Chapter 5 Resource Masters 5 Test, Form 3 .....59 Chapter 5 Extended-Response Test .....61 Standardized Test Practice .....62 Answers

Chapter 5 Resource Masters

CONSUMABLE WORKBOOKS Many of the worksheets contained in the Chapter Resource Masters are available as consumable workbooks in both English and Spanish.

ISBN10 ISBN13

Study Guide and Intervention Workbook 0-07-890848-5 978-0-07-890848-4

Homework Practice Workbook 0-07-890849-3 978-0-07-890849-1

Spanish Version

Homework Pratice Workbook 0-07-890853-1 978-0-07-890853-8

Answers for Workbooks The answers for Chapter 5 of these workbooks can be found in the back of this Chapter Resource Masters booklet.

StudentWorks PlusTM This CD-ROM includes the entire Student Edition test along with the English workbooks listed above.

TeacherWorks PlusTM All of the materials found in this booklet are included for viewing, printing, and editing in this CD-ROM.

Spanish Assessment Masters (ISBN10: 0-07-89085-6, ISBN13: 978-0-07-890856-9) These masters contain a Spanish version of Chapter 5 Test Form 2A and Form 2C.

Copyright © by the McGraw-Hill Companies, Inc. All rights reserved. Permission is granted to reproduce the material contained herein on the condition that such material be reproduced only for classroom use; be provided to students, teachers, and families without charge; and be used solely in conjunction with Glencoe Geometry. Any other reproduction, for use or sale, is prohibited without prior written permission of the publisher.

Send all inquiries to:Glencoe/McGraw-Hill8787 Orion PlaceColumbus, OH 43240 - 4027

ISBN: 978-0-07-890514-8

MHID: 0-07-890514-1

Printed in the United States of America.

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Teacher’s Guide to Using the Chapter 5

Resource Masters .............................................iv

Chapter ResourcesStudent-Built Glossary ....................................... 1

Anticipation Guide (English) .............................. 3

Anticipation Guide (Spanish) ............................. 4

Lesson 5-1Bisectors of Triangles

Study Guide and Intervention ............................ 5

Skills Practice .................................................... 7

Practice .............................................................. 8

Word Problem Practice ..................................... 9

Enrichment ...................................................... 10

Lesson 5-2Medians and Altitudes of Triangles

Study Guide and Intervention .......................... 11

Skills Practice .................................................. 13

Practice ............................................................ 14

Word Problem Practice ................................... 15

Enrichment ...................................................... 16

Lesson 5-3Inequalites in One Triangle


Skills Practice .................................................. 19

Practice ............................................................ 20


Enrichment ...................................................... 22

Cabri Jr. .......................................................... 23

Geometer’s Sketchpad Activity ...................... 24

Lesson 5-4Indirect Proof


Skills Practice .................................................. 27

Practice ............................................................ 28


Enrichment ...................................................... 30

Lesson 5-5The Triangle Inequality


Skills Practice .................................................. 33

Practice ............................................................ 34


Enrichment ...................................................... 36

Lesson 5-6Inequalities Involving Two Triangles


Skills Practice .................................................. 39

Practice ............................................................ 40


Enrichment ...................................................... 42

AssessmentStudent Recording Sheet ................................ 43

Rubric for Extended-Response ....................... 44

Chapter 5 Quizzes 1 and 2 ............................. 45

Chapter 5 Quizzes 3 and 4 ............................. 46

Chapter 5 Mid-Chapter Test ............................ 47

Chapter 5 Vocabulary Test ............................. 48

Chapter 5 Test, Form 1 ................................... 49

Chapter 5 Test, Form 2A ................................. 51

Chapter 5 Test, Form 2B ................................. 53

Chapter 5 Test, Form 2C ................................ 55

Chapter 5 Test, Form 2D ................................ 57

Chapter 5 Test, Form 3 ................................... 59

Chapter 5 Extended-Response Test ............... 61

Standardized Test Practice ............................. 62

Answers ........................................... A1–A29

Contents

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Teacher’s Guide to Using the

Chapter 5 Resource Masters

Chapter Resources

Student-Built Glossary (pages 1–2) These masters are a student study tool that presents up to twenty of the key vocabulary terms from the chapter. Students are to record definitions and/or examples for each term. You may suggest that students highlight or star the terms with which they are not familiar. Give this to students before beginning Lesson 5-1. Encourage them to add these pages to their mathematics study notebooks. Remind them to complete the appropriate words as they study each lesson.

Anticipation Guide (pages 3–4) This master, presented in both English and Spanish, is a survey used before beginning the chapter to pinpoint what students may or may not know about the concepts in the chapter. Students will revisit this survey after they complete the chapter to see if their perceptions have changed.

Lesson Resources

Study Guide and Intervention These masters provide vocabulary, key concepts, additional worked-out examples and Check Your Progress exercises to use as a reteaching activity. It can also be used in conjunction with the Student Edition as an instructional tool for students who have been absent.

Skills Practice This master focuses more on the computational nature of the lesson. Use as an additional practice option or as homework for second-day teaching of the lesson.

Practice This master closely follows the types of problems found in the Exercises section of the Student Edition and includes word problems. Use as an additional practice option or as homework for second-day teaching of the lesson.

Word Problem Practice This master includes additional practice in solving word problems that apply the concepts of the lesson. Use as an additional practice or as homework for second-day teaching of the lesson.

Enrichment These activities may extend the concepts of the lesson, offer a historical or multicultural look at the concepts, or widen students’ perspectives on the mathematics they are learning. They are written for use with all levels of students.

Graphing Calculator, TI-Nspire, or

Spreadsheet Activities These activities present ways in which technology can be used with the concepts in some lessons of this chapter. Use as an alternative approach to some concepts or as an integral part of your lesson presentation.

The Chapter 5 Resource Masters includes the core materials needed for Chapter 5. These materials include worksheets, extensions, and assessment options. The answers for these pages appear at the back of this booklet.

All of the materials found in this booklet are included for viewing and printing on the

TeacherWorks PlusTM CD-ROM.

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Assessment OptionsThe assessment masters in the Chapter 5

Resource Masters offer a wide range of assessment tools for formative (monitoring) assessment and summative (final) assessment.

Student Recording Sheet This master corresponds with the standardized test practice at the end of the chapter.

Extended-Response Rubric This master provides information for teachers and students on how to assess performance on open-ended questions.

Quizzes Four free-response quizzes offer assessment at appropriate intervals in the chapter.

Mid-Chapter Test This 1-page test provides an option to assess the first half of the chapter. It parallels the timing of the Mid-Chapter Quiz in the Student Edition and includes both multiple-choice and free-response questions.

Vocabulary Test This test is suitable for all students. It includes a list of vocabulary words and 10 questions to assess students’ knowledge of those words. This can also be used in conjunction with one of the leveled chapter tests.

Leveled Chapter Tests

• Form 1 contains multiple-choice questions and is intended for use with below grade level students.

• Forms 2A and 2B contain multiple-choice questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Forms 2C and 2D contain free-response questions aimed at on grade level students. These tests are similar in format to offer comparable testing situations.

• Form 3 is a free-response test for use with above grade level students.

All of the above mentioned tests include a free-response Bonus question.

Extended-Response Test Performance assessment tasks are suitable for all students. Sample answers and a scoring rubric are included for evaluation.

Standardized Test Practice These three pages are cumulative in nature. It includes three parts: multiple-choice questions with bubble-in answer format, griddable questions with answer grids, and short-answer free-response questions.

Answers• The answers for the Anticipation Guide

and Lesson Resources are provided as reduced pages.

• Full-size answer keys are provided for the assessment masters.

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Chapter 5 1 Glencoe Geometry

5

This is an alphabetical list of the key vocabulary terms you will learn in Chapter 5.

As you study the chapter, complete each term’s definition or description.

Remember to add the page number where you found the term. Add these pages to

your Geometry Study Notebook to review vocabulary at the end of the chapter.

Student-Built Glossary

Vocabulary TermFound

on PageDefinition/Description/Example

altitude

centroid

circumcenterSUHR·kuhm·sen·tuhr

concurrent lines

incenter

indirect proof

melinda.dezeeuw

Cross-Out

melinda.dezeeuw

Inserted Text

Angle bisector

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5 Student-Built Glossary (continued)

Vocabulary TermFound

on PageDefinition/Description/Example

indirect reasoning

median

orthocenterOHR·thoh·CEN·tuhr

perpendicular bisector

point of concurrency

proof by contradiction

melinda.dezeeuw

Cross-Out

melinda.dezeeuw

Cross-Out

melinda.dezeeuw

Inserted Text

Angle bisector

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5

Before you begin Chapter 5

• Read each statement.

• Decide whether you Agree (A) or Disagree (D) with the statement.

• Write A or D in the first column OR if you are not sure whether you agree or disagree, write NS (Not Sure).

After you complete Chapter 5

• Reread each statement and complete the last column by entering an A or a D.

• Did any of your opinions about the statements change from the first column?

• For those statements that you mark with a D, use a piece of paper to write an example of why you disagree.

Anticipation Guide

Relationships in Triangles

STEP 1 A, D, or NS

StatementSTEP 2A or D

1. Any point that is on the perpendicular bisector of a segment is equidistant from the endpoints of that segment.

2. The circumcenter of a triangle is equidistant from the midpoints of each side of the triangle.

3. The altitudes of a triangle meet at the orthocenter.

4. Three altitudes can be drawn for any one triangle.

5. A median of a triangle is any segment that contains the midpoint of a side of the triangle.

6. The measure of an exterior angle of a triangle is always greater than the measures of either of its corresponding remote interior angles.

7. The longest side in a triangle is opposite the smallest angle in that triangle.

8. To write an indirect proof that two lines are perpendicular, begin by assuming the two lines are not perpendicular.

9. The length of the longest side of a triangle is always greater than the sum of the lengths of the other two sides.

10. In two triangles, if two pairs of sides are congruent, then the measure of the included angles determines which triangle has the longer third side.

Step 1

Step 2

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NOMBRE FECHA PERÍODO

5

Antes de comenzar el Capítulo 5

• Lee cada enunciado.

• Decide si estás de acuerdo (A) o en desacuerdo (D) con el enunciado.

• Escribe A o D en la primera columna O si no estás seguro(a) de la respuesta, escribe NS (No estoy seguro(a).

Después de completar el Capítulo 5

• Vuelve a leer cada enunciado y completa la última columna con una A o una D.

• ¿Cambió cualquiera de tus opiniones sobre los enunciados de la primera columna?

• En una hoja de papel aparte, escribe un ejemplo de por qué estás en desacuerdo con los enunciados que marcaste con una D.

Ejercicios preparatorios

Relaciones en Triángulos

PASO 1A, D o NS

EnunciadoPASO 2 A o D

1. Cualquier punto ubicado sobre la mediatriz de un segmento, equidista de los extremos de dicho segmento.

2. El circuncentro del triángulo equidista de los puntos medios de cada lado del triángulo.

3. Las alturas de un triángulo se unen en el ortocentro.

4. Se pueden dibujar tres alturas para cualquier triángulo.

5. La mediana de un triángulo es un segmento que contiene el punto medio de un lado del triángulo.

6. La medida del ángulo exterior de un triángulo es siempre mayor que las medidas de cualquiera de sus ángulos interiores no adyacentes.

7. El lado más largo en un triángulo está en el lado opuesto al ángulo más pequeño de éste.

8. Para escribir una prueba indirecta de que dos rectas son perpendiculares, comienza por suponer que las dos rectas no son perpendiculares.

9. La longitud del lado más largo de un triángulo es siempre mayor que la suma de los otros dos lados.

10. En dos triángulos, si dos pares de lados son congruentes, entonces la medida de los ángulos inscritos determina qué triángulo tiene el tercer lado más largo.

Paso 1

Paso 2

Capítulo 5 4 Geometrica de Glencoe

Less

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Perpendicular Bisector A perpendicular bisector is a line, segment, or ray that is perpendicular to the given segment and passes through its midpoint. Some theorems deal with perpendicular bisectors.

Perpendicular Bisector

Theorem

If a point is on the perpendicular bisector of a segment, then it is equidistant

from the endpoints of the segment.

Converse of Perpendicular

Bisector Theorem

If a point is equidistant from the endpoints of a segment, then it is on the

perpendicular bisector of the segment.

Circumcenter TheoremThe perpendicular bisectors of the sides of a triangle intersect at a point called

the circumcenter that is equidistant from the vertices of the triangle.

5-1 Study Guide and Intervention

Bisectors of Triangles

Find the measure of FM.

2.8

−− FK is the perpendicular bisector of

−−− GM .

FG = FM

2.8 = FM

−−

BD is the perpendicular

bisector of −−

AC . Find x.

3x + 8

5x - 6

B

C

D

A

AD = DC

3x + 8 = 5x - 6

14 = 2x

7 = x

Example 1 Example 2

Exercises

Find each measure.

1. XW 2. BF

7.5 5

5

4.2

19 19

Point P is the circumcenter of △EMK. List any segment(s) congruent to each segment below.

3. −−−

MY 4. −−

KP

5. −−−

MN 6. −−−

ER

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Angle Bisectors Another special segment, ray, or line is an angle bisector, which divides an angle into two congruent angles.

Angle Bisector

Theorem

If a point is on the bisector of an angle, then it is equidistant from the sides

of the angle.

Converse of Angle

Bisector Theorem

If a point in the interior of an angle if equidistant from the sides of the angle, then

it is on the bisector of the angle.

Incenter TheoremThe angle bisectors of a triangle intersect at a point called the incenter that is

equidistant from the sides of the triangle.

5-1 Study Guide and Intervention (continued)


" MR is the angle bisector of ∠NMP. Find x if m∠1 = 5x + 8

and m∠2 = 8x - 16.

12

N R

PM

! MR is the angle bisector of ∠NMP, so m∠1 = m∠2.

5x + 8 = 8x - 16

24 = 3x

8 = x

Exercises

Find each measure.

1. ∠ABE 2. ∠YBA

43°

47° 8

8

3. MK 4. ∠EWL

3x - 82x + 1

(3x + 21)°(7x + 5)°

Point U is the incenter of △GHY. Find each

measure below.

5. MU 6. ∠UGM

7. ∠PHU 8. HU

Example

28°

21°12

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5-1 Skills Practice

Bisectors of TrianglesFind each measure.

1. FG 2. KL

5x - 1713

13

3x + 1

4.2

3. TU 4. ∠LYF

2x + 24

5x - 30

58°

5. IU 6. ∠MYW

19°

19°

2x + 5

7x

(2x + 5)°

(4x - 1)°

Point P is the circumcenter of △ABC. List any segment(s) congruent to each segment below.

7. −−−

BR

8. −−

CS

9. −−

BP

Point A is the incenter of △PQR. Find each measure below.

10. ∠ARU

11. AU

12. ∠QPK 40°

20°

(4x - 9)°(3x + 2)°

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Find each measure.

1. TP 2. VU

7

9

9

3x+ 10

7x+ 2

3. KN 4. ∠NJZ

3x

x+ 10

I

NZ

J

38°

5. QA 6. ∠MFZ

E

R AQ

3x+ 167x

(2x- 1)°

(x+ 9)°

Point L is the circumcenter of △BKT. List any segment(s) congruent to each segment below.

7. −−−

BN

8. −−

BL

Point A is the incenter of △LYG. Find each measure below.

9. ∠ILA

10. ∠JGA

11. SCULPTURE A triangular entranceway has walls with corner angles of 50, 70, and 60.

The designer wants to place a tall bronze sculpture on a round pedestal in a central

location equidistant from the three walls. How can the designer find where to place the

sculpture?

5-1 Practice


21°

32°

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1. WIND CHIME Joanna has a flat wooden triangular piece as part of a wind chime. The piece is suspended by a wire anchored at a point equidistant from the sides of the triangle. Where is the anchor point located?

2. PICNICS Marsha and Bill are going to the park for a picnic. The park is triangular. One side of the park is bordered by a river and the other two sides are bordered by busy streets. Marsha and Bill want to find a spot that is equally far away from the river and the streets. At what point in the park should they set up their picnic?

3. MOVING Martin has 3 grown children. The figure shows the locations of Martin’s children on a map that has a coordinate plane on it. Martin would like to move to a location that is the same distance from all three of his children. What are the coordinates of the location on the map that is equidistant from all three children?

y

xO

5

5-5

4. NEIGHBORHOOD Amanda is looking at her neighborhood map. She notices that her house along with the homes of her friends, Brian and Cathy, can be the vertices of a triangle. The map is on a coordinate grid. Amanda’s house is at the point (1, 3), Brian’s is at (5, -1), and Cathy’s is at (4, 5). Where would the three friends meet if they each left their houses at the same time and walked to the opposite side of the triangle along the path of shortest distance from their house?

5. PLAYGROUND A concrete company is pouring concrete into a triangular form as the center of a new playground.

a. The foreman measures the triangle and notices that the incenter and the circumcenter are the same. What type of triangle is being created?

b. Suppose the foreman changes the triangular form so that the circumcenter is outside of the triangle but the incenter is inside the triangle. What type of triangle would be created?

5-1 Word Problem Practice


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Inscribed and Circumscribed CirclesThe three angle bisectors of a triangle intersect in a single point called the incenter. This

point is the center of a circle that just touches the three sides of the triangle. Except for the

three points where the circle touches the sides, the circle is inside the triangle. The circle is

said to be inscribed in the triangle.

1. With a compass and a straightedge, construct the inscribed

circle for △PQR by following the steps below.

Step 1 Construct the bisectors of ∠R and ∠Q. Label the point

where the bisectors meet, A.

Step 2 Construct a perpendicular segment from A to −−−

RQ . Use

the letter B to label the point where the perpendicular

segment intersects −−−

RQ .

Step 3 Use a compass to draw the circle with center at A and

radius −−

AB .

Construct the inscribed circle in each triangle.

2. 3.

The three perpendicular bisectors of the sides of a triangle also meet in a single point. This

point is the center of the circumscribed circle, which passes through each vertex of the

triangle. Except for the three points where the circle touches the triangle, the circle is

outside the triangle.

4. Follow the steps below to construct the circumscribed circle

for △FGH.

Step 1 Construct the perpendicular bisectors of −−−

FG and −−−

FH .

Use the letter A to label the point where the

perpendicular bisectors meet.

Step 2 Draw the circle that has center A and radius −−

AF .

Construct the circumscribed circle for each triangle.

5. 6.

F H

G

P

QR

5-1 Enrichment

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Medians A median is a line segment that connects a vertex of a triangle to the midpoint of the opposite side. The three medians of a triangle intersect at the centroid of the triangle. The centroid is located two thirds of the distance from a vertex to the midpoint of the side opposite the vertex on a median.

In △ABC, U is the centroid and

BU = 16. Find UK and BK.

BU = 2 − 3 BK

16 = 2 − 3 BK

24 = BK

BU + UK = BK

16 + UK = 24

UK = 8

Exercises

In △ABC, AU = 16, BU = 12, and CF = 18. Find

each measure.

1. UD 2. EU

3. CU 4. AD

5. UF 6. BE

In △CDE, U is the centroid, UK = 12, EM = 21,

and UD = 9. Find each measure.

7. CU 8. MU

9. CK 10. JU

11. EU 12. JD


Medians and Altitudes of Triangles

Example

16

12

12

9

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Altitudes An altitude of a triangle is a segment from a vertex to the line containing the opposite side meeting at a right angle. Every triangle has three altitudes which meet at a point called the orthocenter.

The vertices of △ABC are A(1, 3), B(7, 7) and C(9, 3). Find the coordinates of the orthocenter of △ABC.

Find the point where two of the three altitudes intersect.

Find the equation of the altitude from

A to −−−

BC .

If −−−

BC has a slope of −2, then the altitude

has a slope of 1 − 2 .

y - y1 = m(x – x

1) Point-slope form

y - 3 = 1 − 2 (x – 1) m = 1 −

2 , (x

1, y

1) = A(1, 3)

y - 3 = 1 − 2 x – 1 −

2 Distributive Property

y = 1 − 2 x +

5 −

2 Simplify.

C to −−

AB .

If −−

AB has a slope of 2 − 3 , then the altitude has a

slope of - 3 − 2 .

y - y1 = m(x - x

1) Point-slope form

y - 3 = - 3 − 2 (x - 9) m = -

3 −

2 , (x

1, y

1) = C(9, 3)

y - 3 = - 3 −

2 x +

27 −

2 Distributive Property

y = - 3 − 2 x +

33 −

2 Simplify.

Solve the system of equations and find where the altitudes meet.

y = 1 − 2 x +

5 −

2 y = - 3 −

2 x +

33 −

2

1 − 2 x +

5 −

2 = - 3 −

2 x +

33 −

2 Subtract 1 −

2 x from each side.

5 −

2 = −2x +

33 −

2 Subtract

33 −

2 from each side.

−14 = −2x Divide both sides by -2.

7 = x

y = 1 − 2 x +

5 −

2 = 1 −

2 (7) +

5 −

2 =

7 −

2 +

5 −

2 = 6

The coordinates of the orthocenter of △ABC is (6, 7).

Exercises

COORDINATE GEOMETRY Find the coordinates of the orthocenter of each triangle.

1. J(1, 0), H(6, 0), I(3, 6) 2. S(1, 0), T(4, 7), U(8, −3)



Example

y

x

(9, 3)(1, 3)

(7, 7)

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5-2

In △PQR, NQ = 6, RK = 3, and PK = 4.

Find each length.

1. KM 2. KQ

3. LK 4. LR

5. NK 6. PM

In △STR, H is the centroid, EH = 6,

DH = 4, and SM = 24. Find each length.

7. SH 8. HM

9. TH 10. HR

11. TD 12. ER

COORDINATE GEOMETRY Find the coordinates of the centroid of each triangle.

13. X(−3, 15) Y(1, 5), Z(5, 10) 14. S(2, 5), T(6, 5), R(10, 0)


15. L(8, 0), M(10, 8), N(14, 0) 16. D(−9, 9), E(−6, 6), F(0, 6)

Skills Practice


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In △ABC, CP = 30, EP = 18, and BF = 39. Find each length.

1. PD 2. FP

3. BP 4. CD

5. PA 6. EA

In △MIV, Z is the centroid, MZ = 6, YI = 18, and NZ = 12.

Find each measure.

7. ZR 8. YZ

9. MR 10. ZV

11. NV 12. IZ

COORDINATE GEOMETRY Find the coordinates of the centroid of each triangle.

13. I(3, 1), J(6, 3), K(3, 5) 14. H(0, 1), U(4, 3), P(2, 5)


15. P(-1, 2), Q(5, 2), R(2, 1) 16. S(0, 0), T(3, 3), U(3, 6)

17. MOBILES Nabuko wants to construct a mobile out of flat triangles so that the surfaces of the triangles hang parallel to the floor when the mobile is suspended. How can Nabuko be certain that she hangs the triangles to achieve this effect?

5-2 Practice


A

C

F

E

D

PB

18 30

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-2


1. BALANCING Johanna balanced a

triangle flat on her finger tip. What point

of the triangle must Johanna be

touching?

2. REFLECTIONS Part of the working space

in Paulette’s loft is partitioned in the

shape of a nearly equilateral triangle

with mirrors hanging on all three

partitions. From which point could

someone see the opposite corner behind

his or her reflection in any of the three

mirrors?

3. DISTANCES For what kind of triangle is

there a point where the distance to each

side is half the distance to each vertex?

Explain.

4. MEDIANS Look at the right triangle

below. What do you notice about the

orthocenter and the vertices of the

triangle?

5. PLAZAS An architect is designing a

triangular plaza. For aesthetic purposes,

the architect pays special attention to the

location of the centroid C and the

circumcenter O.

a. Give an example of a triangular plaza

where C = O. If no such example

exists, state that this is impossible.

b. Give an example of a triangular plaza

where C is inside the plaza and O is

outside the plaza. If no such example


c. Give an example of a triangular plaza

where C is outside the plaza and O is

inside the plaza. If no such example




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Constructing Centroids and OrthocentersThe three medians of a triangle intersect at a single point called the centroid.

You can use a straightedge and compass to find the centroid of a triangle.

1. With a straightedge and compass, construct the

centroid for △STU by following the steps below.

Step 1 Locate the midpoints of sides TU and SU.

Label the midpoints A and B respectively.

Step 2 Draw the segments SA and TB. Use the

letter H to label their point of intersection,

which is the centroid of △STU.

Construct the centroid of each triangle.

2. 3.

The three altitudes of a triangle meet in a single point called the orthocenter of the triangle.

4. Follow the steps below to construct the orthocenter

of △CDE using a straightedge and compass.

Step 1 Extend segments CD and DE past point

D long enough to meet perpendiculars

from E and C as shown.

Step 2 Construct the perpendicular from point C

to the line DE and label the point of

intersection X. Likewise, label the point of

intersection of line CD with the perpendicular

from E as point Z. In this case

both X and Z lie outside △CDE.

Step 3 Label O the point where perpendiculars

! #$ CX and ! #$ EZ intersect. This is the

orthocenter of △CDE.

Construct the orthocenter of each triangle.

5. 6.

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Less

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5-3


Angle Inequalities Properties of inequalities, including the Transitive, Addition, and

Subtraction Properties of Inequality, can be used with measures of angles and segments.

There is also a Comparison Property of Inequality.

For any real numbers a and b, either a < b, a = b, or a > b.

The Exterior Angle Inequality Theorem can be used to prove this inequality involving an

exterior angle.

Exterior Angle

Inequality Theorem

The measure of an exterior angle of a triangle

is greater than the measure of either of its

corresponding remote interior angles.

AC D

1

B

m∠1 > m∠A,

m∠1 > m∠B

List all angles of △EFG whose measures are

less than m∠1.

The measure of an exterior angle is greater than the measure of

either remote interior angle. So m∠3 < m∠1 and m∠4 < m∠1.

Exercises

Use the Exterior Angle Inequality Theorem to list all of

the angles that satisfy the stated condition.

1. measures are less than m∠1

2. measures are greater than m∠3









M J K

3

4 521

L

Exercises 1–2

H E F3

4

21

G


Inequalities in One Triangle

Example

R O

Q

N

P3 4

56

Exercises 9–10

78

21

S

X T W V

3

4

5

67 2 1

U

Exercises 3–8

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Angle-Side Relationships When the sides of triangles are not congruent, there is a relationship between the sides and angles of the triangles.

• If one side of a triangle is longer than another side, then the

angle opposite the longer side has a greater measure than the

angle opposite the shorter side.

• If one angle of a triangle has a greater measure than another

angle, then the side opposite the greater angle is longer than

the side opposite the lesser angle.

B C

A

List the angles in order

from smallest to largest measure.

R T9 cm

6 cm 7 cm

S

∠T, ∠R, ∠S

List the sides in order

from shortest to longest.

A B

C

20°

35°

125°

−−− CB ,

−− AB ,

−− AC

Exercises

List the angles and sides in order from smallest to largest.

1.

T S

R

48 cm

23.7 cm

35 cm

2.

R T

S

60°

80°

40°

3.

A C

B

3.8 4.3

4.0

4.

14

11

5

5.

45

8

6.

20

12

7.

35°

120° 25°

8.

56° 58°

9.

60° 54°



If AC > AB, then m∠B > m∠C.

If m∠A > m∠C, then BC > AB.

Example 1 Example 2

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5-3

Use the Exterior Angle Inequality Theorem to list all of the

angles that satisfy the stated condition.

1. measures less than m∠1

2. measures less than m∠9

3. measures greater than m∠5

4. measures greater than m∠8

List the angles and sides of each triangle in order from smallest to largest.

5.

5

6

2

6. 24°

98°

7.

15

916

8.

38

39

34

9. 10.

1

2 4

6

7

8 93 5

Skills Practice


98°

43°

42°

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Use the figure at the right to determine which angle has the greatest measure.

1. ∠1, ∠3, ∠4 2. ∠4, ∠8, ∠9

3. ∠2, ∠3, ∠7 4. ∠7, ∠8, ∠10

Use the Exterior Angle Inequality Theorem to list

all angles that satisfy the stated condition.





Use the figure at the right to determine the relationship

between the measures of the given angles.

9. m∠QRW, m∠RWQ 10. m∠RTW, m∠TWR

11. m∠RST, m∠TRS 12. m∠WQR, m∠QRW

Use the figure at the right to determine the relationship

between the lengths of the given sides.

13. −−−

DH , −−−

GH 14. −−−

DE , −−−

DG

15. −−−

EG , −−−

FG 16. −−−

DE , −−−

EG

17. SPORTS The figure shows the position of three trees on one

part of a Frisbee™ course. At which tree position is the angle

between the trees the greatest? 53 ft

40 ft

3

2

1

37.5 ft

120°32°

48° 113°

17°H

DE

F

G

3447

45

44

22

14

35

Q

R

S

T

W

12

4 6

78

9

3

5

1

2

46

7

8 9

10

3

5

5-3 Practice


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1. DISTANCE Carl and Rose live on the

same straight road. From their balconies

they can see a flagpole in the distance.

The angle that each person’s line of

sight to the flagpole makes with the

road is the same. How do their distances

from the flagpole compare?

2. OBTUSE TRIANGLES Don notices that

the side opposite the right angle in a

right triangle is always the longest of

the three sides. Is this also true of the

side opposite the obtuse angle in an

obtuse triangle? Explain.

3. STRING Jake built a triangular

structure with three black sticks. He

tied one end of a string to vertex M

and the other end to a point on the

stick opposite M, pulling the string

taut. Prove that the length of the

string cannot exceed the longer of the

two sides of the structure.

string

M

4. SQUARES Matthew has three different

squares. He arranges the squares to

form a triangle as shown. Based on

the information, list the squares in

order from the one with the smallest

perimeter to the one with the largest

perimeter.

54˚47˚

3

1 2

5. CITIES Stella is going

to Texas to visit a friend.

As she was looking at

a map to see where

she might want to go,

she noticed the cities

Austin, Dallas, and Abilene

formed a triangle. She wanted to

determine

how the distances between the cities

were related, so she used a protractor to

measure two angles.

a. Based on the information in the

figure, which of the two cities are

nearest to each other?

b. Based on the information in the

figure, which of the two cities are

farthest apart from each other?

5-3

59˚

64˚

Abilene

Dallas

Austin

Word Problem Practice


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Construction ProblemThe diagram below shows segment AB adjacent to a closed region. The

problem requires that you construct another segment XY to the right of the

closed region such that points A, B, X, and Y are collinear. You are not allowed

to touch or cross the closed region with your compass or straightedge.

Follow these instructions to construct a segment XY so that it is collinear with

segment AB.

1. Construct the perpendicular bisector of −−

AB . Label the midpoint as point C, and the line

as m.

2. Mark two points P and Q on line m that lie well above the closed region. Construct the

perpendicular bisector, n, of −−−

PQ . Label the intersection of lines m and n as point D.

3. Mark points R and S on line n that lie well to the right of the closed region. Construct

the perpendicular bisector, k , of −−

RS . Label the intersection of lines n and k as point E.

4. Mark point X on line k so that X is below line n and so that −−

EX is congruent to −−−

DC .

5. Mark points T and V on line k and on opposite sides of X, so that −−

XT and −−

XV are

congruent. Construct the perpendicular bisector, ℓ, of −−

TV . Call the point where the

line ℓ hits the boundary of the closed region point Y. −−

XY corresponds to the new road.

BA

ExistingRoad

Closed Region(Lake)

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5-3

Cabri Junior can be used to investigate the relationships between angles and sides of a triangle.

Step 1 Use Cabri Junior. to draw and label a triangle. • Select F2 Triangle to draw a triangle. • Move the cursor to where you want the first vertex. Press ENTER .

• Repeat this procedure to determine the next two vertices of the triangle. • Select F5 Alph-num to label each vertex. • Move the cursor to a vertex, press ENTER , enter A, and press ENTER again.

• Repeat this procedure to label vertex B and vertex C.

Step 2 Draw an exterior angle of △ABC. • Select F2 Line to draw a line through

−−− BC .

• Select F2 Point, Point on to draw a point on " #$ BC so that C is between B and the new point.

• Select F5 Alph-num to label the point D.

Step 3 Find the measures of the three interior angles and the exterior angle, ∠ACD. • Select F5 Measure, Angle. • To find the measure of ∠ABC, select points A, B, and

C (with the vertex B as the second point selected). • Repeat to find the remaining angle measures.

Step 4 Find the measure of each side of △ABC. • Select F5 Measure, D. & Length.

• To find the length of −−

AB , select point A and then select point B.

• Repeat this procedure to find the lengths of −−−

BC and −−

CA .

Exercises

Analyze your drawing.

1. What is the relationship between m∠ACD and m∠ABC? m∠ACD and m∠BAC?

2. Make a conjecture about the relationship between the measures of an exterior angle (∠ACD) and its two remote interior angles (∠ABC and ∠BAC).

3. Change the dimensions of the triangle by moving point A. (Press CLEAR so the pointer becomes a black arrow. Move the pointer close to point A until the arrow becomes transparent and point A is blinking. Press ALPHA to change the arrow to a hand. Then move the point.) Is your conjecture still true?

4. Which side of the triangle is the longest? the shortest?

5. Which angle measure (not including the exterior angle) is the greatest? the least?

6. Make a conjecture about where the longest side is in relationship to the greatest angle and where the shortest side is in relationship to the least angle.

Graphing Calculator Activity

Cabri Junior: Inequalities in One Triangle

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5-3

The Geometer’s Sketchpad can be used to investigate the relationships between angles and sides of a triangle.

Step 1 Use The Geometer’s Sketchpad to draw a triangle and one exterior angle. • Construct a ray by selecting the

Ray tool from the toolbar. First, click where you want the first point. Then click a second point to draw the ray.

• Next, select the Segment tool from the toolbar. Use the endpoint of the ray as the first point for the segment and click on a second point to construct the segment.

• Construct another segment joining the second point of the previous segment to a point on the ray.

• Display the labels for each point. Use the Selection Arrow tool to select all four points. Display the labels by selecting Show Label from the Display menu.

Step 2 Find the measures of each angle. • To find the measure of ∠ABC, use the Selection Arrow tool to select points

A, B, and C (with the vertex B as the second point selected). Then, under the Measure menu, select Angle. Use this method to find the remaining angle measures, including the exterior angle, ∠BCD.

Step 3 Find the measures of each side of the triangle. • To find the measure of side AB, select A and then B. Next, under the Measure

menu, select Distance. Use this method to find the length of the other two sides.

Exercises

Analyze your drawing.

1. What is the relationship between m∠BCD and m∠ABC? m∠BCD and m∠BAC?

2. Make a conjecture about the relationship between the measures of an exterior angle (∠BCD) and its two remote interior angles (∠ABC and ∠BAC).

3. Change the dimensions of the triangle by selecting point A with the pointer tool and moving it. Is your conjecture still true?

4. Which side of the triangle is the longest? the shortest?

5. Which angle measure (not including the exterior angle) is the greatest? the least?

6. Make a conjecture about where the longest side is in relationship to the greatest angle and where the shortest side is in relationship to the least angle.

A

B

C D

m/ABC 5 69.29˚

m/BCA 5 55.92˚

m/BAC 5 54.78˚

m/BCD 5 124.08˚

AB 5 2.20 cm

BC 5 2.17 cm

AC 5 2.49 cm

Geometer’s Sketchpad Activity


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Indirect Algebraic Proof One way to prove that a statement is true is to temporarily assume that what you are trying to prove is false. By showing this assumption to be logically impossible, you prove your assumption false and the original conclusion true. This is known as an indirect proof.

Steps for Writing an Indirect Proof

1. Assume that the conclusion is false by assuming the oppposite is true.

2. Show that this assumption leads to a contradiction of the hypothesis or some other fact.

3. Point out that the assumption must be false, and therefore, the conclusion must be true.

Given: 3x + 5 > 8

Prove: x > 1

Step 1 Assume that x is not greater than 1. That is, x = 1 or x < 1.

Step 2 Make a table for several possibilities for x = 1 or x < 1. When x = 1 or x < 1, then 3x + 5 is not greater than 8.

Step 3 This contradicts the given information that 3x + 5 > 8. The assumption that x is not greater than 1 must be false, which means that the statement “x > 1” must be true.

Exercises

State the assumption you would make to start an indirect proof of each statement.

1. If 2x > 14, then x > 7.

2. For all real numbers, if a + b > c, then a > c - b.

Complete the indirect proof.

Given: n is an integer and n2 is even.

Prove: n is even.

3. Assume that

4. Then n can be expressed as 2a + 1 by

5. n2 = Substitution

6. = Multiply.

7. = Simplify.

8. = 2(2a2 + 2a) + 1

9. 2(2a2 + 2a)+ 1 is an odd number. This contradicts the given that n2 is even,

so the assumption must be

10. Therefore,


Indirect Proof

Example

x 3x + 5

1 8

0 5

-1 2

-2 -1

-3 -4

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Indirect Proof with Geometry To write an indirect proof in geometry, you assume that the conclusion is false. Then you show that the assumption leads to a contradiction. The contradiction shows that the conclusion cannot be false, so it must be true.

Given: m∠C = 100

Prove: ∠A is not a right angle.

Step 1 Assume that ∠A is a right angle.

Step 2 Show that this leads to a contradiction. If ∠A is a right angle,

then m∠A = 90 and m∠C + m∠A = 100 + 90 = 190. Thus the

sum of the measures of the angles of △ABC is greater than 180.

Step 3 The conclusion that the sum of the measures of the angles of

△ABC is greater than 180 is a contradiction of a known property.

The assumption that ∠A is a right angle must be false, which

means that the statement “∠A is not a right angle” must be true.

Exercises


1. If m∠A = 90, then m∠B = 45.

2. If −−

AV is not congruent to −−

VE , then △AVE is not isosceles.

Complete the indirect proof.

Given: ∠1 $ ∠2 and −−−

DG is not congruent to −−−

FG .

Prove: −−−

DE is not congruent to −−

FE .

3. Assume that Assume the conclusion is false.

4. −−−

EG $ −−−

EG

5. △EDG $ △EFG

6.

7. This contradicts the given information, so the assumption must

be

8. Therefore,

1

2

D G

F

E

A B

C


Indirect Proof

Example

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5-4 Skills Practice

Indirect Proof


1. m∠ABC < m∠CBA

2. △DEF " △RST

3. Line a is perpendicular to line b.

4. ∠5 is supplementary to ∠6.

Write an indirect proof of each statement.

5. Given: x2 + 8 ≤ 12

Prove: x ≤ 2

6. Given: ∠D ≇ ∠F

Prove: DE ≠ EF

D F

E

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1. −−−

BD bisects ∠ABC.

2. RT = TS

Write an indirect proof of each statement.

3. Given: -4x + 2 < -10

Prove: x > 3

4. Given: m∠2 + m∠3 ≠ 180

Prove: a ∦ b

5. PHYSICS Sound travels through air at about 344 meters per second when the

temperature is 20°C. If Enrique lives 2 kilometers from the fire station and it takes

5 seconds for the sound of the fire station siren to reach him, how can you prove

indirectly that it is not 20°C when Enrique hears the siren?

1

2

3

a

b

5-4 Practice

Indirect Proof

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1. CANOES Thirty-five students went on a canoeing expedition. They rented 17 canoes for the trip. Use an indirect proof to show that at least one canoe had more than two students in it.

2. AREA The area of the United States is about 6,000,000 square miles. The area of Hawaii is about 11,000 square miles. Use an indirect proof to show that at least one of the fifty states has an area greater than 120,000 square miles.

3. CONSECUTIVE NUMBERS David was trying to find a common factor other than 1 between various pairs of consecutive integers. Write an indirect proof to show David that two consecutive integers do not share a common factor other than 1.

4. WORDS The words accomplishment, counterexample, and extemporaneous all have 14 letters. Use an indirect proof to show that any word with 14 letters must use a repeated letter or have two letters that are consecutive in the alphabet.

5. LATTICE TRIANGLES A lattice point is a point whose coordinates are both integers. A lattice triangle is a triangle whose vertices are lattice points. It is a fact that a lattice triangle has an area of at least 0.5 square units.

y

xO

A

B

C

5

5

a. Suppose △ABC has a lattice point in its interior. Show that the lattice triangle can be partitioned into three smaller lattice triangles.

b. Prove indirectly that a lattice triangle with area 0.5 square units contains no lattice point. (Being on the boundary does not count as inside.)


Indirect Proof

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More Counterexamples

Some statements in mathematics can be proven false by counterexamples. Consider the following statement.

For any numbers a and b, a - b = b - a.

You can prove that this statement is false in general if you can fi nd one example for which the statement is false.

Let a = 7 and b = 3. Substitute these values in the equation above.

7 - 3 3 - 7

4 ≠ -4

In general, for any numbers a and b, the statement a - b = b - a is false. You can make the equivalent verbal statement: subtraction is not a commutative operation.

In each of the following exercises a, b, and c are any numbers. Prove that

the statement is false by counterexample.

1. a - (b - c) (a - b) - c 2. a ÷ (b ÷ c) (a ÷ b) ÷ c

3. a ÷ b b ÷ a 4. a ÷ (b + c) (a ÷ b) + (a ÷ c)

5. a + (bc) (a + b)(a + c) 6. a2 + a2 a4

7. Write the verbal equivalents for Exercises 1, 2, and 3.

8. For the Distributive Property, a(b + c) = ab + ac, it is said that multiplication distributes over addition. Exercises 4 and 5 prove that some operations do not distribute. Write a statement for each exercise that indicates this.

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Less

on

5-5


The Triangle Inequality If you take three straws of lengths 8 inches, 5 inches, and 1 inch and try to make a triangle with them, you will find that it is not possible. This illustrates the Triangle Inequality Theorem.

Triangle Inequality

Theorem

The sum of the lengths of any two sides of a

triangle must be greater than the length of the third side. BC

A

a

cb

The measures of two sides of a triangle are 5 and 8. Find a range

for the length of the third side.

By the Triangle Inequality Theorem, all three of the following inequalities must be true.

5 + x > 8 8 + x > 5 5 + 8 > x

x > 3 x > -3 13 > x

Therefore x must be between 3 and 13.

ExercisesIs it possible to form a triangle with the given side lengths? If not, explain

why not.

1. 3, 4, 6 2. 6, 9, 15

3. 8, 8, 8 4. 2, 4, 5

5. 4, 8, 16 6. 1.5, 2.5, 3

Find the range for the measure of the third side of a triangle given the measures

of two sides.

7. 1 cm and 6 cm 8. 12 yd and 18 yd

9. 1.5 ft and 5.5 ft 10. 82 m and 8 m

11. Suppose you have three different positive numbers arranged in order from least to

greatest. What single comparison will let you see if the numbers can be the lengths of

the sides of a triangle?


The Triangle Inequality

Example

a + b > c

b + c > a

a + c > b

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Proofs Using The Triangle Inequality Theorem You can use the Triangle Inequality Theorem as a reason in proofs.

Complete the following proof.

Given: △ABC ! △DEC

Prove: AB + DE > AD − BE

Proof:

Statements

1. △ABC ! △DEC

2. AB + BC > AC

DE + EC > CD

3. AB > AC – BC

DE > CD – EC

4. AB + DE > AC - BC + CD - EC

5. AB + DE > AC + CD - BC - EC

6. AB + DE > AC + CD - (BC + EC)

7. AC + CD = AD

BC + EC = BE

8. AB + DE > AD - BE

Reasons

1. Given

2. Triangle Inequality Theorem

3. Subtraction

4. Addition

5. Commutative

6. Distributive

7. Segment Addition Postulate

8. Substitution

Exercises

PROOF Write a two column proof.

Given: −−

PL ‖ −−−

MT

K is the midpoint of −−

PT .

Prove: PK + KM > PL

Proof:

Statements

1. −− PL ‖ −−− MT

2. ∠P ! ∠T

3. K is the midpoint of −−

PT .

4. PK = KT

5.

6. △PKL ! △TKM

7.

8.

9. PK + KM > PL

Reasons

1.

2.

3. Given

4.

5. Vertical Angles Theorem

6.

7. Triangle Inequality Theorem

8. CPCTC

9.



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Lesso

n 5

-5


5-5 Skills Practice


Is it possible to form a triangle with the given side lengths? If not, explain

why not.

1. 2 ft, 3 ft, 4 ft 2. 5 m, 7 m, 9 m

3. 4 mm, 8 mm, 11 mm 4. 13 in., 13 in., 26 in.

5. 9 cm, 10 cm, 20 cm 6. 15 km, 17 km, 19 km

7. 14 yd, 17 yd, 31 yd 8. 6 m, 7 m, 12 m


of two sides.

9. 5 ft, 9 ft 10. 7 in., 14 in.

11. 8 m, 13 m 12. 10 mm, 12 mm

13. 12 yd, 15 yd 14. 15 km, 27 km

15. 17 cm, 28 cm, 16. 18 ft, 22 ft

17. Proof Complete the proof.

Given: △ABC and △CDE

Prove: AB + BC + CD + DE > AE

Proof:

Statements Reasons

1. AB + BC > AC

CD + DE > CE1.

2. AB + BC + CD + DE > AC + CE 2.

3. 3. Seg. Addition Post

4. 4. Substitution

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Is it possible to form a triangle with the given side lengths? If not explain

why not.

1. 9, 12, 18 2. 8, 9, 17

3. 14, 14, 19 4. 23, 26, 50

5. 32, 41, 63 6. 2.7, 3.1, 4.3

7. 0.7, 1.4, 2.1 8. 12.3, 13.9, 25.2


of two sides.

9. 6 ft and 19 ft 10. 7 km and 29 km

11. 13 in. and 27 in. 12. 18 ft and 23 ft

13. 25 yd and 38 yd 14. 31 cm and 39 cm

15. 42 m and 6 m 16. 54 in. and 7 in.

17. Given: H is the centroid of △EDF

Prove: EY + FY > DE

Proof:

Statements

1. H is the centroid of △EDF

2. −−

EY is a median.

3.

4.

5. EY + DY > DE

6. EY + FY > DE

Reasons

1. Given

2.

3. Definition of median

4. Definition of midpoint

5.

6.

18. GARDENING Ha Poong has 4 lengths of wood from which he plans to make a border

for a triangular-shaped herb garden. The lengths of the wood borders are 8 inches,

10 inches, 12 inches, and 18 inches. How many different triangular borders can

Ha Poong make?

5-5 Practice


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Lesso

n 5

-5


Tanya’s home

Supermarket

Railroad

A B C

1. STICKS Jamila has 5 sticks of lengths 2, 4, 6, 8, and 10 inches. Using three sticks at a time as the sides of triangles, how many triangles can she make?

Use the figure at the

right for Exercises

2 and 3.

2. PATHS To get to the nearest super market, Tanya must walk over a railroad track. There are two places where she can cross the track (points A and B). Which path is longer? Explain.

3. PATHS While out walking one day Tanya finds a third place to cross the railroad tracks. Show that the path through point C is longer than the path through point B.

4. CITIES The distance between New York City and Boston is 187 miles and the distance between New York City and Hartford is 97 miles. Hartford, Boston, and New York City form a triangle on a map. What must the distance between Boston and Hartford be greater than?

5. TRIANGLES The figure shows an equilateral triangle ABC and a point P

outside the triangle.

C

P

B

A

a. Draw the figure that is the result of turning the original figure 60° counterclockwise about A. Denote by P', the image of P under this turn.

b. Note that −−−P'B is congruent to

−−PC . It is

also true that −−−PP' is congruent to

−−PA .

Why?

c. Show that −−PA ,

−−PB , and

−−PC satisfy the

triangle inequalities.



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Constructing Triangles

The measurements of the sides of a triangle are given. If a triangle having sides

with these measurements is not possible, then write impossible. If a triangle is

possible, draw it and measure each angle with a protractor.

1. AR = 5 cm m∠A = 2. PI = 8 cm m∠P =

RT = 3 cm m∠R = IN = 3 cm m∠I =

AT = 6 cm m∠T = PN = 2 cm m∠N =

3. ON = 10 cm m∠O = 4. TW = 6 cm m∠T =

NE = 5.3 cm m∠N = WO = 7 cm m∠W =

OE = 4.6 cm m∠E = TO = 2 cm m∠O =

5. BA = 3.l cm m∠B = 6. AR = 4 cm m∠A =

AT = 8 cm m∠A = RM = 5 cm m∠R =

BT = 5 cm m∠T = AM = 3 cm m∠M =

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Hinge Theorem The following theorem and its converse involve the relationship between the sides of two triangles and an angle in each triangle.

Hinge Theorem

If two sides of a triangle are congruent to two

sides of another triangle and the included

angle of the first is larger than the included

angle of the second, then the third side of

the first triangle is longer than

the third side of the second triangle. RT > AC

Converse of the

Hinge Theorem

If two sides of a triangle are congruent to

two sides of another triangle, and the

third side in the first is longer than the

third side in the second, then the included

angle in the first triangle is greater than

the included angle in the second triangle. m∠M > m∠R

Exercises

Compare the given measures.

1. MR and RP

N

R

P

M

21°

19°

2. AD and CD C

A

DB

22°

38°

3. m∠C and m∠Z 4. m∠XYW and m∠WYZ

Write an inequality for the range of values of x.

5.

115°

120° 24

2440

(4x - 10) 6.

33°

60

60

36

30

(3x - 3)°


Inequalities in Two Triangles

S T80°

R

B C60°

A

3336

TR

SN

M P

Compare the measures

of −−

GF and −−

FE .

H

E

F

G

22°

28°

Two sides of △HGF are congruent to two sides of △HEF, and m∠GHF > m∠EHF. By

the Hinge Theorem, GF > FE.

Compare the measures

of ∠ABD and ∠CBD.

13

16

C

D

A

B

Two sides of △ABD are congruent to two sides of △CBD, and AD > CD. By the

Converse of the Hinge Theorem,

m∠ABD > m∠CBD.

Example 2Example 1

30C

A X

B30

5048

2424

Z Y42

28

ZW

XY

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PROVE RELATIONSHIPS IN TWO TRIANGLES You can use the Hinge Theorem and its converse to prove relationships in two triangles.

Given: RX = XS

m∠SXT = 97

Prove: ST > RT

Proof:

Statements Reasons

1. ∠SXT and ∠RXT are

supplementary

2. m ∠SXT + m∠RXT = 180

3. m∠SXT = 97

4. 97 + m∠RXT = 180

5. m∠RXT = 83

6. 97 > 83

7. m∠SXT > m∠RXT

8. RX = XS

9. TX = TX

10. ST > RT

1. Defn of linear pair

2. Defn of supplementary

3. Given

4. Substitution

5. Subtraction

6. Inequality

7. Substitution

8. Given

9. Reflexive

10. Hinge Theorem

Exercises

Complete the proof.

Given: rectangle AFBC

ED = DC

Prove: AE > FB

Proof:

Statements Reasons

1. rectangle AFBC, ED = DC

2. AD = AD

3. m∠EDA > m∠ADC

4.

5.

6. AE > FB

1. given

2. reflexive

3. exterior angle

4. Hinge Theorem

5. opp sides ! in rectangle.

6. Substitution


Inequalities Involving Two Triangles

Example

97°

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Lesso

n 5

-6


5-6


1. m∠BXA and m∠DXA

2. BC and DC


3. m∠STR and m∠TRU 4. PQ and RQ

31

30

2222

R S

U T

95°

7 7

85°P R

S

Q

5. In the figure, −−

BA , −−−

BD , −−−

BC , and −−−

BE are congruent and AC < DE.

How does m∠1 compare with m∠3? Explain your thinking.

6. PROOF Write a two-column proof.

Given: −−

BA # −−−

DA

BC > DC

Prove: m∠1 > m∠2

1

2

3

B

A

D C

E

Skills Practice

Inequalities Involving Two Triangles

6

98

3

3

B

A C

D

X

1

2

B

A

D

C

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1 2D F

E

G

20 21

R TS

J K

14 14

14

13

12C F

E

D

(x + 3)°(x - 3)°

10 10

R TS

Q

40°

30°

60°A K

M

B


1. AB and BK 2. ST and SR

3. m∠CDF and m∠EDF 4. m∠R and m∠T

5. PROOF Write a two-column proof.

Given: G is the midpoint of −−−

DF .

m∠1 > m∠2

Prove: ED > EF

6. TOOLS Rebecca used a spring clamp to hold together a chair

leg she repaired with wood glue. When she opened the clamp,

she noticed that the angle between the handles of the clamp

decreased as the distance between the handles of the clamp

decreased. At the same time, the distance between the

gripping ends of the clamp increased. When she released the

handles, the distance between the gripping end of the clamp

decreased and the distance between the handles increased.

Is the clamp an example of the Hinge Theorem or its converse?

5-6 Practice


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Lesso

n 5

-6


1. CLOCKS The minute hand of a

grandfather clock is 3 feet long and

the hour hand is 2 feet long. Is the

distance between their ends greater

at 3:00 or at 8:00?

2. FERRIS WHEEL A Ferris wheel has

carriages located at the 10 vertices of

a regular decagon.

12

3

4

5

67

8

9

10

Which carriages are farther away

from carriage number 1 than carriage

number 4?

3. WALKWAY Tyree wants to make two

slightly different triangles for his

walkway. He has three pieces of wood

to construct the frame of his triangles.

After Tyree makes the first concrete

triangle, he adjusts two sides of the

triangle so that the angle they create

is smaller than the angle in the first

triangle. Explain how this changes the

triangle.

4. MOUNTAIN PEAKS Emily lives the

same distance from three mountain

peaks: High Point, Topper, and Cloud

Nine. For a photography class, Emily

must take a photograph from her house

that shows two of the mountain peaks.

Which two peaks would she have the

best chance of capturing in one image?

Emily

CloudNine

HighPoint

Topper

12 miles

9 m

iles

10 miles

5. RUNNERS A photographer is taking

pictures of three track stars, Amy, Noel,

and Beth. The photographer stands on a

track, which is shaped like a rectangle

with semicircles on both ends.

118˚

36˚

146˚

Photographer

Amy

Noel

Beth

a. Based on the information in the

figure, list the runners in order from

nearest to farthest from the

photographer.

b. Explain how to locate the point along

the semicircular curve that the

runners are on that is farthest away

from the photographer.



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Hinge Theorem

The Hinge Theorem that you studied in this section states that if two sides of a

triangle are congruent to two sides of another triangle and the included angle in

one triangle has a greater measure than the included angle in the other, then the

third side of the first triangle is longer than the third side of the second triangle. In

this activity, you will investigate whether the converse, inverse and contrapositive

of the Hinge Theorem are also true.

X

Y

Z

Q

S

R

1 2

Hypothesis: XY = QR, YZ = RS, m∠1 > m∠2

Conclusion: XZ > QS

1. What is the converse of the Hinge Theorem?

2. Can you find any counterexamples to prove that the converse is false?

3. What is the inverse of the Hinge Theorem?

4. Can you find any counterexamples to prove that the inverse is false?

5. What is the contrapositive of the Hinge Theorem?

6. Can you find any counterexamples to prove that the contrapositive is false?

5-6 Enrichment

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Student Recording Sheet

Use this recording sheet with pages 384–385 of the Student Edition.

9.

Read each question. Then fill in the correct answer.

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. F G H J

8. F G H J

Multiple Choice

Record your answer in the blank.

For gridded response questions, also enter your answer in the grid by writing

each number or symbol in a box. Then fill in the corresponding circle for that

number or symbol.

9. (grid in)

10.

11.

12.

13.

14. (grid in)

Extended Response

Short Response/Gridded Response

Record your answers for Question 15 on the back of this paper.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

14.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

. . . . .

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

5

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5 SCORE Rubric for Scoring Extended-Response

General Scoring Guidelines

• If a student gives only a correct numerical answer to a problem but does not show

how he or she arrived at the answer, the student will be awarded only 1 credit.

All extended-response questions require the student to show work.

• A fully correct answer for a multiple-part question requires correct responses for

all parts of the question. For example, if a question has three parts, the correct

response to one or two parts of the question that required work to be shown is not

considered a fully correct response.

• Students who use trial and error to solve a problem must show their method.

Merely showing that the answer checks or is correct is not considered a complete

response for full credit.

Exercise 15 Rubric

Score Specific Criteria

4 The student correctly identifies the number of students that play the

guitar, the number of students that play the piano, and the number of

student that play both guitar and piano based upon the Venn Diagram

provided. The student provides the appropriate computation to show their

work.

3 A generally correct solution, but may contain minor flaws in reasoning

or computation.

2 A partially correct interpretation and/or solution to the problem.

1 A correct solution with no evidence or explanation.

0 An incorrect solution indicating no mathematical understanding of the

concept or task, or no solution is given.

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Assessment

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5 SCORE

1. What is the point called where the perpendicular bisectors of

the sides of a triangle intersect?

2. In △XYZ, point M is the centroid. If

XM = 8, find the length of MA.

3. What is the name of the point that is

two-thirds of the way from each vertex

of a triangle to the midpoint of the

opposite side?

For Questions 4 and 5, use quadrilateral ABCD given

that −−

CD is the perpendicular bisector of −−

AB , and −−

AB is the

perpendicular bisector of −−

CD .

4. Find the value of y.

5. Find the value of x.

1.

2.

3.

4.

5.

1. Determine which angle has the greatest

measure.

For Questions 2 and 3, use quadrilateral PQRS.

2. Find the shortest segment in △PQS.

3. Find the longest segment △QRS.

For Questions 4 and 5, use quadrilateral UVWX.

4. Find the angle with the smallest measure

in △VUW.

5. Find the angle with the greatest measure

in △UWX.

1.

2.

3.

4.

5.

Chapter 5 Quiz 2(Lesson 5-3)

Chapter 5 Quiz 1(Lessons 5-1 and 5-2)

5 SCORE

2

1 3 4

2x 1 7

2y 2 3

y 1 1

A B

C

D

P

QR

S56°63°

61°54° 51°

75°

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5

5

SCORE

SCORE

Chapter 5 Quiz 4(Lesson 5-6)

1. What do you assume in an indirect proof?

For Questions 2 and 3, write the assumption you would make to start an indirect proof of each statement.

2. If 2x + 7 = 19, then x = 6.

3. If △ABC is isosceles with base −−

AC , then −−

AB # −−−

BC .

4. Write an inequality to describe the

possible values of x.

5. MULTIPLE CHOICE Which of the following sets of numbers

can be the lengths of the sides of a triangle?

A 5, 5, 10 B √ %% 39 ,

√ % 8 ,

√ % 5 C 2.5, 3.4, 4.6 D 1, 2, 4

1.

2.

3.

4.

5.

1.

2.

3.

4.

5.

1. Write an inequality 2. Write an inequality relating

relating m∠1 to m∠ 2. relating AB to DE.

3. Write an inequality about the length

of −−−

GH .

For Questions 4 and 5, complete the proof by supplying the missing information for each corresponding location.

Given: AB = DE, and BE > AD

Prove: m∠CAE > m∠CEA

Proof:

Statements Reasons

1. AB = DE, BE > AD 1. Given

2. −−

AB # −−−

DE 2. Def. of # segments

3. (Question 4) 3. Reflexive Prop.

4. m∠CAE > m∠CEA 4. (Question 5)

Chapter 5 Quiz 3(Lessons 5-4 and 5-5)

7

9

x

5

6 12

9

A B D E

FC

75°72°9 95 5

D E

B

C

A

G

I F

H

50°

60° 9

6

67

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5 SCORE Chapter 5 Mid-Chapter Test(Lessons 5-1 through 5-3)

1.

2.

3.

4.

5.

1. Which of the following can intersect outside a triangle?

A angle bisectors C altitudes

B medians D sides

2. What is the name of the point of concurrency of the altitudes of a triangle?

F orthocenter H incenter

G circumcenter J centroid

3. What is the name of the point of concurrency of the medians of a triangle?

A orthocenter C incenter

B circumcenter D centroid

4. Name the longest segment in △ABD.

F −−−

BD H −−−

CD

G −−−

BC J Cannot tell

5. −−

PS is the perpendicular bisector of −−−

QR and −−−

QR is the

perpendicular bisector of −−

PS . If PQ = 2x + 9 and QS = 5x - 12, find x.

A 2 B 3 C 5 D 7

6.

7.

8.

9.

Part II

6. Write an inequality to describe the possible

values of x.

7. List all of the angles that have a measure

greater than ∠1.

8. An advertising company is designing a

corporate flag in the shape of an isosceles

triangle. The right-hand edge of the

company logo will be placed at the centroid,

L, of △ABC. The length of the altitude

CD is 24 inches. How far is the right-hand

edge of the logo from the vertex C?

9. In △XYZ, P is the centroid and YC = 15.

Find the length of YZ.

Part I Write the letter for the correct answer in the blank at the right of each question.

A

B

CD

55°

85°

40°50°

66° 64°

50°

x

C

B

D

A

L

1

43 5 6

7 8

2

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5 SCORE Chapter 5 Vocabulary Test

Write whether each sentence is true or false. If false,

replace the underlined word or number to make a true

sentence.

1. The altitude of a triangle is a segment whose endpoints are a vertex of a triangle and the midpoint of the side opposite the vertex.

2. The centroid of a triangle is the point where the altitudes of the triangle intersect.

Choose the correct term to complete each sentence.

3. The point of concurency of the perpendicular bisectors of a triangle is called the (circumcenter, median).

4. The (incenter, orthocenter) of a triangle is the intersection of the angle bisectors of the triangle.

5. The sum of the measures of any two sides of a triangle is (greater, less) than the measure of the third side.

Choose from the terms above to complete each sentence.

6. A(n) is a segment that joins a vertex of a triangle and is perpendicular to the side opposite to the vertex.

7. Proof by contradiction is a type of .

8. The of a triangle is equidistant from the vertices of the triangle.

Define each term in your own words.

9. concurrent lines

10. median

1.

2.

3.

4.

5.

6.

7.

8.

altitude

centroid

circumcenter

concurrent lines

incenter

indirect proof

indirect reasoning

median

orthocenter

perpendicular bisector

point of concurrency

proof by contradiction

Assessment

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5 SCORE Chapter 5 Test, Form 1

Write the letter for the correct answer in the blank at the right of each question.

For Questions 1–4, refer to the figure at the right.

1. Name an altitude.

A −−−

DE C " #$ GB

B −−

AB D ##$ CF

2. Name a perpendicular bisector.

F −−−

DE G −−

AB H " #$ GB J ##$ CF

3. Name an angle bisector.

A −−−

DE B −−

AB C " #$ GB D ##$ CF

4. Name a median.

F −−−

DE G −−

AB H " #$ GB J ##$ CF

For Questions 5–7, refer to the figure to determine

which is a true statement for the given information.

5. −−

AC is a median.

A m∠ACD = 90 C BC = CD

B ∠BAC & ∠DAC D ∠B & ∠D

6. −−

AC is an angle bisector.

F m∠ACD = 90 G ∠BAC & ∠DAC H BC = CD J ∠B & ∠D

7. −−

AC is an altitude.

A m∠ACD = 90 B ∠BAC & ∠DAC C BC = CD D ∠B & ∠D

8. Name the longest side of △DEF.

F −−−

DE H −−−

DF

G −−

EF J cannot tell

9. Which angle in △ABC has the greatest measure?

A ∠A C ∠C

B ∠B D cannot tell

10. Which theorem compares two sides and the included angle of two triangles?

F Hinge Theorem

G Converse of the Hinge Theorem

H Exterior Angle Inequality Theorem

J Triangle Inequality Theorem

11. Which assumption would you make to indirectly prove x > 5?

A x < 5 B x ≤ 5 C x = 5 D x > 5

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

A

BC D

E F

G

A D

C

B

D62°

10°

108°

F

E

A C

B

9

5 7

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5

12. Find the possible values for m∠1.

F 180 > m∠1 > 62 H 0 < m∠1 < 62

G 90 > m∠1 > 62 J m∠1 = 118


A 5 C 10

B 7 D 15

14. If D is the circumcenter of △ABC and AD = 6, find BD.

F 4 H 9

G 6 J 12

15. Choose the assumption you would make to start an indirect proof of x > 3.

A x < 3 B x ≥ 3 C x ≤ 3 D x = 3

16. Choose the assumption you would make to start an indirect proof. Given: a ∦ b

Prove: ∠1 and ∠2 are not supplementary.

F a ‖ b H ∠1 & ∠2

G ∠1 and ∠2 are supplementary. J ∠1 and ∠2 are complementary.

17. Which of the following sets of numbers can be the lengths of the sides of a triangle?

A 12, 9, 4 B 1, 2, 3 C 5, 5, 10 D √ ) 2 , √ ) 5 , √ )) 18

18. −−

BF is a median of △BEC. If EC = 15, find FC.

F 5 H 10

G 7.5 J 30

For Questions 19 and 20, refer to the figures.

19. Given: −−

AC & −−−

DF , −−

AB & −−−

DE , m∠A > m∠D

Which can be concluded by the Hinge Theorem?

A △ABC & △DEF C BC < EF

B BC = EF D BC > EF

20. Given: −−

AB & −−−

DE , −−−

BC & −−

EF , AC < DF

Which can be concluded by the Converse of the Hinge Theorem?

F m∠B < m∠E H m∠B = m∠E

G m∠B > m∠E J △BAC & △EDF

Bonus −−−

QS is a median of △PQR with point S on −−

PR .

If PS = x2 - 3x and SR = 2x + 6, find the possible

value(s) of x.

A

CBD

E F

A C

B

D E F

A C

B

D

x + 3

5

W X

M

T V

N

15

62° 1 12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

Chapter 5 Test, Form 1 (continued)

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5 SCORE Chapter 5 Test, Form 2A

Write the letter for the correct answer in the blank at the right of each question.

For Questions 1–4, refer to the figure.


A −−

KI B ""# GL C $ ""# JM D −−−

HJ

2. Name a median.

F −−

KI G ""# GL H $ ""# JM J

−−− HJ


A −−

KI B ""# GL C $ ""# JM D

−−− HJ


F −−

KI G ""# GL H $ ""# JM J

−−− HJ



5. −−−

YW is an angle bisector.

A ∠YWZ is a right angle. C XW = WZ

B ∠XYW & ∠ZYW D XY = ZY

6. −−−

YW is an altitude.

F ∠YWZ is a right angle. H XW = WZ

G ∠ XYW & ∠ZYW J XY = ZY

7. −−−

YW is a median.

A ∠YWZ is a right angle. C XW = WZ

B ∠ XYW & ∠ZYW D XY = ZY

8. Name the longest side of △ABC.

F −−

AB H −−

AC

G −−−

BC J cannot tell

9. Name the angle with greatest measure in △DEF.

A ∠D C ∠F

B ∠E D cannot tell

10. Which theorem compares the sides of the same triangle?

F Hinge Theorem H Exterior Angle Inequality Theorem

G Converse of the Hinge Theorem J Triangle Inequality Theorem

11. Tisha wants to plant a garden in the widest corner of her triangular

backyard. The backyard is bordered by the back of the house that is 50 feet

long, fence A that is 27 feet long, and fence B that is 35 feet long. Which

corner has the widest measure?

A corner between fences A and B

B all three corners have the same measure

C corner between the back of the house and fence A

D corner between the back of the house and fence B

3

7

9

F

DE

22° 84°

74°

A C

B

X

Y

Z

W

H

J

MI G

LK

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

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5


F 90 > m∠1 > 74 H 0 < m∠1 < 74

G 180 > m∠1 > 74 J m∠1 = 106


A 9 C 27

B 11 D 32

14. Which is another name for an indirect proof?

F proof by deduction H proof by inverse

G proof by converse J proof by contradiction

15. Choose the assumption you would make to start an indirect proof of x < 2.

A x > 2 B x ≥ 2 C x = 2 D x ≤ 2

16. Choose the assumption you would make to start an indirect proof.

Given: ∠1 is an exterior angle of △ABC.

Prove: m∠1 = m∠B + m∠C

F ∠1 is not an exterior angle of △ABC.

G ∠1 is an interior angle of △ABC.

H m∠1 ≠ m∠B + m∠C

J m∠1 = m∠B

17. Which of the following sets of numbers can be the lengths of the sides of a triangle?

A 6, 6, 12 B 6, 7, 13 C √ ' 2 , √ ' 5 , √ '' 15 D 2.6, 8.1, 10.2

18. What is the relationship between the lengths of −−−

QS and

−− RS ?

F QS = RS H QS > RS

G QS < RS J cannot tell


DC and

−−− AD ?

A DC < AD C DC = AD

B DC > AD D cannot tell

20. What is the relationship between the measures of ∠1

and ∠2?

F m∠1 = m∠2 H m∠1 > m∠2

G m∠1 < m∠2 J cannot tell

Bonus −−−

YW bisects ∠XYZ in △XYZ. Point W is on −−

XZ .

If m∠XYW = 2x + 18 and m∠ZYW = x2 - 5x, find the possible value(s) of x.

8

8

15

131

2

10

10

30°

20°

A

C

D

B

RT

Q S

12.

13.

14.

15.

16.

17.

18.

19.

20.

x + 2

x + 7

A

B

C

D

EF27

174°

Chapter 5 Test, Form 2A (continued)

B:

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5 SCORE Chapter 5 Test, Form 2B

Write the letter for the correct answer in the blank at the right of each

question.

For Questions 1–4, refer to the figure.

1. Name a median.

A −−−

RW C −−−

QT

B " #$ SV D ##$ RU


F −−−

RW G " #$ SV H −−−

QT J ##$ RU


A −−−

RW B " #$ SV C −−−

QT D ##$ RU


F −−−

RW G −−

RP H −−−

QT J ##$ RU



5. −−−

FG is an altitude.

A ∠DGF is a right angle. C DG = GE

B DF = EF D ∠DFG & ∠EFG

6. −−−

FG is a median.

F ∠DGF is a right angle. H DG = GE

G DF = EF J ∠DFG & ∠EFG

7. −−−

FG is an angle bisector.

A ∠DGF is a right angle. C DG = GE

B DF = EF D ∠DFG & ∠EFG

8. Name the longest side of △ABC.

F −−

AB H −−

AC

G −−−

BC J cannot tell

9. Name the angle with the greatest measure in △GHI.

A ∠G C ∠I

B ∠H D cannot tell

10. Two sides of a triangle are congruent to two sides of another triangle and the

included angle in the first triangle has a greater measure than the included

angle in the second triangle. These are the assumptions of which theorem?

F Hinge Theorem H Exterior Angle Inequality Theorem

G Converse of the Hinge Theorem J Triangle Inequality Theorem

5

7

9

H

GI

70° 48°

62°

A C

B

G

D

E F

P

W

R

S

TU V

Q 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

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5

11. Carrie, Maria, and Nayla are friends that

live close to one another. Which two friends

have the shortest distance between them?

A Maria and Nayla

B Carrie and Maria

C Carrie and Nayla

D All three live equal distances from

each other.


F m∠1 = 124 H 90 > m∠1 > 56

G 0 < m∠1 < 56 J 180 > m∠1 > 56

13. Find ST.

A 12 C 23

B 18 D 24

14. Which of the following is the last step in an indirect proof?

F show the assumption true H show the conclusion false

G show the assumption false J contradict the conclusion

15. Choose the assumption you would make to start an indirect proof of x ≤ 1.

A x > 1 B x = 1 C x < 1 D x ≤ 1

16. Choose the assumption you would make to start this indirect proof.

Given: −−

AB bisects ∠CAD.

Prove: ∠ACB ≇ ∠DAB

F −−

AB does not bisect ∠CAD. H −−

AB is a median.

G △ACD is isosceles. J ∠ACB & ∠DAB

17. Which of the following sets of numbers can be the lengths

of the sides of a triangle?

A 12, 9, 2 B 11, 12, 23 C 2, 3, 4 D √ ( 3 , √ ( 5 , √ (( 18


YW and −−

YX ?

F YW = YX H YW > YX

G YW < YX J cannot tell


DG and −−−

GF ?

A DG > GF C DG = GF

B DG < GF D cannot tell

20. What is the relationship between the measures of ∠1 and ∠2?

F m∠1 = m∠2 H m∠1 > m∠2

G m∠1 < m∠2 J cannot tell

Bonus −−−

HJ is an altitude of △GHI with point J on −−

GI .

If m∠GJH = 5x + 30, GH = 3x + 4, HI = 5x - 3,

JI = 4x - 3, and GJ = x + 6, find the perimeter of △GHI.

Nayla

Carrie Maria

(20x + 7)°

(5x + 50)° (5x + 3)°

8

8

30°

20°

D

FG

E

W

Y

X

Z

x + 6

5P

QR

S

T

U2x - 1

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

156°

11

10

10

12

12

M

NK

L


Chapter 5 Test, Form 2B (continued)

B:

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5 SCORE Chapter 5 Test, Form 2C


2. The perimeter of ABCD is 44. Find the value of x. Then describe the

relationship between "# AC and −−−

BD .

3. If point E is the centroid of △ABC, BD = 12, EF = 7, and AG = 15, find ED.

4. The vertices of △XYZ are X(-2, 6), Y(4, 10), and Z(14, 6). Find the coordinates of the centroid of △XYZ.

5. If −−−

PO is an angle bisector of ∠MON, find the value of x.

6. A biker will jump over a ramp, where x and z are measured in feet. Write an inequality relating x and z.

7. List the angles of △GHI in order from smallest to largest measure.

8. List the sides of △PQR in order from shortest to longest.

9. Find the shortest segment.

10. Write the assumption you would make to start an indirect proof of the statement. If 16 is a factor of n, then 4 is a factor

of n.

11. Write the assumption you would make to start an indirect proof of the statement. If

−− AB is an altitude of equilateral

triangle ABC, then −−

AB is a median.

2(z + 3)x5165˚

70˚

55°55°65°

65˚60°

60°

Y

X Z

W

80° 45°

55°

Q

P R

2 in.

1.2 in.3 in.

IG

H

(2x - 10)° (x + 15)°O

PN M

DA C

B

F G

E

2x - 3

2x - 7

x + 5

x + 1

D

B

AC

G

B CD

A

E

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.


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5

C

BAD

5 ft

5 ft

12 ft

11 ft

12

R Q

P

S T U

Y

ZXW

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

12. Write the assumption you would make to start an indirect proof for the following.

Given: −− XY ≇

−− YZ

−−− YW bisects ∠XYZ.

Prove: ∠X ≇ ∠Z

13. The measures of two sides of a triangle are 10 meters and 23 meters. If the measure of the third side is x meters, find the range for the value of x.

14. −−− PU is a median of △PTQ. If UQ = 6, find TQ.

15. If %%& BD bisects ∠ABC, find the value of x.

16. Write an inequality to compare EF and GH.

17. Write an inequality to compare m∠1 and m∠2.

For Questions 18–20, complete the proof below by

supplying the missing information for each corresponding

location.

Given: AD = CB and AC > DB

Prove: m∠ ADC > m∠DCB

Proof:

Statements Reasons

1. AD = CB and AC > DB 1. Given

2. −−− AD '

−−− CB 2. (Question 18)

3. −−− CD '

−−− CD 3. (Question 19)

4. m∠ADC > m∠DCB 4. (Question 20)

Bonus Write an equation in slope-intercept form for the altitude to

−−− BC .

x

y

C(2c, 0)A(0, 0)

B(2a, 2b)

O

2x + 30

3x - 4B

C

A

D


Chapter 5 Test, Form 2C (continued)

23°20°

H

G F

E

6

6

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5 SCORE Chapter 5 Test, Form 2D


2. The perimeter of PRQS is 34. Find the value of x. Then describe the relationship between

"# RS and −−−

PQ .

3. If point N is the centroid of △HIJ, IM = 18, KN = 4, and HL = 15, find JN.

4. The vertices of △DEF are D(4, 12), E(14, 6), and F(-6, 2). Find the coordinates of the circumcenter of △DEF.

5. If −−−

RU is an altitude for △RST, find the value x.

6. A rubber doorstop has a hypotenuse measuring 7z and a height measuring x - 5. Write an inequality relating x and z.

7. List the angles of △TUV in order from smallest to largest measure.

8. List the sides of △FGH in order from shortest to longest.

9. Name the longest segment.

10. Write the assumption you would make to start an indirect proof of the statement. If n is an even number, then n2 is an

even number.

11. Write the assumption you would make to start an indirect proof of the statement. If

−−− AD is an angle bisector of equilateral

triangle ABC, then −−−

AD is an altitude.

50°96°

34°

30° 70°

80°

K

N

L M

82° 41°

57°

H

F

G

1.8

5

3.9

VT

U

7z x-5

150˚

57˚

(5x - 10)°T

U

S

R

MH J

I

K L

N

4x - 10

2x - 3 7

10

R

S

QP

G

H

N

K

JL

I

M

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.


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5

x

y

E(a, 0)C(0, 0)

D(b, c)

O

M

ABK

24°

8

830°

E

D

B

C

7

7

10

9

12

3x - 20

2x + 15Y

Z

X

W

A D E F C

B

S

QPV

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

12. Write the assumption you would make to start an indirect proof for the following.

Given: V is not the midpoint of −−− PQ ; < P " < Q

Prove: −− SV ⊥⁄

−−− PQ .

13. The measures of two sides of a triangle are 14 feet and 29 feet. If the measure of the third side is x feet, find the range for the value of x.

14. −−− BD is a median of △ABE. If AD = 8, find DE.

15. If %%& YW bisects ∠XYZ, find the value of x.

16. Write an inequality relating m∠1 and m∠2.

17. Write an inequality relating BC and ED.

For Questions 18–20, complete the

proof below by supplying the missing

information for each corresponding location.

Given: K is the midpoint of −− AB .

m∠MKB < m∠MKA

Prove: MB < AM

Proof:

Statements Reasons

1. K is the midpoint of −− AB . 1. Given

m∠MKB < m∠MKA

2. −−− BK "

−−− KA 2.

3. −−− MK " −−− MK 3. (Question 19)

4. MB < AM 4. (Question 20)

Bonus Write an equation in slope-intercept form for the perpendicular bisector of

−−− CE .


Chapter 5 Test, Form 2D (continued)

(Question 18)

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5 SCORE Chapter 5 Test, Form 3

1. If point G is the centroid of △ABC, AE = 24, DG = 5, and CG = 14, find DB.

2. The vertices of △EFG are E(2, 4), F(10, -6), and G(-4, -8). Find the coordinates of the orthocenter of △EFG.

3. If −−

JL is a median for △IJK, find the value of x.

4. Write a compound inequality for the possible measures of ∠L.

5. List the angles of △GHI in order from smallest to largest measure.

6. List the sides of △PQR in order from shortest to longest.

7. Name the shortest and the longest segments.

8. Write the assumption you would make to begin an indirect proof of the statement. If 2x + 6 = 12, then x = 3.

9. Determine whether 8, 4, and 2 can be the lengths of the sides of a triangle. Write yes or no. Explain.

10. Write the assumption you would make to begin an indirect proof of the statement. The three angle bisectors of a triangle

are concurrent.

11. Write and solve an inequality for x.

(3x - 4)°

6

10

4

4

(12x - 31)°

53° 64°

63°55°

72°

53°

V

Y

WX

45° 55°

80°

P R

Q

9.6

7 8

I H

G

146°

L

NM

3x + 10 2x + 42LI K

J

FA B

C

DE

G 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.


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5

2x + 23

9x + 6x + 18

7x - 2HE G

F

x

y

F(2a, 0)D(0, 0)

E(2c, 2d)

O

Y

Z W

XK

110°84°

8

866

3x + 10

x + 20

XZ W

T

Y

B

AC

D

E

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

12. If −−−

FH is a median of △EFG, find the

perimeter of △EFG.

13. Write the assumption you would make

to start an indirect proof for the following.

Given: −−

AB ≇ −−−

DE and −−

AC $ −−−

CD

Prove: ∠B ≇ ∠E

14. The measures of two sides of a triangle are 24 inches and

29 inches. If the measure of the third side is x inches, find the

range for the value of x.

15. −−−

YW is the perpendicular bisector of −−

ZT .

If TW = 3, YW = 8, and XZ = 12. Find XY.

16. Write and solve an inequality for

the value of x.

For Questions 17–20, complete the proof below by supplying the missing information for each corresponding location.

Given: XW = YZ, XK > WK, and KZ > KY

Prove: m∠XWZ > m∠YZW

Proof:Statements Reasons

1. XW = YZ, XK > WK, 1. Given

and KZ > KY

2. −−−

XW $ −−

YZ 2. (Question 17)

3. XZ > WY 3. (Question 18)

4. −−−

WZ $ −−−

WZ 4. (Question 19)

5. m∠XWZ > m∠YZW 5. (Question 20)

Bonus Write an equation in slope-intercept

form for the line containing the

median to −−−

DE .


Chapter 5 Test, Form 3 (continued)

Assessment

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NAME DATE PERIOD

5 SCORE Chapter 5 Extended-Response Test

Demonstrate your knowledge by giving a clear, concise solution to

each problem. Be sure to include all relevant drawings and justify

your answers. You may show your solution in more than one way or

investigate beyond the requirements of the problem.

1. Two sticks are bent and connected with a rubber band as shown in the

diagram. Describe what happens to the rubber band as the ends of the

sticks are pulled farther apart. Name the theorem this situation

illustrates.

2. Mary says −−−

HE is a perpendicular bisector of −−−

DG and Ashley says it is not.

Who is correct? Explain your answer.

3. Suppose −−−

BD is drawn on this figure so that point D is on ""# AC and has a

length of 6 centimeters. If the shortest distance from B to ""# AC is

5 centimeters, in how many different places on ""# AC could point D be

located? Explain.

4. Draw a triangle that satisfies each situation.

a. Two of the sides are altitudes.

b. The altitudes intersect outside the triangle.

c. The altitudes intersect inside the triangle.

d. The altitudes are also the medians of the triangle.

10 cm

A

B

C

6 in.4 in.

5 in.

F

J

G

KH

ED


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NAME DATE PERIOD

5 SCORE

35°

35°

3x - 7

x + 5

45°

Standardized Test Practice

(Chapters 1-5)

1. If ∠BXY is a right angle, then which statements are true? (Lesson 1-4)

I m∠BXY = 90

II The measure of an angle vertical to ∠BXY would be 90.

III The measure of an angle supplementary to ∠BXY would be 90.

A I only B I and III C I, II, and III D I and II

2. Which is the contrapositive of the conditional statement If m∠K = 45, then x = 5? (Lesson 2-3)

F If m∠K ≠ 45, then x ≠ 5 H If x = 5, then m∠K = 45

G If x ≠ 5, then m∠K ≠ 45 J If m∠K ≠ 45, then x = 5

3. Find m∠HJK. (Lesson 3-2)

A 33 C 78

B 45 D 147

4. The line y - 5 = -x + 3 satisfies which conditions? (Lesson 3-4)

F m = -1, contains (-5, 3) H m = -1, contains (5, 3)

G m = 1, contains (-5, -3) J m = -1, contains (5, -3)

5. Given D(0, 4), E(2, 4), F(2, 1), A(0, 2), and C(-2, -1), which coordinates for B would make △ABC " △DEF? (Lesson 4-4)

A B(-2, 2) C B(0, 0)

B B(0, 1) D B(-1, 0)

6. In △XYZ, which type of line is ℓ ? (Lesson 5-2) F perpendicular bisector H altitude

G angle bisector J median

7. Which assumption would you make to start an indirect proof of the statement. If 2x - 5 < 17, then x < 11? (Lesson 5-4)

A x < 11 B x ≥ 11 C x > 11 D x ≠ 11

8. Which inequality describes the possible values of x? (Lesson 5-6)

F x > 6 H x ≮ 12

G x < 6 J 6 < x < 12

X Z

Y ℓ

45°33°H K

J

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

Part 1: Multiple Choice

Instructions: Fill in the appropriate circle for the best answer.


Assessment

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NAME DATE PERIOD

5

For Questions 9–11 refer to the figure.

9. What is the measure of ∠E? (Lesson 4-2)

A 18 C 43

B 40 D 81

10. Which of the following could not be the length of −−

EF ? (Lesson 5-5)

F 20 m G 53 m H 75 m J 80 m

11. Which of the following inequalities is true? (Lesson 5-3)

A z = 60 B z = 49 C P = 40 D Q = 39

For Questions 12 and 13

refer to the figure.

12. Which line segment is the shortest? (Lesson 5-3)

F −−−

PQ H −−−

QR

G −−

RS J −−

PS

13. Which line segment is the longest? (Lesson 5-3)

A −−−

PQ B −−−

QR C −−

RS D −−

PS

14. What is the perimeter of △MOP with vertices M(0, 4), O(0, 0), P(3, 0)? (Lesson 3-6)

F -12 G -5 H 5 J 12

15. Which of the following sets of numbers cannot be lengths of the sides of a triangle? (Lesson 5-5)

A 1, 2, 3 B 2, 3, 4 C 3, 4, 5 D 4, 5, 6

120˚

119˚

28˚30˚

SR

QP

123˚

100 m

132 m

(2z + 3)°

42˚D

E

F 9. A B C D

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. If −−−

BD is an altitude of △ABC, find the value of x. (Lesson 5-2)

17. The measures of two sides of △ABC are 19 and 15. The range for measure of the third side n would be 4 < n < . (Lesson 5-5)

16. 17.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

Part 2: Gridded Response

Instructions: Enter your answer by writing each digit of the

answer in a column box and then shading in the appropriate circle

that corresponds to that entry.

(2x + 17)°(3x - 2)°

35°A C

B

D E


?

Standardized Test Practice (continued)

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5

18. Find a counterexample for the statement. Five is the only whole number between 4.5 and 6.1. (Lesson 2-1)

19. What is the length of the side opposite the vertex angle of isosceles △XYZ with vertices at X(-3, 4), Y(8, 6), and Z(3, -4)? (Lesson 4-1)

20. What is the distance between A(-12, 12) and B(-5, -12)? (Lesson 3-6)

21. The vertices of △ABC are A(-2, 3), B(4, 3), and C(-2, -3). Find the coordinates of each point of concurrency of △ABC. (Lesson 5-2)

a. circumcenter

b. centroid

c. orthocenter

22. Given the line y = 5x + 2.

a. What is the equation of the parallel line that intercepts the y-axis at -2? (Lesson 3-5)

b. What is the equation of the perpendicular line that intersects y = 5x + 2 at x = 0? (Lesson 3-6)

c. Find the point of intersection of the lines found in part a and b above? (Lesson 3-6)

18.

19.

20.

21a.

b.

c.

22a.

b.

c.

Part 3: Short Response

Instructions: Write your answer in the space provided.


Standardized Test Practice (continued)

Answers

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Chapter 5 A1 Glencoe Geometry

Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.

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Exam

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Exerc

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Answers (Anticipation Guide and Lesson 5-1)

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8 =

8x -

16

24 =

3x

8 =

x

Exerc

ises

Fin

d e

ach

me

asu

re

.

1

. ∠

AB

E

2.

∠Y

BA

43°

47°

88

3

. M

K

4.

∠E

WL

3x

-8

2x

+1

(3x

+21)°

(7x

+5)°

Po

int

U i

s t

he

in

ce

nte

r o

f △

GH

Y.

Fin

d e

ach

m

ea

su

re

be

low

.

5. M

U

6.

∠U

GM

7

. ∠

PH

U

8. H

U

Exam

ple

28°

21°

12

5

4

3

47

1

9

33

5

21

28

13

Lesson 5-1


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

7

Gle

nc

oe

Ge

om

etr

y

5-1

Sk

ills

Pra

ctic

e

Bis

ecto

rs o

f Tri

an

gle

sF

ind

ea

ch

me

asu

re

.

1

. F

G

2. K

L

5x

- 17

13

13

3x

+ 1

4.2

3

. T

U

4.

∠L

YF

2x

+ 24

5x

- 30

58°

5

. IU

6.

∠M

YW

19°

19°

2x

+ 5

7x

( 2x

+ 5)°

( 4x

- 1)°

Po

int

P i

s t

he

cir

cu

mce

nte

r o

f △

AB

C.

Lis

t a

ny

se

gm

en

t(s)

co

ng

ru

en

t to

ea

ch

se

gm

en

t b

elo

w.

7

. −−

− B

R

8

. −

−

CS

9

. −

−

BP

Po

int

A i

s t

he

in

ce

nte

r o

f △

PQ

R.

Fin

d e

ach

m

ea

su

re

be

low

.

10

. ∠

AR

U

11

. A

U

12

. ∠

QP

K

40°

20°

( 4x

- 9)°

( 3x

+ 2)°

2

8

4.2

6

0

58

7

11

−−

AR

−−

AP

, −

−

CP

40

20

35

−−

AS

Answers (Lesson 5-1)

Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

8

Gle

nc

oe

Ge

om

etr

y

Fin

d e

ach

me

asu

re

.

1

. T

P

2. V

U

7

9

9

3x

+10

7x

+2

3

. K

N

4.

∠N

JZ

3x

x+

10

I

NZ

J

38°

5

. Q

A

6.

∠M

FZ

E

RA

Q

3x

+16

7x

(2x

-1)°

(x+

9)°

Po

int L

is t

he

cir

cu

mce

nte

r o

f △BKT

. L

ist

an

y

se

gm

en

t(s)

co

ng

ru

en

t to

ea

ch

se

gm

en

t b

elo

w.

7.

−−

−

BN

8

. −

−

BL

Po

int A

is t

he

in

ce

nte

r o

f △LYG

. F

ind

ea

ch

m

ea

su

re

be

low

.

9

. ∠

ILA

10

. ∠

JG

A

11

. S

CU

LP

TU

RE

A

tri

an

gu

lar

en

tran

cew

ay h

as

wall

s w

ith

corn

er

an

gle

s of

50,

70,

an

d 6

0.

Th

e d

esi

gn

er

wan

ts t

o p

lace

a t

all

bro

nze s

culp

ture

on

a r

ou

nd

ped

est

al

in a

cen

tral

loca

tion

equ

idis

tan

t fr

om

th

e t

hre

e w

all

s. H

ow

can

th

e d

esi

gn

er

fin

d w

here

to p

lace

th

e

scu

lptu

re?

5-1

Practi

ce

Bis

ecto

rs o

f Tri

an

gle

s

21°

32°

7

16

1

5

38

2

8

19

−−

NT

−−

KL ,

−−

LT

32 37

Fin

d t

he i

ncen

ter,

wh

ere

th

e t

hre

e a

ng

le b

isecto

rs m

eet.

Lesson 5-1


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

9

Gle

nc

oe

Ge

om

etr

y

1

. W

IND

CH

IME

Joan

na h

as

a f

lat

wood

en

tr

ian

gu

lar

pie

ce a

s p

art

of

a w

ind

ch

ime.

Th

e p

iece

is

susp

en

ded

by a

wir

e

an

chore

d a

t a p

oin

t equ

idis

tan

t fr

om

th

e

sid

es

of

the t

rian

gle

. W

here

is

the

an

chor

poin

t lo

cate

d?

2

. P

ICN

ICS

M

ars

ha a

nd

Bil

l are

goin

g

to t

he p

ark

for

a p

icn

ic.

Th

e p

ark

is

tria

ngu

lar.

On

e s

ide o

f th

e p

ark

is

bord

ere

d b

y a

riv

er

an

d t

he o

ther

two

sid

es

are

bord

ere

d b

y b

usy

str

eets

. M

ars

ha a

nd

Bil

l w

an

t to

fin

d a

sp

ot

that

is e

qu

all

y f

ar

aw

ay f

rom

th

e r

iver

an

d

the s

treets

. A

t w

hat

poin

t in

th

e p

ark

sh

ou

ld t

hey s

et

up

th

eir

pic

nic

?

3. M

OV

ING

M

art

in h

as

3 g

row

n c

hil

dre

n.

Th

e f

igu

re s

how

s th

e l

oca

tion

s of

Mart

in’s

ch

ild

ren

on

a m

ap

th

at

has

a

coord

inate

pla

ne o

n i

t. M

art

in w

ou

ld l

ike

to m

ove t

o a

loca

tion

th

at

is t

he s

am

e

dis

tan

ce f

rom

all

th

ree o

f h

is c

hil

dre

n.

Wh

at

are

th

e c

oord

inate

s of

the l

oca

tion

on

th

e m

ap

th

at

is e

qu

idis

tan

t fr

om

all

th

ree c

hil

dre

n?

y

xO5

5-5

4

. N

EIG

HB

OR

HO

OD

A

man

da i

s lo

ok

ing

at

her

neig

hborh

ood

map

. S

he n

oti

ces

that

her

hou

se a

lon

g w

ith

th

e h

om

es

of

her

frie

nd

s, B

rian

an

d C

ath

y,

can

be t

he

vert

ices

of

a t

rian

gle

. T

he m

ap

is

on

a

coord

inate

gri

d.

Am

an

da’s

hou

se i

s at

the p

oin

t (1

, 3),

Bri

an

’s i

s at

(5,

-1),

an

d C

ath

y’s

is

at

(4,

5).

Wh

ere

wou

ld t

he

thre

e f

rien

ds

meet

if t

hey e

ach

left

th

eir

h

ou

ses

at

the s

am

e t

ime a

nd

walk

ed

to

the o

pp

osi

te s

ide o

f th

e t

rian

gle

alo

ng

the p

ath

of

short

est

dis

tan

ce f

rom

th

eir

h

ou

se?

5

. P

LA

YG

RO

UN

D A

con

crete

com

pan

y i

s p

ou

rin

g c

on

crete

in

to a

tri

an

gu

lar

form

as

the c

en

ter

of

a n

ew

pla

ygro

un

d.

a.

Th

e f

ore

man

measu

res

the t

rian

gle

an

d n

oti

ces

that

the i

nce

nte

r an

d t

he

circ

um

cen

ter

are

th

e s

am

e.

Wh

at

typ

e o

f tr

ian

gle

is

bein

g c

reate

d?

b.

Su

pp

ose

th

e f

ore

man

ch

an

ges

the

tria

ngu

lar

form

so t

hat

the

circ

um

cen

ter

is o

uts

ide o

f th

e

tria

ngle

bu

t th

e i

nce

nte

r is

in

sid

e

the t

rian

gle

. W

hat

typ

e o

f tr

ian

gle

w

ou

ld b

e c

reate

d?

5-1

Wo

rd

Pro

ble

m P

racti

ce

Bis

ecto

rs o

f Tri

an

gle

s

the i

ncen

ter,

wh

ich

is w

here

all

thre

e a

ng

le b

isecto

rs i

nte

rsect

a

t th

e i

ncen

ter

of

the p

ark

(

0,

-2)

( 11

−

5 ,

16

−

5 )

eq

uilate

ral

ob

tuse t

rian

gle


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

10

G

len

co

e G

eo

me

try

Inscri

bed

an

d C

ircu

mscri

bed

Cir

cle

sT

he t

hre

e a

ngle

bis

ect

ors

of

a t

rian

gle

in

ters

ect

in

a s

ingle

poin

t ca

lled

th

e i

nce

nte

r. T

his

poin

t is

th

e c

en

ter

of

a c

ircl

e t

hat

just

tou

ches

the t

hre

e s

ides

of

the t

rian

gle

. E

xce

pt

for

the

thre

e p

oin

ts w

here

th

e c

ircl

e t

ou

ches

the s

ides,

th

e c

ircl

e i

s in

sid

e t

he t

rian

gle

. T

he c

ircl

e i

s

said

to b

e i

nscrib

ed

in

th

e t

rian

gle

.

1

. W

ith

a c

om

pass

an

d a

str

aig

hte

dge,

con

stru

ct t

he i

nsc

ribed

circ

le f

or

△P

QR

by f

oll

ow

ing t

he s

tep

s belo

w.

S

tep

1 C

on

stru

ct t

he b

isect

ors

of

∠R

an

d ∠

Q.

Label

the p

oin

t

wh

ere

th

e b

isect

ors

meet,

A.

S

tep

2 C

on

stru

ct a

perp

en

dic

ula

r se

gm

en

t fr

om

A t

o −

−−

RQ

. U

se

the l

ett

er

B t

o l

abel

the p

oin

t w

here

th

e p

erp

en

dic

ula

r

segm

en

t in

ters

ect

s −

−−

RQ

.

S

tep

3 U

se a

com

pass

to d

raw

th

e c

ircl

e w

ith

cen

ter

at

A a

nd

rad

ius

−−

AB

.

Co

nstr

uct

the

in

scrib

ed

cir

cle

in

ea

ch

tria

ng

le.

2

.

3

.

Th

e t

hre

e p

erp

en

dic

ula

r bis

ect

ors

of

the s

ides

of

a t

rian

gle

als

o m

eet

in a

sin

gle

poin

t. T

his

poin

t is

th

e c

en

ter

of

the c

ircu

msc

ribed

cir

cle,

wh

ich

pass

es

thro

ugh

each

vert

ex o

f th

e

tria

ngle

. E

xce

pt

for

the t

hre

e p

oin

ts w

here

th

e c

ircl

e t

ou

ches

the t

rian

gle

, th

e c

ircl

e i

s

ou

tsid

e t

he t

rian

gle

.

4

. F

oll

ow

th

e s

tep

s belo

w t

o c

on

stru

ct t

he c

ircu

msc

ribed

cir

cle

for

△F

GH

.

S

tep

1 C

on

stru

ct t

he p

erp

en

dic

ula

r bis

ect

ors

of

−−

−

FG

an

d −

−−

FH

.

Use

th

e l

ett

er

A t

o l

abel

the p

oin

t w

here

th

e

perp

en

dic

ula

r bis

ect

ors

meet.

S

tep

2 D

raw

th

e c

ircl

e t

hat

has

cen

ter

A a

nd

rad

ius

−−

AF

.

Co

nstr

uct

the

cir

cu

mscrib

ed

cir

cle

fo

r e

ach

tria

ng

le.

5

.

6

.

FH

G

A

P

QR

A B

5-1

En

rich

men

t

Lesson 5-2


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

11

G

len

co

e G

eo

me

try

Me

dia

ns

A m

ed

ian

is

a l

ine s

egm

en

t th

at

con

nect

s a v

ert

ex o

f a t

rian

gle

to t

he m

idp

oin

t of

the o

pp

osi

te s

ide.

Th

e t

hre

e m

ed

ian

s of

a t

rian

gle

in

ters

ect

at

the c

en

tro

id o

f th

e

tria

ngle

. T

he c

en

troid

is

loca

ted

tw

o t

hir

ds

of

the d

ista

nce

fro

m a

vert

ex t

o t

he m

idp

oin

t of

the s

ide o

pp

osi

te t

he v

ert

ex o

n a

med

ian

.

In

△ABC

, U

is t

he

ce

ntr

oid

an

d

BU

= 1

6.

Fin

d UK

an

d BK

.

BU

= 2

−

3 B

K

16 =

2

−

3 B

K

24 =

BK

BU

+ U

K =

BK

16

+ U

K =

24

UK

= 8

Exerc

ises

In △

ABC

, AU

= 1

6, BU

= 1

2,

an

d CF

= 1

8.

Fin

d

ea

ch

me

asu

re

.

1

. U

D

2. E

U

3

. C

U

4

. A

D

5

. U

F

6. B

E

In △

CDE

, U

is t

he

ce

ntr

oid

, UK

= 1

2, EM

= 2

1,

an

d UD

= 9

. F

ind

ea

ch

me

asu

re

.

7. C

U

8. M

U

9. C

K

10

. J

U

11

. E

U

12

. J

D

5-2

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Med

ian

s a

nd

Alt

itu

des o

f Tri

an

gle

s

Exam

ple

16

12 12 9

8 12

6

6 24

18

24

36

14

7 4.5

13.5

Answers (Lesson 5-1 and Lesson 5-2)

Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

12

G

len

co

e G

eo

me

try

Alt

itu

de

s

An

alt

itu

de

of

a t

rian

gle

is

a s

egm

en

t fr

om

a v

ert

ex t

o t

he l

ine c

on

tain

ing t

he

op

posi

te s

ide m

eeti

ng a

t a r

igh

t an

gle

. E

very

tri

an

gle

has

thre

e a

ltit

ud

es

wh

ich

meet

at

a

poin

t ca

lled

th

e o

rth

oce

nte

r.

T

he

ve

rti

ce

s o

f △ABC

are

A(1

, 3

),

B(7

, 7

) a

nd

C(9

, 3

). F

ind

th

e c

oo

rd

ina

tes o

f th

e

orth

oce

nte

r o

f △ABC

.

Fin

d t

he p

oin

t w

here

tw

o o

f th

e t

hre

e a

ltit

ud

es

inte

rsect

.

Fin

d t

he e

qu

ati

on

of

the a

ltit

ud

e f

rom

A t

o −

−−

BC

.

If −

−−

BC

has

a s

lop

e o

f −

2,

then

th

e a

ltit

ud

e

has

a s

lop

e o

f 1

−

2 .

y -

y1 =

m(x

– x

1)

Poin

t-slo

pe f

orm

y -

3 =

1

−

2 (x –

1)

m =

1

−

2 ,

(x1,

y1)

= A

(1,

3)

y -

3 =

1

−

2 x

– 1

−

2

Dis

trib

utive P

ropert

y

y =

1

−

2 x

+ 5

−

2

Sim

plif

y.

C t

o −

−

AB

.

If −

−

AB

has

a s

lop

e o

f 2

−

3 , t

hen

th

e a

ltit

ud

e h

as

a

slop

e o

f -

3

−

2 .

y -

y1 =

m(x

- x

1)

Poin

t-slo

pe f

orm

y -

3 =

- 3

−

2 (

x -

9)

m =

- 3

−

2 ,

(x1,

y1)

= C

(9,

3)

y -

3 =

- 3

−

2 x

+ 2

7

−

2

Dis

trib

utive P

ropert

y

y =

- 3

−

2 x

+ 3

3

−

2

Sim

plif

y.

Solv

e t

he s

yst

em

of

equ

ati

on

s an

d f

ind

wh

ere

th

e a

ltit

ud

es

meet.

y =

1

−

2 x

+ 5

−

2

y =

- 3

−

2 x

+ 3

3

−

2

1

−

2 x

+ 5

−

2 =

- 3

−

2

x +

33

−

2

Subtr

act

1

−

2 x

fro

m e

ach s

ide.

5

−

2 =

−2x +

33

−

2

S

ubtr

act

33

−

2

from

each s

ide.

−

14 =

−2

x

Div

ide b

oth

sid

es b

y -

2.

7 =

x

y =

1

−

2 x

+ 5

−

2 =

1

−

2 (7)

+ 5

−

2 =

7

−

2 +

5

−

2 =

6

Th

e c

oord

inate

s of

the o

rth

oce

nte

r of

△A

BC

is

(6,

7).

Exerc

ises

CO

OR

DIN

AT

E G

EO

ME

TR

Y F

ind

th

e c

oo

rd

ina

tes o

f th

e o

rth

oce

nte

r o

f e

ach

tria

ng

le.

1

. J

(1,

0),

H(6

, 0),

I(3

, 6)

2. S

(1,

0),

T(4

, 7),

U(8

, −

3)

5-2

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Med

ian

s a

nd

Alt

itu

des o

f Tri

an

gle

s

Exam

ple

y

x

(9,

3)

(1,

3)

(7,

7)

(3,

1)

5

−

2 ,

3

−

2

Lesson 5-2


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

13

G

len

co

e G

eo

me

try

5-2

In △

PQR

, NQ

= 6

, RK

= 3

, a

nd

PK

= 4

. F

ind

ea

ch

le

ng

th.

1

. K

M

2.

KQ

2

4

3

. L

K

4.

LR

1.5

4.5

5

. N

K

6.

PM

2

6

In △

STR

, H

is t

he

ce

ntr

oid

, EH

= 6

, DH

= 4

, a

nd

SM

= 2

4.

Fin

d e

ach

le

ng

th.

7

. S

H

8.

HM

16

8

9

. T

H

10

. H

R

8

12

11

. T

D

12

. E

R

12

18

CO

OR

DIN

AT

E G

EO

ME

TR

Y F

ind

th

e c

oo

rd

ina

tes o

f th

e c

en

tro

id o

f e

ach

tria

ng

le.

13

. X

(−3,

15)

Y(1

, 5),

Z(5

, 10)

14

. S

(2,

5),

T(6

, 5),

R(1

0,

0)

(1,

10)

(6

, 3

1

−

3 )

CO

OR

DIN

AT

E G

EO

ME

TR

Y F

ind

th

e c

oo

rd

ina

tes o

f th

e o

rth

oce

nte

r o

f e

ach

tria

ng

le.

15

. L

(8,

0),

M(1

0,

8),

N(1

4,

0)

16

. D

(−9,

9),

E(−

6,

6),

F(0

, 6)

(10,

1)

(-

9,

-3)

Sk

ills

Pra

ctic

e

Med

ian

s a

nd

Alt

itu

des o

f Tri

an

gle

s

3

4


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

14

G

len

co

e G

eo

me

try

In △

ABC

, CP

= 3

0, EP

= 1

8,

an

d BF

= 3

9.

Fin

d e

ach

le

ng

th.

1

. P

D

2. F

P

3

. B

P

4. C

D

5. P

A

6. E

A

In △

MIV

, Z

is t

he

ce

ntr

oid

, MZ

= 6

, YI =

18

, a

nd

NZ

= 1

2.

Fin

d e

ach

me

asu

re

.

7

. Z

R

8. Y

Z

9

. M

R

10

. Z

V

11

. N

V

12. IZ

CO

OR

DIN

AT

E G

EO

ME

TR

Y F

ind

th

e c

oo

rd

ina

tes o

f th

e c

en

tro

id o

f e

ach

tria

ng

le.

13

. I(

3,

1),

J(6

, 3),

K(3

, 5)

14

. H

(0,

1),

U(4

, 3),

P(2

, 5)

CO

OR

DIN

AT

E G

EO

ME

TR

Y F

ind

th

e c

oo

rd

ina

tes o

f th

e o

rth

oce

nte

r o

f e

ach

tria

ng

le.

15

. P

(-1,

2),

Q(5

, 2),

R(2

, 1)

16

. S

(0,

0),

T(3

, 3),

U(3

, 6)

17

. M

OB

ILE

S N

abu

ko w

an

ts t

o c

on

stru

ct a

mobil

e o

ut

of

flat

tria

ngle

s so

th

at

the s

urf

ace

s of

the t

rian

gle

s h

an

g p

ara

llel

to t

he f

loor

wh

en

th

e m

obil

e i

s su

spen

ded

. H

ow

can

N

abu

ko b

e c

ert

ain

th

at

she h

an

gs

the t

rian

gle

s to

ach

ieve t

his

eff

ect

?

5-2

Practi

ce

Med

ian

s a

nd

Alt

itu

des o

f Tri

an

gle

s

AC F

E D

PB

18

30

2

6

45

3

6

54

3

6

9

24

3

6

12

(

4,

3)

(2,

3)

(

2, -

1)

(0,

9)

S

he n

eed

s t

o h

an

g e

ach

tri

an

gle

fro

m i

ts c

en

ter

of

gra

vit

y o

r cen

tro

id,

wh

ich

is t

he p

oin

t at

wh

ich

th

e t

hre

e m

ed

ian

s o

f th

e t

rian

gle

in

ters

ect.

1

5

13


NA

ME

DA

TE

PE

RIO

D

Lesson 5-2

Ch

ap

ter

5

15

G

len

co

e G

eo

me

try

1. B

ALA

NC

ING

Joh

an

na b

ala

nce

d a

tr

ian

gle

fla

t on

her

fin

ger

tip

. W

hat

poin

t of

the t

rian

gle

mu

st J

oh

an

na b

e

tou

chin

g?

2

. R

EFLE

CT

ION

S P

art

of

the w

ork

ing s

pace

in

Pau

lett

e’s

loft

is

part

itio

ned

in

th

e

shap

e o

f a n

earl

y e

qu

ilate

ral

tria

ngle

w

ith

mir

rors

han

gin

g o

n a

ll t

hre

e

part

itio

ns.

Fro

m w

hic

h p

oin

t co

uld

so

meon

e s

ee t

he o

pp

osi

te c

orn

er

beh

ind

h

is o

r h

er

refl

ect

ion

in

an

y o

f th

e t

hre

e

mir

rors

?

3

. D

IST

AN

CE

S F

or

wh

at

kin

d o

f tr

ian

gle

is

there

a p

oin

t w

here

th

e d

ista

nce

to e

ach

si

de i

s h

alf

th

e d

ista

nce

to e

ach

vert

ex?

Exp

lain

.

4

. M

ED

IAN

S L

ook

at

the r

igh

t tr

ian

gle

belo

w.

Wh

at

do y

ou

noti

ce a

bou

t th

e

ort

hoce

nte

r an

d t

he v

ert

ices

of

the

tria

ngle

?

5

. P

LA

ZA

S A

n a

rch

itect

is

desi

gn

ing a

tr

ian

gu

lar

pla

za.

For

aest

heti

c p

urp

ose

s,

the a

rch

itect

pays

speci

al

att

en

tion

to t

he

loca

tion

of

the c

en

troid

C a

nd

th

e

circ

um

cen

ter

O.

a.

Giv

e a

n e

xam

ple

of

a t

rian

gu

lar

pla

za

wh

ere

C =

O.

If n

o s

uch

exam

ple

exis

ts,

state

th

at

this

is

imp

oss

ible

.

b.

Giv

e a

n e

xam

ple

of

a t

rian

gu

lar

pla

za

wh

ere

C i

s in

sid

e t

he p

laza a

nd

O i

s ou

tsid

e t

he p

laza.

If n

o s

uch

exam

ple

exis

ts,

state

th

at

this

is

imp

oss

ible

.

c.

Giv

e a

n e

xam

ple

of

a t

rian

gu

lar

pla

za

wh

ere

C i

s ou

tsid

e t

he p

laza a

nd

O i

s in

sid

e t

he p

laza.

If n

o s

uch

exam

ple

exis

ts,

state

th

at

this

is

imp

oss

ible

.

5-2

Wo

rd

Pro

ble

m P

racti

ce

Med

ian

s a

nd

Alt

itu

des o

f Tri

an

gle

s

cen

tro

id

ort

ho

cen

ter

eq

uilate

ral:

in

cen

ter =

cen

tro

id =

cir

cu

mcen

ter

Th

e o

rth

ocen

ter

co

incid

es w

ith

o

ne o

f th

e v

ert

ices.

an

eq

uilate

ral

tria

ng

le

an

ob

tuse t

rian

gle

imp

ossib

le


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

16

G

len

co

e G

eo

me

try

Co

nstr

ucti

ng

Cen

tro

ids a

nd

Ort

ho

cen

ters

Th

e t

hre

e m

ed

ian

s of

a t

rian

gle

in

ters

ect

at

a s

ingle

poin

t ca

lled

th

e c

en

troid

.

You

can

use

a s

traig

hte

dge a

nd

com

pass

to f

ind

th

e c

en

troid

of

a t

rian

gle

.

1

. W

ith

a s

traig

hte

dge a

nd

com

pass

, co

nst

ruct

th

e

cen

troid

for

△S

TU

by f

oll

ow

ing t

he s

tep

s belo

w.

S

tep

1 L

oca

te t

he m

idp

oin

ts o

f si

des

TU

an

d S

U.

Label

the m

idp

oin

ts A

an

d B

resp

ect

ively

.

S

tep

2 D

raw

th

e s

egm

en

ts S

A a

nd

TB

. U

se t

he

lett

er

H t

o l

abel

their

poin

t of

inte

rsect

ion

,

wh

ich

is

the c

en

troid

of

△S

TU

.

Co

nstr

uct

the

ce

ntr

oid

of

ea

ch

tria

ng

le.

2

.

3.

Th

e t

hre

e a

ltit

ud

es

of

a t

rian

gle

meet

in a

sin

gle

poin

t ca

lled

th

e o

rth

oce

nte

r of

the t

rian

gle

.

4

. F

oll

ow

th

e s

tep

s belo

w t

o c

on

stru

ct t

he o

rth

oce

nte

r

of

△C

DE

usi

ng a

str

aig

hte

dge a

nd

com

pass

.

S

tep

1 E

xte

nd

segm

en

ts C

D a

nd

DE

past

poin

t

D l

on

g e

nou

gh

to m

eet

perp

en

dic

ula

rs

from

E a

nd

C a

s sh

ow

n.

S

tep

2 C

on

stru

ct t

he p

erp

en

dic

ula

r fr

om

poin

t C

to t

he l

ine D

E a

nd

label

the p

oin

t of

inte

rsect

ion

X.

Lik

ew

ise,

label

the p

oin

t of

inte

rsect

ion

of

lin

e C

D w

ith

th

e p

erp

en

dic

ula

r

from

E a

s p

oin

t Z

. In

th

is c

ase

both

X a

nd

Z l

ie o

uts

ide △

CD

E.

S

tep

3 L

abel

O t

he p

oin

t w

here

perp

en

dic

ula

rs

! #$

CX

an

d !

#$

EZ

in

ters

ect

. T

his

is

the

ort

hoce

nte

r of

△C

DE

.

Co

nstr

uct

the

orth

oce

nte

r o

f e

ach

tria

ng

le.

5

.

6.

5-2

En

rich

men

t


NA

ME

DA

TE

PE

RIO

D

Lesson 5-3

Ch

ap

ter

5

17

G

len

co

e G

eo

me

try

An

gle

In

eq

ua

liti

es

Pro

pert

ies

of

inequ

ali

ties,

in

clu

din

g t

he T

ran

siti

ve,

Ad

dit

ion

, an

d

Su

btr

act

ion

Pro

pert

ies

of

Inequ

ali

ty,

can

be u

sed

wit

h m

easu

res

of

an

gle

s an

d s

egm

en

ts.

Th

ere

is

als

o a

Com

pari

son

Pro

pert

y o

f In

equ

ali

ty.

F

or

an

y r

eal

nu

mbers

a a

nd

b,

eit

her

a <

b,

a =

b,

or

a >

b.

Th

e E

xte

rior

An

gle

In

equ

ali

ty T

heore

m c

an

be u

sed

to p

rove t

his

in

equ

ali

ty i

nvolv

ing a

n

exte

rior

an

gle

.

Ex

teri

or

An

gle

Ine

qu

ali

ty T

he

ore

m

Th

e m

ea

su

re o

f a

n e

xte

rio

r a

ng

le o

f a

tria

ng

le

is g

rea

ter

tha

n t

he

me

asu

re o

f e

ith

er

of

its

co

rre

sp

on

din

g r

em

ote

in

terio

r a

ng

les.

AC

D

1

B

m∠

1 >

m∠

A,

m∠

1 >

m∠

B

L

ist

all

an

gle

s o

f △EFG

wh

ose

me

asu

re

s a

re

less t

ha

n m∠

1.

Th

e m

easu

re o

f an

exte

rior

an

gle

is

gre

ate

r th

an

th

e m

easu

re o

f

eit

her

rem

ote

in

teri

or

an

gle

. S

o m

∠3 <

m∠

1 a

nd

m∠

4 <

m∠

1.

Exerc

ises

Use

th

e E

xte

rio

r A

ng

le I

ne

qu

ali

ty T

he

ore

m t

o l

ist

all

of

the

an

gle

s t

ha

t sa

tisfy

th

e s

tate

d c

on

dit

ion

.

1

. m

easu

res

are

less

th

an

m∠

1

2

. m

easu

res

are

gre

ate

r th

an

m∠

3

3

. m

easu

res

are

less

th

an

m∠

1

4

. m

easu

res

are

gre

ate

r th

an

m∠

1

5

. m

easu

res

are

less

th

an

m∠

7

6

. m

easu

res

are

gre

ate

r th

an

m∠

2

7

. m

easu

res

are

gre

ate

r th

an

m∠

5

8

. m

easu

res

are

less

th

an

m∠

4

9

. m

easu

res

are

less

th

an

m∠

1

10

. m

easu

res

are

gre

ate

r th

an

m∠

4

MJ

K

3

45

21L E

xe

rcis

es

1–2

HE

F3

4

21

G

5-3

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Ineq

ualiti

es i

n O

ne T

rian

gle

Exam

ple

RO

QN

P3

456

Ex

erc

ise

s 9

–1

0

78

21

S

XT

WV

3 4

5

67

21

U

Ex

erc

ise

s 3

–8

∠3, ∠

4

∠1, ∠

5

∠5, ∠

6

∠7

∠4

∠1, ∠

7, ∠

TU

V

∠2, ∠

3

∠1, ∠

3, ∠

5, ∠

6, ∠

TU

V

∠4, ∠

5, ∠

7, ∠

NP

R

∠1, ∠

8, ∠

OP

N, ∠

RO

Q


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

18

G

len

co

e G

eo

me

try

An

gle

-Sid

e R

ela

tio

nsh

ips

Wh

en

th

e s

ides

of

tria

ngle

s are

not

con

gru

en

t, t

here

is

a r

ela

tion

ship

betw

een

th

e s

ides

an

d

an

gle

s of

the t

rian

gle

s.

•

If

on

e s

ide o

f a t

rian

gle

is

lon

ger

than

an

oth

er

sid

e,

then

th

e

an

gle

op

posi

te t

he l

on

ger

sid

e h

as

a g

reate

r m

easu

re t

han

th

e

an

gle

op

posi

te t

he s

hort

er

sid

e.

•

If

on

e a

ngle

of

a t

rian

gle

has

a g

reate

r m

easu

re t

han

an

oth

er

an

gle

, th

en

th

e s

ide o

pp

osi

te t

he g

reate

r an

gle

is

lon

ger

than

the s

ide o

pp

osi

te t

he l

ess

er

an

gle

.

BC

A

L

ist

the

an

gle

s i

n o

rd

er

fro

m s

ma

lle

st

to l

arg

est

me

asu

re

.

RT

9 c

m

6 c

m7

cm

S

∠T

, ∠

R,

∠S

L

ist

the

sid

es i

n o

rd

er

fro

m s

ho

rte

st

to l

on

ge

st.

AB

C

20

°

35

°

125

°

−−

−

CB

, −

−

AB

, −

−

AC

Exerc

ises

Lis

t th

e a

ng

les a

nd

sid

es i

n o

rd

er f

ro

m s

ma

lle

st

to l

arg

est.

1

.

TS

R

48

cm

23.7

cm

35

cm

2

.

RT

S

60

°

80

°

40

°

3.

AC

B

3.8

4.3

4.0

4

.

14

11

5

5

.

45

8

6

.

20

12

7

.

35

°

12

0°

25

°

8

.

56

°58

°

9

.

60°

54°

5-3

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Ineq

ualiti

es i

n O

ne T

rian

gle

If A

C >

AB

, th

en m

∠B

> m

∠C

.

If m

∠A

> m

∠C

, th

en B

C >

AB

.

Exam

ple

1Exam

ple

2

∠T,

∠R

, ∠

S

∠

T,

∠R

, ∠

S

∠

C,

∠B

, ∠

A

−−

RS

, −

−

ST ,

−−

RT

−

−

RS

, −

−

ST ,

−−

RT

−

−

AB

, −

−

AC

, −

−

BC

∠S

, ∠

U,

∠T,

∠

B,

∠C

, ∠

A,

∠Q

, ∠

P,

∠R

,

−

−

UT ,

−−

ST ,

−−

SU

−

−

AC

, −

−

BA

, −

−

CB

−

−

PR

, −

−

RQ

, −

−

QP

∠E

, ∠

C, ∠

D,

∠

X,

∠Z

, ∠

Y,

∠

T,

∠S

, ∠

R,

−

−

CD

, −

−

DE

, −

−

CE

−

−

YZ

, −

−

XY

, −

−

XZ

−

−

RS

, −

−

RT ,

−−

ST

Lesson 5-3


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

19

G

len

co

e G

eo

me

try

5-3

Use

th

e E

xte

rio

r A

ng

le I

ne

qu

ali

ty T

he

ore

m t

o l

ist

all

of

the

an

gle

s t

ha

t sa

tisfy

th

e s

tate

d c

on

dit

ion

.

1. m

easu

res

less

th

an

m∠

1

∠2,

∠3,

∠4,

∠5,

∠7,

∠8

2. m

easu

res

less

th

an

m∠

9

∠2,

∠4,

∠6,

∠7

3. m

easu

res

gre

ate

r th

an

m∠

5

∠1,

∠3

4. m

easu

res

gre

ate

r th

an

m∠

8

∠1,

∠3,

∠5

Lis

t th

e a

ng

les a

nd

sid

es o

f e

ach

tria

ng

le i

n o

rd

er f

ro

m s

ma

lle

st

to l

arg

est.

5.

56

2

6

. 2

4°

98°

∠

Q,

∠R

, ∠

S,

−−

RP

, −

−

PQ

, −

−

RQ

∠

K,

∠M

, ∠

L,

−−

ML ,

−−

KL ,

−−

KM

7

.

15

916

8

.

38

39

34

∠

F,

∠H

, ∠

G,

−−

HG

, −

−

FG

, −

−

FH

∠

X,

∠Y

, ∠

Z,

−−

YZ

, −

−

XZ

, −

−

XY

9

.

10

.

∠

A,

∠B

, ∠

C,

−−

BC

, −

−

AC

, −

−

AB

∠

S,

∠U

, ∠

T,

−−

UT ,

−−

ST ,

−−

SU

1

24

6

7

89

35

Sk

ills

Pra

ctic

e

Ineq

ualiti

es i

n O

ne T

rian

gle

98

°

43

°

42

°


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

20

G

len

co

e G

eo

me

try

Use

th

e f

igu

re

at

the

rig

ht

to d

ete

rm

ine

wh

ich

an

gle

ha

s t

he

gre

ate

st

me

asu

re

.

1

. ∠

1,

∠3,

∠4

2.

∠4,

∠8,

∠9

3

. ∠

2,

∠3,

∠7

4.

∠7,

∠8,

∠10

Use

th

e E

xte

rio

r A

ng

le I

ne

qu

ali

ty T

he

ore

m t

o l

ist

all

an

gle

s t

ha

t sa

tisfy

th

e s

tate

d c

on

dit

ion

.

5

. m

easu

res

are

less

th

an

m∠

1

6

. m

easu

res

are

less

th

an

m∠

3

7

. m

easu

res

are

gre

ate

r th

an

m∠

7

8

. m

easu

res

are

gre

ate

r th

an

m∠

2

Use

th

e f

igu

re

at

the

rig

ht

to d

ete

rm

ine

th

e r

ela

tio

nsh

ip

be

twe

en

th

e m

ea

su

re

s o

f th

e g

ive

n a

ng

les.

9

. m

∠Q

RW

, m

∠R

WQ

1

0. m

∠R

TW

, m

∠T

WR

11

. m

∠R

ST

, m

∠T

RS

1

2. m

∠W

QR

, m

∠Q

RW

Use

th

e f

igu

re

at

the

rig

ht

to d

ete

rm

ine

th

e r

ela

tio

nsh

ip

be

twe

en

th

e l

en

gth

s o

f th

e g

ive

n s

ide

s.

13

. −

−−

DH

, −

−−

GH

1

4.

−−

−

DE

, −

−−

DG

15

. −

−−

EG

, −

−−

FG

1

6.

−−

−

DE

, −

−−

EG

17

. S

PO

RT

S T

he f

igu

re s

how

s th

e p

osi

tion

of

thre

e t

rees

on

on

e

part

of

a F

risb

ee™

cou

rse.

At

wh

ich

tre

e p

osi

tion

is

the a

ngle

betw

een

th

e t

rees

the g

reate

st?

53 f

t

40 f

t

3

2

1

37.5

ft

120

°32

°

48

°113

°

17

°H

DE

F

G

34

47

45

44

22

14

35

Q

R

S

T

W

12

46

78

9

3

5

1

2

46

7

891

0

3

5

5-3

Practi

ce

Ineq

ualiti

es i

n O

ne T

rian

gle

∠

1

∠4

∠

7

∠10

∠

3,

∠4,

∠5,

∠7,

∠8

∠

5,

∠7,

∠8

∠

1,

∠3,

∠5,

∠9

∠

6,

∠9

m

∠Q

RW

< m

∠R

WQ

m

∠R

TW

< m

∠TW

R

m

∠R

ST >

m∠

TR

S

m∠

WQ

R <

m∠

QR

W

−−

DH

> −

−

GH

−−

DE

< −

−

DG

−−

EG

< −

−

FG

−−

DE

> −

−

EG

2

Lesson 5-3


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

21

G

len

co

e G

eo

me

try

1

. D

IST

AN

CE

C

arl

an

d R

ose

liv

e o

n t

he

sam

e s

traig

ht

road

. F

rom

th

eir

balc

on

ies

they c

an

see a

fla

gp

ole

in

th

e d

ista

nce

.

Th

e a

ngle

th

at

each

pers

on

’s l

ine o

f

sigh

t to

th

e f

lagp

ole

mak

es

wit

h t

he

road

is

the s

am

e.

How

do t

heir

dis

tan

ces

from

th

e f

lagp

ole

com

pare

?

2. O

BT

US

E T

RIA

NG

LE

S D

on

noti

ces

that

the s

ide o

pp

osi

te t

he r

igh

t an

gle

in

a

righ

t tr

ian

gle

is

alw

ays

the l

on

gest

of

the t

hre

e s

ides.

Is

this

als

o t

rue o

f th

e

sid

e o

pp

osi

te t

he o

btu

se a

ngle

in

an

obtu

se t

rian

gle

? E

xp

lain

.

3

. S

TR

ING

Jak

e b

uil

t a t

rian

gu

lar

stru

ctu

re w

ith

th

ree b

lack

sti

cks.

He

tied

on

e e

nd

of

a s

trin

g t

o v

ert

ex M

an

d t

he o

ther

en

d t

o a

poin

t on

th

e

stic

k o

pp

osi

te M

, p

ull

ing t

he s

trin

g

tau

t. P

rove t

hat

the l

en

gth

of

the

stri

ng c

an

not

exce

ed

th

e l

on

ger

of

the

two s

ides

of

the s

tru

ctu

re.

stri

ng

M

4

. S

QU

AR

ES

M

att

hew

has

thre

e d

iffe

ren

t

squ

are

s. H

e a

rran

ges

the s

qu

are

s to

form

a t

rian

gle

as

show

n.

Base

d o

n

the i

nfo

rmati

on

, li

st t

he s

qu

are

s in

ord

er

from

th

e o

ne w

ith

th

e s

mall

est

peri

mete

r to

th

e o

ne w

ith

th

e l

arg

est

peri

mete

r.

54˚

47˚

3

12

5

. C

ITIE

SS

tell

a i

s goin

g

to T

exas

to v

isit

a f

rien

d.

As

she w

as

look

ing a

t

a m

ap

to s

ee w

here

she m

igh

t w

an

t to

go,

she n

oti

ced

th

e c

itie

s

Au

stin

, D

all

as,

an

d A

bil

en

e

form

ed

a t

rian

gle

. S

he w

an

ted

to

dete

rmin

e

how

th

e d

ista

nce

s betw

een

th

e c

itie

s

were

rela

ted

, so

sh

e u

sed

a p

rotr

act

or

to

measu

re t

wo a

ngle

s.

a.

Base

d o

n t

he i

nfo

rmati

on

in

th

e

figu

re,

wh

ich

of

the t

wo c

itie

s are

neare

st t

o e

ach

oth

er?

b.

Base

d o

n t

he i

nfo

rmati

on

in

th

e

figu

re,

wh

ich

of

the t

wo c

itie

s are

fart

hest

ap

art

fro

m e

ach

oth

er?

5-3

59˚

64˚

Abile

ne

Dal

las

Aust

in

Wo

rd

Pro

ble

m P

racti

ce

Ineq

ualiti

es i

n O

ne T

rian

gle

Th

ey a

re e

qu

al.

Yes.

Sin

ce a

n o

btu

se t

rian

gle

on

ly h

as 1

ob

tuse a

ng

le a

nd

2

acu

te a

ng

les,

the s

ide o

pp

osit

e

the o

btu

se a

ng

le i

s t

he l

on

gest

sid

e.

Sam

ple

an

sw

er:

Th

e s

trin

g

div

ides t

he t

rian

gle

in

tw

o;

on

e o

f

these t

rian

gle

s i

s r

igh

t o

r o

btu

se

becau

se o

ne s

ide o

f th

e s

trin

g

mu

st

make a

rig

ht

or

ob

tuse

an

gle

wit

h t

he s

tick.

In t

his

tria

ng

le,

the s

ide o

pp

osit

e t

he

rig

ht

or

ob

tuse a

ng

le i

s l

on

ger

than

th

e s

trin

g a

nd

th

at

sid

e i

s

als

o a

sid

e o

f th

e t

rian

gle

.

2,

1,

3

Dallas a

nd

Ab

ilen

e

Ab

ilen

e a

nd

Au

sti

n


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

22

G

len

co

e G

eo

me

try

Co

nstr

ucti

on

Pro

ble

mT

he d

iagra

m b

elo

w s

how

s se

gm

en

t A

B a

dja

cen

t to

a c

lose

d r

egio

n.

Th

e

pro

ble

m r

equ

ires

that

you

con

stru

ct a

noth

er

segm

en

t X

Y t

o t

he r

igh

t of

the

close

d r

egio

n s

uch

th

at

poin

ts A

, B

, X

, an

d Y

are

coll

inear.

You

are

not

all

ow

ed

to

tou

ch o

r cr

oss

th

e c

lose

d r

egio

n w

ith

you

r co

mp

ass

or

stra

igh

ted

ge.

Fo

llo

w t

he

se

in

str

ucti

on

s t

o c

on

str

uct

a s

eg

me

nt XY

so

th

at

it i

s c

oll

ine

ar w

ith

se

gm

en

t AB

.

1

. C

on

stru

ct t

he p

erp

en

dic

ula

r bis

ect

or

of

−−

AB

. L

abel

the m

idp

oin

t as

poin

t C

, an

d t

he l

ine

as

m.

2

. M

ark

tw

o p

oin

ts P

an

d Q

on

lin

e m

th

at

lie w

ell

above t

he c

lose

d r

egio

n.

Con

stru

ct t

he

perp

en

dic

ula

r bis

ect

or,

n,

of

−−−

PQ

. L

abel

the i

nte

rsect

ion

of

lin

es

m a

nd

n a

s p

oin

t D

.

3

. M

ark

poin

ts R

an

d S

on

lin

e n

th

at

lie w

ell

to t

he r

igh

t of

the c

lose

d r

egio

n.

Con

stru

ct

the p

erp

en

dic

ula

r bis

ect

or,

k ,

of

−−

RS

. L

abel

the i

nte

rsect

ion

of

lin

es

n an

d k

as

poin

t E

.

4

. M

ark

poin

t X

on

lin

e k

so t

hat

X i

s belo

w l

ine n

an

d s

o t

hat

−−

EX

is

con

gru

en

t to

−−−

DC

.

5

. M

ark

poin

ts T

an

d V

on

lin

e k

an

d o

n o

pp

osi

te s

ides

of

X,

so t

hat

−−

XT

an

d −

−

XV

are

co

ngru

en

t. C

on

stru

ct t

he p

erp

en

dic

ula

r bis

ect

or,

ℓ,

of

−−

TV

. C

all

th

e p

oin

t w

here

th

e

lin

e ℓ

hit

s th

e b

ou

nd

ary

of

the c

lose

d r

egio

n p

oin

t Y

. −

−

XY

corr

esp

on

ds

to t

he n

ew

road

.

Q Pm

k

ℓ

nD

RE

T X V

YB

A

C

S

Exi

stin

gR

oad

Clo

sed R

egio

n(L

ake)

5-3

En

rich

men

t

Lesson 5-3


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

23

G

len

co

e G

eo

me

try

5-3

Cabri

Ju

nio

r ca

n b

e u

sed

to i

nvest

igate

th

e r

ela

tion

ship

s betw

een

an

gle

s an

d s

ides

of

a t

rian

gle

.

Ste

p 1

U

se C

abri

Ju

nio

r. t

o d

raw

an

d l

abel

a t

rian

gle

.

• S

ele

ct F

2 T

ria

ng

le t

o d

raw

a t

rian

gle

.

• M

ove t

he c

urs

or

to w

here

you

wan

t th

e f

irst

vert

ex.

Pre

ss

EN

TE

R.

•

R

ep

eat

this

pro

ced

ure

to d

ete

rmin

e t

he n

ext

two v

ert

ices

of

the t

rian

gle

.

• S

ele

ct F

5 A

lph

-nu

m t

o l

abel

each

vert

ex.

•

M

ove t

he c

urs

or

to a

vert

ex,

pre

ss

EN

TE

R,

en

ter

A,

an

d p

ress

E

NT

ER

again

.

•

R

ep

eat

this

pro

ced

ure

to l

abel

vert

ex B

an

d v

ert

ex C

.

Ste

p 2

D

raw

an

exte

rior

an

gle

of △

AB

C.

•

S

ele

ct F

2 L

ine

to d

raw

a l

ine t

hro

ugh

−−−

BC

.

• S

ele

ct F

2 P

oin

t, P

oin

t o

n t

o d

raw

a p

oin

t on

# $%

BC

so t

hat

C i

s betw

een

B a

nd

th

e n

ew

poin

t.

• S

ele

ct F

5 A

lph

-nu

m t

o l

abel

the p

oin

t D

.

Ste

p 3

F

ind

th

e m

easu

res

of

the t

hre

e i

nte

rior

an

gle

s an

d t

he e

xte

rior

an

gle

, ∠

AC

D.

•

S

ele

ct F

5 M

ea

su

re

, A

ng

le.

• T

o f

ind

th

e m

easu

re o

f ∠

AB

C,

sele

ct p

oin

ts A

, B

, an

d

C (

wit

h t

he v

ert

ex B

as

the s

eco

nd

poin

t se

lect

ed

).

• R

ep

eat

to f

ind

th

e r

em

ain

ing a

ngle

measu

res.

Ste

p 4

F

ind

th

e m

easu

re o

f each

sid

e o

f △

AB

C.

•

S

ele

ct F

5 M

ea

su

re

, D

. &

Le

ng

th.

•

T

o f

ind

th

e l

en

gth

of

−−

AB

, se

lect

poin

t A

an

d t

hen

sele

ct p

oin

t B

.

•

R

ep

eat

this

pro

ced

ure

to f

ind

th

e l

en

gth

s of

−−−

BC

an

d −

−

CA

.

Exercis

es

An

aly

ze

yo

ur d

ra

win

g.

1. W

hat

is t

he r

ela

tion

ship

betw

een

m∠

AC

D a

nd

m∠

AB

C?

m∠

AC

D a

nd

m∠

BA

C?

Sam

ple

an

sw

er:

m∠

AC

D >

m∠

AB

C;

m∠

AC

D >

m∠

BA

C

2

. M

ak

e a

con

ject

ure

abou

t th

e r

ela

tion

ship

betw

een

th

e m

easu

res

of

an

exte

rior

an

gle

(∠

AC

D)

an

d i

ts t

wo r

em

ote

in

teri

or

an

gle

s (∠

AB

C a

nd

∠B

AC

).

T

he m

easu

re o

f an

exte

rio

r an

gle

is e

qu

al

to t

he s

um

of

the m

easu

re o

f

the t

wo

rem

ote

in

teri

or

an

gle

s.

3

. C

han

ge t

he d

imen

sion

s of

the t

rian

gle

by m

ovin

g p

oin

t A

. (P

ress

C

LE

AR

so t

he p

oin

ter

beco

mes

a b

lack

arr

ow

. M

ove t

he p

oin

ter

close

to p

oin

t A

un

til

the a

rrow

beco

mes

tran

spare

nt

an

d p

oin

t A

is

bli

nk

ing.

Pre

ss

ALP

HA

to c

han

ge t

he a

rrow

to a

han

d.

Th

en

m

ove t

he p

oin

t.)

Is y

ou

r co

nje

ctu

re s

till

tru

e?

yes

4

. W

hic

h s

ide o

f th

e t

rian

gle

is

the l

on

gest

? th

e s

hort

est

? S

ee s

tud

en

ts’

wo

rk.

5

. W

hic

h a

ngle

measu

re (

not

incl

ud

ing t

he e

xte

rior

an

gle

) is

th

e g

reate

st?

the l

east

?

See s

tud

en

ts’

wo

rk.

6

. M

ak

e a

con

ject

ure

abou

t w

here

th

e l

on

gest

sid

e i

s in

rela

tion

ship

to t

he g

reate

st a

ngle

an

d w

here

th

e s

hort

est

sid

e i

s in

rela

tion

ship

to t

he l

east

an

gle

.

T

he l

on

gest

sid

e i

s o

pp

osit

e t

he g

reate

st

an

gle

. T

he s

ho

rtest

sid

e i

s

op

po

sit

e t

he l

east

an

gle

.

Gra

ph

ing C

alc

ula

tor

Act

ivit

y

Cab

ri J

un

ior:

In

eq

ualiti

es i

n O

ne T

rian

gle


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

24

G

len

co

e G

eo

me

try

5-3

Th

e G

eom

ete

r’s

Sk

etc

hp

ad

can

be u

sed

to i

nvest

igate

th

e r

ela

tion

ship

s betw

een

an

gle

s an

d

sid

es

of

a t

rian

gle

.

Ste

p 1

U

se T

he G

eom

ete

r’s

Sk

etc

hp

ad

to d

raw

a t

rian

gle

an

d o

ne e

xte

rior

an

gle

.

• C

on

stru

ct a

ray b

y s

ele

ctin

g t

he

Ray t

ool

from

th

e t

oolb

ar.

Fir

st,

clic

k w

here

you

wan

t th

e f

irst

p

oin

t. T

hen

cli

ck a

seco

nd

poin

t to

d

raw

th

e r

ay.

•

N

ext,

sele

ct t

he S

egm

en

t to

ol

from

th

e t

oolb

ar.

Use

th

e e

nd

poin

t of

the r

ay a

s th

e f

irst

poin

t fo

r th

e

segm

en

t an

d c

lick

on

a s

eco

nd

p

oin

t to

con

stru

ct t

he s

egm

en

t.

• C

on

stru

ct a

noth

er

segm

en

t jo

inin

g

the s

eco

nd

poin

t of

the p

revio

us

segm

en

t to

a p

oin

t on

th

e r

ay.

•

D

isp

lay t

he l

abels

for

each

poin

t. U

se t

he S

ele

ctio

n A

rrow

tool

to s

ele

ct a

ll f

ou

r p

oin

ts.

Dis

pla

y t

he l

abels

by s

ele

ctin

g S

ho

w L

ab

el

from

th

e D

isp

lay

men

u.

Ste

p 2

F

ind

th

e m

easu

res

of

each

an

gle

.

• T

o f

ind

th

e m

easu

re o

f ∠ABC

, u

se t

he S

ele

ctio

n A

rrow

tool

to s

ele

ct p

oin

ts

A, B

, an

d C

(w

ith

th

e v

ert

ex B

as

the s

eco

nd

poin

t se

lect

ed

). T

hen

, u

nd

er

the

Me

asu

re

men

u,

sele

ct A

ng

le.

Use

th

is m

eth

od

to f

ind

th

e r

em

ain

ing a

ngle

m

easu

res,

in

clu

din

g t

he e

xte

rior

an

gle

, ∠BCD

.

Ste

p 3

F

ind

th

e m

easu

res

of

each

sid

e o

f th

e t

rian

gle

.

• T

o f

ind

th

e m

easu

re o

f si

de AB

, se

lect

A a

nd

th

en

B.

Next,

un

der

the M

ea

su

re

m

en

u,

sele

ct D

ista

nce

. U

se t

his

meth

od

to f

ind

th

e l

en

gth

of

the o

ther

two

sid

es.

Exerc

ises

An

aly

ze

yo

ur d

ra

win

g.

1

. W

hat

is t

he r

ela

tion

ship

betw

een

m∠BCD

an

d m∠ABC

? m∠BCD

an

d m∠BAC

?

Sam

ple

an

sw

er:

m∠

BC

D >

m∠

AB

C;

m∠

BC

D >

m∠

BA

C

2. M

ak

e a

con

ject

ure

abou

t th

e r

ela

tion

ship

betw

een

th

e m

easu

res

of

an

exte

rior

an

gle

(∠BCD

) an

d i

ts t

wo r

em

ote

in

teri

or

an

gle

s (∠ABC

an

d ∠BAC

).

T

he m

easu

re o

f an

exte

rio

r an

gle

is e

qu

al

to t

he s

um

of

the m

easu

re o

f

the t

wo

rem

ote

in

teri

or

an

gle

s.

3. C

han

ge t

he d

imen

sion

s of

the t

rian

gle

by s

ele

ctin

g p

oin

t A

wit

h t

he p

oin

ter

tool

an

d

movin

g i

t. I

s you

r co

nje

ctu

re s

till

tru

e?

yes

4. W

hic

h s

ide o

f th

e t

rian

gle

is

the l

on

gest

? th

e s

hort

est

? S

ee s

tud

en

ts’

wo

rk.

5. W

hic

h a

ngle

measu

re (

not

incl

ud

ing t

he e

xte

rior

an

gle

) is

th

e g

reate

st?

the l

east

? S

ee s

tud

en

ts’

wo

rk.

6. M

ak

e a

con

ject

ure

abou

t w

here

th

e l

on

gest

sid

e i

s in

rela

tion

ship

to t

he g

reate

st a

ngle

an

d w

here

th

e s

hort

est

sid

e i

s in

rela

tion

ship

to t

he l

east

an

gle

.

T

he l

on

gest

sid

e i

s o

pp

osit

e t

he g

reate

st

an

gle

. T

he s

ho

rtest

sid

e i

s

op

po

sit

e t

he l

east

an

gle

.

A

B

CD

m/ABC

5 6

9.2

9˚

m/BCA

5 5

5.9

2˚

m/BAC

5 5

4.7

8˚

m/BCD

5 1

24

.08˚

AB

5 2.2

0 c

m

BC

5 2.1

7 c

m

AC

5 2.4

9 c

m

Geo

mete

r’s

Sk

etc

hp

ad

Act

ivit

y

Ineq

ualiti

es i

n O

ne T

rian

gle

Lesson 5-4


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

25

G

len

co

e G

eo

me

try

Ind

ire

ct A

lge

bra

ic P

roo

f O

ne w

ay t

o p

rove t

hat

a s

tate

men

t is

tru

e i

s to

tem

pora

rily

ass

um

e t

hat

wh

at

you

are

try

ing t

o p

rove i

s fa

lse.

By s

how

ing t

his

ass

um

pti

on

to b

e

logic

all

y i

mp

oss

ible

, you

pro

ve y

ou

r ass

um

pti

on

fals

e a

nd

th

e o

rigin

al

con

clu

sion

tru

e.

Th

is

is k

now

n a

s an

in

dir

ect

pro

of.

Ste

ps

fo

r W

riti

ng

an

In

dir

ec

t P

roo

f

1.

Assu

me

th

at

the

co

nclu

sio

n is f

als

e b

y a

ssu

min

g t

he

op

pp

osite

is t

rue

.

2.

Sh

ow

th

at

this

assu

mp

tio

n le

ad

s t

o a

co

ntr

ad

ictio

n o

f th

e h

yp

oth

esis

or

so

me

oth

er

fact.

3.

Po

int

ou

t th

at

the

assu

mp

tio

n m

ust

be

fa

lse

, a

nd

th

ere

fore

, th

e c

on

clu

sio

n m

ust

be

tru

e.

G

ive

n:

3x

+ 5

> 8

Pro

ve

: x

> 1

Ste

p 1

A

ssu

me t

hat x i

s n

ot

gre

ate

r th

an

1.

Th

at

is, x =

1 o

r x <

1.

Ste

p 2

M

ak

e a

table

for

severa

l p

oss

ibil

itie

s fo

r x =

1 o

r x <

1.

Wh

en

x =

1 o

r x <

1,

then

3x +

5 i

s n

ot

gre

ate

r th

an

8.

Ste

p 3

T

his

con

trad

icts

th

e g

iven

in

form

ati

on

th

at

3x +

5 >

8.

Th

e

ass

um

pti

on

th

at x i

s n

ot

gre

ate

r th

an

1 m

ust

be f

als

e,

wh

ich

m

ean

s th

at

the s

tate

men

t “x

> 1

” m

ust

be t

rue.

Exerc

ises

Sta

te t

he

assu

mp

tio

n y

ou

wo

uld

ma

ke

to

sta

rt

an

in

dir

ect

pro

of

of

ea

ch

sta

tem

en

t.

1. If

2x >

14,

then

x >

7.

2. F

or

all

real

nu

mbers

, if

a +

b >

c,

then

a >

c -

b.

Co

mp

lete

th

e i

nd

ire

ct

pro

of.

Giv

en

: n

is

an

in

teger

an

d n

2 i

s even

.

Pro

ve

: n

is

even

.

3

. A

ssu

me t

hat

4

. T

hen

n c

an

be e

xp

ress

ed

as

2a

+ 1

by

5

. n

2 =

S

ubst

itu

tion

6.

=

Mu

ltip

ly.

7.

=

Sim

pli

fy.

8.

= 2

(2a

2 +

2a

) +

1

9

. 2(2a

2 +

2a

)+ 1

is

an

od

d n

um

ber.

Th

is c

on

trad

icts

th

e g

iven

th

at n

2 i

s even

,

so

th

e a

ssu

mp

tion

mu

st b

e

10

. T

here

fore

,

x ≤

7

a ≤

c -

b

n i

s n

ot

even

. T

hat

is,

assu

me n

is o

dd

.

the m

ean

ing

of

od

d n

um

ber.

(2a +

1)2

(2a

+ 1

)(2

a +

1)

Dis

trib

uti

ve P

rop

ert

y

fals

e.

n i

s e

ven

.

5-4

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Ind

irect

Pro

of

Exam

ple

x3

x + 5

18

05

-1

2

-2

-1

-3

-4

4a

2 +

4a +

1


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

26

G

len

co

e G

eo

me

try

Ind

ire

ct P

roo

f w

ith

Ge

om

etr

y

To w

rite

an

in

dir

ect

pro

of

in g

eom

etr

y,

you

ass

um

e

that

the c

on

clu

sion

is

fals

e.

Th

en

you

sh

ow

th

at

the a

ssu

mp

tion

lead

s to

a c

on

trad

icti

on

. T

he c

on

trad

icti

on

sh

ow

s th

at

the c

on

clu

sion

can

not

be f

als

e,

so i

t m

ust

be t

rue.

Giv

en

: m

∠C

= 1

00

Pro

ve

: ∠A

is n

ot

a r

igh

t a

ng

le.

Ste

p 1

A

ssu

me t

hat

∠A

is

a r

igh

t an

gle

.

Ste

p 2

S

how

th

at

this

lead

s to

a c

on

trad

icti

on

. If

∠A

is

a r

igh

t an

gle

,

then

m∠

A =

90 a

nd

m∠

C +

m∠

A =

100 +

90

= 1

90.

Th

us

the

sum

of

the m

easu

res

of

the a

ngle

s of

△A

BC

is

gre

ate

r th

an

180.

Ste

p 3

T

he c

on

clu

sion

th

at

the s

um

of

the m

easu

res

of

the a

ngle

s of

△A

BC

is

gre

ate

r th

an

180 i

s a c

on

trad

icti

on

of

a k

now

n p

rop

ert

y.

Th

e a

ssu

mp

tion

th

at

∠A

is

a r

igh

t an

gle

mu

st b

e f

als

e,

wh

ich

mean

s th

at

the s

tate

men

t “∠

A i

s n

ot

a r

igh

t an

gle

” m

ust

be t

rue.

Exerc

ises

Sta

te t

he

assu

mp

tio

n y

ou

wo

uld

ma

ke

to

sta

rt

an

in

dir

ect

pro

of

of

ea

ch

sta

tem

en

t.

1

. If

m∠

A =

90,

then

m∠

B =

45.

2

. If

−−

AV

is

not

con

gru

en

t to

−−

VE

, th

en

△A

VE

is

not

isosc

ele

s.

Co

mp

lete

th

e i

nd

ire

ct

pro

of.

Giv

en

: ∠

1 $

∠2 a

nd

−−

−

DG

is

not

con

gru

en

t to

−−

−

FG

.

Pro

ve

: −

−−

DE

is

not

con

gru

en

t to

−−

FE

.

3

. A

ssu

me t

hat

Ass

um

e t

he c

on

clu

sion

is

fals

e.

4

. −

−−

EG

$ −

−−

EG

5

. △

ED

G $

△E

FG

6

.

7

. T

his

con

trad

icts

th

e g

iven

in

form

ati

on

, so

th

e a

ssu

mp

tion

mu

st

be

8

. T

here

fore

,

1

2

DG

F

E

AB

C

m∠

B ≠

45

△A

VE

is i

so

scele

s.

−−

DE

%

−−

FE

.

Refl

exiv

e P

rop

ert

y

SA

S

−−

DG

%

−−

FG

C

PC

TC

fals

e.

−−

DE

is n

ot

co

ng

ruen

t to

−−

FE

.

5-4

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Ind

irect

Pro

of

Exam

ple

Lesson 5-4


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

27

G

len

co

e G

eo

me

try

5-4

Sk

ills

Pra

ctic

e

Ind

irect

Pro

of

Sta

te t

he a

ssu

mp

tio

n y

ou

wo

uld

ma

ke t

o s

tart

an

in

dir

ect

pro

of

of

ea

ch

sta

tem

en

t.

1.

m∠

AB

C <

m∠

CB

A

m

∠A

BC

≥ m

∠C

BA

2

. △

DE

F $

△R

ST

△D

EF ≇

△R

ST

3.

Lin

e a

is

perp

en

dic

ula

r to

lin

e b

.

L

ine a

is n

ot

perp

en

dic

ula

r to

lin

e b

.

4

. ∠

5 i

s su

pp

lem

en

tary

to ∠

6.

∠

5 i

s n

ot

su

pp

lem

en

tary

to

∠6.

Writ

e a

n i

nd

ire

ct

pro

of

of

ea

ch

sta

tem

en

t.

5

. G

ive

n:

x2 +

8 ≤

12

P

ro

ve

: x ≤

2

P

roo

f:

S

tep

1:

Assu

me x

> 2

.

S

tep

2:

If x

> 2

, th

en

x2 >

4.

Bu

t if

x2 >

4,

it f

ollo

ws t

hat

x2 +

8 >

12.

Th

is

co

ntr

ad

icts

th

e g

iven

fact

that

x2 +

8 ≤

12.

S

tep

3:

Sin

ce t

he a

ssu

mp

tio

n o

f x >

2 l

ead

s t

o a

co

ntr

ad

icti

on

, it

mu

st

be f

als

e.

Th

ere

fore

, x ≤

2 m

ust

be t

rue.

6

. G

ive

n: ∠

D ≇

∠F

P

ro

ve

: D

E ≠

EF

P

roo

f:

S

tep

1:

Assu

me D

E =

EF

.

S

tep

2:

If D

E =

EF,

then

−−

DE

% −

−

EF b

y t

he d

efi

nit

ion

of

co

ng

ruen

t seg

men

ts.

Bu

t if

−−

DE

% −

−

EF ,

then

∠D

% ∠

F b

y t

he I

so

scele

s

Tri

an

gle

Th

eo

rem

. T

his

co

ntr

ad

icts

th

e g

iven

in

form

ati

on

th

at

∠D

≇ ∠

F.

S

tep

3:

Sin

ce t

he a

ssu

mp

tio

n t

hat

DE

= E

F l

ead

s t

o a

co

ntr

ad

icti

on

, it

m

ust

be f

als

e.

Th

ere

fore

, it

mu

st

be t

rue t

hat

DE

≠ E

F.

DF

E


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

28

G

len

co

e G

eo

me

try

Sta

te t

he

assu

mp

tio

n y

ou

wo

uld

ma

ke

to

sta

rt

an

in

dir

ect

pro

of

of

ea

ch

sta

tem

en

t.

1

. −

−−

BD

bis

ect

s ∠

AB

C.

2

. R

T =

TS

Writ

e a

n i

nd

ire

ct

pro

of

of

ea

ch

sta

tem

en

t.

3

. G

ive

n:

-4

x +

2 <

-10

P

ro

ve

: x >

3

4

. G

ive

n:

m∠

2 +

m∠

3 ≠

180

P

ro

ve

: a ∦

b

5

. P

HY

SIC

S S

ou

nd

tra

vels

th

rou

gh

air

at

abou

t 344 m

ete

rs p

er

seco

nd

wh

en

th

e

tem

pera

ture

is

20

°C.

If E

nri

qu

e l

ives

2 k

ilom

ete

rs f

rom

th

e f

ire s

tati

on

an

d i

t ta

kes

5 s

eco

nd

s fo

r th

e s

ou

nd

of

the f

ire s

tati

on

sir

en

to r

each

him

, h

ow

can

you

pro

ve

ind

irect

ly t

hat

it i

s n

ot

20

°C w

hen

En

riqu

e h

ears

th

e s

iren

?

1 2

3

a

b

−−

BD

do

es n

ot

bis

ect

∠A

BC

.

RT

≠ T

S

P

roo

f:

Ste

p 1

Assu

me x

≤ 3

.

Ste

p 2

If

x ≤

3,

then

-4x ≥

-12.

Bu

t -

4x ≥

-12 i

mp

lies t

hat

-4x +

2 ≥

-10,

wh

ich

co

ntr

ad

icts

th

e g

iven

in

eq

uality

.

Ste

p 3

Sin

ce t

he a

ssu

mp

tio

n t

hat

x ≤

3 l

ead

s t

o a

co

ntr

ad

icti

on

,

it m

ust

be t

rue t

hat

x >

3.

P

roo

f:

Ste

p 1

Assu

me a

|| b.

Ste

p 2

If a

|| b,

then

th

e c

on

secu

tive i

nte

rio

r an

gle

s ∠

2 a

nd

∠3 a

re

su

pp

lem

en

tary

. T

hu

s m

∠2 +

m∠

3 =

180.

Th

is c

on

trad

icts

the g

iven

sta

tem

en

t th

at

m∠

2 +

m∠

3 ≠

180.

Ste

p 3

Sin

ce t

he a

ssu

mp

tio

n l

ead

s t

o a

co

ntr

ad

icti

on

, th

e s

tate

men

t

a |

| b m

ust

be f

als

e.

Th

ere

fore

, a ∦

b m

ust

be t

rue.

Assu

me t

hat

it i

s 2

0°C

wh

en

En

riq

ue h

ears

th

e s

iren

, th

en

sh

ow

th

at

at

this

tem

pera

ture

it

will

take m

ore

th

an

5 s

eco

nd

s f

or

the s

ou

nd

of

the

sir

en

to

reach

him

. S

ince t

he a

ssu

mp

tio

n i

s f

als

e,

yo

u w

ill

have p

roved

that

it i

s n

ot

20

°C w

hen

En

riq

ue h

ears

th

e s

iren

.

5-4

Practi

ce

Ind

irect

Pro

of

Lesson 5-4


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

29

G

len

co

e G

eo

me

try

1

. C

AN

OE

S T

hir

ty-f

ive s

tud

en

ts w

en

t on

a c

an

oein

g e

xp

ed

itio

n.

Th

ey r

en

ted

17 c

an

oes

for

the t

rip

. U

se a

n i

nd

irect

p

roof

to s

how

th

at

at

least

on

e c

an

oe

had

more

th

an

tw

o s

tud

en

ts i

n i

t.

2

. A

RE

A T

he a

rea o

f th

e U

nit

ed

Sta

tes

is

abou

t 6,0

00,0

00 s

qu

are

mil

es.

Th

e a

rea

of

Haw

aii

is

abou

t 11,0

00 s

qu

are

mil

es.

U

se a

n i

nd

irect

pro

of

to s

how

th

at

at

least

on

e o

f th

e f

ifty

sta

tes

has

an

are

a

gre

ate

r th

an

120,0

00 s

qu

are

mil

es.

3

. C

ON

SE

CU

TIV

E N

UM

BE

RS

D

avid

w

as

tryin

g t

o f

ind

a c

om

mon

fact

or

oth

er

than

1 b

etw

een

vari

ou

s p

air

s of

con

secu

tive i

nte

gers

. W

rite

an

in

dir

ect

pro

of

to s

how

David

th

at

two c

on

secu

tive i

nte

gers

do n

ot

share

a c

om

mon

fact

or

oth

er

than

1.

4

. W

OR

DS

T

he w

ord

s a

ccom

pli

shm

ent,

co

un

tere

xa

mp

le,

an

d e

xte

mp

ora

neo

us

all

h

ave 1

4 l

ett

ers

. U

se a

n i

nd

irect

pro

of

to

show

th

at

an

y w

ord

wit

h 1

4 l

ett

ers

mu

st

use

a r

ep

eate

d l

ett

er

or

have t

wo l

ett

ers

th

at

are

con

secu

tive i

n t

he a

lph

abet.

S

up

po

se t

he l

ett

ers

are

dis

tin

ct

an

d n

on

co

nsecu

tive.

Th

en

th

e

alp

hab

et

mu

st

have a

t le

ast

14 +

13 o

r 27 l

ett

ers

, a

co

ntr

ad

icti

on

.

5

. LA

TT

ICE

TR

IAN

GLE

S A

la

ttic

e p

oin

t is

a p

oin

t w

hose

coord

inate

s are

both

in

tegers

. A

latt

ice t

rian

gle

is

a t

rian

gle

w

hose

vert

ices

are

latt

ice p

oin

ts.

It i

s a

fact

th

at

a l

att

ice t

rian

gle

has

an

are

a

of

at

least

0.5

squ

are

un

its.

y

xO

A

B

C

5

5

a.

Su

pp

ose

△A

BC

has

a l

att

ice p

oin

t in

it

s in

teri

or.

Sh

ow

th

at

the l

att

ice

tria

ngle

can

be p

art

itio

ned

in

to t

hre

e

small

er

latt

ice t

rian

gle

s.

b.

Pro

ve i

nd

irect

ly t

hat

a l

att

ice t

rian

gle

w

ith

are

a 0

.5 s

qu

are

un

its

con

tain

s n

o l

att

ice p

oin

t. (

Bein

g o

n t

he

bou

nd

ary

does

not

cou

nt

as

insi

de.)

Sam

ple

an

sw

er:

Su

pp

ose a

ll

can

oes h

ad

≤ 2

stu

den

ts,

then

th

e t

ota

l w

ou

ld b

e l

ess

than

or

eq

ual

to 1

7 ×

2 =

34,

a c

on

trad

icti

on

.

Sam

ple

an

sw

er:

Su

pp

ose n

o

sta

te h

as a

rea >

120,0

00 m

i2.

Th

en

th

e t

ota

l are

a c

ou

ld n

ot

exceed

120,0

00 ×

49 +

11,0

00 =

5,8

91,0

00,

a c

on

trad

icti

on

.

Sam

ple

an

sw

er

in d

iag

ram

ab

ove.

5-4

Wo

rd

Pro

ble

m P

racti

ce

Ind

irect

Pro

of

Sam

ple

an

sw

er:

Fro

m E

xerc

ise

5a,

the l

att

ice t

rian

gle

co

nta

ins

3 s

maller

latt

ice t

rian

gle

s,

each

of

wh

ich

has a

rea a

t le

ast

0.5

sq

uare

un

its.

Th

e o

rig

inal

wo

uld

th

en

have a

rea a

t le

ast

1.5

sq

uare

un

its,

a

co

ntr

ad

icti

on

.

S

am

ple

an

sw

er:

Assu

me x

an

d y

are

in

teg

ers

wit

h a

co

mm

on

facto

r g

reate

r th

an

1. F

or

co

nsecu

tive

inte

gers

on

e is e

ven

an

d t

he o

ther

is

od

d, so

x =

2a a

nd

y =

2a +

1, fo

r an

in

teg

er

a. L

et

n b

e t

he c

om

mo

n

facto

r g

reate

r th

at

1.

Th

ere

fore

x

−

n =

2a

−

n

is a

n i

nte

ger

an

d y

−

n =

2a+

1

−

n

i

s a

lso

an

in

teg

er.

Bu

t 2

a+

1

−

n

= 2

a

−

n

+

1

−

n a

nd

1

−

n i

s n

ot

an

in

teg

er

un

less

n

= 1

, a c

on

trad

icti

on

.


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

30

G

len

co

e G

eo

me

try

Mo

re C

ou

nte

rexam

ple

s

Som

e s

tate

men

ts i

n m

ath

em

ati

cs c

an

be p

roven

fals

e b

y counterexamples.

Con

sid

er

the f

oll

ow

ing s

tate

men

t.

For

an

y n

um

bers

a a

nd

b, a

- b

= b

- a

.

You

can

pro

ve t

hat

this

sta

tem

en

t is

fals

e i

n g

en

era

l if

you

can

fi n

d o

ne

exam

ple

for

wh

ich

th

e s

tate

men

t is

fals

e.

Let a

= 7

an

d b

= 3

. S

ubst

itu

te t

hese

valu

es

in t

he e

qu

ati

on

above.

7 -

3

3 -

7

4 ≠

-4

In g

en

era

l, f

or

an

y n

um

bers

a a

nd

b, th

e s

tate

men

t a

- b

= b

- a

is

fals

e.

You

can

mak

e t

he e

qu

ivale

nt

verb

al

state

men

t: s

ubtr

act

ion

is not

a

com

mu

tati

ve o

pera

tion

.

In e

ach

of

the

fo

llo

win

g e

xe

rcis

es a

, b

, a

nd

c a

re

an

y n

um

be

rs.

Pro

ve

th

at

the

sta

tem

en

t is

fa

lse

by

co

un

tere

xa

mp

le.

1

. a

- (b

- c

)

(a

- b

) -

c

2. a

÷ (b ÷

c)

(a

÷ b

) ÷

c

3

. a

÷ b

b

÷ a

4. a

÷ (b +

c)

(a

÷ b

) +

(a

÷ c

)

5

. a

+ (bc)

(a

+ b

)(a

+ c

) 6. a

2 +

a2

a4

7

. W

rite

th

e v

erb

al

equ

ivale

nts

for

Exerc

ises

1,

2,

an

d 3

.

8

. F

or

the D

istr

ibu

tive P

rop

ert

y, a

(b +

c)

= ab +

ac,

it

is s

aid

th

at

mu

ltip

lica

tion

d

istr

ibu

tes

over

ad

dit

ion

. E

xerc

ises

4 a

nd

5 p

rove t

hat

som

e o

pera

tion

s d

o n

ot

dis

trib

ute

. W

rite

a s

tate

men

t fo

r each

exerc

ise t

hat

ind

icate

s th

is.

Sam

ple

an

sw

ers

are

giv

en

.

6 -

(4 -

2)

(

6 -

4)

- 2

6 ÷

(4 ÷

2)

(

6 ÷

4)

÷ 2

6 -

2

2 -

2

6

−

2

1.5

−

2

4 ≠

0

3 ≠

0.7

5

1

. S

ub

tracti

on

is n

ot

an

asso

cia

tive o

pera

tio

n.

2.

Div

isio

n i

s n

ot

an

asso

cia

tive o

pera

tio

n.

3.

Div

isio

n i

s n

ot

a c

om

mu

tati

ve o

pera

tio

n.

4

. D

ivis

ion

do

es n

ot

dis

trib

ute

over

ad

dit

ion

.5.

Ad

dit

ion

do

es n

ot

dis

trib

ute

over

mu

ltip

licati

on

.

5-4

En

rich

men

t

6 ÷

4

4 ÷

6

6 ÷

(4 +

2)

(

6 ÷

4)

+(6

÷ 2

)

3

−

2 ≠

2

−

3

6 ÷

6

1.5

+ 3

1 ≠

4.5

6 +

(4 .

2)

(

6 +

4)

(6 +

2)

62 +

62

64

6 +

8

(10)

(8)

36 +

36

1296

1

4

80

72 ≠

1296


NA

ME

DA

TE

PE

RIO

D

Lesson 5-5

Ch

ap

ter

5

31

G

len

co

e G

eo

me

try

Th

e T

ria

ng

le I

ne

qu

ali

ty

If y

ou

tak

e t

hre

e s

traw

s of

len

gth

s 8 i

nch

es,

5 i

nch

es,

an

d

1 i

nch

an

d t

ry t

o m

ak

e a

tri

an

gle

wit

h t

hem

, you

wil

l fi

nd

th

at

it i

s n

ot

poss

ible

. T

his

il

lust

rate

s th

e T

rian

gle

In

equ

ali

ty T

heore

m.

Tri

an

gle

In

eq

ua

lity

Th

eo

rem

Th

e s

um

of

the

le

ng

ths o

f a

ny t

wo

sid

es o

f a

tria

ng

le m

ust

be

gre

ate

r th

an

th

e le

ng

th o

f th

e t

hird

sid

e.

BC

A

a

cb

Th

e m

ea

su

re

s o

f tw

o s

ide

s o

f a

tria

ng

le a

re

5 a

nd

8.

Fin

d a

ra

ng

e

for t

he

le

ng

th o

f th

e t

hir

d s

ide

.

By t

he T

rian

gle

In

equ

ali

ty T

heore

m,

all

th

ree o

f th

e f

oll

ow

ing i

nequ

ali

ties

mu

st b

e t

rue.

5 +

x >

8

8 +

x >

5

5 +

8 >

x

x >

3

x >

-3

13 >

x

Th

ere

fore

x m

ust

be b

etw

een

3 a

nd

13.

Exerc

ises

Is i

t p

ossib

le t

o f

orm

a t

ria

ng

le w

ith

th

e g

ive

n s

ide

le

ng

ths? I

f n

ot,

ex

pla

in

wh

y n

ot.

1

. 3,

4,

6

2. 6,

9,

15

3

. 8,

8,

8

4. 2,

4,

5

5

. 4,

8,

16

6. 1.5

, 2.5

, 3

Fin

d t

he

ra

ng

e f

or t

he

me

asu

re

of

the

th

ird

sid

e o

f a

tria

ng

le g

ive

n t

he

me

asu

re

s

of

two

sid

es.

7

. 1 c

m a

nd

6 c

m

8. 12 y

d a

nd

18 y

d

9. 1.5

ft a

nd

5.5

ft

10

. 82 m

an

d 8

m

11

. S

up

pose

you

have t

hre

e d

iffe

ren

t p

osi

tive n

um

bers

arr

an

ged

in

ord

er

from

least

to

gre

ate

st.

Wh

at

sin

gle

com

pari

son

wil

l le

t you

see i

f th

e n

um

bers

can

be t

he l

en

gth

s of

the s

ides

of

a t

rian

gle

?

yes

no

; 6 +

9 =

15

yes

yes

no

; 4 +

8 <

16

yes

5 c

m <

n <

7 c

m6 y

d <

n <

30 y

d

4 ft

< n

< 7

ft74 m

< n

< 9

0 m

Fin

d t

he s

um

of

the t

wo

sm

aller

nu

mb

ers

. If

th

at

su

m i

s g

reate

r th

an

th

e

larg

est

nu

mb

er,

th

en

th

e t

hre

e n

um

bers

can

be t

he l

en

gth

s o

f th

e s

ides

of

a t

rian

gle

.

5-5

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Th

e T

rian

gle

In

eq

uality

Exam

ple

a +

b >

c

b +

c >

a

a +

c >

b


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

32

G

len

co

e G

eo

me

try

Pro

ofs

Usi

ng

Th

e T

ria

ng

le I

ne

qu

ali

ty T

he

ore

m

You

can

use

th

e T

rian

gle

In

equ

ali

ty T

heore

m a

s a r

easo

n i

n p

roofs

.

Co

mp

lete

th

e f

oll

ow

ing

pro

of.

Giv

en

: △

AB

C !

△D

EC

Pro

ve

: A

B +

DE

> A

D −

BE

Pro

of:

Sta

tem

en

ts

1.

△A

BC

! △

DE

C

2.

AB

+ B

C >

AC

DE

+ E

C >

CD

3.

AB

> A

C –

BC

DE

> C

D –

EC

4.

AB

+ D

E >

AC

- B

C +

CD

- E

C

5.

AB

+ D

E >

AC

+ C

D -

BC

- E

C

6.

AB

+ D

E >

AC

+ C

D -

(B

C +

EC

)

7.

AC

+ C

D =

AD

BC

+ E

C =

BE

8.

AB

+ D

E >

AD

- B

E

Re

aso

ns

1.

Giv

en

2.

Tri

an

gle

In

equ

ali

ty T

heore

m

3.

Su

btr

act

ion

4.

Ad

dit

ion

5.

Com

mu

tati

ve

6.

Dis

trib

uti

ve

7.

Segm

en

t A

dd

itio

n P

ost

ula

te

8.

Su

bst

itu

tion

Exercis

es

PR

OO

F W

rit

e a

tw

o c

olu

mn

pro

of.

Giv

en

: −

−

PL

‖ −

−−

MT

K i

s th

e m

idp

oin

t of

−−

PT

.

Pro

ve

: P

K +

KM

> P

L

Pro

of:

Sta

tem

en

ts

1.

−−

PL

‖ −

−−

MT

2.

∠P

! ∠

T

3.

K i

s th

e m

idp

oin

t of

−−

PT

.

4.

PK

= K

T

5.

6.

△P

KL

! △

TK

M

7.

8.

9.

PK

+ K

M >

PL

Re

aso

ns

1.

Giv

en

2.

Alt

ern

ate

In

teri

or

An

gle

s T

heo

rem

3.

Giv

en

4.

Defi

nit

ion

of

mid

po

int

5.

Vert

ical

An

gle

s T

heore

m

6.

AS

A

7.

Tri

an

gle

In

equ

ali

ty T

heore

m

8.

CP

CT

C

9.

Su

bsti

tuti

on

5-5

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Th

e T

rian

gle

In

eq

uality

∠P

KL !

∠M

KT

PK

+ K

L >

PL

KL

= K

M


NA

ME

DA

TE

PE

RIO

D

Lesson 5-5

Ch

ap

ter

5

33

G

len

co

e G

eo

me

try

5-5

Sk

ills

Pra

ctic

e

Th

e T

rian

gle

In

eq

uality

Is i

t p

ossib

le t

o f

orm

a t

ria

ng

le w

ith

th

e g

ive

n s

ide

le

ng

ths? I

f n

ot,

ex

pla

in

wh

y n

ot.

1. 2 f

t, 3

ft,

4 f

t 2

. 5 m

, 7 m

, 9 m

3

. 4 m

m,

8 m

m,

11 m

m

4. 13 i

n., 1

3 i

n., 2

6 i

n.

5

. 9 c

m,

10 c

m,

20 c

m

6. 15 k

m,

17 k

m,

19 k

m

7

. 14 y

d,

17 y

d,

31 y

d

8. 6 m

, 7 m

, 12 m

Fin

d t

he

ra

ng

e f

or t

he

me

asu

re

of

the

th

ird

sid

e o

f a

tria

ng

le g

ive

n t

he

me

asu

re

s

of

two

sid

es.

9

. 5 f

t, 9

ft

1

0. 7 i

n., 1

4 i

n.

11

. 8 m

, 13 m

12

. 10 m

m,

12 m

m

13

. 12 y

d,

15 y

d

14

. 15 k

m,

27 k

m

15

. 17 c

m,

28 c

m,

1

6. 18 f

t, 2

2 f

t

17

. P

ro

of

Com

ple

te t

he p

roof.

G

ive

n:

△A

BC

an

d △

CD

E

P

ro

ve

: A

B +

BC

+ C

D +

DE

> A

E

Pro

of:

Sta

tem

en

tsR

ea

so

ns

1.

AB

+ B

C >

AC

CD

+ D

E >

CE

1.

Tri

an

gle

In

eq

uality

Th

eo

rem

2.

AB

+ B

C +

CD

+ D

E >

AC

+ C

E2

. A

dd

itio

n P

rop

ert

y o

f E

qu

ality

3.

AC

+ C

E =

AE

3.

Seg.

Ad

dit

ion

Post

4.

AB

+ B

C +

CD

+ D

E >

AE

4.

Su

bst

itu

tion

yes

yes

yes

no

; 13 +

13 ≯

26

yes

yes

no

; 9 +

10 ≯

20

no

; 14 +

17 ≯

31

4 f

t <

n <

14 f

t

5 m

< n

< 2

1 m

3 y

d <

n <

27 y

d

11 c

m <

n <

45 c

m

7 i

n. <

n <

21 i

n.

2 m

m <

n <

22 m

m

12 k

m <

n <

42 k

m

4 f

t <

n <

40 f

t


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

34

G

len

co

e G

eo

me

try

Is i

t p

ossib

le t

o f

orm

a t

ria

ng

le w

ith

th

e g

ive

n s

ide

le

ng

ths? I

f n

ot

ex

pla

in

wh

y n

ot.

1

. 9,

12,

18

2. 8,

9,

17

3

. 14,

14,

19

4. 23,

26,

50

5

. 32,

41,

63

6. 2.7

, 3.1

, 4.3

7

. 0.7

, 1.4

, 2.1

8. 12.3

, 13.9

, 25.2

Fin

d t

he

ra

ng

e f

or t

he

me

asu

re

of

the

th

ird

sid

e o

f a

tria

ng

le g

ive

n t

he

me

asu

re

s

of

two

sid

es.

9. 6 ft

an

d 1

9 ft

10

. 7 k

m a

nd

29 k

m

11

. 13 in

. an

d 2

7 in

.

12

. 18 ft

an

d 2

3 ft

13

. 25 y

d a

nd

38 y

d

14

. 31 c

m a

nd

39 c

m

15

. 42 m

an

d 6

m

16

. 54 in

. an

d 7

in.

17

. G

ive

n:

H i

s th

e c

en

troid

of

△E

DF

P

ro

ve

: E

Y +

FY

> D

E

Pro

of:

Sta

tem

en

ts

1.

H i

s th

e c

en

troid

of

△E

DF

2.

−−

EY

is

a m

ed

ian

.

3.

Y i

s t

he m

idp

oin

t o

f −

−

DF

4. D

Y =

FY

5.

EY

+ D

Y >

DE

6.

EY

+ F

Y >

DE

Re

aso

ns

1.

Giv

en

2.

Defi

nit

ion

of

cen

tro

id

3.

Defi

nit

ion

of

med

ian

4.

Defi

nit

ion

of

mid

poin

t

5.

Tri

an

gle

In

eq

uality

Th

eo

rem

6.

Su

bsti

tuti

on

18

. G

AR

DE

NIN

G H

a P

oon

g h

as

4 l

en

gth

s of

wood

fro

m w

hic

h h

e p

lan

s to

mak

e a

bord

er

for

a t

rian

gu

lar-

shap

ed

herb

gard

en

. T

he l

en

gth

s of

the w

ood

bord

ers

are

8 i

nch

es,

10 i

nch

es,

12 i

nch

es,

an

d 1

8 i

nch

es.

How

man

y d

iffe

ren

t tr

ian

gu

lar

bord

ers

can

H

a P

oon

g m

ak

e?

yes

no

; 8 +

9 =

17

yes

no

; 23 +

26 <

50

yes

yes

no

; 0.7

+ 1

.4 =

2.1

yes

13ft

< n

< 2

5ft

22 k

m <

n <

36 k

m

14 in

. <

n <

40 in

.5 ft

< n

< 4

1 ft

13 y

d <

n <

63 y

d8 c

m <

n <

70 c

m

36 m

< n

< 4

8 m

47 in

. <

n <

61 in

.

3

5-5

Practi

ce

Th

e T

rian

gle

In

eq

uality


NA

ME

DA

TE

PE

RIO

D

Lesson 5-5

Ch

ap

ter

5

35

G

len

co

e G

eo

me

try

Tan

ya’s

ho

me

Su

per

mar

ket

Rai

lro

ad AB

C

1

. S

TIC

KS

Jam

ila h

as

5 s

tick

s of

len

gth

s 2,

4,

6,

8,

an

d 1

0 i

nch

es.

Usi

ng t

hre

e

stic

ks

at

a t

ime a

s th

e s

ides

of

tria

ngle

s,

how

man

y t

rian

gle

s ca

n s

he m

ak

e?

Use

th

e f

igu

re

at

the

rig

ht

for E

xe

rcis

es

2 a

nd

3.

2

. P

AT

HS

T

o g

et

to t

he n

eare

st

sup

er m

ark

et,

T

an

ya m

ust

walk

over

a r

ail

road

tr

ack

. T

here

are

tw

o p

lace

s w

here

sh

e c

an

cro

ss t

he

track

(p

oin

ts A

an

d B

). W

hic

h p

ath

is

lon

ger?

Exp

lain

.

3

. P

AT

HS

W

hil

e o

ut

walk

ing o

ne d

ay

Tan

ya f

ind

s a t

hir

d p

lace

to c

ross

th

e

rail

road

tra

cks.

Sh

ow

th

at

the p

ath

th

rou

gh

poin

t C

is

lon

ger

than

th

e

path

th

rou

gh

poin

t B

.

4. C

ITIE

S T

he d

ista

nce

betw

een

New

York

C

ity a

nd

Bost

on

is

187 m

iles

an

d t

he

dis

tan

ce b

etw

een

New

York

Cit

y a

nd

H

art

ford

is

97 m

iles.

Hart

ford

, B

ost

on

, an

d N

ew

York

Cit

y f

orm

a t

rian

gle

on

a

map

. W

hat

mu

st t

he d

ista

nce

betw

een

B

ost

on

an

d H

art

ford

be g

reate

r th

an

?

5

. T

RIA

NG

LE

ST

he f

igu

re s

how

s an

equ

ilate

ral

tria

ngle

AB

C a

nd

a p

oin

tP

ou

tsid

e t

he t

rian

gle

.

B´

A´

C´

C

P´

P

B

A

a.

Dra

w t

he f

igu

re t

hat

is t

he r

esu

lt

of

turn

ing t

he o

rigin

al

figu

re 6

0°

cou

nte

rclo

ckw

ise a

bou

t A

. D

en

ote

by

P', t

he i

mage o

f P

un

der

this

tu

rn.

b.

Note

th

at

−−

−P

'B i

s co

ngru

en

t to

−−

PC

. It

is

als

o t

rue t

hat

−−

−P

P' is

con

gru

en

t to

−−

PA

. W

hy?

c.

Sh

ow

th

at

−−

PA

, −

−P

B ,

an

d −

−P

C s

ati

sfy t

he

tria

ngle

in

equ

ali

ties.

3

By t

he T

rian

gle

In

eq

uality

Th

eo

rem

, th

e d

ista

nce f

rom

Tan

ya’s

ho

me t

o p

oin

t B

an

d o

n

to t

he s

up

erm

ark

et

is g

reate

r th

an

the s

traig

ht

dis

tan

ce f

rom

Tan

ya’s

ho

me t

o t

he S

up

erm

ark

et.

Sam

ple

an

sw

er:

Let

S b

e t

he

Su

perm

ark

et

an

d T

be T

an

ya’s

ho

me.

Becau

se ∠

SA

B i

s 9

0,

m∠

SB

A <

90,

so

m∠

SB

C >

90,

makin

g S

C >

SB

. S

imilarl

y,

CT

> B

T.

Th

ere

fore

CT +

CS

> B

T +

BS

.

90 m

i

See f

igu

re.

Sam

ple

an

sw

er:

△P

'PB

is a

tria

ng

le w

ith

sid

e l

en

gth

s

eq

ual

to P

A,

PB

, an

d P

C.

5-5

Wo

rd

Pro

ble

m P

racti

ce

Th

e T

rian

gle

In

eq

uality

Sam

ple

an

sw

er:

PA

is

co

ng

ruen

t to

P'A

an

d m

∠P

AP

' is

60

°, S

o b

y S

AS

, tr

ian

gle

P

P'A

is e

qu

ilate

ral. T

hu

s,

PP

' =

PA


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

36

G

len

co

e G

eo

me

try

Co

nstr

ucti

ng

Tri

an

gle

s

Th

e m

ea

su

re

me

nts

of

the

sid

es o

f a

tria

ng

le a

re

giv

en

. If

a t

ria

ng

le h

av

ing

sid

es

wit

h t

he

se

me

asu

re

me

nts

is n

ot

po

ssib

le,

the

n w

rit

e i

mp

ossib

le.

If a

tria

ng

le i

s

po

ssib

le,

dra

w i

t a

nd

me

asu

re

ea

ch

an

gle

wit

h a

pro

tra

cto

r.

1

. AR

= 5

cm

m

∠A

=

2. PI =

8 c

m

m∠P

=

RT

= 3

cm

m

∠R

=

IN

= 3

cm

m

∠I =

AT

= 6

cm

m

∠T

=

PN

= 2

cm

m

∠N

=

3

. ON

= 1

0 c

m

m∠O

=

4. TW

= 6

cm

m

∠T

=

NE

= 5

.3 c

m

m

∠N

=

WO

= 7

cm

m

∠W

=

OE

= 4

.6 c

m

m

∠E

=

TO

= 2

cm

m

∠O

=

5

. BA

= 3

.l c

m

m

∠B

=

6. AR

= 4

cm

m

∠A

=

AT

= 8

cm

m

∠A

=

RM

= 5

cm

m

∠R

=

BT

= 5

cm

m

∠T

=

AM

= 3

cm

m

∠M

=

M

RA

T

BA

W

T

O

AR T

30

94

56

imp

ossib

le

112

15 53

imp

ossib

le

162

90

11

37

753

5-5

En

rich

men

t

Lesson 5-5


NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

37

G

len

co

e G

eo

me

try

Hin

ge

Th

eo

rem

T

he f

oll

ow

ing t

heore

m a

nd

its

con

vers

e i

nvolv

e t

he r

ela

tion

ship

betw

een

th

e s

ides

of

two t

rian

gle

s an

d a

n a

ngle

in

each

tri

an

gle

.

Hin

ge

Th

eo

rem

If t

wo

sid

es o

f a

tria

ng

le a

re c

on

gru

en

t to

tw

o

sid

es o

f a

no

the

r tr

ian

gle

an

d t

he

in

clu

de

d

an

gle

of

the

first

is la

rge

r th

an

th

e in

clu

de

d

an

gle

of

the

se

co

nd

, th

en

th

e t

hird

sid

e o

f

the

first

tria

ng

le is lo

ng

er

tha

n

the

th

ird

sid

e o

f th

e s

eco

nd

tria

ng

le.

RT

> A

C

Co

nv

ers

e o

f th

e

Hin

ge

Th

eo

rem

If t

wo

sid

es o

f a

tria

ng

le a

re c

on

gru

en

t to

two

sid

es o

f a

no

the

r tr

ian

gle

, a

nd

th

e

third

sid

e in

th

e f

irst

is lo

ng

er

tha

n t

he

third

sid

e in

th

e s

eco

nd

, th

en

th

e in

clu

de

d

an

gle

in

th

e f

irst

tria

ng

le is g

rea

ter

tha

n

the

in

clu

de

d a

ng

le in

th

e s

eco

nd

tria

ng

le.

m∠

M >

m∠

R

Exerc

ises

Co

mp

are

th

e g

ive

n m

ea

su

re

s.

1

. MR

an

d RP

N

R

P

M

21

°

19

°

2

. AD

an

d CD

C ADB

22

°

38

°

M

R >

RP

A

D >

CD

3

. m

∠C

an

d m

∠Z

4

. m

∠XYW

an

d m

∠WYZ

m∠

C <

m∠

Z

m

∠X

YW

< m

∠W

YZ

Writ

e a

n i

ne

qu

ali

ty f

or t

he

ra

ng

e o

f v

alu

es o

f x.

5.

115

°

120

°24

24

40

( 4x

- 1

0)

6

.

33

°

60

60

36

30

( 3x

- 3

) °

5-6

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

Ineq

ualiti

es i

n T

wo

Tri

an

gle

s

ST

80

°

R

BC

60

°A

33

36

TR

SN

MP

C

om

pa

re

th

e m

ea

su

re

s

of

−−

GF

an

d −

−

FE

.

H

EFG

22°

28°

Tw

o s

ides

of

△HGF

are

co

ng

rue

nt

to t

wo

si

de

s o

f △HEF

, a

nd

m∠GHF

> m

∠EHF

. B

y

the H

inge T

heore

m, GF

> FE.

C

om

pa

re

th

e m

ea

su

re

s

of

∠A

BD

an

d ∠

CB

D.

13

16

C D A

B

Tw

o s

ides

of

△ABD

are

co

ng

rue

nt

to

two

sid

es

of

△CBD

, a

nd

AD

> CD

. B

y t

he

Con

vers

e o

f th

e H

inge T

heore

m,

m∠ABD

> m

∠CBD.

Exam

ple

2Exam

ple

1

x

> 1

2.5

x <

12

30

C

AX

B3

0

50

48

24

24

ZY

42

28

ZW

XY


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

38

G

len

co

e G

eo

me

try

PR

OV

E R

ELA

TIO

NS

HIP

S I

N T

WO

TR

IAN

GLE

S

You

can

use

th

e H

inge T

heore

m

an

d i

ts c

on

vers

e t

o p

rove r

ela

tion

ship

s in

tw

o t

rian

gle

s.

Giv

en

: R

X =

XS

∠

SX

T =

97°

Pro

ve

: S

T >

RT

Pro

of:

Sta

tem

en

tsR

ea

so

ns

1

. ∠

SX

T a

nd

∠R

XT

are

sup

ple

men

tary

2

. m

∠S

XT

+ m

∠R

XT

= 1

80

°

3

. m

∠S

XT

= 9

7°

4

. 9

7 +

m∠

RX

T =

180

5

. m

∠R

XT

= 8

3

6

. 9

7 >

83

7

. m

∠S

XT

> m

∠R

XT

8

. R

X =

XS

9

. T

X =

TX

10

. S

T >

RT

1

. D

efn

of

lin

ear

pair

2

. D

efn

of

sup

ple

men

tary

3

. G

iven

4

. S

ubst

itu

tion

5

. S

ubtr

act

ion

6

. In

equ

ali

ty

7

. S

ubst

itu

tion

8

. G

iven

9

. R

efl

exiv

e

10

. H

inge T

heore

m

Exerc

ises

Co

mp

lete

th

e p

ro

of.

Giv

en

: re

ctan

gle

AF

BC

ED

= D

C

Pro

ve

: A

E >

FB

Pro

of:

Sta

tem

en

tsR

ea

so

ns

1.

rect

an

gle

AF

BC

, E

D =

DC

2.

AD

= A

D

3.

m∠

ED

A >

m∠

AD

C

4. A

E >

AC

5. A

C =

FB

6.

AE

> F

B

1.

giv

en

2.

refl

exiv

e

3.

exte

rior

an

gle

4.

Hin

ge T

heore

m

5.

op

p s

ides

" i

n r

ect

an

gle

.

6.

Su

bst

itu

tion

5-6

Stu

dy

Gu

ide a

nd

In

terv

en

tio

n

(con

tin

ued

)

Ineq

ualiti

es I

nvo

lvin

g T

wo

Tri

an

gle

s

Exam

ple

97°


NA

ME

DA

TE

PE

RIO

D

Lesson 5-6

Ch

ap

ter

5

39

G

len

co

e G

eo

me

try

5-6

Co

mp

are

th

e g

ive

n m

ea

su

re

s.

1

. m

∠B

XA

an

d m

∠D

XA

m∠

BX

A <

m∠

DX

A

2

. B

C a

nd

DC

B

C >

DC

Co

mp

are

th

e g

ive

n m

ea

su

re

s.

3

. m∠

ST

R a

nd

m∠

TR

U

4. P

Q a

nd

RQ

31

30

22

22

RS

UT

95°

77

85°

PR

SQ

m∠

STR

< m∠

TR

U

P

Q >

RQ

5

. In

th

e f

igu

re,

−−

BA

, −

−−

BD

, −

−−

BC

, an

d −

−−

BE

are

con

gru

en

t an

d A

C <

DE

.

How

does

m∠

1 c

om

pare

wit

h m∠

3?

Exp

lain

you

r th

ink

ing.

m∠

1 <

m∠

3;

Fro

m t

he g

iven

in

form

ati

on

an

d t

he

SS

S I

neq

uality

Th

eo

rem

, it

fo

llo

ws t

hat

in △

AB

C

an

d △

DB

E w

e h

ave m∠

AB

C <

m∠

DB

E.

Sin

ce

m∠

AB

C =

m∠

1 +

m∠

2 a

nd

m∠

DB

E =

m∠

3 +

m∠

2,

it f

ollo

ws t

hat

m∠

1 +

m∠

2 <

m∠

3 +

m∠

2.

Su

btr

act

m∠

2 f

rom

each

sid

e o

f th

e l

ast

ineq

uality

to

get

m∠

1 <

m∠

3.

6

. P

RO

OF

W

rite

a t

wo-c

olu

mn

pro

of.

G

ive

n:

−−

BA

" −

−−

DA

BC

> D

C

P

ro

ve

: m∠

1 >

m∠

2

P

roo

f:

S

tate

me

nts

R

ea

so

ns

1

. B

A "

DA

1.

Giv

en

2.

BC

> D

C

2.

Giv

en

3.

AC

" A

C

3.

Refl

exiv

e P

rop

ert

y

4.

m∠

1 >

m∠

3

4.

SS

S I

nequ

ali

ty

1

2

3

B

A

DC

E

Sk

ills

Pra

ctic

e

Ineq

ualiti

es I

nvo

lvin

g T

wo

Tri

an

gle

s

6

98

3

3

B

AC

D

X

1 2

B

A

D

C


Answers

Co

pyri

gh

t ©

Gle

nco

e/M

cG

raw

-Hill, a

div

isio

n o

f T

he

Mc

Gra

w-H

ill C

om

pa

nie

s,

Inc

.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

40

G

len

co

e G

eo

me

try

12

DF

E

G

20

21

RT

S

JK

14

14

14

13

12

CF

E

D

( x+

3) °

( x-

3) °

10

10

RT

S

Q

40

°

30

°

60

°A

KM

B

Co

mp

are

th

e g

ive

n m

ea

su

re

s.

1

. A

B a

nd

BK

2. S

T a

nd

SR

3. m

∠C

DF

an

d m

∠E

DF

4. m

∠R

an

d m

∠T

5. P

RO

OF W

rite

a t

wo-c

olu

mn

pro

of.

G

ive

n:

G i

s th

e m

idp

oin

t of

−−

−

DF

.

m∠

1 >

m∠

2

P

ro

ve

: E

D >

EF

Pro

of:

Sta

tem

en

tsR

easo

ns

1.

G i

s t

he m

idp

oin

t o

f −

−

DF .

1.

Giv

en

2.

−−

DG

# −

−

FG

2.

Defi

nit

ion

of

mid

po

int

3.

−−

EG

# −

−

EG

3.

Refl

exiv

e P

rop

ert

y

4.

m∠

1 >

m∠

24.

Giv

en

5.

ED

> E

F5.

Hin

ge T

heo

rem

6

. T

OO

LS

R

ebecc

a u

sed

a s

pri

ng c

lam

p t

o h

old

togeth

er

a c

hair

leg s

he r

ep

air

ed

wit

h w

ood

glu

e.

Wh

en

sh

e o

pen

ed

th

e c

lam

p,

she n

oti

ced

th

at

the a

ngle

betw

een

th

e h

an

dle

s of

the c

lam

p

decr

ease

d a

s th

e d

ista

nce

betw

een

th

e h

an

dle

s of

the c

lam

p

decr

ease

d.

At

the s

am

e t

ime,

the d

ista

nce

betw

een

th

e

gri

pp

ing e

nd

s of

the c

lam

p i

ncr

ease

d.

Wh

en

sh

e r

ele

ase

d t

he

han

dle

s, t

he d

ista

nce

betw

een

th

e g

rip

pin

g e

nd

of

the c

lam

p

decr

ease

d a

nd

th

e d

ista

nce

betw

een

th

e h

an

dle

s in

crease

d.

Is t

he c

lam

p a

n e

xam

ple

of

the H

inge T

heore

m o

r it

s co

nvers

e?

AB

> B

KS

T >

SR

m∠

CD

F <

m∠

ED

Fm

∠R

< m

∠T

Hin

ge T

heo

rem

5-6

Practi

ce

Ineq

ualiti

es i

n T

wo

Tri

an

gle

s


NA

ME

DA

TE

PE

RIO

D

Lesson 5-6

Ch

ap

ter

5

41

G

len

co

e G

eo

me

try

1

. C

LO

CK

S T

he m

inu

te h

an

d o

f a

gra

nd

fath

er

clock

is

3 f

eet

lon

g a

nd

the h

ou

r h

an

d i

s 2 f

eet

lon

g.

Is t

he

dis

tan

ce b

etw

een

th

eir

en

ds

gre

ate

r

at

3:0

0 o

r at

8:0

0?

2

. FE

RR

IS W

HE

EL A

Ferr

is w

heel

has

carr

iages

loca

ted

at

the 1

0 v

ert

ices

of

a r

egu

lar

deca

gon

.

12

3 4

5

67

89

10

Wh

ich

carr

iages

are

fart

her

aw

ay

from

carr

iage n

um

ber

1 t

han

carr

iage

nu

mber

4?

3

. W

ALK

WA

Y T

yre

e w

an

ts t

o m

ak

e t

wo

slig

htl

y d

iffe

ren

t tr

ian

gle

s fo

r h

is

walk

way.

He h

as

thre

e p

iece

s of

wood

to c

on

stru

ct t

he f

ram

e o

f h

is t

rian

gle

s.

Aft

er

Tyre

e m

ak

es

the f

irst

con

crete

tria

ngle

, h

e a

dju

sts

two s

ides

of

the

tria

ngle

so t

hat

the a

ngle

th

ey c

reate

is s

mall

er

than

th

e a

ngle

in

th

e f

irst

tria

ngle

. E

xp

lain

how

th

is c

han

ges

the

tria

ngle

.

4. M

OU

NT

AIN

PE

AK

S E

mil

y l

ives

the

sam

e d

ista

nce

fro

m t

hre

e m

ou

nta

in

peak

s: H

igh

Poin

t, T

op

per,

an

d C

lou

d

Nin

e.

For

a p

hoto

gra

ph

y c

lass

, E

mil

y

mu

st t

ak

e a

ph

oto

gra

ph

fro

m h

er

hou

se

that

show

s tw

o o

f th

e m

ou

nta

in p

eak

s.

Wh

ich

tw

o p

eak

s w

ou

ld s

he h

ave t

he

best

ch

an

ce o

f ca

ptu

rin

g i

n o

ne i

mage?

Em

ily

Clo

ud

Nin

e

Hig

hP

oin

t

Top

per

12 m

iles

9 miles

10 m

iles

5

. R

UN

NE

RS

A

ph

oto

gra

ph

er

is t

ak

ing

pic

ture

s of

thre

e t

rack

sta

rs,

Am

y,

Noel,

an

d B

eth

. T

he p

hoto

gra

ph

er

stan

ds

on

a

track

, w

hic

h i

s sh

ap

ed

lik

e a

rect

an

gle

wit

h s

em

icir

cles

on

both

en

ds.

11

8˚

36˚

14

6˚

Ph

oto

gra

ph

er

Am

y

No

el

Bet

h

a.

Base

d o

n t

he i

nfo

rmati

on

in

th

e

figu

re,

list

th

e r

un

ners

in

ord

er

from

neare

st t

o f

art

hest

fro

m t

he

ph

oto

gra

ph

er.

b.

Exp

lain

how

to l

oca

te t

he p

oin

t alo

ng

the s

em

icir

cula

r cu

rve t

hat

the

run

ners

are

on

th

at

is f

art

hest

aw

ay

from

th

e p

hoto

gra

ph

er.

8:0

0

5,

6,

an

d 7

Sam

ple

an

sw

er:

By t

he H

ing

e

Th

eo

rem

, th

e t

hir

d s

ide o

pp

osit

e

the a

ng

le t

hat

was m

ad

e s

maller

is

no

w s

ho

rter

than

th

e t

hir

d s

ide o

f

the f

irst

tria

ng

le.

Am

y,

Beth

, N

oel

5-6

Wo

rd

Pro

ble

m P

racti

ce

Ineq

ualiti

es i

n T

wo

Tri

an

gle

s To

pp

er

an

d C

lou

d N

ine

Exte

nd

th

e l

ine t

hro

ug

h t

he

ph

oto

gra

ph

er

an

d t

he c

en

ter

of

the s

em

icir

cle

to

wh

ere

it

inte

rsects

th

e s

em

icir

cu

lar

track.


Co

pyrig

ht ©

Gle

nc

oe

/Mc

Gra

w-H

ill, a d

ivis

ion

of T

he

Mc

Gra

w-H

ill Co

mp

an

ies, In

c.



NA

ME

DA

TE

PE

RIO

D

Ch

ap

ter

5

42

G

len

co

e G

eo

me

try

Hin

ge T

heo

rem

Th

e H

inge T

heore

m t

hat

you

stu

die

d i

n t

his

sect

ion

sta

tes

that

if t

wo s

ides

of

a

tria

ngle

are

con

gru

en

t to

tw

o s

ides

of

an

oth

er

tria

ngle

an

d t

he i

ncl

ud

ed

an

gle

in

on

e t

rian

gle

has

a g

reate

r m

easu

re t

han

th

e i

ncl

ud

ed

an

gle

in

th

e o

ther,

th

en

th

e

thir

d s

ide o

f th

e f

irst

tri

an

gle

is

lon

ger

than

th

e t

hir

d s

ide o

f th

e s

eco

nd

tri

an

gle

. In

this

act

ivit

y,

you

wil

l in

vest

igate

wh

eth

er

the c

on

vers

e,

invers

e a

nd

con

trap

osi

tive

of

the H

inge T

heore

m a

re a

lso t

rue.

X

Y

Z

Q

S

R

1 2

Hyp

oth

esi

s: XY

= QR

, YZ

= RS

, m

∠1

> m

∠2

Con

clu

sion

: XZ

> QS

1

. W

hat

is t

he c

on

vers

e o

f th

e H

inge T

heore

m?

2

. C

an

you

fin

d a

ny c

ou

nte

rexam

ple

s to

pro

ve t

hat

the c

on

vers

e i

s fa

lse?

3

. W

hat

is t

he i

nvers

e o

f th

e H

inge T

heore

m?

4

. C

an

you

fin

d a

ny c

ou

nte

rexam

ple

s to

pro

ve t

hat

the i

nvers

e i

s fa

lse?

5

. W

hat

is t

he c

on

trap

osi

tive o

f th

e H

inge T

heore

m?

6

. C

an

you

fin

d a

ny c

ou

nte

rexam

ple

s to

pro

ve t

hat

the c

on

trap

osi

tive i

s fa

lse?

I

f tw

o s

ides o

f o

ne t

rian

gle

are

co

ng

ruen

t to

tw

o s

ides o

f an

oth

er

tria

ng

le,

an

d t

he t

hir

d s

ide o

f th

e f

irst

is l

on

ger

than

th

e t

hir

d s

ide

of

the s

eco

nd

, th

en

th

e i

nclu

ded

an

gle

of

the f

irst

is l

arg

er

than

th

e

inclu

ded

an

gle

of

the s

eco

nd

.

N

o,

it a

pp

ears

to

be t

rue.

I

f tw

o s

ides o

f a t

rian

gle

are

no

t co

ng

ruen

t to

tw

o s

ides o

f an

oth

er

tria

ng

le o

r th

e i

nclu

ded

an

gle

in

on

e t

rian

gle

do

es n

ot

have a

gre

ate

r

measu

re t

han

th

e i

nclu

ded

an

gle

in

th

e o

ther,

th

en

th

e t

hir

d s

ide o

f th

e

firs

t tr

ian

gle

is n

ot

lon

ger

than

th

e t

hir

d s

ide o

f th

e s

eco

nd

tri

an

gle

.

N

o,

it a

pp

ears

to

be t

rue.

I

f th

e t

hir

d s

ide o

f th

e f

irst

tria

ng

le i

s n

ot

lon

ger

than

th

e t

hir

d s

ide o

f th

e

seco

nd

tri

an

gle

, th

en

th

e o

ther

two

sid

es a

re n

ot

co

ng

ruen

t o

r th

e

inclu

ded

an

gle

do

es n

ot

have a

gre

ate

r m

easu

re.

N

o,

it a

pp

ears

to

be t

rue.

5-6

En

ric

hm

en

t


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Chapter 5 Assessment Answer KeyQuiz 1 (Lessons 5-1 and 5-2) Quiz 3 (Lessons 5-4 and 5-5) Mid-Chapter Test

Page 45 Page 46 Page 47

Quiz 2 (Lesson 5-3) Quiz 4 (Lesson 5-6)

Page 45 Page 46

1.

2.

3.

4.

5.

circumcenter

4

centroid

4

-1

1.

2.

3.

4.

5.

∠4

−−

PQ

−−

QR

∠VUW

∠X

1.

2.

3.

4.

5.

The conclusion

is false.

x ≠ 6

−−

AB ≇ −−

BC

2 < x < 16

C

1.

2.

3.

4.

5.

m∠1 < m∠2

AB < DE

GH > 7

−−

AE % −−

AE

SSS Inequality

1.

2.

3.

4.

5.

C

F

D

G

D

6.

7.

8.

9.

180 > x > 50

∠7 and ∠5

16 in.

30

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Chapter 5 Assessment Answer KeyVocabulary Test Form 1

Page 48 Page 49 Page 50

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

false, median

false, orthocenter

circumcenter

incenter

greater

altitude

indirect proof

circumcenter

three or more lines

intersecting at a

common point

segment with endpoints

at a vertex and the

midpoint of the side

opposite to the vertex

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

A

H

D

G

C

G

A

H

B

F

B

12.

13.

14.

15.

16.

17.

18.

19.

20.

F

B

G

C

G

A

G

D

F

B: 6, -1

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Chapter 5 Assessment Answer KeyForm 2A Form 2B

Page 51 Page 52 Page 53 Page 54

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

B

J

A

H

B

F

C

F

A

J

A

12.

13.

14.

15.

16.

17.

18.

19.

20.

G

A

J

B

H

D

H

B

H

B: 9, -2

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

C

J

B

F

A

H

D

G

C

F

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

C

J

B

G

A

J

C

G

A

H

B: 160

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Chapter 5 Assessment Answer KeyForm 2C

Page 55 Page 56

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

"# AD

x = 8; "# AC is the ⊥

bisector of −−

BD .

4

(

16

− 3 , 22

− 3 )

25

2(z + 3) > x − 5

∠I, ∠H, ∠G

−−

PQ , −−

PR , −−

QR

−−

XY

4 is not a factor of n.

−−

AB is not a median.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

∠X $ ∠Z

13 m < x < 33 m

12

34

EF < GH

m∠1 > m∠2

Definition of $ segments

Reflexive Prop.

Converse of Hinge Th.

y = c - a

− x x

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Chapter 5 Assessment Answer KeyForm 2D

Page 57 Page 58

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

"# LM

x = 5; "# RS is the ⊥

bisector of −−− PQ .

8

( 9 −

2 ,

3 −

2 )

20

7z > x - 5

∠T, ∠V, ∠U

−−

FH , −−

GH , −−

GF

−−

LM

n2 is not an even number.

−−

AD is not an altitude.

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

−−

SV ⊥ −−

PQ

15 ft < x < 43 ft

8

35

m∠1 < m∠2

BC < ED

Midpoint Theorem

Reflexive Prop.

Hinge Th.

x = 1 − 2 a

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Chapter 5 Assessment Answer KeyForm 3

Page 59 Page 60

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

15

( 38

− 13

, - 32 −

12 )

32

146 > m∠L > 0

∠H, ∠I, ∠G

−−

QR , −−−

PQ , −−−

PR

shortest: −−

VY ;

longest: −−

VW

x ≠ 3

no; 2 + 4 < 8

The ∠ bisectors are

not concurrent.

12x - 31 > 3x - 4; x > 3

12.

13.

14.

15.

16.

17.

18.

19.

20.

B:

140

∠B $ ∠E

5 in. < x < 53 in.

17

3x + 10 > x + 20;

x > 5

Def. of $ segments

Addition Prop. of Inequality

Reflexive Prop.

Converse of Hinge Th.

y = d −

c - 2a x - 2ad

− c - 2a

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Score General Description Specific Criteria

4 Superior

A correct solution that is supported by

well-developed, accurate explanations

• Shows thorough understanding of the concepts of

bisectors, medians, altitudes, inequalities in triangles,

indirect proof, the Triangle Inequality, Hinge Theorem and

its Converse.

• Uses appropriate strategies to solve problems.

• Computations are correct.

• Written explanations are exemplary.

• Figures are accurate and appropriate.

• Goes beyond requirements of some or all problems.

3 Satisfactory

A generally correct solution, but may

contain minor flaws in reasoning or

computation

• Shows an understanding of the concepts of bisectors,

medians, altitudes, inequalities in triangles, indirect proof,

the Triangle Inequality, Hinge Theorem and its Converse.

• Uses appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are effective.

• Figures are mostly accurate and appropriate.

• Satisfies all requirements of problems.

2 Nearly Satisfactory

A partially correct interpretation and/or

solution to the problem

• Shows an understanding of most of the concepts of

bisectors, medians, altitudes, inequalities in triangles,


its Converse.

• May not use appropriate strategies to solve problems.

• Computations are mostly correct.

• Written explanations are satisfactory.

• Figures are mostly accurate.

• Satisfies the requirements of most of the problems.

1 Nearly Unsatisfactory

A correct solution with no supporting

evidence or explanation

• Final computation is correct.

• No written explanations or work shown to substantiate the

final computation.

• Figures may be accurate but lack detail or explanation.

• Satisfies minimal requirements of some of the problems.

0 Unsatisfactory

An incorrect solution indicating no

mathematical understanding of the

concept or task, or no solution is given

• Shows little or no understanding of most of the concepts

of bisectors, medians, altitudes, inequalities in triangles,


its Converse.

• Does not use appropriate strategies to solve problems.

• Computations are incorrect.

• Written explanations are unsatisfactory.

• Figures are inaccurate or inappropriate.

• Does not satisfy requirements of problems.

• No answer given.

Chapter 5 Assessment Answer KeyExtended-Response Test, Page 61

Scoring Rubric

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Chapter 5 Assessment Answer KeyExtended-Response Test, Page 61

Sample Answers

1. As the sticks are pulled apart the angle

gets greater and the rubber band will

be stretched and become longer. This

situation illustrates the Hinge Theorem.

2. Ashley is correct, any point on a

perpendicular bisector is equidistant

from the endpoints of the segment it is

bisecting. The distance from D to H is

6 inches. The distance from H to G is

5 inches.

3. The segment from B to AC##$ could

intersect AC##$ in two different points

because the length of the segment, 6, is

more than the perpendicular distance

from B to AC##$, 5, and less than the length

of AwBw, 10. BwDw can either slant in towards

A or out towards C as shown in this

figure.

1066 5

A

B

CD

4. a. The student should draw a right

triangle.

altitudes

b. The student should draw an obtuse

triangle.

c. The student should draw an acute

triangle.

d. The student should draw an

equilateral triangle.

In addition to the scoring rubric found on page A27, the following sample answers may be used as guidance in evaluating open-ended assessment items.

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Chapter 5 Assessment Answer KeyStandardized Test Practice

Page 62 Page 63

1. A B C D

2. F G H J

3. A B C D

4. F G H J

5. A B C D

6. F G H J

7. A B C D

8. F G H J

9. A B C D

10. F G H J

11. A B C D

12. F G H J

13. A B C D

14. F G H J

15. A B C D

16. 17.

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

8

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

9

8

7

6

5

4

3

2

1

0

3 4

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Chapter 5 Assessment Answer KeyStandardized Test Practice (continued)

Page 64

18.

19.

20.

21a.

b.

c.

22a.

b.

c.

6

10

25

(1, 0)

(0, 1)

(-2, 3)

y = 5x - 2

y = - x − 5 + 2

( 10

− 13

, 24 −

13 )

Documents

Chapter 5 Resource Masters 5 Test, Form 3 .....59 Chapter 5 Extended-Response Test .....61 Standardized Test Practice .....62 Answers