Tutor in a Book Geometry 9780978639020

Citation preview

UmrIrH'ffi Jo Greiu

GEOMETRY By Jo Greig

TUTOR IN A BOOK. Berkeley, California

Compositor and Design: Jo Greig Mathematics Editor: James R. Shilleto, Ph.D Editor: Christine E. McGowan futwork Ka-Wai Lui Problems Editor: D. \Tilliam McPhee, Jr.

AII rights reserved. No part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying, recording, or by any information storage and retrieval system, without written permission from the publisher, except for the inclusion of brief quotations in a review. International Standard Book Number: 978-0-9786390-2-0

Copyright @ 2006 byJo Greig Published by Tutor In a Book' 1476 University Avenue, Suite 100 Berkeley, California 947 02

Contents r. PoINTS, LINES AND PLANES Points Lines

2

Segments

3

Length and Distance Getting R."dy for Proofs, First Postulate

3

I

5

Ryt

5

Planes

5

Postulates and Theorems about Points, Lines and Planes

8

2. ANGLES Angles Special Pairs ofAngles Postulates and Theorems about Angles Getdng Started with Proofs Properties and How to Use them in Proofs Perpendicular Lines

3. LOGIC If -Then Statements The Law of Detachment Contrapositives, Converses and Inverses The Law of Syllogism Indirect Proofs

9

r3 r9 20

))

25 )-7

27 28 30

3r

4. PARALLEL LINES Lines, Thansversals and More Special Pairs of Angles Parallel Lines and Planes Postulates and Theorems about Parallel Lines Proving Lines are Parallel

5. TRIANGLES Tiiangles Exterior Angles of Tiiangles Parts ofTliangles Congruency and Congruent Figures Congruent Thiangles Postulates and Theorems about Congruent Tliangles Getting Ready for Proofs about Congruent Tiiangles Proving Tliangles are Congruent Hypotenuse Leg Theorem for Right Thiangles Only Using CongruentThiangles to Prove Other Things Perpendicular Bisectors and Angle Bisectors Isosceles Triangles and the Isosceles Thiangle Theorem lll

33 35 36 42

44 48

5r 52 53 53 56 57 58 60

6t 64

6. POLYGONS Polygons Regular Polygons Finding the Measure of the Interior Angles of Polygons Finding the Measure of the Exterior Angles of Polygons Finding the Measure of the Exterior Angles of Polygons

65

66 67 70 70

7. QUADRIIAIERALS Parallelograms Proving the Properties of Paralellograms

-

A Very Close Look At Proofs

Proving a Quadrilateral Is a Parallelogram Rectangles, Rhombuses and Squares Specid Parallelograms TLapezoids

-

Miscellaneous Theorems about Parallel Lines Kites

8. INEQUALITIES Properties of Inequalities The Exterior Angle Inequaliry Theorem Thiangle Inequaliry Theorem Theorems about Inequalities for One Tbiangle teorems about Inequalities for Two Tliangles 9. SIMILAR FIGURES Ratios and Proportions

Similarity Postulates and Theorems which Prove Tiiangles Are Similar Similariry Versus Congruency

10. RIGHT TRIANGLES Radicals Radical Rules Geometric Mean Theorems about Special Properties of Right Thiangles The Pythagorean Theorem The Converse of the Pythagorean Theorem and Related Theorems Special RightTliangles - 45"-45"-90" and 30'-60"-90'

Tiigonometry Solving Right Thiangles Practical Applications of Thigonometry Solving General Thiangles - The Law of Sines, The Law of Cosines Three Important Ideas

lv

72 73 75 78 82 84 86

87 88 89 90 93

97 99 103

t07 108 109

110 111 115 116 120

r25

r3r t32 r33 136

11. CIRCLES

Circle Terms and Definitions

r37

Spheres

138

Thngents and Theorems about Thngents Inscribed and Circumscribed Angles, Circles and Polygons Arcs, Central Angles and Semi-Circles Chords Circle Theorems and Proofs Angles and the Arcs They Intercept

r39

Putting It All Together - Angles and fucs Segments and Circles 12.

r59

AREA Area of Squares fuea of Rectangles Area of Parallelograms Area of Isosceles, Equilateral and General tiangles Putting k All Together - Finding the Area ofTliangles fuea of Rhombuses Area ofThapezoids Area of Regular Polygons Area of Circles, Sectors and Circle Segments Geometric Probability A Simple Counting Argument

13.

141 142 145 145 150 158

162 163 165 169 174

175 177 179

t84 192 195

VoLUMEAND AREA OF SOLrD OBJECTS Prisms

Pyramids Cylinders Cones Spheres

Similar Solids

ANSWER SECTION

r97 204

2tt 216 219 224 229

in mathematics for more than 30 years. A good tutor can help a struggling student pass a class that the student might otherwise ftil. A good tutor can help a strong student do much better in a class than he or she might otherwise do. I have written this book with the hope of giving all students the advantage of a good tutor. I have attempted to include every explanation, every drawing, every hint, every memory tool, every problem that students always seem to struggle with and every bit of enthusiasm that I try hard to impart to my private tutoring students. Good luckwithTutor in a Book's Geometry! Jo Greig, July 1, 2006

Author's Note

-

I

have tutored students

A Note to Parents on How to Use this Book If the semester is just beginning: The student should read the next dayt assignment in this book (check in the textbook for the correct topic) prior to each school day. The student should be sure to do the exercises in this book to reinforce his or her understanding of the topic.

If the school year is well under way and the student is really struggling: He or she needs to start at the beginning of this book and read through to the point where his or her class is at the present time. This is important because geometry is sequential and you cannot build understanding on a shalry foundation. This book is much shorter than a textbook and much more informal and accessible. Be sure the student reads each section, carefully studies the example problems and then does the exercises. In addition to reinforcing the review and proper learning of the material, doing the exercises successfully is very empowering to the student. Step-by-step solutions to each exercise with appropriate explanations and illustrations are included in the Answer Section at the back of the book.

If the school year is we[[ under way and the student understands

some material: He or she can use this book like a cookbook, carefully going over the chapters that cover the troublesome topics. However, rhe student should study any necessary chapters in their entirety, paylng careful attention to the examples, and complete all exercises in those chapters.

Suggestions for AII Students Make a fashcard for every symbol, term, definition, theorem, postulate and corollary as each is introduced in your textbook. Memorize the information. Test yourself on your fashcards everyday. Only use those postulates, theorems, corollaries and definitions that are introduced in your textbook. Always read your class's next days topic in this book the night before class. Carefully study the example problems or proofs. Do the related exercises in this book and be sure to check your answers. The time you spend will end up saving you study time in the long run. In class, try to sit in the front of the room. Listen to your teachert lecture and take careful notes including the problems done in class.

Carefully read your textbook and do the class work and homework when it is assigned. Dont fall behind. Geometry cannot be crammed in the night before an exam. Be diligent and stay positive. I ve had students who received low grades on the first exams but who worked diligently and never gave up and ended up with As for both semesters. vl

General Notes to Parents

-

Here are some hints on how to help your child succeed in geometry:

If you were good in geometry and are trying to help your son or daughter but are just put offby the 700+ page textbook, this is the book for you. The language is simple and the illustrations and review charts tie related information together to quickly refresh your memory. The standard problems (in both senses of the word) complete with fully explained and illustrated solutions are here.

If you didnt like geometry

itt

probably better not to mention this to your child. After all, you would never tell your child how much you hated to read! In fact, doing so gives your child permission to fail. Instead, encourage your child to follow the many tips in this book, to be respectful to his or her teacher and to do all required assignments when they are assigned. Always ask to see each days completed homework. You will be able to tell if each problem was attempted. Meet your child's teacher and let the teacher know that you support him or her. (Most will sincerely appreciate it).

A major reason that students do poorly in geometry is they dont realize that geometry is like a combination of a foreign language and a mathematics course. Students need to learn the basics (the symbols, how to name an angle, etc. ), and then the vocabulary, all of the theorems and so forth, in order to be successful. To compound the problem, textbooks give a piece of information one time, then expect the student to recall the information for the rest of the school year. Some teachers allow students to make and refer to notes or theorem sheets, but this is rarely helpful because the student has no idea where to look or what to look for. Experience has shown, that all of this information needs to be learned and learned thoroughly. In this book, key information is stated, stressed and restated. You can help your child by insisting on fashcards and then quizzing him or her on the information. Regarding proofs, over the years, proofs have gained an unfair reputation for difficulry. However, proofs follow fairly predictable patterns and once a student learns the patterns and check steps, proofs become much easier. This book includes dozens of thoroughly explained proofs including many visuals and tips to help the student recognize and learn the patterns. An interesring fact is that beginning proofs are very subtle and therefore more difficulu later proofs are actually easier. Unless your school provides an integrated algebra/geometry course, successful completion of firstyear algebra is the prerequisite for geometry. This is necessary because there is a good deal of algebra in a geometry course, especially in the second semester. If you want your child to accelerate, ask the

school counselor if the school will allow geometry and second-year algebra to be taken together. Sometimes, a student feels that he or she just does not click with a particular teacher and/or the textbook. In my experience, switching teachers is rarely helpful. However, if your school will allow a change, go ahead, but then insist that your child take a new approach with the course (being very attentive to this "better" teacher, doing all homework promptly and so forth). As for textbooks, they do differ in qualiry and sometimes a concept that is perfectly clear to a team of authors is beyond the reach of a teenager. I have tried to address this issue in this book by using less formal teaching methods ones that I ve developed over the years and which are teen-age tested and approved.

-

vll

Tirtor in a Book't GeometqP Jo Greig

vlll

1. POINTS. L INES AND PLANES Points are the building blocla of geometry. In fact, all of the objects that we learn about in geometry are made up of points. Butwe know 1

') 3

a

lot about them:

h^urAEachpointb. Points-f Points

4 \We draw a c 5 Points are

-

.A

.C

.B Lines are important objects in geometry.

Butwe know

G

a

lot about them

l. A line is

are 3.Eacht* 2. Lines

.

4. Lines have a

/d-

of a, ). lo 6.

All lines go on forever, that is, infnitely, in two directions.

7. To show that a line is infinite, we pur an arrowhead on each end.

Lines and points together

-

\fle say a line includts or contains the points through which it passes. 'We say the points are on or in or captured by or contained in the line. Two Ways to Name Lines

1. A line may be named by a single n

lowercase letter.

d

oa,o

line n

Iine

d

2. A line may also be named by using the names (the letters) of any tuto points through which the line passes, together with a double arrowhead on top of the two letters. 'We

may name the two points in any orderwe choose. Heret an example

The line on the right can be called any of the following:

fr,8e, fr,68,ffi", 1

d

QuickReview-,\7heneveryouseethewordlineingeometryitIt Lines

*

The measure of anyline

Answer true or false: A line is defined as a straight, infinite collection of points. '(paugap lou sr oull

v) irsp{

:ra rsuv

,r-_ . 1, ,. U ll

_

Collinear_tf

anyffiTharis,

s

Points L and M are collinear. Points L andNare collinear. Points M and.A/are collinear Points L, M and l/are not collineer.

L

lr

Any two points are collinear. Points D, E and.Fare collinear. Points D, G and E are not collinear. Points D, G, and F are not collinear. D Points D, G, and.Fand E are not collinear.

Here are the same 3 points and the lines which capture each pair of collinear points. Since lines are straight, no singb line could capture all 3 points.

M N

F D

G

Here are the same 4 points and the lines which capture each collection of collinear points.

Between

1.G 2.--When

problem says that a point is between two points you are being given both conditions. \7hen you are trying to decide if a point is between two points you must check for both conditions. Here are some examples: a

Is S between R and Q? a a

Is Ubetween Vand T?

R

a

U

a

V

,S

Qbut

is collinear S is

and Q.

with R and

not in between R

?

tw'

tT

'e

.9

No.

Is X betwe en Y and W

o

tYx

No. Uis not collinear withVand Yes. Xis collinear with Izand WandXis in bet'ween YandW Talthough [/is in between 7 and

T 2

segment

Segments

s

A segment is: I 2

The symbol for a segment is the names of the two endpoints (the capital letters in either order) with a bar over the top. Heret an example:

B

The segment at right is named AB or BA.

A \What doesABinclude? l. AB includes its two endpoints, A and B, and 2. Nl of the points between (collinear with and in between) A and B. Note that BA includes the same collection of points. That's why AB is the."-. .Br4 .

",

Here are more segments and the symbols that could be used to name them:

E

D CD or DC

l/ EF or FE

F

MN or NM

M

Length and Distance

i5 +R.S+

1. The symbol for distance is the names of the two endpoints (the capital

F{ R

letters in either order). So "plairi' RS means the length of the segmenr.

s

RS is R.9 units long!

2. Length is always positive, so its measure, rhar is, the distance between two points, is always p ositiue. For example, you cannot be minus 5t 6" tall!

VW WV= VW= 4

Sometimes a segment is placed on a number line. 'Ihe distanrs between any rwo points is the number of units from the first point to the second point. Remember distance isnt about which side of zero either point is on. Itt about how far you would have to go to get from one point to the other. Is distance always a whole number? No

A

B

C

D

E

F

-2

I

0

1

2

3

5

I

A

B

6

-2

1

0

t23aLs56 \-\r-r,

HA=7

PD = 3.5

3

How Do You Find Distance?

Three'ways:

-11;1

t.

o 7 2 3 4 ttlf

: KL=

5.7 --1.8 = 5.7 +7.8 =7.5 or, KL = 15.7--1.81 = 7.5, or,

z. a.

Equality and Congruency A segmentt

size is measured by its length.

If we say two objects

If two segments have the same size, their lengths are equal.

aHn

w.Yx

are congruent, we are saying

l,w=QP

2 things:

Shape _-.-r^t, is IJ: The symbol for congruency & a \.: 2. Same Size 1. Same

Same ShaPe

6. Same Size

Yet all segments have the same shape. This means, if two segments have the same length, they are equal and congruent. In fact, for segments, equaliry and congruence can be used interchangeably:

s

/"2N, 3 . MN=OP 'M o P MN=op

.SI = QR

SI=

Midpoints of segments

QR

5

(/) mean the These two segments are conBruent'

x,5,

Ylnrr=xy

W

VW=XY

Segments are finite; they have a beginning and an end. Therefore, segments have a midpoint. In fact, a segment is the only geometric object that can have a midpoint.

'

Know your definitions. If a proof gives: -Ay'is the midpoint of FG, you are supposed to think and respond with Fltf ffG Other facts are Afi, = t ' ffue, but they are not the definition of a midpoint.

Finding

a

FN=NG

I ABC DE FGH -2 -1 0 t 2 3 4 5 6

Particular Point and Its Coordinate

tract 4 from 6 (or count 4 units back from 6), which equds 2. Since the number 2 is paired with point E, themidpoint of IA is E and2 is its coordinate./* *Note: In this book r,/means, this is the correct answer.

NowYouTryIt-Usingthenumberlineontherightfind:

t.

2.

WP

3.

EA

4.

GA

4 lA A D E f EWry I=I 2 3 44.) -2-t'.2 .+_5 -15 -t--t 0 1

>./

5.The midpoint and the coordinate of the midpoint of HB. Check your answers in the back of the book. 4

Getting Ready for Proofs

C

TnB SncrranNr AoorrroN Posrurann: If B h between (collinear utith and in benaeen) A and C, then + BC = AC use the segment Addition Postulate in 2 *^ylB 1. In a proofwhen you need to show that a segment is equal to the sum of its parts, or that the sum its parts is equal to the (entire) segment. Remember, r4,B means the of AlB.

2. In algebra problems like this one: If MO = 24, find x. +6

M R"yr

lr

of

O 4x+ 6 +2x=24 6x+5=24,6x=18,x=3,1

Point r4 is the endpoint. Point .B is another point through which this ray passes.

B

l. 2. 3.

A

---->

AB

Using a single arrowhead to name a ray makes sense because the ray goes on forever, but only in one direction. Here are some more examples of rays and their names:

qJ

\r \

---->

Either is correct and both name the same ray, that is, the same collection of Iooints.

KL is a difFerent

--->

HG.Theendpoint is ll No matterwhichway

JK orJL.

ffi

(D ID DC.'Ihe endpoint in this case) is always

the ray itself is going,

/,

the arowhead on the symbol goes to the right.

ray.

named first and the arrowhead on the symbol goes to the right.

DprrNrrroN or e Rey:

\\\

\\ \ \\\

\ \\

\\ \

Opposite Rays If 2 rays share a common endpoint, and that endpoint is between (collinear with and in benveen) another point on eaclt ray, the 2 rays are "opposite" rays. Here are some examples:

U

NowYou Try It

-->

1.AB

---->------>

V

VU andVW

.R

are

opposite rays.

-

M >

LKand

Give the definition of each symbol shown below 2.

e

AB

3. AB 5

LM are

opposite rays.

4. AtB

Planes

-

\7e draw figures

these

to represent

namrng

with a single capital letter:

Plane

Plane R

M

Plane

A

M Another way to name a plane is by naming any three (or more) non-collinear points which are included in the plane. For example, in the figure on the E right, plane EFG includes the small rectangle EFGH that we see in the drawing, but it goes on forever. Other names for plane EFG are EFGH, EFH, FGH, HFG and so on. There are five other planes shown in the I figure. Tiy to name them. It is also important to imagine planes that are not shown, for example, plane HEJK(see Figure 1 below). Tiy to sketch in plane GHA in Figure 2. F F G

F

K

G

E Fig. I

K

K

Fig.2

I

Finding and Naming the Intersection Recall that the intersection of two or more objects means the

that are shared by the o b ects. 2 lines, 2 possibilities

A line and

A/

a

plane,

> No intersection.

#of

-Line Shared poirrl. r

0

J

I

t

oo

Name

Description The line misses the plane. The line passes through the plan6. The (entire) line is contained in the plane.

lr,,.r3J.rio, none

2 planes, 2 poisibilities. Planes U andV no intersection.

v/

Planes V and'W intersection is a line, m.

t 6

G

Intersect in I point, U.

Here are some examples of the intersections of points, lines and planes:

C

A 1. Name a line that's

Here are a few, 2.

not shown.

fr,fii,ffi,

Frt,

fr,tE

""d

I

ffi. E

Vhat

is the intersection of planes ABC and DCG? (Remember that 3 noncollinear points name and determine a unique plane.)

\Thenever two planes intersecr, they intersect in a line, in this

H Questions 1-6

."r.,6d.

3. Name the intersection of planes ABCD, DCGH and BCGF. An intersection means the point(s) shared by all of the named objects. The answer is point C. Look up at the corner of your room to see another example of this intersection. 4. Name 3 planes that are not shown. Here are fow ACGE, BCHE, CDEF, ADGF. Recall that ACG narnes the same plane In fact, any three non-collinear points narne a plane. 5. \Mhat is the intersection of planes ABGH and DCGH? 'Whenever nvo planes intersect, they intersect in a line, in this

as

ACGE.

,ur,6fi.

td

6. \7.hat is the intersectio n of and /? \7hat is the intersectio n of ABGH and 1? The intersection of a point and another object is either the point or nothing. In the case of tdthrans\Mer is point L In the case of A*GH, theanswer (and the interseltion) is nothing. Some textbooks and teachers will say @ or { } or the "null set" to describe no intersection. Segment Bisectors

I a bisectar,

that is, they can do the

DsrrNrrroN oF A S Given AB wirh midpoint M: R

A

B

A

3

n

Line n bisecting segment /4,8.

n"y

@

bisecting segment,,4B.

Plane P bisecting

segmenr/,8.

7

B

Segment CDbisecting segment AB, but not being bisectedby AB.

Postulates and Theorems about Points, Lines and Planes Your textbook might have a number of postulates and theorems about points, Iines and planes and it is important that you study, understand and memorize them. As you read the postulates and theorems below, remember that each point is distinct (different) from any other point; each line is distinct from any other line; each plane is distinct from any other plane. The following are some of the postulates and theorems that your textbook might include:

Posrurern:

Two points determine

a (unique) line

Study the drawing. Points A and B which are on opposite corners of the box are fixed. Lines are straight and cannot move. Only one line can capture both point A and point.B. Given any two points, one (unique) line is determined.

PosruranB: Through aryt three points there is at least one Plane and through an! three noncollinear points there is a unique plane. If three points happen to be on the same line (that is, if they

are collinear), then

an infinite number of planes include the line and therefore, the three points. The figure at right shows three (of the infinite number) of planes going through the line which contains points A, B and C.

If three points do not lie on the same line (that is, if they

not collinear), then only one plane can capture all three points. The figure at right shows the unique plane through noncollinear points G, H and I.

TnBonnu:

are

I

I

Through a line and a point not 0n the line there is a unique phne.

H

.c

A

Choose any rwo points on the line and then visualize the plane through those two points and the third point which is not on the line. See the 2 figures on the right.

THronsu: If two lines intersect, tlten one and only one plane contains

them.

A good way to visualize this theorem is to imagine 2 Pick Up Sticks, cross them and then imagine a piece of paper that is laid on top of the cross. The paper represents the unique plane that contains both lines. "Two points determine a unique line" means two things: that the line exists and that only one such line exists. Other ways to state these two ideas would be to say one and only one, or exacdy one.

8

2. ANGLES Angles

\7hat makes an angle?

l. The sides of an angle are 2 rays. 2.'Ihe2 ruys share a common endpoint. 3. The common endpoint is the aertex of the angle. 4. Since the vertex is a point, it is named by a single capital letter. 5. The angle also includes the infinite "interior region" bet'ween the rays.

F --> ray EF

E

i\''

Sometimes a figure will show segments as an angle's sides, however, we understand that the anglet actual sides are rays which go on forever. Nevertheless, rhe lengths of the segments can be important pieces of informarion for the

--->

tay ED

vertex

D 10

problem we are solving.

r4

E

vertex

NamingAngles In naming

-

angles,

t

.

The angle on the Ieft can be called ZABC or /.CBA.

ZABC or ICBA or ZB

C

A

ZB

would be no confusion angle we had in mind.

B

since there

as to

which

Look at the angles on the left. \7e could call

C

IABC

or

ICBA

X

ZCBD

oT

ZDBC

&

D B

ZABD orZoae

any angle lB would be confusing. Similarly, the right most angle could be called ZCBD

d

or ^^1 L'-^

t -, ".^- ^-rt^

--

2

Sometimes your textbook will use numbers to name angles, especially in complicated figures where many angles are shown.

9

1

5 4

3

NowYou Try It

-

Give as many correct names as you can for the angles below:

)

I A

4

3

C

K

E

I H

Y

F

B

MeasuringAngles The

\(hat

the size of an angle measures is rotation. The vertex of an angle is the cenrer of the angle. The vertex is the point around which the rotation takes place. An anglet measure is the amount of rotation around the vertex beginning at one ray (sometimes called the initial rafl) and ending at the other ray (sometimes called the terminal or terminating ray*).

Angles are measured in degrees. The symbol for degree is '. Each of the angles at right measures 56". A complete rotation (as in a circle) measures 360".It is traditional in beginning geometry, when the rotation is more than 180', to measure in the other direction. So, all angles studied will be 180" or less.

M

l/

"m"

means degree meAsure.

This is ZABC. wIABC = 56

wIABC = 56. wIABC is a number.

IMNO. wIMNO = 43

This is

"plaii'ZABC

is the angle itself.

B

* If your textbook uses these terms, remember that it doesn't matter which is the initial ray and which is the terminal ray. One side of the angle is initial and the other side is terminal (or terminating). 10

Measuring Angles With a Protractor How To Use a Protractor Task Measre ZDEF on the right. Before you start, be sure that your protractor is face up. The numbers should be oriented as shown in the figure below.

F E

Step one:

the bottom of the protracor. Center the circle over the veftex, point 4 of the angle.

Step two:

a

Tirrn up

the protracror so that rhis with the right side of the angle being careful to keep the circle centered over the vertex. Now read the measure on the inside row of numbers where the second side of the angle .E .rorr., ,h. protractor (or would cross it if you were to extend the ray).

ZDEF

measures approximately 113".

\When you always things in a particular way, youte less likely to make mistakes.

11

Pictures of angles do not always show the points through which the rays pass. In fact, there are many different ways in which your teacher and your textbook might draw and label angles. Severd of these are shown in the measuring exercises below.

NowYou Try It

-

Using a protractor:

1.

Measure

4-

ZB

Measrre

Zl

I

B

2.

Mexure

A

5.

Measure IABL. (Remember, the vertex is the middle

lH. H

3.

Measure

ZDEF. E

6.

Measure

C

B

letter.)

ZGHI.

I

G

H

F PuttingAngles into Categories

Angles are pur into categories based on their sizes. The names of the categories below are used in problems and it is important that you memorize them.

Acute angle: Right angle: Obtuse angle: Straightangle:

# For example, an

An angle measuring 89.9999" is acute. An angle measuring 90.0001" is obtuse.

NowYou Try It

-

IdentiS, each of the following 2

1 1

5

as an acute,

right, obtuse or straight angle. 4

3

o

90

180'

6.

7 1

t2

gg.g'

More about Angles

C 3 10

31"

I

A 1

Because their degree measure is the same, the four angles above are equal. Angles of equal degree measure have the same shape (remember, although it's not shown in the figures, the sides of each angle are rays, and all rays go on forever). Therefore, every angle of say, 31', is the same size. So

for angles, congruence is the same as equality. Therefore, we can write:

ZA

=

ZB

=

ZC

=

ZD or

mZ.A = mZB = rnZC = mZD

tWhen anglCI in figures are marked with the same number of marks, ir means rhe anglCI are congruenr.

Special Pairs ofAngles Vertical Angles One of the few things we can assume in geometry is that lines that appear straight are straight. two lines cross, two pairs of equal, opposite angles are formed. Pairs of opposite angles are called uertical angles. Remember, there is no such thing as a verticd, angle, they always come in pairs. Here are a few examples: 'S7henever

90'

90'

90"

90"

t2 60'

152" 152"

80

40"

140"

740"

In

each case, opposite angles are equal, and together, they make

a

pair of vertical angles.

The Vertical Angle Theorem

*This

is often a piece of information that you need in order to solve the problem. T3

All'

More Special Pairs ofAngles

Complementdry 90"

ComplementaryAngles

Two angles whose measures total 90" are clmplementary. The angles may or may not be adjacent angles or even near each other. Look at lA and lB below. Since their measures total 90", they are complemenrary angles. \7e say 2 angles are clm?lernentary or that one angle czmPlernenls the other or that rwo angles are com?lements.In each case we mean the same thing: the two angles add up to 90'.

In the figure on the right, ao

)

2

B

A

mzA+mzB-90" supprementarvAngre'

angles

MNO

and ONP are complementary. In this case they are also adjacent angles and therefore, we know that ZMNP is a right angle.

P 39

l'/

o

51"

M

SuPPlgmentary 180"

Two angles whose measures total 180" are supplementary. The angles may or may not be adjacent angles or even near each other. Look at ZG and lH below. Since their measures total 180", they are supplementary angles.'We say 2 angles are supplementary or that one angle supplements the other or thar two angles are supplements.In each case we mean the same thing: the two angles add up to 180".

mzG + l%zH 26'

G

-

180

In the figure on the right, angles FDE and E EDC are supplementary. In this case, o they are also adjacent I

o

angles and, therefore, we

154"

F

know that Z-FDC is a straight angle.

H

*MEMORY HINTS* Complemenrary and Supplementary sound alike. How do you keep them straight??

180

Heretoneway:

S= with C, which

comes before S in the al phabet. a Here's another: Complementary starts 90 comes before 180 on the number line. AB

1.-9

\

180-zoo

t4

C

More Special Pairs ofAngles AdjacentAngles

lt ,_.r^f1" "^-;ru 1. The two angles have the same verrex.

2. One of the anglet terminal side is the other angle's initial side. In other words, they share a common side, and 3. The angles do not overlap.

I and 2 are examples of pairs of adjacent angles: ZABC and ICBD ZMNO and ZONP

Figures

B Fig.

C

o

l/

M

1

P

Fig.2

Here are3 examples of angles that are NOTadiacent:

I In Figure 3, ZFGH

IJGI

nor ad.iacent angles. The angles share point G as their vertex and do not overlap, however, they do not share common side. They are breaking condition 2. and

are

H

l

a

In Figure 4, Z. KLO (the larger surrounding angle) and./.MLN are not adjacent angles. They do share

G

Fig. 3

F

a

co{nmon vertex L and they do share a common side LO - (which is the same as LN), however, the angles

L

overlap. They are breaking condition 3.

If

Fis.4

o In Figure 5, ZABC and ZCDE are nlt adjacent angles. Although the angles do not overlap, they do not share a common vertex and do not share a common side since a segment DZ is not includ.ed ----> "f in BC. The angles are breaking conditions 1 and 2.

.s7hen

asked

if 2 angles

are adjacent, be sure 15

A

C

B

Fig.

D

E

to check for all 3 conditions!

5

As,

Special Pairs ofAngles continued

Note: Check in the index of your textbook. If "linear pair" and theorems in

linear pairs are not listed, skip this definition.

n'rru-mnrlrnu4"hWUe

wwryjE,@ *Memory Hint:

the word line hidden in linear? The initial ray of one angle and the terminating ray of the other, form a (straight) [ine.

Here are rwo examples:

Pa

See

l. IABC and ( -->----->

CBD are adjacent angles, and 2. BA and BD are opposite rays.

s

l.lPqRand

ZRQS are adjacent angles, and .-->----> 2.QP and QSare opposite rays.

Therefore,

IABC

and

ICBD form a linear

pair

Therefore,

lPQRand IRQS form a linear

pair.

IMNO

I

Quick Angle Review

1.

{

er.

IMNP IPNM

2."&l4-M

or

Z ONP or

I PNO

A

"Plain"ZABC means the angle itself. wIABC is a number. wIABC = 25.

B

C 3

a,cljlce

0'

gi ht

10" 20" 30" 40" 50" 60" 70'80"

OBTT'SE 100" 1 10" 120' 130" 140" I 50' 160' 170" 80'

lA. = lB.2r mlA = mlB

4. Congruent angles are also equal angles. Equal angles are also congruent angles. 5. Special pairs of angles. Wrtical Angles are formed by mo intersecting lines.

IEBD

Complementary Angles

Supplementary Angles:

Adjacent Angles:

2 angles that total 90.

2 angles that total 180

2 angles that have l.

Same vertex.

c--"'""]ia.. Y overlap a RT Given is the angle bisector of ZQRS, we know that r = 47.t/ bisector of ZGHI. Note: Angle bisecSuaightAngles in Problems and Proofs tors are always rays.

_z x ir

--M r/

Given the figure above, you are allowed to assume that ruZMNP = 180". In an algebra rype of problem, this assumption is enough to allow you ro solve. For example: 158+x=180 ,r*" ,P x = 22t/ But in a proof, if you must show that the two angles are supplemenrary, it takes 2 steps:

ffi

First'Way - If your textbook gives this second part of the Angle Addition Postulate:

Second \Vry

If ZMNP is4graight angle and point O is not on ME then

Angles

wZMNO+wZPNO=180.

l. wIMNO + wZPNO=180.

If your textbook defines linear pairs

and gives this theorem:

thatform a linear pair are supplernentar!

t/ N

Given the above figure, these are the 2 steps you should use to show 2 angles are supplemenrary:

Statement

-

Given the above figure, these are the 2 steps you should use to show 2 angles are supplemenrary:

Statement

Reason

l.Angle

2. ZMNO ETZPNO are suppl. 2. Def, of suppl.

Reason

I.ZMNOUIPNO are a lin. pr. 1.De[ of lin. pair. 2.ZMNOAIPNO are suppl. 2.2's forminga

angles.

lin. pair L9

are suppl.

Trrnonru:

Complements of the sArne or congruent angles are congruent.

lA

/-B

and

(20 +70

+ +

C

I

are

complementary

=90'/)

B

90 therefore

70"

90

A of

Pay attention to the details

ZB are complementary (20+70=90'/)

C and

this theorem:

ZA

and

lC

are

congruent

(NOT complementary). Trrnonrrvr: Supplernents of the same or congruent

ZD

and

ZE

are

angles are

congrasnt.

supplementary

(116+ 64=180'/) 116

o

+

D F 116'

I 16'

180 64"

E

D 1

+ lF

180

and

lE

are

supplementary

(115+ 54=180'/) Getting Started with Proofs

-

Pay attention

/

5.

aP

Statements 134 and mZR = 46. mZQ= mZQ+ mlR = 180. lQand R are supplementary. and are supplementary. Z-P.

lR

=

lP

to the details of

ZD

and

lF

are

congruent

(NOT supplementary).

An Example

about its definition and use the definition to

I

16'

this theorem:

Given: lQissupplementary to ZP, mZ.Q = 734 and mlR = 46. Prove: lR ZP. = Sketch in the given information. Supplementary is part of the specid

7. 2. 3. 4. lQ

F

therefore,

P

R

of geometry. Think (< )r. Prove ls true.

Reasons

1. Given. . in Stmt. 1 together.) 2. Addition Prop. (Added 2 angles. (Stmt. 2). Definition of supplementary 3. 4. Given. 5. Supplements of the same angle are congruent.

.q

20

What Is

Proof? A proof is a problem in which you are given some information and then asked to reach a certain conclusion (the prove) in a logical way and according to certain rules. a

Given: AE = CE, EB = ED The Two Parts of a Two-Column Proof Prove: CD = AB. 1. The Statements The statements are where you state your case. The statements conrain information about this particular problem. Tlte last statement is always the proue! 2.Ihe Reasons For each statement, you must say why, the reason (the "because"). And in geometry the reasons are limited to: 1. \W.hat is given 2. Definitions 3. Postulates (and the Properties) 4. Theorems (and

sometimes mini-theorems called Corollaries). Some textbooks will call these by other names such as conjectures and facts or arguments or rules, but you'll recognize them; they are the reasons that you are allowed to give in a proof, And remember, you are only allowed to use those that are listed in

your textbook or given byyour teacher!

C'""

Statements

G

D

efinitions

These are the GooD PoinTs

Reasons

p*,utrtes

P

T

(+ Properties)

of Geometry!

!.o..ms

(+ Corollaries)

\When you're searching for reasons, the GooD PoinTs are where you search.

Thinking Through

€r6*il,l,(r*?

a

Proof

1. Draw a sketch. Add the information from rhe "given". 2.The prove is always true. Make notes as you rry ro figure out why it is true 3. Jot down your reasoning. The path of your thoughts zi the proof. 4. Add any new information that you were able to figure our ro your sketch.

;$erAr*''oo&

Memorize the definitions, postulates, theorems and corollaries as they are introduced. tff/hat good is a theorem sheet (which some teachers allow) if you have no idea what the theorems are? - Homework proofs emphasize the postulates, theorems and definitions that were introduced in the current section; proofs on exams emphasize the new information from the chapter being tested. given or prove, you will usually use the definition of the term - If any defined terms are used in the "biconditionds". as one of the reasons. Definitions are This means that you can use them in 2 ways: A right angle measures 90". An angle that measures 90' is a right angle. The beginning part of each sentence is what you have already shown. The ending part is what you are trying to show.

-

Proofs Versus Other Problems To do a proof you need to know the DPTs so that you can recognize relationships, think logically and name the DPTs that supporr your staremenrs. For other problems, you need to know the DPTs in order to recognize relationships and form the correct equation(s). 21

The Properties and How to Use Them in Proofs Use the Equaliry Properties with segment lengths and angle measures since both are real numbers.

AoonroN Pnoponrr or Equeurv: lf x = y and A = b,then x a 4 = y + b.

DrvrsroN Pnoprnrv or Equeurr:

This property allows you to add two equations together, left side to left side, right side to right side For example:

MurtrpuceuoN Pnopnnrr or Equer-Irv: lf x =7, then cx = c!. The Multiplication Properry allows you to multiply both sides of an equationby the

IfAB=DEandBC=EE thenAB+BC=DE+EF

Ifx=yandc*0,thenf=*

same number. DrsrmsutrvE, PnopERTY:

D

Remember, AB is a length, that is, a number. SunrnecrroN Propenrv or Equerrv: If x = y and d. = dtthen x - a = ! - a, for example: If wZABC + ruICBD = wIDBE + wZCBD and wZCBD = wICBD (Refexive),

then ruIABC = mlDBE. Remember,

rnlABCis

a number.

SussrrrurroN Pnoprnrv or Equaurr: If a = b then either a or b can be substituted for the other in any equation. For example,

lf AB+BC=DE+EF

F

andAB+BC=AC, DE+EF=DF then, AC = DF

a(b+c)=Ab+d.c or Equeurr:

SvraNasrRrc Pnopnnrv

lf

x=!,

theny=v.

For example: lf AB = CD, then, CD = AB. Symmetric is also a congruence properry so: If lB, then ZB = Z-A and = 'tf AE = CD then CD = AB The last statement of a proof must match the prove exacdy. Use the Symmetric Property to "fip" an equation if necessary.

lA

Rrrupxvs Pnoppnrv: x = xt AB = AB, m/.D = mZD. Refexive is also a congruence properry, so:

Substitution is calling the same quantiry by a different name. It's as if using the information in the 2nd line, you went back to the lst line and crossed out the quantities, replacing them with their other narnes:

Ni%c'=Di%r

ZA = ZA, and BC = BC Before you add or subtract the measure of an angle or a segment from or to both sides of an equation, stop and say that the measure of the object is equal to itself by "Reflexive Properry". Notice that this creates the second equation that you need in order to use the Addition or Subtraction Properry.

TneNsrrrys Pnopnnrv: Equaliry is transitive: lf AB = CD and CD = EF then AB = EF. Congruence is transitive: lf lA ZB and lB lC then ZA lC. = = = Same eye color is transitive: If my eyes are the same color as your eyes and your eyes are the same color as your mother's eyes then my eyes are the same color as your mother's eyes. Some characteristics "transfer" and some

dont. In proofs, if a quantiry or object is skipped over you are probably looking at the Tlansitive Property. (CD was skipped over in the equaliry example, /.8 wx skipped over in the congruency example and you were skipped over in the eye color example.)

Know the properties and know how to use them! 22

Srcuprvrs

DEFINITION

ANcrrs

vs.

Midpoint of a Segment: The t divides a segment into 2 congruent parts.

Angles do not have midpoints.

DEFINITION

DEFINITION

Segment Bisector: A taft plane or segment that intersects a segment at its midpoint.

n'l

Angle Bisector: A ray which divides ^tl ,/o angle into 2 congruentadjacentangles. m

THEOREM

THEOREM

POSTUTAIE

POSTULAIE

D

*Midpoint Theorem: *A1gfe Bisector Theorem : If M is the midpoint ofB, then If BD is the bisector of IABC, then AM = YIAB and MB = YzAB ZABD I/z/-ABC and I DBC I/zZABC = = *Only learn and use these theorems if your textbook includes them. If they do, pay attention to the details: The definitions of a midpoint and an angle bisector state the 2 parts are congruent. The theorems state the 2 parts areYz of the whole. Definition5 ) parts =. Theoremr * p".tr@whole.

is

Segment Addition Postulate: If B between A and C then AB + BC = AC.

{

-4C U"

nZ.JOK *

K

Postulate: If point region of ZJOL thenmZ.tfOL =

liSf-;a:,fition lies in the interior

'

"4 mlJOL. l

Proofs about Segments For proofs about segments try the Addition Property rogerher with the Segment Addition Postulate. Heret an example: Given: AE = CE, EB = ED. C Prove: AB = 6P.

D

Before you start, ask yourself, "-Why is the prove true?" (The prove is always true and figuring out why is what a proof is all about.) Sketch the figure adding the given information. Now study the figure. If why the prove is true still isnt obvious to you, try assigning lengths to the segmenrs. For example, let

AE and CE = 2, and EB and ED = 5. Studying the sketch, CD = 7 and AB = 7,so yes, AB = CD. This is not a proof, but it shows that if the parts are equal, the wholes are equal and suggests the idea of adding the parts together. The shaded letters are to help you track the segments: Reasons

Statements

l. AE = CE,

the 2

from I together Using the Segment tion Post. to state that the sum the parts equals the whole. This brings and CD into the

1. Given.

EB = ED, 2. AE + EB = CE + 3. AE + EB = AB, CE + ED = CD. AB = CD.

ED.

ls

statement the Confused by substitution? Try remembering when to use ir, like this:

23

2. Addition Property of Equaliry. 3. Segment Addition Postulate.

4. Substitution Stmts.2& (from the same thing Stmt. 2) by a t name (the one determined in Stmt. 3 the left sides of two equations are equal: c-- - nlrAE + EB =AB, + ED = nD.

)tmt''\cr f.x;5

cD.

;r*,ffi:L;'"' "'

t5D>

Now You Try

It

A

Given: AC = EC and AB = ED Prove: BC = DC

D

B

C Study the figure and ask yourself "\(hy is the prove true?" Mark A E to is asking us information. This proof up the sketch with the given B show that if the wholes are equal and one pair of parts are equal, then €egA.tbl,(r*? the other pair of parts must also be equal. This suggests taking something SwDIr(rry? away, or subtraction. And in order to subtract something, we'll need to show that the whole is equal to its parts and the Segment Addition Postulate does that. As you think about the proof jot down the properties or facts you will probably use. Check your answer in the back of the book.

Statements

Reasons

1.

1

2.

2

3.

)

4.

4

5.

BC = DC.

5

Proofs aboutAngles For proofs about angles rry the Addition Property together with the Angle Addition Postulate. Here's an

example:

A

ml

BAE = wIABD and Prove: wZ-BAC = m^/.ABC

Given:

ml

EAC =

B

wIDBC A

Study the figure. \7e are given thar- 2 pairs of parts are equal and asked to prove that the wholes are equal. In fact, this proof is logically identical to the proof about segments onpage23. Reasons

Statements

Proofl

l. wIBAE = wABD

2 eouatlons from ^l Statement 1 to

ml

1. Given.

wZEAC = wZDBC. 2. wZBAE + wZEAC

Using the Post. to state that the sum of the parts equals the whole. This brings ml BAC and into the

and

2. Addition Properry of Equaliry.

=

wZABD+ mlDBC.

ruZBAE+ nZEAC = mlBAC,3 m IABD + ml DBC = wZABC.

Be

This

Stop is about

(not

4. Substitution (Stmts. 2 &.3)

mZBAC =mlABC. is al

AdditionPostulate.

a the same thing (from Stmt. 2) (the name one determined in Stmt. 3)

statement the 24

)

Perpendicular Lines and Proofs

DrrrNrrroN: Perpendicular

lines are lines that intersect

Tnnonru: If ruo lines form

congruent adjacent angles, the lines are perpendicular.

Given:

to

form right

angles.

IABC = ZCBD

;;, tr;fi;iz-\JL,) \4 * \A lBV

,}(=,X

X-'X

ilu\^u

Proor:

To prove that some object has a particular property, show that the object meers the definition of that properry. In fact, if you are stuck on a proof you might try working backwards, srarting with the definition. Below is an example, using arrows to indicate the fow of thought: The last statemenr + is always the p rove

e

e

CEt

AD

ZABC

is a

Def,

of

perpendicular lines. (r lines form rt. Z's.)

efinition of right angles. (Rt. l

ruIABC = 90 .....'....-'

's = 90".)

)

At this point we need to study the problem in order to decide how we can show that ruZABC = 90, but at least we know what it is that we need to show. Here's the completed proof going "forwards":

Statements l. ZABC= ZCBD, or

Reasons 1. Given.

wIABC

= wZCBD. 2. ruZABC + wZ-CBD

=180. 2. Angle Addition Postulate*. 3.2(m/-ABC) = 180. 3. Substitution (Statements lg.D. it: 4. Division Properry of Equaliry. 4. ryZABC = 90. +srxte 5.ZABC it angle. ;f;jfl +5. Definition oiright (k.Z= ^right "n$Ls. 6..aE t Ab \ t t',fi'#t',/_:_^ ,_ F. Definition of perpendicular lines.

\

The last statement is always the prove.

use it /

90")

g lin., form rt. z's)

* If your text defines linear pairs, do the following steps in place of Step 2:

Statements ZABC

and

ICBD

are a linear

Reasons

pair.

Definition of linear pairs.

ZABC and /.CBD are supplementary. Linear pairs are supplementary. mz.ABC + mlcBD = 180. Definition of supplemenrary.

&to'

On an exam, put down something. Never leave a proof blank. At least put in the given and thJ prove. Never be afraid to try (teachers appreciate effort). If you think you can recall a patrern, try it. Most Important Tip: Beginning proofs take getting used to and many students find them hard. This is because you are learning to think in a new way. Proofs will become easier. Be patient! 25

NowYou Try

It D

A

1. Given: AB = CD. Prove: AC = DB.

Problems

I

and,2

2. Given: wIAEB = rnZDEC, Prove: ruZAEC = mlDEB.

3. Prove the following:

Tnnonsrvr: If nto lines are perpendicuhr they forrn congruent adjacent angles. Create an appropriate figure and state the given and the Prove. The theorem ar rhe top of the previous page and this one have a close connection. They are called "converses". Can you see what the connection is?

Be sure

to check in the back of the book to

see

how you did! 26

3. LOGIC Here are some examples of conditional or "if-tlten" statements: If you live in Malibu, then you live in California. If this is alternative rock, then it is music. If you are a soloist in the choir, then you have a good voice. In symbols we write: If p then q. p is called the premise (or hypothesis) and q is called the conclusion. Premise

if

orh

Conclusion you live in California.

ls

you live in Malibu, this is alternative rock, you are a soloist in the choit

rhen v-- it

is music. you have a good voice.

The statements above are examples of true if-then sratemenrs. Look at Figure l. Since Malibu is in California, you cannor live in Malibu without also living in California. A point cannor be in the small circle

without

also being

C

A L

in the surrounding figure.

I

%

\we could also say, "You musr live in california if you live in Malibu" or "You live in Malibu, only if you live in california." Even though the phrases have been switched around, living in Malibu is still the hypothesis and living ln California is still the conclusion. Be careful, "onb ,f" actually means "then".

R

N IA

Fig. I

The circles in Figure 2 are an example of a Wnn Diagram, a logical model which helps us to picture how conditional srarements work. You can see thar any points in the small circle are automatically in the larger surrounding circle. Figure 3 demonstrates a parrern. Note that the premise, thar is, the "if" part of the statemenr, is the petite (smaller) inner circle. The conclusion, the "then" part, is the larger circle.

Truthvalue of If-Then statements and the Law of Detachment True: An example of a true if-then sratement is, if x = 2, rhen 2x = 4.In

Fig. 3

a true

if-then statement, given that the premise, that is, the "if" part of the statement is true, then the conclusion, that is, the "then" part of the srarement is also true. This is the Lew oF DETACHMENT. False: An if-then statement isfalse if you can show that given a rrue premise, that is, given that the part of the statement is true, there is et)en one example of the conclusion, that is the "then" part of the statement, being false. Heret an example of a false if-then staremenr:

"if"

Statement: If a person is a star, then he or she lives in Hollywood. Counterexample: Kelly Clarkson lives in Malibu. Since Kelly Clarkson is a star, (that is, the premise is true) and she does notlive in Hollywood (that is, the conclusion is not true) Kelly is a counterexample that proves that the if-then staremenr is false. 27

Another example of a false if-then statement: Statement: 7f xz = 25, then x = 5 Counterexample: x = -5 = 2J, x = -5 is a valid counterexample and disproves the if-then statement. Notice that in this case, -5 is the only cowterexample, but one counterexample is all that we need to disprove an

Since

(-5)'

hypothesis. an if-then statement:

statements that are related to and based

There are other lo

If-then statement If p, then

q

First comes the p, then comes the q. The positionsof

p

and

q

have been swapped

Contrapositive

If not !1, then not p.

Converse

If ![ then p.

Inverse

If not p, then not ![ The positions remain the same; both p and (made negative). Q havebeen negated

Example 1. If-then statement: Contrapositive: Converse: Inverse:

andboth have been negated (made negative). The positions of

p

and

q

have been swapped.

P-9

m-s 9-P @-cD

If you are an Olympic gymnast, then you are an excellent athlete. If you are not an excellent athlete, then you are not an Olympic gymnast. If you are an excellent athlete, then you are an Olympic gymnast. If you are not an Olympic gymnast, then you are not an excellent athlete.

In this example, the if-then sratement and its contrapositive are both true. The converse is false since many excellent athletes are not Olympic gymnasts and the inverse is false for the same reason. Notice that the words Example 2. If-then statement: Contrapositive: Converse: Inverse:

"if"

and "then" stay put. They are not part of the premise or the conclusion.

If x>

l,

then x>

lf x)5, thenx}' If x>5, thenr>

Example 3. If-then statement: Contrapositive: Converse: Inverse:

5 1 1

Ifxl l, thenx\ 5

Ifx=7,then2x=14 If 2x * 14, then x *7 lf2x=14,thenx=7 Ifx*7,then2x*14

In this example, the if-then statement and

In this example, the if-then statement, its

its contrapositive are both false, the converse and the inverse are both true.

contrapositive, converse and inverse are all true. In other words allfour statements are true.

Although there seem to be many different combinations of true and false possibilities, in fact, two pairs of the four conditional statements are logically connected. 28

contra

f - t hen

OS I tive

The if-then statement and its contrapositive are loeicallv () J linked: both are true or both are false. If you live in Malibu, then If you dont live in California, rhen you live in California. T you dont live in Malibu. T I

lnverse

converse Tte

conuerse of a statement

and the inuerse of a sraremenr are logically linked: both are rrue or both are false. If yo.u live in California, then If you dont live in Malibu, then you you live in Malibu. F dont live in California. F

*Memory Hint: Ttte"uerses" are logically linked.

The Venn diagrams of the four statements, togerher with their "Tluth Value" is below:

IF-THEN If you live in Malibu, then you live in California. TRUE

CONTRAPOSITIVE If you dont live in California, then you don't live in Malibu. Study the figure. Since the conclusion is negated, the object (you!) is outside the larger circle and therefore could not possibly be in the smaller inner circle.

CONVERSE If you live in California, then you live in Malibu. Since we can find a counterexample (Hollywood for Maybe you live in Malib example), the statement is false

TRUE

FALSE

Or maybe you live in Hollywood.

INVERSE If you dont live in Malibu, then you dont live in California. Since we can find a counrerexample, Hollywood, the statement is

false.

you might live in 29

FALSE

Tnn

Lerurr

If Hypothesis

or SvrroGISM

If C.ondrsion

1 then C.,onclrston 1.

I

then Conclusion 2.

Syllogisms are combinations of rwo conditionals in which the conclusion of the first statement is the hypothesis of the second. Heret an example,

) th. conclusion of the first I .rr,.r,,..rt must be the

If you live in Malibu,

f

you live in the United States

Lnurr

op

hypo,h.sis of the

J second statement.

Synocrsu If both

conditionals Are true, then you may conclud.e correctly, that giuen Hypothesisl, Conclusion 2 is true.

This means that the first hypothesis leads directly to the second conclusion. In symbols we can write:

9+ f or more simplyP * f. Using the example above, since both statements Are true, you may conclude that if p * gand

q+

rwhich can be combined to form: P *

you live in Malibu, then you live in the United States. Here's the Venn diagram of our example.

NowYou Try

States

f

It

1. List the four logical conditionals and using symbol shorthand give an example of each. State the two pairs that are logically connected.

2. Given the statement: If you are in a top band, then you are famous. a. Form the contraposirive, converse and inverse and state the truth value of each of the 4 statements

b. If your cousin Sally is not famous, what can you conclude about her? c. If your cousin Sam is famous, what can you conclude about him? 3. Given the statem ent: If * > 16, then x > 4. Form the contrapositive, the converse and the inverse and determine the truth value of each of the four statements.

4. Explain the terms syllogism and the Law of Syllogism. Give an original example of each.

30

Indirect Proofs Certain proofs are much easier if done indirectly Regular Proofs (also called Direct P-ort) \We are given some information (the "given") and asked to prove a certain conclusion (the "prove"). tWe start with the given and add other information based on theorems, postulates, and definitions until we reach the conclusion.

Indirect Proofs

\We are given some

information (the "given") and asked to prove a certain conclusion \(/e start by temporarily assuming that the prove (neuer the given) is negative. \We then add other information based on the GooD PoiriTs until we conGooD Poinlls tradict either the given or some other known fact (the GooD Poirils Qiven again). Since rhis is impossible, (because the GooD Poinls are true), the pefinitions postulates(+ Properties) contradiction shows that our temporary assumprion was false. Therefore, (+ "the Corollaries) Jh.o.e-s prove" must be true. (Don't worry, indirect proofs are easier than they

(the "prove").

sound.)

Heret the pattern for Indirect Proofs: temporarily that the prove is false, this means negare (make negative) rhe proue. 2. Study the problem and figure out the effect that the negarive of the prove has on the problem. 3. Give statements and reasons until you contradict either the given or some other known fact (the GooD Poinfs again!). 4. State that since something that is true has been contradicted, your temp orary assumprion must be false and therefore, the prove musr be ffue. 1. Assume

To get started we need to learn how to make the prove negative. Here are some examples:

If the prove is: the negatiue of the prove is:

AB=MN AB+MN

If the prove is: mlC the negatiue of the prove is: mZC If the prove

is:

the negatiur of the prove is:

I mlD =

mZD

AABD is an isosceles triangle. LABD is not an isosceles triangle.

is a < b the negatiue of the prove is a Z b If the prove

(You must allow for a to be greater than b or equal to b.)

3t

Heret an example*:

* mZB, and /-A is complementary to lC proof, ml C I mlD indirectly using a

Given: mZA Prove:

and

ZB is complementary to lD.

"Assume

mlC= mlD. Since lAis complementary to lC and ZB is complemenfus ume tempo rary to lD (Given), mZA + mlC = 90 and mZB + mZD = 90 (definition of complementary angles). Therefore, mlA + mZ C = mZB + mlD (Substitution). Since we have assumed temporarily that mZ C = mZD we can subtract them (Refexive) from the equation. Doing so would mean , so our tem mlA = rnlB (Subtraction). But this contradicts the Given which is assumPuon was false,

andmlCt mZD

Restate the

4. Contradict a Good Point.

NowYou Try It - Hint, follow the steps. temPorary assumption was false" Given: mZI#wZJ, Prove: I I and ZJ are not both right angles. Do the proof indirectly and use a paragraph proof,

Indirect proofs are a lot like the contrapositive form of an if-then statement. Heret an example:

If-then srarement: If the speed limit is 65 miles per hour and you arrived at a concert which is 65 miles away in 45 minutes, then you were speeding. Contrapositive: If you did not speed, then you could not be at a concert which is 65 miles away if you have been traveling for 45 minutes. Using an indirect proof prove: Given that the speed limit is 55 miles per hour and that you arrived at a concert which is 65 miles away in 45 minutes, prove that you have been speeding. Proof: First, assume tem?orarily thatyou were not speeding. But if you were not speeding and have traveled for 45 minutes at the top legal speed of 65 miles per hour you should only be 48.75 miles away

(65mph x 45160 hrs). But you are 65 miles away (the Given). Therefore, since we have contradicted the Given, you must have been speeding!

*Note:

For additional examples of Indirect Proofs see page 235 in the Answer Section.

32

4

PAFIAI I

FI

I

INES

Lines, Transversals and More Special Pairs ofAngles c

r

M*w rn

Special Pairs

Eight angles are formed by the transversal and the lines it crosses. There are special terms for pairs of angles in certain relatiue positions. It is important to understand and memorize these terms.

I

rn5

)-

1

2 4

6 8

of Angles:

Corresponding Angles

Alternate Interior Angles

@enu ndm.

srynddmnn**r {mrylp

Corresponding angles dtFH

@ @

rmnlilnr;frrr.

*nelnift*af,tlftCtm*)

ns.

Irook at the 2 angles marked Look at the 2 angles marked with the black't's. The sides with a single J.lhe sides of each Pairs of Pairs of of both angles are formed angle are formed by the ffansCorresponding Alternate Interior by the tran-sversal and one versal and one of the lines. Both Angles Angles of th.lines. The angles are angles are on the same side (the on opposite (alternate) sides Ieft side in this example) of the of the ffansversal within the transversal and both angles are above the line that forms one of interior region determined by the lines. its sides. *Consecutive) Same Side (or Interior Angles 1{.[ternate Exterior Angles

W*Mor @s.

Exreruon

interior angles always refer to a pair of angles in particular positions. Same side

@

*inrmahr

Alternate exterior angles refer to a pair of angles.

Look at the 2 angles marked a black *. The sides RrcroN Look at the 2 angles marked of each angle is formed by with the black't. The sides of the transversal and one of each angle is formed by the pairs of the lines. The angles are on Pairs of transversal and one of the lines. Same Side Alternate Exterior opposite (alternate) sides of Interior Angles Both angles are on the same side Angles rhe transversal in the exterior of the ffansversal and within the regions created by the lines. interior region between the lines. *lf your textbook does not define alternate xUse the term your textbook uses. exterior angles, skip this definition.

with

33

Consider lines i and j in the illustration below. Lines k and /are transversals from the point of view lines i and j. However, if we are considering lines k and /, then lines i and are the transversals.

j

of

Example: Classify the following pairs of angles as corresponding angles, alternate interior angles, same side interior angles or none of these. If the angles do form a special pair, name the lines and the transversal that form the 2 angles.

l. Z7 and Z9 - Corresponding angles, lines i and j, transversal *. 2. ll and 15 - Corresponding angles, lines * and l, transversdT. 3. /5 and Zl3 - Corresponding angles, lines i andj, transversal /. 4. ll2 and 113 - Alternate interior angles, lines 2 and l, transversal i. 5. Zt and Zl2 - Nternate exterior angles, lines i and j, transversal A. exterior angles, lines * and l, transversalT. 6. Zl and Z8 -Alternate 7. Zl2 and 213 - Alternate interior angles, lines I and l, transversal l. 8. Z4

and

110

-

14 and ll3 lO. Z4 and 26 Now You Try It 9.

Same side

J 1,

1

9 1

5

t2

8

t5

4

interior (or consecutive) angles, lines I

and j, transversal

[.

None (no transversal is shared by both angles). None (this pair of angles are not in any of the correct relative positions).

Using the figure above, classify the following pairs of angles as corresponding angles, alternate interior angles, same side interior angles or none of these. If the angles do form one of the listed special pairs, name the lines and the ffansversal that form the 2 angles.

1. 13 2.

l4

and and

3. Z15

17

4.

l2

25

5.

23 and Zl0

and

Z6

6.

and

lll

Z5

and

7.

27

and

173

8.Z4andZl3

Zl0

9.

lll

and

18

A Closer Look The angles that make up the special pairs defined on the previous page must both be formed by the same transversal: ./-7 and 22 are not coffesponding angles because their sides are formed by 4 (entirely different) lines. 8

)

Z l0

Z5

not samc side interior angles because their sides are formed by 4 (entirely different) lines. and

34

are

Parallel Lines

-

C

lines.

B

Coplanar means that a single plane can caprure both Lines that are coplanar in the figure on the right include:

AB andDC, AB

V

-

and

HG, AC

and

BD, FH

G

H

BD. ri",gl."

and Uy

Non-coplanar, oru iiilin r.*rrroib..ri,"r.a pl^rr! Lines that are skew in the figure on the left include: MN and W QM andNo, PS and(JV, MP and OT Do not intersect means that the lines share no points L.S.r i" on the- right that do not intersect include:

K

f3firy.

7fr and,ffi,

ee N""dTi, jfr and KO, IK and MO.

o

M

Lines which meet both conditions (coplanar and do not intersect) are parallel. Pairs of parallel lines in the figure on the right include:

iB

^nd.bd,

VE

^nd,Fd,

Ffi

^nd,EB,

7B

The symbol for parall.l is ll. For example: -

c

,ll a

*ll

€ll

, n->

"ll,

"

+>

AB II DC €rr€ EA II HD

Arrowheads at the end of lines do not mean the lines are parallel

k?t Parallelism is transitive. Here's an example: If

And never assume that lines are parallel!

Fdll

Af

,r,d

@ llfri

th.r, Fd

ll

ft,

U

.t

T Parallel Planes

-

Planes are Parallel

In the figure on the right: Plane GINK ll G

tttl,tt

Plane GHLKIIltl,tw Plane HIJGII

tu\r

if They Meet

1

Condition: Parallel Planes Do Not Intersect

In the,figure on the right, plane ABC ll DEE There are no other parallel planes in the figure. Planes may not appear to intersect even when they do. Planes PRQ

and SUT do intersect. 35

s

B

D

a

Eight AnglesTwo Measures Interesting things happen when wo parallel lines are cut by a transversal. Examine the figure at right. As we learned in the prior section, eight angles are formed. However, because the lines are parallel, the measures of the angles are connected.

If

the two lines are parallel, any rwo of the eight angles are either equal or supplementary. Here are a few examples:

4

3

6

5

8

7

1

15

115 5 11

5

9

125 9

125

1

1

5

15"

11

125 125 5

but only two angle sizes. In fact, in the special case of the transversal being perpendicular to the lines, all 8 angles will measure 90" and be both equal and supplementary, but this is only in the specid case. Putting the new information together with the definitions about certain pairs of angles leads to important postulates and theorems.

In

each of the above figures, there are eight angles

The figures below show each pair of corresponding angles from the first figure:

15"

11

15"

11 5

if the mro lines

are pardlel. formed by parallellines. This means that not all pairs of corresponding angles are equal, only those

It is imporranr ro understand that corresponding

angles are equal only

36

Investigating further, the figures below show each pair of alternate interior angles from the first 6gure.

11 5

15"

Now look at the third special pairs of angles, same side (or consecutive) interior angles. Study the two drawings. The connection between the angles is different than the connecrion of the previous special pairs. Each pair of angles is supplementary. Itt importanr to stop and think when using this theorem, think supplementary, not equal.

1

15'

15'

The figures below show each pair of alternate exterior angles from the first figure.

1

15'

115 5

*Not all textbooks define exterior or state this theorem. Although is true, do not use any theorem in proofs if your textbook does not state it. (But this onet still a good fact to know!) Given &rallel lines

Givenffirallel ffilternate $nterior are

Givenprallellines #ame #ide flnterior are

37

are

)tMemory Hints*

ffi WP*andffi as abbreviations and memory tools. However, on homework thit your teacher taught you for each postulate and theorem.

This book uses and tests, use tfi"e name

Hints for RecognizingN7hich Typ" of Special Pair ofAngles Are in a Problem: \y/ith corresponding -. ncs \With alternate interior -r!ffN, With same side interior "Qry "

make;S angles, the lines make, *ffi,ro' "./-- C(or3). r aZ(orS).

,.,r't.*-Sl anF(orllorEorC). "f

angles, the lines

angles, the lines

PROJECT Draw two parallel lines. Recall that parallel lines hJve the same slope. Then draw a third line (the rransversal). Using your protractor, carefully measure the eight angles formed. Now, draw a transversal with a different slant. Measure the angles formed. Do your findings agree with our new postulate and theorems?

Angle measures for first transversal:

Angle measures for second transversal:

Review Solve:

-

on Both Sides

Equations

5x+15=2x+75 tff/hat You Do

5x+15=2x+75 2x -2x 3x+15=75

-

15

-r5

Why You Do It 'Write The Equation. Subtract 2x from each side. (You are adding the opposite of +2x) Subtract 15 from each side. (You are adding the op

te of

+

l5)

3x=60 3x =50

33

x=2Ol

NowYou Try It

-

Divide by the coefficient of x. Done! (Because we have a single positive x allby itself on one side of the equation .)

Solve for x:

l.7x-5=4x+31

3. 4x + 7 = 5x

- 10.5 (The answer is not an integer.)

4.20x+4=l3x+39

2.3x+24=4x-B 38

Algebra Review- Equations withVariables on Both Sides (continued) Solve: 4x + l0 = 180 - (2x -10)

\7hat You Do 4x + l0 = 180 - (2x -10) 4x + l0 = 180 - l(2x) -l (-10)

4x+10=180-2x+10

+2x

+ 2x

-10

-

6x =

180

6x 66

180

x

=

30t/

NowYou Try It

-

10

\ThyYou Do It

\Write the equation.

Distribute the formerly "invisible" negative one. Remember: (-) (*) = - and (-) (-) = * Add2x to both sides. (Adding the opposite of -2x). Subtract 10. (You are adding the opposite of +10).

Divide by the coefficient of x. Done! (Because we have a single positive x all by itself on one side of the equation.)

Solve for x

2. 180-(2x+3)=6x-3

l. 10r+5=180-(4x-7) Putting Algebra to'Work with Parallel Line Problems Example

1.

\Mhat to Do Study the drawing. The arrowheads tell us the lines are parallel. Since the two given expressions are for corresponding angles (note that the angles form an {), we can apply PCC and set the two expressions equal to each other:

Solve for x: o

(7x +10

(10r - 38

7x+10=10x-38

o

-7x + 38 -7x

n

Example

2.

+

38 Add opposites.

48=3x

Solve

15

48 33

for z:

16 =

-

xt/

Divide by the coefficient of x Done.

*ffi

rVhat to Do The lines are parallel and the given expressions are for same side interior angles (note that they form a f,). PSSIS, tells us the angles are supplementary:

z +10) (202

3x

5

l5z+10+202-5 = 180 352+5 = 180

-5

-5

352 = 175

z=5t/ 39

Tirrn English into algebra. Like terms collected. Adding opposites. Now divideby coef. of z. Done.

As, \7hen yousee

,orll think#$€W,B..& %

More Parallel Line Problems Example:

At first glance, this figure doesnt look like

Solve for x.

Fig. I

the figures we've been seeing, however, if you extend the lines, it does. Now, referring to the 4 methods above, PSSIS tells us that the 2 angles are supplementary:

Fig.2

2x+7 +x+20 =180 3x+27=180

x = 5lJ

Example: Solve for x, y and

6x-17)"

z.

The key to this sort of problem is to use one of the methods at the top the page to create an equation with one variable. Study the diagram. PSSIS tells us that the 53" angle must be supplementary to the angle measuring 6x -17. This means we can form the following equation:

53

6x-17+53=180 (3

y +28)"

(7

6x = 144

z+l)"

x=24J

Now solve for7. Since we know that x = 24, we can substitue 24 for x in the original expression. By PCC and substitution we have: 6(24) - 17 = 3y + 28

127=3y+28 99=31 And finally, solve for

z,

alsoby PCC and substitution:

33

=

y'/

3(33)+28=72+l 127=72+7 126=72 18= zJ Notice that there were other ways to solve this problem. Also, the problem asked for the values of x, y and z. Sometimes you might be asked for the measure of each angle Always be careful to answer the question you were asked to answer.

NowYou Try

It

1. Solve for a, b and

c.

2. Solve for u, u and

Qu+5

40

w.

10

N'S.* the$l$ht quQi}ion.

of

Proofs and Parallel Lines

ll

I

a

3

5

Fig.

1

is supplementary to 17. Givenp llq, prove that None of the 4 methods deal with angles in the relative positions of Zl and 27, so we need to use other methods to get started. Study Figure l. Vertical V & Z4 6 a Angle Theorem (VAT) tells us that mll = rnl4 ) and ml6 = ml7.Itt a good strategy to mark up the Fig.2 drawing while you're trying to do a proof which we've done in Figure 2. Studying Figure 2, we know that 24 afi and 16 are supplementary 6y PSSIS). But then it follows that Zl is supplementary to 27 (substitution?). Saying what we have figured out in a more formal way is what a proof is. First, figure out why the conclusion is true, then do the proof,

Given: pll q. Prove:

Zl

is supplementary to 2.7.

Proofi

Statements

t. pllq.

mll = ml4, m/7 = mZ6. 3. /.4 is supplementary to 26. 4. mZ4 + mZ6 = 180. 5. mZl + mZ7 = 180. 6. ll is supplementary to 27. 2.

Reasons

1. Given. 2. YNf.

3. PSSIS. 4. Definition of supplementary. 5. Substitution Properry (Statements 2 A q. 6. Definition of supplementary.

'Sf.[ry

couldnt we have substituted right after step 3? Question: Answer: Substitution is a properry of equaliry and congruence. You cant use substitution with other types of statements.

NowYou Try Given:

It

pllq.

1. Prove:

12 is supplementary

to 2.5 4

1

3

2. Prove:

mZl

=

ml8

6

without using PAEAC. Problems

4r

I

k- 2.

Proving Lines are Parallel In the previous section, we learned that if 2 parallel lines were cut by a transversal, the pairs of angles which were formed were either congruent and/or supplementary. The logic sketch for the 4 methods in the previous section look like this:

+

PAIAC:

PSSIS:

]|(-ft,'0"

pAEAC:

+-#

In each case we start lut knowing that the rwo lines are parallel (the given). Then we can conclude information about a special pair of angles (the conclusion). In this section we go backrvards. \fle start out with rwo coplanar lines (about which we know nothing), cut by a transversal. We are given that a particular pair of angles formed are equal or supplementary. Based upon this Act, we conclude that the lines are parallel.

It is enough to know that a single pair of corresponding

angles are equal to conclude that the lines are

parallel.

Here is the logic sketch for "CCP"

CCP:

-*+

Notice that we start out knowing that a special pair of angles are equal (the given). Then we conclude that the two lines are parallel (the conclusion). Here is a comparison of the logic sketches for PCC and its converse, CCP:

Here are 3 other new methods that are the converses of the theorems shown at the top of this page:

*Although this theorem is true, do not use any theorem in proofs, if your textbook does not state it 42

-:p

\7hen you are doing a proof or a problem about lines and transversals, A$, it is important to choose the correct method. In each case, stop and ask K yourself, Am I starting out knowing (being given) that the lines are parallel? * If so, you want a method that begins with a p (for parallel). Or, are you being asked to show (conclude) that the lines are parallel? If so, you want a method endswith a p (also for parallel). Once you decide which is the right group to choose from, add additional markings to the figure; this should help you select the correct method within that group.

$

ry *lp

*ry

Here's an example of how to use one of the methods that ends with a p in a proof,

I

Given: 12 is supplementary to 28. Prove: llln. I n Since we are being asked to conclude that 2 lines are parallel, 4 1 vaT we definitely want to use one of the new methods (the ones )1 6 that end with a p). But none of the methods deal with 4 angles in the relative positions of 12 and 28, so we need to use other methods to get started. Study Figure 1. AT Fig. 1 Z l5 The Vertical Angle Theorem (VAT) tells us that mZ2 = ml3 and mZ8 = mZ5. Since 12 is supplementary to Z-8, l3 Fig.2 must be supplementary to 15.1his, in turn, would prove that the two lines are parallel by SSISP. As you think about a proof mark up the figure (see Figure 2). Your "notes" will help you when you do the proof itself. z3*25 =18A"

Proof:

Statements l. 12 is supplementary to 18. 2. ml2 + mZ8 = 180. 3. mZ2 = mZ3 and mZ8 = mZ5. 4. mZ3 + mZ5 = 180. 5. m/3 is supplementary to ml5. 6.

tll".

*Note: \fe've

used "SSISP"

Reasons

1. Given. 2. Definition of supplementary. 3. Vertical Angle Theorem.

4. Substitution Properry (Statements 2 e 5. Definition of supplementary. 6. ssrsP.*

3).

but be sure to name all methods in the way that your teacher requires.

Heret an example of a problem that uses the methods in this section and is the geometry version of aqpe of problem which appears on many exams and standardized tests: m Using the figure, find the values of x, y and zthat makep ll n *ll o. ^nd Solution: The key to solving problems like this one is to assume that whatever n condition you are "making" is already ffue. Then, given that the condition is ffue, how does this affect the problem? Given: plln and *ll o,what do we 0 know? PSSIS tells us that same side interior angles are supplementary. This would mean: (2x+16)+(x+14) = 180, x+76)+(31'11)=180 3y-ll)+z=180 + 3x+30 = 180, 2 105 180 3(25)-11+a=180 = )+16*3y-ll=3! 3x=l50so x=50 3y=75so!=25Vl z = ll6J 43

5. TRIANGLES

Tnronru:

The

interior angles of a triangle total 180".

This theorem is easy to prove to yourself. Thke any paper triangle, first cut off the three angles, then place the points together. The result will be a straight (180) angle. Although this is not a formal proof it does demonstrate this most basic property of all triangles.

d

^& Many problems are based on this simple and well-known theorem. Here are a few examples:

l.

Since the three angles total 180, translate the picture into this equation

Find x.

x+(2x+7)+(3x-31)=180 (3x

6x+7-31=180 6x-24=180

- 3l)"

r

2x+

2.

24

+24

6x = 204 204 6x = -6-6x =34,/

Find x

The single arc marks on each of the lower angles tell us that the two angles are equal. Since there are 180" in all triangles, the correct equation will be:

x+x+96=180 2x+96=180 -96 -96

2x= 84 2x-84

22 x = 42t/

3. Find the values of athroughf

This is apuzzle problem. A good strategy for this rype of problem is to find the angles in the same order that the text has labeled them. Usually this is a good hint about the logical order in which the problem should be solved.

As,

Letter

Many problems that look complicated arent. AII of the angles in the figure above can be found by using: VAI, the fact that a straight angle measures 180", and the fact that the angles of a triangle add up to 180'. Complete the table at right by finding d, e andf Be sure to check in the answer section at the back of the book to see how you did. 44

a b c

d e

f

Equation

Reason

Straisht

l=

180

(180 - r03)

VAI AnslesofaA=

Answer

77 103

180

(1

80-lo3-58)

t9

Seeing

All the Triangles

Many problems in geometry require careful analysis of figures. Seeing shapes within shapes is a skill that you will learn with practice.

Lett re-examine the triangle in Example 3 on the previous page How many triangles are in the figure?

'We'll count them

45

Putting Triangles into Categories Tiiangles are categorized in two different ways. One way is by the number of sides of equal length and the other way is by the sizes of the angles of the triangle. Memorize these terms because they will be used in problems and proofs.

Comparing the Lengths of

I.

If the sides of a triangle

a

Thiangle's Sides

are all different lengths,

it is a scalene triangle.

7

Here are some examples of scalene triangles 8

16

16 2

7

r3

r4

r3

13.2

8.25

4.875

II. If (at least) two of the sides of a triangle

are the same length,

it is an isosceles triangle.

Here are some examples of isosceles triangles:

l2

12

SOSCELES

10

10

20

6.5

11

6.5

10

11

3

r9 The triangle on the left has three equal sides. This meets the definition of isosceles: at least wo equal sides.

III. If all three sides of a triangle are equal, the triangle

is an equilateral triangle. means Since equilateral equal sides, this definition should be easy to remember. Below are examples of equilateral triangles. Notice that although they are different sizes, they all

look very similar.

10

10

t6

1

1l 11

9 11

9

r6

10

ALOGIC DIAGRAM

The entire figure represents all triangles. The dark gray triangle represents all isosceles triangles which include all equilateral triangles. All equilateral triangles are isosceles. Is the converse true?

46

Grouping Triangles by the Size of Their Angles

I.

If all the angles in a triangle are less than 90', that is, if all the angles are acute, the triangl e is acute Think of it as "a cute little triangle". Below are examples of acute triangles. Remember that an angle may be very close to 90' and still be acute.

u

o

10"

T

ts

II.

right angle, it is a right triangle. The side opposite the right angle is called the hypotenuse of the triangle. The hypotenuse is always the longest side of a right triangle. Here are some examples of right triangles:

If

a triangle has a

This sym

means the two legs are perpendicular (r)

R

90"

I

Perpendicular lines form 90' angles.

G

11'

H

T

Rules for Right tiangles:

A right triangle is the only triangle that can have a hypotenuse. 2. Since the three interior angles of a triangle total 180", if one angle measures 90', there is only 90" left for the other two angles to share. Both angles' measures are positive and together add to 90'. This is why a triangb can ltaue at mlst zne 1.

right angle and, the acute angles of a right niangle are complementary. 3. Never assume a triangle is a right triangle just because it "loola like one". 4. Not all right triangles have two 45" angles. In fact, most dont.

III. If a triangle has an obruse angle, the triangle

is an obtuse triangle. Here are some examples

obtuse triangles

r9

5"

of

OBTUS

3

6"

o

\!)

Note that a triangle can haae at mlst one obtuse angle. This makes sense because the three angles of a triangle total 180" and if the obtuse angle uses up more than 90", there are less than 90" left for the other two angles to share. +/

Exterior Angles of Triangles Thiangles and other polygons have steps to

interior angles but they also have exterior angles. Here are the

find them:

2.'Ihe angle formed by the dashed line

1. Extend one side of the triangle.

and the triangle is an exterior angle.

Exterior angle

#l 3. Moving in a counter clockrise direction, extend the next side of the triangle. (Tirrning the piece of paper around as you work makes this easier.) The angle formed by the dashed line and the triangle is another exterior angle,

4. Again, moving counter clockwise, extend the third side of the triangle and, once again, measure back to the triangle. The angle formed by the dashed line and the triangle is the third exterior angle.

Measure back to the triangle.

Exterior angle frL1 .t+

Exterior angle

#3 Allowing one at each vertex, each triangle has three exterior angles, which is the same as the number of its interior angles. By the way, if we had gone in the other direction (clockwise) we would have ended up with 3 exterior angles having the same measures as those of the exterior angles found above. Remote Interior Angles

point of view, the two interior angles on the other side of the triangle are far away. That is why they are called remote interior angbs. Itt important to remember that what is considered a remote interior angle

Remote means far away. From the exterior anglet

Rcmotc

lnterior Anglc

Exterior

depends entirely on which exterior angle is being considered. IntaiorAngflc Rcmotc

Exterior Angle #2

InuiorAnglc Exterior #7

Rcmoe

Andc

\

Rcmoclnulor

Rcmotc

Andc

IncriorAngh Rcmoe Intcrior

48

Exterior Angle #3

ExteriorAngle Theorem \7e already know that The interior angles of a triangle total 180'

and a straight angle measures 180.'

An exterior angle of a triangle and its adjacent angle make a straight angle. That is, they total 180".

,/,/

a,A\

+,/ = 180"

Something interesting happens when we put these ideas together. Study the 3 figures below. Do you see a

pattern?

r

1

18"

III

46"

108"

,'1 In

72"

o

each case we have:

Two Third Remote + Angle Angles

Third Exterior + Angle Angle

180

180

Since the right sides of the equations are equal, the left sides must also be equal. This allows us to set the left sides equal to each other. Notice how our reasoning follows the pattern of a formal proof,

tfl,k

+

ft*

= 'i:;,':'. ffi*

SUBSTITUTION PROPERTY

By the reflexive property, the third angle is equal to itself, so we can subtract above equation and therefore: Two Exterior Remote Angle Angles

Exrrnron ANcrs

TnBonsNa

:

it from each side of the SUBTRACTION PROPERTY

Tlte rneAsure of an exterior angle of a triangle is equal to the sum the measures of the two remote interior angles.

4

,/\

,/ \ - -\-& 49

of

Notice how perfectly the two remote interior angles fit in the place of the exterio"r angle.

Example: Find the measures of ZA, ZB, ZBCA and ZBCD. Although we could solve this problem in several ways, we will use the Exterior Angle Theorem: n

r.u

(8x

-

8x-18 = (4x-5)+3x 8x-18 = 7x-5

18)

-7x + 18 -7x

you've finished a

problem,

+18

x =13

mlA=3(13) = 39r/ mlB=4(13) ruZBCA=6(13)+16=94'/ A and m/-BCD = 180 - 94 = 86'/

B

-When

5

=47\/

As'

always go back and read the question to make sure you've answered the question(s) that were asked.

Example: Find the measures of ZMON, ZNOE lM and lN. A straight angle equals 180", which means ruZMON + wZNOP = 180

N

+Start with this equation because 180 it has a single variable, 7.

(3y + 5) + (51 + 15) = 180 81t + 20 =

8/=160, !=20

Therefore, ruZMON = 3! + 5 = 3(20) + 5 = 65'/ 2INOP = 5! + 15 =5(20) + 15 = ( 5r* t 5)" ZNOP is an exterior angle of L MNO and therefore equal to ..+ the sum of its 2 remote interior angles, ZM and ZN:

ll5'/

P

mZM + mZN = 115

(2x-25)+2x=ll5

- 25 = 115, 4x = 140,x =35 mZM = 2x - 25 = 2(35) - 25 = 45'/ mZN = 2x = 2(35) =70'/

4x

NowYou Try

It u-6)"

Find the measures of ISRQ ISRT ZQand (Hinc Form a "system" of equations.)

lS.

\(/hen a problem gives a hint, always take it.

A$,

.t

a Vocabulary Checkup

1. Scalene triangle.

2. Obtuse

Give the definition and sketch an example for each of the following terms 4. Acute triangle.

triangle.

3. Equilateral

5. Right triangle.

triangle.

6. Isosceles triangle 50

.

Parts of Triangles

-

Be Sure to Memorize These Terms!

ASp

MBomN: A median is a line segrnent that goes from a uertex of a triangh Median -+ Middle to the middle of the opposite side. One Thiangle

Rules for Medians:

-

Three Medians

1. All triangles have three medians. 2. Medians stay insidr the triangle. 3. Each median divides the opposite side into two equal Parts 4. A median does not necessarily divide the triangle into two ll congruent triangles (it depends on the triangle). 5. A median does not necessarily divide the vertex angle in any particular way (it depends on the triangle.) 6. The medians of a triangle are not necessarily equal in length (it depends on rhe triangle). 7.The 3 medians of a triangle intersect in one point.

Aruruon:

2

un2

E

D

AE =6 BD =8 EC =13

Altitude --- Pelpendicular An abitude of a triangle is the petpendicular line segment that goes from a uertex to the line that includes the opposite side.

Rules for Altitudes: 1. All triangles have three altitudes. 2. An altitude doesnt necessarily divide the side into cwo 2 congruent parts or the triangle into 2 congruent triangles (it depends on the triangle).

AAAI:ilr,l,*i:;

un2

3

4 ,6

\When working with altitudes, pay attention to whether the triangle is acute, right or obtuse: Acute Tiiangles:

All 3 altitudes are in the interior of

F

the triangle.

Right Thiangles: The two legs of a right The 3 altitudes of a right triangle: triangle are ular each leg is the altitude to the other leg.

Obtuse Thiangles:

The 3 altitudes of an obtuse triangle.

The definition of an altitude is the perpendicular segment t0 the line that includes the opposite side. This is necessary for obtuse triangles because an altitude is

perpendicular. First extend the side of the triangle and then draw the altitude to that line.

51

Remembering that each leg of a right triangle is an altitude, will make some problems much easier.

i$o

Notice that one altitude is inside and two altitudes are outside the triangle.

Theword Congruent \7hat

does

it mean for two figures to be congruent? The symbol for congruency

the two parts of the symbol:

,., The squiggly sign on top

:

is

=.

Pay attention to

means that the figures have the same shape.

The equal sign on the bottom means that the figures have the same size.

Putting this together tells us that congruent figures have the same shape and the same size.

If mo figures

have the same shape and the same size,

it means that one is a perfect copy of the other,

That is, they are exactb alike,

-When

two figures are congruent, their corresponding parts and angles will be equal. But itt important to orient the figures in the same way because, when figures are oriented properly, you are less likely to compare parts and angles that are not corresponding and are not meant to be set equal to each other. For example, study the C figures on the left. The congruency sign tells us that the two triangles are congruent. However, it is best to take a few seconds to re-draw the second 6gure so that both triangles are oriented similarly. This is especially true for figures that are easy to sketch, such as triangles. Be B sure to carefully place identifying letters in the correct AD positions. See the re-drawn figures on the right. B E Order Counts A D

;[so

Thiangle ABC is congruent to triangle

DEF. In symbols, we write:

\ABCYADEF This simple congruenrr is giving us 6pieces of information:

ZA=ID

IB=IE

ZC=IF

AB=DE gC=EF

Now study the order of the original congruence:

CA

FD

L(N = A.r^rA Order counts! ---J

By the way, it would be equally correct to say that LBCA= LEFD, LCAB LFDE, LACB = =LDFE, LCBA = LFED, LBAC = LEDE and so forth. The important idea is that corresponding parts must be listed in corresponding positions. 52

Congruent Triangles parts: three sides and three angles. tWhen we say wo triangles are congruent, we mean that each of the six parts of one triangle is equal to the corresponding part of the other triangle. And remember, order counts.

A triangle

has six

C

F 31

The two triangles on the left are certainly congruent. Each corresponding part is equal. Stated in symbols:

3L

25

25

)7

B

A

)7

D

LABC = LDEF

E

\7e know that the above congruence means:

/.A=ID IB=IE IC=IF AB=DE NC=BT CA=FO A lot of problems in geometry involve trying to figure out if two triangles

are congruent.

If you

can

show that each of the six pairs of corresponding parts are congruent, you've accomplished the task. However, usually you are not given that much information for a problem and, fortunately, you dont need to check all six pairs of parts to prove that the triangles are congruent. There are 4

dffirent

metbods you can use to prove that rwo triangles are congruent.

I

SSS

= SroB SrpB SroB CoNGRUENCy PosruranB If three sides of one triangle are congruent to three sides of another niangle, the triangles are congruent. Side Side Side Congruency Postulate is true because if the sides of two triangles are the same lengths, it turns out that their corresponding angles are also equal. Since all six parts are congruent, this means the two triangles are congruent.'Sfhy is this so? Because, if you are given three lengths and

told to make a triangle, there is, at most, one triangle you can make. Heret an example: Given three segments with lengths 2, 3 and 4 respectively, construct as many triangles as possible:

2

w 4

4

You can put the sides in any order, fip the triangles over, rotate them, it doesnt matter, you always get exacdy the same triangle, only the orientation is different. All of the triangles on the right are congruent by SSS=.

53

4

4

4

3

"lJsing" SSS= means that if you can show that two triangles have three pairs of congruent sides, you can state that the triangles are congruent based on SSS = Postulate. Here are some examples of pairs of triangles that are congruent based on SSS =

t2

5

t2

13

1

t2

+-+

13

1

\7here is the third pair of congruent t2 sides in each figure?

lq parer{s aPrs uoruruof JrlI :Ja.&\suv 'say8uern qloq

Each letter in SSS = stands for one pair

of consruent corresponding

Itt

A

sides.

Think of it like this:

SSS

C

habit to check off each letter of the postulate as you find the pair of equal parts that the letter stands for:

D

AB =DE S,/ BC =EF S,/ CA =FD Sy

Naming the Congruency: Order Counts! You must name the congruency by pairing up congruent parts in the two triangles. Choose a path in one triangle and follow the same path in the other: A

B

a good

26

26

C

t4 34

34

ACBA = AFED

D

to "check offthe letters" on the following postulates and theorems that prove that triangles are congruent. This is especially true when you are working on complicated proofs. Adjacent means "right next to each other".If any three adjacent parts of one triangle are congruent ro three adjacent parts of another triangle, the two triangles are congruent. This fact leads to rwo additional congruency postulates:

It

also makes sense

I Sro E

SroB CoNcRUENCY

SAS !I

II

Posrulern

If nro sides and

the included angle of one triangle are czngruent to two sides and the included angle anotber niangle, the tu.n triangles are congruent.

of

Study the figures above. Thiangle I has a side, the adjacent angle and the immediate next side equal to the corresponding parts of tiangle II. \Mhenever you see this relationship in a problem, you know that the triangles are congruent by SAS Congruency Postulate. Although it's not part of the name of the postulate, remember the important word "included". The angle mustbe berween the rwo sides that make up the congruent pairs (of sides) for this postulate to work. Here are some examples of pairs of triangles that are congruent based on SAS=:

t3 t2

r3

t2

5

Each letter in SAS = srands for one pair of congruent corresponding parts. 54

R

I

T

U

m ANcrB SroB ANcre CoNcnusNcv Postur-arn

If

ASA=

two angles and the included side of one triangle are clngruent t0 two angles and the includ.ed side of another

triangle, then the two triangbs are congruent.

Here are some examples of triangles that are congruent by ASA

=:

Mark the second pair ofcongruent angles.

Each letter in ASA= stands for one

pair of congruent corresponding parts ln the introducdon to this postulate. And yet it is a You may have noticed that there is no part of the postulate, so it must be important.'S7hy is it missing? To find out, we need to re-examine the way triangles work. If rwo angles of one triangle are equal to two angles of a second triangle, what else must be true? Think of it like this: the interior angles of each triangle total 180". Subtrac the measure of the two angles that are in both triangles and what is left? For example:

AI

^ So the

ll

? = 180o -

43"- 36o = 101"

? =180o-43"

- 36o = 101"

third pair of angles are also equal. This leads to the next theorem.

Tnnonru: If tuo angles of two tiangles

are equal, the third angles ltaue no choice but to be equal.

It might help you to remember this theorem if you think of it

or the no r:, the letters ASA stand for: angle, choice theorem. Now, re-examining our newest postulate, ASA adjacent side, adjacent angle in that order. However, given these conditions, we now know that the third pair of angles are also equal,

MW

+

as the third angle theorem

bM

which in turn leads to the fourth method for showing that rwo triangles are congruent.

ry ANcrr ANcrn Sron CoNcRUENCy TirronBlut: If two arugles and a non-includrd sid.e of one *iangle

AAS dre congraent to the conesponding two angles and

conesponding non-included side of a second niangle, then the tuto triangbs are clngruent. 55

=

Here are some examples of triangles that are congruent by AAS non-included sides are congruent.

Each letter in AAS = stands for one pair of congruent corresponding parts.

:. Notice that the corresponding

\Where is the

-

4

side?

Matching the Problem to the Correct Theorem - ASA ! and AAS are like first cousins, however, = in a proof you need to name the correct theorem, so we'll compare them: Tiiangle ABC is congruent to DCB based on ASA ! because thatt what's in the picture: each triangle has a 62 angle (A y), then side ,BC (S l), and then a T6 angle (A /). This means, the picture

B

A

D we were given and the postulate that we have chosen, match.

C

O

tiangle MNO is congruent to triangle POlVbased on AAS

=.

Remembet AAS means equal pairs of: P an angle, an adjacent angle and then a side M Studying the figure, in each triangle we have N a34" angle, then an 81' angle and then side NO A (a common side), that is, A y A / S /. Being sure that the reason we provide (in this case AAS !) matches the information in the problem is a very important part of doing proofs correctly. Naming the congruency, we could say LMIVO A?Ol/by AAS = =. This correctly orders corresponding equal parts and names the correct theorem.

B

D C

ASA

o

Yefsus

AAS

P

N

And now to answer the question: \7hy isn't it important to stress that the side must be included? if the side is not included, the triangles are still congruent based on AAS = (not ASA =). This is not true for SAS =. There is no SSA = theorem (because it doesn't work for all triangles which a qeneral . congruency theorem must do). \We will give a counterexample later in the book which disproves ffi =

Because,

Getting Ready for Proofs about Congruent Triangles Example: Decide if the two triangles below are congruent. If the triangles are congruent, name the congruency and state the method which proves they are congruent.

AD 1

17

I

17

B 7 C E7 F

Studying the problem, we see the three pairs of equal sides so we know the triangles are congruent based on SSS =:

Remember, each

AABC > ADEF lerter of SSS represents one pair of equal corresponding

sides: an oe = 56

(ss

sc

=

EF s(s

ca

=

ro

ss(

Proving Triangles are Congruent But to use any of the 4 ASA or AAS To prove triangles are congruent you can use SSS SAS =. =, =, =, methods you must first show that pairs of corresponding parts of the triangles are equal. To find pairs of equal angles and sides, always look for: Recurring Patterns

-

l.

Shared sides,2. Vertical angles,3. Information from the parallel line methods (especially PAI.ACJ, and, 4.Information from definitions.

** l*'p

\7hen you've added as much information as you can find, study what you've shown and pick the right method from the four, the one that matches what you see. How You Do It

S7hat You Do

1. Find pairs of eq :ual Z's and sides Mark up the figure. \What do you

1. By the GooD Poirils. Choose the method that matches the picture.

see?

2.By

2.StateA=A.

SSS, SAS,

ASAoTAAS=.

Now You Try It Search the triangles below for additional pairs of equal parts (state your reasons). Then, name the postulate or theorem which proves that the pair of triangles is congruent. 1

w

2

&

3

@

Formal Proof

-

uVu

4

RS Iis the midpoint of RV.

Moving to formal proofs is not difficult. Heret an example:

N.

tWO Given: MOIIPQ = ""d AQ /P. Prove: LMNO = \We need to prove that pairs of parts of the 2 triangles are equal. The 7 prove, in this case LMNO = AQN'4 is always true and conveniendy lists for us, exacdywhich pairs of parts are equal (lM lQ, IMNO = = Z-QNP, ZO lP and so forth). As to how to prove that the equal parts are = a equal, the arrowheads should make us think about the parallel line methods 2 and, in fact, PAIAC tells us: ZM = lQand, ZO = ZP Nso,we were given tttO = pQ 17 Herc's the figure marked up to show the congruent parts. Compare the figure to the 4 congruency methods. \X4rich do you see? The method that matches is ASA=.

P

t7

&$,

o

I o

6

5.

t/rt/r/

tM = ZQ ASA MO = PQ ASA tM = lQ

Proof: Statements t. MollPa. 2.lMs lQand ZO x lP. ,Q. 3. Mo = AQI/P 4. LMNO:

Reasons

1. Given.

2. PAIAC. 3. Given. 4,

ASA 57

(Stmts. 2 and 3.)

ASA

Remember, you cant use one of the 4 r methods unless you've accounted for each of the 3 letters in the statements of the proof.

Now You Try

It

A

Given: ABIIDE Cis the midpoint of ,4E "nd Prove: LACB LECD. =

tWhen you are given Cis the midpoint of AE, you are pposed to respond with:

-

ac=ca

C

Proof: Statements

Reasons

E

e[ ofa

'Ihe 4 methods for establishing congruency work for all triangles. But, for right triangles only, there is an additional theorem which proves that two (right) triangles are congruent:

HL

N

HvporBNusE LEG Tsronrrra : If the hypotenuse and leg of one right triangle are czngruent t0 the hypotenuse and leg of a second right triangle, the two triangles are congruent.

HL is the usual abbreviation for this theorem. Flere congruent based on HL:

2

are some examples

of right triangles that

are

2

8

\Where are the r: hypotenuses ? 'aPrs

\7here are the rg legs? 'aPIs uotuuof, eql :Ja./tasuv

uoluruof, aql :Jaasuv

Remember what the name HL stands for Hypotenuse Leg. Rules for using HL: 1. Only use HL for right triangles (only right triangles have hypotenuses). 2. ln a proof, you must be given or show that the triangles are right, in order to use 3. \Mhen a proof involves right triangles, try using HL first. 4. 'Ihe two hypotenuses must be equal to each other or you cant use HL. For example: These two triangles are congruent but the

-

congruency is by SAS >, notHL.

58

HL

It

Now You Try 1

.

BD L AC . Given: AA = CA Prove: LABD LCBD = ^rd B Statements Reasons

ADC 2. Given: tWO

l/

=

o

PR,

ZM = ZP, NO r MO

and

@

t

pA.

Statements

Hint you are given that -\Mhen gD t AC , you are supposed to respond with ZADB is a rt.Zk ICDB ts a rt.l(Def. of r lines.)

Prove: Reasons

LMNO = APQR

2

M 5 O R 5 Quick Review

l.

-

P

To prove that two triangles are congruent:

Use one of the 4 Congruency Methods for all triangles: SSS

r:, SAS ri, ASA !:, SAS ry or

a

fifth, HL, for right triangles only.

2. Use other right triangle congruency theorems only if they are given by your teacher or textbook. Please

.Ways

note: Do not make up theorems. There are no other congrueng/ theorems that work for all triangles. Not to prove Congruency:

AAA

\Mhy not? These two triangles have three pairs of equal corresponding angles and are definitely not congtluent.

n

f

SSA

\Mhy not ? These two triangles have equal pairs of sides, the adjacent sides and the adjacent angles, and yet the triangles are defi nitely not congruent.

10 8

*

8

0

And remember, although we could provide an infinite number of counterexamples in both of these cases, even one counterexample disproves an hypothesis. 59

Using Congruent Triangles to Prove Other Things Here are 5 important ways in which congruent triangles are used in theorems, problems and proofs. Pay attention to the pattern: 1. Prove triangles are congruent. 2. Use the definition of congruent trianglestoprove other facts.

I. A DnrrNrrroN or CoNcnuBNt TnreNcrrs: CPCT Corresponding Parts of Congruent Triangles are Congruent By definition, if 2 triangles are congruent, every corresponding pair of parts of the 2 triangles is congruent. For example:

t. lA

=ZD 2.lB =lE YZF

+

3.

FA

ZC

4. AB 5.

N

BC

6. CA

N

DE EF FD

F C D CD This means that if you can show that2 triangles are congruent, you can then state that aryt pair of corresponding parts of the triangles are congruent, including those that werent mentioned in the proof of the congruency of ,h. triangles. Heret an example Given: AC I EC and BC DC. A

Prove:

=

AA=OE.

n r o Proofl Statements I. AC O EC, BC = DC.

A

D

Reasons

1. Given.

2. ZACB = ZECD. 3. AACB AECD. = 4. AB DE.

2.VNf .

=

B

E

C

B 3. SAS =. 4. Corresponding parts of congruent triangles are congruent.(CPCT)

Use

this to A

LACB

=

LECD

A

D

AB>DE

B

D

B F-

E

You cannot use CPCT until AFTER you have proved that the 2 triangles are congruent. This means that you must first prove A ! A, then you may use CPCT. Learn the pattern:

WhatYou Do 1. Find pairs of equal Mark up the figure. \What do you

HowYou Do

l.

and sides.

see?

2. StateA=A. 3. State side ri side or

It

Use the GooD Poinls. Choose the method = that matches the picture

2.By SSS,SAS,ASA AAS

l> Z

3.By CPCT. 60

=

or HL.

E

II.

Proofs Using More Than One Pair of Congruent Triangles.

Given Prove:

AB= AC, BD = CD aE= CE

E

First think through the proofi The goal is to get one of the pairs of triangles that

I.

"See" the pairs of triangles that you think are congruent:

include BE^nd "into" the proof.

2. Mark up the figure with the given information and anything else you can figure out

3. Note thar

LABD and

LACD share parts with LABE and LACE. E This will help us to

@O = 1D, Reflexive)

Studying the figure, we see that we can prove: LABD = AACD by SSS =

prove that the larger triangles are congruent

C

Cf f .ll

B

E C

5. Now "see" only the A larger triangles,

4. Look for parts shared by both pairs of triangles, AB = AC, (Given) and IBAD ZCAD, (by CPCT). C

E

AE =,,4E(Reflexive), so LABE LACE =

E

C =). Note that in these problems, you work your 6. Finally, we are ready way from one pair of triangles to another pair, to state, gE CE by = E the ones that contain the parts you need to A CPCT. pfove are congruent. C Learn the pattern First prove triangles are congruent, then use CPCT

=

-

+ A=A AABD = AACD

(by SAS

CPCT ---->

proofi

D

B

A C

A

E

A

E C

Statements

Reasons

1. AB

CD. = 49, BD = P AD AD. 2. 3. AABD = AACD. 4. /.BAD= ZCAD.

Given. 2. Reflexive Prop. (Note, we could have 3. SSS 1.

4.

5.8=AE.

5.

6. AABE = AACE,

6.

=

BE ! CE E

A

A

7. BE

CPCT

IBAD = Z.CAD AABE = AACE B A

Here's the formal

€

A=A

CE,

=. CpCT. Refexive. SAS =.

5. CPCT. 6t

proved LBDE LCDE, = but to do so would have taken more work.)

Perpendicular Bisectors and Angle Bisectors Proofs of these important theorems are exercises on the following page.

III.

P E R P E N D

Perpendicular Bisectors

I

DerrNrrroN: A perpendicular bisector

is a segment, ray or

c

line

U

L

A

uhich is perpendicular to a segment at its midpoint. (Remember, a midpoint divides a segment into 2 congruent pieces.)

A

I

S E C

Rr

There arc 2 important theorems about perpendicular bisectors:

TlrBonru: EVERY ?oint on the perpendicukr is

B

B

I

T

o R

bisector of a segment

equidistantfom the endpoints of the segment. For example, given that segment RS is a perpendicular bisector of MN, euery point on RS is equally distant from point M and point.A[ For example point C is 25 units from point M and 25 units from point N.

M

Ar

s.

And the converse

THeoREL,r: Any point that is equidistantfrom the endpoints 0f a segment lies on the perpendicular bisector of the segment. For example, since Wis equally distant from points A and B, W lies on the perpendicular bisector of AB. A

* A

IV. Angle Bisectors

DrrrNrrroN: An

angle bisector is a ray that diuidrs an angle into two congruent adjacent angles

There are 2 important theorems about angle bisectors: Ti.rnoRErrr: then the

For example, since

s

If apoint is on the bisector of an anglr,

point

is

equidistant*from the sidts of the angb.

32

And the converse:

Tnr,onnrra If a point is equidistant*from the sides of an angle, then the point lies on the bisector

s

ZRSU ZTSU, = Uis equally distant from SR and ST. For example, since ooint Uis eouallv L L-----S J distant from .lR ---->

of the angle.

and

SI

ZRSU *Remember, the distance from a line to a point not on the line, means the 62

J-

=

ITSU.

distance:

Fp

NowYou Try It

Every proof on this page can be done using congruent triangles and/or the definition of congruency.

-

To Prove 7$Congruent

& HL for rt N's = l. Co.-onparts. ^

SSS, SAS, ASA, AAS

only.

/ ll t z.Vertical angles. Nl o :. Information from parallel line V f merhods (especially PAIAC). 4. Information from definitions.

1. Given: AC

EC and ZBAC ZDEC. =

=

Prove: AA

=

2. Given: FH and

Prove:

Of

Statements

Reasons

Statements

Reasons

Statements

Reasons

Statements

Reasons

Statements

Reasons

Statements

Reasons

B

.

=JH, GI=

IO, HI

t

GK

I

IFHG z 4HK.

ec=lf. G

3. Given: P is on the perpendicular

P

bisector of @-R. Prove: PQ= PR.

Hint-\7hen Pis on the

r

you are given that bisector of Qn-&

respond with: QS = RS, ,rrd ZQSP is a rt. (Def. of r bis.)

IK

as

R

I

4. Given: SU = SV Prove: S lies on the perpendicular bisector of UV.

U 5. Given: )zlies on the bisector of llWZ, YWI)(W, YZ TXZ. Prove: YW = YZ. Hint-'!7hen you are given that Izlies on the bisector of ZWXZ, respond with: ZIFXY = (Def of Z bis.)

V

I I

a

V/

X

ZZY

6. Given: YW = YZ, yW

W

tXW,

ardyZLXZ. Prove: ZIX/XY

=

ZZXY. 63

Using Congruent Thiangles

-

continued

V. Theorems about Isosceles Triangles

A triangle with at least two equal

) a

. .

Atp

Become an expert with isosceles triangles. They are a powerful tool in geometry!

sides is an isosceles triangle. The

A

angle between two equal sides is

4

Isosceles

tiangle ABC

called the uertex angle.

attention to the technique used to prove the next theorem. By drawing the altitude from the vertex angle of the isosceles triangle, we divide the original triangle into 2 congruent right uiangles. Pay special

* Trrn IsoscELES TRTANGLE Tnponnu: ITT If a triangle has two congruent sides, the angles opposite those sides are also congruent. Although we call it ITI you must call each theorem by the name given by your teacher or textbook. To prove C

ITT

use congruent triangles:

of altitude

Given

ADB

k

Reflexive

ADB

CPCT

D B

A DD

B

+ Trrs CoNVERsE oF rrrB IsoscELEs TRIANGLE TrrponEkr: If a triangb has two congruent angles, the sides opposite those angles are also congruent, To prove this theorem use congruent triangles: Given

Def. of altitude

Refexive

B

Name the right theorem! If slldes are congruent,

If

itt

aB8gles are congruent,

'Working with Isosceles Triangles

IffT.

itt

CPCT

AAS

ADD 4\. -->

D

+

the CoBf,verse of ITT.

Due to the symmetry of isosceles triangles, segment CD (which is drawn from the vertex angle) is an altitude, a median and an angle bisector. And segment CD divides LABC into A D B 2 congruent right niangles. Right triangles are easy to Isosceles Tiiangle ABC work with so knowing this will help you to solveA with AC = BC many different types of problems

As, See an IsoscelesTliangle

+

Equrranrxer Tnreucrrs 3 TnBonsN,Is l. Equikteral triangles are equiangular.

:

2. Equiangular tiangles are equilateral. 3. Each angle of an equikteral triangle equab 60". 64

DD +

-Ih'nk 2 congruent right triangles.

€

B

€

B

6. POLYGONS These are not polygons:

These are polygons:

Er*w

Polygons are: 1. Closed figures. 2.Made up of (straight, line) segments.

3. The segments intersect at their endpoints (they are not allowed to cross one another). 4. Each segment must intersect with exactly two other segments. '$7hen

you

see a

figure, you are expected to know whether or not it is a polryon.

Polygons come in 2 varieties, cont)ex and non-conuex (also called concaue). You are expected to know the difFerence berween the two types. Heret how:

ConvexorNon-Convex? Imaginewrapping arubberbandaroundapolygon. If therubberband touches every one of the polygon's vertices, then the polygon is conuex.To demonstrate, we will test from above each of the six

aex

coruaex

Although you do need to be able to recognize a non-convex polygon, geometry problems are usually about convex polygons. tWhen a problem uses the word polygon, assume it's convex unless the figure or given information tells you it's non-convex. B

Naming Polygons

D

Polygons are named by their consecutive vertices which A are points, and points are named by capital letters. Polygon The letters are usually named in alphabetical order. 65

E

ABCDEF

Z Polygon

XYZ

Polygon PQRS

Special Names for Special Polygons You are responsible for memorizing these terms as well as any additional ones that your teacher or textbook think are important:

Number of Sides

Name

)a

Tliangle

4

Quadrilateral

5

Pentagon

6

Hexagon

8

Octagon

10

Decagon

72

Dodecagon

Example

n-gofis: Another way to name polygons is "number of sides -gon". For example, a triangle is a 3-gon, an octagon is an 8-gon and, speaking in general, we say an"n-gori', which means a polygon with an unstated number of sides. Remember, "z" stands for the number of sides. Diagonals of a Polygon: A diagaonal of a polygon is a segment that goes from one vertex to a non-adjacent (not next to) vertex.

All the diagonals of a pentagon.

DpprNrrroN oF A Rp,cuun PorycoN: A polygon with equal sides and equal angles is called a regular polygon.lhar is, a regular polygon is both eq

ui latera

I

and e qu i angu lar.

Here are some examples of regular polygons: 20

24 24

I

6

24

l

0

t4 4

t4

r6

1

I

14

I

I

) 73

16

.l

)

1

11

3

4

t3

3

1l

Note that if a triangle is either equiangular or equilateral, then it is both. That is, it is regular. This is not true about any other polygon. For example, shown below are polygons that are equiangular or equilateral but not both (and not regtiar): 2 I

42

I

90'

t9

t9 90'

5

B

7

50'1t 17

1

2 8

1

5

66

17

Finding the Sum of the Interior Angles of Polygons The three interior angles of any triangle add up to

I7e can use this fact to figure out the

formula for the sum of the measures of the lntenor angles of polygons. Since the interior angles Drawing a diagonal of each triangle add up to Beginning with q divides the 180", the interior angles a quadrilaterd: eral into two triangles. of the quadrilateral must

180"

total: 2x180'=360". In the example above, we used an oddly shaped quadrilateral, but we would have had the same result with any quadrilateral Since the interior angles Drawing the diagonals 180" of each triangle add up to Now we'll try a from a single vertex 180', the interior angles pentagon: divides the pentagon into three triangles.

of the pentagon must total:

3x180"=540". Since the interior angles of each triangle add up to 180", the interior angles of the hexagon must total:

Drawing the diagonals from a single vertex divides the hexagon into four triangles

Now we'lI choose a hexagon:

4 x 180' =720". Heret the pattern: Each increase in the number of sides produces one additional triangle, which adds 180" to the total measure of the polygont interior angles. Heret a table organizing our findings: Total

Number of Number of Sides

Calculation

Tiiangles

IxiUO 2x180 3x180 4x 180

I 2 1

Finish the last four rows of the table. It will help you to understand and remember the pattern.

(

4

of

Interior Ansles 180

360 740

720

7 9 10

In mathematics, we like to use formulas that are concise and correct for all possible cases. And a formula should include a variable to represent the quantiry that is changing from case to case. In our example, it is the number of sides of the polygon that is changing. Notice that the number of triangles is always 2 less than the number of sides. So the correct formula is: (number of sides -2) (180)

Letting n stand for the number of sides, the formula becomes (n

-

2) (180) = the sum of the measures of the interior 67

ofa

with n sides.

Heret an example: Find the sum of the measures of the interior angles of

a dodecagon

A dodecagon is 12 sided, so n = 12. (n

-

t2

(/-

2)(180) = the sum of the measure of its interior angles. 2)(180) = (10)(180) =

1800'\,/

B

Heret another example: Find the sum of the interior angles

of

rectangle ABCD:

-

(n

- 2)(t80) = (tr- 2)(180) = (2)(1so) =,

Since the sides are peqpendicular (which means they meet to form 90' angles), we can confirm our solution: 4x90 = 360

2)(180) = the sum of the measure of its interior angles.

4B

D

A

A rectangle is a quadrilateral (4 sided), so n = 4 (n

C

)o

C

360"'/

D

- 2)(180) might be used in a problem: of ZM,IN,ZO,ZP and lQ

Heret an example of how the formula (n Find the measure Step

l.

Count the number of sides (find n).

Step 2. Use the formula (n

- 2)(l8q

x o

6

o

x+ 13)" (7x+

rl=5

()

= - 2)(r8o) = (3)(180) = 540. Step 3. Since 540 is the total sum of the interior angles, set the angles equal to 540: 5x + (8x + 4) + 6x + (7x + 13) + (7x + 13) = 540 Step 4. Collect like terms and solve for x:

34x+30=540,34x=510,

x=15

Af,,

Step 5. Go back and read the question and answer the question you were asked,o

ZM = 6x, =6(15) =90'/,

lN

ZP=7x+3= 7(15)+13= NowYou Try It l.

,rr?-Bt'

= 8x + 4= 8(15) + 4 = 124'/, Z.O= 6(15) =90'/, 118 =7x+13= 7(15)+13= 118/.

'/,andlQ

Find the total sum of the measure of the interior angles of a 100-gon

2. Find the measure of the smallest and largest angle of polygon PQRSTUVW.

R.t (1

13)"

(9x+

06'

0xff P

W 68

l1

Y

02x+

V

T U

Going Backwards: Finding z W'hen you Knowthe Sum of the Interior Angles Example: How many sides does a polygon have if the interior angles @al1440"?

In this problem, the answer to (n - 2)(180) is given. Carefully distribute

So set (n

(n-2)(180)=t440 each term of the expression (n - 2): n(180) - 2(180)= t440 l80n-360=1440 Balance the equation:

2)(180) equal to the answer:

+ 360 + 360

l80n = 1800 l0 Divide bvJ the coefrcient of n: !fin -fiOA JJ W-W

n=10

Finding the Measure of One InteriorAngle of a Regular Polygon Remember that the word regular means equal sides and equal angles. Since the angles are equal, divide the total measure of the interior angles by the number of angles to find the measure of a single angle.

Heret a compact formula to find the value of 1 interior angle of a regular polygon (n-2)r80 n

Heret an example: Find the measure of each interior angle of a regular pentagon First, find the total of the measures of the interior angles Then divide the total by the number of angles:

(5

-Z) 180 = 540

ff=rca"

The following example combines two concepts: How large is one interior angle of a regular polygon if the sum of its interior angles is 1800?

In this problem, the answer to (n - 2)(180) is given, so set (n - 2)(180) equal to the answer: (n-2)(180) = 1800 l80n - 360 = 1800, l80n = 1800 + 360

l80n=2160,n=12

Since n = 12, and the polygon (which we now know is a dodecagon) is regular, we know that the 12 angles share the 1800'equally, which means the measure of each angle = 150.r/

is: ,iry

NowYou Try

It -

If the sum of the interior

angles of a regular polygon

one exterior angle of the polygon.

69

isl260, find the measure of

Exterior Angles of Polygons

Quick Review - Exterior angles of a polygon are formed by extending a side of the polygon. Start with one side, extend it and measure back to the polygon. If you need more than one exterior angle, work your way around the polygon in one direction. Draw only one exterior angle for each vertex. Exterior angle

#1=747"

Exterior angle

#2=t23' Exterior angle

#3=90

ForAll Polygons: nlm ber of sides = zumber of interior

0

For Regular Polygons:

@

angles = number of exterior angles

SIDES ARE EQUAL

rNrunroR ANGLEs ARE EeuAL

@nxrrruon ANGLES ARE EqUAL Finding the Sum of the ExteriorAngles The interior angles of a triangle total 180". The exterior angles of the triangle above total:

147"+123"+ 90" = 360" The interior angles of a quadrilateral rctal (n exterior angles? Here's an

example:

-

2)180, or (4-2)180 = 360". Butwhat about the r

10

r 88o

74" + 107" +91" + 88" = 360" So the exterior angles of the above quadrilateral total 360'. The interior angles of a pentagon total (5 - Z1 x 180 = 540" . Checking the exterior angles:

57"+ 99" + 42+ 95" + 67" = 360" So the sum of the exterior angles of the pentagon above equals 360".

In fact, the sum of the exterior angles of any polygon is always 360"1 Note: As z increases, the sum of the interior angles u But, as n increases,

nt nl

the sum of the exterior angles stays the same.

70

t

then interiorZ rr'rm thrn exterior sum= 360o

I

Tnnonrrr:

7be Sum of the Exterior Angles of a Polygon =

360"

The above theorem is especially helpful with problems involving regular polygons. Since the exterior angles of a regularpolygon are equal, they share the 360' equally:

360"

n

-

369" exterior , =' tr'. , '*,of a regular polygon ^;;f;;i;",r"s"iiip-ity[i'i, angle

rhe measure of one

An understanding of exterior angles can help solve many polygon problems. Here

are some examples:

1. An exterior angle of a regular polygon measures 30". Find the measure of an interior angle. a single vertex of the polygon drawing in the exrerior angle. t ' Since the interior angle is supplementary to the adjacent exrerior angle: sAS,{*Sk.rch

30'+xo=180" x = l50r/

2. An exterior

angle of a regular polygon measures 30'. How many sides does the polygon have? Since the sum of the exterior angles of a polygon is always 360', and since all of the exrerior angles of a regular polygon are equal, the 350" is shared equally among the exterior angles. So, if each exterior angle is

30'the polygon

has:

= 12 sidesr/

# 3. An interior

angle of a regular polygon measures 140'. How many sides does the polygon have? Sketch a single vertex of the polygon drawing in the interior angle. Since the interior angle is supplementary to the adjacent exterior angle:

140"+x"=180"

x=40

xo

Since the sum of the exterior angles of a polygon is always 360', and since all of the exrerior angles of a regular polygon are equal, the 360' is shared equally among the exterior an ifeach exterior angle is 40" the polygon has:

Now You Try 1.

It

W

If an exterior angle of a regular polygon

= 9 sidesr/

measures

72",what is the total measure of the interior

angles of the polygon?

2.If

the interior angles of a regular polygon total 1520", find the measure of one exrerior angle.

3. Polygon ABCDEFGH is regular. Find the measure of

A

71

lI.

7. Quadrilateral Parallelograms B

C

Like all quadrilaterals, parallelograms are named by their vertices. The symbol for a parallelogram is E. The figure at right is EABCD.

4

F^rrrYou Need To Know About Parallelograms:

D

A

1. DrrrNrtroN: Both pairs of opposite sides are parallel. 2. TrrBonrvr: Both pairs of opposite sidrs are equal. 3. Tnronnu: Both pairs of opposite angles are equal. 4. TnsoR-Elur: The diagonals diuid.e each other in half

8

ffi

Thke a close look at the last theorem. Diagonals of parallelograms arent necessarily equal. rrue is that the diagonals break each other in half, that is, they bisect each other.

-What

is

a certain type of object, we say those Properties - In mathematics, when things are true about things are the ?ro?erties of that object. The definition is not a property, the definition is what defines the object. The theorems are the properties of the object. Given a parallelogram (a quadrilateral with rwo pairs of parallel sides), the other three facts (the properties) are always true.

In a problem, you will know a figure is a parallelogram either from the 6gure itselfl or by the word parallelogram or by the symbol o. Remember, if a figure is a parallelogram, you know a lot about it. Always think "b"ut}r* 4 facts and add any information to the figure to show those facts that are connected to a problem. For example:

@

EMNOP,

on the left, we can add the information shown on the right to the original figure

Given

M|OPMIOP How did we figure out that ZM and ZO = 50"? There are two ways: l. Like all quadrilaterals, the interior angles of z MNOP add up to 350" (Remember, (n- 2)180). This tells us that ZM and lOtogether share 100' (360" - 130" - 130" = 100").Since opposite angles of parallelograms are tWhen a transversal cuts two pardlel lines, equal, lO and lM must both equal 50'. 2. **--/

uinssls,samesideinterior",gl.,aresupplementary.ffirao. 4fi, So, consecutive angles of a parallelogram are supplementary! It Given EGHIJ on the right, add the correct measurements for the following: HI HG wIGHI

NowYou T"y

ruZ.HIJ

ruZ.HGJ

7')

GI

I G

Proving the Properties of Parallelograms

If a quadrilateral

-

A Detailed Look at Proof

two pairs of parallel sides, the other parallelogram properties are always presenr. And since those properties were given as theorems, we must be able to prove each of them based only on the figuret two pairs of parallel sides. Lett investigate: has

\7e will "use" congruent triangles to prove all three theorems

Given: ERSTU. Prove: The opposite sides of

a parallelogram are congruenr.

W

and

57 = RU.

(FiS. 1).

Our

reason is,

For the given figure we need to prove that S? 'We

T

.t

will begin by drawing diagonal

nf

=

R

u

T

two points determine a line.

Sf ll

AUand S? llTu,our rransversal theorems should .o* to mind. There are no angles on the outside of the figure so we should think about alternate interior angles. Studying Figure l: ruISTR = wZURT and m ISRT = wIUTR by PAIAC (Fig. 2). Since

Fig.

1

U

T Studying Fig.2, we see that the diagonal creares 2 triangles. the triangles are congruent, we'll have our "prove". That is: .9R

3

=

TU andS7 = nU

If R

Ay CPCT.

But in order to prove congruency wernust use one of the four triangle congruency methods. Since RZ is a side common ro 2 both triangles, we can state that it is equal to itself by the Refexive Property and mark it. Now, comparing Fig. 3 to the four R congruency methods, we see that ARIS LTRU by ASA

Fig.3

R

Fig.4

Remember, you must first prove the triangles are congruent.

cpcT --+Then you may use CPCT.

PAIAC will help solve many parallelogram proofs and problems.

73

U

T T

Having proved that the two lrrr"g]S are congruent, we can now say that SR TU and S7= RU because corresponding = parts of congruent triangles are congruent (CPCT) (Fig. a).

@m_

U

=.

=

4

Fig.2

U

Given:

DRSTU. Prove:

T

s

Opposite angles of a parallelogram are congruent.

ruZSTU. Using the R U LTRU, same reasoning (the same steps) that we used in the previous proof we arrive at ARIS = which proves that ml6R = mlRUTby CPCT This completes one-half of the proof. Continuing with the steps of the previous proof we also found that: mZffit = wZWIR and

'We

need to prove ruITSR = ruZR(JT and ruZSRU =

ZUtrf = wZSE'N, The addition properry of equaliry allows us to add the left sides of two equations R and set them equal to the sum of the right sides of the two equations:

mlWt +mZURi =mZ$W+m Keeping in mind that angle to make the

rs

which

T

s + R

mISRU

to

SRUand

the two pieces of each

ISTU

.T

T

R

U

R

T

T

+

=

R

anm we

=

U

R

U

U

mlWR + ml$iW = wISTU mlSffi,jt + mlUR'J = wZSRU The reason for the last statement is the Angle Addition Postulate. And finally, since we have already shown that the left sides of the above two equations are equal, we can state that the right sides must also be equal, mlSR[J = mlSTU, by the Substitution Property. Notice that we also could have completed the last half of the proof by simply arguing that since mZTSR = ruIRUT that is, one pair of opposite angles were proved equal in an arbitrary parallelogram, any pair of opposite angles are equal. NowYou Try

It

Supply the reasons to prove the diagonals of a parallelogram bisect each other. ABCD. Prove: ^BD bisects AC; AC bisects BD. Given:

E

^:f81',

Statements

Reasons

I.

D

ABCD.

1.

2.

IAEB = IDEC.

.,

AE oC. nym' 4.t. BCllao'AEllDC.

3.

=

5.

IDBA= ZCDB.

LAEB = LCED. 7. AE = EC; BE = ED. A , -rv?.vc 8. E is the midpoin t of AZ;

Kffi'

"'t\r--,

4. 5.

6.

6.

7. E is

8.

the midpoint of BD.

zB>S&iY,c 9. BDbisects AZ;

AT bisects BD.

9. 74

C

A

Proving a Quadrilateral is a Parallelogram

A quadrilateral with two pairs of parallel

-

AVery Close Look at Proof

But if certain other conditions are met, could a quadrilateral be a parallelogram even though we aren't given that the sides are parallel? Notice that we are going backwards, we are starting with a quadrilateral and asking, sides is, by

definition,

a parallelogram.

is the figure a parallelogram?

\Mhat if we are given a quadrilateral with two pairs of equal opposite sides? Is the figure a parallelogram and, if so, can we prove it? That is, can we prove that both pairs of opposite sides are parallel as well as being equal?

l/

o

M

P Fig. I

If we

are going to prove that lines are parallel, we'll need the transversal theorems since they are our only theorems that deal

with parallel lines. But which of the rwo groups of transversal

s tr ? 2 =180o

theorems do we want?'We need to get the order right. That is, given certain conditions, we want to be able to prove that pairs of lines are parallel. Recall that the transversal theorems that endwith a "p" are the ones that do this. Now we need to findout exactly which one of the group we need.

z

The angles in Fig. 1 are in the interior of the figure, which makes CCP unlikely; the problem gives no angle measures which eliminates SSISP This leaves CAIAB which says that if mo lines are cut by a transversal and alternate interior angles are congruent, then the lines are parallel. Diagonal UO 1fig.2) serves as a transversal. Looking at Fig. 2, we see that we have created two triangles, and that they share MO . We mark MO eqtal to itself by the Reflexive Properry (Fig. 3). Figure 3 has rwo triangles that are congruent (by SSS and, therefore, three =) pairs of corresponding angles congruent by CPCT (Fig. 4). a NMO ZPOM = tells us MN ll ro Ay CAIAP. lNoM lPMo tells us ,n/o ll Mp, also by = CAIAP These two statements prove that MNOP is a parallelogram based on rhe definition of a parallelogram, a quadrilateral with two pairs of parallel sides.

Tnsonrn: A quadrilateral with two pairs of equal opposite sides is a parallelogram. Here are two examples 18

_+

1

7-->7 18

7

r8

Remember, once a theorem is given, you can use it in any subsequent (later) proofs.

75

P

o Fig. 3

P

o Fig.4

Here is the theorem we just proved:

&ff,

Fig.2

P

There are 3 other sets of conditions that guarantee that a quadrilateral is a parallelogram. Trrnoruu: If both pairs of opposite angles of a quadrilateral are + congruent the quadrilateral is a parallelogram.

Given: Quadrilateral GHIJwith

mlG = mll

and

mZH

=

ml]

Prove: GHU is a parallelogram.

G

Statements \. mlG = mll and m/.H = mZJ. 2. mZG + mZI + mZH + mlJ = 360". 3. Z(nZG) + 2( mZH) = 360";

2(mlG) * 2( *ln = 350'. 4. mlG + mZH = 180'; mlG 5.

qll

rur; GH ll

tr

+

mlJ

Reasons

1. Given.

2. Z's of a quadrilateral total360". 3. Substitution Properry of Equaliry

= 180'.

of Equality.

5. ssISP. 6. Definition of a parallelogram.

6. GHIJ is a parallelogram.

Now You Try It

4. Division Property

For the next two theorems, we've given the statements; Trrnoruu: If both diagonals of a quadrilateral bisect eachother the quadrilateral is a parallelogram B Given: Quadrilateral ABCD with AE = EC and BE = ED. A Prove: ABCD is a parallelogram. Reasons Statements

t. AE = {C, ar

-

=

ro.

pair of opposite

3. 4. 5.

quadrilateral are congruent and parallel, the quadrilateral is a parallelogram. sides of a

Special Note: Be careful with this theorem. For this theorem to work, the same pair of sides must be both congruent and parallel.

Given: Quadrilate ral MNOP with MN = pO Prove: MNOP is a parallelogram.

9

and MN ll pO.

o

Reasons

1. Construrt MO.

z. MNllPo. 3. INMO = IPOM. 4. MN = PO.

2.

=

tWO.

5. LMNO = LOPM. 7. NO = UP.

B. MNOP is a parallelogram

130"

50'

1.

tWO

----->

B

Statements

5.

D

2.

5. ABCD is a parallelogram. one

----->

1.

2. IBEA = ICED; IAED = IBEC. 3. LAEB = LCED; LBEC = LDEA. 4. AE = DC, AD = BC. Throns\a: If

fill in the reasons.

a ).

4. 5

6. 7. 8. 76

Two points determine a line

NowYou Try It

- For problems I - 4, find any parallelograms in the drawings. Be sure to thoroughly explain your conclusions. ,O B 3.e

2

1

72

2

IL 5. Find the values of x and y that make MNOP a parallelogram

B)"

Qt +2)"

A

Y

Z (5y+30)'

R

UT

6. Find the values of a and b that make FGHI

(9x-28)'

P

M

4

a

parallelogram.

(a+9)

Gb -4)

(b* 12)

Quick Review- Properties of Parallelograms If you're given a parallelogram, heret what you're supposed to know about the figure: 1. Both pairs of opposite sides are parallel

2. Both pairs of opposite sides are congruent.

C

A

D

ABCD

D

+

'ii";

3. Both pairs of opposite angles are equal 4. The diagonals break each other in half.

Remember, whenever you see the E symbol, you know all of the above facts. Another fact about parallelograms, is that same side interior (consecutive) angles always add up to 180".

W'hen is a Quadrilateral a Parallelogram?

If you can show euen zne of the following conditions, /ou have proved that the quadrilateral is a parallelogram. If you're given a quadrilateral that satisfies euen one of the following conditions you are supposed to know that the quadrilateral is a parallelogram

l.

E

If Both pairs of opposite sides are parallel or

2. Both pairs of opposite sides are congruenr ar 3. Both pairs of opposite angles are eqlal or 4. Both diagonals bisect each other ar 5. One pair of opposite sides are both equal and parallel,

then the quadrilateral is a parallelogram. 77

Special Parallelograms The following figures are parallelograms, and it's important to remember that they have all of the properties of parallelograms, but they also have special properties that are all their own. First we'll define them:

m rln

DrrrNrrroN

ffi-55-

oF A

RrcreNcrn A rectangle is a quadrilateral

with four right angles.

DBrrNruoN oF A Rnol,rsus: A rhombus is a quadrilateral with four congruent sides. Note: The word rhombus has two plural forms, rhombuses and rhombi (rhornbT)

b

uffiu

DrrrNrrroN

oF A

Squeru: A square

is a quadrilateral with

four right

6

it

Based on their definitions,

congruent sides and

angles.

is easy to prove that squares, rectangles and

--+

Use (one of) these theorems:

four

rhombi are parallelograms

---->

The logic diagram below shows the relationship between the three special parallelograms:

All squares are rectangles. Not all rectangles are squares. All squares are rhombi. Not all rhombi are squares. Remember, euery figure in the drawing has all of the properties of parallelograms.

Rectangles

ThBonru:

The diagonak of

Given: Rectangle

GHfl.

Construct Ct U

2

HG = IJ,

il

) . GJ=GJ. 4 . ZIGH rs rt., IGJI a

5

. wIJGH

6.

7

= mlGJI.

LIGH = LGI.

. GI=H]

Prove: Ct

=

I

Al

---->

Reasons

Statements 1

a rectangle are equal.

Two points determine a line. G 2. Opposite sides of a D= H 3. Reflexive. 4. Definition of a rectangle 5. All right angles xre =. G

G

1.

is rt

6

SAS

GI=HJ N

Sketch the triangles separately

=. 7. CPCT.

Here is an important rheorem about right triangles, which follows from the previous theorem: each aertex. the midpoint of the hypotenuse is ThBonBv: In a right Many simple calculation AB=CD problems are based yg "* f x=5 on this theorem. Jt'P 2 N's form a rectangle. Diagonals are equal. Diag. ol /-/ blsect. -1.

78

Rhombuses

THBonBu:7he diagonals of a rhombus

are ?erpendicular.

---->

Proof: The proof of this theorem demonstrates why knowing your definitions and properties is so important in geometry. It is also an example of using congruent triangles and then CPCT in a proof

Given: Rhombus ABCD. Prove: BO L AZ.

C

D

A

Fig. I

Looking at Fig. I above, we might not have any idea how to begin, excepr we drawing. Now we'll add some knowledge to the picture: B By its definition, a rhombus has four congruenr sides:

A Since a rhombus is a parallelogram, the diagonals must bisect each other (Properry of a

four triangles in the C

Fig.2

B

D): A

Studying Fig. 3, clearly the four triangles are congruenr by

see

D SSS

Fig. 3

=.

Our goal is to show nO t AZ, and there are theorems which show rwo lines are perpendicular. However, itt probably easier, especially on an exam, to use the definition of perpendicular lines: rwo lines that intersect to form right angles. So we have to figure our a way to show that the cenrer angles are right. \Working on IAEB and ICEB, we see straight IAEC, which we know equals 180'. Since B the two triangles are congruent, CPCT tells us that ZAEB > ZCEB.If the measures of rwo equal angles total 180' (either as a linear pair or using the Angle Addition E Postulate, depending on which method your rext uses), then each angle must measure 90'by the division properry of equaliry. If an angle equals 90", it is a right angle (definition of right angles), which then proves that the diagonals are perpendicular (the definition perpendicular lines). QED (which means what was to have been proven is proved)!

TuBonnn:

The angles of

a rhombus dre bisected by its diagonak.

Here's the proof for two pairs of angles. The same argument proves rhe pairs Given: Rhombus MNOP. Prove: ZQMN = ZqMP and IMNQ= IONQ Statements Reasons

l. MNOP

is a rhombus.

+ of

are equal

o

1. Given.

2.MN=NO=OP=PM. l.tWQ = OQ, lrQ = rQ. 4. LMNQ= LMPQ AOI/Q. = 5. IQMN= /.QMP, /MNQ= ZONQ.

2. Definition of a rhombus. f 3. Diagonals ol e l_J blsect -1. 4. SSS =. 5. CPCT. 79

of

.r.htth.r.

Due to the symmetry of rhombuses, there are other facts that your textbook may introduce theorems. For example:

as

A rhombus is divided into 4 congruent rigbt triangles by its diagonals.

If

2 consecutiue (right next to each other) sides of a parallelogram are congruent, the figure is a rhombus. (Which property of parallelograms proves this?)

+

5

Notice that the 4 angles of a rhombus aren't necessarily equal, If the angles are equal as well, then the figure is a square.

itt

5

the sides that are equal:

The definition of a square is a quadrilateral with 4 congruent sides and 4 right angles Squares So a square is both a rectangle and a rhombus! Everything that is true about a rectangle is true about a square and everything that is true about a rhombus is true about a square. The square is the ultimate parallelogram! Be sure you know the information in the chart below:

Putting It All Together 1. 4

(=)

-

Special Parallelograms

RncreNcrr Fecrs:

E-H

right angles.

2. Diagonals are =GC = BD). 3. Diagonals bisecteach other.

4

Opposite sides are s.

Opposite sid.r rre ll. it will 6 If a Ehasl right angle, have 4 right angles (and it's a rectangle!). 5

A

(So all 4 halves are equal.)

RnoMsus Fecrs:

1. 4 congruent

5. OPPosite angles are rr.

sides.

2. Diagonals are r 3. Diagonals bisect the angles. ffi 4. Diagonals form 4 = right triangles 1. 4 congruent

.

.

4 (congruent) right angles. Diagonals are r.

7. Opposite rid.r 8. A

E

with2

rr.ll

consec.

=

sides is a rhombus.

Squenn Fecrs: 5. Diagonals form 4=righttrirngles.

sides.

Diagonals are congrurnt (AC

5. Diagonals bisect each other.

B_C

=

BD)

,X,

X 80

X

rt.l's, into 45"2's.ffi 7. Diagonals bisect (f,k'fl each other.ffi 8. Opposite sides are ll. u

6. Diagonals bisect the

Now You Try

It

1. Find the perimeter of

2. a. Find x, ! and z in

B

DABCD, wIAED

b. LGEF

5

and ruZEAD.

=which

D

FGHI.

triangle?

A

F

E

3. Given: find NM.

4. Find the perimeter of square MNOP, wZMQN, mlPOQand name any triangles that are congruent rc LOQP. P

JKLM,

I

The diagonals divide any parallelogram into 2 pairs of congruent triangles. Heret why:

II rw

r!

ry by SSS.

E are E

Opposite sides of a and, diagonals of a = bisect each other.

!

The diagonals divide a rltombus into 4congruent

right triangles. Heret why:

II

r!

rw

All

sides of a rhombus are

=

4 rt. A

.

Diagonals of any ^E bisect each other. Diagonals of a rhombus are

I. 81

by SSS.

!

by SSS (or HL)

.\J

t

t/ o

Trapezoids

Quick Review - A quadrilateral is a four-sided polygon. Quad means four and lateral means side, so it makes sense that a quadrilateral is a four (straight) sided figure. It's easy to figure out that the interior angles of a quadrilateral total 360". To do this we can use the formula for polygons, (" - 2) 180' or (4 - 2) 180" = 360". However, itt better just to memorize this important fact: The interior angles of a quadrilateral total 360".

DErrNrnoN: ,4 napezoid

is a

quadrihteral with exactly one pair ofparallel

sides.

One pair of base

The parts a

of

trapezoid:

L ne pair of

t base

"rrgl.J

A base of auapezoid is one of Bases of Trapezoids its parallel sides. It doesnt mafter whether the base is on the side or on the top. If it's a base, itt one of the parallel sides, if it's a parallel side, itt a base.

b

t.

b

e

The bases are the parallel sides

e

Like all polygons, a trapezoid is named by its vertices. Many texts use the small trapezoid figure a to indicate a trapezoid. Here are some examples of the different ways trapezoids may be oresented in '12 The problems C rhe figuqe D,MNOP figure on the right.

on the right.

on the right.

11

In the fourth figure, the same side interior angles between the top and bottom sides of the figure are supplementary, which means (by SSISP) that the top and bottom sides are parallel. Since the ll sides are unequal (which means the figure is not a parallelogram), the figure is a trapezoid. trapezoid is a traPezoid with both legs congruent. Many of the trapezoids you work with in geometry are isosceles trapezoids. Note: the bases of avapezoid can never be equd. \Why not? (1uur3o1a11ercd e eq PFo./rr 1r uaql esnrcag)

DBrrNrrroN: An

Tunonrvr: This

isosceles

The base angles of an isosceles trapezoid are congruent.

lrq uals this

1

\tirz

equals this

------)

Base angles come in pairs. Each trapezoid has 2 pairs of base angles. Each pair shares one of the bases. Notice how the top side of the trapezoid, a 4 base, is a side of both of the | 32" angles. Therefore, the 132" angles make up one pair of base angles of this trapezoid . The pair of 48" angles are the other pair of base angles.

Z/

82

Median of a Trapezoid A trapezoid

with exacdy 2 parallel sides called the bases. The two non-parallel sides of a trapezoid are called the legs. is a quadrilateral

DnrINIuoN:

%

tltat connects the midpoints of the legs.

Zbe median of a trapezoid is the segrnent

33

Here are some examples B

C 8

e

d

F

i

e

n

D

median EF

Tnronru:

7be length of the median of a napezoid is equal to the aaerage length of its bases.

hare

[

base 2

MM= Example

-

4fi,

Know where

hare n .]).,

M=

2s+4

)

x*

5

2

2. Collect like terms: 12

5x+

TneonsM: Tlte median of a trapezoid is parallel to the

-

p

2

28 2

=uJ

zhz=7x+to\2

\

7x+10 2

2l

24 = 7x + 10, 14 =7x

x=2r/

bases.

MN)P with median QR,

#

frnd wZNMP and r. The base angles of an isosceles trapezoid are equal. Therefore, ZNMP is also equal to (2x - 20)".Since rhe median of a trapezoid is parallel to its bases, PCC tells us thar ruINMP equals ruZNQR So, ruZNMP = 48/and2x -20 = 48,2x = 68, x = iq.'/

Given isosceles trapezoid

O

=

+22

3. Use "fraction busters" to cancel the 2 in the denominator

lhllrbase2

1. Know where each value goes!

5

Example

2

3&=

each value goes!

2

Solve for x.

M

83

Now You Try

l.ls ABCD

It

a trapezoid?

2. Given D,MNOP with ,A/O and

C

N

A

bases

3. Given GHIJ, find u and

u.

MP, frnd x and y. o

M

4. Given aUVWXwith median YZ, find a, b and

a QRSI find r. R 3x+B .t 26

5. Given c.

6. Given aIJKL with median MN, find z.

10

I

(3r+B) 7x+2 (4a

-

2b+16

Other Parallel Line Theorems + The median of a trapezoid equals hare n harez

But what happens to rhe median when the top base of a trapezoid gets smaller and smaller?

20

20

12+20

22 =2

Trrsonru

=16

8+20 28

22

20

20

4+20 24

1/.

22

The segrnent connecting the midpoints of

1. Half as hng as the third side, and 2. Parallel to the third side.

A

tuo

1.

sides of a

ry=T=ro triangle is:

DE = YzAC DE II AC

+ A

For triangles, this special segment is called a midsegmentbecause a median of a triangle is something else:

Examples using the new theorem: 1. Find

x

M 28

and y.

Since PQ joins the midpoints of two sides of AMNO, PQis Yz as long as and ll to th. third side,

! =14, o byPCC, x=55.'/ MO.lherefore,

and

B4

2. Find a and

b.

Study L FHJ. The new theorem tells us that a = 12. Now study aEGIK. Since a is the median of EGIK, 12 = (6+ b) *2, 24 = 5 + b, b = 18./

a

Now You Try 1. Find

It

z and rnl DEC. B

2. Solve for x, y and mZ HJI.

l5x-2 A

F

L

Segment Lengths and Parallel Lines

the

distance between two parallel lines is the perpendicular distance.

Trreonura:

If 3 or more parallel lines cut offequal segments on A transuersal, they are the sarne distance dpart. If 3 or more ?arallel lines are the sAme distance apart, they cut off equal segments on all transaersall

9

5

9

on even one transversal, the lines are the

off congruent Example 1. Find

segmenfr

on

eaeryl

5

9S

transversal.

ea.

Since the

7w-2 4w+/

AllO \7hen you see 3 or more parallel lines - * in a problem, think of this theorem.

3ll

lines cut offequal parts on one transversal, they cut off equal parts on the other transversal, so set the two expressions equal:

Example

2.

Find x.

7

7w-2=4w+7, 3w=9, ut=3. I Now You Try It l. Given A = b, find z.

x=2.t/

2.Tre three lines in the figure are equally far apart. Solve for

x +12

.r,

Since the 3 ll 1i.,., offequal parts on line m, they cut offequal parts on line n, so 3x = 6, and

and y.

(Hint: Think about

which segment equals which segmenr.)

B5

3t

-7

2x +4

KITES

If your geometry course doesn't include kites, wanr to skip this

with exactly two pairs of consecutive congruent sides. Here

A kite is a 4-sided examples:

6.

7

5

are some

3

7

) a

4 8

6

Kites have interesting properties that we can discover by using what we know about congruent and isosceles triangles. \We'll investigate using the first figure:

7

Add the longer diagonal, which will always be between the pairs of equal sides.

7

7

7

7

The longer diagonal divides the kite into rwo congruent triangles by SSS

=.

By CPCT, the angles between the unequal sides are equal and the angles between the equal sides are bisected.

7

7

7

6

Now add the shorter diagonal, which will always be between the pairs of unequal sides.

Facts about kites:

The shorter diagonal divides the kite into two isosceles triangles.

l.

The longer diagonal divides each isosceles triangle into two congruent right triangles.

The opposite sides are never equal. 2.The angles berween the equal sides are never equal. 3. The diagonals are never equal. 4. the diagonals are perpendicular. 5. The longer diagonal bisects the shorter diagonal. 86

8. INEOUALITI S In most problems, we work with equalities, that is, expressions that are equal to each other or figures that are congruent to each other. This section deals with geometric objects rhat are not equal to each other. The goal is to compare two (or more) objects and to figure out which is larger. Properties of Inequalities Here are the Properties of Inequalities that are used most often in geometry:

I. lf a > b and. c> d, then a + c > b + d. Here are some examples:

8, 3 and 2 =2,so 8 + 2 > 3 + 2. 2) 5, 2 and4 > 3,so 5 + 4 > 2 + 3. 1)

3) AB >

CD, BC =BCso AB + BC > CD + BC.

ABCD 4) MN > Ptrland NO > l\/Qso MN + NO > PN + NQ.

P

5)

*Zl

>

mZ3

a

l/

and mZZ =

m/2,

o

so

mZ.l + rn./.2 > ml3 + mZ2.

1

ll. lf a > b and, b, c, then a > c. Here's an example:

If EF > FG

and

FG > GH, then EF > GH,

E

GH

F

III. If

a = b+ cand b > Oand c > 0, then a > b and. a > c, (The whole thing is bigger than either one of its parts.) Here are some examples:

1) l0 =2+ 8 so l0 >2and l0 > 8. 2) IK = IJ + JK (Seg. Add. Post.), so IK

> IJ and

IK > JK.

Il

K C

3) lf ml ABD = wZABC + wZCBD (Angle Add. Post.), then rnZ ABD > mZ. ABC and mZ ABD > mlCBD. B 87

An Important Theorem Used in Many Inequality Problems Exterior Angle

Quick Review - An exterior angle of a triangle is formed by extending a side of a triangle and measuring back to the triangle.

Exrnnron ANcrB TnsoRErur: The measure of an exterior angle of a triangle is equal to the sum the measures of the ttt)o remote interior angles.

Remote Interior Angles

1

Now, recall Properry

Sum ofTwo Remote Interior Angles

Exterior Angle

Exterior Angle

of

46"

08"

+ 62"

III of Inequalities: If a = b + c, and b > 0 and c > O, then a > b and A > c. (The whole thing is bigger than either one of its parts.)

Using the previous problem, 108'= 46" + 62", so 108" > 45" and 108" > 62'. The above properry leads to a theorem, which is the key to solving many inequaliry problems:

Exrnnron ANcrB INrquerrry Tnnoner\a:

B

IDCB

-G

A

C

measure of an exterilr angle of a triangle is larger than tlte measure of either of the two remote interior angles. TVte

is an exterior angle of LABC.

For the reason in a proof,, write: "Exterior Angle Inequality Theoreni' (If you dont say Inequaliry, itt wrong!)

ruIDCB > mlA wIDCB > mlB

D

l/

tffhen you're doing inequaliry problems, always look for the exterior angles in the figure. For example, in the figure on the right, 13 is an exterior angle of LNPO and Z4 is an exterior angle

NowYou Try

It

of LNPM.

o

1. How many exterior angles are named in the figure? List them. Can you find any exterior angles that are shown but not named?

2. Using the Exterior Angle Inequaliry Theorem, list as many inequalities as you can. Problems 1-3

> ml4 (never assume that what else can you conclude?

3.If ml3

88

it

is based on the drawing)

Triangle Inequality Theorem 5

7

You cant make a triangle with sides of just any lengths. Heret an example

t6

No matter how fat we make the figure, the two smaller sides just arent long enough to meet and form a triangle. This is so because in order to make a triangle, the sum of the rwo smaller sides must be larger than the third side.

6 16

TnreucrB INrqueuryTnnon-nu In a triangle, the sum of the lengths of the two smaller sides is larger than the length of the third side. There are 2 types of Thiangle Inequaliry problems:

lst Type of Problem Given three lengths, can you make a triangle? To check: l. Add the two smaller lengths together. 2. Is their sum than the largest length?

Examples

-

No{ L-:2

Can you make a triangle with the given 3 lengths? 5, 3 and 2? No. 2 +3 = 5 and 5 is not larger than 6. 2,6 and 4? No. 2+4 / (. "Equal to" isn't good enough. The sum must be larger! 2, 4.01and 6? Yes. 2+4.01 = 6.01 which is larger than 6. (Even .01 larger is enough.) , 3 and 5.8 ? Yes. 3+3 6 and 6 is larger than 5.8 , 5 and 5 ? Yes. 2+5 = 7 and 7 is larger than !. , 4 and 4? Yes. 4+4 = 8 and 8 is larger than 4. (Any equilateral triangle "works".)

Isosceles

Equilateral

Gm2nd,TypeofProblem-Giventhelengthsof2sidesofa) '

I i

3

,riangle, find i.e lower and upper .length of third side < limits of the length of the third side.J the lower and upper limits of the Example - If the lengths of rwo sides of a triangle are 9 and 4, find

lD

F

In other words, in a triangle, a bigger side is opposite a bigger angle,

w4l.(

and a smaller side is opposite a smaller angle.

\7ith

Use the above theorem when you are given or can figure out how the sizes of two (or more) sides of a triangle compare and you're trying to find out

these problems it-s a

good idea to -rrk sides and ansles.

un,h.fifip

@ x^ndrfol uig, f stands for small and M stands for medium.

how the sizes of the angles opposite those sides comPare Example: Find the largest angle in the triangle below.

x

Important Matltematics Note : Physical quantities (like the side of a triangle, for example) are positive, so the expression x - 2 must be positive. This tells us that r is positive and in fact that x > 2.

,,\

x+3 Step l. Label the sides of the triangle. x

Step 2. Label the angles site the sides

st

lr

x+

3

\

oppo-

Step 3. Studying the figure in Step 2, we see that:

12

is the largest angle in the triangle.

x

3

The converse of this theorem is also true: As the sides of an angle move further apart, the larger the angle becomes. This idea leads to the second theorem. 90

Tnnonsu: Within a singh triangle, if one angle is larger than a second angle, then the side oPPosite the first angle is larger than the side opposite the second angh. E

F

Since

lD>lF,

EF > DE

F

Notice that EF is not the largesr side in the triangle but it islarger thanDE. This is all that the theorem is claiming. Use this theorem when you can compare the sizes of angles in a triangle

and you're

asked to compare the lengths of its sides.

Remember

-

Use the 1st theorem when you have information about the

sides. T*

Use the 2nd theorem when you have information about the angles.

Example: Put the side lengths of the two triangles at right in order from smallest to largest.

Since we are given the sizes of the angles and we are asked to compare the lengths of the sides, we should use the second theorem.

C

A

;-\

4 * S

D

Step l. Figure out how big the third angles are in each triangle and add that information to the figure.

Step 2.

to indicate their relative sizes A then mark the sides opposite each angle with the corresponding letter.

side lengths of the two triangles in order

B

Confused? Think of it like this: The biggest side in LDCB is BD, 60'

:

.OC.BO.en.AD'/

Example: Find the largest side of triangleTW

each triangle, mark the

angles

Step 3. Now study the drawing. Since the biggest side of ADCB, is the same size as the smallest side of ADr4,B, we are able to put the

BC

\Tithin

but BD is the smallest side of LDAB. This means that the other sides of ADr4B are all larger than every side of ADCB.

Step 1. This is an isosceles triangle, so mark ZT and Z-Vequal (by ITT). This means ZV is also equal to 58" and Z Uis equal to:

180"-58"-58" = 64"

V

Step 2. Add the angle measures and markings

V

showing ZU is the largest angle. This means that 77 must be the largest side of the triangle. ,/ 9r

V

Here is another example of the use of these theorems:

Given: MN = PN = OP. Prove: Ol/is the largest side of APIVO and ON > MN. Proof: 1.

ll - 13, 25 = 16.* (ITT) *Note: Although MN PN OP, ll and 13 = = are not necessarily equal to ZS ana, Z-e .

M

o

P

M

2.

P

24 > Z2 and Z4 > 11. (Ext. Zlneq. Thm.)

Z4 > 13. (Subst. since lt = 13, Step l.) 4. 23 > 15 and 13 > 15. (Ext. Z Ineq. Thm.) 5. 14 > Z5 and 14 > 26. (Tlansitive Properry of Ineq. Steps 3&4.) 3.

APIVO

k ffi

6. ON is the largest side of (From Step 5 and the 2nd theorem in this section -Within a single triangle, the largest angle is opposite the largest side.) 7.

Zl > 16. (Subst. Steps l&4.)

8.

ON

> MN. (\Within a single triangle, LMNO, a larger angle is opposite a larger side.)

$?o

o

Check List for Inequality Proofs

l. Look for isosceles triangles.Mark

opposite angles equal.

4

2. Look for exterior angles. They help you put the angles in size 3. Combine what you learned

A$,

otdrr.l,

>Z 4, t>Ze in Steps I A 2.If 25 > Z3 then 25 >26 and so forth.

ofthe indiuidual trianglesin the problem and try to apply one of the two theorems in this section For example, in the above proof there are 3 separate trianghs, and each needs to be checked.

4. Lookat

each

Now You Try It Given: BD =BC list everything you can

P

discover about the angles, sides and triangles in the figure below:

E

-A

D 92

M

4\* 4 €/'\ * {#

Inequalities for 2 Triangles Even when two triangles are not congruent, if they have certain parts in common, we can make comparisons of some of their other parts.

22

71

27

1

The triangles above have two equal sides and an unequal included angle. (The included angle in this case, is the angle between the sides measuring l5 and 24.) As the size of the included angle increases, the length of the opposite side also increases. This idea leads to the following theorem: word there or

itt

" has to be not correct

SAS SroB crn Srpn INnquerrrv Turonsu (Thn HrNcr Turonrvr): If two pairs of sides of two niangles are equal and the included angle of the first triangle is smaller than the included angle of the second triarugle, then the third side of the frst triangle is smaller than the third side of the second triangle. i:\ ,i Sound Conftrsing? Read on. =>=

{,}

SAS Ineq. -Vhat we know: Each of the two St in SAS Ineq. stands for one pair of congruent sides. The A stands for one pair of unequal included angles. B

tVhat we find out (the conclusion): \Mhich of the sides opposite the unequal angles is larger and

B

E

)

3

At2

which is smaller.

AB KL

H #

L

G

T

t2

2

U

I

V

Y

+

X

U

93

W


xolzo=(5ozr)(4]

*

7200 = (5}n)mZKOL

7200 50n

fr
lP, wO t l4O and qA t =

Proofl

l/

Statements

M 5 O R 5

l. P

N M

OR

t

MO, QR L PR.

2.ZO isart. 3. 4.

P

NO

Reasons

l, lRisart.Z

mlO = 90", mlR ZO = lR.

= 90"

5.ZM = lP, UO = Pn 6. LMNO APQR. = 240

1. Given.

r

2. lines meet to form rt. angles. 3. Def. of rt. angles. 4. Definition of congruency. 5. Given. 6. ASA. (Steps 4 and 5.)

Page63 NowYou T"y It Every proof on this page can be done using congruent triangles and/or the definition of congruency. Statements 1. Given: AC

1. ZACB = ZECD. 2. AC =EC,

EC and ZBAC ZDEC = B Prove: BZ DE =

VAT. 2 Given. 1

ZBAC = IDEC. 3. LACB = LECD. 4. BA=DE.

=

=JE,Gr = KI, HI L GK and ZFHG IJHK = Prove: FG=JK

2. Given: FH

l

that

"See" the 2 pairs of triangles

G

l.

Co--o,

p^rrr. 11/

2. Vertical angles. f; 3. Information from parallel line methods (especially PAIAC).

6

? 4. Information from definitions.

KG

I A D.

4. CPCT

angles, and = 90"

ZGIH = ZKIH. HI HI. = 5. LGIH = LIOH. 6. GH= KH. 7. FH =JE, ZFHG = IJHK. 8. LFGH LIKH. = 9. FG=IK.

3. 4.

you rhink are congruent.

! f

3. ASA=. (Stmts

Reasons Statements 1. GI=KI, At tCX.t Given. 2. IGIH, ZIOH arc 2. Def. of r lines and

rt.

1H

Reasons

def of right angles. Def. of congruency. 3. 4. Refexive. 5. SAS=.(Stmts 1, 3 Aq. 6. CPCT. 7. Given. 8. SAS=. (Stmts 9. CPCT.

6 A7).

Start with the pair of A's that yov can Prove are congruent.

Statements

3. Given: P is on the perpendicular bisector of QR Prove: PQ= PR

1. P is on the

r bisector l. Given.

of QR-R.

as

R

2. S is the midpoint

Z.RSPisart. Z. 4. ZQSP LRSP = .t

R

,. QS= Rs. 6.N=PT.

First A A, 7. LQSP = ARS? = then CPCT { 8.PQ = PR.

24t

of

2. Definition of l-

@, Ps r

QR. ZQSP is a rt. Z, 3.

sAs

a

Reasons

bisector.

3. Definition of perpendicular lines. 4.

Nl

righr Z's are

=. 5. Def. of midpoint. 6. Refexive Properry. 7. SAS (Stmts.4,5&6). =.

8. CPCT.

page

63 (continued)

Statements l. Construct S7 L W

4. Given: SU = SV Prove: S lies on the perpendicular bisector of UV

U

V

P

l, l.

2. ZUPS is a rt. ZWS is a rt. 3. LSUP & ASIZP are right triangles.

HL

I I

+ I

V

P

tA=

4. SU=SVSU=SV. 5. 57 = .tZ'. 6. LSUP

=

ASIz?

Reasons 1. There is exactly one

segment between a line and a point not on the line. 2. Definition of perpendicular lines.

3. Def. of rt. A's. 4. Given.(Definitionof =.) 5. Refexive. 6. Hl.(Statements 4 & 5).

7.UP=W.

7. CPCT, 8. Pis the midpt. of W.8. Def. of midpt. 9. S lies on the l- bisector 9. Def of ..t- bisector.

of UV 5. Given: )zlies on the bisector

l.

W

W AAS

Y

Z 6. Given: YW = YZ, YlV L XW and YZ t XZ. W

lrst A=A,

YlV

t

l.

Statements YrX/ L XW , YZ L

Reasons

XZ.l.

IXWYis a rt. Z,

ZXZYis a rt. Z. 3. LXWY k XZY are W

HL

Reasons

of perpenare rt. l's and = 90". dicular lines & rt. Z's. 3. ZXWY > ZXZY 3. Defintion of =. 4. Ylies on the bisector 4. Given. of llWZ. 5. IWXY IZXY 5. Defintion of I bisect. = 6.XY=XY. 6. Reflexive Prop. LXWY= LXZY 7. 7. AAS=. (Steps 3,5 U 6). 8. YW=YZ. 8. CPCT.

2.

Z\WY = ZZXY

I & 8).

t XW, YZ L XZ. 1. Given. 2. IXWY and ZXZY 2. Definition

Prove: YW = YZ

Prove:

(Statements

Statements

ofZIVXZ,WtXlXt, YZ T XZ,

r

Given.

2. Definition of perpendicular lines. 3. Def of rt. A's.

rt. A's. 4. Y'W= YZ, YW= YZ. 4. Given.(Definitionof=.) 5' Reflexive' 5. XY= XY. 5. HL. [e . xxwv=

LxZY lt. zwxY=tzxY 7'3PCT' 242

Polygon Chart page67

Number of Number of

Calculation

Sides

Triangles

7

I

Ix

180

180

4

2

6

4

2x180 3 x 180 4x180 5 Xtro

360 540 720

1

5

7

9 'to

Ansles

900

6X/rO

/o80

7

7X/KO

o

K

/260 /440

6

page

Total of Interior

Xlgo

68 NowYou T"f It

1. Find the

totd sum of the measure of the interior

Use the formula, (n

-

angles of a 100-gon.

2)(180) = (t092)(180) = (98)(180)

=

17,640",/

2. Find the measure of the smallest and largest angle of polygon PQRSTTIVW.

RS

7

(1

(9x+

06'

0xl)' P

11

(l2x+

V

W

5

gtep 1. Counting the sides of the polygon, there are eight, which T means that the total npasure of the interior angles is: (n - 2)(180) = z)(rso) = (6)(180) = 1080'

ff-

U

Step 2. Now set the total of the values of each angle equal to 1080":

(10x- 7)" +106"+(I4x- l3)'+ 12x" + (9x+1)"+ 1lx" +(12x+13)" +(l3x+8)" = 1080"

81x+108=1080, 8\x=972, x=t2

Step 3. Go back and read the question and answer the questions that were asked: The smallest angle is ZQwhich equals 106', the largest is lWwhich equals 164".'/

page69

NowYou Try

It -

If the sum of the interior

angles of a regular polygon is 1260,

find the measure of

one exterior angle of the polygon.

This problem is giving the answer to the formula (n

-

2)(180) so set the formula equal to the answer:

(n-2)(180)=1260" Carefully distribute each term of the expression (n - 2): Balance the equation:

Divide by the coefficient of n:

n(180) - 2(t80)= 1260 180n - 360= 1260 + 360 + 360 1520

l80n

n=9

180 2) 180 for the measure of one interior angle, (9 180

Using the compact formula (n

Now, draw one corner of the polygon:

-

n9

I

Since a straight angle equals 180", the exterior angle must

equal 180-40 = 40.1 243

-

2) 180

= 140".'/

pageTl Now You Try 1.

It

If an exterior angle of a regular polygon is72",what is the total measure of the interior

the polygon? Since the exterior angles of a regular polygon are equal, use the formula: 360" =fi the measure of one exterior 72" angle of a regular polygon

angles

of

t

(

Since n = 5, the total measure of the interior angles = (fr -2)(l8o) = 540"

'/

2.lf

the interior angles of a regular polygon total 1620", 6nd the measure of one exterior angle. The problem is giving the answer to the formula (n-2)180 so set the formula equal to the answer:

Use the

(n-2)180 = 1620. Now distribute: 180n - 360 = 1620, 18on = 1980, n =

formula: 360" the r_neas-ure of one exterior 360" = 150" x 32.7" / n - angle of a regular polygon n 1l

3. Polygon ABCDEFGH is regular. Find the measure of

A

11

ZI.

There are lots of ways to do this problem, heret one of them: Polygon ABCDEFGH is a regular octagon. Each exterior angle measures: ,6qr, "

= 3Q =

45"

Sketching one vertex of the octagon helps you to see that each interior angle equals 180"

- 45" = 135". Now study the original figure. Polygon EFGI is a

quadrilateral so its interior angles total (n-2)180 = 360": Z.FEI + IIGF + (360" - Z.EF1 +

Now substitute

, ,fr 45"

+

. ,,rE;.(360" 45"

lI

=360"

,W. tI = 360"

+ (225) + lI = 360",

lI

= 360"- 315" = 45"1

page72 NowYou T"y It Given E GHIJ at right, add the correct measurements for the following:

ruZGHI, mlHIJ, HG, ruZHGJ, HI Gf . ^"d t/ , ruIHIJ = 180'- 133" = 47" (consecutiveG ruIGHI = ruZGJI = 133 (E) angles of n are supplementar/), ftC tJ = 5 (#), ruIHGJ = rnZ-HIJ = =

47"(

),HI=GJ=g

, GI=6+6=12(&).'/ 244

I 8

I

page74 NowYou Try

It

Supply the reasons to prove the diagonals of a parallelogram bisect each other. B

Given:

E ABCD

Prove: BD bisects AC and

Proof:

1Z

^:/w'

l.

ABCD. 2. ZAEB ZDEC. = t. AE oC. 4. BC

A

5.

= ll aDt

3. Opposite sides of

ABllDC.

4. Definition of

IDBA= ZCDB.

,'iFQ,-,' 9. BDbisects AC, AC bisects ^BD

E.

9. Definition of segment bisector.

It

If both diagonals of a quadrilateral

bisect each other, the

quadrihteral is a parallelogram. Given: Quadrilaterd, ABCD with AE = EC and BE Prove: ABCD is a pardlelogram. Statements

EC, Af = ZO. 2. IBEA= ICED; IAED = IBEC. 3. LAEB = LCED; LBEC = LDEA. 4. AE = DC, AD = BC,

ED D

1. Given.

2.YNl. 3. SAS. 4. CPCT.

5. ABCD is a parallelogram.

pair of opposite

=

Reasons

=

TrrBoruu: If

areo

7. CPCT. 8. Definition of a midpoint.

t of AC; E is BD. the midpoint of

Proof:

E

6. AAS.

ro,:n, 8. E is the midpoin ",-r$.GXc

TnBon-ENr:

a

a

5. PAIAC.

LAEB v LCED. 7.AE = EC; BE = ED.

page76 NowYou Try

Given.

2.YNr.

6.

A

D

A Reasons

E

^ffi'

AE

BD

Statements I.

l.

bisects

C

5.

A Quad. with 2

prs. of

=

opp. sides is a

E. o

quadrilateral are congruent parallel, and the quadrilateml is a parallelogram. one

Given: Quadrilate

Proof

ral

MNOP with MN

Statements

1. Construct MO . 2. A,INllPo. 3. INIUIO_: ZPOM.

4. MN = PO. 5. MO =tWO. 6. LMNO = LOPM. 7. NO = tWP.

sides of a

8. MNOP is a parallelogram.

=

PO

and

MN llPZ Pror., MNOP is a parallelogram Reasons

Two points determine a line. 2. Given. 3. PATAC. 4. Given. 5. Refexive Property. 6. SAS=. 7. CPCT. 8. A Quad. with 2 prs. of opp. sides is a D. 1.

=

245

page77 Now You T.y

It

For problems I clusions.

- 4, find any parallelograms in the drawings.

Be sure to thoroughly explain your conB

2

1

I

PSSIS tells us

3

X

and we can see IJ = LK but no theorem tells us that this is a E so the correct answer is "No Conclusion'.'.r/ Had one pair of sides been both llr"d IJKL =,

.ZF

wouldbeaEby:

A

Y

Z

i ll lk

4.

9

s

R

V

2

L

0

D

UT

IZXY of LZY mexures 180 - 44 - 30 = 106.5o 2

The diagonals bisect each other

CCP tells u, AC ll ED, ITT tells us AB = AC= 9, ZXWis 180 - 106 = 74. I so RS7Vis a D VZX of Quad. VZXW mea- so, AC= ED. Therefore, t/. H.r. is the CDEisaEbecause sures 360 - 74 - 74 - 106 = logic sketchof one pair of sides is both 106. Since opposite angles the supporting and =. t/ Here is the logic theorem: are equal, Q""d. VZXWis a sketch of the supporting E.Here is the logic sketch rheorem: of the supporting theorem: I

D-&

5. Find the values of x and y that make MNOP a parallelogram.

o 30)'

g)"

(9x

(7y+2)

M

Step l. Assume tempo6x+8 = 9x -28 3x=36, x=12 rarily that MNOP is a l-l. )teP z. Lreate equa- 7y+2 = 5y+30 tions based on the facts !=14 x=12,!=14t/ that you know about a D, in this case thar opposite angles are equal.

2!=28,

p

6. Find the values of a and b that make FGHI a parallelogram. Step 1.

fusume that

FGHI

is a

ti

sides are equal.

- 4)

+

b-4=a+9 i#n*.y.,

Step 2. Create equations

you know about a E, in this case that opposite

(3b

-2 = b+l

D.

based on the facts that

+9)

equatlons.

lst eq. I =b Using lsolate , 3(4a r4) - 4=a*f!!!2 r l2a -42 - 4= a+9

lla=55,

a = JSolvefora. 4(5) -14 = b = 6 Solve forb.

page 81

NowYou T.y

a.=5,b=5/

It

l.

Find the perimeter of EABCD, m/.AED B 56' and ruZEAD.

A

C ABCD is a rhombus because BA= AD.

D

(A

o

with 2 consecutive sides is a rhombus.) Therefore, the perimeter is 4xl0 = 40.1 = Diagonals of a rhombus are t so,mlAED = 90"J (def. of r). wIBAD = 68o (ZABC = 2x56= 112", so ZBAD = 180" - ll2"= 68" (Consec. Z's of a are suppl.), ZEAD = 68"* 2 = 34".1 (Diagonals of a rhombus bisect the l's.) 246

page

8l

(Continued)

2. a.Find x, y and zin E FGHI is not a rhombus (2ZGEF = 7B), nor is it a rectangl, Gf * f]l E FGHI. b. LGEF | ? so only the regular properties of parallelograms can be used to solve the

H

aD

problem: ! = 5, z= 12 (diagonals of bisecteach other),wZFEI = 102 (180 - 78),wZEFI =21 (lB0 - 102 -57,31's of aA= 180"), sox=21, (PAIAC). ,/ b. LGEF LIEH (Since GE = IE, IGEF = IIEH, f7 fE.)

G

=

4. Find the perimeter of square MNOP, wZMQN, wZPOQ

F 3. Given D JKLM,find, NM.

I

is a rectangle (a/- with I rt. is a rectangle). JN> NL (diagonals

IKLM

Z

=

of a E bisect each other), so JL= 32. Diagonals of a rectangle are .r, so Kfu[=32, and NM = 16. (A second proof; LJLM is a right triangle, and l/is the midpoint of JL, so NM =16 because the midpoint of the hypotenuse is equidistant from each vertex.)

and name any triangles A OQP. = A square has four sides and four P o = = (rt.) angles. The length of one side of the square is r. In this case, we cannot solve for (find the value of) t, and that means the answer must be "in terms of" t; the perimeter is 4xt or, 4t. J To frnd mlMQN, recall that the diagonals of a square (since a square is a rhombus) are t, so wIMQN = 90 r/(def. of r lines). The diagonals of a square (since it's a rhombus) bisect the vertices, so

wIPOQ=

Since order counts, there are 7 triangles

=

90 +2 =

45/.

to A OQP: LPQO,

LOQN, AlrQO, LMQN LNQM, LMQP,

LPuM.,/

page84 NowYouT"yIt

l.

ls

ABCD a trapezoid? C

A

2. Given D,MNOP with bases l/Oand MP, frrd x andy.

N

o

M

3. Given

GHA find u and u.

To be a trapezoid, two sides (and only mo) must be ll which would then mean that by PSSIS, two adjacent angles (two angles that are next to each other) must be supplementary. But 101 + 78 = 179, and78 + 93 = l7l. So no, ABCD is not atrapezoid.t/ Since NO and MP bases, they must b. ll. Th.refore, by ^r, PSSIS, IONM and IPMN must be supplementary, so set

180, l4x + 12 = 180 14x=168, x=12.t/ For the same reason, ZMPO and INOP must also be supplementary. Substitute 12 for x in the equation: 4(12) +7 + y = 180, 55 + y = 180, y =125.'/ their sum equal to 180: 9x+8+5x+ 4

=

This is an isosceles a so by the Isosceles Tiapezoid Theorem, each pair of base angles is equal. But first, we need to find the value of either u or u. ZIHG and ZJGH are supplementary by PSSIS so set their sum equal to 180: 123 + 5u + 2 = 180, 125 + 5u = 180, 5u = 55. u = ll./ Nowsubstitute: u = 7l into (5u + 2)andset = /. 5(11) + 2 = u.

u=57t/ 247

page 84 (Continued)

4. Given a, UVlWwith median YZ, find a, b and l0

Finding a and c. Since YZ ,t the median of the A, YandZ are midpoints of the sides. This means:

c.

(3c+B)

(4a

-

4a- 6 =24,

2a=6,a=31

6)

2b+15

and,4c=3c+8

Findins

&:

base

-.di,i-

1+ base2 2

r0 + (2b +t6) 2

zbz=rc+2b+t6\tr

\21

44 = 2b + 26, 18 = 2b,9 = b.t/

c=8t/ 5. Given

a

26

QRSTfind r.

Because

aa:

median- basel+base2

UR

2

andST-VT, UV

R 3x+8 S

5x-3=

is a median.

22

2

7x+2

k.-

(3x+8)t(7x+Z)

) I 0x +

3

10

4

L

x

l2x - $ = l}x +10,2x = 16, x =8t/ 6. Given D,IJKL with median , find z.

MN

a

is parallel to both bases. PCC tells us The median of a that IJMN= ZMIL so set the corresponding expressions

equal:

5z

*

13 = 4z +

I Top of page 85

-

2,

2. Solve for x, y and mZ HJI. First concentrate on L,GIK,

I

NowYouTty It

z =15.1

5x+ 4 = Vz(l1x -2) =7.5 x -l Find zandmZ DEC. Since DE joins 5=2.5x, x=2t/ the mid points of two sides of AABC, DE mIHJI= 36"(PCC) is Yz as long as and ll to, th. third side, (HJ is llto GKe.FL) 68 48 AC. So,2z +14 = Vz(68), z =10 r/and F study ttFHJL. L A wZDEC = 180 - 48 = 132 (PSSIS). GKis the median of trapezoid FHJL so Bottom of page 85 NowYouTiw It substitute x = 2 into the median formula: Since th. a'llh.r.r cut offequal 1 + 1 + 1. Given a = b, find cut they on one transversal, p"rtr '' 2 off equal parts on the other trans+ +4 I 2 8= 14 l3y 5z versal, so set the two expressions I a ,r,(r +12 equal: 5z = 2z +12,32 = 12, z = 4. t/ r/ 56 = l3y +18, l3y = 38, y =2

l.

yk=

2.Ihe three

lines in the figure are

equally far apart. Solve for x and y. (Hint: Think about which segment equals which segment.)

Parallel lines that are equally far apart, cut offequal segments on every transversal. Therefore, we can form these equatlons: 2y +5 = 4x |This is a system (2 or Jmore) of equations

'

(1) -I (2) 3y-7=2x+4

Vorking on equation (1), +5

4x -1

3v

-7

2x +4

isolate

y:

2y

=4x-

Now substitute into equation (2): 3(2x 3)-7=2x+4 Distribute: 6x - 9 - 7= 2x + 4

4x=20,x=5t/

Substitute x = 5 into equation 248

(l):

2y +5 = 4(5)

-

I = 19, I =7 t/

page 88 Now Yo uTryIt

How many exterior angles are named in the figure? List them. Can you find any exterior angles that are shown but not named? (1) 5 (2) 13 (to 2 different triangles), 24, 18, 110 (to 2 different tri-

1.

angles), and t2

(l)

Zl2.

Yes

Problems l-3

2. Using the Exterior Angle Inequality Theorem, list as many inequalities as you can. ml3> ml5, ml3 > ml6, ml3 > mZ7, mZ3> mZ9, mZ4> mll, ml4> mZZ, ml8> m ll, ml8 > ml2 (Notice that mZ8 = mZ4, by VAT), mZ.lO > mZ7, mZl) > ml9, mZ-10 > ml5, mZl\ > mZ6, (Notice that mll} = ml3 by VAT), mll2 > ml7, mZl2 > m211.

3.If ml3

ml4

(never assume that it is based on the drawing) what else can you conclude? and ml4 > ml2, rhen ml3 > ml1 and ml3 > ml2 as Since ml3 > mZ4 and ml4 \We could not have known this otherwise. well.

>

) mll

page 89

NowYou T"y

It

1. Can you make a triangle with sides equal to a) 4,4, and 8? No, 4 + 4 = 8 * g. "Equal to" isn't good enough. The sum must be larger.

b) 4.01, I and3? No. 1+ 3 =4* 4.01. Rememberalways addthe wo smallernumbers together c) 2, 7.9 and 5? Yes, 2 + 6 = 8 and 8 >7.9 (ever so slightly bigger, but thatt enough).

2.lhreesides ofa triangle

aZWXV 250

Reasons

1. Given. 2. Refexive. 3. SSS Ineq.

Step

I

Step3.

4. Ext. Z Ineq.Thm. 6. Properry of Ineq.

Y Step 4.

Y

Page 98 NowYou Tiy

It

1. Name the 6 properties of proportion and give an example of each.

w=Y If xz

and

3

6

Z

8

(1) Cross multiply: uz (3) Swap the means

=

lt

lx =

and (3)(8) =(4)(6)

x

yz

(5)Bring up the denominato,

and

34 68

2.

tf

j

h

ll)

=

z

v

48 36

and

z

=y XU

and

86

43

* =y ?# xz 3+6 9 3 6

l:a xz =

48

{

(4) Swap the extremes:

u*x

3+4 g

(2) Flip the ratios: '

(6) Add them vp,

6+8

=

4+8124

8

, which of the following are equivalent proportions?

1 ") i =

.I7hrt

happened? gand j got switched. Since swapping the extremes is one of the

properties, the answer is yes, this is an equivalent proportion.

b.\! 'g

= !i

d+b !i 3.rfy=a xz 4.If

t=,

\ifhat happened? Compare b to the original proportion; each ratio has been fipped. Since fipping (inverting) the ratios is one of the properties, the answer is yes, this is an equivalent proportion.

tVhat happened? Compare c to the origind proportion; the right ratio has been fipped (inverted), but not the left. None of the properties of proportion allow only one side to be fipped. Conclusion: No, this is not an equivalent proportion. then

then,

z

Compare to the original, the ratios were fipped and then both the

v

means and extremes were swapped, to b

? Co-p"re to the original, the denominator of the left side got "brought up" and then the ratio was flipped (inverted) so do

-a+b

exactly the same to the right side: page

. t/ ; =;

h= *.

/

102 NowYouTryIt

1. \Write

four congruencies and the extended proportion given by this similarity: MNOP

ZM=ZQ ./.N=ZR ZO=lS ZP=Z T

MN NO OP

QR RS .t7 25r

PM TQ

*

QRSTI

order counts!

- QRST t&@if

MNOP

page 102 (continued)

2.1he two quadrilaterals shown

G J

are similar.

D

2 ,-t

a. Name two similar quadrilaterals.

t

A

2t 3

93" 111"

t2

C

Answer: Any similariry where the corresponding 9 angles are matched up as shown below, is correct. J GHIJ,-, BADC. (Match up the corresponding angles) Re-sketched .; b. CBAD is similar to what quadrilateral? GHIJ so that rt,

B +

JGHr

is oriented like tVhat is the scale factor of the two figures? BADC. Answer: Studying the two figures, the problem gives a length of two corresponding sides, .Br4 and GH , the sides berween the 125" and 3l " angles in each figure. Since the figures were glven in small to big order, that is the order for this problem, small to big. The scale factor which reduces to 2 Always write the scale factor on the figures. It helps you to keep 12 ) track of the scale factor and the order of the p roblem. d. Find r, s, and t. +

c.

ir9

22 3r 2.t

39

2t

32t

J

(2)(r) = (3)(2) =

6,

(2)(9) = 18 = (3)G),

r = 3'/ s

=

J q

6'/

(2)(2r) = 42 = (3)(t), t =

14

I

D r

c

2t 93"

t2

9

page 106

NowYou Try

3

111'

B

It

In problems 1-4, decide if rwo triangles are similar. If they are similar, explain why and name the similarity. If not, explain why not. 1

Here are the ) Arr{ways you rho*!, SASrwo triangles I SSSare

Since no lengths are given in the problem, the only way we can prove LGHI * LFHJ is by AA-.

similar )

Draw the two triangles separately.

F

2-Study the lettersr. to help find

H

shared parts,

Look for straight angles

lHisin

\

I

G

F

I 180-108

both As.

7)

F

ZG=ZFandlH=ZH LGHI* LFHI byAA-./ 252

page 106 (continued)

2.6 M

second triangle gives two angles, so the first step is to find its third angle:

180-86"-48"=46". 8

L

The first triangle has a 33'angle and the angles of the second triangle are 86", 4B', and 46.The two triangles can't be similar because all three corresponding angles of similar triangles are congruent and triangle NOP doesnt have a 33"

1

l/

angle. 1

P

3

ZY = ZV

and

ZX

=

lU

(and

IYWX = lVWXby

VAT)

In naming the similariry, follow the order of the pairs of

W

W

congruent angles. =z_

I

4.a

V

U P

-z_

I

A

-z_ by

AA-.'/

Both triangles shows the size of 2 angles so we can find the third. ARQP; 180 - 26" - 28" = 126" LMNO:180 - 26" - 126" = 28" Therefore, the three angles of both triangles are 26" , 28" , and 126" . The two triangles are similar by AA- because they share pairs of congruent angles. To name the similariry correctly, follow the same angle measure path in each triangle 26",28", 126 APRQ - LMNO.\/

R

M

V

I

o

5. Using the figures below, name two similar triangles and find g and i.

I

H

G The problem is telling us that the 2 triangles are similar so the first thing to do is to re-orient the second triangle (triangle lfi{ so both triangles are turned the same 5.6 way. Do this by drawing a larger copy of FGH. You know where the angles go (in the same relative locaH tions as in LFGH). Then carefully transfer the letters and other numbers by referring to the original figure.

K

F o

7

6

5 J

Now it's easy to name 2 similar triangles: /rFGH - LIIOJ Find the scale factor by forming the ratio of two corresponding sides that

havelength, 6. Using the figure below, prove

AC BC *Notice

L =.g= 7 5

5=6

AE BD

that once youve proved the triangles are similar, you can state that the ratios of any pair ofcorresponding sides must be in proportion.

4 = g s=4\/ 4 - 8 i=l0r/ 5 i 5 5o

Statements

r. ABIIBD 2,

IEAC = ZDBC

ZAEC = ZBDC 3. AEAC - ADBC ,* AC AE

BC

253

BD

Reasons

1. Given. 2. PCC. 3.

AA-

4. Definition of Similariry. (Corr. sides are in nronortion-)

Page 109

NowYouTty It

Complete the indicated operation and if necessary simplif, (finish) the following radical expressions.

-

Answers:

t.

(3)(2^fr) = 6 \/T

=

6^M

2. 6^fr) o^fr) =a^R

\a ^3 (fl (4Xl) 3.m ,G

4

^12

ts

'/ -=

page 111

6)G)(^h) = 18 ^6,t

=

=8^M

= (8)(4)

(^R)=32\b'/

,/5v/

---;4

.G ,l ts

F9

NowYou T"y It Given right LGHIwith altitudr-

(@; HI

z6

(2)(3x5)

15

Note: You can't rr..l

,/

ffi

Think of the radical sign as protecung lts contents

drawn from the right angle, name three pairs of similar triangles.

Remember, order counts! tWhat this problem is testing is if you know how to name the similariry; that means, Iisting the angles of each triangle in the correct (corresponding) order. 'We know the three triangles are similar so they dl have the same shape. Draw wo smaller triangles that are the same shape as the original triangle (GHI) and that are oriented (turned) like GHI. And then

H

carefully place each letter on the correct vertex, beginning with the right angle, and then the acute angle from which the dtitude was not drawn. ++

J

++

J

Now

itt

easy

to name three similar triangles:

l. LGHI

2. LGJH 3. There are lots of other correct answers, for example, LGIH * LGHJ

LHI

- LHII.

pagell4 NowYou Try

l.

It

Based on the figure below,

fill in the blanks:

a

a is the geometric mean between & is the geometric mean between c is the seometric mean between b

254

d and @A. e and @A e and d

page

l14 (continued)

A3

AC=AD+DC=3a9=12J AB 3 Findinsl,B: OAB12 -

2. Based on the figure, find AC, AB, DB and CB:

9C

D

(AB)'=36, AB=6J

FindinsDB: ADBg3 B

A

3

D

(DB)' = )/, DB

C

9

9 Findins.BC ABCO

J'

6

=

(BCy=1o8,BC=

DB

-

il@O)= 3J5 = 5.2J BC

-

(3X-6)(6) 6 Jj=10.4'/

B

page 116 NowYou Try It l. Find z in the right triangle below. Answer: In the problem, you are given that this is a right triangle so you know that the lengths of the sides of the triangle are related (Leg1)'z + (Leg2)2 = (Hypotenuse)2 by the Pphagorean Theorem: 1- ,1 3z-2 LLi' (22 + 2)2 + * = (32 - 2)z Foil the binomials.

(4*+82+4)+*=9*-l2z+4

/,

4i-202=0, 4z(z-5)=0 4z =O,rXO (i-possible), (z -5) = 0, z = 5 r/ 2. Is triangle PQR a right triangle? Answer: The converse 6f the Pythagorean Theorem says that the lengths of a right triangle are related by the formula: (Leg l)'z+ (Leg 2)2 = (Leg3)2' then the triangle is a righltriangle. Always let the largest side of the triangle be Leg 3: 32 +

22 ?-

q'

3

9+4

* 16

So triangle PQR is not

a righrtriangle.

118 NowYouTrylt-Given: a2 + b2 < P Prove: LABC is obtuse. Hints: Construct right LTUZ with right lV andlegs a and b. Now recall the inequality

if 4

P

R

page

B d

Statements l. a2 + b2 < P. U 2. a2 + b2 = ri. a 3. ti< *, u< c. v 4. ZC > lV.

theorems!

Reasons 1. Given.

^%.

2. Pythagorean Theorem. 3. Substitution. 4. SSS Ineq. Thm. b 5. ZC A LABC are obtuse. 5. Def. of Obtuse Z's &. A's. 1. Is triangle BCD acute, right or obtuse? Prove it. page l19 NowYou Try It To use the three theorems, we need to know the lengths of all three sides. Since triangle ABC is a right triangle, we can use the Pythagorean Theorem to find CB:

A

bC

D 8

A4

B

82 + 42

JBo

=

=

(CB)z,

ru

80 = (CB)2

(s )(4)(4)=

4\5 =8.94=CB

Knowing all three sides we can now form a test equation to test the triangle: 62 + 72 ? (4\/-5)'z Since 415 is the largest leg, we placed it by itself on the right. Left side: 62 + 72 = 36 + 49 = 85 Right side: (4t/j)' = (4)' 6/r)'= (16) (5) = 80 Since 85 > 80, that is, the left side of the test equation is larger than the right side, we know the triangle is acute. r/ 255

page

121-

Now You Try

length of a diagonal of the square below.

1. Find the

It

A diagonal divides a Draw the 45"-45"-90" Since "rGis the length of square into two congruent model so that it is ori- a leg, the hypotenuse of 45"- 45"-90" triangles. ented like the problem. the triangle (which is the diagonal of the square)='s x

2. Find the length of a side and the perimeter of the square below.

{60

x

The diagonal divides the Draw the 45"- 45"-90" square into two congruent model so that it is ori45"- 45"-90" triangles ented like the problem. with hypotenuse = 1.8

="h1=2",13

*3.46J

Since 1.8 is the length of the diagonal (the hypotenuse of each triangle),

set 1.8 =

*J1

1.8*-rt = x 1.27 = r =side,,/

.8

I

Perimeter = 4(1.27)

page 124

NowYou T.y Find a. d

It

= 5.08r/

Answer: To solve a right triangle problem we need one side and one acute angle. At first glance, it doesn't look like we have enough information about either triangle to get started. But the lower triangle is isosceles, and since it's a right triangle, LU must equal 45 Draw the 45"- 45"-90" model so that it is oriented x like the lower triangle

x

5

@ Comparing the lower triangle to the model, the hypotenuse of the

a

@But the hypotenuse of the lower 5J1 triangle is the left side of the upper triangle, and since p = {J, )qt = lower triangle must equal5fi. 5 90. Now we know that the upper Lower Tiiangle triangle is a 30"- 60"-90'triangle. d

@Co-p"ring

2x

the

x^15

x

@ Draw the 30"- 60"-90" SO like that it is oriented the upper triangle in the problem.

model,

5Ji

the same

isin

position

10 as xy3 soset v'6 them ..._..._-1**.. eoual.

\-=

5# -_ 3

256

2

Upper

tiangle

5J, = xt6

upper triangle to the

5

5

10 ="A ^h

^"6

=x

Divide

bvlj

Rationalize denomin.

(^fr^fr=3)

10\/6 = 4.08 so 2x = a = 3

= 8.16

\/

Page 131

NowYou T"y Find r and y:

It

acute angle.

S

o

H C A H

8

8

v

m

30"

v Remember, if one angle is 30", 60", or 45", then you can use the special right triangle formulas instead of trig to solve for the 2. Find the measure angle

M:

of

lol

d

I

a different

),= 8\/3

O

n l/ 4M

uC

cross

.58

8

1

v

mubiplying

from p:ru:alculznr

A

H

T

diuide both

13.8

sides by

A

mahe the

3. Find the length of the base of the triangle.

H C

A H

lol lal

S

o

4

H

EI IAI

ln

4

T

A/

o A

cos62o

1(8)

diutde both sides b1 .5

x t2

/. M

a

4

= '44 4

=

I6t/

o@@

)

or

@@@

mZM = 71.6J

m

ratio

8

(.5)x =

.58

tan

side a

multiplying

cross

S

o

lf

.5= lx

=13.8J

o This is a "going backwards" t2 problem, use the second key: 4

+

.i.r'30" =

o

(5a7, x 1(8)

t

+

=

sin 30"

loi

mahe the left side a ratio

modeltriangle

g x=16,_

+

&--+

variable.

a

tst

from^yoycalculator

The 30-60-90

with

=

tan 30"

lA

60" 0"

FindingT

FindingT

Finding the

cross

mubiplying

(.47)4 o *1,6

from yogr calculator

*b*'+ L/

4

make tbe left side a ratio

1.88

* *1,5

orl,6x2

ro base

1.88x

Y

2 x3.76J

47

4. Solve triangle

ABC:

A

a,/-, 1 - 4" I Finding AC; _ 3 Finding AB:AB

Remember, "solve" A tan 41" = means find all 1ficos47"=+E 6.5 lel-,,t1r-A missing vour calculator o"-li'y''**':'- "#om:-.ta ry measurements AB B of the triangle. 6.5 Label the triangle from mahe tbe left side a ratio mahe the left side a ratxo

t

He+_#la #"o

5

B

lC's

point of

Finding mlA: mZA= 90 - ml C = 90

X

(W'z*

(4.25)')

- (3.33)2

(W'z x 6.97 4V4

Y

ros muhip[ing

x 6'5 - 41= 49,/ ('75)AC AC x 8.67

+

257

AB 6.5

cros mubiplying

(.87)6.5 x AB

Y3

'/

coslY = 3Y' =.78 4V4

mZY = 38'/

\Wx2.6/

r

5.7 =AB This is a "going backwards" pro blem, use the second key:

X

(3.33)'*(W'z x(4.25)' W

%

#'*#

mZY:

To find WX, use PythagoreanTheorem:

5. Solve triangle WXY

Ji.*

mZW

o

H C

A

IJ

T

Finding ? mlW= 90 - m./.Y = 90 -38 = 52'/

pageI35 NowYou Try It In problems l-5, name a method that could be used to solve each triangle, then solve the triangle l. x Trig can be used to Fi ndingT 2. This is a Finding x S solve this problem, o cos 32" 45-45"-90" s tan 32" = 16tz so label the triangle H t2 triangle. C frorrt vour calculator H f'rom your calculator " -:/ from the 32 angle's A ' Therefore, one .tJZ T il " point of view: o re*TN method to A x

= lr? +85-: t') t rx-T"+

make-the ufr

ro/" o *rrT

make the hfr side

v .62

-32/o, r,

I

t

1

cros

(.85)! x

cross

I(r2)

diuide both sides by

x

special

triangles although trig or the Law Of Sines could also be used. Comparing the problem to the model:

mubiplying

(.62)

.85

solve is

12

1

multiplying

d

t2

ns 1(x)

X

=7.44r/

t =14J2\/

3r/,

n,-5 , Since 52 + 122 = 132, /VML LMN is a right triangle 12W13 (Converse of Pythag. Th-.) with ZN = 90",/. To find the other 2 angles, use trig, ("go bachvards", 5 LS o use the second key).

W

4 'When

you are given 2 angles of a triangle, always find the

C

fl

H

r3

C

A

H

ranzM=

+

fom your calcut"*,

H

right triangle, the Law of Cosines would have worked. 5

b c

mZS=180-69-65=46

Now that we have a pair, we can use the Law of Sines.

r R

9@

mZM x 22.6/ mZL= 90 - 22.6 x 67.4'/ If you didnt norice that this was a

third:

t

lh

T2

x4tfr.=

sin

46

5

=

srn

sin69"

t

7;r

\.93

-*#tr = @ t .72

t

t sin65'

5r 7;r ' \.gt Affitr =@ 5-r from your calculator

from vour calculator

'

4e

5

t x 6.46t/

.72r=5(.91),rx6.32/

Now that we have a pair, use This is not a right triangle and we do not have the Law of Sines: apair, so the Law of Cosines is the method to use. Since we are given a, b, and c use: y,o,y''53 s,,ffi lo= sin ZB calculator 5 9 Az = bz + c2 - 2bc(cos A) cos A is our .53(9) = 5(sinZB), sinZB = .95, yll unhnoun 52 = 92 + 62 - Z(9)(6)(cos A) .95 @ @, mZB x71.8 25=81 +36-108(cos,4) mZC = 180 - 31.8 -71.8 x76.4 B Zut B must be the biggest angle 9 ls the 108(cos,4) -92= biggest side) So B I 80 7 1 8 1 0 8 .2 (-gZ)* (- 108) = cosA,.85 = cos-,4 and mZC = 1 80 3 1 8 I 08 2 40 .85€@ ZAx3l.B''/ *See the bottom of 133 more

I

@,

258

page 135 (continued)

6.The light beam falls 100 ft. from the bottom of the thouse. If the lighthouse is 70 ft. tall, what is the angle of depression of the light

1. Find the right trian in the picture: S

angle

'tsh

of

o H C

2. Use trig (go backrvards, use the second key) to find the angle of elevation (which equals the angle of depression, by PAIAC.)

A

tan

Z! =

70

m_0

Qfi (@ mlelevation x 35 so the mZdeoression es 35 r/

fom your calcukn,

l0

beam?

page 138

NowYou T"y It Study the illustration on the right and identify each object. (Hint, some are duplicates and for this exercise, objects that appear tangenr are tangent.) Answers: G center A oointoftansencv

B C

D E F

tansent

H

diameter

secant

I

secant

K

chord tansent

I

chord chord diameter

K

2. Explain what you know about AB? The converse of the Theorem tells us that triangle ABC is 2 a right triangle and that 13 angle ,B is a right angle. Therefore, AB is tangent rc circle C.t/

It

l.

Find x. Since all tangents from a given point are equal, the middle unnamed tangent equals

6and

B

Z ooint of tansencv

pagel4O

NowYou Try

A

x=6/.

page 144

NowYou Thy It (5x+20\ l. FindT;

) a

2. Find z:

4x 3. \Mhat can you_ conclude about AB and DE? Explain.

-,"

Both of the chords go through the center so both are diameters.

(2x+20f

(5x+201

" 3*+5x+20=180 8rr= 160, x=20/ 3y=90=yt/

Both chords go through the center and therefore both pairs of opposite angles are cenffal angles and by VAT, congruent. This means we can set the two expressions equal to each other,

D

IACB = Z ECD by VAI

and since all radiuses of a circle are equal, AC = BC = DC = EC andthe two triangles are congruent by SAS. Therefore, by CPCT, AB DE =

259

'/

4x=2x+20 2x=20,x=10 4(10)+z=180 z = l40r/ D

pagel(T NowYouTrylt same or congruent , congruent chords are equally -In distant from the center of the circle(s).

Using the drawings

hints, prove the above theorem for a single circle. Prove: In a O chords are equidistant from the center. = E E E E F F F

as

G

H

l.

EF

=

HG.

G

Statements

4.

Reasons

Given.

2. All radii of a circle

lH=lF.

5. Construct altitudes AD and BD. 5. Z DAF A ZDBH are right angles.

LHBD = LFAD. 8. BD =AD, BD = AD.

7.

page

G

G

1.

2. ED=FD=UO=CO. 3. LHDG = LFDE.

F

are equal.

3. SSS congruency postulate. 4. CPCT. 5. Thru a pt. not on a line there is ex.l line r to the given 6. Definition of altitude and perpendicular lines. 7. AAS = Theorem. 8. CPCT.

149 Now You Try It tst

Find the length of a diameter of circle C.

k-)l

1.

sin 55"

HI

A

H

110" Let x = the radius x

110"

3

T

o

\$

+

.82*3x

C

x

6

=

82

A

3

lx

5

.82x

0-Ev0.tv

"3

x

3

xx

3.55 Since the diameter equals two times the radius, the diameter is x 7.32/ tst

2.The radius of circle Wis 4, find x

xo

x" is equal to the

lol HI

C

'7)

measure of the central angle.

xo

A H

3.6

tv f.

3.6

T

o A

y(

3.6 h,SY

3.6 9 4 going backwards, sin

Vz

x"

=

Cg@@ or@@CGl Vzx

x

260

x 64.2

x 128.4'/

Pagel52

NowYou Try It l. Given the anglerand arc measures shown in the figure and that BA and DC are tangents, find all of the numbered measurements. (The point at 4 is the center of the circle in the figure.)

1. 20" ^ = central I 2. 10" inscribed I =Yz^ 3. l0o inscribed I =Yz^ 4. 160" straightl = 180" 5. 10" ITT 6. 20" ^ = (2) (inscribed Z) 7. 50" Z tankchord = Yz^

A

8. 60" semicircle = 180" 9. 30" inscribed I =Yz10. 90" ^ =2(ltan&chord) ll. 70" Semicircle = 180" 12. 35" inscribed I = Yz^ 11 13. 45o inscribed Z = Yz^ 14. 100"

l'sofaA=

1

1

180"

+

2. Giye{r BC is tangent to the circle at right, mAB = (4x + 32)' and wIABC = (x + 30)o, find r.

pagel54 NowYou Try 1. Find x:

kt I

formed by a tangent and a chord is equal to Vz the arc of the chord: (r +30) = Vz(4x +32)" x +30 = 2x +16 x = l4'/

A

k Use the relationshi

3x+6

2x+4

p:

I

=

G+G 2

Substitute in the given expressions: 2x+14 = (3x+6) + (2x+ 4) 2 (2x+74) = (3x+5) + (2x+ 4) Make the left side a ratio:

l2

Cross Multiply:

Distribute: Combine like terms and solve

2.

Find

* 5d Use the relationship:

2(2x+14)

= l((3x+6) + (2x+ 4))

4x+28 =3x+6+2x+4 4x+28 =5x+10, x =l8t/

a- G*G 2

= 8l t- DC 86 - $ + 5d Cross Multiolv: LJ -T-

A

Make the left side a ratio: 86

81'

'2

-2--

B

Distribute and solve: 2(86) = 1(81 * 5d1,172 = 8l + 5d, 9l = 5a 26t

/

page 157 NowYou T"y 1. Find the value of r in the figure on the right:

It

Z

Use the formula

multiply:

2

(2x+4)"

(2x+4) (4

(2x +4)) 2(4x 1 ((8r+ 20) - 8x -2)= (8x+20) -(2x+4) Now, -4 = 8x

x- 2)"

carefully distribute the invisible negative one.

-4 = 8x+20 -2x- 4 = 6x+16 2x=20, x=70t/

Z - 6E 2- G] Let x = Small then 360- r = Big ,j60 -:r\ - (x\ 79 =

2. Given tangents ,D arrd C-D, and mZD =70, find. * 7C-, and m

EE-Gt (8x+20)'

(8r+20) . F rorm a proporuon: Gx-2\ = 2 Cross

=

Use the formula:

IEa.

70

-1-

A

_ Q60-x)-(x) 2

B

2(70)= 1((360-x)-(r))

C

140=360-x-x=360-2x -220=-2x divide by (-2) and

,o page 1.

*fr

= I lO"and

solve:

110 = x t/,

*iEd

= 360"

-

110' = 250"r/.

159 NowYouT"yIt

Findy: The product of the chord

segments are equal (2)(6) = (3)(y)

12 =3!

4=ll

2. Find the length of the shorter chord. Since the products of each chord's parts are equal, set them equal to each other:

(r +8)(x) = (5)(x +2) Now distribute. terms to

of the equation and then like terms, we end up with a which has a quadratic (r2x), a linear term (8r) and a constant term (- l0). we need to factor and then use the zero produc property to solve.

*+8x=5x+10 +3x - 10=0 (x + 5)(x

- 2) = 0

One or both expressions must equal0 so take turns setting each expression = 0 Physical quanti ties must be POSltrve so throw out the ,r * 5 answer. The shorter chord is 5 + (2 + 2) = 9 /

(r+5)=0 (x-2)=0

262

Page 16r

NowYou Try

It

l. Findr:

(x)(x+5) - (8X20) *+ 6x = 160

2. MP is a tangent,

frndy

Findingy: (3)(3+9) =(2)(2+y) 36 = 4+2!

and z:

*+6x -160 = 0 (x+16)(x-10) = g

8

32=2y,!=16

M

(x+16)=0, x=X6

O

2

(physical quantities are positive)

P

(r-10)=0, x=10/

Finding z: (3)(3+9) = *

36=* z =6t/

3. Find AB: Let x = AB. Then BC = 10.8

10.8

- x.

10.8(10.8 -x)=15(4.5) 115.54 - 10.8r =67.5

4.5 c

-10.8r=-49.14 x = 4.55 AB = 4.55'/ page 168

NowYou T"y

It

1. Find the perimeter of square UVWX. LetT equal the length of one side. By the Pphagorean Theorem,

l'*f-2f=142=195 Z1f =195 98 -z z

!2=98

I = 70 *

9.9, perimeter

^,

(4X9.9) = 39.6\/

U

t

2.The area of a rectangle is 180. The ratio of the base to the height is 4:5. Find the perimeter and the diagonal of the rectangle. Draw a rectangle that roughly matches the given information The ratio is 4:5, so you must supply the x's,4x:5x. Now, mark up the drawing, showing all the information that you know. Since

/5

A = bh

180 = (ai6x) = 20x2, * both sides by 20, then 9 = x2,3 = x. The perimeter equds 4x + 5x + 4x + 5x = l8x = (18)(3) = 54 '/ By the Pythagorean Theorem, the diagonal squared equals (12)'z+(15)2 = 369 The diagonal = x 19.2'/

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263

X

page 168 (continued) 3. Find the area of the

,1,*

Mark up the drawing, breaking the figure into

below:

r6

28

-

14

rectangular pieces.

x

Heret another way ro divide the figure: Note

x

28

50

50

20

20

I

I

Findthemissingdimensions, find x: 50 = x + x +2x+ find7: ! = 20 + 28 + 16 + 14 =78, ! =78 r4

,,/l-

11

x

14

11

x

30=330

6,

6,

50 = 4x +

44 = 4x, x =

11

= t54

30

//

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16+14 :t