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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
Study Guide and Intervention .......................... 17
Skills Practice .................................................. 19
Practice ............................................................ 20
Word Problem Practice ................................... 21
Enrichment ...................................................... 22
Cabri Jr. .......................................................... 23
Geometer’s Sketchpad Activity ...................... 24
Lesson 5-4Indirect Proof
Study Guide and Intervention .......................... 25
Skills Practice .................................................. 27
Practice ............................................................ 28
Word Problem Practice ................................... 29
Enrichment ...................................................... 30
Lesson 5-5The Triangle Inequality
Study Guide and Intervention .......................... 31
Skills Practice .................................................. 33
Practice ............................................................ 34
Word Problem Practice ................................... 35
Enrichment ...................................................... 36
Lesson 5-6Inequalities Involving Two Triangles
Study Guide and Intervention .......................... 37
Skills Practice .................................................. 39
Practice ............................................................ 40
Word Problem Practice ................................... 41
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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NAME DATE PERIOD
Ch
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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
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Chapter 5 2 Glencoe Geometry
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
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Chapter 5 3 Glencoe Geometry
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
on
5-1
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Chapter 5 5 Glencoe Geometry
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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Chapter 5 6 Glencoe Geometry
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)
Bisectors of Triangles
" 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
5
Lesso
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-1
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Chapter 5 7 Glencoe Geometry
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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Chapter 5 8 Glencoe Geometry
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
Bisectors of Triangles
21°
32°
Lesso
n 5
-1
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Chapter 5 9 Glencoe Geometry
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
Bisectors of Triangles
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Chapter 5 10 Glencoe Geometry
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
Lesso
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Chapter 5 11 Glencoe Geometry
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
5-2 Study Guide and Intervention
Medians and Altitudes of Triangles
Example
16
12
12
9
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Chapter 5 12 Glencoe Geometry
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)
5-2 Study Guide and Intervention (continued)
Medians and Altitudes of Triangles
Example
y
x
(9, 3)(1, 3)
(7, 7)
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Chapter 5 13 Glencoe Geometry
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)
COORDINATE GEOMETRY Find the coordinates of the orthocenter of each triangle.
15. L(8, 0), M(10, 8), N(14, 0) 16. D(−9, 9), E(−6, 6), F(0, 6)
Skills Practice
Medians and Altitudes of Triangles
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Chapter 5 14 Glencoe Geometry
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)
COORDINATE GEOMETRY Find the coordinates of the orthocenter of each triangle.
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
Medians and Altitudes of Triangles
A
C
F
E
D
PB
18 30
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Lesso
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-2
Chapter 5 15 Glencoe Geometry
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
exists, state that this is impossible.
c. Give an example of a triangular plaza
where C is outside the plaza and O is
inside the plaza. If no such example
exists, state that this is impossible.
5-2 Word Problem Practice
Medians and Altitudes of Triangles
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Chapter 5 16 Glencoe Geometry
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.
5-2 Enrichment
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Less
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Chapter 5 17 Glencoe Geometry
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
3. measures are less than m∠1
4. measures are greater than m∠1
5. measures are less than m∠7
6. measures are greater than m∠2
7. measures are greater than m∠5
8. measures are less than m∠4
9. measures are less than m∠1
10. measures are greater than m∠4
M J K
3
4 521
L
Exercises 1–2
H E F3
4
21
G
5-3 Study Guide and Intervention
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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Chapter 5 18 Glencoe Geometry
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°
5-3 Study Guide and Intervention (continued)
Inequalities in One Triangle
If AC > AB, then m∠B > m∠C.
If m∠A > m∠C, then BC > AB.
Example 1 Example 2
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Chapter 5 19 Glencoe Geometry
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
Inequalities in One Triangle
98°
43°
42°
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Chapter 5 20 Glencoe Geometry
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.
5. measures are less than m∠1
6. measures are less than m∠3
7. measures are greater than m∠7
8. measures are greater than m∠2
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
Inequalities in One Triangle
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Chapter 5 21 Glencoe Geometry
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
Inequalities in One Triangle
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Chapter 5 22 Glencoe Geometry
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)
5-3 Enrichment
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Chapter 5 23 Glencoe Geometry
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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Chapter 5 24 Glencoe Geometry
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
Inequalities in One Triangle
Less
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Chapter 5 25 Glencoe Geometry
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,
5-4 Study Guide and Intervention
Indirect Proof
Example
x 3x + 5
1 8
0 5
-1 2
-2 -1
-3 -4
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Chapter 5 26 Glencoe Geometry
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
State the assumption you would make to start an indirect proof of each statement.
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
5-4 Study Guide and Intervention (continued)
Indirect Proof
Example
Lesso
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Chapter 5 27 Glencoe Geometry
5-4 Skills Practice
Indirect Proof
State the assumption you would make to start an indirect proof of each statement.
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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Chapter 5 28 Glencoe Geometry
State the assumption you would make to start an indirect proof of each statement.
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
Lesso
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Chapter 5 29 Glencoe Geometry
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.)
5-4 Word Problem Practice
Indirect Proof
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Chapter 5 30 Glencoe Geometry
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.
5-4 Enrichment
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Less
on
5-5
Chapter 5 31 Glencoe Geometry
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?
5-5 Study Guide and Intervention
The Triangle Inequality
Example
a + b > c
b + c > a
a + c > b
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Chapter 5 32 Glencoe Geometry
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.
5-5 Study Guide and Intervention (continued)
The Triangle Inequality
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Lesso
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-5
Chapter 5 33 Glencoe Geometry
5-5 Skills Practice
The Triangle Inequality
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
Find the range for the measure of the third side of a triangle given the measures
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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Chapter 5 34 Glencoe Geometry
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
Find the range for the measure of the third side of a triangle given the measures
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
The Triangle Inequality
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Lesso
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-5
Chapter 5 35 Glencoe Geometry
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.
5-5 Word Problem Practice
The Triangle Inequality
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Chapter 5 36 Glencoe Geometry
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 =
5-5 Enrichment
Less
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Chapter 5 37 Glencoe Geometry
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)°
5-6 Study Guide and Intervention
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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Chapter 5 38 Glencoe Geometry
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
5-6 Study Guide and Intervention (continued)
Inequalities Involving Two Triangles
Example
97°
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Lesso
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-6
Chapter 5 39 Glencoe Geometry
5-6
Compare the given measures.
1. m∠BXA and m∠DXA
2. BC and DC
Compare the given measures.
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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Chapter 5 40 Glencoe Geometry
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
Compare the given measures.
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
Inequalities in Two Triangles
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Lesso
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-6
Chapter 5 41 Glencoe Geometry
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.
5-6 Word Problem Practice
Inequalities in Two Triangles
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Chapter 5 42 Glencoe Geometry
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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Chapter 5 43 Glencoe Geometry
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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Chapter 5 44 Glencoe Geometry
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
NAME DATE PERIOD
NAME DATE PERIOD
Chapter 5 45 Glencoe Geometry
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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NAME DATE PERIOD
Chapter 5 46 Glencoe Geometry
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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Chapter 5 47 Glencoe Geometry
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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Chapter 5 48 Glencoe Geometry
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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Chapter 5 49 Glencoe Geometry
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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Chapter 5 50 Glencoe Geometry
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
13. Find the value of x.
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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Chapter 5 51 Glencoe Geometry
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.
1. Name an angle bisector.
A −−
KI B ""# GL C $ ""# JM D −−−
HJ
2. Name a median.
F −−
KI G ""# GL H $ ""# JM J
−−− HJ
3. Name an altitude.
A −−
KI B ""# GL C $ ""# JM D
−−− HJ
4. Name a perpendicular bisector.
F −−
KI G ""# GL H $ ""# JM J
−−− HJ
For Questions 5–7, refer to the figure to determine
which is a true statement for the given information.
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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Chapter 5 52 Glencoe Geometry
5
12. Find the possible values for m∠1.
F 90 > m∠1 > 74 H 0 < m∠1 < 74
G 180 > m∠1 > 74 J m∠1 = 106
13. Find the value of x.
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
19. What is the relationship between the lengths of −−−
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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Chapter 5 53 Glencoe Geometry
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
2. Name an angle bisector.
F −−−
RW G " #$ SV H −−−
QT J ##$ RU
3. Name a perpendicular bisector.
A −−−
RW B " #$ SV C −−−
QT D ##$ RU
4. Name an altitude.
F −−−
RW G −−
RP H −−−
QT J ##$ RU
For Questions 5–7, refer to the figure to determine
which is a true statement for the given information.
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.
12. Find the possible values for m∠1.
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
18. What is the relationship between the lengths of −−−
YW and −−
YX ?
F YW = YX H YW > YX
G YW < YX J cannot tell
19. What is the relationship between the lengths of −−−
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 54 Glencoe Geometry
Chapter 5 Test, Form 2B (continued)
B:
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NAME DATE PERIOD
5 SCORE Chapter 5 Test, Form 2C
1. Name an angle bisector.
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.
Chapter 5 55 Glencoe Geometry
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NAME DATE PERIOD
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 56 Glencoe Geometry
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
1. Name a perpendicular bisector.
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.
Chapter 5 57 Glencoe Geometry
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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 58 Glencoe Geometry
Chapter 5 Test, Form 2D (continued)
(Question 18)
Assessment
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NAME DATE PERIOD
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.
Chapter 5 59 Glencoe Geometry
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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 60 Glencoe Geometry
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
Chapter 5 61 Glencoe Geometry
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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.
Chapter 5 62 Glencoe Geometry
Assessment
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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
Chapter 5 63 Glencoe Geometry
?
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.
Chapter 5 64 Glencoe Geometry
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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Chapter Resources
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Lesson 5-1
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
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Ch
ap
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5
5
Gle
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Ge
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3
x +
8 =
5x -
6
14 =
2x
7 =
x
Exam
ple
1Exam
ple
2
Exerc
ises
Fin
d e
ach
me
asu
re
.
1
. X
W
2. B
F
7.5
5
5
4.2
19
19
Po
int
P i
s t
he
cir
cu
mce
nte
r o
f △
EM
K.
Lis
t a
ny
se
gm
en
t(s)
co
ng
ru
en
t to
ea
ch
se
gm
en
t b
elo
w.
3
. −
−−
MY
4. −
−
KP
5
. −
−−
MN
6. −
−−
ER
7
.5
4.2
−−
YE
−−
NK
−
−R
K
−−
MP
, −
−
EP
Answers (Anticipation Guide and Lesson 5-1)
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.
Chapter 5 A2 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
PE
RIO
D
Ch
ap
ter
5
6
Gle
nc
oe
Ge
om
etr
y
An
gle
Bis
ect
ors
A
noth
er
speci
al
segm
en
t, r
ay,
or
lin
e i
s an
an
gle
bis
ect
or,
wh
ich
d
ivid
es
an
an
gle
in
to t
wo c
on
gru
en
t an
gle
s.
An
gle
Bis
ec
tor
Th
eo
rem
If a
po
int
is o
n t
he
bis
ecto
r o
f a
n a
ng
le,
the
n it
is e
qu
idis
tan
t fr
om
th
e s
ide
s
of
the
an
gle
.
Co
nv
ers
e o
f A
ng
le
Bis
ec
tor
Th
eo
rem
If a
po
int
in t
he
in
terio
r o
f a
n a
ng
le if
eq
uid
ista
nt
fro
m t
he
sid
es o
f th
e a
ng
le,
the
n
it is o
n t
he
bis
ecto
r o
f th
e a
ng
le.
Inc
en
ter
Th
eo
rem
Th
e a
ng
le b
ise
cto
rs o
f a
tria
ng
le in
ters
ect
at
a p
oin
t ca
lled
th
e in
ce
nte
r th
at
is
eq
uid
ista
nt
fro
m t
he
sid
es o
f th
e t
ria
ng
le.
5-1
Stu
dy
Gu
ide a
nd
In
terv
en
tio
n
(con
tin
ued
)
Bis
ecto
rs o
f Tri
an
gle
s
"
MR
is t
he
an
gle
bis
ecto
r o
f ∠
NM
P.
Fin
d x
if
m∠
1 =
5x
+ 8
a
nd
m∠
2 =
8x -
16.
12
NR
PM
! M
R i
s th
e a
ngle
bis
ect
or
of
∠N
MP
, so
m∠
1 =
m∠
2.
5x +
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
.
Chapter 5 A3 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-1)
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.
Chapter 5 A4 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
.
Chapter 5 A5 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-2)
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.
Chapter 5 A6 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (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
.
Chapter 5 A7 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-2 and Lesson 5-3)
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.
Chapter 5 A8 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (Lesson 5-3)
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
.
Chapter 5 A9 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-3)
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.
Chapter 5 A10 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (Lesson 5-3)
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
.
Chapter 5 A11 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-3 and Lesson 5-4)
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.
Chapter 5 A12 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (Lesson 5-4)
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
.
Chapter 5 A13 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
.
Answers (Lesson 5-4)
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.
Chapter 5 A14 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (Lesson 5-4 and Lesson 5-5)
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
.
Chapter 5 A15 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-5)
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.
Chapter 5 A16 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (Lesson 5-5)
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
.
Chapter 5 A17 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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
Answers (Lesson 5-5 and Lesson 5-6)
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.
Chapter 5 A18 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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°
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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 (Lesson 5-6)
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
.
Chapter 5 A19 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, 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
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
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.
Answers (Lesson 5-6)
Co
pyrig
ht ©
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/Mc
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ill, a d
ivis
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Mc
Gra
w-H
ill Co
mp
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ies, In
c.
Chapter 5 A20 Glencoe Geometry
Copyright © Glencoe/McGraw-Hill, a division of The McGraw-Hill Companies, Inc.
NA
ME
DA
TE
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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
Answers (Lesson 5-6)
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-Hill
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Answers
Chapter 5 A21 Glencoe Geometry
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 A22 Glencoe Geometry
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 A23 Glencoe Geometry
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 A24 Glencoe Geometry
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 A25 Glencoe Geometry
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 A26 Glencoe Geometry
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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Chapter 5 A27 Glencoe Geometry
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,
indirect proof, the Triangle Inequality, Hinge Theorem and
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,
indirect proof, the Triangle Inequality, Hinge Theorem and
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 A28 Glencoe Geometry
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 A29 Glencoe Geometry
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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c.
Chapter 5 A30 Glencoe Geometry
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 )
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