . . . . . . . . 1Vocabulary Builder . . . . . . . . . . . . . . . . . . 21-1 Points, Lines, and Planes . . . . . . . . . 41-2 Linear Measure and Precision . . . . . 81-3 Distance and Midpoints. . . . . . . . . 111-4 Angle Measure . . . . . . . . . . . . . . . . 131-5 Angle Relationships . . . . . . . . . . . . 161-6 Two-Dimensional Figures. . . . . . . . 191-7 Three-Dimensional Figures . . . . . . 23Study Guide . . . . . . . . . . . . . . . . . . . . . . . 26Foldables. . . . . . . . . . . . . . . . . . . . . . . . . . 31Vocabulary Builder . . . . . . . . . . . . . . . . . 322-1 Inductive Reasoning andConjecture. . . . . . . . . . . . . . . . . . . . 342-2 Logic . . . . . . . . . . . . . . . . . . . . . . . . 362-3 Conditional Statements . . . . . . . . . 392-4 Deductive Reasoning . . . . . . . . . . . 422-5 Postulates and ParagraphProofs . . . . . . . . . . . . . . . . . . . . . . . 452-6 Algebraic Proof . . . . . . . . . . . . . . . 482-7 Proving Segment Relationships. . . 512-8 Proving Angle Relationships . . . . . 53Study Guide . . . . . . . . . . . . . . . . . . . . . . . 57Foldables. . . . . . . . . . . . . . . . . . . . . . . . . . 61Vocabulary Builder . . . . . . . . . . . . . . . . . 623-1 Parallel Lines and Transversals. . . . 643-2 Angles and Parallel Lines. . . . . . . . 673-3 Slopes of Lines . . . . . . . . . . . . . . . . 693-4 Equations of Lines . . . . . . . . . . . . . 723-5 Proving Lines Parallel. . . . . . . . . . . . 753-6 Perpendiculars and Distance . . . . . . 79Study Guide . . . . . . . . . . . . . . . . . . . . . . . 82Foldables. . . . . . . . . . . . . . . . . . . . . . . . . . 85Vocabulary Builder . . . . . . . . . . . . . . . . . 864-1 Classifying Triangles . . . . . . . . . . . . . 884-2 Angles of Triangles . . . . . . . . . . . . . . 914-3 Congruent Triangles . . . . . . . . . . . . . 954-4 Proving Congruence—SSS, SAS . . . . 984-5 Proving Congruence—ASA, AAS. . 1024-6 Isosceles Triangles . . . . . . . . . . . . . 1054-7 Triangles and Coordinate Proof . . 108Study Guide . . . . . . . . . . . . . . . . . . . . . . 111Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 115Vocabulary Builder . . . . . . . . . . . . . . . . 1165-1 Bisectors, Medians, andAltitudes . . . . . . . . . . . . . . . . . . . . 1185-2 Inequalities and Triangles . . . . . . 1225-3 Indirect Proof . . . . . . . . . . . . . . . . 1255-4 The Triangle Inequality . . . . . . . . 1285-5 Inequalities Involving TwoTriangles . . . . . . . . . . . . . . . . . . . . 130Study Guide . . . . . . . . . . . . . . . . . . . . . . 133Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 137Vocabulary Builder . . . . . . . . . . . . . . . . 1386-1 Angles of Polygons. . . . . . . . . . . . 1396-2 Parallelograms . . . . . . . . . . . . . . . 1426-3 Tests for Parallelograms. . . . . . . . 1446-4 Rectangles. . . . . . . . . . . . . . . . . . . 1476-5 Rhombi and Squares . . . . . . . . . . 1526-6 Trapezoids. . . . . . . . . . . . . . . . . . . 1556-7 Coordinate Proof withQuadrilaterals. . . . . . . . . . . . . . . . 159Study Guide . . . . . . . . . . . . . . . . . . . . . . 162Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 167Vocabulary Builder . . . . . . . . . . . . . . . . 1687-1 Proportions . . . . . . . . . . . . . . . . . . 1707-2 Similar Polygons . . . . . . . . . . . . . . 1727-3 Similar Triangles . . . . . . . . . . . . . . 1757-4 Parallel Lines and ProportionalParts. . . . . . . . . . . . . . . . . . . . . . . . 1787-5 Parts of Similar Triangles . . . . . . . 181Study Guide . . . . . . . . . . . . . . . . . . . . . . 184
  • 4. iv Glencoe GeometryContentsFoldables. . . . . . . . . . . . . . . . . . . . . . . . . 187Vocabulary Builder . . . . . . . . . . . . . . . . 1888-1 Geometric Mean. . . . . . . . . . . . . . 1908-2 The Pythagorean Theorem andIts Converse. . . . . . . . . . . . . . . . . . 1948-3 Special Right Triangles. . . . . . . . . 1988-4 Trigonometry . . . . . . . . . . . . . . . . 2018-5 Angles of Elevation andDepression . . . . . . . . . . . . . . . . . . 2048-6 The Law of Sines . . . . . . . . . . . . . 2078-7 The Law of Cosines . . . . . . . . . . . 211Study Guide . . . . . . . . . . . . . . . . . . . . . . 214Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 219Vocabulary Builder . . . . . . . . . . . . . . . . 2209-1 Reflections . . . . . . . . . . . . . . . . . . 2229-2 Translations. . . . . . . . . . . . . . . . . . 2269-3 Rotations. . . . . . . . . . . . . . . . . . . . 2299-4 Tessellations . . . . . . . . . . . . . . . . . 2319-5 Dilations . . . . . . . . . . . . . . . . . . . . 2339-6 Vectors . . . . . . . . . . . . . . . . . . . . . 236Study Guide . . . . . . . . . . . . . . . . . . . . . . 239Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 243Vocabulary Builder . . . . . . . . . . . . . . . . 24410-1 Circles and Circumference . . . . . . 24610-2 Measuring Angles and Arcs. . . . . 24910-3 Arcs and Chords . . . . . . . . . . . . . . 25210-4 Inscribed Angles . . . . . . . . . . . . . . 25610-5 Tangents . . . . . . . . . . . . . . . . . . . . 26010-6 Secants, Tangents, and AngleMeasures. . . . . . . . . . . . . . . . . . . . 26210-7 Special Segments in a Circle . . . . 26510-8 Equations of Circles . . . . . . . . . . . 267Study Guide . . . . . . . . . . . . . . . . . . . . . . 269Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 273Vocabulary Builder . . . . . . . . . . . . . . . . 27411-1 Areas of Parallelograms. . . . . . . . 27511-2 Areas of Triangles, Trapezoids,and Rhombi . . . . . . . . . . . . . . . . . 27711-3 Areas of Regular Polygons andCircles . . . . . . . . . . . . . . . . . . . . . . 28011-4 Areas of Composite Figures. . . . . 28311-5 Geometric Probability andAreas of Sectors . . . . . . . . . . . . . . 285Study Guide . . . . . . . . . . . . . . . . . . . . . . 288Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 291Vocabulary Builder . . . . . . . . . . . . . . . . 29212-1 Representations ofThree-Dimensional Figures . . . . . 29412-2 Surface Areas of Prisms. . . . . . . . . 29812-3 Surface Areas of Cylinders. . . . . . 30012-4 Surface Areas of Pyramids. . . . . . 30212-5 Surface Areas of Cones . . . . . . . . 30512-6 Surface Areas of Spheres. . . . . . . 307Study Guide . . . . . . . . . . . . . . . . . . . . . . 310Foldables. . . . . . . . . . . . . . . . . . . . . . . . . 315Vocabulary Builder . . . . . . . . . . . . . . . . 31613-1 Volumes of Prisms andCylinders . . . . . . . . . . . . . . . . . . . . 31713-2 Volumes of Pyramids andCones. . . . . . . . . . . . . . . . . . . . . . . 32013-3 Volumes of Spheres . . . . . . . . . . . 32313-4 Congruent and Similar Solids . . . 32513-5 Coordinates in Space . . . . . . . . . . 328Study Guide . . . . . . . . . . . . . . . . . . . . . . 330
  • 5. Glencoe Geometry vOrganizing Your FoldablesCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Make this Foldable to help you organize andstore your chapter Foldables. Begin with onesheet of 11" × 17" paper.FoldFold the paper in half lengthwise. Then unfold.Fold and GlueFold the paper in half widthwise and glue all of the edges.Glue and LabelGlue the left, right, and bottom edges of the Foldableto the inside back cover of your Noteables notebook.Foldables OrganizerReading and Taking Notes As you read and study each chapter, recordnotes in your chapter Foldable. Then store your chapter Foldables insidethis Foldable organizer.
  • 6. vi Glencoe GeometryThis note-taking guide is designed to help you succeed in Geometry.Each chapter includes:Glencoe Geometry 187Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter8Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.Right Triangles and TrigonometryNOTE-TAKING TIP: When you take notes, drawa visual (graph, diagram, picture, chart) thatpresents the information introduced in the lessonin a concise, easy-to-study format.Stack the sheets. Foldthe top right cornerto the bottom edgeto form a square.Fold the rectangularpart in half.Staple the sheetsalong the fold infour places.Label each sheet witha lesson number andthe rectangular partwith the chapter title.Begin with seven sheets of grid paper.C H A P T E R8188 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 8.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.C H A P T E R8Vocabulary TermFoundDefinitionDescription oron Page Exampleangle of depressionangle of elevationcosinegeometric meanLaw of CosinesLaw of SinesThe Chapter Openercontains instructions andillustrations on how to makea Foldable that will help youto organize your notes.A Note-Taking Tipprovides a helpfulhint you can usewhen taking notes.The Build Your Vocabularytable allows you to writedefinitions and examplesof important vocabularyterms together in oneconvenient place.Within each chapter,Build Your Vocabularyboxes will remind youto fill in this table.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.
  • 7. STUDY GUIDE214 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERUse your Chapter 8 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 8, go to:glencoe.comYou can use your completedVocabulary Builder(pages 188–189) to help yousolve the puzzle.฀฀ ฀฀Find the geometric mean between each pair of numbers.1. 4 and 9 2. 20 and 303. Find x and y.4. For the figure shown, which statements are true?a. m2 + n2 = p2 b. n2 = m2 + p2c. m2 = n2 + p2 d. m2 = p2 - n2e. p2 = n2 - m2 f. n2 - p2 = m2g. n = √m2 + p2 h. p = √m2 - n2Which of the following are Pythagorean triples? Write yes or no.5. 10, 24, 26 6. √2, √2, 2 7. 10, 6, 88-1Geometric Mean8-2The Pythagorean Theorem and Its ConverseC H A P T E R8Glencoe Geometry 203Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT8–4Use Trigonometric Ratios to Find a LengthEXERCISING A fitness trainer sets the incline on atreadmill to 7°. The walking surface is 5 feet long.Approximately how many inches did the trainer raisethe end of the treadmill from the floor?Let y be the height of the treadmill from the floor in inches.The length of the treadmill is feet, or inches.sin 7° = sin =leg opposite_hypotenusesin 7° = y Multiply each side by .The treadmill is about inches high.Check Your Progress The bottom of a handicap ramp is15 feet from the entrance of a building. If the angle of the rampis about 4.8°, how high does the ramp rise off the ground to thenearest inch?Glencoe Geometry viiExamples parallel theexamples in your textbook.Bringing It All TogetherStudy Guide reviews themain ideas and keyconcepts from each lesson.Check Your ProgressExercises allow you tosolve similar exerciseson your own.Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.฀ ฀฀฀฀฀฀฀฀฀฀฀ ฀ ฀ ฀ ฀ ฀ ฀฀ ฀ ฀ ฀ ฀฀฀฀฀ ฀ ฀ ฀ ฀฀฀฀ ฀ ฀ ฀฀ ฀ ฀฀฀฀ ฀฀฀฀ ฀ ฀฀ ฀ ฀ ฀฀ ฀ ฀ ฀ ฀ ฀ ฀฀ ฀ ฀ ฀ ฀ ฀ ฀฀Foldables featurereminds you to takenotes in your Foldable.48 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.A two-column proof, or formal proof, containsand organized intwo columns.Algebraic Proof2–6• Use algebra to writetwo-column proofs.• Use properties ofequality in geometryproofs.MAIN IDEASVerify Algebraic RelationshipsSolve 2(5 - 3a) - 4(a + 7) = 92.Algebraic Steps Properties2(5 - 3a) - 4(a + 7) = 92 Original equation10 - 6a - 4a - 28 = 92 Property-18 - 10a = 92 Property= Addition Propertya = PropertyCheck Your Progress Solve -3(a + 3) + 5(3 - a) = -50.BUILD YOUR VOCABULARY (pages 32–33)Write a two-columnproof for the statementif 5(2x - 4) = 20, thenx = 4 under the tab forLesson 2-6.ORGANIZE ITFoldables featurereminds you to takenotes in your Foldable.Lessons cover the content ofthe lessons in your textbook.As your teacher discusses eachexample, follow along andcomplete the fill-in boxes.Take notes as appropriate.
  • 8. viii Glencoe GeometryYour notes are a reminder of what you learned in class. Taking good notescan help you succeed in mathematics. The following tips will help you takebetter classroom notes.• Before class, ask what your teacher will be discussing in class. Reviewmentally what you already know about the concept.• Be an active listener. Focus on what your teacher is saying. Listen forimportant concepts. Pay attention to words, examples, and/or diagramsyour teacher emphasizes.• Write your notes as clear and concise as possible. The following symbolsand abbreviations may be helpful in your note-taking.• Use a symbol such as a star (#) or an asterisk (*) to emphasize importantconcepts. Place a question mark (?) next to anything that you do notunderstand.• Ask questions and participate in class discussion.• Draw and label pictures or diagrams to help clarify a concept.• When working out an example, write what you are doing to solve theproblem next to each step. Be sure to use your own words.• Review your notes as soon as possible after class. During this time, organizeand summarize new concepts and clarify misunderstandings.Note-Taking Don’ts• Don’t write every word. Concentrate on the main ideas and concepts.• Don’t use someone else’s notes as they may not make sense.• Don’t doodle. It distracts you from listening actively.• Don’t lose focus or you will become lost in your note-taking.NOTE-TAKING TIPSWord or PhraseSymbol orAbbreviationWord or PhraseSymbol orAbbreviationfor example e.g. not equal ≠such as i.e. approximately ≈with w/ therefore ∴without w/o versus vsand + angle ∠Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.
  • 9. Glencoe Geometry 1Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter1C H A P T E R1Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.Tools of GeometryNOTE-TAKING TIP: When you take notes, listen orread for main ideas. Then record what you knowand apply these concepts by drawing, measuring,and writing about the process.Begin with a sheet of 11" × 17" paper.Fold the short sidesto meet in the middle.Fold the top to thebottom.Open. Cut flaps alongsecond fold to makefour tabs.Label the tabs as shown.Points, Lines,PlanesLength andPerimeterAnglesAngleMeasure
  • 10. 2 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 1.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.Vocabulary TermFoundDefinitionDescription oron Page Exampleadjacent angles[uh-JAY-suhnt]angleangle bisectorcollinear[koh-LIN-ee-uhr]complementary anglescongruent[kuhn-GROO-uhnt]coplanar[koh-PLAY-nuhr]degreelineline segmentlinear pairC H A P T E R1
  • 11. Glencoe Geometry 3Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY1Vocabulary TermFoundDefinitionDescription oron Page Examplemidpointperpendicularplanepointpolygon[PAHL-ee-gahn]polyhedronprecisionraysegment bisectorsidessupplementary anglesvertexvertical angles
  • 12. 4 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–1Points on the same are collinear.Points that lie on the same are coplanar.Points, Lines, and PlanesName Lines and PlanesUse the figure to name each ofthe following.a. a line containing point KThe line can be named as line .There are three points on the line. Any two of the pointscan be used to name the line. The possible names are.b. a plane containing point LThe plane can be named as plane .You can also use the letters of any threepoints to name the plane.plane plane planeCheck Your Progress Use the figure to name each ofthe following.a. a line containing point Xb. a plane containing point Z• Identify and modelpoints, lines, andplanes.• Identify collinear andcoplanar points andintersecting lines andplanes in space.MAIN IDEASPoint A point hasneither shape nor size.Line There is exactlyone line through anytwo points.Plane There is exactlyone plane throughany three noncollinearpoints.KEY CONCEPTS(page 2)BUILD YOUR VOCABULARY
  • 13. Glencoe Geometry 5Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–1Model Points, Lines, and PlanesName the geometric term modeled by each object.a. the long hand on a clockThe long hand on a clock models a .b. a 10 × 12 patioThe patio models a .c. a water glass on the tableThis models a .Check Your Progress Name the geometric shapemodeled by each object.a. a colored dot on a map usedto mark the location of a cityb. the ceiling of your classroomDraw Geometric FiguresDraw and label a figure for each situation.a. ALGEBRA Plane R contains lines AB and DE, whichintersect at point P. Add point C on plane R so that itis not collinear with AB or DE.Draw a surface to representand label it.Draw a line anywhere on the planeand draw dots on the line for pointsA and B. Label the points.Draw a line intersectingand draw dots on theline for points D and E.Label the points.Label the intersection of thetwo lines as .Draw and label a pointP, a line AB, and a planeXYZ under the Points,Lines, and Planes tab.Points, Lines,PlanesLength andPerimeterAnglesAngleMeasureORGANIZE IT
  • 14. 6 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–1Draw a dot for point C in plane Rsuch that it will not lie on AB orDE. Label the point.b. QR on a coordinate plane contains Q(-2, 4) andR(4, -4). Add point T so that T is collinear withthese points.Graph each point and draw QR.There are an infinite number of points that are collinearwith Q and R. In the graph, one such point is .Check Your Progress Draw and label a figure foreach relationship.a. Plane D contains line a, line m, and line t, with all threelines intersecting at point Z. Add point F on plane D so thatit is not collinear with any of the three given lines.b. BA on a coordinate plane contains B(-3, -2) and A(3, 2).Add point M so that M is collinear with these points.REMEMBER ITThe prefix co-means together. So,collinear means lyingtogether on the sameline. Coplanar meanslying together in thesame plane.
  • 15. Glencoe Geometry 7Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT1–1Interpret DrawingsUse the figure for parts a–d.a. How many planes appear in thisfigure?b. Name three points that are collinear.Points , , and are collinear.c. Are points A, B, C, and D coplanar? Explain.Points A, B, C, and D all lie in , so theyare coplanar.d. At what point do DB and CA intersect?The two lines intersect at point .Check Your Progress Refer tothe figure.a. How many planes appear inthis figure?b. Name three points thatare collinear.c. Are points X, O, and R coplanar?Explain.d. At what point do BN and XO intersect?Explain the differentways of naming a plane.WRITE IT
  • 16. 8 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–2A line segment can be measured because it hastwo .Linear Measure and PrecisionLength in Metric UnitsFind the length of−−−LM using the ruler.The long marks indicate , and theshorter marks indicate .L−−−LM is about millimeters long.Check Your Progressa. Find the length of−−AB.Length in Customary UnitsFind the length of each segment using each ruler.a.−−DEEach inch is divided into .−−−DE is aboutinches long.• Measure segments anddetermine accuracy ofmeasurement.• Compute with measures.MAIN IDEAS(page 2)BUILD YOUR VOCABULARY
  • 17. Glencoe Geometry 9Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.b.−−FG−−−FG is about inches long.Check Your Progressa. Find the length of−−AZ.b. Find the length of−−IX.PrecisionFind the precision for each measurement. Explainits meaning.a. 323_4inchesThe measuring tool is divided into increments.Thus, the measurement is precise to within 1_2(1_4) or 1_8inch.The measurement could be to inches.b. 15 millimetersThe measuring tool is divided into millimeters. Thus themeasurement is precise to within 1_2of 1 millimeter. Themeasurement could be 14.5 to 15.5 millimeters.Check Your ProgressFind the precision for eachmeasurement.Explain how to findthe precision of ameasurement. Writethis under the Lengthand Perimeter tab.Points, Lines,PlanesLength andPerimeterAnglesAngleMeasureORGANIZE IT1–2
  • 18. 10 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT1–2Find MeasurementsFind the measurement of each segment.a.−−−LMLM + MN = LN ฀LM + = SubstitutionLM + - = - Subtract.LM = Simplify.−−−LM is centimeters long.b. x and ST if T is between S and U, ST = 7x, SU = 45,and TU = 5x - 3.ST + TU = SU7x + = Substitute known values.7x + 5x - 3 + 3 = 45 + 3 Add 3 to each side.12x = 48 Simplify.12x_12= 48_12Divide each side by 12.= Simplify.ST = 7x Given= 7(4) x = 4= Thus, x = 4, ST = .Check Your Progressa. Find SE.b. Find a and AB if AB = 4a + 10, BC = 3a - 5, and AC = 19.
  • 19. Glencoe Geometry 11Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–3 Distance and MidpointsFind Distance on a Number LineUse the number line to find QR.The coordinates of Q and Rare and .QR = |-6 - (-3)| Distance Formula= |-3| or 3 Simplify.Check Your Progress Use the number line to find AX.Find Distance on a Coordinate PlaneFind the distance betweenE(-4, 1) and F(3, -1).Method 1 Pythagorean Theorem(EF)2 = (ED)2 + (DF)2(EF)2 = 2+ 2(EF)2 = 53EF = √53Method 2 Distance Formulad = √(x2 - x1)2 + (y2 - y1)2 Distance FormulaEF = √[3 - (-4)]2 + (-1 - 1)2 (x1, y1) = (-4, 1) and(x2, y2) = (3, -1)EF = ( )2+ ( )2Simplify.EF = √53 Simplify.• Find the distancebetween two points.• Find the midpoint of asegment.MAIN IDEASDistance FormulasNumber Line If P and Qhave coordinates a and b,respectively, then thedistance betweenP and Q is givenby PQ = |b - a|or |a - b|.Coordinate Plane Thedistance d between twopoints with coordinates(x1, y1) and (x2, y2) isgiven by d =√(x2 - x1)2 + (y2 - y1)2.KEY CONCEPT
  • 20. 12 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT1–3Check Your Progress Find thedistance between A(-3, 4) and M(1, 2).Find Coordinate of EndpointFind the coordinates of D if E(-6, 4) is the midpointof−−DF and F has coordinates (-5, -3).Let F be (x2, y2) in the Midpoint Formula.E(-6, 4) = E(x1 + (-5)_2,y1 + (-3)_2 ) (x2, y2) = (-5, -3)Write and solve two equations to find the coordinates of D.-6 =x1 + (-5)_24 =y1 + ( )__2= 8 = y1 - 3= x1 = y1The coordinates of D are .Check Your Progress Find the coordinates of R if N(8, -3)is the midpoint of−−RS and S has coordinates (-1, 5).The midpoint of a segment is the point halfwaybetween the of the segment.Any segment, line, or plane that intersects a segment atits is called a segment bisector.Draw a coordinatesystem and show howto find the distancebetween two pointsunder the Length andPerimeter tab.Points, Lines,PlanesLength andPerimeterAnglesAngleMeasureORGANIZE ITMidpoint The midpointM of−−PQ is the pointbetween P and Q suchthat PM = MQ.KEY CONCEPT(page 3)BUILD YOUR VOCABULARY
  • 21. Glencoe Geometry 13Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–4 Angle MeasureAngles and Their PartsRefer to the figure.a. Name all angles that have B as a vertex.b. Name the sides of ∠5.and or are the sides of ∠5.c. Write another name for ∠6., , , and areother names for ∠6.Check Your Progressa. Name all angles that have X as a vertex.b. Name the sides of ∠3.c. Write another name for ∠3.• Measure and classifyangles.• Identify and usecongruent anglesand the bisector ofan angle.MAIN IDEASAngle An angle isformed by twononcollinear rays thathave a commonendpoint.KEY CONCEPTWhen can you use asingle letter to namean angle?WRITE ITA ray is a part of a . A ray starts at aon the line and extends endlessly in .An angle is formed by two rays thathave a common endpoint.BUILD YOUR VOCABULARY (pages 2–3)
  • 22. 14 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–4Measure and Classify AnglesMeasure each angle named and classify it as right,acute, or obtuse.a. ∠XYV∠XYV is marked with a rightangle symbol, so measuring isnot necessary. m∠XYV = ,so ∠XYV is a(n) .b. ∠WYTUse a protractorto find thatm∠WYT = 130.180 > m∠WYT > 90,so ∠WYT is a(n).c. ∠TYUUse a protractorto find thatm∠TYU = 45.45 < 90, so ∠TYU isa(n) .Check Your Progress Measure each angle named andclassify it as right, acute, or obtuse.a. ∠CZDb. ∠CZEc. ∠DZXClassify AnglesA right angle has adegree measure of 90.An acute angle has adegree measure lessthan 90. An obtuseangle has a degreemeasure greater than 90and less than 180.Congruent AnglesAngles that have thesame measure arecongruent angles.KEY CONCEPTSDraw and label ∠RSTthat measures 70° underthe Angle Measure tab.Classify ∠RST as acute,right, or obtuse.Points, Lines,PlanesLength andPerimeterAnglesAngleMeasureORGANIZE IT
  • 23. Glencoe Geometry 15Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTA that divides an angle into congruentis called an angle bisector.1–4Use Algebra to Find Angle MeasuresINTERIOR DESIGN Wall stickers of standardshapes are often used to provide a stimulatingenvironment for a young child’s room.A five-pointed star sticker is shown withvertices labeled. Find m∠GBH and m∠HCIif ∠GBH ∠HCI, m∠GBH = 2x + 5, andm∠HCI = 3x - 10.∠GBH ∠HCI Givenm∠GBH = m∠HCI2x + 5 = 3x - 10 Substitution= Add 10 to each side.15 = x Subtract 2x from each side.Use the value of x to find the measure of one angle.m∠GBH = 2x + 5 Given= 2(15) + 5 x == + or 35 Simplify.Since m∠GBH = m∠HCI, m∠HCI = .Check Your Progress A railroadcrossing sign forms congruent angles.In the figure, m∠WVX m∠ZVY.If m∠WVX = 7a + 13 andm∠ZVY = 10a - 20, find the actualmeasurements of m∠WVX and m∠ZVY.REMEMBER ITIf angles arecongruent, then theirmeasures are equal.BUILD YOUR VOCABULARY (page 2)
  • 24. 16 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–5Identify Angle PairsRefer to the figure. Name two acute vertical angles.There are four acute anglesshown. There is one pair ofvertical angles. The acutevertical angles are.Check Your Progress Name an angle pair that satisfieseach condition.a. two angles that form a linear pairb. two adjacent angles whose measureshave a sum that is less than 90Angle MeasureALGEBRA Find the measures of two supplementaryangles if the measure of one angle is 6 less thanfive times the other angle.The sum of the measures of supplementary angles is .Draw two figures to represent the angles.If m∠A = x, then m∠B = .• Identify and use specialpairs of angles.• Identify perpendicularlines.MAIN IDEASAngle RelationshipsAngle PairsAdjacent angles aretwo angles that lie inthe same plane, havea common vertex, anda common side, but nocommon interior points.Vertical angles aretwo nonadjacentangles formed bytwo intersecting lines.A linear pair is a pair ofadjacent angles whosenoncommon sides areopposite rays.KEY CONCEPTSDraw andlabel examples under theAngles tab.
  • 25. Glencoe Geometry 17Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Lines that form are perpendicular.1–5m∠A + m∠B = 180 Given+ ( )= 180m∠A = x andm∠B = 5x - 66x - 6 = 180 Simplify.6x = 186 Add to each side.x = 31 Divide each side by 6.m∠A = x m∠B =m∠A = 31 m∠B = 5(31) - 6 or 149Check Your Progress Find the measures of twocomplementary angles if one angle measures six degrees lessthan five times the measure of the other.Perpendicular LinesALGEBRA Find x so that KO ⊥ HM.If KO ⊥ HM, then m∠KJH = .m∠KJH = m∠KJI + m∠IJH90 = (3x + 6) + 9x Substitution90 = Add.84 = 12x Subtract 6 from each side.= x Divide each side by 12.Angle RelationshipsComplementary anglesare two angles whosemeasures have asum of 90°.Supplementary anglesare two angles whosemeasures have asum of 180°.KEY CONCEPTPerpendicular linesintersect to formfour right angles.Perpendicular linesintersect to formcongruent adjacentangles. Segments andrays can be perpendicularto lines or to other linesegments and rays.The right angle symbolis used in figures toindicate that lines,rays, or segmentsare perpendicular.KEY CONCEPTBUILD YOUR VOCABULARY (page 3)
  • 26. 18 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT1–5Check Your Progress Find xand y so that AD and CE areperpendicular.Interpret FiguresDetermine whether each statement can be assumedfrom the figure below.a. m∠VYT = 90The diagram is marked to show VY ⊥ XT.From the definition of perpendicular,perpendicular lines intersect to formcongruent adjacent angles.; VY and TX are perpendicular.b. ∠TYW and ∠TYU are supplementary.Yes; they form a of angles.c. ∠VYW and ∠TYS are adjacent angles.No; they do not share a .Check Your Progress Determine whether eachstatement can be assumed from the figure below.Explain.a. m∠XAY = 90b. ∠TAU and ∠UAY arecomplementary.c. ∠UAX and ∠UXA are adjacent.
  • 27. Glencoe Geometry 19Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–6 Two-Dimensional FiguresIdentify PolygonsName each polygon by the number of sides. Thenclassify it as convex or concave, regular or irregular.a. There are 4 sides, so this isa .No line containing any of the sides will pass through theinterior of the quadrilateral, so it is . The sidesare not congruent, so it is .b. There are 9 sides, so this is a.A line containing a side will pass through the interior of thenonagon, so it is .The sides are not congruent, so it is .Check Your Progress Name each polygon by thenumber of sides. Then classify it as convex or concave,regular or irregular.a. b.• Identify and namepolygons.• Find perimeter orcircumference and areaof two-dimensionalfigures.MAIN IDEASA polygon is a figure whose areall segments.BUILD YOUR VOCABULARY (page 3)Polygon A polygon is aclosed figure formed bya finite number ofcoplanar segments suchthat (1) the sides thathave a commonendpoint arenoncollinear, and(2) each side intersectsexactly two other sides,but only at theirendpoints.KEY CONCEPT
  • 28. 20 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–6Find Perimeter and AreaFind the perimeter or circumference and area ofthe figure.C = 2πr Circumference of a circleC = 2π( ) r =C = Multiply.C = Use a calculator.A = πr2 Area of a circleA = π( 2) r =A = Simplify.A = Use a calculator.The circumference of the circle is about inches and thearea is about square inches.Check Your Progress Find the perimeter or circumferenceand area of the figure.Perimeter The perimeterP of a polygon is the sumof the lengths of thesides of a polygon.Area The area is thenumber of square unitsneeded to cover asurface.KEY CONCEPTDraw a polygon andexplain how to find theperimeter. Place thisunder the Length andPerimeter tab.Points, Lines,PlanesLength andPerimeterAnglesAngleMeasureORGANIZE IT
  • 29. Glencoe Geometry 21Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–6Largest AreaTEST EXAMPLE Terri has 19 feet of tape to make an areain the classroom where the students can read. Which ofthese shapes has a perimeter or circumference thatwould use most or all of the tape?A square with side length of 5 feetB circle with the radius of 3 feetC right triangle with each leg length of 6 feetD rectangle with a length of 8 feet and a width of 3 feetRead the Test ItemYou are asked to compare the perimeters of four differentshapes.Solve the Test ItemFind the perimeter of each shape.Square Circle RectangleP = 4s C = 2πr P = 2ℓ + 2wP = 4(5) C = 2π(3) P = 2(8) + 2(3)P = ft C = 6π or about ft P = ftRight TriangleUse the Pythagorean Theorem to find the length of thehypotenuse.c2 = a2 + b2c2 = 62 + 62c2 = 72c = 6√2P = 6 + 6 + 6√2P = or about ftThe shape that uses the most of the tape is the circle. Theanswer is .Check Your Progress Jason has 20 feet of fencing tomake a pen for his dog. Which of these shapes enclosesthe largest area?A square with a side length of 5 feetB circle with radius of 3 feetC right triangle with each leg about 6 feetD rectangle with length of 4 feet and width of 6 feet
  • 30. 22 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT1–6Perimeter and Area on the Coordinate PlaneFind the perimeter of pentagonABCDE with A(0, 4), B(4, 0),C(3, -4), D(-3, -4), and E(-3, 1).Since−−−DE is a vertical line segment,we can count the squares onthe grid. The length of−−−DE isunits. Likewise, since−−−CD is ahorizontal line segment, count the squares to find that thelength is units.To find AE, AB, and BC, use the distance formula.AE = √(0 - (-4))2 + (4 - 1)2 SubstitutionAE =2+2Subtract.AE = or 5 Simplify.AB = √(0 - 4)2 + (4 - 0)2 SubstitutionAB = ( )2+2Subtract.AB = √32 or about Simplify.BC = √(4 - 3)2 + (0 - (-4))2 SubstitutionBC =2+2Subtract.BC = or about 4.1 Simplify.To find the perimeter, add the lengths of each side.P = AB + BC + CD + DE + AEP ≈ 5.7 + 4.1 + 6 + 5 + 5P ≈The perimeter is approximately units.Check Your Progress Find the perimeter of quadrilateralWXYZ with W(2, 4), X(-3, 3), Y(-1, 0) and Z(3, -1).
  • 31. Glencoe Geometry 23Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–7A solid with all that enclose a singleregion of space is called a polyhedron.A prism is a polyhedron with congruentfaces called bases.A regular prism is a prism with that are regularpolygons.A polyhedron with all faces (except for one) intersecting atis a pyramid.A polyhedron is a regular polyhedron if all of its faces areand all of theare congruent.A cylinder is a solid with congruentin a pair of parallel planes.A cone has a base and a .A sphere is a set of in space that are a givendistance from a given point.Three-Dimensional Figures• Identify three-dimensional figures.• Find surface area andvolume.MAIN IDEASBUILD YOUR VOCABULARY (page 3)
  • 32. 24 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.1–7Identify SolidsIdentify each solid. Name the bases, faces, edges, andvertices.a.The bases and faces are rectangles. This is a rectangularprism.Bases: rectangles andFaces:Edges:Vertices:b.This figure has two faces that are hexagons. Therefore, it isa hexagonal prism.Bases: hexagons andFaces: rectangles EFLK, FGML, GHMN, HNOI, IOPJ, andJPKEEdges:−−FL,−−−GM,−−−HN,−−IO,−−JP,−−−EK,−−EF,−−−FG,−−−GH,−−HI,−−IJ,−−JE,−−KL,−−−LM,−−−MN,−−−NO,−−−OP, and−−PKVertices: E, F, G, H, I, J, K, L, M, N, O, Pc.The base of the solid is a circle and the figure comes to apoint. Therefore, it is a cone.Base: Vertex:There are faces or edges.
  • 33. Glencoe Geometry 25Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT1–7Check Your Progress Identify the solid.Name the bases, faces, edges, and vertices.Surface Area and VolumeRefer to the cone with r = 3 cm, h = 4 cm, and ℓ = 5 cm.a. Find the surface area of the cone.T = πrℓ + πr2 Surface area of a coneT = -π( )( )+ π( 2) SubstitutionT = 15π + 9π Simplify.T ≈ cm2 Use a calculator.b. Find the volume of the cone.V = 1_3πr2h Volume of a coneV = 1_3π( 2)( ) SubstitutionV = 12π Simplify.V ≈ cm3 Use a calculator.Check Your Progress Find the surfacearea and volume of the triangular prism.
  • 34. 26 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.STUDY GUIDEBRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R1Refer to the figure.1. Name a point contained in line n.2. Name the plane containing lines n and p.3. Draw a model for therelationship AK and CGintersect at point Min plane T.Find the measure of each segment.4.−−−AD 5.−−−WX1-1Points, Lines, and PlanesUse your Chapter 1 Foldableto help you study for yourchapter test.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 1, go to:glencoe.comYou can use your completedVocabulary Builder(pages 2–3) to help you solvethe puzzle.1-2Linear Measure and Precision
  • 35. Glencoe Geometry 27Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER16. CARPENTRY Jorge used the figure at the right to make apattern for a mosaic he plans to inlay on a tabletop. Name allof the congruent segments in the figure.Use the number line to find each measure.7. LN 8. JLFind the distance between each pair of points.9. F(-3, -2), G(1, 1) 10. Y(-6, 0), P(2, 6)Find the coordinates of the midpoint of a segment havingthe given endpoints.11. A(3, 1), B(5, 3) 12. T(-4, 9), U(7, 5)For Exercises 13–16, refer to the figure at the right.13. Name a right angle.14. Name an obtuse angle.15. Name a point in the interior of ∠EBC.16. What is the angle bisector of ∠EBC?1-3Distance and Midpoints1-4Angle Measure
  • 36. 28 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER1In the figure, CB and CD are opposite rays, CE bisects ∠DCF,and CG bisects ∠FCB.17. If m∠DCE = 4x + 15 and m∠ECF = 6x - 5, find m∠DCE.18. If m∠FCG = 9x + 3 and m∠GCB = 13x - 9, find m∠GCB.19. TRAFFIC SIGNS The diagram shows a sign usedto warn drivers of a school zone or crossing. Measure andclassify each numbered angle.For Exercises 20–23, use the figure at the right.20. Name two acute vertical angles.21. Name a linear pair.22. Name two acute adjacent angles.23. Name an angle supplementary to ∠FKG.24. Find x so that−−TR ⊥−−TSif m∠RTS = 8x + 18.25. Find m∠PTQ if−−TR ⊥−−TSand m∠PTQ = m∠RTQ - 18.1-5Angle Relationships
  • 37. Glencoe Geometry 29Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER1Determine whether each statement can be assumed from the figure.Explain.26. ∠YZU and ∠UZV are supplementary.27. ∠VZU is adjacent to ∠YZX.Name each polygon by its number of sides and then classify it as convexor concave and regular or irregular.28. 29.30. The length of a rectangle is 8 inches less than six times itswidth. The perimeter is 26 inches. Find the length of each side.Find the surface area and volume of each solid.31. 32. 33.1-6Two-Dimensional Figures1-7Three-Dimensional Figures
  • 38. 30 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistC H A P T E R1Visit glencoe.com toaccess your textbook,more examples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 1.ARE YOU READY FORTHE CHAPTER TEST?Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 1 Practice Test on page 73of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 1 Study Guide and Reviewon pages 68–72 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 1 Practice Test onpage 73 of your textbook.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 1 Foldable.• Then complete the Chapter 1 Study Guide and Review onpages 68–72 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 1 Practice Test onpage 73 of your textbook.
  • 39. Glencoe Geometry 31Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter2Reasoning and ProofC H A P T E R2NOTE-TAKING TIP: When you take notes, listen orread for main ideas. Then record concepts, defineterms, write statement in if-then form, and writeparagraph proofs.Stack the sheets ofpaper with edges3_4-inch apart. Foldthe bottom edges upto create equal tabs.Staple along the fold.Label the top tab withthe chapter title. Labelthe next 8 tabs withlesson numbers. Thelast tab is forKey Vocabulary.Begin with five sheets of 81_2" × 11" plain paper.Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.
  • 40. 32 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R2Vocabulary TermFoundDefinitionDescription oron Page ExampleThis is an alphabetical list of new vocabulary terms you will learn in Chapter 2.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.conclusionconditional statementconjecture[kuhn-JEK-chur]conjunctioncontrapositiveconversecounterexampledeductive argumentdeductive reasoningdisjunctionhypothesisif-then statement
  • 41. Chapter BUILD YOUR VOCABULARY2Glencoe Geometry 33Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Vocabulary TermFoundDefinitionDescription oron Page Exampleinductive reasoninginversenegationparagraph proofpostulaterelated proofconditionalsstatementtheoremtruth tabletruth valuetwo-column proof
  • 42. 34 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.A conjecture is an educated based on knowninformation.Inductive reasoning is reasoning that uses a numberof specific examples to arrive at a plausible generalizationor .Inductive Reasoning and ConjecturePatterns and ConjectureMake a conjecture about the next number based on thepattern 2, 4, 12, 48, 240.Find a Pattern: 2 4 12 48 240Conjecture: The next number will be multiplied by 6.So, it will be · 240 or .Check Your Progress Make a conjecture about the nextnumber based on the pattern 1, 1_4, 1_9, 1_16, 1_25.Geometric ConjectureFor points L, M, and N, LM = 20, MN = 6, and LN = 14.Make a conjecture and draw a figure to illustrate yourconjecture.Given: Points L, M, and N; LM = 20, MN = 6, and LN = 14.Since LN + MN = LM, the points can be collinearwith point N between points L and M.Conjecture: L, M and N are .REMEMBER ITWhen looking forpatterns in a sequence ofnumbers, test allfundamental operations.Sometimes twooperations can be used.(pages 32–33)BUILD YOUR VOCABULARY• Make conjectures basedon inductive reasoning.• Find counterexamples.MAIN IDEAS2–1
  • 43. Glencoe Geometry 35Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTA counterexample is one false example showing that aconjecture is not true.BUILD YOUR VOCABULARY (pages 32–33)Check Your Progress ACE is a right triangle withAC = CE. Make a conjecture and draw a figure to illustrateyour conjecture.Find a CounterexampleUNEMPLOYMENT Refer to the table. Find acounterexample for the following statement.The unemployment rate is highest in the cities withthe most people.City Population RateArmstrong 2163 3.7%Cameron 371,825 7.2%El Paso 713,126 7.0%Hopkins 33,201 4.3%Maverick 50,436 11.3%Mitchell 9402 6.1%Maverick has a population of people, and it hasa higher rate of unemployment than El Paso, which has apopulation of people.Check Your Progress Refer to the table. Find acounterexample for the following statement.The unemployment rate is lowest in the cities with theleast people.2–1Write definitions for aconjecture and inductivereasoning under the tabfor Lesson 2-1. Design apattern and state aconjecture about thepattern.ORGANIZE IT
  • 44. 36 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–2 Logic• Determine truth valuesof conjunctions anddisjunctions.• Construct truth tables.MAIN IDEAS A statement is any sentence that is either true or false,but not . The truth or falsity of a statement iscalled its truth value.(pages 32–33)BUILD YOUR VOCABULARYTruth Values of ConjunctionsUse the following statements to write a compoundstatement for each conjunction. Then find its truth value.p: One foot is 14 inches.q: September has 30 days.r: A plane is defined by three noncollinear points.a. p and qOne foot is 14 inches, and September has 30 days. p and qis because is false and q is .b. ∼p ∧ rA foot is not 14 inches, and a plane is defined by threenoncollinear points. ∼p ∧ r is , because ∼p isand r is .Check Your Progress Use the following statements towrite a compound statement for each conjunction. Thenfind its truth value.p: June is the sixth month of the year.q: A square has five sides.r: A turtle is a bird.a. p and rb. ∼q ∧ ∼rNegation If a statementis represented by p, thennot p is the negationof the statement.Conjunction Aconjunction is acompound statementformed by joining twoor more statements withthe word and.KEY CONCEPTS
  • 45. Glencoe Geometry 37Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.A convenient method for organizing the truth values ofstatements is to use a truth table.Disjunction A disjunctionis a compound statementformed by joining twoor more statements withthe word or.KEY CONCEPTTruth Values of DisjunctionsUse the following statements to write a compoundstatement for each disjunction. Then find its truth value.p:−−AB is proper notation for “segment AB.”q: Centimeters are metric units.r: 9 is a prime number.a. p or q−−AB is proper notation for “segment AB,” or centimeters aremetric units. p or q is because q is .It does not matter that is false.b. q ∨ rCentimeters are metric units, or 9 is a prime number. q ∨ ris because q is . It does not matterthat is false.Check Your Progress Use the following statements towrite a compound statement for each disjunction. Thenfind its truth value.p: 6 is an even number.q: A cow has 12 legs.r: A triangle has three sides.a. p or rb. ∼q ∨ ∼rBUILD YOUR VOCABULARY (pages 32–33)2–2
  • 46. 38 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTConstruct a truth tablefor the compoundstatement ∼q ∨ p andwrite it under the tabfor Lesson 2-2.ORGANIZE ITUse Venn DiagramsDANCING The Venn diagram shows the number ofstudents enrolled in Monique’s Dance School for tap,jazz, and ballet classes.Tap2817 2513 Jazz4329Ballet9a. How many students are enrolled in all three classes?The number of students that are enrolled in allclasses is represented by the of the threecircles. There are students enrolled in all three classes.b. How many students are enrolled in tap or ballet?The number of students enrolled in tap or ballet isrepresented by the union of the two sets. There areor studentsenrolled in tap or ballet.c. How many students are enrolled in jazz and ballet,but not tap?The number of students enrolled in and ballet butnot tap is represented by the intersection of the jazz andsets. There are students enrolled inonly.Check Your Progress How many students are enrolled inballet and tap, but not jazz?2–2
  • 47. Glencoe Geometry 39Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–3 Conditional StatementsWrite a Conditional in If-Then FormIdentify the hypothesis and conclusion of eachstatement. Then write each statement inif-then form.a. Distance is positive.Hypothesis:Conclusion:If a distance is measured, then it is positive.b. A five-sided polygon is a pentagon.Hypothesis:Conclusion:If a polygon has , then it is a .Check Your Progress Identify the hypothesis and conclusionof the statement.To find the distance between two points, you can use theDistance Formula.• Analyze statements inif-then form.• Write the converse,inverse, andcontrapositive ofif-then statements.MAIN IDEASIf-Then Statement Anif-then statement iswritten in the form if p,then q. The phraseimmediately followingthe word if is called thehypothesis, and thephrase immediatelyfollowing the word thenis called the conclusion.KEY CONCEPT
  • 48. 40 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Truth Values of ConditionalsDetermine the truth value of the following statementfor each set of conditions. If Yukon rests for 10 days, hisankle will heal.a. Yukon rests for 10 days, and he still has a hurt ankle.The hypothesis is , since he rested for 10 days. Theconclusion is since his ankle will heal. Thus,the conditional statement is .b. Yukon rests for 10 days, and he does not have a hurtankle anymore.The hypothesis is since Yukon rested for 10 days,and the conclusion is because he does not have ahurt ankle. Since what was stated is true, the conditionalstatement is .c. Yukon rests for 7 days, and he does not have a hurtankle anymore.The hypothesis is , and the conclusion is .The statement does not say what happens if Yukon onlyrests for 7 days. In this case, we cannot say that thestatement is false. Thus, the statement is .Check Your Progress Determine the truth value ofthe following statement for each set of conditions. If itrains today, then Michael will not go skiing.a. It rains today; Michael does not go skiing.b. It rains today; Michael goes skiing.Under the tab forLesson 2-3, explain howto determine the truthvalue of a conditional.Be sure to include anexample.ORGANIZE IT2–3
  • 49. Glencoe Geometry 41Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTRelated ConditionalsWrite the converse, inverse, and contrapositive of thestatement All squares are rectangles. Determinewhether each statement is true or false. If a statementis false, give a counterexample.Conditional: If a shape is a square, then it is a rectangle.The conditional statement is true.Write the converse by switching the andconclusion of the conditional.Converse: If a shape is a rectangle, then it is a square.The converse is . A rectangle with ℓ = 2and w = 4 is not a square.Inverse: If a shape is not a square, then it is not arectangle. The inverse is . A rectanglewith side lengths 2, 2, 4, and 4 is not a square.The contrapositive is the of the hypothesis andconclusion of the converse.Contrapositive: If a shape is not a rectangle, then it is nota square. The contrapositive is .Check Your Progress Write the converse, inverse, andcontrapositive of the statement The sum of the measures oftwo complementary angles is 90. Determine whether eachstatement is true or false. If a statement is false, give acounterexample.Related ConditionalsConditional Formed bygiven hypothesis andconclusionConverse Formed byexchanging thehypothesis andconclusion of theconditionalInverse Formed bynegating both thehypothesis andconclusion of theconditionalContrapositive Formedby negating both thehypothesis andconclusion of theconverse statementKEY CONCEPTS2–3
  • 50. 42 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Deductive reasoning uses facts, rules, , orto reach logical conclusions.Deductive Reasoning• Use the Law ofDetachment.• Use the Law ofSyllogism.MAIN IDEASDetermine Valid ConclusionsThe following is a true conditional. Determinewhether the conclusion is valid based on the giveninformation. Explain your reasoning.If two segments are congruent and the second segmentis congruent to a third segment, then the first segmentis also congruent to the third segment.a. Given:−−−WX−−UV;−−UV−−RTConclusion:−−−WX−−RTThe hypothesis states that W andU . Since the conditional is true and thehypothesis is true, the conclusion is .b. Given:−−UV,−−−WX−−RTConclusion:−−−WX−−UV and−−UV−−RTThe hypothesis states that W and .Not enough information is provided to reach the conclusion.The conclusion is .Check Your Progress The following is a true conditional.Determine whether the conclusion is valid based on the giveninformation. Explain your reasoning.If a polygon is a convex quadrilateral, then the sum of themeasures of the interior angles is 360°.BUILD YOUR VOCABULARY (pages 32–33)Law of Detachment Ifp → q is true and p istrue, then q is also true.KEY CONCEPT2–4
  • 51. Glencoe Geometry 43Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Given: m∠X + m∠N + m∠O = 360°Conclusion: If you connect X, N, and O with segments, thefigure will be a convex quadrilateral.Determine Valid Conclusions From Two ConditionalsPROM Use the Law of Syllogism to determine whethera valid conclusion can be reached from each set ofstatements.a. (1) If Salline attends the prom, she will go with Mark.(2) If Salline goes with Mark, Donna will go with Albert.The conclusion of the first statement is theof the second statement. Thus, if Salline attends the prom,Donna will go with .b. (1) If Mel and his date eat at the Peddler Steakhouse beforegoing to the prom, they will miss the senior march.(2) The Peddler Steakhouse stays open until 10 P.M.There is . While both statementsmay be true, the conclusion of one statement is not used asthe of the other statement.Check Your Progress Use the Law of Syllogism todetermine whether a valid conclusion can be reachedfrom each set of statements.a. (1) If you ride a bus, then you attend school.(2) If you ride a bus, then you go to work.b. (1) If your alarm clock goes off in the morning, then youwill get out of bed.(2) You will eat breakfast, if you get out of bed.Law of Syllogism Ifp → q and q → r aretrue, then p → r isalso true.KEY CONCEPTUse symbols and wordsto write the Law ofDetachment and theLaw of Syllogism underthe tab for Lesson 2-4.ORGANIZE IT2–4
  • 52. 44 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTAnalyze ConclusionsDetermine whether statement (3) follows from statements(1) and (2) by the Law of Detachment or the Law ofSyllogism. If it does, state which law was used. If it doesnot, write invalid.(1) If the sum of the squares of two sides of a triangleis equal to the square of the third side, then thetriangle is a right triangle.(2) For △XYZ, (XY)2 + (YZ)2 = (ZX)2.(3) △XYZ is a right triangle.p: The of the squares of the lengths of the two sidesof a is equal to the square of the length ofthe side.q: the triangle is a triangle.By the Law of , if p → q is true andp is true, then q is also true. Statement (3) is aconclusion by the Law of Detachment.Check Your Progress Determine whether statement(3) follows from statements (1) and (2) by the Law ofDetachment or the Law of Syllogism. If it does, state whichlaw was used. If it does not, write invalid.(1) If a children’s movie is playing on Saturday, Janine willtake her little sister Jill to the movie.(2) Janine always buys Jill popcorn at the movies.(3) If a children’s movie is playing on Saturday, Jill willget popcorn.2–4
  • 53. Glencoe Geometry 45Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–5A postulate is a statement that describes a fundamentalrelationship between the basic terms of .Postulates are accepted as .Postulates and Paragraph Proofs• Identify and use basicpostulates about points,lines, and planes.• Write paragraph proofs.MAIN IDEASPoints and LinesSNOW CRYSTALS Some snow crystals are shaped likeregular hexagons. How many lines must be drawn tointerconnect all vertices of a hexagonal snow crystal?Draw a diagram of a hexagon toillustrate the solution. Connect eachpoint with every other point. Then,count the number of segments.Between every two points there isexactly segment.Be sure to include the sides of thehexagon. For the six points,segments can be drawn.Check Your Progress Jodi is making a string art design.She has positioned ten nails, similar to the vertices of adecagon, onto a board. How many strings will she need tointerconnect all vertices of the design?BUILD YOUR VOCABULARY (pages 32–33)Postulate 2.1Through any two points, there is exactly one line.Postulate 2.2Through any three points not on the same line, there isexactly one plane.
  • 54. 46 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Use PostulatesDetermine whether the statement is always, sometimes,or never true. Explain.a. If plane T contains EF and EF contains point G, thenplane T contains point G.; Postulate 2.5 states that if twolie in a plane, then the entire containing thosepoints lies in the plane.b. For XY, if X lies in plane Q and Y lies in plane R, thenplane Q intersects plane R.; planes Q and R can both be ,and can intersect both planes.c. GH contains three noncollinear points.; noncollinear points do not lie on the sameby definition.Check Your Progress Determine whether the statementis always, sometimes, or never true. Explain.Plane A and plane B intersect in one point.Postulate 2.3A line contains at least two points.Postulate 2.4A plane contains at least three points not on the same line.Postulate 2.5If two points lie on a plane, then the entire line containingthose points lies in that plane.Postulate 2.6If two lines intersect, then their intersection is exactlyone point.Postulate 2.7If two planes intersect, then their intersection is a line.Collinear means lyingon the same line, sononcollinear means notlying on the same line.(Lesson 1-1)REVIEW IT2–5
  • 55. Glencoe Geometry 47Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT2–5Write a Paragraph ProofGiven AC intersects CD, write a paragraph proof toshow that A, C, and D determine a plane.Given: AC intersects CD.Prove: A, C, and D determine a plane.AC and CD must intersect at C because if linesintersect, then their intersection is exactly point.Point A is on AC and point D is on CD. Therefore, pointsA and D are . Therefore, A, C, and Ddetermine a plane.Check Your Progress Given−−RT−−TY, S is the midpoint of−−RT,and X is the midpoint of−−TY, writea paragraph proof to show that−−ST−−TX.Theorem 2.1 Midpoint TheoremIf M is the midpoint of−−AB, then−−−AM−−−MB.Proofs Five essentialparts of a good proof:• State the theorem orconjecture to beproven.• List the giveninformation.• If possible, draw adiagram to illustratethe given information.• State what is to beproved.• Develop a system ofdeductive reasoning.CopyExample 3 under the tabfor Lesson 2-5 as anexample of a paragraphproof.KEY CONCEPT
  • 56. 48 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.A two-column proof, or formal proof, containsand organized intwo columns.Algebraic Proof2–6• Use algebra to writetwo-column proofs.• Use properties ofequality in geometryproofs.MAIN IDEASVerify Algebraic RelationshipsSolve 2(5 - 3a) - 4(a + 7) = 92.Algebraic Steps Properties2(5 - 3a) - 4(a + 7) = 92 Original equation10 - 6a - 4a - 28 = 92 Property-18 - 10a = 92 Property= Addition Propertya = PropertyCheck Your Progress Solve -3(a + 3) + 5(3 - a) = -50.BUILD YOUR VOCABULARY (pages 32–33)Write a two-columnproof for the statementif 5(2x - 4) = 20, thenx = 4 under the tab forLesson 2-6.ORGANIZE IT
  • 57. Glencoe Geometry 49Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–6Write a Two-Column ProofWrite a two-column proof to show that if7d + 3_4= 6,then d = 3.Statements Reasons1. 7d + 3_4= 6 1. Given2. 4(7d + 3_4 ) = 4(6) 2.3. 7d + 3 = 24 3. Substitution4. 7d = 21 4.5. 5.Check Your Progress Write a two-column proof for thefollowing.Given: 10 - 8n_3= -2Proof: n = 2What are the fiveessential parts of a goodproof? (Lesson 2-5)WRITE IT
  • 58. 50 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT2–6Geometric ProofSEA LIFE A starfish has five arms. If the length of arm 1is 22 centimeters, and arm 1 is congruent to arm 2, andarm 2 is congruent to arm 3, prove that arm 3 has alength of 22 centimeters.Given: arm 1 arm 2arm 2 arm 3m arm 1 = 22 cmProve: m arm 3 = 22 cmProof:Statements Reasons1. arm 1 arm 2; 1. Givenarm 2 arm 32. arm 1 arm 3 2.3. m arm 1 = m arm 3 3. Definition of congruence4. 4. Given5. m arm 3 = 22 cm 5. Transitive PropertyCheck Your Progress A stopsign as shown at right is a regularoctagon. If the measure of angle Ais 135 and angle A is congruent toangle G, prove that the measureof angle G is 135.
  • 59. Glencoe Geometry 51Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–7• Write proofs involvingsegment addition.• Write proofs involvingsegment congruence.MAIN IDEASProving Segment RelationshipsProof with Segment AdditionProve the following.Given: PR = QSProve: PQ = RSStatements Reasons1. PR = QS 1. Given2. PR - QR = QS - QR 2. Subtraction Property3. PR - QR = PQ; 3. Segment AdditionQS - QR = RS Postulate4. PQ = RS 4. SubstitutionCheck Your Progress Prove the following.Given: AC = ABAB = BXCY = XDProve: AY = BDPostulate 2.8 Ruler PostulateThe points on any line or line segment can be paired withreal numbers so that, given any two points A and B on aline, A corresponds to zero, and B corresponds to a positivereal number.Postulate 2.9 Segment Addition PostulateIf B is between A and C, then AB + BC = AC.If AB + BC = AC, then B is between A and C.Under the tab forLesson 2-7, write theSegment AdditionPostulate, draw anexample, and writean equation for yourexample.ORGANIZE IT
  • 60. 52 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT2–7Transitive Property of CongruenceProve the following.Given: WY = YZ,−−YZ−−XZ−−XZ−−−WYProve:−−−WX−−−WYStatements Reasons1. WY = YZ 1. Given2.−−−WY−−YZ 2. Definition ofsegments3.−−YZ−−XZ;−−XZ−−−WX 3. Given4. 4. Transitive Property5.−−−WX−−−WY 5.Check Your Progress Prove the following.Given:−−−GD−−−BC,−−−BC−−−FH,−−−FH−−AEProve:−−AE−−−GDTheorem 2.2Congruence of segments is reflexive, symmetric, andtransitive.
  • 61. Glencoe Geometry 53Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–8 Proving Angle RelationshipsAngle AdditionTIME At 4 o’clock, the angle betweenthe hour and minute hands of a clockis 120°. If the second hand stopswhere it bisects the angle betweenthe hour and minute hands, what arethe measures of the angles betweenthe minute and second hands andbetween the second and hour hands?If the second hand stops where the angle is bisected, then theangle between the minute and second hands isthe measure of the angle formed by the hour and minute hands,or (120) = .By the Angle Addition Postulate, the sum of the two angles is, so the angle between the second and hour hands isalso .Check Your Progress The diagramshows one square for a particular quiltpattern. If m∠BAC = m∠DAE = 20°,and ∠BAE is a right angle, find m∠CAD.Postulate 2.10 Protractor PostulateGiven AB and a number r between 0 and 180, there isexactly one ray with endpoint A, extending on either sideof AB, such that the measure of the angle formed is r.Postulate 2.11 Angle Addition PostulateIf R is in the interior of ∠PQS, then m∠PQR + m∠RQS =m∠PQS. If m∠PQR + m∠RQS = m∠PQS, then R is in theinterior of ∠PQS.• Write proofs involvingsupplementary andcomplementary angles.• Write proofs involvingcongruent and rightangles.MAIN IDEAS
  • 62. 54 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–8The angles of a linearpair are alwayssupplementary, butsupplementary anglesneed not form a linearpair. (Lesson 1-5)REVIEW ITSupplementary AnglesIf ∠1 and ∠2 form a linear pair and m∠2 = 166, find m∠1.m∠1 + m∠2 = 180m∠1 + = 180 m∠2 =m∠1 = 14 Subtraction PropertyCheck Your Progress If ∠1 and ∠2 are complementaryangles and m∠1 = 62, find m∠2.Theorem 2.5Congruence of angles is reflexive, symmetric, and transitive.Theorem 2.6Angles supplementary to the same angle or to congruentangles are congruent.Theorem 2.7Angles complementary to the same angle or to congruentangles are congruent.Theorem 2.3 Supplement TheoremIf two angles form a linear pair, then they are supplementaryangles.Theorem 2.4 Complement TheoremIf the noncommon sides of two adjacent angles form aright angle, then the angles are complementary angles.
  • 63. Glencoe Geometry 55Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.2–8Use Supplementary AnglesIn the figure, ∠1 and ∠4 form a linear pair, andm∠3 + m∠1 = 180. Prove that ∠3 and ∠4 are congruent.Given: ∠1 and ∠4 form a linear pair.m∠3 + m∠1 = 180Prove: ∠3 ∠4Proof:StatementsReasons1. m∠3 + m∠1 = 180; ∠1 1. Givenand ∠4 form a linear pair.2. ∠1 and ∠4 are 2.supplementary.3. ∠3 and ∠1 are 3. Definition of supplementarysupplementary. angles.4. ∠3 ∠4 4.Check Your ProgressIn the figure, ∠NYR and∠RYA form a linear pair,∠AXY and ∠AXZ form alinear pair, and ∠RYAand ∠AXZ are congruent.Prove that ∠RYN and∠AXY are congruent.Under the tab forLesson 2-8, copyTheorem 2.12: If twoangles are congruentand supplementary, theneach angle is a rightangle. Illustrate thistheorem with a diagram.ORGANIZE IT
  • 64. 56 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT2–8Vertical AnglesIf ∠1 and ∠2 are vertical angles and m∠1 = d - 32 andm∠2 = 175 - 2d, find m∠1 and m∠2.∠1 ∠2 Theoremm∠1 = m∠2 Definition of congruent angles= Substitution3d = 207 Addition Propertyd = 69 Divide each side by 3.m∠1 = m∠2 = 175 - 2d= - 32 = 175 - 2( )= =Check Your Progress If ∠A and ∠Z are vertical angles andm∠A = 3b - 23 and m∠Z = 152 - 4b, find m∠A and m∠Z.Theorem 2.8 Vertical Angle TheoremIf two angles are vertical angles, then they are congruent.Theorem 2.9Perpendicular lines intersect to form four right angles.Theorem 2.10All right angles are congruent.Theorem 2.11Perpendicular lines form congruent adjacent angles.Theorem 2.12If two angles are congruent and supplementary, then eachangle is a right angle.Theorem 2.13If two congruent angles form a linear pair, then they areright angles.REMEMBER ITBe sure to readproblems carefully inorder to provide theinformation requested.Often the value of thevariable is used to findthe answer.
  • 65. STUDY GUIDEGlencoe Geometry 57Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERUse your Chapter 2 Foldableto help you study for yourchapter test.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 2, go to:glencoe.comYou can use your completedVocabulary Builder(pages 32–33) to help you solvethe puzzle.Make a conjecture about the next number in the pattern.1. -6, -3, 0, 3, 6 2. 4, -2, 1, -1_2, 1_43. Make a conjecture based on the given information.Points A, B, and C are collinear. D is between A and B.Use the statements p: -1 + 4 = 3, q: A pentagon has 5 sides,and r: 5 + 3 > 8 to find the truth value of each statement.4. p ∨ r 5. q ∧ r 6. (p ∨ q) ∧ r7. Construct a truth table for the compound statement p ∨ (∽p ∧ q).2-1Inductive Reasoning and Conjecture2-2LogicC H A P T E R2 BRINGING IT ALL TOGETHER
  • 66. Chapter BRINGING IT ALL TOGETHER258 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8. Identify the hypothesis and conclusion of the statement.If x - 7 = 4, then x = 11.9. Write the statement in if-then form.Supplementary angles have measures with a sum of 180°.10. Use the Law of Detachment to tell whether the followingreasoning is valid. Write valid or not valid.If a car has air bags it is safe. Maria’s carhas air bags. Therefore, Maria’s car is safe.11. Use the Law of Syllogism to write a valid conclusion fromthis set of statements.(1) All squares are rectangles.(2) All rectangles are parallelograms.12. Determine the number of line segments that canbe drawn connecting each pair of points.Name the definition, property, postulate, or theorem thatjustifies each statement.13.−−−CD−−−CD14. If A is the midpoint of−−−CD, then−−CA−−−AD.2-3Conditional Statements2-4Deductive Reasoning2-5Postulates and Paragraph Proofs
  • 67. Chapter BRINGING IT ALL TOGETHER2Glencoe Geometry 59Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Name the definition, property, postulate, or theorem thatjustifies each statement.15. If x - 7 = 9, then x = 1616. 5(x + 8) = 5x + 4017. If−−−CD−−EF and−−EF−−−GH, then−−−CD−−−GH.Name the definition, property, postulate, or theorem thatjustifies each statement.18. If−−AB−−−CD, then−−−CD−−AB.19. If AB + BD = AD, then B is between A and D.Name the definition, property, postulate, or theorem thatjustifies each statement.20. If m∠1 + m∠2 = 90, and m∠2 + m∠3 = 90, then m∠1 = m∠3.21. If ∠A and ∠B are vertical angles, then ∠A ∠B.2-7Proving Segment Relationships2-6Algebraic Proof2-8Proving Angle Relationships
  • 68. 60 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureC H A P T E R2ChecklistCheck the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 2 Practice Test onpage 137 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 2 Study Guide and Reviewon pages 132–136 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 2 Practice Test onpage 137 of your textbook.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 2 Foldable.• Then complete the Chapter 2 Study Guide and Review onpages 132–136 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 2 Practice Test onpage 137.ARE YOU READY FORTHE CHAPTER TEST?Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 2.
  • 69. Chapter3Glencoe Geometry 61Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.C H A P T E R3 Parallel and Perpendicular LinesNOTE-TAKING TIP: When taking notes, writing aparagraph that describes the concepts, thecomputational skills, and the graphics will helpyou to understand the math in the lesson.Fold in half matchingthe short sides.Unfold and fold thelong side up 2 inchesto form a pocket.Staple or glue theouter edges tocomplete the pocket.Label each side asshown. Use indexcards to recordexamples.Begin with one sheet of 81_2" × 11" paper.Parallel Perpendicular
  • 70. 62 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R3This is an alphabetical list of new vocabulary terms you will learn in Chapter 3.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.Vocabulary TermFoundDefinitionDescription oron Page Examplealternate exterioranglesalternate interioranglesconsecutive interioranglescorresponding anglesequidistant[ee-kwuh-DIS-tuhnt]non-Euclidean geometry[yoo-KLID-ee-yuhn]parallel lines
  • 71. Chapter BUILD YOUR VOCABULARY3Glencoe Geometry 63Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Vocabulary TermFoundDefinitionDescription oron Page Exampleparallel planespoint-slope formrate of changeskew linesslopeslope-intercept formtransversal
  • 72. 64 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–1 Parallel Lines and TransversalsIdentify RelationshipsUse the figure shown at the right.a. Name all planes that areparallel to plane AEF.b. Name all segments thatintersect−−AF.c. Name all segments that are skew to−−AD.Check Your Progress Use the figure shown.a. Name all planes that are parallelto plane RST.Coplanar lines that do not are calledparallel lines.Two that do not are calledparallel planes.Lines that do not intersect and are not arecalled skew lines.A line that intersects or more lines in a planeat different points is called a transversal.BUILD YOUR VOCABULARY (pages 62–63)• Identify therelationships betweentwo lines or two planes.• Name angles formed bya pair of lines and atransversal.MAIN IDEAS
  • 73. Glencoe Geometry 65Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–1Identify TransversalsBUS STATION Some of a bus station’s driveways areshown. Identify the sets of lines to which each givenline is a transversal.a. line v If the lines are extended, line v intersects lines.b. line yc. line uCheck Your Progress A groupof nature trails is shown. Identifythe set of lines to which eachgiven line is a transversal.a. line ab. line bc. line cd. line db. Name all segments that intersect−−YZ.c. Name all segments that are skew to−−TZ.
  • 74. 66 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–1Page(s):Exercises:HOMEWORKASSIGNMENTconsecutive interior angles:∠4 and ∠5, ∠3 and ∠6alternate interior angles:∠4 and ∠6, ∠3 and ∠5alternate exterior angles:∠1 and ∠7, ∠2 and ∠8corresponding angles: ∠1 and ∠5, ∠2 and ∠6, ∠3 and ∠7,∠4 and ∠8BUILD YOUR VOCABULARY (page 62)1 23 4658710911 12nᐉ mpIdentify Angle RelationshipsIdentify each pair of angles as alternate interior, alternateexterior, corresponding, or consecutive interior angles.a. ∠7 and ∠3b. ∠8 and ∠2c. ∠4 and ∠11 d. ∠7 and ∠1e. ∠3 and ∠9 f. ∠7 and ∠10Check Your Progress Identify each pair of angles asalternate interior, alternate exterior, corresponding,or consecutive interior angles.a. ∠4 and ∠5b. ∠7 and ∠9c. ∠4 and ∠7 d. ∠2 and ∠11Using a separate cardfor each type of anglepair, draw two parallellines cut by a transversaland identify consecutiveinterior angles,alternate exteriorangles, alternateinterior angles, andcorresponding angles.Place your cards in theParallel Lines pocket.Parallel PerpendicularORGANIZE IT
  • 75. Glencoe Geometry 67Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–2 Angles and Parallel LinesSketch two parallel linesand a transversal on acard. Below the sketch,write a brief note toexplain how knowingone of the anglemeasures allows you tofind the measures of allthe other angles. Placethe card in the ParallelLines pocket.Parallel PerpendicularORGANIZE ITDetermine Angle MeasuresIn the figure x y and m∠11 = 51. Find m∠16.∠11 ∠15 Corresponding Angles PostulateVertical Angles Theorem∠11 ∠16 Transitive Propertym∠11 = m∠16 Definition of congruent angles= m∠16 SubstitutionCheck Your Progress In the figure, m∠18 = 42. Find m∠25.Postulate 3-1 Corresponding Angles PostulateIf two parallel lines are cut by a transversal, then each pairof corresponding angles is congruent.Theorem 3.1 Alternate Interior Angles TheoremIf two parallel lines are cut by a transversal, then each pairof alternate interior angles is congruent.Theorem 3.2 Consecutive Interior Angles TheoremIf two parallel lines are cut by a transversal, then each pairof consecutive interior angles is supplementary.• Use the propertiesof parallel lines todetermine congruentangles.• Use algebra to findangle measures.MAIN IDEAS
  • 76. 68 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTWhen two lines intersect,what types of anglepairs are congruent?(Lesson 2-8)REVIEW ITFind Values of VariablesALGEBRA If m∠5 = 2x - 10,m∠6 = 4(y - 25), andm∠7 = x + 15, find x and y.Find x. Since p q, m∠5 m∠7by the Corresponding AnglesPostulate.m∠5 = m∠7 Definition of congruent angles= Substitutionx = Subtract x from each side andadd 10 to each side.Find y. Since m n, m∠5 m∠6 by the Alternate ExteriorAngles Theorem.m∠5 = m∠6Definition of congruentangles= Substitution= x = 25= y Add 100 to each side anddivide each side by 4.Check Your ProgressIf m∠1 = 9x + 6,m∠2 = 2(5x - 3), andm∠3 = 5y + 14, find x and y.Theorem 3.3 Alternate Exterior Angles TheoremIf two parallel lines are cut by a transversal, then each pairof alternate exterior angles is congruent.Theorem 3.4 Perpendicular Transversal TheoremIn a plane, if a line is perpendicular to one of two parallellines, then it is perpendicular to the other.3–2
  • 77. Glencoe Geometry 69Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–3 Slope of LinesFind the Slope of a Linea. Find the slope of the line.฀฀ ฀Use the rise_run method. From(-3, 7) to (-1, -1), go downunits and 2 units.rise_run = orb. Find the slope of the line.Use the slope formula.Let (0, 4) be (x1, y1) and(0, -3) be (x2, y2).m =y2 - y1_x2 - x1=--0 0or-7_0, which is .c. Find the slope of the line.฀m =y2 - y1_x2 - x1==The slope of a line is the ratio of the vertical rise to itshorizontal run.BUILD YOUR VOCABULARY (page 63)• Find slopes of lines.• Use slope to identifyparallel andperpendicular lines.MAIN IDEASSlope The slope m ofa line containing twopoints with coordinates(x1, y1) and (x2, y2) isgiven by the formulam =y2 - y1_x2 - x1, wherex1 ≠ x2.KEY CONCEPT-6 - (-2)
  • 78. 70 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.The rate of change describes how a quantity isover .Postulate 3.2Two nonvertical lines have the same slope if and only ifthey are parallel.Postulate 3.3Two nonvertical lines are perpendicular if and only if theproduct of their slopes is -1.3–3d. Find the slope of the line.m =y2 - y1_x2 - x1== orCheck Your Progress Find the slope of each line.a. b.c. d.REMEMBER ITLines with positiveslopes rise as you movefrom left to right, whilelines with negativeslopes fall as you movefrom left to right.BUILD YOUR (page 63)
  • 79. Glencoe Geometry 71Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTDetermine Line RelationshipsDetermine whether FG and HJ are parallel,perpendicular, or neither.a. F(1, -3), G(-2, -1), H(5, 0), J(6, 3)Find the slopes of FG and HJ.slope of FG =--= or -2_3slope of HJ =--= 3_1orThe slopes are not the same, so FG and HJ are. The product of the slopes is . SoFG and HJ are neither nor .b. F(4, 2), G(6, -3), H(-1, 5), J(-3, 10)slope of FG =--= or -5_2slope of HJ =--= or -5_2The slopes are the same, so FG and HJ are .Check Your Progress Determine whether AB and CDare parallel, perpendicular, or neither.a. A(-2, -1), B(4, 5), C(6, 1), D(9, -2)b. A(7, -3), B(1, -2), C(4, 0), D(-3, 1)On a study card, writethe equation of twolines that are paralleland explain what is trueabout their slopes. Placethis card in the ParallelLines pocket.Parallel PerpendicularORGANIZE IT3–3
  • 80. 72 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–4The slope-intercept form of a linear equation is, where m is the of theline and b is the y-intercept.The point-slope form is , where(x1, y1) are the coordinates of any point on the line and mis the slope of the line.Equations of LinesSlope and y-interceptWrite an equation in slope-intercept form of the linewith slope of 6 and y-intercept of -3.y = mx + b Slope-intercept formy = x + m = 6, b = -3The slope-intercept form of the equation is .Check Your Progress Write an equation in slope-interceptform of the line with slope of -1 and y-intercept of 4.Slope and a PointWrite an equation in point-slope form of the line whoseslope is -3_5and contains (-10, 8).y - y1 = m(x - x1) Point-slope formy - = -3_5[x - ] m = -3_5, (x1, y1) = (-10, 8)y - 8 = -3_5(x + 10) Simplify.• Write an equation of aline given informationabout its graph.• Solve problems bywriting equations.MAIN IDEASBUILD YOUR VOCABULARY (page 63)REMEMBER ITNote that the point-slope form of anequation is different foreach point used.However, the slope-intercept form of anequation is unique.
  • 81. Glencoe Geometry 73Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Check Your Progress Write an equation in point-slopeform of the line whose slope is 1_3and contains (6, -3).Two PointsWrite an equation in slope-intercept form for a linecontaining (4, 9) and (-2, 0).Find the slope of the line.m =y2 - y1_x2 - x1Slope formula= 0 - 9_-2 - 4x1 = 4, x2 = -2,y1 = 9, y2 = 0= or Simplify.Use the point-slope form to write an equation.Using (4, 9):y - y1 = m(x - x1) Point-slope formy - = (x - 4) m = , (x1, y1) = (4, 9)y = 3_2x + 3 Distributive Property andadd 9 to both sides.Using (-2, 0):y - y1 = m(x - x1) Point-slope formy - = [x - (-2)] m = , (x1, y1) = (-2, 0)y = Distributive PropertyCheck Your Progress Write an equation in slope-interceptform for a line containing (3, 2) and (6, 8).3–4
  • 82. 74 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTOne Point and an EquationWrite an equation in slope-intercept form for a linecontaining (1, 7) that is perpendicular to the liney = -1_2x + 1.Since the slope of the line is , the slopeof a line perpendicular to it is .y - y1 = m(x - x1) Point-slope formy - = (x - ) m = 2, (x1, y1) = (1, 7)y = Distributive Property andadd 7 to each side.Check Your Progress Write an equation in slope-interceptform for a line containing (-3, 4) that is perpendicular to theline y = 3_5x - 4.On a study card, writethe slope-interceptequations of two linesthat are perpendicular,and explain what istrue about their slopes.Place this card in thePerpendicular Linespocket.Parallel PerpendicularORGANIZE IT3–4
  • 83. Glencoe Geometry 75Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–5 Proving Lines ParallelIdentify Parallel LinesDetermine which lines,if any, are parallel.Since ∠RQT and ∠SQP areangles,m∠SQP = .Since m∠UPQ + m∠SQP = + or ,consecutive interior angles are supplementary. So, a b.Since m∠TQR + m∠VRQ = 77 + 100 or 177, consecutiveinterior angles are . So, c is notparallel to a or b.• Recognize angleconditions that occurwith parallel lines.• Prove that two lines areparallel based on givenangle relationships.MAIN IDEASPostulate 3.4If two lines in a plane are cut by a transversal so thatcorresponding angles are congruent, then the lines areparallel.Postulate 3.5 Parallel PostulateIf there is a line and a point not on the line, then thereexists exactly one line through the point that is parallel tothe given line.Theorem 3.5If two lines in a plane are cut by a transversal so that a pairof alternate exterior angles is congruent, then the two linesare parallel.Theorem 3.6If two lines in a plane are cut by a transversal so that a pairof consecutive interior angles is supplementary, then thelines are parallel.Theorem 3.7If two lines in a plane are cut by a transversal so that a pairof alternate interior angles is congruent, then the lines areparallel.Theorem 3.8In a plane, if two lines are perpendicular to the same line,then they are parallel.
  • 84. 76 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–5Check Your Progress Determine which lines, if any,are parallel.Solve Problems with Parallel LinesALGEBRA Find x and m∠ZYN so that PQ MN.m∠WXP = m∠ZYN Alternate exterior angles= Substitution4x Ϫ 25 = 35 Subract 7x from each side.4x = 60 Add 25 to each side.x = 15 Divide each side by 4.Now use the value of x to find m∠ZYN.m∠ZYN = 7x + 35 Original equation= 7΂ ΃+ 35 x = 15= 140 Simplify.So, x = 15 and m∠ZYN = 140.Check Your ProgressFind x and m∠GBAso that GH RS.On a card, write thebiconditional statementthat you obtain bycombining Postulates 3.1and 3.4. Place the cardin the Parallel Linespocket.Parallel PerpendicularORGANIZE IT
  • 85. Glencoe Geometry 77Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–5Prove Lines ParallelGiven: ℓ m∠4 ∠7Prove: r sWrite how you can useangles formed by twolines and a transversalto decide whether thetwo lines are parallel ornot parallel.WRITE ITStatements Reasons1. ℓ m, ∠4 ∠7 1. Given2. ∠4 and ∠6 are suppl. 2. Consecutive Interior ∠Thm.3. ∠4 + ∠6 = 180 3.4. = m∠7 4. Def. of congruent ∠s5. m∠7 + = 180 5.6. ∠7 and ∠6 are suppl. 6. Def. of suppl. Єs7. r s 7. If cons. int. Єs are suppl.,then lines are parallel.Slope and Parallel LinesDetermine whether p q.slope of p: m ==slope of q: m ϭϭSince the slopes are , p q.--ϪϪ΂ ΃
  • 86. 78 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTCheck Your Progressa. Prove the following lines parallel.Given: x y∠1 ∠4Prove: a bb. Determine whether r s.3–5
  • 87. Glencoe Geometry 79Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–6Equidistant means that the between twolines measured along a to thelines is always the same.Perpendiculars and DistanceDistance from a Point to a LineDraw the segment that represents the distance fromA to BP.Since the distance from a line to a point not on the line is theof the segment to the linefrom the point, extend and draw so that.Check Your Progress Draw the segment that representsthe distance from R to−−XY.• Find the distancebetween a point anda line.• Find the distancebetween parallel lines.MAIN IDEASTheorem 3.9In a plane, if two lines are each equidistant from a thirdline, then the two lines are parallel to each other.BUILD YOUR VOCABULARY (page 62)Distance Between aPoint and a Line Thedistance from a line toa point not on the lineis the length of thesegment perpendicularto the line from thepoint.On a card,describe how to find thedistance from a pointin the coordinate planeto a line that does notpass through the point.Place the card in thePerpendicular Linespocket.KEY CONCEPT
  • 88. 80 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3–6Distance Between LinesFind the distance between parallel lines a and b whoseequations are y = 2x + 3 and y = 2x - 3, respectively.• First, write an equation of a line p perpendicular toa and b. The slope of p is the oppositeof 2, or . Use the y-intercept of line a, (0, 3), asone of the endpoints of the perpendicular segment.y - y1 = m(x - x1) Point-slope formy - = (x - ) x1 = , y1 = ,m =y = -1_2x + 3 Simplify and add 3 toeach side.• Next, use a system of equations to determine the point ofintersection of line b and p.= Substitutefor y in the secondequation.2x + 1_2x = 3 + 3 Group like terms on eachside.x = 2.4 Divide each side 5_2.y = ϭ Substitute 2.4 for x in theequation for p.The point of intersection is .Distance BetweenParallel Lines Thedistance between twoparallel lines is thedistance between one ofthe lines and any pointon the other line.KEY CONCEPT
  • 89. Glencoe Geometry 81Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTThen, use the Distance Formula to determine the distancebetween (0, 3) and (2.4, 1.8).d = Distance Formula=x2 = 2.4, x1 = 0,y2 = 1.8, y1 = 3=The distance between the lines is or aboutunits.Check Your Progress Find the distance between theparallel lines a and b whose equations are y = 1_3x + 1 andy = 1_3x - 2, respectively.What is the point-slopeform of a linearequation?WRITE IT3–6
  • 90. 82 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.STUDY GUIDEBRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R3Use your Chapter 3 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumble puzzleof the vocabulary words inChapter 3, go to:glencoe.comYou can use your completedVocabulary Builder (pages 62–63)to help you solve the puzzle.Refer to the figure at the right.1. Name all planes that are parallel to plane ABC.2. Name all segments that are parallel to FෆGෆ.In the figure, m∠5 = 100. Find the measure of each angle.3. ∠1 14. ∠35. ∠4 16. ∠77. ∠6 18. ∠2Determine the slope of the line that contains the given points.9. F(-6, 2), M(7, 9) 10. Z(1, 10), L(5, -3)3-1Parallel Lines and Transversals3-2Angles and Parallel Lines3-3Slope of Lines
  • 91. Chapter BRINGING IT ALL TOGETHER3Glencoe Geometry 83Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11. Determine whether−−EF and−−−PQ are parallel, perpendicular, orneither. E(0, 4), F(2, 3), P(-3, 5), Q(1, 3)12. Write an equation in slope-intercept form of the line with slope-2 that contains (2, 5).13. Write an equation in slope-intercept form of the line thatcontains (-4, -2) and (-1, 7).Given the following information, determine which lines,if any, are parallel. State the postulate or theorem thatjustifies your answer.14. ∠1 ∠1515. ∠9 ∠1116. ∠2 ∠617. Draw the segment that representsthe distance from m to NP.Find the distance between each pair of parallel lines.18. x = 3 19. y = x + 5x = -5 y = x - 53-4Equations of lines3-5Proving Lines Parallel3-6Perpendiculars and Distance
  • 92. 84 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureC H A P T E R3ChecklistCheck the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 3 Practice Test onpage 195 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 3 Study Guide and Reviewon pages 191–194 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 3 Practice Test onpage 195.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 3 Foldable.• Then complete the Chapter 3 Study Guide and Review onpages 191–194 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 3 Practice Test onpage 195.Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 3.ARE YOU READY FORTHE CHAPTER TEST?
  • 93. Glencoe Geometry 85Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter4C H A P T E R4 Congruent TrianglesUse the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: Before each lesson, skimthrough the lesson and write any questions thatcome to mind in your notes. As you work throughthe lesson, record the answer to your question.Stack the grid paperon the constructionpaper. Fold diagonallyto form a triangle andcut off the excess.Staple the edge toform a booklet. Writethe chapter title onthe front and labeleach page with alesson number and title.Begin with two sheets of grid paper and onesheet of construction paper.
  • 94. 86 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R4This is an alphabetical list of new vocabulary terms you will learn in Chapter 4.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.Vocabulary TermFoundDefinitionDescription oron Page Exampleacute trianglebase anglescongruencetransformation[kuhn-GROO-uhns]congruent trianglescoordinate proofcorollaryequiangular triangleequilateral triangleexterior angle
  • 95. Glencoe Geometry 87Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY4Vocabulary TermFoundDefinitionDescription oron Page Exampleflow proofincluded angleincluded sideisosceles triangleobtuse triangleremote interior anglesright trianglescalene triangle[SKAY-leen]vertex angle
  • 96. 88 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–1Classify Triangles by AnglesARCHITECTURE The triangular truss below is modeledfor steel construction. Classify △JMN, △JKO, and△OLN as acute, equiangular, obtuse, or right.△JMN has one angle with measuregreater than , so it is antriangle.△JKO has one angle with measureequal to , so it is a triangle.△OLN is an triangle with all anglescongruent, so it is an triangle.Check Your Progress The frame of this window designis made up of many triangles. Classify △ABC, △ACD, and△ADE as acute, equiangular, obtuse, or right.Classify Triangles by SidesIdentify the indicated triangles inthe figure if−−UV−−VX−−UX.a. isosceles trianglesIsosceles triangles have at leasttwo sides congruent. Triangle andare .• Identify and classifytriangles by angles.• Identify and classifytriangles by sides.MAIN IDEASClassifying Triangles byAngles In an acutetriangle, all of the anglesare acute. In an obtusetriangle, one angle isobtuse. In a righttriangle, one angleis right.Classifying Triangles bySides No two sides of ascalene triangle arecongruent. At least twosides of an isoscelestriangle are congruent.All of the sides of anequilateral triangleare congruent.KEY CONCEPTSClassifying Triangles
  • 97. Glencoe Geometry 89Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–1b. scalene trianglesScalene triangles have no sides.The scalene triangles are.Check Your Progress Identifythe indicated triangles inthe figure.a. isosceles trianglesb. scalene trianglesFind Missing ValuesALGEBRA Find d and the measure of each side ofequilateral triangle KLM if KL = d + 2, LM = 12 - d, andKM = 4d - 13.Since △KLM is equilateral, eachside has the same length.So = .= Substitution2 = 3d - 13 Subtract d from each side.15 = 3d .= Divide each side by 3.Next, substitute to find the length of each side.KL = d + 2 LM = 12 - d KM = 4d - 13= + 2 = 2 - = 4( )- 13= = =For △KLM, d = and the measure of each side is .On the page forLesson 4-1, draw the6 different types oftriangles and describeeach one.ORGANIZE ITExplain why anequilateral triangle is aspecial kind of isoscelestriangle.WRITE IT
  • 98. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.90 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT4–1Check Your Progress Find x and the measure of each sideof equilateral triangle ABC if AB = 6x - 8, BC = 7 + x, andAC = 13 - x.Use the Distance FormulaCOORDINATE GEOMETRY Find the measures of thesides of △RST. Classify the triangle by sides.Use the Distance Formula to find the length of each side.RS == √25 + 49=ST == √(16 + 25)=RT == √81 + 4=RS = , ST = , RT = . Since all3 sides have different lengths, △RST is .Check Your Progress Find themeasures of the sides of △ABC.Classify the triangle by its sides.REVIEW ITWrite the DistanceFormula. (Lesson 1-3)
  • 99. Glencoe Geometry 91Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–2 Angles of TrianglesInterior AnglesFind the missing angle measures.Find m∠1 first because themeasure of two angles of the triangleare known.m∠1 + 43 + 74 = 180 Angle Sum Theoremm∠1 + = 180 Simplify.m∠1 = Subtract 117 from each side.∠1 and ∠2 are congruent angles.So, m∠2 = .m∠3 + + = 180 Angle Sum Theoremm∠3 + 142 = 180 Simplify.m∠3 = Subtract 142 fromeach side.So, m∠1 = , m∠2 = , and m∠3 = .Check Your Progress Find the missing angle measures.• Apply the Angle SumTheorem.• Apply the ExteriorAngle Theorem.MAIN IDEASTheorem 4.1 Angle Sum TheoremThe sum of the measures of the angles of a triangle is 180.
  • 100. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.92 Glencoe Geometry4–2An exterior angle of a triangle is formed by one side ofa and the of another side.The angles of the triangle notto a given exterior angle are called remote interior anglesof the exterior angle.Exterior AnglesFind the measure of eachnumbered angle in the figure.m∠1 = + Exterior Angle Theorem= Simplify.m∠1 + m∠2 = 180 Linear pairs aresupplementary.+ m∠2 = 180 Substitutionm∠2 = Subtract 70 fromeach side.m∠2 = m∠3 + 64 Exterior Angle Theorem= m∠3 + 64 Substitution= m∠3 Subtract 64 fromeach side.REMEMBER ITTwo angles aresupplementary if thesum of the angles is 180.(pages 86–87)On the page forLesson 4-2, draw anacute triangle andidentify an exteriorangle and its tworemote interior angles.ORGANIZE ITTheorem 4.2 Third Angle TheoremIf two angles of one triangle are congruent to two anglesof a second triangle, then the third angles of the trianglesare congruent.Theorem 4.3 Exterior Angle TheoremThe measure of an exterior angle of a triangle is equal tothe sum of the measures of the two remote interior angles.BUILD YOUR VOCABULARY
  • 101. Glencoe Geometry 93Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–2A that can be easily proved using ais often called a corollary.Corollary 4.1The acute angles of a right triangle are complementary.Corollary 4.1There can be at most one right or obtuse angle in atriangle.m∠4 + (m∠3 + 32) = 180 Linear pairs aresupplementary.m∠4 + ( + 32) = 180 Substitutionm∠4 = Simplify and subtract 78from each side.m∠5 + m∠4 + 41 = 180 Angle Sum Theoremm∠5 + 102 + 41 = 180 Substitutionm∠5 = Simplify and subtract143 from each side.So, m∠1 = , m∠2 = , m∠3 = ,m∠4 = , and m∠5 = .Check Your Progress Find the measure of each numberedangle in the figure.(page 86)BUILD YOUR VOCABULARY
  • 102. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.94 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT4–2Right AnglesGARDENING The flower bed shown is in the shape of aright triangle. Find m∠A if m∠C is 20.+ = Corollary 4.1m∠A + = Substitutionm∠A = Subtract fromeach side.Check Your Progress The piece of quilt fabric is in theshape of a right triangle. Find m∠A if m∠C is 62.
  • 103. Glencoe Geometry 95Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–3 Congruent TrianglesTriangles that are the same and arecongruent triangles.If you slide, flip, or turn a triangle, the size anddo not change. These three transformations are calledcongruence transformations.Corresponding Congruent PartsARCHITECTURE A drawing of a tower’s roof iscomposed of congruent triangles all converging ata point at the top.a. Name the corresponding congruent angles and sidesof △HIJ and △LIK.Since corresponding parts of congruent triangles arecongruent, ∠HJI ∠LKI, ∠ILK ∠IHJ,∠HIJ ,−−HI ,−−−HJand−−JI−−KI.b. Name the congruent triangles.Name the triangles in the order of their correspondingcongruent parts. So, △HIJ .• Name and labelcorresponding parts ofcongruent triangles.• Identify congruencetransformations.MAIN IDEASDefinition of CongruentTriangles (CPCTC) Twotriangles are congruentif and only if theircorresponding partsare congruent.Write thisdefinition in your notes.Be sure to include adiagram.KEY CONCEPTBUILD YOUR VOCABULARY (page 86)
  • 104. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.96 Glencoe Geometry4–3Check Your ProgressThe support beams onthe fence formcongruent triangles.a. Name the corresponding congruent angles and sides of△ABC and △DEF.b. Name the congruent triangles.REMEMBER ITWhen namingcongruent triangles,always list the name inthe order of congruentparts.Theorem 4.4Congruence of triangles is reflexive, symmetric, andtransitive.Transformations in the Coordinate Plana. Verify that △RST △R′S′T′.Use the Distance Formula to find thelength of each side of the triangles.RS == √9 + 25 orR′S′ == √9 + 25 orST = √(0 - 1)2 + (5 - 1)2= √1 + 16 orS′T′ = √(0 - (-1))2 + (5 - (-1))2= √1 + 16 or
  • 105. Glencoe Geometry 97Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT4–3TR = √(1 - (-3))2 + (1 - 0)2= √16 + 1 or √17T′R′ = √(-1 - 3)2 + (-1 - 0)2= √16 + 1 or √17The lengths of the corresponding sides of two trianglesare equal. Therefore, by the definition of congruence,−−RS ,−−ST−−TR .In conclusion, because−−RS R−−−R′S′,−−ST S−−−S′T′, T−−TR−−−T′R′,∠R , ∠S , and ∠T′,△RST .b. Name the congruence transformation for △RST and△R′S′T′.△R′S′T′ is a of .Check Your Progress The vertices of △ABC areA(-5, 5), B(0, 3), and C(-4, 1). The vertices of △A′B′C′are A′(5, -5), B′(0, -3), and C′(4, -1).a. Verify that △ABC △A′B′C′.b. Name the congruence transformation for △ABC and△A′B′C′.
  • 106. 98 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–4 Proving Congruence – SSS, SAS• Use the SSS Postulate totest for trianglecongruence.• Use the SAS Postulateto test for trianglecongruence.MAIN IDEASPostulate 4.1 Side-Side-Side Congruence (SSS)If the sides of one triangle are congruent to the sidesof a second triangle, then the triangles are congruent.Statements Reasons1. 1.2. 2.3. 3.Statements Reasons1.−−FE−−HI; G is the midpoint of−−EI; 1.G is the midpoint of−−−FH.2. 2. Midpoint Theorem3. 3.Use SSS in ProofsENTOMOLOGY The wings of one type of moth formtwo triangles. Write a two-column proof to prove that△FEG △HIG if−−EI−−FH,−−FE−−HI, and G is themidpoint of both−−EI and−−FH.Given:−−EI−−−FH;−−FE−−HI;G is the midpoint ofboth−−EI and−−−FH.Prove: △FEG △HIGProof:Check Your Progress Write atwo-column proof to prove that△ABC △GBC if−−−GB−−ABand−−AC−−−GC.Proof:
  • 107. Glencoe Geometry 99Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–4SSS on the Coordinate PlaneCOORDINATE GEOMETRY Determine whether△WDV △MLP for D(-5, -1), V(-1, -2), W(-7, -4),L(1, -5), P(2, -1), and M(4, -7). Explain.Use the Distance Formula toshow that the correspondingsides are congruent.WD = √(-7 - (-5))2 + (-4 - (-1))2= or √13ML = √(4 - 1)2 + (-7 - (-5))2= or √13DV == √16 + 1 or √17LP = √(1 - 2)2 + (-5 - (-1))2= or √17VW = √(-1 - (-7))2 + (-2 - (-4))2= √36 + 4 orPM = √(2 - 4)2 + (-1 - (-7))2= orWD = , DV = , and VW = .By definition of congruent segments, all correspondingsegments are congruent. Therefore, △WDVby .
  • 108. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.100 Glencoe Geometry4–4Check Your ProgressDetermine whether△ABC △DEF for A(-5, 5),B(0, 3), C(-4, 1), D(6, -3),E(1, -1), and F(5, 1). Explain.Use SAS in ProofsWrite a flow proof.Given:−−−RQ−−TS−−−RQ−−TSProve: △QRT △STRProof:Postulate 4.2 Side-Angle-Side Congruence (SAS)If two sides and the included angle of one triangle arecongruent to two sides and the included angle of anothertriangle, then the triangles are congruent.In a triangle, the formed by isthe included angle for those two sides.BUILD YOUR VOCABULARY (page 87)
  • 109. Glencoe Geometry 101Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT4–4Identify Congruent TrianglesDetermine which postulate can be used to prove thatthe triangles are congruent. If it is not possible to provethat they are congruent, write not possible.a. Two sides of each triangle are. The included anglesare congruent. Therefore, the trianglesare congruent by .b. Each pair of sides arecongruent. Two are given and the third iscongruent by the Property.So the triangles are congruent by .Check Your Progress Determine which postulate canbe used to prove that the triangles are congruent. If itis not possible to prove that they are congruent, writenot possible.a. b.Check Your Progress Write aflow proof.Given: C is the midpoint of−−−DB;∠ACB ∠ACD.Prove: △ABC △ADCProof:On the page forLesson 4-4, copy twosample flow proofs toshow how to use SSSand SAS to provetriangles congruent.ORGANIZE IT
  • 110. 102 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–5 Proving Congruence – ASA, AASThe side of a triangle that is contained in each of twoangles is called the included side for the two angles.Use ASA in ProofsWrite a paragraph proof.Given: L is the midppoint of−−−WE; W−−−WR−−−ED.Prove: △WRL △EDLProof:because anglesare congruent. By the Midpoint Theorem, .Since vertical angles are congruent, .△WRL by ASA.Check Your Progress Writea paragraph proof.Given:−−−DE−−−CB; A is themidpoint of−−−EC.Prove: △DEA △CBAProof:• Use the ASA Postulateto test for trianglecongruence.• Use the AAS Theoremto test for trianglecongruence.MAIN IDEASPostulate 4.3 Angle-Side-Angle Congruence (ASA)If two angles and the included side of one triangle arecongruent to two angles and the included side of anothertriangle, the triangles are congruent.BUILD YOUR VOCABULARY (page 87)
  • 111. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Glencoe Geometry 1034–5Use AAS in ProofsWrite a flow proof.Given: ∠NKL ∠NJM−−KL−−−JMProve:−−−LN−−−MNProof:Check Your ProgressWrite a flow proof.Given: ∠ADB ∠ACE−−−EC−−−BDProve: ∠AEC ∠ABDProof:REMEMBER ITWhen trianglesoverlap, it may behelpful to draw eachtriangle separately.Theorem 4.5 Angle-Angle-Side Congruence (AAS)If two angles and a nonincluded side of one triangle arecongruent to the corresponding two angles and side of asecond triangle, then the two triangles are congruent.
  • 112. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.104 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT4–5Determine if Triangles are CongruentSTANCES When Ms. Gomez puts her handson her hips, she forms two triangles withher upper body and arms. Suppose herarm lengths AB and DE measure 9 inches,and AC and EF measure 11 inches. Alsosuppose that you are given that−−BC−−DF.Determine whether △ABC △EDF.Justify your answer.Since = DE = ,−−AB . Likewise,AC = EF = ,−−AC . We are given−−−BC−−−DF.Check each possibility using the five methods you know.We are given information about . Since allthree pairs of corresponding sides of the triangles arecongruent, △ABC △EDF by .Check Your Progress Thecurtain decorating the windowforms 2 triangles at the top. B isthe midpoint of−−AC. AE = 13 inchesand CD = 13 inches. BE and BDeach use the same amount ofmaterial, 17 inches. Determinewhether △ABE △CBD. Justifyyour answer.On the page forLesson 4-5, draw apicture to show howASA is different fromAAS.ORGANIZE IT
  • 113. Glencoe Geometry 105Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–6 Isosceles TrianglesState the four ways toprove triangles arecongruent.WRITE IT• Use properties ofisosceles triangles.• Use properties ofequilateral triangles.MAIN IDEASFind the Measure of a Missing AngleTEST EXAMPLE If−−DE−−CD,−−BC−−AC, and m∠CDE = 120,what is the measure of ∠BAC?A 45.5 C 68.5B 57.5 D 75Read the Test Item△CDE is isosceles with base−−−CE. Likewise, △CBA is isosceleswith base−−BA. The base angles of △CDE are congruent.Solve the Test ItemLet x = m∠DEC = m∠DCE.m∠DEC + m∠DCE + m∠CDE = 180 Angle SumTheorem+ + = 180 Substitution+ = 180 Add.Theorem 4.9 Isosceles Triangle TheoremIf two sides of a triangle are congruent, then the anglesopposite those sides are congruent.In an triangle, the angle formed by thecongruent sides is called the vertex angle.The two angles formed by the base and one of thesides are called base angles.BUILD YOUR VOCABULARY (pages 86–87)
  • 114. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.106 Glencoe Geometry= Subtract 120 fromeach side.x = Divide each sideby 2.So, m∠DEC = m∠DCE = .∠DCE and ∠BCA are vertical angles, so they haveequal measures.m∠DCE = m∠BCA Definition of vertical angles= m∠BCA SubstitutionThe base angles of △CBA are congruent.Let y = m∠CBA = m∠BAC.m∠CBA + m∠BAC + m∠BCA = 180 Angle SumTheoremy + y + 30 = 180 Substitution+ 30 = 180 Add.2y =Subtract 30 fromeach side.y =Divide each sideby 2.The measure of ∠BAC is . The answer is D.Check Your ProgressIf−−AB−−−BC,−−AC−−−CD, andm∠ABC = 80, what is themeasure of ∠ADC?Theorem 4.10If two angles of a triangle are congruent, then the sidesopposite those angles are congruent.Corollary 4.3A triangle is equilateral if and only if it is equiangular.Corollary 4.4Each angle of an equilateral triangle measures 60°.4–6
  • 115. Glencoe Geometry 107Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–6Use Properties of Equilateral Triangles△EFG is equilateral, and−−−EH bisects∠E.−−EJ bisects ∠HEG.a. Find m∠HEJ and m∠EJH.Each angle of an equilateral trianglemeasures .So, m∠FEH + = . Since the anglewas , m∠FEH = m∠HEG = .EJ bisects , so m∠HEJ = m∠JEG = .m∠EJH = m∠GEJ + m∠EGJ Exterior Angle Theorem= + Substitution= Add.So, m∠HEJ = , m∠EJH = .b. Find m∠EJG.∠EJH and ∠EJG form a .∠EJH and ∠EJG = Def. of linear pairs+ m∠EJG = 180 Substitution.m∠EJG = Subtract.Check Your Progress △ABC is an equilateral triangle.−−AD bisects ∠BAC.a. Find x.b. Find m∠ADB.Page(s):Exercises:HOMEWORKASSIGNMENTOn the page forLesson 4-6, list theproperties and theoremsyou learned about eachtype of triangle.ORGANIZE IT
  • 116. 108 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.4–7Coordinate proof uses figures in theand to provegeometric concepts.Triangles and Coordinate ProofPlacing Figures on theCoordinate Plane1. Use the origin as avertex or center of thefigure.2. Place at least one sideof a polygon on anaxis.3. Keep the figurewithin the firstquadrant if possible.4. Use coordinates thatmake computations assimple as possible.KEY CONCEPT• Position and labeltriangles for use incoordinate proofs.• Write coordinateproofs.MAIN IDEASPosition and Label a TrianglePosition and label right triangle XYZ with leg−−XZd units long on the coordinate plane.Use the as vertex Xof the triangle. Place the base ofthe triangle along the positive.Position the triangle in the quadrant.Since Z is on the x-axis, its y-coordinate is .Its x-coordinate is because the base is d units long.Since triangle XYZ is a right triangle the x-coordinate of Y is. We cannot determine the y-coordinate so call it .Check Your Progress Position and label equilateraltriangle ABC with side−−BC w units long on the coordinate plane.BUILD YOUR VOCABULARY (page 86)
  • 117. Glencoe Geometry 109Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Find the Missing CoordinatesName the missing coordinates ofisosceles right triangle QRS.Q is on the origin, so its coordinatesare . The x-coordinate ofS is .The y-coordinate for S is the distance from R to S. Since △QRSis an isosceles right triangle, .The distance from Q to R is units. The distance from Rto S must be the same. So, the coordinates of S are .Check Your Progress Namethe missing coordinates of isoscelesright △ABC.Coordinate ProofWrite a coordinate proof to prove that the segment thatjoins the vertex angle of an isosceles triangle to themidpoint of its base is perpendicular to the base.Given: △XYZ is isosceles.−−−XW−−−WZProve:−−−YW ⊥−−XZProof:By the Formula, the coordinates of W, themidpoint of−−XZ, are (0 + 2a_2,0 + 0_2 ) or .The slope of−−−YW is (0 - b_a - a) or .The slope of−−XZ is ( 0 - 0_0 - 2a) or , therefore, .REVIEW ITThe vertex angle of anisosceles triangle is theangle formed by thecongruent sides.(Lesson 4-6)4–7
  • 118. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.110 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT4–7Check Your Progress Write acoordinate proof to prove that thesegment drawn from the right angleto the midpoint of the hypotenuseof an isosceles right triangle isperpendicular to the hypotenuse.Proof:Classify TrianglesDRAFTING Write a coordinate proof toprove that the outside of this drafter’stool is shaped like a right triangle. Thelength of one side is 10 inches and thelength of another side is 5.75 inches.Proof:The slope of−−−ED is (10 - 0_0 - 0 ) or . The slope of−−DFis ( 0 - 0_0 - 5.75) or , therefore . △DEF is atriangle. The drafter’s tool is shaped like a righttriangle.Check Your Progress Writea coordinate proof to prove thisflag is shaped like an isoscelestriangle. The length is 16 inchesand the height is 10 inches.Proof:Suppose a triangleis in the coordinateplane. Explain how youcan classify the triangleby its sides if you knowthe coordinates of allthree vertices. Includethe explanation on thepage for Lesson 4-7.ORGANIZE IT
  • 119. Glencoe Geometry 111Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.STUDY GUIDEBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R4Use your Chapter 4 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 4, go to:glencoe.comYou can use your completedVocabulary Builder(pages 86–87) to help you solvethe puzzle.Find x and the measure of each side of the triangle.1. △ABC is equilateral with AB = 3x - 15, BC = 2x - 4,and CA = x + 7.2. △DEF is isosceles, ∠D is the vertex angle, DE = x + 5,DF = 5x - 7 and EF = 2x - 1.3. Find the measures of the sides of △RST and classify the triangleby its sides. Triangle RST has vertices R(2, -2), S(0, 1), and T(2, 4).Find the measure of each angle.4. m∠15. m∠34-1Classifying TrianglesBRINGING IT ALL TOGETHER4-2Angles of Triangles
  • 120. 112 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER4Find the measure of each angle withoutusing a protractor.6. ∠DBC 7. ∠ABC8. ∠ACF 9. ∠EABComplete each congruence statement if △TSR △WVU.10. ∠R 11. W 12. ∠S13.−−RT 14.−−−VU 15.−−−WV16. In quadrilateral RSTU,−−RS−−TS and−−−SUbisects ∠RST. Name the postulate thatcould be used to prove △RSU △TSU.17. In quadrilateral PQRS,−−−QR−−SP and−−−PQ−−RS.Name the postulate that could be used to prove△PQR △RSP.Determine whether you have enough information to prove that thetwo triangles in each figure are congruent. If so, write a congruencestatement and name the congruence postulate or theorem that youwould use. If not, write not possible.18. 19. T is the midpoint of−−−RU.4-5Proving Congruence – ASA, AAS4-3Congruent Triangles4-4Proving Congruence – SSS, SAS
  • 121. Glencoe Geometry 113Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER4Refer to the figure.20. What kind of triangle is △QRS?21. Name the legs of △QRS.22. Name the base of △QRS.23. Name the vertex angle of △QRS.24. Name the base angles of △QRS.△ABF is equilateral, and−−AE ⊥−−CF.Find each measure.25. m∠CFD 26. m∠BFC27. m∠ABF 28. m∠AFind the missing coordinates of each triangle.29. 30.4-6Isosceles Triangle4-7Triangles and Coordinate Proof
  • 122. 114 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureC H A P T E R4ARE YOU READY FORTHE CHAPTER TEST?Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 4.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 4 Practice Test onpage 261 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 4 Study Guide and Reviewon pages 256–260 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 4 Practice Test onpage 261 of your textbook.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 4 Foldable.• Then complete the Chapter 4 Study Guide and Review onpages 256–260 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 4 Practice Test onpage 261 of your textbook.ChecklistCheck the one that applies. Suggestions to help you study aregiven with each item.
  • 123. Glencoe Geometry 115Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter5C H A P T E R5 Relationships in TrianglesUse the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: When you take notes, writeconcise definitions in your own words. Addexamples that illustrate the concepts.Fold lengthwise tothe holes.Cut five tabs.Label the edge.Then label the tabsusing lesson numbers.Begin with one sheet of notebook paper.
  • 124. 116 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 5.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.Vocabulary TermFoundDefinitionDescription oron Page Examplealtitudecentroidcircumcenter[SUHR-kuhm-SEN-tuhr]concurrent linesincenterindirect proofC H A P T E R5
  • 125. Glencoe Geometry 117Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY5Vocabulary TermFoundDefinitionDescription oron Page Exampleindirect reasoningmedianorthocenter[OHR-thoh-CEN-tuhr]perpendicular bisectorpoint of concurrencyproof by contradiction
  • 126. 118 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–1 Bisectors, Medians, and AltitudesA perpendicular bisector of a side of a triangle is a line,segment, or ray that passes through the midpoint of theside and is to that side.When three or more lines intersect at a commonpoint, the lines are called concurrent lines, and theirpoint of is called the point ofconcurrency. The point of concurrency of thebisectors of a triangle iscalled the circumcenter.Use Angle BisectorsGiven: m∠F = 80 and m∠E = 30−−−DG bisects ∠EDFProve: m∠DGE = 115Proof:Statements Reasons1. m∠F = 80, m∠E = 30, 1. Givenand−−−DG bisects ∠EDF.2. m∠EDF + m∠E + 2.m∠F = 180• Identify and useperpendicular bisectorsand angle bisectors intriangles.• Identify and usemedians and altitudesin triangles.MAIN IDEASBUILD YOUR VOCABULARY (pages 116–117)Theorem 5.1Any point on the perpendicular bisector of a segment isequidistant from the endpoints of the segment.Theorem 5.2Any point equidistant from the endpoints of a segment lieson the perpendicular bisector of the segment.Theorem 5.3 Circumcenter Theorem The circumcenter ofa triangle is equidistant from the vertices of the triangle.State the Angle SumTheorem. (Lesson 4-2)WRITE IT
  • 127. Glencoe Geometry 119Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.3. 3. Substitution4. m∠EDF = 180 - 110 = 70 4. Subtraction Property5. m∠GDE = 35 5. Definition of angle bisector6. m∠GDE + m∠E + 6. Angle Sum Theoremm∠DGE = 1807. 35 + 30 + m∠DGE = 180 7. Substitution8. 8. Subtraction PropertyCheck Your ProgressGiven: m∠B = 66 and m∠C = 50−−−AD bisects ∠BAC.Prove: m∠ADC = 98Proof:Statements Reasons1. m∠B = 66, m∠C = 50, 1. Givenand−−−AD bisects ∠BAC.2. 2.3. 3.4. 4.5. 5.6. 6.7. 7.8. 8.5–1
  • 128. 120 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.The angle bisectors of a triangle are concurrent, and theirpoint of concurrency is called the incenter of a triangle.A median is a segment whose endpoints are a vertex of atriangle and the of the side opposite the. The point of concurrency for the mediansof a triangle is called a centroid.5–1Segment MeasuresALGEBRA Points U, V, and W aremidpoints of−−YZ,−−ZX, and−−XY,respectively. Find a, b, and c.Find a.VY = 2a + 7.4 Segment Addition Postulate= 2_3VY Centroid Theorem7.4 = 2_3(2a + 7.4) Substitution= 4a + Multiply each side by 3and simplify.= a Subtract 14.8 from eachside and divide by 4.Theorem 5.4Any point on the angle bisector is equidistant from thesides of the angle.Theorem 5.5Any point equidistant from the sides of an angle lies on theangle bisector.Theorem 5.6 Incenter TheoremThe incenter of a triangle is equidistant from each side ofthe triangle.Theorem 5.7 Centroid TheoremThe centroid of a triangle is located two-thirds of thedistance from a vertex to the midpoint of the side oppositethe vertex on a median.Under the tab forLesson 5-1, drawseparate pictures thatshow the centroid,circumcenter, incenter,and orthocenter. Write adescription for each.ORGANIZE ITBUILD YOUR VOCABULARY (pages 116–117)
  • 129. Glencoe Geometry 121Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTFind b.XU = 3b + 2 + 8.7 Segment Addition Postulate3b + 2 = 2_3XU }23b + 2 = 2_3(3b + 2 + 8.7) Substitution= Multiply each side by 3and simplify.b ≈ Solve for b.Find c.WZ = + Segment AdditionPostulate15.2 = 2_3WZ Centroid Theorem15.2 = 2_3(5c + 15.2) Substitution= c Simplify and solve for c.Check Your Progress Points T, H, and G are themidpoints of−−−NM,−−−MK, and−−−NK, respectively. Find w, x, and y.An altitude of a triangle is a segment from ato the line containing the opposite sideand to the line containing that side.The intersection point of the of a triangle iscalled the orthocenter.BUILD YOUR VOCABULARY (pages 116–117)5–1
  • 130. 122 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–2 Inequalities and TrianglesCompare Angle MeasuresDetermine which angle hasthe greatest measure.Compare m∠3 to m∠1.m∠1 = Exterior Angle Theoremm∠1 m∠3 Definition of InequalityCompare m∠4 to m∠1.m∠1 = Exterior Angle Theoremm∠1 m∠4 Definition of Inequality∠4 ∠5m∠4 = m∠5 Definition of ∠sm∠1 > m∠5Compare to .By the Exterior Angle Theorem, m∠5 = m∠2 + m∠3.By the definition of inequality, m∠5 m∠2.Since we know that m∠1 > m∠5, by the Transitive Property,m∠1 m∠2. Therefore, has the greatestmeasure.Check Your ProgressDetermine which anglehas the greatest measure.• Recognize and applyproperties ofinequalities to themeasures of anglesof a triangle.• Recognize and applyproperties ofinequalities to therelationships betweenangles and sides of atriangle.MAIN IDEAS
  • 131. Glencoe Geometry 123Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Exterior AnglesUse the Exterior AngleInequality Theorem tolist all of the anglesthat satisfy the statedcondition.a. measures less than m∠14By the Exterior Angle Inequality Theorem, m∠14 > m∠4,m∠14 > m∠11, m∠14 > m∠2, and m∠14 >+ .Since ∠11 and ∠9 are angles, they haveequal measures, so m∠14 > m∠9. > m∠9 > m∠6 andm∠9 > m∠7, so m∠14 > m∠6 and m∠14 > m∠7.Thus, the measures ofare all less than m∠14.b. measures greater than m∠5By the Exterior Angle Inequality Theorem, m∠5 < m∠10,m∠5 < m∠16, m∠5 < m∠12, m∠5 < m∠15, and m∠5 < m∠17.Thus, the measures ofare each greater than m∠5.Check Your Progress Use the Exterior Angle InequalityTheorem to list all angles whose measures are greater than m∠8.Theorem 5.8 Exterior Angle Inequality TheoremIf an angle is an exterior angle of a triangle, then itsmeasure is greater than the measure of either of itscorresponding remote interior angles.Under the tab forLesson 5-2, summarizethe proof of Theorem5.9 using your ownwords in paragraphform.ORGANIZE IT5–2
  • 132. 124 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT5–2REMEMBER ITThe longer andshorter sides must beopposite the largerand smaller anglesrespectively, notadjacent to them.Side-Angle RelationshipsDetermine the relationship between the measures ofthe given angles.a. ∠RSU and ∠SURThe side opposite ∠RSU is than the sideopposite ∠SUR, so m∠RSU > .b. ∠TSV and ∠STVThe side opposite ∠TSV is shorter than the side opposite, so m∠TSV m∠STV.Check Your Progress Determine the relationshipbetween the measures of the given angles.a. ∠ABD, ∠DABb. ∠AED, ∠EADTheorem 5.10If one angle of a triangle has a greater measure thananother angle, then the side opposite the greater angle islonger than the side opposite the lesser angle.Theorem 5.9If one side of a triangle is longer than another side, thenthe angle opposite the longer side has a greater measurethan the angle opposite the shorter side.
  • 133. Glencoe Geometry 125Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–3When using indirect reasoning, you assume that theis false and then show that thisassumption leads to a contradiction of the, or some other accepted fact, such as adefinition, postulate, theorem, or corollary. A proof of thistype is called an indirect proof or proof by contradiction.Stating ConclusionsState the assumption you would make to start anindirect proof of each statement.a.−−EF is not a perpendicular bisector.b. 3x = 4y + 1c. If B is the midpoint of−−LH and−−LH = 26, then−−−BH iscongruent to−−LB.Check Your Progress State the assumption you wouldmake to start an indirect proof of each statement.a.−−AB is not an altitude. b. a = 1_2b - 4Algebraic ProofWrite an indirect proof.Given: 1_2y + 4= 20Prove: y ≠ -2• Use indirect proofwith algebra.• Use indirect proofwith geometry.MAIN IDEASBUILD YOUR VOCABULARY (pages 116–117)Steps for Indirect Proof:(1) Assume that theconclusion is false.(2) Show that thisassumption leads toa contradiction ofthe hypothesis or ofsome other fact.(3) Point out that theoriginal conclusionmust be true, since thisis the only way to avoidthe contradiction.KEY CONCEPTIndirect Proof
  • 134. 126 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–3Indirect Proof:STEP 1 Assume that .STEP 2 Substitute -2 for y in the equation 1_2y + 4= 20.1_2y + 4= 20= 20 Substitution1_-4 + 4= 20 Multiply.1_0= 20 Add.This is a contradiction because thecannot be 0.STEP 3 The assumption leads to a contradiction. Therefore,the assumption that y = -2 must be ,which means that y ≠ -2 must be .Check Your Progress Write an indirect proof.Given: 1_2a + 6≤ 12Prove: a ≠ -3Indirect Proof:STEP 1STEP 2Under the tab forLesson 5-3, list thesteps for writing anindirect proof. Then,give an example ofan indirect proof.ORGANIZE IT
  • 135. Glencoe Geometry 127Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT5–3STEP 3Geometry ProofWrite an indirect proof.Given: △JKL with side lengths 5, 7,and 8 as shown.Prove: m∠K < m∠LIndirect Proof:STEP 1 Assume that .STEP 2 By angle-side relationships, JL ≥ JK.By substitution, .This inequality is a false statement.STEP 3 This contradicts the given side lengths, sothe assumption must be false. Therefore,.Check Your Progress Write an indirectproof.Given: △ABC with side lengths 8,10, and 12 as shown.Prove: m∠C > m∠AIndirect Proof:STEP 1STEP 2STEP 35–3
  • 136. 128 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–4Identify Sides of a TriangleDetermine whether the given measures can be lengthsof the sides of a triangle.a. 61_2, 61_2, and 141_2Because the of two measures is not greaterthan the length of the side, the sides cannotform a triangle.61_2+ 61_2141_2≯b. 6.8, 7.2, and 5.1Check each inequality.6.8 + 7.2 5.1 6.8 + 5.1 7.2 5.1 + 7.2 6.8> 5.1 > 7.2 > 6.8All of the inequalities are , so 6.8, 7.2, and 5.1 canbe the lengths of the sides of a triangle.Check Your Progress Determine whether the givenmeasures can be lengths of the sides of a triangle.a. 6, 9, 16b. 14, 16, 27• Apply the TriangleInequality Theorem.• Determine the shortestdistance between apoint and a line.MAIN IDEASTheorem 5.11 Triangle Inequality TheoremThe sum of the lengths of any two sides of a triangle isgreater than the length of the third side.The Triangle Inequality
  • 137. Glencoe Geometry 129Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT5–4Determine Possible Side LengthTEST EXAMPLE In △PQR, PQ = 7.2 and QR = 5.2. Whichmeasure cannot be PR?A 7 B 9 C 11 D 13Read the Test ItemLet PR = n. Solve each inequality to determine the range ofvalues for PR.Solve the Test ItemPQ + QR > PR7.2 + > n> n or n <PQ + PR > QR PR + QR > PQ7.2 + > 5.2 n + > 7.2n > n >The range of values that fits all three inequalities is< n < . The answer is .Check Your Progress In △XYZ, XY = 6 and YZ = 9. Findthe range of possible values for XZ.Theorem 5.12The perpendicular segment from a point to a line is theshortest segment from the point to the line.Corollary 5.1The perpendicular segment from a point to a plane is theshortest segment from the point to the plane.Under the tab forLesson 5-4, write theTriangle InequalityTheorem in your ownwords. Draw and labela triangle to showwhy the theoremmakes sense.ORGANIZE ITCorollary 5.1The perpendicular segment from a point to a plane is theshortest segment from the point to the plane.
  • 138. 130 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–5Use SAS Inequality in a ProofWrite a two-column proof.Given:−−KL−−−JH, JK = HLm∠JKH + m∠HKL < m∠JHK + m∠KHLProve: JH < KLProof:Statements Reasons1. m∠JKH + m∠HKL < 1. Givenm∠JHK + m∠KHL2. m∠HKL = m∠JHK 2.3. m∠JKH + m∠JHK < 3. Substitutionm∠JHK + m∠KHL4. m∠JKH < m∠KHL 4. Subtraction Property5. 5. Given6. HK = HK 6. Reflexive Property7. 7.• Apply the SASInequality.• Apply the SSSInequality.MAIN IDEASTheorem 5.13 SAS Inequality/Hinge TheoremIf two sides of a triangle are congruent to two sides ofanother triangle and the included angle in one triangle hasa greater measure than the included angle in the other,then the third side of the first triangle is longer than thethird side of the second triangle.Theorem 5.14 SSS InequalityIf two sides of a triangle are congruent to two sides ofanother triangle and the third side in one triangle is longerthan the third side in the other, then the angle betweenthe pair of congruent sides in the first triangle is greaterthan the corresponding angle in the second triangle.Explain why the SASInequality Theorem isalso called the HingeTheorem.WRITE ITInequalities Involving Two Triangles
  • 139. Glencoe Geometry 131Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.5–5Prove Triangle RelationshipsGiven: ST = PQSR = QRST = 2_3SPProve: m∠SRP > m∠PRQProof:Statements Reasons1. SR = QR 1. Given2. PR = PR 2.3. ST = PQ 3. Given4. ST = 2_3SP; SP > ST 4. Given5. SP > PQ 5. Substitution6. 6.Check Your ProgressGiven: X is the midpoint of−−−MB.△MCX is isosceles.CB > CMProve: m∠CXB > m∠CMXProof:Statements Reasons1. X is the midpoint of−−−MB. 1. Given2. 2.3. 3.4. 4.5. 5.6. 6.7. 7.
  • 140. 132 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTRelationships Between Two TrianglesWrite an inequality using theinformation in the figure.a. Compare m∠LDM andm∠MDN.In △MDL and △MDN,−−−LD−−−ND,−−−MD−−−MD,and ML > MN.The allows us to conclude that> .b. Find the range of values containing a.By the SSS Inequality, m∠LDM > m∠MDN orm∠MDN < m∠LDM.m∠MDN < m∠LDM9a + 15 < 141 Substitution< Subtract 15 from each side anddivide each side by 9.Also, recall that the measure of any angle is always greaterthan 0.9a + 15 > 09a > -15 Subtract 15 from each side.> or -5_3Divide each side by 9.The two inequalities can be written as the compoundinequality .Check Your Progress Write aninequality using the informationin the figure.a. Compare m∠WYX and m∠ZYW.b. Find the range of values containing n.Under the tab forLesson 5-5, draw twosets of two triangleseach with two pairsof congruent sides.Measure and markthe sides and angles toillustrate the SASInequality Theoremand the SSS InequalityTheorem.ORGANIZE IT5–5
  • 141. Glencoe Geometry 133Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.STUDY GUIDEBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R5 BRINGING IT ALL TOGETHERUse your Chapter 5 Foldableto help you study for yourchapter test.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 5, go to:glencoe.comYou can use your completedVocabulary Builder(pages 116–117) to help yousolve the puzzle.Fill in the correct word or phrase to complete each sentence.1. A(n) of a triangle is a segment drawn from avertex of the triangle perpendicular to the line containingthe opposite side.2. The point of concurrency of the three perpendicular bisectorsof a triangle is called the .3. Any point in the interior of an angle that is equidistant fromthe sides of that angle lies on the .4. The vertices of △PQR are P(0, 0), Q(2, 6), and R(6, 4). Find thecoordinates of the orthocenter of △PQR.5. If−−XZ is the perpendicular bisectorof−−−WY and−−−WY is the perpendicularbisector of−−XZ, find x and y.5-1Bisectors, Medians, and Altitudes
  • 142. 134 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER5Determine which angle has the greatest measure.6. ∠1, ∠3, ∠47. ∠5, ∠6, ∠8Determine the relationship between the measures of the given angles.8. m∠ABD, m∠BDA9. m∠DBC, m∠BCDWrite the assumption you would make to start an indirect proof ofeach statement.10. If 3x - 8 = 10, then x = 6.11. If m∠1 = 100 and m∠2 = 80, then ∠1 and ∠2 aresupplementary.Write the assumption you would make to start an indirect proof.12. Given: 5x + 4 < 14Prove: x < 213. Given: △EFG is a right triangle.∠G is a right angle.Prove: m∠E < 905-2Inequalities and Triangles5-3Indirect Proof
  • 143. Glencoe Geometry 135Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER5Determine whether the given measures can be the lengths of atriangle. Write yes or no.14. 5, 6, 7 15. 6, 8, 1016. 10, 10, 21 17. 12, 12, 12Refer to the figure. Determine whether each statement is true or false.18. DE > EF + FD19. EG = EF + FG20. The shortest distance from Dto EG is DF.21. The shortest distance from D to EG is DG.Write an inequality relating the given pair of segment measures.22. 23.MR RP AD CD5-4The Triangle Inequality5-5Inequalities Involving Two Triangles
  • 144. 136 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureC H A P T E R5ChecklistCheck the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 5 Practice Test onpage 313 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 5 Study Guide and Reviewon pages 310–312 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 5 Practice Test onpage 313.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 5 Foldable.• Then complete the Chapter 5 Study Guide and Review onpages 310–312 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 5 Practice Test onpage 313.ARE YOU READY FORTHE CHAPTER TEST?Visit glencoe.com toaccess your textbook,more examples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 5.
  • 145. Chapter6Glencoe Geometry 137Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R6Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: When taking notes, alwayswrite clear and concise notes so they can be easilyread when studying for a quiz or exam.Fold lengthwise tothe left margin.Cut four tabs.Label the tabs usingthe lesson concepts.Begin with a sheet of notebook paper.฀Quadrilaterals
  • 146. 138 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R6Vocabulary TermFoundDefinitionDescription oron Page Examplediagonalisosceles trapezoidkitemedianparallelogramrectanglerhombussquaretrapezoidThis is an alphabetical list of new vocabulary terms you will learn in Chapter 6.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.
  • 147. Glencoe Geometry 139Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–1The diagonals of a polygon are segments that connect anytwo nonconsecutive .• Find the sum of themeasures of the interiorangles of a polygon.• Find the sum of themeasures of theexterior angles of apolygon.MAIN IDEASAngles of PolygonsInterior Angles of Regular PolygonsARCHITECTURE A mall is designed sothat five walkways meet at a food courtthat is in the shape of a regularpentagon. Find the sum of measures ofthe interior angles of the pentagon.A pentagon is a convex polygon. Use the Angle Sum Theorem.S = 180(n - 2) Interior Angle Sum Theorem= 180( - 2) n == 180( )or Simplify.The sum of the measures of the angles is .Check Your Progress A decorativewindow is designed to have the shapeof a regular octagon. Find the sum ofthe measures of the interior anglesof the octagon.BUILD YOUR VOCABULARY (page 138)Theorem 6.1 Interior Angle Sum TheoremIf a convex polygon has n sides and S is the sum of themeasures of its interior angles, then S = 180(n - 2).
  • 148. 140 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–1Sides of a PolygonThe measure of an interior angle of a regular polygonis 135. Find the number of sides in the polygon.Use the Interior Angle Sum Theorem.S = 180(n - 2) Interior Angle Sum Theorem(135)n = 180(n - 2) S = 135n= 180n - Distributive Property0 = 45n - 360 Subtract.= Add.= n The polygon has sides.Interior AnglesFind the measure of each interior angle.Since n= 4, the sum of the measures ofthe interior angles is 180(4 - 2) or 360.Write an equation to express the sum ofthe measures of the interior angles of the polygon.360 = m∠R + m∠S + m∠T + m∠U360 = + + +360 ===Use the value of x to find the measure of each angle.m∠R = orm∠S = orm∠T = orm∠U = orCan you use the methodin Example 2 if thepolygon is not regular?Explain why or why not.WRITE ITSketch a quadrilateralwith two pairs ofparallel sides. Writethe sum of the anglemeasures below thefigure. Include thesketch under theparallelograms tab ofyour Foldable.ORGANIZE IT
  • 149. Glencoe Geometry 141Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–1Page(s):Exercises:HOMEWORKASSIGNMENTCheck Your Progressa. The measure of an interior angle of a regular polygonis 144. Find the number of sides in the polygon.b. Refer to the figure shown. Find themeasure of each interior angle.Exterior AnglesFind the measures of an exterior angle and an interiorangle of convex regular nonagon ABCDEFGHJ.At each vertex, extend a side to formone exterior angle. The sum of themeasures of the exterior angles is 360.A convex regular nonagon has 9 congruentexterior angles.9n = 360n = Divide each side by 9.Since each exterior angle and its corresponding interior angleform a linear pair, the measure of the interior angle is180 - or .Check Your Progress Findthe measures of an exterior angleand an interior angle of convexregular hexagon ABCDEF.Theorem 6.2 Exterior Angle Sum TheoremIf a polygon is convex, then the sum of the measures of theexterior angles, one at each vertex, is 360.REVIEW ITWhat is an exteriorangle? (Lesson 4-2)
  • 150. 142 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–2A quadrilateral with opposite sides is calleda parallelogram.ParallelogramsProperties of ParallelogramsQuadrilateral RSTU is aparallelogram. Find m∠URT,m∠RST, and y.First, find m∠URT.∠URT ∠STR You know that if parallel lines arecut by a transversal, alternate interior∠s are congruent.m∠URT = m∠STR Congruent anglesm∠URT = SubstitutionNow, find m∠RST.m∠STU = + or Angle AdditionPostulatem∠RST + m∠STU = Consecutive ∠s inare supplementary.m∠RST + 58 = Substitutionm∠RST = Subtract.−−RS Opposite sidesof are .BUILD YOUR VOCABULARY (page 138)Theorem 6.3Opposite sides of a parallelogram are congruent.Theorem 6.4Opposite angles in a parallelogram are congruent.Theorem 6.5Consecutive angles in a parallelogram are supplementary.Theorem 6.6If a parallelogram has one right angle, it has four rightangles.• Recognize and applyproperties of the sidesand angles ofparallelograms.• Recognize and applyproperties of thediagonals ofparallelograms.MAIN IDEASWrite the properties ofparallelograms underthe parallelograms tab.Parallelogram Aparallelogram is aquadrilateral with bothpairs of opposite sidesparallel.KEY CONCEPT
  • 151. Glencoe Geometry 143Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT6–2RS = TU Congruent segments3y = , so y = Substitution; Divide.m∠URT = , m∠RST = , y =Check Your ProgressABCD is a parallelogram. Findm∠BDC, m∠BCD, and x.Diagonals of a ParallelogramTEST EXAMPLE What are the coordinates of theintersection of the diagonals of parallelogram MNPR,with vertices M(-3, 0), N(-1, 3), P(5, 4), and R(3, 1)?A (2, 4) B (9_2, 5_2) C (1, 2) D (-2, 3_2)Read the Test Item Since the diagonals of a parallelogrambisect each other, the intersection point is the midpointof−−−MP and−−−NR.Solve the Test Item Find the midpoint of−−−MP.(x1 + x2_2,y1 + y2_2 )= (_______________2, _____________2)=The answer is .Check Your Progress What are the coordinates of theintersection of the diagonals of parallelogram LMNO, withvertices L(0, -3), M(-2, 1), N(1, 5), O(3, 1)?Theorem 6.7The diagonals of a parallelogram bisect each other.Theorem 6.8The diagonal of a parallelogram separates theparallelogram into two congruent triangles.Write what it means fordiagonals to bisect eachother.WRITE IT
  • 152. 144 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–3 Tests for Parallelograms• Recognize theconditions that ensurea quadrilateral is aparallelogram.• Prove that a set ofpoints forms aparallelogram in thecoordinate plane.MAIN IDEASProperties of ParallelogramsSome of the shapes in this Bavariancrest appear to be parallelograms.Describe the information needed todetermine whether the shapes areparallelograms.If both pairs of opposite sides are the same length or if onepair of opposite sides is , thequadrilateral is a parallelogram. If both pairs of oppositeangles are or if the bisecteach other, the quadrilateral is a parallelogram.Check Your Progress The shapes in the vest pictured hereappear to be parallelograms. Describe the information neededto determine whether the shapes are parallelograms.Theorem 6.9 If both pairs of opposite sides of aquadrilateral are congruent, then the quadrilateral is aparallelogram.Theorem 6.10 If both pairs of opposite angles of aquadrilateral are congruent, then the quadrilateral is aparallelogram.Theorem 6.11 If the diagonals of a quadrilateral bisecteach other, then the quadrilateral is a parallelogram.Theorem 6.12 If one pair of opposite sides of aquadrilateral is both parallel and congruent, then thequadrilateral is a parallelogram.
  • 153. Glencoe Geometry 145Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–3Properties of ParallelogramsDetermine whether the quadrilateral isa parallelogram. Justify your answer.Each pair of opposite sides have the samemeasure. Therefore, they are .If both pairs of opposite sides of a quadrilateralare , the quadrilateral is aparallelogram.Check Your Progress Determine whether thequadrilateral is a parallelogram. Justify your answer.Find MeasuresFind x so that the quadrilateral isa parallelogram.Opposite sides of a parallelogramare congruent.−−AB−−−DC Opposite sides of a are .AB = DC Definition of segments.= Substitution= Distributive Property= Subtract 3x from each side.x = Add 1 to each side.When x is , ABCD is a parallelogram.Record the tests forparallelograms underthe parallelograms tab.ORGANIZE ITWhat does it mean fortwo segments to becongruent? (Lesson 1-2)REVIEW IT
  • 154. 146 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT6–3Check Your Progress Find m so thatthe quadrilateral is a parallelogram.Use Slope and DistanceCOORDINATE GEOMETRYDetermine whether ABCD withvertices A(-3, 0), B(-1, 3),C(3, 2), and D(1, -1) is aparallelogram. Use the SlopeFormula.If the opposite sides of a quadrilateralare parallel, then it is a parallelogram.slope of−−AB = 3 - 0_-1 - (-3)orslope of−−−CD = -1 - 2_1 - 3orslope of−−−AD = -1 - 0_1 - (-3)orslope of−−−BC = 3 - 2_-1 - 3orSince opposite sides have the same slope,−−AB−−−CD and−−−AD−−−BC.Therefore, ABCD is a parallelogram by definition.Check Your Progress Determine whether the figurewith the given vertices is a parallelogram. Use the methodindicated.A(-1, -2), B(-3, 1), C(1, 2), D(3, -1); Slope Formula
  • 155. Glencoe Geometry 147Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–4A rectangle is a quadrilateral with four .• Recognize and applyproperties of rectangles.• Determine whetherparallelograms arerectangles.MAIN IDEASRectanglesDiagonals of a RectangleQuadrilateral RSTU is a rectangle.If RT = 6x + 4 and SU = 7x - 4, find x.The diagonals of a rectangle are congruent,so−−RT−−−SU.−−RT Diagonals of a rectangleare .RT = Definition of congruentsegments= Substitution= Subtract 6x from each side.= x Add 4 to each side.Check Your Progress Quadrilateral EFGH is a rectangle.If FH = 5x + 4 and GE = 7x - 6, find x.BUILD YOUR VOCABULARYTheorem 6.13 If a parallelogram is a rectangle, then thediagonals are congruent.(page 138)
  • 156. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.148 Glencoe Geometry6–4Angles of a RectangleQuadrilateral LMNP is a rectangle.Find x.∠MLP is a right angle, som∠MLP = 90.m∠MLN + m∠NLP = m∠MLP Angle AdditionPostulate5x + 8 + 3x + 2 = Substitution.= Simplify.= Subtract.x = Divide each side by 8.Check Your Progress Quadrilateral EFGH is arectangle.a. Find x.b. Find y.Properties of aRectangle1. Opposite sides arecongruent andparallel.2. Opposite angles arecongruent.3. Consecutive angles aresupplementary.4. Diagonals arecongruent and bisecteach other.5. All four angles areright angles.KEY CONCEPTDiagonals of a ParallelogramKyle is building a barn for his horse.He measures the diagonals of the dooropening to make sure that they bisect eachother and they are congruent. How does heknow that the measure of each corner is 90?We know that−−AC−−−BD. A parallelogram withdiagonals is a rectangle. Therefore, the corners areangles.Theorem 6.14If the diagonals of a parallelogram are congruent, then theparallelogram is a rectangle.
  • 157. Glencoe Geometry 149Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–4Check Your Progress Max is building a swimming poolin his backyard. He measures the length and width of thepool so that opposite sides are parallel. He also measures thediagonals of the pool to make sure that they are congruent.How does he know that the measure of each corner is 90?Rectangle on a Coordinate PlaneQuadrilateral ABCD has vertices A(-2, 1), B(4, 3),C(5, 0), and D(-1, -2). Determine whether ABCD is arectangle.Method 1Use the Slope Formula, m =y2 - y1_x2 - x1, to see ifconsecutive sides are perpendicular.slope of−−AB = 3 - 1_4 - (-2)orslope of−−−CD = -2 - 0_-1 - 5orslope of−−−BC = 0 - 3_5 - 4orslope of−−−AD =1 - (-2)_-2 - (-1)orRecord similarities anddifferences betweenrectangles and othertypes of parallelogramsunder the rectanglestab.ORGANIZE IT
  • 158. 150 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–4Because−−AB−−−CD and−−−BC−−−AD, quadrilateral ABCD is a. The product of the slopes of consecutivesides is . This means that−−AB⊥−−−BC,−−AB⊥−−−AD,−−−AD⊥−−−CD,and−−−BC⊥−−−CD.The perpendicular segments create four right angles.Therefore, by definition ABCD is a .Method 2Use the Distance Formula, d = √(x2 - x1)2 + (y2 - y1)2,to determine whether opposite sides are congruent.AB DC= √[4 - (-2)]2 + (3 - 1)2 = √[5 - (-1)]2 + [0 - (-2)]2= == =AD BC= √[-1 - (-2)]2 + (-2 - 1)2 = √(5 - 4)2 + (0 - 3)2= == =Since each pair of opposite sides of the quadrilateral have thesame measure, they are congruent. Quadrilateral ABCD is aparallelogram.What do you knowabout the slopes of twoperpendicular lines?(Lesson 3-3)REVIEW IT
  • 159. Glencoe Geometry 151Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–4Page(s):Exercises:HOMEWORKASSIGNMENTFind the length of the diagonals.AC = √[-5 - (-2)]2 + (0 - 1)2==DB = √[4 - (-1)]2 + [3 - (-2)]2==The length of each diagonal is .Since the diagonals are congruent, ABCD is a rectangle.Check Your Progress Quadrilateral WXYZ has verticesW(-2, 1), X(-1, 3), Y(3, 1), and Z(2, -1). Determine whetherWXYZ is a rectangle using the Distance Formula.
  • 160. 152 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–5A rhombus is a quadrilateral with all four sides congruent.Rhombi and Squares• Recognize and applythe properties ofrhombi.• Recognize and applythe properties ofsquares.MAIN IDEASMeasures of a RhombusUse rhombus LMNP and the giveninformation to find the value ofeach variable.a. Find y if m∠1 = y2 - 54.m∠1 = The diagonals of arhombus are perpendicular.= Substitutiony2 = 144 Add 54 to each side.y = Take the square root ofeach side.The value of y can be .b. Find m∠PNL if m∠MLP = 64.m∠PNM = Opposite angles arecongruent.m∠PNM = SubstitutionThe diagonals of a rhombus bisect the angles.So, m∠PNL = or .BUILD YOUR VOCABULARY (page 138)Theorem 6.15 The diagonals of a rhombus areperpendicular.Theorem 6.16 If the diagonals of a parallelogram areperpendicular, then the parallelogram is a rhombus.Theorem 6.17 Each diagonal of a rhombus bisects a pairof opposite angles.
  • 161. Glencoe Geometry 153Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.If a quadrilateral is both a and a rectangle,then it is a square.6–5Check Your Progress Use rhombus ABCD and thegiven information to find the value of each variable.a. Find x if m∠1 = 2x2 - 38.b. Find m∠CDB if m∠ABC = 126.SquaresDetermine whether parallelogram ABCD is a rhombus,a rectangle, or a square for A(-2, -1), B(-1, 3), C(3, 2),and D(2, -2). List all that apply. Explain.Use the Distance Formula to compare the lengths of thediagonals.DB = √(-1 - 2)2 + [3 - (-2)]2==AC = √[3 - (-2)]2 + [2 - (-1)]2==BUILD YOUR VOCABULARY (page 138)REMEMBER ITA square is arhombus, but a rhombusis not necessarily asquare.
  • 162. 154 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT6–5Use slope to determine whether the diagonals areperpendicular.slope of−−−DB =3 - (-2)_-1 - 2orslope of−−AC =2 - (-1)_3 - (-2)orSince the slope of−−AC is the negative of theslope of−−−DB the diagonals are . The lengthsof−−−DB and−−AC are the same so the diagonals are congruent.ABCD is a .Check Your Progress Determine whether parallelogramEFGH is a rhombus, a rectangle, or a square for E(0, -2),F(-3, 0), G(-1, 3), and H(2, 1). List all that apply. Explain.Record the conceptsabout squares andrhombi, includingtheir similarities anddifferences, under thesquares and rhombi tab.ORGANIZE IT
  • 163. Glencoe Geometry 155Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–6A trapezoid is a quadrilateral with exactly one pair ofsides.If the legs are , then the trapezoid is anisosceles trapezoid.• Recognize and applythe properties oftrapezoids.• Solve problemsinvolving the mediansof trapezoids.MAIN IDEASTrapezoidsIdentify Isosceles TrapezoidsThe top of this work station appears to be two adjacenttrapezoids. Determine if they are isosceles trapezoids.Each pair of base angles is , so the legs arethe same length. Both trapezoids are .Check Your Progress The sides of a picture frame appearto be two adjacent trapezoids. Determine if they are isoscelestrapezoids.BUILD YOUR VOCABULARY (page 138)Theorem 6.18 Both pairs of base angles of an isoscelestrapezoid are congruent.Theorem 6.19 The diagonals of an isosceles trapezoid arecongruent.
  • 164. 156 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–6Identify TrapezoidsABCD is a quadrilateral with vertices A(5, 1), B(-3, -1),C(-2, 3), and D(2, 4).a. Verify that ABCD is a trapezoid.A quadrilateral is a trapezoid ifexactly one pair of opposite sidesare parallel. Use the Slope Formula.slope of−−AB = -1 - 1_-3 - 5slope of−−−CD = 4 - 3_2 - (-2)= =slope of−−−DA = 1 - 4_5 - 2slope of−−−BC =3 - (-1)_-2 - (-3)= =Exactly one pair of opposite sides are parallel,and . So, ABCD is a trapezoid.b. Determine whether ABCD is an isosceles trapezoid.Explain.First use the Distance Formula to determine whether thelegs are congruent.DA = √(2 - 5)2 + (4 - 1)2==BC = √[-2 - (-3)]2 + [3 - (-1)]2==Since the legs are not , ABCD is not anisosceles trapezoid.฀฀฀฀Draw an isoscelestrapezoid under thetrapezoids tab. Includelabels on the drawing toshow congruent angles,congruent diagonals,and parallel sides.ORGANIZE IT
  • 165. Glencoe Geometry 157Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.The segment that joins of theof a trapezoid is the median.6–6Check Your Progress QRST is a quadrilateral withvertices Q(-3, -2), R(-2, 2), S(1, 4), and T(6, 4).a. Verify that QRST is a trapezoid.b. Determine whether QRST is an isosceles trapezoid. Explain.Median of a TrapezoidDEFG is an isosceles trapezoid withmedian−−−MN.a. Find DG if EF = 20 and MN = 30.MN = 1_2(EF + DG) Theorem 6.20= 1_2( + DG) Substitution= + DG Multiply each side by 2.= Subtract 20 from each side.BUILD YOUR VOCABULARY (page 138)Theorem 6.20The median of a trapezoid is parallel to the bases, and itsmeasure is one-half the sum of the measures of the bases.
  • 166. 158 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT6–6b. Find m∠1, m∠2, m∠3, and m∠4 if m∠1 = 3x + 5 andm∠3 = 6x - 5.Since−−EF−−−DG, ∠1 and ∠3 are supplementary. Because thisis an isosceles trapezoid, ∠1 ∠2 and ∠3 ∠4.m∠1 + m∠3 = Consecutive InteriorAngles Theorem+ = Substitution= Combine like terms.x = Divide each side by 9.If x = , then m∠1 = and m∠3 = .Because ∠1 ∠2 and ∠3 ∠4, m∠2 = andm∠4 = .Check Your Progress WXYZ is an isosceles trapezoidwith median−−JK.a. Find XY if JK = 18 and WZ = 25.b. Find m∠1, m∠2, m∠3, and m∠4 ifm∠2 = 2x - 25 and m∠4 = 3x + 35.
  • 167. Glencoe Geometry 159Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–7• Position and labelquadrilaterals for usein coordinate proofs.• Prove theorems usingcoordinate proofs.MAIN IDEASCoordinate Proof With QuadrilateralsPositioning a RectanglePosition and label a rectangle with sides a and b unitslong on the coordinate plane.• Let A, B, C, and D be vertices of a rectangle with sides−−ABand−−−CD a units long, and sides−−−BC and−−−AD b units long.• Place the rectangle with vertex A at the ,−−AB alongthe positive , and−−−AD along the .Label the vertices A, B, C, and D.• The y-coordinate of B is because the vertex is on thex-axis. Since the side length is a, the x-coordinate is .• D is on the y-axis so the x-coordinate is . Since the sidelength is b, the y-coordinate is .• The x-coordinate of C is also . The y-coordinate is0 + b or b because the side−−−BC is b units long.Check Your Progress Position and label a parallelogramwith sides a and c units long on the coordinate plane.
  • 168. 160 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.6–7Coordinate ProofPlace a rhombus on the coordinate plane. Label themidpoints of the sides M, N, P, and Q. Write acoordinate proof to prove that MNPQ is a rectangle.The first step is to position arhombus on the coordinateplane so that the origin isthe midpoint of the diagonalsand the diagonals are on theaxes, as shown. Label thevertices to make computationsas simple as possible.Given: ABCD is a rhombus as labeled. M, N, P, Q aremidpoints.Prove: MNPQ is a rectangle.Proof: By the Midpoint Formula, the coordinates of M, N, P,and Q are as follows.M(-2a + 0_2, 0 - 2b_2 ) =N(0 - 2a_2,2b + 0_2 ) =P(2a + 0_2,0 + 2b_2 ) =Q(0 + 2a_2,-2b + 0_2 ) =Find the slopes of−−−QP,−−−MN,−−−QM, and−−−PN.slope of−−−QP =-__a - aorslope of−−−MN =b - (-b)__-orslope of−−−QM =- (-b)__-a - aorslope of−−−PN =b -_-a - aorMake sketches toshow how each typeof quadrilateral in thischapter can be placedin the coordinateplane to have thesimplest coordinatesfor the vertices. Labelthe vertices with theircoordinates. Includeeach sketch under theappropriate tab.ORGANIZE IT
  • 169. Glencoe Geometry 161Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT6–7When proving theoremsabout quadrilaterals,why is it convenient toplace a figure on thecoordinate plane withone side parallel to anaxis and one vertex at(0, 0)?WRITE ITA segment with slope 0 is perpendicular to a segment withslope. Therefore, consecutive sides of thisquadrilateral are . Since consecutive sidesare perpendicular, MNPQ is, by definition, a .Check Your Progress Write a coordinate proof.Given: ABCD is an isoscelestrapezoid. M, N, P, andQ are midpoints.Prove: MNPQ is a rhombus.Proof:
  • 170. STUDY GUIDE162 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R6Use your Chapter 6 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 6, go to:glencoe.comYou can use your completedVocabulary Builder (page 138)to help you solve the puzzle.Give the measure of an interior angle and the measure of anexterior angle of each polygon.1. equilateral triangle2. regular hexagon3. Find the sum of the measures of the interior angles of a convex 20-gon.For Exercises 4–7, let ABCD be a parallelogram with AB ≠ BCand with no right angles.4. Sketch a parallelogram that matches thedescription above and draw diagonal−−−BD.Complete each sentence.5.−−AB and−−AD .6.−−AB and−−−BC .7. ∠A and ∠ABC .6-1Angles of Polygons6-2Parallelograms
  • 171. Chapter BRINGING IT ALL TOGETHER6Glencoe Geometry 163Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8. Which of the following conditions guarantee that aquadrilateral is a parallelogram?a. Two sides are parallel.b. Both pairs of opposite sides are congruent.c. A pair of opposite sides is both parallel and congruent.d. There are two right angles.e. All four sides are congruent.f. Both pairs of opposite angles are congruent.g. The diagonals bisect each other.h. All four angles are right angles.Determine whether there is enough given information toknow that each figure is a parallelogram. If so, state thedefinition or theorem that justifies your conclusion.9. 10.11. Find m∠1, m∠2, and m∠3 in the rectangle shown.ABCD is a rectangle with AD > AB. Name each of thefollowing in this figure.12. all segments that are congruent to−−−BE13. all angles congruent to ∠114. two pairs of congruent triangles6-3Tests for Parallelograms6-4Rectangles
  • 172. Chapter BRINGING IT ALL TOGETHER6164 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Sketch each of the following.15. a quadrilateral with 16. a quadrilateral withperpendicular diagonals congruent diagonalsthat is not a rhombus that is not a rectangleList all of the special quadrilaterals that have each listedproperty: parallelogram, rectangle, rhombus, square.17. Opposite sides are congruent.18. The diagonals are perpendicular.19. The quadrilateral is equilateral.20. The quadrilateral is equiangular.21. The diagonals are perpendicular and congruent.Complete each sentence.22. A quadrilateral with only one pair of opposite sides paralleland the other pair of opposite sides congruent is a(n).23. The segment joining the midpoints of the nonparallel sides ofa trapezoid is called the .24. A quadrilateral with only one pair of opposite sides parallelis a(n) .6-5Rhombi and Squares6-6Trapezoids
  • 173. Chapter BRINGING IT ALL TOGETHER6Glencoe Geometry 165Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.EFGH is a trapezoid, I is the midpoint of−−FE, and J is themidpoint of−−−GH. Identify each of the following segmentsor angles.25. the bases of trapezoid EFGH26. the legs of trapezoid EFGH27. the median of trapezoid EFGH28. Find the missing coordinates in the figure. Then write thecoordinates of the four vertices of the quadrilateral.Refer to quadrilateral EFGH.29. Find the slope of each side.30. Find the length of each side.31. Find the slope of each diagonal.32. Find the length of each diagonal.33. What can you conclude about the sides of EFGH?34. What can you conclude about the diagonals of EFGH?6-7Coordinate Proof with Quadrilaterals
  • 174. 166 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureC H A P T E R6ChecklistARE YOU READY FORTHE CHAPTER TEST?Visit glencoe.com toaccess your textbook,more examples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 6.Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 6 Practice Test onpage 373 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 6 Study Guide and Reviewon pages 369–372 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 6 Practice Test onpage 373.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 6 Foldable.• Then complete the Chapter 6 Study Guide and Review onpages 369–372 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 6 Practice Test onpage 373.
  • 175. Glencoe Geometry 167Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter7C H A P T E R7 Proportions and SimilarityUse the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: A visual study guide like theFoldable shown above helps you organize what youknow and remember what you have learned. Youcan use them to review main ideas or keywords.Fold widthwise.Leave space to punchholes so it can beplaced in your binder.Open the flap anddraw lines to dividethe inside into sixequal parts.Label each partusing the lessonnumbers.Write the nameof the chapteron the front flap.Begin with one sheet of 11" × 17" paper.
  • 176. 168 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 7.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.C H A P T E R7Vocabulary TermFoundDefinitionDescription oron Page Examplecross productsextremesmeans
  • 177. Glencoe Geometry 169Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY7Vocabulary TermFoundDefinitionDescription oron Page Examplemidsegmentproportionratioscale factorsimilar polygons
  • 178. 170 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7–1 ProportionsA ratio is a comparison of two quantities. The ratio ofa to b can be expressed , where b is not zero.An equation stating that two ratios are is calleda proportion.Write a RatioThe total number of students who participate in sportsprograms at Central High School is 520. The totalnumber of students in the school is 1850. Find theathlete-to-student ratio to the nearest tenth.To find this ratio, divide the number of athletes by the totalnumber of students.number of athletes___total number of students= or aboutThe athlete-to-student ratio is athletes for eachstudent in the school.Check Your Progress The country with the longest schoolyear is China with 251 days. Find the ratio of school days tototal days in a year for China to the nearest tenth. (Use 365 asthe number of days in a year.)Solve Proportions by Using Cross ProductsSolve each proportion.a. 6_18.2= 9_y.6_18.2= 9_yOriginal proportion= Cross productsy = Divide each side by 6.• Write ratios.• Use properties ofproportions.MAIN IDEASBUILD YOUR VOCABULARY (pages 168–169)Property of ProportionsFor any numbers a andc and any nonzeronumbers b and d, a_b= c_dif and only if ad = bc.KEY CONCEPT
  • 179. Glencoe Geometry 171Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT7–1As you skim the lesson,write down questionsthat you have on thesection for Lesson 7-1.Then write the answersnext to each question.ORGANIZE ITb. 4x - 5_3= -26_64x - 5_3= -26_6Original proportion6 = 3 Cross multiply.24x - 30 = -78 Multiply.24x = -48 Add 30 to each side.x = Divide.Check Your Progress Solve each proportion.a. 13.5_42= b_14b. 7n - 1_8= 15.5_2Solve Problems Using ProportionsTRAINS A boxcar on a train has a length of 40 feet andawidth of 9 feet. A scale model is made with a length of16 inches. Find the width of the model.Write and solve a proportion.boxcar’s length (ft)__model’s length (in.)=boxcar’s width (ft)__model’s width (in.)= Substitution40x = 16(9) Cross products40x = 144 Multiply.x = Divide each side by 40.The width of the model is inches.Check Your Progress Two large cylindrical containers arein proportion. The height of the larger container is 25 meterswith a diameter of 8 meters. The height of the smaller containeris 7 meters. Find the diameter of the smaller container.
  • 180. 172 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7–2 Similar PolygonsSimilar PolygonsDetermine whether the pair of figures is similar.Justify your answer.Since m∠B = m∠S, .The m∠C = 40 and m∠R = 60.So, ∠C ∠T and .Thus, all the corresponding angles are congruent.Now determine whether corresponding sides are proportional.AC_RT= 8_6or 1.−3 AB_RS= or 1.−3 BC_ST= or 1.−3The ratio of the measures of the corresponding sides areequal and the corresponding angles are ,so △ABC ∼ △RST.Check Your Progress Determine whetherthe pair of figures is similar. Justifyyour answer.When polygons have the same shape but may be differentin , they are called similar polygons.When you compare the lengths ofsides of similar figures, you usually get a numerical ratio.This ratio is called the scale factor for the two figures.BUILD YOUR VOCABULARY (page 169)• Identify similar figures.• Solve problemsinvolving scale factors.MAIN IDEASSimilar polygons Twopolygons are similar ifand only if theircorresponding anglesare congruent and themeasures of theircorresponding sidesare proportional.KEY CONCEPT
  • 181. Glencoe Geometry 173Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7–2Explain why twocongruent polygonsmust also be similar.WRITE ITScale FactorARCHITECTURE An architect prepared a 12-inch modelof a skyscraper to look like an actual 1100-foot building.What is the scale factor of the model compared to theactual building?Before finding the scale factor you must make sure that bothmeasurements use the same unit of measure.1100(12) = 13,200 inchesheight of model___height of actual building==The ratio comparing the two heights is or: . The scale factor is , whichmeans that the model is the height of theactual skyscraper.Check Your Progress A space shuttle is about 122 feet inlength. The Science Club plans to make a model of the spaceshuttle with a length of 24 inches. What is the scale factor ofthe model compared to the real space shuttle?Proportional Parts and Scale FactorThe two polygons are similar.a. Write a similarity statement.Then find x, y, and UV.Use the congruent anglesto write the correspondingvertices in order.polygon ∼ polygon
  • 182. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.174 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT7–2Write a description ofthe information youwould include in adiagram of twopolygons to enable afriend to decide that thepolygons are similar.Record your descriptionon the section forLesson 7-2.ORGANIZE ITNow write propotions to find x and y.To find x:AB_RS= CD_TUSimilarity proportion4=3AB = 6, RS = 4, CD = x, TU = 318 = 4x Cross products= x Divide each side by 4.To find y:AB_RS= DE_UVSimilarity proportion6=8AB = 6, RS = 4, DE = 8,UV = y + 16y + 6 = 32 Cross products6y = 26 Subtract 6 from each side.y = Divide each side by 6 and simplify.UV = y + 1, so UV = + 1 or .b. Find the scale factor of polygon ABCDE topolygon RSTUV.The scale factor is the ratio of the lengths of any twocorresponding sides.AB_RS=4or 3_2Check Your Progress The twopolygons are similar.a. Write a similarity statement. Then finda, b, and ZO.b. Find the scale factor of polygon TRAPto polygon ZOLD.
  • 183. Glencoe Geometry 175Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7–3 Similar Triangles• Identify similartriangles.• Use similar triangles tosolve problems.MAIN IDEASDetermine Whether Triangles are SimilarIn the figure,−−AB−−DC, BE = 27, DE = 45, AE = 21,and CE = 35. Determine which triangles in thefigure are similar.Since−−−AB−−−DC, ∠BAC by the Alternate InteriorAngles Theorem.Vertical angles are congruent, so .Therefore, by the AA Similarity Theorem, .Check Your Progress In the figure, OW = 7, BW = 9,WT = 17.5, and WI = 22.5. Determine which triangles in thefigure are similar.Postulate 7.1 Angle-Angle (AA) SimilarityIf the two angles of one triangle are congruent to twoangles of another triangle, then the triangles are similar.Theorem 7.1 Side-Side-Side (SSS) SimilarityIf the measures of the corresponding sides of two trianglesare proportional, then the triangles are similar.Theorem 7.2 Side-Angle-Side (SAS) SimilarityIf the measures of two sides of a triangle are proportionalto the measures of two corresponding sides of anothertriangle and the included angles are congruent, then thetriangles are similar.฀
  • 184. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.176 Glencoe Geometry7–3Parts of Similar TrianglesALGEBRA Given−−RS−−UT, RS = 4, RQ = x + 3,QT = 2x + 10, UT = 10, find RQ and QT.Since−−RS U−−−UT, ∠SRQ and TUQbecause they are alternate interior angles. By AA Similarity,. Using the definition of similarpolygons, = .4_10=x + 3_2x + 10Substitution4(2x + 10) = 10(x + 3) Cross products8x + 40 = 10x + 30 Distributive Property= x Simplify.Now find RQ and QT.RQ = QT = 2x + 10= + or = 2( )+ 10 orCheck Your Progress Given−−−AB−−−DE, AB = 38.5, DE = 11,AC = 3x + 8, and CE = x + 2,find AC and CE.Theorem 7.3Similarity of triangles is reflexive, symmeteric, and transitive.Write a short paragraphto describe how youcould apply thepostulate and theoremsin this lesson to helpyou construct similartriangles. Include yourparagraph on thesection for Lesson 7-3.ORGANIZE IT
  • 185. Glencoe Geometry 177Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTFind a MeasurementINDIRECT MEASUREMENT Josh wantedto measure the height of the SearsTower in Chicago. He used a 12-footlight pole and measured its shadowat 1 P.M. The length of the shadow was2 feet. Then he measured the length ofthe Sears Tower’s shadow and it was242 feet at the time. What is theheight of the Sears Tower?Assuming that the sun’s rays form similar triangles, thefollowing proportion can be written.height ofSears Tower (ft)=light pole shadowlength (ft)Now substitute the known values and let x be the height of theSears Tower.= Substitutionx · 2 = 242(12) Cross productsx = Simplify and divide eachside by 2.The Sears Tower is feet tall.Check Your Progress On hertrip along the East coast, Jenniestops to look at the tallestlighthouse in the U.S. located atCape Hatteras, North Carolina.Jennie measures her shadow tobe 1 feet 6 inches in length andthe length of the shadow of thelighthouse to be 53 feet 6 inches.Jennie’s height is 5 feet 6 inches.What is the height of the CapeHatteras lighthouse to the nearest foot?REMEMBER ITShadows and similartriangles are commonlyused for indirectlymeasuring the heightsof objects that areotherwise too tall tomeasure.7–3
  • 186. 178 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7–4 Parallel Lines and Proportional PartsFind the Length of a SideIn △RST,−−RT V−−VU, SV = 3,VR = 8, and UT = 12.Find SU.From the Triangle Proportionality Theorem, SV_VR= .Substitute the known measures.=3(12) = 8x Cross products36 = 8x Multiply.36_8= x Divide each side by 8.= x Simplify.Check Your Progress In △ABC,−−AC−−XY, AX = 4,XB = 10.5, and CY = 6. Find BY.• Use proportional partsof triangles.• Divide a segment intoparts.MAIN IDEASTheorem 7.4 Triangle Proportionality TheoremIf a line is parallel to one side of a triangle and intersectsthe other two sides in two distinct points, then it separatesthese sides into segments of proportional lengths.Theorem 7.5 Converse of the Triangle Proportionality If aline intersects two sides of a triangle and separates thesides into corresponding segments of proportionallengths, then the line is parallel to the third side.
  • 187. Glencoe Geometry 179Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.A midsegment of a triangle is a segment whose endpointsare the of the two sides of the triangle.7–4Determine Parallel LinesIn △DEF, DH = 18, HE = 36, andDG = 1_2GF. Determine whether−−−GH−−FE. Explain.In order to show that−−−GH−−FE, we must show that DG_GF= DH_HE.DG = 1_2GF, so DG_GF= 1_2.Since, DG_GF= and DH_HE= or , the sideshave lengths. Since the segments haveproportional lengths, .Check Your Progress In △WXZ, XY = 15, YZ = 25,WA = 18, and AZ = 32. Determine whether−−−WX−−AY. Explain.Sketch a pentagon thatis not regular. Write abrief explanation ofhow you could use thetheorems in this lessonto construct a pentagonsimilar to the givenpentagon. Include thesketch and explanationon the section forLesson 7-4.ORGANIZE ITWrite the MidpointFormula.WRITE ITBUILD YOUR VOCABULARY (page 135)Theorem 7.6 Triangle Midsegment TheoremA midsegment of a triangle is parallel to one side of thetriangle, and its length is one-half of that side.
  • 188. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.180 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT7–4Proportional SegmentsMAPS In the figure, Larch, Maple, and Nuthatch Streetsare all parallel. The figure shows the distances inbetween city blocks. Find x.From Corollary 7.1, if three or more parallel lines intersecttwo transversals, then they cut off the transversals. Write a proportion.26_13= x_16Triangle Proportionality Theorem26(16) = 13x .= 13x Multiply.= x Divide.Check Your Progress In the figure, Davis, Broad, andMain Streets are all parallel. The figure shows the distances incity blocks that the streets are apart. Find x.A 4 B 5 C 6 D 7
  • 189. Glencoe Geometry 181Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.7–5 Parts of Similar Triangles• Recognize and useproportionalrelationships ofcorrespondingperimeters of similartriangles.• Recognize and useproportionalrelationships ofcorresponding anglebisectors, altitudes,and medians of similartriangles.MAIN IDEASTheorem 7.7 Proportional Perimeters TheoremIf two triangles are similar, then the perimeters areproportional to the measures of corresponding sides.Perimeters of Similar TrianglesIf △ABC ∼ △XYZ, AC = 32, AB = 16,BC = 16√5, and XY = 24, find theperimeter of △XYZ.Let x represent the perimeter of △XYZ.The perimeter of△ABC = 16 + 16√5 + 32 or + 16√5.XY_AB=perimeter ofperimeter ofProportionalPerimeterTheorem= Substitution24(48 + 16√5) = 16x Cross products1152 + 384√5 = 16x Multiply.+ 24√5 = x Divide.The perimeter of △XYZ is units.Check Your Progress If △PNO∼ △XQR, PN = 6, XQ = 20,QR = 20√2, and RX = 20, find the perimeter of △PNO.
  • 190. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.182 Glencoe Geometry7–5Theorem 7.8If two triangles are similar, then the measures of thecorresponding altitudes are proportional to the measuresof the corresponding sides.Theorem 7.9If two triangles are similar, then the measures of thecorresponding angle bisectors of the triangles areproportional to the measures of the corresponding sides.Theorem 7.10If two triangles are similar, then the measures of thecorresponding medians are proportional to the measuresof the corresponding sides.Write a ProofWrite a paragraph proof.Given: △JKL ∼ △QRS−−−MK is a median of △JKL.−−TR is a median of △QRS.Prove: △JKM ∼ △QRTProof: By , JK_QR= JL_QS.Since similar triangles have correspondingproportional to the corresponding sides, JL_QS= KM_RT. Bysubstitution, JK_QR= KM_RT. Since−−−MK and−−TR areof △JKL and △QRS, respectively, M and T are midpoints of−−JL and−−−QS. By , 2JM = JLand 2QT = QS. Substitute in the equation JK_QR= JL_QStoget JK_QR= 2JM_2QT. Simplify to find JK_QR= JM_QT. Therefore,△JKM ∼ △QRT by .Use a pair of similarisosceles triangles toillustrate all of thetheorems in this lesson.Include a sketch andexplanation on thesection for Lesson 7-5.ORGANIZE IT
  • 191. Glencoe Geometry 183Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTMedians of Similar TrianglesIn the figure, △EFD ∼ △JKI.−−EGis a median of △EFD, and−−JL isa median of △JKI. Find JL ifEF = 36, EG = 18, and JK = 56.EG_JL= EF_JKWrite a proportion.18_x= 36_56EG = 18, JL = x, EF = 36, and JK = 561008 = 36x Cross products= x Divide each side by 36.Thus, .Check Your Progressa. △EFG ∼ △MSY and EF = 5_4MS. Find the ratio of thelength of an altitude of △EFG to the length of an altitudeof △MSY.b. In the figure, △ABD ∼ △MNP.−−AC is a median of △ABDand−−−MO is a median of △MNP. Find x if AC = 5, AB = 7,and MO = 12.5.REVIEW ITAn altitude of a triangleis a segment from avertex to the linecontaining the oppositeside and perpendicularto the line containingthe side. (Lesson 5-1)7–5฀฀฀Theorem 7.11Angle Bisector TheoremAn angle bisector in a triangle separates the oppositeside into segments that have the same ratio as the othertwo sides.
  • 192. STUDY GUIDE184 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R7Use your Chapter 7 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 7, go to:glencoe.comYou can use your completedVocabulary Builder(pages 168–169) to help yousolve the puzzle.1. ADVERTISEMENT A poster measures 10 inches by 14 inches.If it is enlarged to have a width of 60 inches, how tall will thenew poster be?Solve each proportion.2. 3_8= x_403. 9_11= 15_x4.x + 2_5= 4_3Determine whether each pair of figures is similar. If so, writethe appropriate similarity statement.5.7-2Similar Polygons7-1Proportions
  • 193. Glencoe Geometry 185Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER76.If △HIJ ∼ △MKJ, find x and the scale factor of △HIJ to △MKJ.7. ฀ ฀฀ ฀8.9. SHADOWS A tree casts a 60 foot shadow. At the same time,a 6-foot tall man casts a shadow that is 2 feet long. How tallis the tree?Determine whether−−BC−−DE.10. AD = 16, DB = 12, AE = 14, and EC = 10.511. BD = 5, BA = 20, and CE is one third of EA12. Find FG if △RST ∼ △EFG,−−−SH is an altitude of △RST,−−FJ isan altitude of △EFG, ST = 10, SH = 8, and FJ = 10.฀ ฀ ฀ ฀7-3Similar Triangles7-4Parallel Lines and Proportional Parts7-5Parts of Similar Triangles
  • 194. 186 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistC H A P T E R7Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 7.Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 7 Practice Test onpage 427 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 7 Study Guide and Reviewon pages 424–426 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 7 Practice Test onpage 427 of your textbook.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 7 Foldable.• Then complete the Chapter 7 Study Guide and Review onpages 424–426 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 7 Practice Test onpage 427 of your textbook.ARE YOU READY FORTHE CHAPTER TEST?
  • 195. Glencoe Geometry 187Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter8Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.Right Triangles and TrigonometryNOTE-TAKING TIP: When you take notes, drawa visual (graph, diagram, picture, chart) thatpresents the information introduced in the lessonin a concise, easy-to-study format.Stack the sheets. Foldthe top right cornerto the bottom edgeto form a square.Fold the rectangularpart in half.Staple the sheetsalong the fold infour places.Label each sheet witha lesson number andthe rectangular partwith the chapter title.Begin with seven sheets of grid paper.C H A P T E R8
  • 196. 188 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 8.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.C H A P T E R8Vocabulary TermFoundDefinitionDescription oron Page Exampleangle of depressionangle of elevationcosinegeometric meanLaw of CosinesLaw of Sines
  • 197. Glencoe Geometry 189Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY8Vocabulary TermFoundDefinitionDescription oron Page ExamplePythagorean triplesinesolving a triangletangenttrigonometric ratiotrigonometry
  • 198. 190 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–1The geometric mean between two numbers is thesquare root of their .Geometric MeanFind the geometric mean between 25 and 7.16x= Definition of geometric meanx2 = 175 Cross multiplyx = Take the positive squareroot of each side.x ≈ Use a calculator.Check Your Progress Find the geometric meanbetween each pair of numbers.a. 3 and 12 b. 25 and 7• Find the geometricmean between twonumbers.• Solve problemsinvolving relationshipsbetween parts of aright triangle andthe altitude to itshypotenuse.MAIN IDEASGeometric Mean Fortwo positive numbers aand b, the geometricmean is the positivenumber x where theproportion a : x = x : bis true. This proportioncan be written usingfractions as a_x=x_borwith cross products asx2 = ab or x = √ab.KEY CONCEPTBUILD YOUR VOCABULARY (page 188)Theorem 8.1If the altitude is drawn from the vertex of the right angleof a right triangle to its hypotenuse, then the two trianglesformed are similar to the given triangle and to each other.Theorem 8.2The measure of an altitude drawn from the vertex of theright angle of a right triangle to its hypotenuse is thegeometric mean between the measures of the twosegments of the hypotenuse.Geometric Mean
  • 199. Glencoe Geometry 191Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Why do youautomatically discardthe negative root whenfinding the altitude orgeometric mean?WRITE ITAltitude and Segments of the HypotenuseIn △ABC, BD = 6 and AD = 27. Find CD.Let x = CD.== BD = , AD = ,and CD = x.x2 = Cross productsx = 9√2 Take the positive square root ofeach side.x ≈ Use a calculator.CD is about .Check Your Progress In △DEF, FG = 18, and DG = 4.Find EG.Altitude and Length of the HypotenuseKITES Ms. Alspach constructeda kite for her son. She had toarrange perpendicularly twosupport rods, the shorter ofwhich was 27 inches long. Ifshe had to place the short rod7.25 inches from one end of thelong rod in order to form tworight triangles with the kitefabric, what was the length ofthe long rod?8–1CDCDxx
  • 200. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.192 Glencoe Geometry8–1Draw a diagram of one ofthe right triangles formed.Let XY be the altitudedrawn from the right angleof △WYZ.WX_YX= YX_ZX= WX = and YX =7.25ZX = Cross productsZX ≈ Divide each side by .The length of the long rod is 7.25 + , or aboutinches long.Check Your ProgressA jetliner has a wingspan, BD,of 211 feet. The segment drawnfrom the front of the planeto the tail,−−AC, intersects−−−BD at point E. If AE is163 feet, what is thelength of the aircraft?ZXTheorem 8.3If the altitude is drawn from the vertex of the right angleof a right triangle to its hypotenuse, then the measure of aleg of the triangle is the geometric mean between themeasures of the hypotenuse and the segment of thehypotenuse adjacent to that leg.
  • 201. Glencoe Geometry 193Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT8–1Hypotenuse and Segment of HypotenuseFind c and d in JKL.−−−KM is the altitude of right triangle JKL. Use Theorem 8.2 towrite a proportion.JM_KM= KM_LM= JM = , KM = , and LM =100 = 5c Cross products= c Divide each side by .−−JK is a leg of right triangle JKL. Use Theorem 8.3 to writea proportion.JL_JK= JK_JM= JL = , JM = 5, and JK = d= Cross productsd = √125 Take the .d = Simplify.d ≈ 11.2 Use a calculator.Check Your Progress Find e and f.On the Lesson 8-1Foldable, includedrawings to show howto solve problemsinvolving therelationships betweenparts of a right triangleand the altitude to itshypotenuse.ORGANIZE ITdd
  • 202. 194 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–2฀The Pythagorean Theorem and Its ConverseFind the Length of the HypotenuseLONGITUDE AND LATITUDECarson City, Nevada, islocated at about 120 degreeslongitude and 39 degreeslatitude; NASA Ames islocated at about 122 degreeslongitude and 37 degreeslatitude. Use the lines oflongitude and latitude tofind the degree distance tothe nearest tenth degree if youwere to travel directly from NASA Ames to Carson City.The change in longitude between NASA Ames and Carson Cityis 119 - 122 or 3 degrees. Let this distance be a.The change in latitude is 39 – 37 or 2 degrees.Let this distance be b.Use the Pythagorean Theorem to find the distance from NASAAmes to Carson City.a2 + b2 = c2 Pythagorean Theorem32 + 22 = c2 a = 3, b = 2+ = c2 Simplify.13 = c2 Add.= c Take the square root of each side.≈ c Use a calculator.The degree distance between NASA Ames and Carson City isabout degrees.Theorem 8.4 Pythagorean TheoremIn a right triangle, the sum of the squares of the measuresof the legs equals the square of the measure of thehypotenuse.• Use the PythagoreanTheorem.• Use the converse of thePythagorean Theorem.MAIN IDEAS
  • 203. Glencoe Geometry 195Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–2Check Your Progress Carson City, Nevada, is locatedat about 120 degrees longitude and 39 degrees latitude.NASA Dryden is located about 117 degrees longitude and34 degrees latitude. Use the lines of longitude and latitudeto find the degree distance to the nearest tenth degree if youwere to travel directly from NASA Dryden to Carson City.Find the Length of a LegFind d.(PQ)2 + (QR)2 = (PR)2 PythagoreanTheorem32 + d2 = 62 PQ = 3, PR = 69 + d2 = 36 Simplify.d2 = Subtract from each side.d = Take the square root of each side.d ≈ Use a calculator.Check Your Progress Find x.Verify a Triangle is a Right TriangleCOORDINATE GEOMETRYVerify that △ABC is aright triangle.Use the Distance Formulato determine the lengths ofthe sides.Theorem 8.5 Converse of the Pythagorean TheoremIf the sum of the squares of the measures of two sides of atriangle equals the square of the measure of the longestside, then the triangle is a right triangle.On the Lesson 8-2Foldable, write thePythagorean Theoremand its converse.ORGANIZE IT
  • 204. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.196 Glencoe Geometry8–2AB = √[1 - (-9)]2 + [-1 - (-3)]2 x1 = -9, y1 = -3,x2 = 1, y2 = -1= √2+2Subtract.= Simplify.BC = √(-3 - 1)2 + [-7 - (-1)]2 x1 = 1, y1 = -1,x2 = -3, y2 = -7= √(-4)2 + (-6)2 Subtract.= Simplify.AC = √[-3 - (-9)]2 + [-7 - (-3)]2 x1 = -9, y1 = -3,x2 = -3, y2 = -7= √62 + (-4)2 Subtract.= Simplify.By the converse of the Pythagorean Theorem, if the sum of thesquares of the measures of two sides of a triangle equals thesquare of the measure of the longest side, then the triangle isa right triangle.(BC)2 + (AC)2 = (AB)2 Converse of thePythagorean Theorem( )2+ ( )2= ( )2+ = 104 Simplify.= 104 Add.Since the sum of the squares of two sides equals the square ofthe , △ABC is a right triangle.Check Your Progress Verify that△RST is a right triangle.
  • 205. Glencoe Geometry 197Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTA Pythagorean triple is three whole numbers that satisfythe equation , where c is thegreatest number.8–2Pythagorean TriplesDetermine whether each set of measures can be thesides of a right triangle. Then state whether they form aPythagorean triple.a. 9, 12, 15Since the measure of the longest side is 15, 15 must be c.Let a and b be 9 and 12.a2 + b2 = c2 Pythagorean Theorem92 + 122 152 a = 9, b = 12, c = 15+ Simplify.= Add.These segments form the sides of a right triangle since theysatisfy the Pythagorean Theorem. The measures are wholenumbers and form a Pythagorean triple.b. 4√3, 4, and 8Since the measure of the longest side is 8, let c = 8.a2 + b2 = c2 Pythagorean Theorem(4√3)2 + 42 82 Substitution+ Simplify.= Add.Since these measures satisfy the Pythagorean Theorem,they form a right triangle. Since the measures are not allwhole numbers, they do not form a Pythagorean triple.Check Your Progress Determine whether each set ofmeasures are the sides of a right triangle. Then statewhether they form a Pythagorean triple.a. 5, 8, 9 b. 3, √5, √14BUILD YOUR VOCABULARY (page 189)Write examples ofPythagorean triplesunder the Lesson 8-2tab.ORGANIZE IT
  • 206. 198 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–3• Use properties of45°-45°-90° triangles.• Use properties of30°-60°-90° triangles.MAIN IDEASREMEMBER ITTo rationalize adenominator, multiplythe fraction by 1 in theform of a radical overitself so that theproduct in thedenominator is arational number.Find the Measure of the LegsFind a.The length of the hypotenuse of a 45°-45°-90° triangle is √2times as long as a leg of the triangle.RS = (ST)√2= √2 RS = , ST =8_√2= a Divide each side by √2.8_√2·√2_√2= a Rationalize the denominator.= a Divide.Check Your Progress Find b.Theorem 8.6In a 45°-45°-90° triangle, the length of the hypotenuse is√2 times the length of a leg.Theorem 8.7In a 30°-60°-90° triangle, the length of the hypotenuse istwice the length of the shorter leg, and the length of thelonger leg is √3 times the length of the shorter leg.Special Right Triangles
  • 207. Glencoe Geometry 199Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.30°-60°-90° TrianglesFind QR.−−PR is the longer leg,−−−PQ is the shorter leg, and−−−QR is thehypotenuse.PQ = 1_2(QR)= 1_2(QR) PQ == QR Multiply each side by 2.The length of QR is centimeters.Check Your Progress Find BC.Special Triangles in a Coordinate PlaneCOORDINATE GEOMETRY △WXY is a 30°-60°-90°triangle with right angle X and−−−WX as the longer leg.Graph points X(-2, 7) and Y(-7, 7) and locate point Win Quadrant III.Graph X and Y.−−XY lies on ahorizontal gridline of thecoordinate plane. Since−−−WXwill be perpendicular to−−XY, itlies on a vertical gridline. Findthe length of−−XY.Explain what it meansfor a triangle to beclassified as a righttriangle. (Lesson 4-1)REVIEW ITOn the Lesson 8-3Foldable, includedrawings of a45°-45°-90° triangleand of a 30°-60°-90°triangle. Show how touse the PythagoreanTheorem and propertiesof these special trianglesto find missing parts.ORGANIZE IT8–3
  • 208. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.200 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT8–3XY = ß-7 - (-2)ß = ß-7 + 2ß =−−XY is the shorter leg.−−−WX is the longer leg. So, WX = √3(XY).Use XY to find WX.WX = √3(XY)= √3( ) XY ==Point W has the same x-coordinate as X. W is locatedunits below X. So, the coordinates of W are(-2, 7 - 5√3) or about .Check Your Progress △RST is a 30°-60°-90° triangle withright angle R and−−RT as the shorter leg. Graph points T(3, -3)and R(3, -6) and locate point S in Quadrant III.
  • 209. Glencoe Geometry 201Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–4A ratio of the of sides of a right triangle iscalled a trigonometric ratio.The three most common trigonometric ratios are sine,cosine, and tangent.Find Sine, Cosine, and Tangent RatiosFind sin L, cos L, tan L, sin N, cos N, and tan N. Expresseach ratio as a fraction and as a decimal.sin L =opposite leg__hypotenusesin N =opposite leg__hypotenuse= MN_LN= LM_LN= or 0.47 = or 0.88cos L =adjacent leg__hypotenusecos N =adjacent leg__hypotenuse= LM_LN= MN_LN= or 0.88 = or 0.47tan L =opposite leg__adjacent legtan N =opposite leg__adjacent leg= MN_LM= LM_MN= or 0.53 = or 1.88• Find trigonometricratios using righttriangles.• Solve problems usingtrigonometric ratios.MAIN IDEASBUILD YOUR VOCABULARY (page 189)Trigonometric Ratiossin A = BC_ABsin B = AC_AB฀ ฀cos A = AC_ABcos B = BC_ABtan A = BC_ACtan B = AC_BCKEY CONCEPTTrigonometry
  • 210. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.202 Glencoe GeometryCheck Your Progress Find sin A, cos A, tan A, sin B, cos B,and tan B. Express each ratio as a fraction and as a decimal.Evaluate ExpressionsUse a calculator to find each value to the nearestten-thousandth.a. tan 56°TAN ENTER 1.482560969tan 56° ≈b. cos 90°COS ENTER 0cos 90° ≈Check Your Progress Use a calculator to find eachvalue to the nearest ten-thousandth.a. sin 48° b. cos 85°On your Lesson 8-4Foldable, include stepsto follow when findingtrigonometric ratios ona calculator. Be sure tonote that the calculatorshould be in degreemode rather thanradian mode.ORGANIZE IT8–4
  • 211. Glencoe Geometry 203Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT8–4Use Trigonometric Ratios to Find a LengthEXERCISING A fitness trainer sets the incline on atreadmill to 7°. The walking surface is 5 feet long.Approximately how many inches did the trainer raisethe end of the treadmill from the floor?Let y be the height of the treadmill from the floor in inches.The length of the treadmill is feet, or inches.sin 7° = sin =leg opposite_hypotenusesin 7° = y Multiply each side by .The treadmill is about inches high.Check Your Progress The bottom of a handicap ramp is15 feet from the entrance of a building. If the angle of the rampis about 4.8°, how high does the ramp rise off the ground to thenearest inch?
  • 212. 204 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–5An angle of elevation is the angle between the line of sightand the horizontal when an observer looks .Angles of Elevation and DepressionAngle of ElevationCIRCUS ACTS At the circus, a person in the audiencewatches the high-wire routine. A 5-foot-6-inch tall acrobatis standing on a platform that is 25 feet off the ground.How far is the audience member from the base of theplatform, if the angle of elevation from the audiencemember’s line of sight to the top of the acrobat is 27°?Make a drawing.Since QR is 25 feet and RS is 5 feet 6 inches or feet,QS is 30.5 feet. Let x represent PQ.tan 27° = tan =opposite_adjacenttan 27° = 30.5_xQS = 30.5, PQ = xx tan 27° = 30.5 Multiply each side by x.x = 30.5_tan 27°Divide each side by tan 27°.x ≈ Simplify.The audience member is about feet from the base ofthe platform.• Solve problemsinvolving angles ofelevation.• Solve problemsinvolving angles ofdepression.MAIN IDEASBUILD YOUR VOCABULARY (page 188)REVIEW ITWhat is true about thesum of the measures ofthe acute angles of aright triangle?(Lesson 4-2)
  • 213. Glencoe Geometry 205Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.An angle of depression is the angle between the lineof sight when an observer looks , andthe horizontal.8–5Check Your Progress At a diving competition, a 6-foot-talldiver stands atop the 32-foot platform. The front edge of theplatform projects 5 feet beyond the end of the pool. The poolitself is 50 feet in length. A camera is set up at the opposite endof the pool even with the pool’s edge. If the camera is angled sothat its line of sight extends to the top of the diver’s head, whatis the camera’s angle of elevation to the nearest degree?On the Lesson 8-5Foldable, include adrawing that illustratesangle of elevation andone that illustratesangle of depression.ORGANIZE ITBUILD YOUR VOCABULARY (page 188)REMEMBER ITThere may be morethan one way to solve aproblem. Refer to page465 of your textbook foranother method youcould use to solveExample 2.Angle of DepressionTEST EXAMPLE A wheelchair ramp is 3 meters longand inclines at 6°. Find the height of the ramp to thenearest tenth centimeter.A 0.3 cm B 31.4 cm C 31.5 cm D 298.4 cmRead the Test ItemThe angle of depression between the ramp and the horizontalis . Use trigonometry to find the height of the ramp.Solve the Test ItemThe ground and the horizontal level with the platform towhich the ramp extends are . Therefore,m∠ZYX = m∠WXY since they areangles.
  • 214. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.206 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT8–5sin 6° = WY sin =opposite_hypotenusesin 6° = WY XY =sin 6° = WY Multiply each side by 3.≈ WY Simplify.The height of the ramp is about meters, or0.314(100) = 31.4 centimeters. The answer is .Check Your Progress A roller coaster car is at one of itshighest points. It drops at a 63° angle for 320 feet. How highwas the roller coaster car to the nearest foot before it beganits fall?
  • 215. Glencoe Geometry 207Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–6In trigonometry, the Law of Sines can be used to findmissing parts of triangles that are not right triangles.Use the Law of Sinesa. Find p. Round to the nearest tenth.sin P_p= sin R_rLaw of Sinessin 17°_p= sin 29°_8m∠P = 17, m∠R = 29, r = 88 sin = p sin Cross products8 sinsin= p Divide.4.8 ≈ p Use a calculator.b. Find m∠L to the nearest degree in △LMN if n = 7,ℓ = 9, and m∠N = 43.sin L_ℓ= sin N_nLaw of Sinessin L=9 sin 43ℓ = ,m∠N=43,n =sin L = sin 43 Cross Productssin L = 9 sin 43_7Divide.L = sin-1(9 sin 43_7 ) Solve for L.L ≈ Use a calculator.• Use the Law of Sines tosolve triangles.• Solve problems by usingthe Law of Sines.MAIN IDEASLaw of Sines Let △ABCbe any triangle with a, b,and c representing themeasures of the sidesopposite the angleswith measures A, B,and C, respectively. Thensin A_a= sin B_b= sin C_c.KEY CONCEPTThe Law of SinesBUILD YOUR VOCABULARY (page 188)
  • 216. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.208 Glencoe GeometrySolving a triangle means finding the measures of all of theand of a triangle.8–6On the Lesson 8-6Foldable, write anexample of how theLaw of Sines is used.ORGANIZE ITCheck Your Progressa. Find c.b. Find m∠T to the nearest degree in △RST if r = 12, t = 7,and m∠R = 76.Solve Trianglesa. Solve △DEF if m∠D = 112, m∠F = 8,and f = 2. Round angle measuresto the nearest degree and sidemeasures to the nearesttenth.We know the measures oftwo angles of the triangle. Use the Angle Sum Theorem tofind m∠E.m∠D + m∠E + m∠F = 180 Angle Sum Theorem+ m∠E + = 180 m∠D = ,m∠F =+ m∠E = 180 Add.m∠E = Subtract 120 fromeach side.Since we know m∠F and f, use proportions involving sin F_f.(page 189)BUILD YOUR VOCABULARY
  • 217. Glencoe Geometry 209Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–6To find d:sin F_f= sin d_dLaw of Sinessin 8°_2= sindSubstitute.d sin 8° = 2 sin Cross products.d =2 sinsin 8°Divide each side by sin 8°.d ≈ Use a calculator.To find e:sin F_f= sin E_eLaw of Sinessin 8° = sin 60°_eSubstitute.e sin 8° = sin 60° Cross products.e = sin 60°sin 8°Divide each side by sin 8°.e ≈ Use a calculator.Therefore, m∠E = , d ≈ , and e ≈ .Check Your Progress Solve △RST if m∠R = 43,m∠T = 103, and r = 14. Round angle measures to thenearest degree and side measures to the nearest tenth.
  • 218. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.210 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT8–6b. Solve △HJK if m∠J = 32, h = 30, and j = 16. Roundangle measures to the nearest degree and sidemeasures to the nearest tenth.We know the measure of two sides and an angle oppositeone of the sides. Use the Law of Sines.sin J_j= sin H_hLaw of Sinessin 32°_16= sin H_30m∠J = 32, j = 16, h = 30sin 32° = sin H Cross products30 sin 32°_16= sin H Divide.sin-1(30 sin 32°_16 ) = H Solve for H.≈ H Use a calculator.m∠H + m∠J = m∠K = 180 Angle Sum Theorem+ + m∠K = 180 Substitute.m∠K ≈ Subtract 116 fromeach side.sin J_j= sin K_kLaw of Sinessin 32°_16= sin 64°_km∠J = 32,m∠K = 64, j = 16k sin 32° = 16 sin 64° Cross productsk = 16 sin 64°_sin 32°Divide.k ≈ Use a calculator.So, m∠H ≈ , m∠K ≈ , and k ≈ .Check Your Progress Solve △TUV if m∠T = 54, t = 12,and v = 9. Round angle measures to the nearest degree andside measures to the nearest tenth.REMEMBER ITIf you round beforethe final answer, yourresults may differfrom results in whichrounding was not doneuntil the final answer.
  • 219. Glencoe Geometry 211Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.8–7The Law of Cosines allows us to solve a triangle when thecannot be used.The Law of CosinesTwo Sides and the Included AngleFind x if y = 11, z = 25, and m∠X = 45.Use the Law of Cosines since the measures of two sides andthe included angle are known.x2 = y2 + z2 - 2yz cos Law of Cosinesx2 = 112 + 252 - 2(11)(25) cos y = 11, z = 25,m∠ =x2 = 746 - 550 cos Simplify.x = 746 - 550 cos Take the square rootof each side.x ≈ Use a calculator.Check Your Progress Find r if s = 15, t = 32, andm∠R = 40.• Use the Law of Cosinesto solve triangles.• Solve problems by usingthe Law of Cosines.MAIN IDEASBUILD YOUR VOCABULARY (page 188)Law of Cosines Let△ABC be any trianglewith a, b, and crepresenting themeasures of sidesopposite angles withmeasures A, B, and C,respectively. Then thefollowing equationsare true.a2 = b2 + c2 - 2bc cos Ab2 = a2 + c2 - 2ac cos Bc2 = a2 + b2 - 2ab cos CKEY CONCEPT
  • 220. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.212 Glencoe Geometry8–7On the Lesson 8-7Foldable, try to includeyour own example ofa problem that canbe solved using theLaw of Cosines. Showhow you solved yourproblem.ORGANIZE ITName two cases whenyou would use the Lawof Cosines to solve atriangle and two caseswhen you would use theLaw of Sines to solve atriangle.WRITE ITThree SidesFind m∠L.ℓ2 = m2 + n2 - 2mn cos L Law of Cosines= 272 + 52 - 2(27)(5) cos L Replace ℓ, m, and n.= 754 - 270 cos L Simplify.-178 = -270 cos L Subtract.-178_-270= L Divide.cos-1(178_270) = L Solve for L.≈ L Use a calculator.Check Your Progress Find m∠F.Select a StrategySolve △DEF. Round anglemeasures to the nearest degreeand side measures to thenearest tenth.Since we know the measures of twosides and the included angle,use the Law of Cosines.f 2 = d2 + e2 - 2de cos F Law of Cosinesf 2 = 322 + 172 - 2(32)(17) cos 145° d = 32, e = 17,m∠F = 145f = - 1088 cos 145° Take the squareroot of each side.f ≈ Use a calculator.
  • 221. Glencoe Geometry 213Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT8–7Next, we can find m∠D or m∠E. If we decide to find m∠D, wecan use either the Law of Sines or the Law of Cosines to findthis value.sin F_f= sin D_dLaw of Sinessin F=sin Df = , d = ,m∠F =32 sin 145° = 46.9 sin D Cross products32 sin 145°_46.9= sin D Divide each side by 46.9.sin-1(32 sin 145°_46.9 ) = D Take the inverse of each side.≈ D Use a calculator.Use the Angle Sum Theorem to find m∠E.m∠D + m∠E + m∠F = 180 Angle Sum Theorem+ m∠E + ≈ 180 m∠D ≈ 23,m∠F = 145m∠E + 168 ≈ 180m∠E ≈ Subtract fromeach side.Therefore, f ≈ , and m∠D ≈ , m∠E ≈ .Check Your Progress Determine whether the Law of Sinesor the Law of Cosines should be used first to solve △ABC.Then solve △ABC. Round angle measures to the nearestdegree and side measures to the nearest tenth.
  • 222. STUDY GUIDE214 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERUse your Chapter 8 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 8, go to:glencoe.comYou can use your completedVocabulary Builder(pages 188–189) to help yousolve the puzzle.Find the geometric mean between each pair of numbers.1. 4 and 9 2. 20 and 303. Find x and y.4. For the figure shown, which statements are true?a. m2 + n2 = p2 b. n2 = m2 + p2c. m2 = n2 + p2 d. m2 = p2 - n2e. p2 = n2 - m2 f. n2 - p2 = m2g. n = √m2 + p2 h. p = √m2 - n2Which of the following are Pythagorean triples? Write yes or no.5. 10, 24, 26 6. √2, √2, 2 7. 10, 6, 88-1Geometric Mean8-2The Pythagorean Theorem and Its ConverseC H A P T E R8
  • 223. Glencoe Geometry 215Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER8Complete each statement.8. In a 45°-45°-90° triangle, to find the length of thehypotenuse, multiply the length of a leg by .9. In a 30°-60°-90° triangle, to find the length of thehypotenuse, multiply the length of the shorter leg by .10. In a 30°-60°-90° triangle, to find the length of the longer leg,multiply the length of the shorter leg by .Indicate whether each statement is always, sometimes,or never true.11. The lengths of the three sides of an isosceles triangle satisfythe Pythagorean Theorem.12. The lengths of the sides of a 30°-60°-90° triangle form aPythagorean triple.13. The lengths of all three sides of a 30°-60°-90° triangle arepositive integers.Write a ratio using the side lengths inthe figure to represent each of thefollowing trigonometric ratios.14. sin N 15. cos N16. tan N 17. tan M18. sin M 19. cos M8-3Special Right Triangles8-4Trigonometry
  • 224. 216 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER8Refer to the figure. The two observersare looking at one another. Select thecorrect choice for each question.20. What is the line of sight?a. line RS b. line ST c. line RT d. line TU21. What is the angle of elevation?a. ∠RST b. ∠SRT c. ∠RTS d. ∠UTR22. What is the angle of depression?a. ∠RST b. ∠SRT c. ∠RTS d. ∠UTR23. A tree that is 12 meters tall casts a shadow that is 15 meterslong. What is the angle of elevation of the sun?24. Refer to the figure. According to the Law of Sines,which of the following are correct statements?a. m_sin M= n_sin N=p_sin Pb. sin m_M= sin n_N=sin p_Pc. cos M_m= cos N_n= cos P_pd. sin M_m+ sin N_n= sin P_pe. (sin M)2 + (sin N)2 = (sin P)2 f. sin P_p= sin M_m= sin N_n25. Solve △ABC if m∠A = 50, m∠B = 65, and a = 12. Roundangle measures to the nearest degree and side measures tothe nearest tenth.8-5Angles of Elevation and Depression8-6The Law of Sines
  • 225. Glencoe Geometry 217Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER8Write true or false for each statement. If the statement is false,explain why.26. The Law of Cosines applies to right triangles.27. The Law of Cosines is used to find the third side of a trianglewhen you are given the measures of two sides and thenonincluded angle.28. Refer to the figure. According to theLaw of Cosines, which statements arecorrect for △DEF?a. d2 = e2 + f2 - ef cos D b. e2 = d2 + f2 - 2df cos Ec. d2 = e2 + f2 + 2ef cos D d. f 2 = d2 + e2 - 2ef cos Fe. f2 = d2 + e2 - 2de cos F f. d2 = e2 + f2g. sin D_d= sin E_e= sin F_fh. d2 = e2 + f 2 - 2ef cos D29. Solve △DEF if m∠F = 37, d = 3, and e = 7. Round anglemeasures to the nearest degree and side measures to thenearest tenth.8-7The Law of Cosines
  • 226. 218 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistC H A P T E R8Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 8 Practice Test onpage 491 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 8 Study Guide and Reviewon pages 486–490 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 8 Practice Test onpage 491 of your textbook.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 8 Foldable.• Then complete the Chapter 8 Study Guide and Review onpages 486–490 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 8 Practice Test onpage 491 of your textbook.Visit glencoe.com toaccess your textbook,more examples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 8.ARE YOU READY FORTHE CHAPTER TEST?
  • 227. Chapter9Glencoe Geometry 219Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R9 TransformationsUse the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: In addition to writingimportant definitions in your notes, be sure toinclude your own examples of the conceptspresented.Fold a sheet of notebookpaper in half lengthwise.Cut on every third lineto create 8 tabs.Label each tab with avocabulary word fromthis chapter.Begin with one sheet of notebook paper.reflectiontranslationrotationdilation
  • 228. 220 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.C H A P T E R9Vocabulary TermFoundDefinitionDescription oron Page ExampleThis is an alphabetical list of new vocabulary terms you will learn in Chapter 9.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.angle of rotationcenter of rotationcomponent formcompositiondilationdirectioninvariant pointsisometryline of reflectionline of symmetrymagnitudepoint of symmetryreflection
  • 229. Chapter BUILD YOUR VOCABULARY9Glencoe Geometry 221Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Vocabulary TermFoundDefinitionDescription oron Page Exampleregular tessellationresultantrotationrotational symmetryscalarscalar multiplicationsemi-regular tessellationsimilarity transformationstandard positiontessellationtranslationuniformvector
  • 230. 222 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–1 Reflections• Draw reflected images.• Recognize and drawlines of symmetry andpoints of symmetry.MAIN IDEASReflecting a Figure in a LineDraw the reflected image of quadrilateral WXYZ inline p.STEP 1 Draw segments perpendicular toline p from each point W, X, Y,and Z.STEP 2 Locate W′, X′, Y′, and Z′ so that linep is the bisectorof−−−WW′,−−XX′,−−YY′, and−−ZZ′. Points W′, X′, Y′,and Z′ are the respective images of.STEP 3 Connect vertices W′, X′, Y′, and Z′.Check Your Progress Draw the reflected image ofquadrilateral ABCD in line n.A reflection is a transformation representing aof a figure.The segment connecting a point and its image isto a line m, and isby line m. Line m is called the line of reflection.A reflection is a congruence transformation, or an isometry.BUILD YOUR VOCABULARY (pages 220–221)
  • 231. Glencoe Geometry 223Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–1Write the definition ofreflection under thereflection tab. Includea sketch to illustrate areflection.ORGANIZE ITRelection in the x-axisCOORDINATE GEOMETRY Quadrilateral ABCD hasvertices A(1, 1), B(3, 2), C(4, -1), and D(2, -3).a. Graph ABCD and its image under reflection in thex-axis. Compare the coordinates of each vertex withthe coordinates of its image.Use the vertical grid lines to find the corresponding pointfor each vertex so that the x-axis is equidistant from eachvertex and its image.A(1, 1) A′B(3, 2) B′C(4, -1) C′D(2, -3) D′Plot the reflected vertices and connect to form the imageA′B′C′D′. The x-coordinates stay the same, but they-coordinates are opposite. That is, (a, b) (a, -b).b. Graph ABCD and its image under reflection in theorigin. Compare the coordinates of each vertex withthe coordinates of its image.Use the horizontal and vertical distances. From A to theorigin is 2 units down and 1 unit left. So, A′ is located byrepeating that pattern from the origin.A(1, 1) A′B(3, 2) B′C(4, -1) C′D(2, -3) D′Plot the reflected vertices and connectto form the image A′B′C′D′. Both thex-coordinates and y-coordinates areopposite. That is, (a, b) (-a, -b).
  • 232. 224 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–1c. Graph ABCD and its image under reflection in theline y = x. Compare the coordinates of each vertexwith the coordinates of its image.The slope of y = x is 1.−−−C′C is perpendicular to y = x, soits slope is -1. From C, move up 5 units and to the left5 units to C′.A(1, 1) A′B(3, 2) B′C(4, -1) C′D(2, -3) D′Plot the reflected vertices and connect. Comparingcoordinates shows that (a, b) (b, a).Check Your Progressa. Quadrilateral LMNP has vertices L(-1, 1), M(5, 1),N(4, -1), and P(0, -1). Graph LMNP and its image underreflection in the x-axis.b. Quadrilateral LMNP has vertices L(-1, 1), M(5, 1),N(4, -1), and P(0, -1). Graph LMNP and its image underreflection in the origin.
  • 233. Glencoe Geometry 225Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–1Page(s):Exercises:HOMEWORKASSIGNMENTSome figures can be folded so that the two halves. The fold is a line of reflection calleda line of symmetry.For some figures, a point can be found that is a commonpoint of reflection for all points on a figure. This commonpoint of reflection is called a point of symmetry.Draw Lines of SummetryDetermine how many lines of symmetry a regularpentagon has. Then determine whether a regularpentagon has a point of symmetry.A regular pentagon has lines of symmetry.A point of symmetry is a point that is acommon point of reflection for all points onthe figure. There is not one point ofsymmetry in a regular pentagon.Check Your Progress Determine how many lines ofsymmetry an equilateral triangle has. Then determinewhether an equilateral triangle has a point of symmetry.BUILD YOUR VOCABULARY (pages 220–221)c. Quadrilateral LMNP has vertices L(-1, 1), M(5, 1),N(4, -1), and P(0, -1). Graph LMNP and its image underreflection in the line y = x.
  • 234. 226 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–2A translation is a transformation that moves all points of afigure the same distance in the same .Translations• Draw translated imagesusing coordinates.• Draw translated imagesby using repeatedreflections.MAIN IDEASTranslations in the Coordinate PlaneCOORDINATE GEOMETRYParallelogram TUVW has verticesT(-1, 4), U(2, 5), V(4, 3), andW(1, 2). Graph TUVW and itsimage for the translation(x, y) (x - 4, y - 5).This translation moved every pointof the preimage 4 units left and5 units down.T(-1, 4) T′ or T′U(2, 5) U′ or U′V(4, 3) V′ or V′W(1, 2) W′ or W′Plot and then connect the translated vertices T′U′V′ and W′.Check Your Progress Parallelogram LMNP has verticesL(-1, 2), M(1, 4), N(3, 2), and P(1, 0). Graph LMNP and itsimage for the translation (x, y) (x + 3, y - 4).BUILD YOUR VOCABULARY (pages 220–221)
  • 235. Glencoe Geometry 227Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–2Write the definition oftranslation under thetranslation tab. Drawa figure and its imagefor a translation on acoordinate plane.ORGANIZE ITREMEMBER ITWhen you translatethe point at (x, y) to thepoint at (x + a, y + b),the number a tells howfar to move left or right.The number b tells howfar to move up or down.Repeated TranslationsANIMATION The graph showsrepeated translations that resultin the animation of a raindrop.Find the translation that movesraindrop 2 to raindrop 3, andthen the translation that movesraindrop 3 to raindrop 4.To find the translation from raindrop2 to raindrop 3, use the coordinatesat the top of each raindrop. Use thecoordinates (1, 2) and (-1, -1) inthe formula.(x, y) (x + a, y + b).(1, 2) (-1, -1)x + a =+ a = x = 1a = Subtract 1 from each side.y + b =+ b = y = 2b = Subtract 2 from each side.The translation is ( , )from raindrop 2to raindrop 3. Use the coordinates (-1, -1) and (-1, -4) tofind the translation from raindrop 3 to raindrop 4.(x, y) (x + a, y + b)(-1, -1) (-1, -4)x + a = -1 y + b = -4+ a = + b =a = b =The translation is from raindrop 3 to raindrop 4.
  • 236. Glencoe Geometry 229Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–3A rotation is a transformation that turns every point of apreimage through a specified andabout a fixed point.• Draw rotated imagesusing the angle ofrotation.• Identify figures withrotational symmetry.MAIN IDEASRotationsBUILD YOUR VOCABULARY (pages 220–221)Reflections in Intersection LinesFind the image of parallelogram WXYZ underreflections in line p and then line q.First reflect parallelogram WXYZ in line . Then labelthe image W′X′Y′Z′.Next, reflect the image in line . Then label the imageW!X!Y!Z!.Parallelogram W!X!Y!Z! is the image of parallelogramunder reflections in lines p and q.Postulate 9.1In a given rotation, if A is the preimage, A′ is the image,and P is the center of rotation, then the measure of theangle of rotation, ∠APAЈ is twice the measure of the acuteor right angle formed by the intersecting lines of reflection.Corollary 9.1Reflecting an image successively in two perpendicular linesresults in a 180° rotation.
  • 237. Glencoe Geometry 229Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–3A rotation is a transformation that turns every point of apreimage through a specified andabout a fixed point.• Draw rotated imagesusing the angle ofrotation.• Identify figures withrotational symmetry.MAIN IDEASRotationsBUILD YOUR VOCABULARY (pages 220–221)Reflections in Intersection LinesFind the image of parallelogram WXYZ underreflections in line p and then line q.First reflect parallelogram WXYZ in line . Then labelthe image W′X′Y′Z′.Next, reflect the image in line . Then label the imageW!X!Y!Z!.Parallelogram W!X!Y!Z! is the image of parallelogramunder reflections in lines p and q.Postulate 9.1In a given rotation, if A is the preimage, A′ is the image,and P is the center of rotation, then the measure of theangle of rotation, ∠APAЈ is twice the measure of the acuteor right angle formed by the intersecting lines of reflection.Corollary 9.1Reflecting an image successively in two perpendicular linesresults in a 180° rotation.
  • 238. 230 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT9–3Check Your Progress Find the image of △ABC underreflections in line m and then line n.Identifying Rotational SymmetryQUILTS Using the quilt in Example 3 in the StudentEdition, identify the order and magnitude of thesymmetry in each part of the quilt.a. medium star directly to the left of the large star inthe center of the quiltThe medium-sized star has points. So it has arotational symmetry of order 16. To find the magnitude,divide by 16 to find that the magnitude is .b. tiny star above the medium sized star in part aThe tiny star has 8 points, so the order is . Divide360° by 8 to find that the magnitude is .Check Your Progress Identify the order andmagnitude of the rotational symmetry for each regularpolygon.a. nonagonb. 18-gonWrite the definitionof rotation under therotation tab. Include asketch that indicates theangle, direction, andcenter of rotation.ORGANIZE ITHow many degrees arein a half-turn rotation?a full-turn rotation?WRITE IT
  • 239. Glencoe Geometry 231Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–4A pattern that a plane bythe same figure or set of figures so that there are nooverlapping or empty spaces is called a tessellation.A regular tessellation is a tessellation formed by only onetype of regular polygon.Tessellations containing the same arrangement of shapesand at each vertex are called uniform.A uniform tessellation formed using two or more regularis called a semi-regular tessellation.• Identify regulartessellations.• Create tessellationswith specific attributes.MAIN IDEASTessellationsWrite the definitionsof tessellation, regulartessellation, anduniform tessellationunder the appropriatetabs. In each case,include a sketch thatillustrates the definition.reflectiontranslationrotationdilationORGANIZE ITRegular PolygonsDetermine whether a regular 16-gon tessellates theplane. Explain.Use the Interior Angle Theorem. Let ∠1 represent one interiorangle of a regular 16-gon.m∠1 =180(n - 2)_n=180(16 - 2)_16orSince is not a factor of 360, a 16-gon will nottessellate the plane.Check Your Progress Determine whether a regular20-gon tessellates the plane. Explain.Semi-Regular TessellationDetermine whether a semi-regular tessellation can becreated from regular nonagons and squares, all havingsides 1 unit long.Each interior angle of a regular nonagon measures 140°. Eachangle of a square measures 90°. Find whole-number values forBUILD YOUR VOCABULARY (pages 220–221)
  • 240. 232 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT9–4REVIEW ITWhat is the sum of theinterior angles of aregular pentagon?(Lesson 8-1)n and s such that 140n + 90s = 360. All whole numbersgreater than 3 will result in a negative value for s.Let n = 1.Let n = 2.140(1) + 90s = 360 140(2) + 90s = 360+ 90s = 360 + 90s = 360= 220 = 80s = s =There are no whole number values for n and s so that140n + 90s = 360.Check Your Progress Determine whether a semi-regulartessellation can be created from regular hexagon and squares,all having sides 1 unit long. Explain.Classify TessellationsSTAINED GLASS Determinewhether the pattern is atessellation. If so, describeit as uniform, regular,semi-regular, or not uniform.The pattern is a tessellation because atthe different verticesthe sum of the angles is . Thetessellation is notuniform because each vertex does not have the samearrangement of shapes and .Check Your Progress Determine whether the pattern is atessellation. If so, describe it as uniform, regular, semi-regular,or not uniform.
  • 241. Glencoe Geometry 233Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–5A dilation is a transformation that changes theof a figure.A dilation is a similarity transformation; that is, dilationsproduce figures.• Determine whethera dilation is anenlargement, areduction, or acongruencetransformation.• Determine the scalefactor for a givendilation.MAIN IDEASDilationsDilation• If ßrß > 1, the dilationis an enlargement.• If 0 < ßrß < 1, thedilation is a reduction.• If ßrß = 1, the dilationis a congruencetransformation.• If r > 0, P′ lies on CPand CP = r · CP.If r < 0, P′ lies on CP′the ray opposite CP,and CP′ = ßrß · CP. Thecenter of a dilation isalways its own image.KEY CONCEPTBUILD YOUR VOCABULARY (pages 220–221)Theorem 9.1If a dilation with center C and a scale factor of r transformsA to E and B to D, then ED = ßrß(AB).Theorem 9.2If P(x, y) is the preimage of a dilation centered at the originwith a scale factor r, then the image is P′(rx, ry).Determine Measures Under DilationsFind the measure of the dilation image or the preimageof−−CD using the given scale factor.a. CD = 15, r = 3Since ßrß > 1, the dilation is an enlargement.C′D′ = ßrß(CD) Dilation Theorem= ßrß = 3, CD = 15= Multiply.b. C′D′ = 7, r = -2_3Since 0 < ßrß < 1, the dilation is a reduction.C′D′ = ßrß(CD) Dilation Theorem= (CD) ßrß = 2_3, C′D′ = 7= CD Multiply each side by 3_2.
  • 242. 234 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–5Check Your Progress Find the measure of thedilation image or the preimage of−−AB using the givenscale factor.a. AB = 16, r = 22 b. A′B′ = 24, r = 2_3Dilations in the Coordinate PlaneCOORDINATE GEOMETRY Trapezoid EFGH has verticesE(-8, 4), F(-4, 8), G(8, 4) and H(-4, -8). Find the imageof trapezoid EFGH after a dilation centered at theorigin with a scale factor of 1_4. Sketch the preimage andthe image. Name the vertices of the image.Preimage (x, y) Image (1_4x, 1_4y)E(-8, 4) E′F(-4, 8) F′G(8, 4) G′H(-4, -8) H′Check Your Progress Triangle ABC has vertices A(-1, 1),B(2, -2), and C(-1, -2). Find the image of △ABC after adilation centered at the origin with a scale factor of 2. Sketchthe preimage and the image.Write the definition ofa dilation under thedilation tab. Thenshow with figures howdilations can result ina larger figure and asmaller figure than theoriginal.ORGANIZE ITWhat image is producedwhen a dilation has ascale factor of r = 1?WRITE IT
  • 243. Glencoe Geometry 235Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT9–5Identify Scale FactorDetermine the scale factor used for each dilation withcenter C. Determine whether the dilation is anenlargement, reduction, or congruence transformation.a. scale factor =image length__preimage length=← image length← preimage length= Simplify.Since the scale factor is less than 1, the dilation is a.b. scale factor =image length__preimage length== Simplify.Since the image falls on the opposite side of the center, C, thanthe preimage, the scale factor is . So the scalefactor is ß-1ß. The absolute value of the scale factor equals 1,so the dilation is a transformation.Check Your Progress Determine the scale factor usedfor each dilation with center C. Determine whether thedilation is an enlargement, reduction, or congruencetransformation.a. b.← imagelength← preimagelength
  • 244. 236 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–6A vector in standard position has its initial point at the.The representation of a vector is calledthe component form of the vector.VectorsVectors A vector is aquantity that has bothmagnitude, or length,and direction, and isrepresented by a directedsegment.KEY CONCEPTWrite Vectors in Component FormWrite the component form of AB.Find the change of x-values and thecorresponding change in y-values.AB = 〈x2 - x1, y2 - y1〉 Component form of vector= 〈 - ( ), x1 = ,y1 = ,- ( )〉 x2 = ,y2 == 〈 , 〉 Simplify.Because the magnitude and direction of a vector are notchanged by translation, the vector representsthe same vector as AB.Check Your Progress Write the component form of AB.BUILD YOUR VOCABULARY (pages 220–221)• Find magnitudes anddirections of vectors.• Perform translationswith vectors.MAIN IDEAS
  • 245. Glencoe Geometry 237Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9–6Write the definition ofvector under the vectortab. Include a sketch toillustrate the definition.ORGANIZE ITEqual Vectors Twovectors are equal if andonly if they have thesame magnitude anddirection.Parallel Vectors Twovectors are parallel ifand only if they have thesame or oppositedirection.KEY CONCEPTSMagnitude and Direction of a VectorFind the magnitude and direction of ST for S(-3, -2)and T(4, -7).Find the magnitude.ßSTß = √(x2 - x1)2 + (y2 - y1)2 Distance Formula= x1 = -3,y1 = -2,x2 = -4,y2 = -7= Simplify.≈ Use a calculator.Graph ST to determine how to findthe direction. Draw a right trianglethat has ST as its hypotenuse andan acute angle at S.tan S =y2 - y1_x2 - x1tan =length of opposite side__length of adjacent side=--Substitution.=So, m∠S = tan-1( ) Simplify.≈ Use a calculator.A vector in standard position that is equal to ST forms aangle with the positive x-axis in the fourthquadrant. So it forms a 360 + (-35.5) or anglewith the positive x-axis.
  • 246. 238 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT9–6Check Your Progress Find the magnitude and directionof AB for A(2, 5) and B(-2, 1).Translations with VectorsGraph the image of quadrilateral HJLK with verticesH(-4, 4), J(-2, 4), L(-1, 2) and K(-3, 1) under thetranslation of v = 〈5, -5〉.First graph quadrilateral HJLK.Next translate each vertex by v,units andunits .Connect the vertices for quadrilateralH′J′L′K′.Check Your Progress Graph the image of triangle ABCwith vertices A(7, 6), B(6, 2), and C(2, 3) under the translationof v 〈-3, -4〉.
  • 247. STUDY GUIDEGlencoe Geometry 239Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R9Use your Chapter 9 Foldableto help you study for yourchapter test.To make a crossword puzzle,word search, or jumble puzzleof the vocabulary words inChapter 9, go to:glencoe.comYou can use your completedVocabulary Builder(pages 220–221) to help yousolve the puzzle.1. Draw the reflected image for areflection of pentagon ABCDEin the origin. Label the imageof ABCDE as A′B′C′D′E′.Determine the number of lines of symmetry for each figuredescribed below. Then determine whether the figure haspoint symmetry and indicate this by writing yes or no.2. a square 3. an isosceles trapezoid4. the letter E 5. a regular hexagonFind the image of each preimage under the indicatedtranslation.6. (x, y); 5 units right and 3 units up7. (x, y); 2 units left and 4 units down8. (-7, 5); 7 units right and 5 units down9-1Reflections9-2Translations
  • 248. Chapter BRINGING IT ALL TOGETHER9240 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.9. △RST has vertices R(-3, 3), S(0, -2),and T(2, 1). Graph △RST and its image△R′S′T′ under the translation(x, y) (x + 3, y - 2). List thecoordinates of the vertices of the image.List all of the following types of transformations thatsatisfy each description: reflection, translation, rotation.10. The transformation is also called a slide.11. The transformation is also called a flip.12. The transformation is also called a turn.Determine the order and magnitude of the rotationalsymmetry for each figure.13. 14.Underline the correct word, phrase, or number to form atrue statement.15. A tessellation that uses only one type of regular polygon iscalled a (uniform/regular/semi-regular) tessellation.16. The sum of the measures of the angles at any vertex in anytessellation is (90/180/360).Write all of the following words that describe eachtessellation: uniform, non-uniform, regular, semi-regular.17. 18.9-3Rotations9-4Tessellations
  • 249. Chapter BRINGING IT ALL TOGETHER9Glencoe Geometry 241Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.19. △XYZ has vertices X(-4, 3), Y(6, 2), and Z(8, -3). Find thecoordinates of the image of △XYZ after a dilation centered atthe origin with a scale factor of 2.Each value of r represents the scale factor for a dilation.In each case, determine whether the dilation is anenlargement, a reduction, or a congruence transformation.20. r = 321. r = 0.522. r = -123. Find the magnitude and direction to the nearest degree ofx = 〈2, -5〉.Write each vector described below in component form.24. a vector with initial point (a, b) and endpoint (c, d)25. a vector in standard position with endpoint (-3, 5)26. a vector with initial point (2, -3) and endpoint (6, -8)9-5Dilations9-6Vectors
  • 250. 242 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureC H A P T E R9ChecklistARE YOU READY FORTHE CHAPTER TEST?Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 9.Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 9 Practice Test onpage 547 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 9 Study Guide and Reviewon pages 543–546 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 9 Practice Test onpage 547.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 9 Foldable.• Then complete the Chapter 9 Study Guide and Review onpages 543–546 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 9 Practice Test onpage 547.
  • 251. Glencoe Geometry 243Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter10Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.C H A P T E R10 CirclesFold two of the circles inhalf and cut one-inch slitsat each end of the folds.Fold the remaining threecircles in half and cut a slitin the middle of the fold.Slide the two circles withslits on the ends throughthe large slit of the othercircles.Fold to make a booklet.Label the cover withthe title of the chapterand each sheet with alesson number.Begin with five sheets of plain 81_2" × 11"paper, and cut out five large circles that arethe same size.10-1CirclesNOTE-TAKING TIP: Take notes in such a mannerthat someone who did not understand the topicwill understand after reading what you havewritten.
  • 252. BUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.244 Glencoe GeometryC H A P T E R10This is an alphabetical list of new vocabulary terms you will learn in Chapter 10.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.Vocabulary TermFoundDefinitionDescription oron Page Examplearccentercentral anglechordcirclecircumferencecircumscribeddiameterinscribed
  • 253. Glencoe Geometry 245Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY10Vocabulary TermFoundDefinitionDescription oron Page Exampleinterceptedmajor arcminor arcpi (π)point of tangencyradiussecantsemicircletangent
  • 254. 246 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–1A circle is the locus of all points in a planefrom a given point called the center of the circle.Any segment with that are on the circle isa chord of the circle.A chord that passes through the is a diameterof the circle.Any segment with endpoints that are the anda point on the circle is a radius.Circles and Circumference• Identify and use partsof circles.• Solve problemsinvolving thecircumference ofa circle.MAIN IDEASREMEMBER ITThe letters d and rare usually used torepresent diameter andradius. So, d = 2r andr = d_2or r = 1_2d.Find Radius and DiameterCircle R has diameters−−ST and−−−QM.a. If ST = 18, find RS.r = 1_2dr = 1_2or Substituteand simplify.b. If RN = 2, find RP.Since all radii are congruent, RN = RP.So, RP = .Check Your Progress Circle M hasdiameters−−BG and−−−DN.a. If BG = 25, find MG.b. If MF = 8.5, find MG.(pages 244–245)BUILD YOUR VOCABULARY
  • 255. Glencoe Geometry 247Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–1Find Measures in Intersecting CirclesThe diameters of ⊙X, ⊙Y, and ⊙Z are 22 millimeters,16 millimeters, and 10 millimeters, respectively. Find EZ.Since the diameter of ⊙X is, EF = .Since the diameter of ⊙Z is, FZ = .−−FZ is part of−−EZ.EF + FZ = EZ Segment Addition Postulate+ = EZ Substitution= EZ Simplify.So, the measure of−−EZ is millimeters.Check Your Progress The diameters of ⊙D, ⊙B, and⊙A are 5 inches, 9 inches, and 18 inches respectively.a. Find AC.b. Find EB.The circumference of a circle is the arounda circle.The ratio is an irrational number called pi (π).BUILD YOUR VOCABULARYUnder the tab forLesson 10-1, sketch acircle and label its parts.List the parts of thecircle and theirdefinitions.10-1CirclesORGANIZE IT(pages 244–245)
  • 256. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.248 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENTFind Circumference, Diameter, and RadiusFind the indicated measure for each circle.a. Find C if r = 13 inches.C = 2πr Circumference formula= 2π Substitution.= π or aboutb. Find C if d = 6 millimeters.C = πd= π or aboutc. Find d and r to the nearest hundredth if C = 65.4 feet.C = πd Circumference formula= πd Substitution65.4_π = d Divide each side by π.r = 1_2d Radius formular = 1_2( ) Substitution.r ≈ Use a calculator.The diameter is approximately feet and the radiusis approximately feet.Check Your Progress Find the indicated measure foreach circle.a. Find C if d = 3 feet.b. Find d and r to the nearest hundredth if C = 16.8 meters.Circumference For acircumference of C unitsand a diameter of d unitsor a radius of r units,C = πd or C = 2πr.KEY CONCEPT10–1
  • 257. Glencoe Geometry 249Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–2A central angle has the of the circle as itsvertex, and its sides contain two radii of the circle.Angles and ArcsMeasures of Central AnglesALGEBRA−−RV is a diameter of ⊙T.Find m∠RTS.The sum of the measures of ∠RTS,∠STU, and ∠UTV is 180.m∠RTS + m∠STU + m∠UTV = 180(8x - 4) + + = 180= 180 Simplify.= 182 Add 2.x = Divide.Use the value of x to find m∠RTS.m∠RTS = 8x - 4 Given= 8 - 4 or SubstitutionCheck Your Progress Find m∠CZD.• Recognize major arcs,minor arcs, semicircles,and central angles andtheir measures.• Find arc length.MAIN IDEASBUILD YOUR VOCABULARYSum of Central AnglesThe sum of the measuresof the central angles of acircle with no interiorpoints in common is 360.Include thisconcept in your notes.KEY CONCEPT(pages 244–245)A central angle separates the circle into two parts, each ofwhich is an arc.BUILD YOUR VOCABULARY (pages 244–245)
  • 258. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.250 Glencoe Geometry10–2Measures of ArcsIn ⊙P, m∠NPM = 46,−−PL bisects ∠KPM, and−−OP ⊥−−−KN.a. Find m⁀OK.⁀OK is a minor arc, so mO⁀OK = m∠KPO.⁀KON is a semicircle.m⁀ON = m∠NPO= ∠NPO is a right angle.m⁀KON = m⁀OK + m⁀ON Arc Addition Postulate= m⁀OK + Substitution= m⁀OK Subtract.b. Find m⁀LM.m⁀LM = 1_2⁀KM since−−PL bisects ∠KPM.⁀KMN is a semicircle.m⁀KM + m⁀MN = m⁀KMN Arc Addition Postulatem⁀KM + = m⁀MN = m∠NPM = 46m⁀KM = Subtract.m⁀LM = 1_2or 67c. m⁀JKO⁀JKO is a major arc.m⁀JKO = m⁀JLM + m⁀MN + m⁀NO Arc Addition Postulatem⁀JKO = 180 + + Substitutionm⁀JKO =Arcs of a CircleA minor arc can benamed by its endpointsand has a measure lessthan 180.A major arc can benamed by its endpointsand another point onthe arc, and its measureis 360 minus the measureof the related minor arc.A semicircle can benamed by its endpointsand another point onthe arc, and its measureis 180.KEY CONCEPTSTheorem 10.1Two arcs are congruent if and only if their correspondingcentral angles are congruent.Postulate 10.1 Arc Addition PostulateThe measure of an arc formed by two adjacent arcs is thesum of the measures of two arcs.
  • 259. Glencoe Geometry 251Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT10–2Check Your Progress In ⊙B,−−XP andand−−YN are diameters, m∠XBN = 108,and−−BZ bisects ∠YBP. Find eachmeasure.a. m⁀YZb. m⁀XY c. m⁀XNZArc LengthIn ⊙B, AC = 9 and m∠ABD = 40. Find the length of ⁀AD.In ⊙B, AC = 9 so C = π(9) or 9π andA BCDm⁀AD = m∠ABD or 40. Write a proportionto compare each part to its whole.degree measure of arc arc lengthdegree measure ofcircumferencewhole circleNow solve the proportion for ℓ.40_360= ℓ Multiply each side by 9π.= ℓ Simplify.The length of ⁀AD is units or about units.Check Your Progress In ⊙A, AY = 21 and m∠XAY = 45.Find the length of ⁀WX.Arc Length Suppose acircle has radius r andcircumference C. If an arcof the circle has degreemeasure A andlength ℓ, then A_360= ℓ_2πrand A_360· C = ℓ.KEY CONCEPT==
  • 260. 252 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–3 Arcs and ChordsProve TheoremsPROOF Write a proof.Given:−−DE−−FG,m∠EBF = 24⁀DFG is a semicircle.Prove: m∠FBG = 78Proof:• Recognize and userelationships betweenarcs and chords.• Recognize and userelationships betweenchords and diameters.MAIN IDEASStatements Reasons1.−−DE−−FG; m∠EBF = 24; 1. Given⁀DFG is a semicircle.2. m⁀DFG = 2. Def. of semicircle3. = 3. In a circle, if 2 chords are ,corr. minor arcs are .4. m⁀DE = m⁀FG 4. Def. of arcs5. m⁀EF = 5. Def. of arc measure6. m⁀ED + m⁀EF + m⁀FG 6. Arc Addition Postulate= m⁀DFG7. m⁀FG + + m⁀FG 7.= 1808. 2(m⁀FG) = 156 8. Subtraction Property andsimplify9. m⁀FG = 9. Division Property10. m⁀FG = ∠mFBG 10.11. m∠FBG = 78 11.Theorem 10.2In a circle or in congruent circles, two minor arcs arecongruent if and only if their corresponding chords arecongruent.
  • 261. Glencoe Geometry 253Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.What is a biconditionalstatement? Include anexample in yourexplanation. (Lesson 2-3)REVIEW IT10–3Check Your Progress Write a proof.Given: ⁀AB ⁀EF−−AB−−−CDProve: ⁀CD ⁀EFProof:A figure is considered inscribed if all of its vertices lie onthe circle.A circle is considered circumscribed about a polygon if itcontains all the vertices of the polygon.BUILD YOUR VOCABULARY (pages 244–245)Theorem 10.3In a circle, if a diameter (or radius) is perpendicular to achord, then it bisects the chord and its arc.Radius Perpendicular to a ChordCircle W has a radius of 10 centimeters.Radius−−−WL is perpendicular to chord−−−HK, which is 16 centimeters long.a. If m⁀HL = 53, find m⁀MK.You know that ⁀HL ⁀LK.m⁀MK + m⁀KL = m⁀MLm⁀MK + = m⁀ML Substitutionm⁀MK + = Substitutionm⁀MK = Subtract.
  • 262. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.254 Glencoe Geometry10–3b. Find JL.Draw radius−−−WK. △WJK is a right triangle.WK = r =−−−HL bisects HK. A radius perpendicular toa chord bisects it.JK = 1_2(HK) Definition of segment bisector.= or HK =Use the Pythagorean Theorem to find WJ.(WJ)2 + (JK)2 = (WK)2 Pythagorean Theorem(WJ)2 + = JK = 8, WK = 10(WJ)2 + = Simplify.(WJ)2 = Subtract 64 from each side.WJ =Take the square root of eachside.WJ + JL = WL Segment addition+ JL = WJ = , WL =JL = Subtract 6 from each side.Check Your Progress Circle O hasa radius of 25 units. Radius−−OC isperpendicular to chord−−AE, whichis 40 units long.a. If m⁀MG = 35, find m⁀CG.b. Find CH.
  • 263. Glencoe Geometry 255Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT10–3Chords Equidistant from CenterChords−−EF and−−−GH are equidistantfrom the center. If the radius of⊙P is 15 and EF = 24, find PR and RH.EෆFෆ and GෆHෆ are equidistant from P,so EෆFෆ GෆHෆ.QF = 1_2EF, so QF = or 12RH = 1_2GH, so RH = or 12Draw−−−PH to form a right triangle. Use the PythagoreanTheorem.(PR)2 + (RH)2 = (PH)2 Pythagorean Theorem(PR)2 + = RH = , PH =(PR)2 + = Simplify.(PR)2 = Subtract.PR = Take the square root ofeach side.Check Your Progress Chords SෆZෆ and UෆVෆ are equidistantfrom the center of ⊙X. If TX is 39 and XY is 15, find WZand UV.Summarize what youhave learned in thislesson about arcs andchords of a circle.Include sketches thatillustrate the importantfacts. Include thissummary under the tabfor Lesson 10-3.10-1CirclesORGANIZE ITTheorem 10.4In a circle or in congruent circles, two chords are congruentif and only if they are equidistant from the center.
  • 264. 256 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–4 Inscribed AnglesTheorem 10.5If an angle is inscribed in a circle, then the measure of theangle equals one-half the measure of its intercepted arc (orthe measure of the intercepted arc is twice the measure ofthe inscribed angle).Measures of Inscribed AnglesIn ⊙F, m⁀WX = 20, m⁀XY = 40,m⁀UZ = 108, and m⁀UW = m⁀YZ.Find the measures of thenumbered angles.First determine m⁀YZ and m⁀UW.Use the Arc Addition Thereom.m⁀WX + m⁀XY + m⁀YZ + m⁀UZ + m⁀UW = 36020 + 40 + m⁀YZ + 108 + m⁀YZ = 360+ 2(m⁀YZ) = Simplify.2(m⁀YZ) = Subtract.m⁀YZ = Divide.So, m⁀YZ = and m⁀UW = . Now find the measuresof the numbered angles.m∠1 = 1_2m⁀UW m∠2 = 1_2m⁀XY= 1_2or = 1_2orm∠3 = 1_2m⁀UZ m∠5 = 1_2m⁀UZ= 1_2or = 1_2orm∠2 + m∠3 + m∠4 = 180+ + m∠4 = 180m∠4 =• Find measures ofinscribed angles.• Find measures of anglesof inscribed polygons.MAIN IDEAS
  • 265. Glencoe Geometry 257Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–4Check Your Progress In ⊙A, m⁀XY = 60, m⁀YZ = 80, andm⁀WX = m⁀WZ. Find the measures of the numbered angles.Theorem 10.6If two inscribed angles of a circle (or congruent circles)intercept congruent arcs or the same arc, then the anglesare congruent.Theorem 10.7If an inscribed angle intercepts a semicircle, the angle is aright angle.Angles of an Inscribed TriangleALGEBRA Triangles TVU and TSUare inscribed in ⊙P with ⁀VU ⁀SU.Find the measure of each numberedangle if m∠2 = x + 9 and m∠4 = 2x + 6.△UVT and △UST are right triangles.m∠1 = m∠2 since they intercept congruentarcs. Then the third angles of the trianglesare also congruent, so m∠3 = m∠4.m∠2 + m∠4 + m∠S = 180+ + = 180= 180 Simplify.= Subtract.x = Divide.Use the value of x to find the measures of ∠1, ∠2, ∠3, and ∠4.m∠2 = x + 9 m∠4 = 2x + 6= + 9 or 34 = 2 + 6 or 56m∠1 = m∠2 = m∠3 = m∠4 =Explain how to find themeasure of an inscribedangle in a circle if youknow the measure ofthe intercepted arc.Include your explanationunder the tab forLesson 10-4.10-1CirclesORGANIZE IT
  • 266. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.258 Glencoe Geometry10–4Angles of an Inscribed QuadrilateralQuadrilateral QRST is inscribed in ⊙M. If m∠Q = 87and m∠R = 102, find m∠S and m∠T.Draw a sketch of this situation.To find m∠S, we need to know m⁀RQT.To find m⁀RQT, first find m⁀RST.m⁀RST = 2(m∠Q) Inscribed Angle Theorem= 2΂ ΃or 174 m∠Q =m⁀RST + m⁀RQT = 360 Sum of angles = 360+ m⁀RQT = 360 m⁀RST =m⁀RQT = Subtract.m⁀RQT = 2(m∠S) Inscribed Angle Theorem= 2(m∠S) Substitution= (m∠S) Divide.To find m∠T, we need to know m⁀QRS, but first find m⁀STQ.m⁀STQ = 2(m∠R) Inscribed Angle Theorem= 2΂ ΃or 204 m∠R =m⁀STQ + m⁀QRS = 360 Sum of angles in circle = 360+ m⁀QRS = 360 m⁀STQ =m⁀QRS = Subtract.m⁀QRS = 2(m∠T) Inscribed Angle Theorem= 2(m∠T) m⁀QRS == m∠T Divide.
  • 267. Glencoe Geometry 259Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT10–4Check Your Progressa. Triangles MNO and MPO are inscribedin ⊙D with ⁀MN ⁀OP. Find the measureof each numbered angle if m∠2 = 4x - 8and m∠3 = 3x + 9.b. BCDE is inscribed in ⊙X. If m∠B = 99 and m∠C = 76,find m∠D and m∠E.Theorem 10.8If a quadrilateral is inscribed in a circle, then its oppositeangles are supplementary.
  • 268. 260 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–5 TangentsFind LengthsALGEBRA−−RS is tangent to ⊙Q atpoint R. Find y.Use the Pythagorean Theorem to find QR,which is one-half the length y.(SR)2 + (QR)2 = (SQ)2 Pythagorean Theorem+ (QR)2 = SR = , SQ =+ (QR)2 = Simplify.(QR)2 = 144 Subtract from each side.QR = Take the square root of each side.Because y is the length of the diameter, ignore the negativeresult. Thus, y is twice QR or y = .Check Your Progress CෆDෆ is a tangentto ⊙B at point D. Find a.• Use properties oftangents.• Solve problemsinvolving circumscribedpolygons.MAIN IDEASTheorem 10.9If a line is tangent to a circle, then it is perpendicular to theradius drawn to the point of tangency.Theorem 10.10If a line is perpendicular to a radius of a circle at its endpointon the circle, then the line is a tangent to the circle.Theorem 10.11If two segments from the same exterior point are tangentto a circle, then they are congruent.A ray is tangent to a circle if the line containing the rayintersects the circle in exactly one point. This point is calledthe point of tangency.BUILD YOUR VOCABULARY (pages 244–245)
  • 269. Glencoe Geometry 261Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT10–5Congruent TangentsALGEBRA Find y. Assume thatsegments that appear tangentto circles are tangent.−−−DE and−−−DF are each drawn fromthe same exterior point and areboth tangent to ⊙S, so−−−DE−−−DF.Likewise,−−−DG and−−−DH are bothdrawn from D and are tangentto ⊙T, so−−−DG−−−DH. From theSegment Addition Postulate,DG = DE + EG andDH = DF + FH.DG = DH Definition of congruent segmentsDE + EG = DF + FH Substitution+ y - 5 = y + DE = , EG = y - 5,DF = y, FH =y + 5 = 2y + 4 Simplify.5 = y + 4 Subtract y from each side.= y Subtract 4 from each side.Triangles Circumscribed About a CircleTriangle HJK is circumscribedLJMKHGN1618about ⊙G. Find the perimeterof △HJK if NK = JL + 29.Use Theorem 10.10. JM = JL = ,LH = HN = , and NK = .Since NK = JL + 29, NK = 16 + 29 or 45. Then MK = 45.P = JM + MK + HN + NK + JL + LH= 16 + 45 + 18 + 45 + 16 + 18 or unitsCheck Your Progress TriangleNOT is circumscribed about ⊙M.Find the perimeter of △NOT ifCT = NC - 28.Explain what a tangentto a circle is. Provide asketch to illustrate theexplanation. Includeyour explanation andsketch under the tab forLesson 10-5.10-1CirclesORGANIZE ITWrite the PythagoreanTheorem in words andin symbols. Include adiagram. (Lesson 8-2)REVIEW IT
  • 270. 262 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–6 Secants, Tangents, and Angle MeasuresSecant-Secant AngleFind m∠4 if m⁀FG = 88 and m⁀EH = 76.Method 1m∠3 = 1_2΂m⁀FG + m⁀EH΃= 1_2( + )orm∠4 = 180 - m∠3= - orMethod 2Find m⁀EF + m⁀GH.m⁀EF + m⁀GH = 360 - ΂m⁀FG + m⁀EH΃= 360 - ( + )= 360 - orm∠4 = 1_2΂m⁀EF + m⁀GH΃= 1_2΂ ΃orCheck Your Progress Find m∠5 ifm⁀AC = 63 and m⁀XY = 21.• Find measures ofangles formed bylines intersecting onor inside a circle.• Find measures ofangles formed bylines intersectingoutside the circle.MAIN IDEASTheorem 10.12 If two secants intersect in the interior ofa circle, then the measure of an angle formed is one-halfthe sum of the measure of the arcs intercepted by theangle and its vertical angle.A line that intersects a circle in exactly two points is calleda secant.BUILD YOUR VOCABULARY (pages 244–245)
  • 271. Glencoe Geometry 263Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–6Theorem 10.13If a secant and a tangent intersect at the point of tangency,then the measure of each angle formed is one-half themeasure of its intercepted arc.Theorem 10.14If two secants, a secant and a tangent, or two tangentsintersect in the exterior of a circle, then the measure of theangle formed is one-half the positive difference of themeasures of the intercepted arcs.Secant-Tangent AngleFind m∠RPS if m⁀PT = 114 and m⁀TS = 136.m⁀PS = 360 - m⁀PTS= 360 - ΂ + ΃=m∠RPS = 1_2m⁀PS= 1_2΂ ΃orCheck Your Progress Find m∠FEGif m⁀HF = 80 and m⁀HE = 164.Secant-Secant AngleFind x.= 1_2(141 - x) Theorem10.14= 141 - x Multiplyeach side by 2.= 141 Add x to each side.x = Subtract 124 from each side.
  • 272. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.264 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT10–6Check Your Progress Find x.Secant-Tangent AngleFind x.55 = 1_2(6x - 40)= - Multiply each side by 2.= Add to each side.= x Divide each sideby .Check Your Progress Find x.Devise a chart thatsummarizes how themeasure of an angleformed by chords,secants, or tangents isrelated to the measuresof intercepted arcs.Include your chart underthe tab for Lesson 10-6.10-1CirclesORGANIZE IT
  • 273. Glencoe Geometry 265Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–7 Special Segments in a CircleIntersection of Two ChordsFind x.To find x, use Theorem 10.15.· = ·8x = 108 Multiply.x = Divide each side by 8.Theorem 10.15If two chords intersect in a circle, then the products of themeasures of the segments of the chords are equal.Theorem 10.16If two secant segments are drawn to a circle from anexterior point, then the product of the measures of onesecant segment and its external secant segment is equal tothe product of the measures of the other secant segmentand its external secant segment.Intersection of Two SecantsFind x if EF = 10, EH = 8,and FG = 24.EH · EI = EF · EG Secant Segment Products· (8 + x) = · (10 + 24) Substitution64 + 8x = Distributive Property8x = 276 Subtract 64 from each side.x = Divide each side by .• Find measures ofsegments that intersectin the interior of acircle.• Find measures ofsegments that intersectin the exterior of acircle.MAIN IDEAS
  • 274. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.266 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENTCheck Your Progressa. Find x. b. Find x if GO = 27,OM = 25, and KI = 24.Intersection of Secant and a TangentFind x. Assume that segments thatappear to be tangent are tangent.(x + 4)2 = x(x + x + 2)x2 + 8x + 16 = x(2x + 2)x2 + 8x + 16 = 2x2 + 2x0 =0 =x - 8 = 0 x + 2 = 0x = x = Disregard.Check Your Progress Find x.Assume that segments that appearto be tangent are tangent.Theorem 10.17 If a tangent segment and a secantsegment are drawn to a circle from an exterior point, thenthe square of the measure of the tangent segment is equalto the product of the measures of the secant segment andits external secant segment.10–7Write a paragraphto describe what youhave learned about therelationships betweenthe lengths of specialsegments in a circle.Include your paragraphunder the tab forLesson 10-7.10-1CirclesORGANIZE IT
  • 275. Glencoe Geometry 267Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.10–8 Equations of CirclesEquation of a CircleWrite an equation for the circle with center at (3, -3),d = 12.If d = 12, r = 6.(x - h)2 + (y - k)2 = r2 Equation ofa circle΂ ΃2+ [y - (-3)]2 = (h, k) = (3, -3),r = 6= Simplify.Check Your Progress Write an equation foreach circle.a. center at (0, -5), d = 18b. center at (7, 0), r = 20• Write the equation of acircle.• Graph a circle on thecoordinate plane.MAIN IDEASStandard Equation of aCircle An equation for acircle with center at(h, k) and radius ofr units is(x - h)2 + (y - k)2 = r2.KEY CONCEPT
  • 276. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.268 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENTGraph a CircleGraph (x - 2)2 + (y + 3)2 = 4.Compare each expression in the equation to the standard form.(x - h)2 = (y - k)2 =x - h = y - k =-h = -k =h = k =r2 = , so r = .The center is at , andthe radius is .Graph the center. Use a compassset to a width of grid squares todraw the circle.Check Your Progress Graph x2 + (y - 5)2 = 25.10–8
  • 277. Glencoe Geometry 271Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER1022. Find m∠2. 23. Find x.Supply the missing length to completeeach equation.24. BH · HD = FH ·25. AD · AE = AB · 26. AF · AC =΂ ΃2Write an equation for each circle.27. center at origin, r = 828. center at (3, 9), r = 1Write an equation for each circle.29. 30.10-6Secants, Tangents, and Angle Measures10-7Special Segments in a Circle
  • 278. Glencoe Geometry 269Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.STUDY GUIDEBRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R10Use your Chapter 10 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 10, go to:glencoe.comYou can use your completedVocabulary Builder(pages 244–245) to help yousolve the puzzle.1. In ⊙A, if BD = 18, find AE.Refer to the figure.2. Name four radii of the circle.3. Name two chords of the circle.Refer to ⊙P. Indicate whether each statement is true or false.4. ⁀DAB is a major arc.5. ⁀ADC is a semicircle.6. ⁀AD ⁀CD7. ⁀DA and ⁀AB are adjacent arcs.Refer to ⊙P. Give each of the following arc measures.8. m⁀AB 9. m⁀BC10. m⁀DAB 11. m⁀DAC10-1Circles and Circumference10-2Angles and Arcs
  • 279. 270 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER10If ⊙P has a diameter 40 centimeters long, andAC = FD = 24 centimeters, find each measure.12. PA 13. HEIn ⊙Q, RS = VW and m⁀RS = 70. Find eachmeasure.14. m⁀RT 15. m⁀VWRefer to the figure. Find each measure.16. m∠ABC 17. m∠⁀AD18. m∠BCA 19. m∠⁀BCD20. Two segments from P are tangent to ⊙O. If m∠P = 120 and theradius of ⊙O is 8 feet, find the length of each tangent segment.21. Each side of a circumscribed equilateral triangle is 10 meters.Find the radius of the circle.10-3Arcs and Chords10-4Inscribed Angles10-5Tangents
  • 280. Glencoe Geometry 271Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER1022. Find m∠2. 23. Find x.Supply the missing length to completeeach equation.24. BH · HD = FH ·25. AD · AE = AB · 26. AF · AC =΂ ΃2Write an equation for each circle.27. center at origin, r = 828. center at (3, 9), r = 1Write an equation for each circle.29. 30.10-6Secants, Tangents, and Angle Measures10-7Special Segments in a Circle
  • 281. 272 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistC H A P T E R10ARE YOU READY FORTHE CHAPTER TEST?Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 10 Practice Test onpage 625 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 10 Study Guide and Reviewon pages 620–624 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 10 Practice Test onpage 625.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 10 Foldable.• Then complete the Chapter 10 Study Guide and Review onpages 620–624 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 10 Practice Test onpage 625.Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 10.
  • 282. Glencoe Geometry 273Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter11Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: When you take notes, write asummary of the lesson, or write in your ownwords what the lesson was about.Stack 4 of the 5 sheetsof notebook paperas illustrated.Cut in about 1 inchalong the heading lineon the top sheet of paper.Cut the marginsoff along theright edge.Stack in order of cuts,placing the uncut fifthsheet at the back. Labelthe tabs as shown. Stapleedge to form a book.Begin with five sheets of notebook paper.11-111-211-311-411-5C H A P T E R11 Areas of Polygons and Circles
  • 283. 274 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 11.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.apothemcomposite figuregeometric probabilityheight of a parallelogramsectorsegmentVocabulary TermFoundDefinitionDescription oron Page ExampleC H A P T E R11
  • 284. Glencoe Geometry 275Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–1Perimeter and Area of a ParallelogramFind the perimeter and area of RSTU.Base and Side:Each base is inches long, andeach side is inches long.Perimeter:The perimeter of RSTU is 2( )+ 2( )orinches.Height:Use a 30°-60°-90° triangle to find the height. Recall that if themeasure of the leg opposite the 30° angle is x, then the lengthof the hypotenuse is , and the length of the leg oppositethe 60° angle is x√3.24 = 2x Substitute 24 for the hypotenuse.12 = x Divide each side by 2.So, the height is x√3 or inches.Area:A = bh== or about square inches.Check Your ProgressFind the perimeter andarea of DEFG.• Find perimeters andareas of parallelograms.• Determine whetherpoints on a coordinateplane define aparallelogram.MAIN IDEASArea of a ParallelogramIf a parallelogram has anarea of A square units, abase of b units, and aheight of h units, thenA = bh.KEY CONCEPTAreas of Parallelograms
  • 285. 276 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT11–1Area of a ParallelogramThe Kanes want to sod a portion of their yard. Find thenumber of square yards of grass needed to sod theshaded region in the diagram.The area of the shaded region is the sum of two rectangles.The dimensions of the first rectangle are 50 feet by 150 - 40or 110 feet. The dimensions of the second rectangle are150 - 60 or 90 feet and 50 + 100 or 150 feet.Area of shaded region = Area of Rectangle 1 + Area ofRectangle 2= += += square feetNext, change square feet to square yards.19,000 ft2 ×1 yd2_9 ft2≈The Kanes need approximately square yardsof sod.Under the tab forLesson 11-1, make asketch to show howa parallelogram canbe cut apart andreassembled to forma rectangle. Write theformula for the areaof the parallelogram.11-111-211-311-411-5ORGANIZE IT
  • 286. Glencoe Geometry 277Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–2 Areas of Triangles, Trapezoids, and RhombiArea of TrianglesFind the area of quadrilateral ABCD ifAC = 35, BF = 18, and DE = 10.area of ABCD= area of △ABC + area of △ADC= 1_2bh + 1_2bh Area formula= + Substitution= square units Simplify.Check Your Progress Find thearea of quadrilateral HIJK if IK = 16,HL = 5, and JM = 9.Area of a Trapezoid on the Coordinate PlaneFind the area of trapezoid RSTUwith vertices R(4, 2), S(6, -1),T(-2, -1) and U(-1, 2).Bases:Since−−−UR and−−TS are horizontal, findtheir length by subtracting thex-coordinates of their endpoints.UR =  -  TS =  - =  or =  orHeight:Because the bases are horizontal segments, the distancebetween them can be measured on a vertical line. So,subtract the y-coordinates to find the trapezoid’s height.• Find areas of triangles.• Find areas of trapezoidsand rhombi.MAIN IDEASArea of a Triangle If atriangle has an areaof A square units, abase of b units, and acorresponding height ofh units, then A = 1_2bh.Area of a Trapezoid If atrapezoid has an area ofA square units, bases ofb1 units and b2 units anda height of h units, thenA = 1_2h(b1 + b2).KEY CONCEPTS
  • 287. 278 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–2h = 2 - ( ) orArea:A = 1_2h(b1 + b2)= 1_2( )( )+ ( )or square unitsArea of a Rhombus on the Coordinate PlaneFind the area of rhombus MNPRwith vertices at M(0, 1), N(4, 2),P(3, -2), and R(-1, -3).Let−−−MP be d1 and−−−NR be d2.Use the Distance Formula to find MP.d1 = √(x2 - x1)2 + (y2 - y1)2= (0 - 3)2 + ΄ ΅2= √18 orUse the Distance Formula to find NR.d2 = ΄ ΅2+ ΄ ΅2= orA = 1_2d1d2 Area of a rhombus= 1_2( )( )or square unitsCheck Your Progress Find the area of each figure.a. b.Area of a Rhombus If arhombus has an area ofA square units anddiagonals of d1 and d2units, then A = 1_2d1d2.Solve threeproblems under the tabfor Lesson 11-2: find thearea of a triangle, findthe area of a trapezoid,and find the area of arhombus.KEY CONCEPT
  • 288. Glencoe Geometry 279Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTAlgebra: Find Missing MeasuresRhombus RSTU has an area of64 square inches. Find US ifRT = 8 inches.Use the formula for the area of arhombus and solve for d2.A = 1_2d1d2 Area of rhombus= 1_2( )(d2) A = , d1 =64 = 4(d2) Multiply.= d2−−−US is inches long.Check Your Progressa. Rhombus ABCD has an area of81 square centimeters. Find BDif AC = 6 centimeters.b. Trapezoid QRST hasan area of 210 squareyards. Find the heightof QRST.11–2
  • 289. 280 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–3 Areas of Regular Polygons and CirclesArea of a Regular PolygonFind the area of a regularpentagon with a perimeterof 90 meters.Apothem:The central angles of a regularpentagon are all congruent.Therefore, the measure of eachangle is 360_5or 72.−−−GF is anapothem of pentagon ABCDE. Itbisects ∠EGD and is a perpendicular bisector of−−−ED. So,m∠DGF = 1_2(72) or 36. Since the perimeter is 90 meters, eachside is 18 meters and FD = 9 meters.Write a trigonometric ratio to find the length of−−GF.tan∠DGF = DF_GFtan θ =length of opposite side__length of adjacent sidetan = _GFm∠DGF = ,DF =(GF)tan = Multiply each side by GF.GF = __tanDivide each side by tan.GF ≈ Use a calculator.• Find areas of regularpolygons.• Find areas of circles.MAIN IDEASAn apothem is a segment that is drawn from theof a regular polygon to a side of thepolygon.BUILD YOUR VOCABULARY (page 278)Area of a RegularPolygon If a regularpolygon has an areaof A square units, aperimeter of P units,and an apothem of aunits, then A = 1_2Pa.Write theformula for the area of aregular polygon underthe tab for Lesson 11-3.KEY CONCEPT
  • 290. Glencoe Geometry 281Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–3Area: A = 1_2Pa Area of a regular polygon≈ P = , a ≈≈ 558 Simplify.The area of the pentagon is about square meters.Check Your Progress Find the areaof a regular pentagon with a perimeterof 120 inches.Use Area of a Circle to Solve a Real-World ProblemMANUFACTURING An outdooraccessories companymanufactures circular coversfor outdoor umbrellas. If thecover is 8 inches longer thanthe umbrella on each side,find the area of the cover insquare yards.The diameter of the umbrella is 72 inches, and the cover mustextend 8 inches in each direction. So the diameter of the coveris + + or inches. Divide by 2 tofindthat the radius is 44 inches.A = πr2 Area of a circle= π Substitution≈ Use a calculator.The area of the cover is square inches. Toconvert to square yards, divide by 1296. The area of thecover is square yards to the nearest tenth.Check Your Progress A swimming poolcompany manufactures circular coversfor above-ground pools. If the cover is10 inches longer than the pool on each side,find the area of the cover in square yards.Area of a Circle If a circlehas an area of A squareunits and a radius ofr units, then A = πr2.KEY CONCEPT
  • 291. 282 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT11–3Area of an Inscribed PolygonFind the area of the shaded region.Assume that the triangle is equilateral.The area of the shaded region is thedifference between the area of the circleand the area of the triangle. First, findthe area of the circle.A = πr2 Area of a circle= π Substitution≈ Use a calculator.To find the area of the triangle, useproperties of 30°-60°-90° triangles.First, find the length of the base. Thehypotenuse of △RSZ is 7, so RS is 3.5and SZ is 3.5√3. Since YZ = 2(SZ),YZ = 7√3.Next, find the height of the triangle,XS. Since m∠XZY is 60,XS = 3.5√3 (√3) or 10.5.Use the formula to find the area of the triangle.A = 1_2bh= 1_2( )( )≈The area of the shaded region isor approximately squarecentimeters to the nearest tenth.Check Your Progress Find the area ofthe shaded region. Assume that the triangleis equilateral. Round to the nearest tenth.REVIEW ITDraw a 30°-60°-90°triangle with the shorterleg labeled 5 meterslong. Label the anglesand the remaining sides.(Lesson 8-3)
  • 292. Glencoe Geometry 283Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–4A composite figure is a figure that be classifiedinto the specific shapes that we have studied.Areas of Composite FiguresFind the Area of a Composite FigureA rectangular rose garden is centered in a border oflawn. Find the area of the lawn around the garden insquare feet.One method to find the area of the lawn around the garden isto find the total area and then subtract the area of the garden.The overall length of the lawn and garden is 25 + 100 + 25or feet. The overall width of the lawn and garden is25 + 20 + 25 or feet.Area of lawn = Area of lawn and garden - Area of garden= ℓ1w1 - ℓ2w2 Area formulas= (150)(70) - (100)(20) ℓ1 = 150, w1 = 70,ℓ2 = 100, w2 = 20= 10,500 - 2000 Multiply.= Subtract.The area of the lawn around the garden is squarefeet.BUILD YOUR VOCABULARY (page 274)• Find areas of compositefigures.• Find areas of compositefigures on thecoordinate plane.MAIN IDEASExplain how to find thearea of a compositefigure. Include theexplanation under thetab for Lesson 11-4. Alsoinclude an example toshow how to find thearea of such a figure.11-111-211-311-411-5ORGANIZE ITPostulate 11.2The area of a region is the sum of the areas of all of itsnonoverlapping parts.
  • 293. 284 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT11–4Coordinate PlaneFind the area of polygonMNPQR.First, separate the figure intoregions. Draw an auxiliary lineperpendicular to−−−QR from M(we will call this point ofintersection S) and an auxiliaryline from N to the x-axis (we willcall this point of intersection K).This divides the figure into triangle MRS, triangle NKM,trapezoid POKN, and trapezoid PQSO.Now, find the area of each of the figures. Find the differencebetween x-coordinates to find the lengths of the bases of thetriangles and the lengths of the bases of the trapezoids. Findthe difference between y-coordinates to find the heights of thetriangles and trapezoids.area of MNPQR= area of △MRS + area of △NKM + area of trapezoid POKN+ area of trapezoid PQSO= 1_2bh + 1_2bh + 1_2h(b1 + b2) + 1_2h(b1 + b2)= + ++=The area of polygon MNPQR is square units.Check Your Progress Find the areaof polygon ABCDE.REMEMBER ITEstimate the area ofthe figure by countingthe unit squares. Use theestimate to determineif your answer isreasonable.
  • 294. Glencoe Geometry 285Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.11–5Probability that involves a geometric measure such aslength or area is called geometric probability.A sector of a circle is a region of a circle bounded by aand its .Probability with SectorsRefer to the figure.a. Find the total area of theshaded sectors.The shaded sectors have degreemeasures of 45 and 35 or 80° total. Usethe formula to find the total area of theshaded sectors.A = N_360πr2 Area of a sector= _________360π( )2N = , r == Simplify.b. Find the probability that a point chosen at randomlies in the shaded region.To find the probability, divide the area of the shaded sectorsby the area of the circle. The area of the circle is πr2 with aradius of 9.P(shaded) = area of sectors__area of circleGeometric probability formula= _________π ·area of circle =≈ Use a calculator.The probability that a random point is in the shadedsectors is about .• Solve problemsinvolving geometricprobability.• Solve problemsinvolving sectors andsegments of circles.MAIN IDEASProbability and AreaIf a point in region A ischosen at random, thenthe probability P(B) thatthe point is in region B,which is in the interiorof region A, isP(B) =area of region B__area of region A.Area of a SectorIf a sector of a circle hasan area of A squareunits, a central anglemeasuring N°, and aradius of r units, thenA = N_360πr2.KEY CONCEPTSBUILD YOUR VOCABULARY(page 274)Geometric Probabilityarea of circle =
  • 295. 286 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.The region of a circle bounded by an arc and ais called a segment of a circle.11–5Probability with SegmentsA regular hexagon is inscribed in a circlewith a diameter of 12.a. Find the area of the shaded regions.Area of a sector:A = N_360πr2 Area of a sector= _360π( 2) N = , r == or about Simplify. Then usea calculator.Area of a triangle:Since the hexagon was inscribed inthe circle, the triangle is equilateral,with each side 6 units long. Useproperties of 30°-60°-90° trianglesto find the apothem. The value of xis 3 and the apothem is x√3 or3√3, which is approximately 5.20.BUILD YOUR VOCABULARY (page 274)Check Your Progress Find thearea of the shaded sectors. Then findthe probability that a point chosen atrandom lies in the shaded regions.
  • 296. Glencoe Geometry 287Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT11–5Next, use the formula for the area of a triangle.A = 1_2bh Area of a triangle≈ 1_2b = , h ≈≈ Simplify.Area of segment:area of one segment = area of sector - area of triangle≈ - Substitution≈ Simplify.Since three segments are shaded, we will multiply this by 3.3( )=The area of the shaded regions about squareunits.b. Find the probability that a point chosen at randomlies in the shaded regions.Divide the area of the shaded regions by the area of the circleto find the probability. First, find the area of the circle. Theradius is 6, so the area is or about 113.10 square units.P(shaded) =area of shaded region__area of circle≈ or aboutThe probability that a random point is on the shaded regionis about or %.Check Your Progress A regularhexagon is inscribed in a circle witha diameter of 18. Find the area of theshaded regions. Find the probabilitythat a point chosen at random liesin the shaded regions.Write a paragraph tosummarize what youhave learned aboutfinding geometricprobabilities. Illustrateyour remarks withsketches. Include yourparagraph and sketchesunder the tab forLesson 11-5.11-111-211-311-411-5ORGANIZE IT
  • 297. STUDY GUIDE288 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R11Use your Chapter 11 Foldableto help you study for yourchapter test.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 11, go to:glencoe.comYou can use your completedVocabulary Builder (page 274)to help you solve the puzzle.Refer to the figure. Determine whether each statement istrue or false. If the statement is false, explain why.1.−−AB is an altitude of the parallelogram.2.−−−CD is a base of parallelogram ABCD.3. Find the perimeter and area of the parallelogram.Round to the nearest tenth if necessary.Find the area of each quadrilateral.4. 5.11-1Areas of Parallelograms11-2Area of Triangles, Trapezoids, and Rhombi
  • 298. Glencoe Geometry 289Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER116. Find the area of a regular pentagon with a perimeter of100 meters.7. Find the area of the shaded region to the nearesttenth. Assume that the polygon is regular.Find the area of the shaded region of each figure to thenearest tenth.8. 19.10. Suppose you are playing a game of darts with a targetlike the one shown at the right. If your dart landsinside equilateral △UVW, you get a point. Assumethat every dart will land on the target. The radius ofthe circle is 1. Find the probability of getting a point.Round to the nearest thousandth.11-3Areas of Regular Polygons and Circles11-5Geometric Probability11-4Areas of Composite Figures
  • 299. 290 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistC H A P T E R11ARE YOU READY FORTHE CHAPTER TEST?Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 11 Practice Test onpage 675 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 11 Study Guide and Reviewon pages 672–674 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 11 Practice Test onpage 675.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 11 Foldable.• Then complete the Chapter 11 Study Guide and Review onpages 672–674 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 11 Practice Test onpage 675.Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 11.
  • 300. Glencoe Geometry 291Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter12NOTE-TAKING TIP: When taking notes, place aquestion mark next to anything you do notunderstand. Then be sure to ask questions beforeany quizzes or tests.Fold lengthwiseleaving a two-inchtab.Fold the paperinto five sections.Open. Cut alongeach fold to makefive tabs.Label as shown.Begin with a sheet of 11" × 17" paper.Surface AreaPrismsCylindersPyramidsConesSpheresExtending Surface AreaC H A P T E R12Use the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.
  • 301. 292 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Vocabulary TermFoundDefinitionDescription oron Page Exampleaxisgreat circlehemispherelateral arealateral edgeslateral facesThis is an alphabetical list of new vocabulary terms you will learn in Chapter 12.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.C H A P T E R12
  • 302. Glencoe Geometry 293Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BUILD YOUR VOCABULARY12Vocabulary TermFoundDefinitionDescription oron Page Exampleoblique conereflection symmetryregular pyramidright coneright cylinderright prism[PRIZ-uhm]slant height
  • 303. 294 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–1 Representations of Three-Dimensional FiguresDraw a SolidSketch a triangular prism 6 units high with bases thatare right triangles with legs 6 units and 4 units long.STEP 1 Draw thecorner ofthe solid:4 units up,6 units to theleft, and 6 unitsto the right.STEP 2 Draw aparallelogramfor the backof the solid.Draw thehypotenuse ofone base.STEP 3 Draw a dashedline 6 unitslong.STEP 4 Connect thecorrespondingvertices. Usedashed linesfor the hiddenedges.• Draw isometric views ofthree-dimensionalfigures.• Investigate crosssections of three-dimensional figures.MAIN IDEAS
  • 304. Glencoe Geometry 295Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–1Check Your ProgressSketch a rectangular prism1 unit high, 5 units long, and2 units wide.Use Orthographic DrawingsDraw the corner view of the figure given itsorthographic drawing.• The top view indicates one row of different heights andone column in the front right.• The front view indicates that there arestanding columns. The first column to the left isblocks high, the second column is blocks high,the third column is blocks high, and the fourthcolumn to the far right is block high. The darksegments indicate breaks in the surface.• The right view indicates that the front right column isonly block high. The dark segments indicatebreaks in the surface.• The columnshould be visible.Connect the dots onthe isometric dotpaper to representthe solid.What views are includedin an orthographicdrawing? Is each viewthe same shape?WRITE IT
  • 305. 296 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Check Your Progress Draw the corner view of the figuregiven its orthogonal drawing.BAKERY A customer ordered a two-layer sheet cake.Describe and draw the possible cross sections ofthe cake.If the cake is cut , the cross section will be a.If the cake is cut , the cross section will be a. Sketch the cross sections.Horizontal: Vertical:12–1Sketch a prism, pyramid,cylinder, cone, andsphere. Write the nameof each figure belowthe sketch. Includeeach sketch under theappropriate tab.Surface AreaPrismsCylindersPyramidsConesSpheresORGANIZE IT
  • 306. Glencoe Geometry 297Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–1Page(s):Exercises:HOMEWORKASSIGNMENTCheck Your Progress ARCHITECTUREAn architect is building a scale model ofthe Great Pyramids of Egypt. Describethe possible cross sections of the model.If the pyramid is cut verticallythe cross section is a triangle.If the pyramid is cut horizontallythe cross section is a square.
  • 307. 298 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–2 Surface Areas of PrismsLateral Area of a Hexagonal PrismFind the lateral area of the regular hexagonal prism.The bases are regular hexagons.So the perimeter of one base is6(5) or 30 centimeters.L = Ph= ( )( )= square centimetersCheck Your Progress Findthe lateral area of the regularoctagonal prism.• Find lateral areas ofprisms.• Find surface areas ofprisms.MAIN IDEASLateral Area of a PrismIf a right prism has alateral area of L squareunits, a height of h units,and each base has aperimeter of P units,then L = Ph.KEY CONCEPTIn a prism, the faces that are not are calledlateral faces.The lateral faces intersect at the lateral edges. Lateraledges are segments.A prism with lateral edges that are alsois called a right prism.The lateral area L is the sum of the of thelateral faces.BUILD YOUR VOCABULARY (pages 292–293)
  • 308. Glencoe Geometry 299Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT12–2Surface Area of a Square PrismFind the surface area of the square prism.T = L + 2B Surface area of a prism= + L = Ph= + Substitution= Simplify.The surface area is square centimeters.Check Your Progress Find the surface area of thetriangular prism.Surface Area of a PrismIf the surface area of aright prism is T squareunits, its height is hunits, and each base hasan area of B square unitsand a perimeter of Punits, then T = L + 2B.KEY CONCEPTSketch a rightprism. Below the sketch,explain how finding theperimeter of the basecan help you calculatethe surface area of theprism. Include the sketchand explanation underthe tab for Prisms.
  • 309. 300 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–3 Surface Areas of CylindersLateral Area of a CylinderMANUFACTURING A fruit juice can is cylindrical withaluminum sides and bases. The can is 12 centimeterstall, and the diameter of the can is 6.3 centimeters. Howmany square centimeters of aluminum are used to makethe sides of the can?The aluminum sides of the can represent thearea of the cylinder. If the diameter of the can is 6.3centimeters, then the radius is centimeters.The height is 12 centimeters. Use the formula to find thelateral area.L = 2πrh Lateral area of a cylinder= 2π( )( ) r = , h =≈ Use a calculator.About square centimeters of aluminum are usedto make the sides of the can.• Find lateral areas ofcylinders.• Find surface areas ofcylinders.MAIN IDEASLateral Area of aCylinder If a rightcylinder has a lateralarea of L square units,a height of h units, andthe bases have radii ofr units, then L = 2πrh.Surface Area of aCylinder If a rightcylinder has a surfacearea of T square units,a height of h units,and the bases haveradii of r units, thenT = 2πrh + 2πr2.KEY CONCEPTSThe axis of a cylinder is the segment with endpoints thatare of the circular bases.An altitude of a cylinder is a segment that isto the bases of the cylinder and hasits endpoints on the bases.If the axis is also the , then the cylinder iscalled a right cylinder. Otherwise, the cylinder is an obliquecylinder.BUILD YOUR VOCABULARY (pages 292–293)Take notesabout cylinders underthe Cylinders tab.
  • 310. Glencoe Geometry 301Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT12–3Check Your Progress A set of toy blocks are sold in acylindrical shape container. A product label wraps around allsides of the container without any overlaps or gaps. How muchpaper is used to make the label the appropriate size if thediameter of the container is 12 inches and the height is18 inches?Surface Area of a CylinderFind the surface area of the cylinder.The radius of the base and theheight of the cylinder aregiven. Substitute these valuesin the formula to find thesurface area.T = 2πrh + 2πr2 Surface area ofa cylinder= 2π( )( )+ 2π( )2r = ,h =≈ Use a calculator.The surface area is approximately square feet.Check Your Progress Find the surface area ofthe cylinder.REVIEW ITWhat is the area of acircle with a diameterof 7 centimeters?(Lesson 11-3)
  • 311. 302 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–4 Surface Areas of PyramidsUse Lateral Area to Solve a ProblemCANDLES A candle store offers a pyramidal candle thatburns for 20 hours. The square base is 6 centimeterson a side and the slant height of the candle is 22centimeters. Find the lateral area of the candle.The sides of the base measure 6 centimeters, so the perimeteris 4(6) or 24 centimeters.L = 1_2Pℓ Lateral area of a regular pyramid= P = , ℓ == Multiply.The lateral area of the candle is square centimeters.Check Your Progress A pyramidal shaped tent is putup by two campers. The square base is 7 feet on a side andthe slant height of the tent is 7.4 feet. Find the lateral areaof the tent.Surface Area of a Square PyramidFind the surface area of the square pyramid. Round tothe nearest tenth if necessary.To find the surface area, first find theslant height of the pyramid. The slantheight is the hypotenuse of a righttriangle with legs that are the altitudeand a segment with a length that isone-half the side measure of the base.c2 = a2 + b2 Pythagorean Theorem= Replace a, b, and ℓ.ℓ ≈ Use a calculator.• Find lateral areas ofregular pyramids.• Find surface areas ofregular pyramids.MAIN IDEASLateral Area of aRegular Pyramid If aregular pyramid has alateral area of L squareunits, a slant height of ℓunits, and its base has aperimeter of P units,then L = 1_2Pℓ.Surface Area of aRegular Pyramid If aregular pyramid has asurface area of T squareunits, a slant height of ℓunits, and its base has aperimeter of P units, andan area of B squareunits, thenT = 1_2Pℓ + B.KEY CONCEPTSDefine thebasic properties ofpyramids under thePyramids tab. Includea sketch of a pyramidwith the parts of apyramid labeled.
  • 312. Glencoe Geometry 303Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–4Now find the surface area of the regular pyramid. The perimeterof the base is 4(8) or 32 meters, and the area of the base is 82 or64 square meters.T = 1_2Pℓ + B Surface area of a regularpyramidT ≈ 1_2( )( )+ Replace P, ℓ, and B.T ≈ Use a calculator.The surface area is square meters to the nearest tenth.Check Your Progress Find thesurface area of the regular pyramidto the nearest tenth.Surface Area of a Pentagonal PyramidFind the surface area of the regularpyramid. Round to the nearest tenth.The altitude, slant height, and apothem forma right triangle. Use the PythagoreanTheorem to find the apothem. Let xrepresent the length of the apothem.c2 = a2 + b2 Pythagorean Theorem= + b = , c == a Simplify.Now find the length of the sides of the base. The central angleof the hexagon measures 360°_6or 60°. Let a represent themeasure of the angle formed by a radius and the apothem.Then a = 60_2or 30.
  • 313. Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.304 Glencoe GeometryPage(s):Exercises:HOMEWORKASSIGNMENT12–4Use trigonometry to find the length ofthe sides.tan 30° = _______ tan A =opposite_adjacent9(tan 30°) = Multiply each side by 9.(tan 30°) = s Multiply each side by 2.≈ s Use a calculator.Next, find the perimeter and area of the base.P = 6s≈ 6( )orB = 1_2Pa≈ 1_2 ( )( )orFinally, find the surface area.T = 1_2Pℓ + B Surface area of aregular pyramid≈ 1_2 ( )( )+ Replace P, ℓ, and B.≈ Simplify.The surface area is approximately squarecentimeters.Check Your Progress Find thesurface area of the regular pyramid.
  • 314. Glencoe Geometry 305Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–5 Surface Areas of ConesLateral Area of a ConeICE CREAM A sugar cone has an altitudeof 8 inches and a diameter of 21_2inches.Find the lateral area of the sugar cone.Use the Pythagorean Theorem. Write anequation and solve for ℓ.ℓ2 =2+2ℓ2 =ℓ ≈Next, use the formula for the lateral area of aright circular cone.L = πrℓ≈ π( )( )≈The lateral area is approximatelysquare inches.Check Your Progress A hat fora child’s birthday party has a conicalshape with an altitude of 9 inchesand a diameter of 5 inches. Find thelateral area of the birthday hat.• Find lateral areas ofcones.• Find surface areas ofcones.MAIN IDEASLateral Area of a ConeIf a right circular conehas a lateral area of Lsquare units, a slantheight of ℓ units, andthe radius of the base isr units, then L = πrℓ.KEY CONCEPTThe shape of a tepee suggests a circular cone.BUILD YOUR VOCABULARY (pages 292–293)
  • 315. 306 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT12–5Surface Area of a ConeFind the surface area of the cone. Round to thenearest tenth.T = πrℓ + πr2 Surface area of a cone= π( )( )+ π( )2r = ,ℓ =≈ Use a calculator.The surface area is approximately squarecentimeters.Check Your Progress Find the surface area of the cone.Round to the nearest tenth.Surface Area of a ConeIf a right circular conehas a surface area of Tsquare units, a slantheight of ℓ units, andthe radius of the base isr units, thenT = πrℓ + πr2.KEY CONCEPTSketch a rightcircular cone. Use theletters r, h, and ℓ toindicate the radius,height and slant height,respectively. Write theformula for the surfacearea. Include all thisunder the tab for Cones.
  • 316. Glencoe Geometry 307Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–6 Surface Areas of SpheresSpheres and CirclesIn the figure, O is the centerof the sphere, and plane Pintersects the sphere in thecircle R. If OR = 6 centimetersand OS = 14 centimeters,find RS.The radius of circle R is segment , and S is a point oncircle R and on sphere O. Use the Pythagorean Theorem forright triangle ORS to solve for RS.OS2 = RS2 + OR2 Pythagorean Theorem= RS2 + OS = ,OR == RS2 + Simplify.= RS2 Subtract fromeach side.≈ RS Use a calculator.RS is approximately centimeters.• Recognize and definebasic properties ofspheres.• Find surface areas ofspheres.MAIN IDEAS When a plane intersects a sphere so that it contains theof the sphere, the intersection is called agreat circle.Each great circle separates a sphere into, each called a hemisphere.BUILD YOUR VOCABULARY (pages 292–293)
  • 317. 308 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.12–6Surface Areaa. Find the surface area of the sphere, given a greatcircle with an area of approximately 907.9 squarecentimeters.The surface area of a sphere is four times the area of thegreat circle.T = 4πr2 Surface area of a sphere= 4( ) πr 2 ≈= Multiply.The surface area is approximately squarecentimeters.b. Find the surface area of a hemisphere with a radiusof 3.8 inches.A hemisphere is half of a sphere. To find the surface area,find half of the surface area of the sphere and add the areaof the great circle.T = 1_2(4πr2) + πr2 Surface area of ahemisphere= 1_2[4π( )2] + π( )2Substitution≈ Use a calculator.The surface area is approximately squareinches.Check Your Progress In the figure, O is the center ofthe sphere, and plane U intersects the sphere in circle L. IfOL = 3 inches and LM = 8 inches, find OM.Surface Area of a SphereIf a sphere has a surfacearea of T square unitsand a radius of r units,then T = 4πr2.KEY CONCEPTDefine theterms great circle andhemisphere under theSpheres tab. Also,include the formula forfinding the surface areaof a sphere.
  • 318. Glencoe Geometry 309Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENTREVIEW ITWhat is thecircumference of acircle with a radius of6 centimeters?(Lesson 10-1)Check Your Progressa. Find the surface area of the sphere, given a great circlewith an area of approximately 91.6 square centimeters.b. Find the surface area of a hemisphere with a radius of6.4 inches.A ball is a sphere with a circumference of 24 inches.Find the approximate surface area of the ball to thenearest tenth of a square inch.To find the surface area, first find the radius of the sphere.C = 2πr Circumference of a circle= 2πr C =_______ = r Divide each side by .= r Simplify.Next, find the surface area of the sphere.T = 4πr2 Surface area of a sphere.≈ 4π( )2r =≈Use a calculator.The surface area is approximately square inches.Check Your Progress Find the approximate surface areaof a ball with a circumference of 18 inches to the nearest tenthof a square inch.12–6
  • 319. 310 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.STUDY GUIDEBRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERC H A P T E R12Identify each solid. Name the bases, faces, edges, and vertices.1.2.12-1Representations of Three-Dimensional FiguresUse your Chapter 12 Foldable tohelp you study for your chaptertest.To make a crossword puzzle,word search, or jumblepuzzle of the vocabulary wordsin Chapter 12, go to:glencoe.comYou can use your completedVocabulary Builder(pages 292–293) to help yousolve the puzzle.
  • 320. Glencoe Geometry 311Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER12Refer to the figure.3. Name this solid with as specific a name as possible.4. Name the bases of the solid.5. Name the lateral faces.6. Name the edges.7. Name an altitude of the solid.8. The lateral area of a prism is 90 square inches andthe perimeters of its base is 15 inches. Find the height.Underline the correct word or phrase to form a true statement.9. The bases of a cylinder are (rectangles/regular polygons/circles).10. The (axis/radius/diameter) of a cylinder is the segment with endpointsthat are the centers of the bases.11. The net of a cylinder is composed of two congruent (rectangles/circles) andone (rectangle/semicircle).12. In a right cylinder, the axis of the cylinders is also a(n) (base/lateraledge/altitude).13. A cylinder that is not a right cylinder is called an (acute/obtuse/oblique)cylinder.12-2Surface Areas of Prisms12-3Surface Areas of Cylinders
  • 321. 312 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER1214. Find the lateral area andsurface area of the cylinder.Round to the nearest tenth.In the figure, ABCDE has congruent sides and congruent angles.15. Use the figure to name the base of this pyramid.16. Describe the base of the pyramid.17. Name the vertex of the pyramid.18. Name the altitude of the pyramid.19. Write an expression for the height of the pyramid.20. Write an expression for the slant height of the pyramid.21. Find the lateral area and surface area of theregular figure. Round to the nearest tenth.A right circular cone has a radius of 7 meters and aslant height of 13 meters.22. Find the lateral area to the nearest tenth.23. Find the surface area to the nearest tenth.24. Suppose you have a right cone with radius r, diameter d,height h, and slant height ℓ. Which of the followingrelationships involving these lengths are correct?a. r = 2d b. r + h = ℓ c. r2 + h2 = ℓ2d. r2 + ℓ2 = h2 e. r = √ℓ2 - h2 f. h = ±√ℓ2 - r212-4Surface Areas of Pyramids12-5Surface Areas of Cones
  • 322. Glencoe Geometry 313Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER12In the figure, P is the center of the sphere. Nameeach of the following in the figure.25. two chords of the sphere26. a great circle of the sphere27. a tangent to the sphere28. Complete: A sphere is separated by a great circle into two congruenthalves, each called a(n) .Determine whether each sentence is sometimes, always, or never true.29. If a sphere and a plane intersect in more than one point, theirintersection will be a great circle.30. A great circle has the same center as the sphere.31. A chord of a sphere is a diameter of the sphere.32. T = πrℓ + πr2 a. regular pyramid33. T = PH + 2B b. hemisphere34. T = 4πr2 c. cylinder35. T = 1_2Pℓ + B d. prism36. T = 2πrh + 2πr2 e. sphere37. T = 3πr2 f. cone38. A sphere has a radius that is 28 inches long. Find the surface area to thenearest tenth.39. The radius of a sphere is doubled. How is the surface area changed?12-6Surface Areas of Spheres
  • 323. 314 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistC H A P T E R12ARE YOU READY FORTHE CHAPTER TEST?Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 12 Practice Test onpage 723 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 12 Study Guide and Reviewon pages 719–722 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 12 Practice Test onpage 723 of your textbook.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 12 Foldable.• Then complete the Chapter 12 Study Guide and Review onpages 719–722 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 12 Practice Test onpage 723 of your textbook.Visit glencoe.com to accessyour textbook, moreexamples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 12.
  • 324. Glencoe Geometry 315Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter13C H A P T E R13 Extending VolumeUse the instructions below to make a Foldable to help youorganize your notes as you study the chapter. You will seeFoldable reminders in the margin of this Interactive StudyNotebook to help you in taking notes.NOTE-TAKING TIP: When you take notes ingeometry, be sure to make comparisons amongthe different formulas and concepts. For example,how are pyramids and cones similar? different?This will help you learn the material.Fold in thirds.Fold inhalflengthwise.Label asshown.Unfold book. Drawlines along folds andlabel as shown.Begin with one sheet of 11" × 17" paper.
  • 325. 316 Glencoe GeometryBUILD YOUR VOCABULARYCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.This is an alphabetical list of new vocabulary terms you will learn in Chapter 13.As you complete the study notes for the chapter, you will see Build YourVocabulary reminders to complete each term’s definition or description onthese pages. Remember to add the textbook page number in the secondcolumn for reference when you study.C H A P T E R13Vocabulary TermFoundDefinitionDescription oron Page Examplecongruent solidsordered triplesimilar solids
  • 326. Glencoe Geometry 317Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–1 Volumes of Prisms and CylindersVolume of a Triangular PrismFind the volume of thetriangular prism.V = Bh= 1_2(10)( )( )=The volume of the prism is cubic centimeters.Check Your Progress Findthe volume of the triangular prism.Volume of a Rectangular PrismThe weight of water is 0.036 pound times the volumeof water in cubic inches. How many pounds of waterwould fit into a rectangular child’s pool that is12 inches deep, 3 feet wide, and 4 feet long?First, convert feet to inches.3 feet = 3 × or inches4 feet = 4 × or inches• Find volumes of prisms.• Find volumes ofcylinders.MAIN IDEASVolume of a Prism If aprism has a volume of Vcubic units, a height of hunits, and each base hasan area of B squareunits, then V = Bh.KEY CONCEPT
  • 327. 318 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–1To find the pounds of water that would fit into the child’s pool,find the volume of the pool.V = Bh Volume of a prism= 36(48)( ) B = 36(48), h ==Now multiply the volume by 0.036.× 0.036 ≈ Simplify.A rectangular child’s pool that is 12 inches deep, 3 feet wide,and 4 feet long, will hold about pounds of water.Check Your Progress The weight of water is 62.4 poundsper cubic foot. How many pounds of water would fit into a backyard pond that is rectangular prism 3 feet deep, 7 feet wide, and12 feet long?Volume of a CylinderFind the volume of the cylinder.The height h is centimeters,and the radius r is centimeters.V = πr2h Volume of a cylinder= π( )2( ) r = , h =≈ Use a calculator.The volume is approximately cubic centimeters.Volume of a CylinderIf a cylinder has a volumeof V cubic units, a heightof h units, and the baseshave radii of r units,then V = Bh or V = πr2h.Explain howto find the volume of aprism and a cylinder. Putthe explanations undertheir respective tabs inthe Foldable.KEY CONCEPT
  • 328. Glencoe Geometry 319Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT13–1Check Your Progress Find the volume of eachcylinder to the nearest tenth.a. b.Volume of an Oblique CylinderFind the volume of the obliquecylinder to the nearest tenth.To find the volume, use the formulafor a right cylinder.V = πr2h Volume of a cylinder= π( )2( ) r = , h =≈ Use a calculator.The volume is approximately cubic feet.Check Your Progress Find thevolume of the oblique cylinder to thenearest tenth.Cavalieri’s PrincipleIf two solids have thesame height and thesame cross-sectional areaat every level, then theyhave the same volume.KEY CONCEPTREMEMBER ITThe diameter of thebase, the diagonal,and the lateral edgeof the cylinder forma right triangle.
  • 329. 320 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–2 Volumes of Pyramids and ConesVolume of a PyramidCLOCKS Teofilo has a solid clock that is in the shape ofa square pyramid. The clock has a base of 3 inches anda height of 7 inches. Find the volume of the clock.V = 1_3Bh Volume of a pyramid= 1_3s2h B = s2= 1_3( )2( ) s = , h == Multiply.The volume of the clock is cubic inches.Check Your Progress Brad isbuilding a model pyramid for a socialstudies project. The model is a squarepyramid with a base edge of 8 feetand a height of 6.5 feet. Find thevolume of the pyramid.• Find volumes ofpyramids.• Find volumes ofcones.MAIN IDEASVolume of a PyramidIf a pyramid has avolume of V cubic units,a height of h units, anda base with an area ofB square units, thenV = 1_3Bh.KEY CONCEPT
  • 330. Glencoe Geometry 321Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–2Volumes of ConesFind the volume of the cone to the nearest tenth.V = 1_3πr2h Volume of a cone= 1_3π( )2( ) r = , h =≈ Use a calculator.The volume is approximately cubic inches.Check Your Progress Find the volume of each cone tothe nearest tenth.a. b.Volume of an Oblique ConeFind the volume of the obliquecone to the nearest tenth.V = 1_3Bh Volume of a cone= 1_3h B == 1_3π( )2( ) r ≈ , h =≈ Use a calculator.The volume is approximately cubic inches.Volume of a Cone If aright circular cone has avolume of V cubic units,a height of h units, andthe base has a radius of runits, thenV = 1_3Bh or V = 1_3πr2h.KEY CONCEPTExplain how to find thevolume of a pyramidand a cone. Put theexplanations under theirrespective tabs in theFoldable.ORGANIZE IT
  • 331. 322 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT13–2Check Your Progress Findthe volume of the oblique cone tothe nearest tenth.
  • 332. Glencoe Geometry 323Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–3 Volumes of Spheres• Find volumes ofspheres.• Solve problemsinvolving volumesof spheres.MAIN IDEASVolume of a Sphere If asphere has a volume of Vcubic units and a radiusof r units, then V = 4_3πr3.KEY CONCEPTVolumes of SpheresFind the volume of each sphere. Round to thenearest tenth.a.V = 4_3πr3 Volume of a sphere= 4_3π(3) r =≈ Use a calculator.The volume is approximately cubiccentimeters.b.First find the radius of the sphere.C = 2πr Circumference of a circle= 2πr C == r Solve for r.Now find the volume.V = 4_3πr3 Volume of a sphere= 4_3π( )3r =≈ Use a calculator.The volume is approximately cubic centimeters.
  • 333. 324 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT13–3Check Your Progress Find the volume of each sphereto the nearest tenth.a. b.Volume of a HemisphereFind the volume of the hemisphere.The volume of a hemisphere is one-halfthe volume of the sphere.V = 1_2(4_3πr3) Volume of a hemisphere= π(3)3 r = 3≈ Use a calculator.The volume of the hemisphere is approximatelycubic feet.Check Your Progress Find thevolume of the hemisphere.Write a paragraph thatincludes formulas forthe volume of a sphere,cylinder, and cone.Then describe how thevolumes of these threesolids are related. Putyour paragraph underthe Spheres tab.ORGANIZE IT
  • 334. Glencoe Geometry 325Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–4Similar solids are solids that have exactly the samebut not necessarily the same .Congruent solids are exactly the same andexactly the same .Congruent and Similar Solids• Identify congruent orsimilar solids.• State the properties ofsimilar solids.MAIN IDEASCongruent Solids Twosolids are congruent ifthe correspondingangles are congruent,the corresponding edgesare congruent, thecorresponding faces arecongruent, and thevolumes are equal.KEY CONCEPTBUILD YOUR VOCABULARY (page 316)Similar and Congruent SolidsDetermine whether each pair of solids is similar,congruent, or neither.a. Find the ratiosbetween thecorrespondingparts of thesquare pyramids.base edge of larger pyramid___base edge of smaller pyramid=5√5_2√5Substitution= Simplify.height of larger pyramid___height of smaller pyramid=5√7_2_√7Substitution= Simplify.lateral edge of larger pyramid___lateral edge of smaller pyramid=5√3_2√3Substitution= Simplify.The ratios of the measures are equal, so we can concludethat the pyramids are similar. Since the scale factor is not1, the solids are not congruent.
  • 335. 326 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–4b. Compare the ratios betweenthe corresponding parts ofthe cones.radius of larger cone__radius of smaller cone= ______ or Use substitution.Then simplify.height of larger cone__height of smaller cone= _______ SubstitutionSince the ratios are not the same, the cones are neithersimilar nor congruent.Check Your Progress Determine whether each pair ofsolids is similar, congruent, or neither.a. b.Describe the differencesbetween congruentand similar solids underthe Similar tab in theFoldables.ORGANIZE ITREVIEW ITAn architect drewblueprints of a house,where the 44-footlength of the housewas represented by20 inches. What is thescale factor of theblueprints comparedto the real house?(Lesson 7-2)SportsSOFTBALL Softballs and baseballs฀are both used to play a game witha bat. A softball has a diameter of3.8 inches, while a baseball hasa diameter of 2.9 inches.a. Find the scale factor of the two balls.Write the ratio of the radii. The scale factor of the two ballsis 3.8 : 2.9 or about : .Theorem 13.1If two solids are similar with a scale factor of a:b, then thesurface areas have a ratio of a2:b2, and the volumes have aratio of a3:b3.REMEMBER ITAll spheres aresimilar, just as all circlesare similar.
  • 336. Glencoe Geometry 327Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT13–4b. Find the ratio of the surface areas of the two balls.If the scale factor is a:b, then the ratio of the surface areasis a2:b2.surface area of the larger ball___surface area of the smaller ball= a2_b2Theorem 13.1≈1.32_______2≈The ratio of the surface areas is about .c. Find the ratio of the volumes of the two balls.If the scale factor is a:b, then the ratio of the volumesis a3:b3.volume of the larger ball___volume of the smaller ball= a3_b3Theorem 13.1≈3_______13≈The ratio of the volume of the two balls is about .Check Your Progress Twosizes of balloons are beingused for decorating at aparty. When fully inflated,the balloons are spheres.The first balloon has adiameter of 18 inches whilethe second balloon has aradius of 7 inches.a. Find the scale factor of the two balloons.b. Find the ratio of the surface areas of the two balloons.c. Find the ratio of the volumes of the two balloons.
  • 337. 328 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.13–5A point in space is represented by an ordered triple of realnumbers (x, y, z).Coordinates in SpaceGraph a Rectangular SolidGraph the rectangular solidthat contains the orderedtriple A(-3, 1, 2) and theorigin as vertices. Label thecoordinates of each vertex.• Plot the x-coordinate first. Draw a segment from theorigin units in the negative direction.• To plot the y-coordinate, draw a segment unit in thepositive direction.• Next, to plot the z-coordinate, draw a segment unitslong in the positive direction.• Label the coordinate A.• Draw the rectangular prism and label each vertex.Check Your Progress Graph the rectangular solid thatcontains the ordered triple N(1, 2, -3) and the origin. Label thecoordinates of each vertex.• Graph solids in space.• Use the Distance andMidpoint Formulas forpoints in space.MAIN IDEASBUILD YOUR VOCABULARY (page 316)REMEMBER ITThe three planesdetermined by the axesof a three-dimensionalcoordinate systemseparate space intoeight regions. Theseregions are calledoctants.Write a short paragraphto explain how to grapha rectagular solid.Record your paragraphunder the Prisms tab.ORGANIZE IT
  • 338. Glencoe Geometry 329Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Page(s):Exercises:HOMEWORKASSIGNMENT13–5Distance and Midpoint Formulas in Spacea. Determine the distance between F(4, 0, 0) andG(-2, 3, -1).FG = √(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2 DistanceFormula inSpace= + +Substitution= Simplify.b. Determine the midpoint M of−−FG.M = (x1 + x2_2,y1 + y2_2,z1 + z2_2 ) MidpointFormula inSpace= ( , , )Substitution= ( ) Simpliify.Check Your Progressa. Determine the distance between A(0, -5, 0) andB(1, -2, -3).b. Determine the midpoint M of−−AB.Distance Formula inSpace Given two pointsA(x1, y1, z1) and B(x2, y2,z2) in space, the distancebetween A and B is givenby the followingequation.d = (x2 - x1)2 +(y2 - y1)2 + (z2 - z1)2Midpoint Formula inSpace Given two pointsA(x1, y1, z1) andB(x2, y2, z2) in space, themidpoint of−−AB is at(x1 + x2_2,y1 + y2_2,z1 + z2_2 ).KEY CONCEPTS
  • 339. STUDY GUIDE330 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.BRINGING IT ALL TOGETHERBUILD YOURVOCABULARYVOCABULARYPUZZLEMAKERUse your Chapter 13 Foldableto help you study for yourchapter test.To make a crossword puzzle,word search, or jumble puzzleof the vocabulary words inChapter 13, go to:glencoe.comYou can use your completedVocabulary Builder (page 316) tohelp you solve the puzzle.C H A P T E R13In each case, write a formula for the volume V of the solid interms of the given variables.1. a rectangular box with length a, width b, and height c2. a cylinder with height h whose bases each have diameter d3. The volume of a rectangular prism is 224 cubic centimeters, thelength is 7 centimeters, and the height is 8 centimeters. Findthe width.4. Find the volume to the nearest tenth.13-1Volumes of Prisms and Cylinders13-2Volumes of Pyramids and Cones
  • 340. Glencoe Geometry 331Copyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Chapter BRINGING IT ALL TOGETHER13Let r represent the radius and d represent the diameter of asphere. Determine whether each formula below can be usedto find the volume of a sphere, a hemisphere, or neither.5. V = 2πr3_36. V = 1_3πr37. V = 1_6πd38. Determine whether thecylinders are congruent,similar, or neither.9. If the ratio of the diameters of two spheres is 3:1, then the ratioof their surface areas is , and the ratio of theirvolumes is .For the following questions, A(-4, 3, 12) and B(7, 2, 8).10. Find the coordinates of the midpoint of−−AB.11. Find the distance between A and B.12. If B is the midpoint of−−AC, find the coordinates of C.13-3Volumes of Spheres13-4Congruent and Similar Solids13-5Coordinates in Space
  • 341. 332 Glencoe GeometryCopyright©Glencoe/McGraw-Hill,adivisionofTheMcGraw-HillCompanies,Inc.Student Signature Parent/Guardian SignatureTeacher SignatureChecklistARE YOU READY FORTHE CHAPTER TEST?Visit glencoe.com toaccess your textbook,more examples, self-checkquizzes, and practice teststo help you study theconcepts in Chapter 13.Check the one that applies. Suggestions to help you study aregiven with each item.I completed the review of all or most lessons without usingmy notes or asking for help.• You are probably ready for the Chapter Test.• You may want to take the Chapter 13 Practice Test onpage 769 of your textbook as a final check.I used my Foldable or Study Notebook to complete the reviewof all or most lessons.• You should complete the Chapter 13 Study Guide and Reviewon pages 765–768 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 13 Practice Test onpage 769.I asked for help from someone else to complete the review ofall or most lessons.• You should review the examples and concepts in your StudyNotebook and Chapter 13 Foldable.• Then complete the Chapter 13 Study Guide and Review onpages 765–768 of your textbook.• If you are unsure of any concepts or skills, refer back to thespecific lesson(s).• You may also want to take the Chapter 13 Practice Test onpage 769.C H A P T E R13