geometry: a complete course - videotextthese course notes are replications of the essential content...

Written by: Larry E. Collins

Geometry:A Complete Course

(with Trigonometry)

Module E – Instructor's Guidewith Detailed Solutions for

Progress Tests

RobbinsCreative

Errata March 2015

Geometry: A Complete Course (with Trigonometry)Module E -Instructor's Guide with Detailed Solutions for Progress Tests

Copyright © 2014 by VideotextInteractive

Send all inquiries to:VideotextInteractiveP.O. Box 19761Indianapolis, IN 46219

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted,in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the priorpermission of the publisher, Printed in the United States of America.

ISBN 1-59676-113-X1 2 3 4 5 6 7 8 9 10 - RPInc - 18 17 16 15 14

Letter from the author . . .Well, are you ready to start wrapping up this study of Plane Geometry? You really have accomplished a lot.You have developed quite an extensive list of fundamental terms and spatial relationships, and have appliedthem, just as extensively, to the investigation of triangles and their parts.

Of course, in the process, I’m sure you have become much more proficient, and confident, in the demonstra-tions of “if-then” statements. As well, it doesn’t hurt, does it, that the proofs themselves have become moreconcise and more efficient. In fact, I suspect you have not had to call the help-line as much as you did at thebeginning, right? You are getting better and better. Thanks again for all of your hard work.

So, what will we encounter in this Module? Actually, we will cover two units. The first, Unit V, is a generalinvestigation of polygons other than triangles. That makes sense, when you consider that there are many “sim-ple closed plane curves, made up of straight line segments”, and we have focused only on triangles. However,you will immediately begin to notice how important it is, to have a thorough understanding of triangle relation-ships, as you examine other polygons.

Then, in Unit VI, we will take this study “to the limit” by considering the ultimate polygon, the one with aninfinite number of sides, all of which are infinitesimally small. That’s right. In Unit VI, we will be exploringrelationships with circles. Trust me, it won’t be terribly difficult, but we will encounter quite a few new terms,concepts, and applications. I’m sure you will do fine.

And, one more time, our daily work approach will be to do as many of the application exercises as possible,while exercising some selectivity when it comes to doing proofs.

So there you are. We wish you the best with these two units. And don’t hesitate to contact us if you have anydifficulty.

Thomas E. Clark, Author

Table of ContentsInstructional Aids

Program Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .ivScope and Sequence Rationale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .vi

Detailed Solutions for Progress TestsUnit V - Other PolygonsPart A - Properties of Polygons

LESSON 1 - Basic TermsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5

LESSON 2 - ParallelogramsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13

LESSON 3 - Special Parallelograms (Rectangle, Rhombus, Square)Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21

LESSON 4 - TrapezoidsLESSON 5 - Kites

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29

LESSON 6 - MidsegmentsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .35

LESSON 7 - General PolygonsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .39

Part B - Areas of PolygonsLESSON 1 - Postulate 14 - AreaLESSON 2 - Triangles

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .43

LESSON 3 - ParallelogramsLESSON 4 - Trapezoids

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .47

LESSON 5 - Regular PolygonsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .49Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .51

Module E - Table of Contents i

combined

combined

combined

Part C - ApplicationsLESSON 1 - Using Areas in ProofsLESSON 2 - Schedules

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57

Unit V Test - Form A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61Unit V Test - Form B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .71

Unit VI - CirclesPart A - Fundamental Terms

LESSON 1 - Lines and SegmentsLESSON 2 - Arcs and AnglesLESSON 3 - Circle Relationships

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .83Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .85

Part B - Angle and Arc RelationshipsLESSON 1 - Theorem 65 - “If, in the same circle, or in congruent circles, two

central angles are congruent, then their intercepted minor arcs arecongruent.”Theorem 66 -”If, in the same circle, or in congruent circles, twominor arcs are congruent, then the central angles which interceptthose minor arcs are congruent.”

LESSON 2 - Theorem 67 - “If you have an inscribed angle of a circle, then themeasure of that angle, is one-half the measure of its intercepted arc.”

LESSON 3 - Theorem 68 - “If, in a circle, you have an angle formed by a secantray, and a tangent ray, both drawn from a point on the circle, then themeasure of that angle, is one-half the measure of the intercepted arc.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .87Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .91

LESSON 4 - Theorem 69 -“If, for a circle, two secant lines intersect inside the circle,then the measure of an angle formed by the two secant lines,(or its vertical angle), is equal to one-half the sum of the measures of the arcs intercepted by the angle, and its vertical angle.”Theorem 70 - “If, for a circle, two secant lines intersect outside the circle, then the measure of an angle formed by the two secant lines, (or its vertical angle), is equal to one-half the difference of the measures of the arcs intercepted by the angle.”

LESSON 5 - Theorem 71 - “If, for a circle, a secant line and a tangent line intersectoutside a circle, then the measure of the angle formed, is equal to one-halfthe difference of the measures of the arcs intercepted by the angle.”Theorem 72 - “If, for a circle, two tangent lines intersect outside the circle,then the measure of the angle formed, is equal to one-half the difference ofthe measures of the arcs intercepted by the angle.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .95Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .99

ii Module E - Table of Contents

combined

combined

combined

combined

Module E - Table of Contents iii

Part C - Line and Segment RelationshipsLESSON 1 - Theorem 73 - “If a diameter of a circle is perpendicular to a chord

of that circle, then that diameter bisects that chord.”LESSON 2 - Theorem 74 - “If a diameter of a circle bisects a chord of the circle which

is not a diameter of the circle, then that diameter is perpendicular to thatchord.”Theorem 75 - “If a chord of a circle is a perpendicular bisector of anotherchord of that circle, then the original chord must be a diameter of the circle.”

LESSON 3 - Theorem 76 - “If two chords intersect within a circle, then the productof the lengths of the segments of one chord, is equal to the product of the lengths of the segments of the other chord.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .103Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .105

LESSON 4 - Theorem 77 - “If two secant segments are drawn to a circle from asingle point outside the circle, the product of the lengths of one secant segment and its external segment, is equal to the product of the lengths of the other secant segment and its external segment.”Theorem 78 - “If a secant segment and a tangent segment are drawn to a circle, from a single point outside the circle, then the length of that tangent segment is the mean proportional between the length of the secant segment,and the length of its external segment.”

LESSON 5 - Theorem 79 - “If a line is perpendicular to a diameter of a circle at one ofits endpoints, then the line must be tangent to the circle, at that endpoint.”

LESSON 6 - Theorem 80 - “If two tangent segments are drawn to a circle from the same point outside the circle, then those tangent segments are congruent.”

LESSON 7 - Theorem 81 - “If two chords of a circle are congruent, then theirintercepted minor arcs are congruent.”Theorem 82 - “If two minor arcs of a circle are congruent, then thechords which intercept them are congruent.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .107Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .111

Part D - Circles and ConcurrencyLESSON 1 - Theorem 83 - “If you have a triangle, then that triangle is cyclic.”LESSON 2 - Theorem 84 - “If the opposite angles of a quadrilateral are supplementary,

then the quadrilateral is cyclic.”Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .115Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .117

Unit VI Test - Form A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .119Unit VI Test - Form B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .125Unit I-VI Cumulative Review - Form A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .131Unit I-VI Cumulative Review - Form B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .137

combined

combined

combined

iv

Program OverviewThe VideoTextInteractive Geometry program addresses two of the most important aspects of math-ematics instruction. First, the inquiry-based video format contributes to the engaging of studentsmore personally in the concept development process. Through the frequent use of the pause button,you, as the instructor, can virtually require interaction and dialogue on the part of your student. As well,students who work on their own, can “simulate” having an instructor present by pausing the lesson everytime a question is asked, and trying to answer it correctly before continuing. Of course, the student mayanswer incorrectly, but the narrator will be sure to give the right answer when the play button is pressedto resume the lesson. Right or wrong, however, the student is regularly engaging in analytical and criti-cal thinking, and that is a healthy exercise, in and of itself. Second, each incremental concept isexplored in detail, using no shortcuts, tricks, rules, or formulas, and no step in the process isignored. As such, the logic and the continuity of the development assure students that they understandcompletely. Subsequently, learning is more efficient, and all of the required concepts (topics) of the sub-ject can be covered with mastery. Of course, the benefits of these efforts can be seen even more clearlyin a description of a typical session, as follows:

After a brief 2 or 3 sentence introduction of the concept to be considered, usually by examining thedescription, and the objective given at the beginning of the video lesson, you and your student can begin.You should pause the lesson frequently, usually every 15-20 seconds (or more often if appropriate), toengage your student in discussion. This means that, for a 5-10 minute VideoText lesson, it may take 10-15 minutes to finish developing the concept. Dialogue is a cornerstone. In addition, during this time,your student should probably not be allowed to take notes. Students should not have their attentiondivided, or they risk missing important links. Neither should you be dividing your attention, by lookingat notes, or writing on a pad, or an overhead projector. Everyone should be concentrating on conceptdevelopment and understanding. Please understand that a student who is accustomed to workingalone, or can be motivated to study independently, has, with the VideoText, a powerful resource toexplore and master mathematical concepts by simulating the dialogue normally encountered with a“live” instructor. And, because of the extensive detail of the explanations, along with the computer gen-erated graphics, and animation, students are never shortchanged when it comes to the insight necessaryto fully comprehend.

Once the concept is developed, and the VideoText lesson is completed, you can then employ the CourseNotes to review, reinforce, or to check on your student's comprehension. These Course Notes arereplications of the essential content that was viewed in the VideoText lesson, illustrating the same terms,diagrams, problems, numbers, and logical sequences. In fact, at this time, if your student needs a littlemore help, he or she can use the Course Notes while viewing the lesson again, using them as a guide, tore-examine the concept. The key here is that students concentrate on understanding first, and takecare of documentation later.

Please understand that it is not the intent of the program to let the VideoText lesson completely take theplace of personal instruction or interaction. Actually, the video should never tell your students any-thing that hasn't been considered or discussed (while the lesson is paused), and it should neveranswer questions that have not already been considered and resolved. As such, it becomes a “newbreed” of chalkboard or overhead projector, whereby you, as the teacher, or your student working alone,can “write”, simply by pressing the “play” button. This is a critical point to be understood, and should

v

serve to help you examine all of the materials and strategies from the proper perspective.Next, your student can begin to do some work independently, either by your personal introduction ofadditional examples from the WorkText, or by the student immediately going to the WorkText on his orher own. The primary feature of the WorkText, beside providing problem banks with which studentscan work on mastery, is that objectives are restated, important terms are reviewed, and additionalexamples are considered, in noticeable detail, taking students, once again, through the logic of theconcept development process. The premise here is simple. When students work with an instructor,whether doing exercises on their own, or working through them with other students, they are usuallyconcentrating more on “how to do” the problems. Then, when they leave the instructor, they simplydon't take the discussion of the concept with them. The goal of this program is to provide a resourcewhich will help students “re-live” the concept development on their own, whether for review, or foradditional help. That is the focus of the Student WorkText.

Having completed the exercises for the lesson being considered, your student is now ready to use thedetail in the Solutions Manual to check work and engage in error analysis. Again, it is essential to astudent's understanding that he or she find mistakes, correct them, and be required to give some explana-tion, either verbal or in writing, to you as the instructor. In fact, at this stage, you might even considergrading your student only on the completion of the work, not on its accuracy. Remember, this is thefirst time the student has tried to demonstrate understanding of a concept, and he or she may still needsome fine-tuning. So, because this is part of the initial learning process, the focus should be on a care-ful analysis of the logic behind the work, not just the answers. Finally, it is time to assess your stu-dent's mastery of the concept behind the work. Just be sure you are not testing on the same day theexercises were completed. Short-term memory can trick you into thinking that you “have it”, when, infact, you are just remembering what you did moments before. A more accurate evaluation can be madeon the next day, before moving on to the next lesson. Further, the quizzes and tests in the program oftenutilize open-response questions which will require your student to state, in writing, his or herunderstanding of the concept. This often reveals much more about a student's understanding than justchecking to see if an answer on a test is correct. Remember too, that there are two versions of everyquiz and test, allowing you to retest, if necessary, in order to make sure that your student has masteredthe concept.

Of course, just as with the WorkText, there are detailed solutions for all of the quiz and test problems,in the Instructor's Guide. Again, your student should be required to analyze problems that weremissed, and explain why the problem should have been done differently. It is simply a fact that one ofthe most powerful and effective teaching tools you can employ, is to ask your students to “articulate”to you what their thinking was, as they worked toward a given answer.

As you can see, the highly interactive quality of this program, affords students a much greater opportuni-ty than usual to grow mathematically, at a personal level, and develop confidence in their ability. Thatcan have a tremendous impact on a student's future pursuits, especially in an age where applications ofmathematics are so important.

vi

Scope and Sequence RationaleThere are two basic premises which drive concept development in Geometry, and these two essentialsshape the logical scope and sequence of geometric content.

First, it is generally understood that Geometry is the study of spatial relations. In the same way thatAlgebra is the study of numerical relations (equations and inequalities), and Calculus is concerned prima-rily with rates of change, Geometry is a comprehensive exploration of “shapes” (as sets of points), themeasurements associated with those shapes, and the relationships that can be established between thoseshapes. As such, no treatment of Geometry should ever investigate those relationships only individually,or in isolation. This is especially noticeable with traditional textbooks, which generally use a formatwhich addresses them in different “chapters”. In the VideoText Interactive Geometry course, conceptsare discussed from a “Unit” perspective, pursuing and connecting, in an exhaustive way, all of theoutcomes associated with various possibilities for a specific relationship. Of course, as much as ispossible, students need to “see” those relationships, and experience the “motion”, or “transformation”,necessary to clearly illustrate the concept. It really is impossible to put a value on the benefits of visuali-zation, in life in general, and in Geometry in particular. So, in the VideoText Interactive Geometry pro-gram, computer-generated graphics are used extensively, along with animation and color-sequencing,in order that students can actually see the relationships develop.

The second premise is that geometric concepts should be studied utilizing all of the power and convic-tion that both inductive and deductive reasoning can bring to the table. In other words, it is alwaysdesirable, and helpful, for students to “experiment”, inductively, with a geometric relationship, in aneffort to come to some general conclusion. Once that general conclusion has been arrived at, however, itis even more convincing if the student is able to “prove”, deductively, that the conclusion absolutely mustfollow, logically, from the given information. No, formal proof is not often asked for in everyday life.On the other hand, the exercise of developing that kind of thinking is invaluable, not only in some specif-ic job-related activities, but, more generally, in the daily problem-solving situations that confront us. TheVideoText Interactive Geometry program is formatted in such a way that formal proof is a cornerstone.

Unit I, then, focuses on a complete preparation for students to begin a formal study of Geometry by“re-teaching” of all of the basic geometric concepts for which students have simply memorized theappropriate term, definition, or formula. That means we must re-establish that Mathematics in general,and Geometry in particular, is a language, with parts of speech and sentence structure. We mustdevelop, in detail, the concepts associated with building geometric shapes. We must investigate, againin detail, the concepts dealing with the measurement of those shapes. Finally, we must thoroughlydevelop the principles of inductive and deductive reasoning, giving significant attention to the dynamicsof mathematical deductive logic, which are the building blocks that students will use to construct formalproofs.

In Unit II, we begin the actual study of “Plane Geometry” by developing all of the necessary terms,definitions, and assumptions we will be using as a basis for studying geometric relationships. Inother words, we draw on the analogy that studying any area of Mathematics is like “playing a game”.We must first determine which basic elements will be “undefined” in our Geometry, or accepted with-out definition. We must then determine which basic elements can be formally defined, using those unde-

fined terms. Finally, we must build a list of “postulates”, or conditional assumptions which will serveas the “rules of the game”, guiding us through the investigation of relationships, in our Geometry. It isimportant to note, at this point, that every Plane Geometry study will, in certain ways, be unique to thephilosophy of the instructor, depending on the acceptance of these fundamental terms. In other words,while the prevailing context will always be that of classical Euclidean Geometry, the lists of definitionsand postulates may differ from person to person. The key, however, is that each study will rely on itsown particular list of Essential Elements to prove the rest of the relationships to be investigated.

So, in Unit III, we use the Fundamental Terms developed in Unit II, to prove FundamentalTheorems related to points, lines, rays, segments, and angles. These theorems will be foundational tothe study of Simple Closed Plane Curves, which are the primary backdrop of all studies of PlaneGeometry.

At this point, since we have put in place the “rules of the game”, we can begin, and, for all practical pur-poses, complete, a methodic investigation of the geometric relationships associated with Triangles(Unit IV), Other Polygons (Unit V), and Circles (Unit VI). That then allows us to conclude our studyby the investigation of several applications, internal to the study of Geometry.

First, in Unit VII, we will engage in the classic geometric exploration of “Construction”. Thismeans that, with the use of only a straight edge (to construct lines, rays, and segments), and a compass(to construct circles, and arcs of circles), we will attempt to use our knowledge of geometric relationshipsto “build”, and “operate on”, various geometric shapes. Included will be the replication and division ofline segments and angles, the building of polygons to desired specifications, and the generation of circlesto desired specifications.

Second, in Unit VIII, we will examine, in significant detail, the relationships between the variouscomponents of triangles. This is, of course, the study of Trigonometry, from the Greek, meaning “tri-angle-measure”. Included are the basic relationships of sine, cosine, and tangent, as well as appli-cations involving the Pythagorean Theorem, the Laws of Sines and Cosines, and several other ambiguouscases.

Please understand that the organizational argument presented here is not meant to stifle the creativity of the instructor. Neither should it prohibit the instructor from utilizing a modular approach to conceptdevelopment. It does, however, serve to remedy the fragmented, isolated topic, “chapter” approach to a subject which has been traditionally presented to us in “textbooks”, without that element of develop-mental continuity. To that end, it speaks loudly to the curricular issues which all instructors face, and the attitudinal issues students deal with when they are presented with a new and differentMathematics course.

vii

© 2014 VideoTextInteractive Geometry: A Complete Course2

NameUnit V, Part A, Lessons 1, Quiz Form A—Continued—

2. Match each statement in column I with a phrase in column II.

Column I Column II

Rectangle _________

Diagonal of a polygon _________

Polygon _________

Convex Polygon _________

Square _________

Parallelogram _________

Trapezoid _________

Vertex of a polygon _________

Quadrilateral _________

Rhombus _________

3. A list of properties found in the group of seven special quadrilaterals is given below. Write the name of the special quadrilateral(s) beside the given property for which that property is always present.

a. Both pairs of opposite sides are parallel. _________________________________________

b. Exactly one pair of opposite sides are parallel. _________________________________________

c. Both pairs of opposite sides are congruent. _________________________________________

d. Exactly one pair of oppsite sides are congruent. _________________________________________

e. All sides are congruent. _________________________________________

f. All angles are congruent. _________________________________________

a) An equilateral parallelogram

b) A parallelogram that has one right angle

c) A closed “path” of four segments that doesnot cross itself

d) A quadrilateral that has exactly one pair ofparallel sides

e) An end point of a side of a polygon

f) A polygon in which any diagonal liesinside the polygon

g) A quadrilateral with opposite sides parallel

h) A simple closed curve made up entirelyof line segments

i) A segment whose endpoints are two non-consecutive vertices of a polygon

j) An equilangular equilateral quadrilateral

b

i

h

f

j

g

d

e

c

a

Parallelogram, Rectangle, Rhombus, Square

Trapezoid, Isosceles Trapezoid

Parallelogram, Rectangle, Rhombus, Square

Isosceles Trapezoid

Square, Rhombus

Rectangle, Square

© 2014 VideoTextInteractive Geometry: A Complete Course 3


4. Indicate whether each of the following is true or false.

a) Every square is a rhombus. _________

b) Every rhombus is a square. _________

c) Every square is a kite. _________

d) Every rhombus is a kite. _________

e) If a quadrilateral has three sides of equallength, then it is a kite. _________

f) Every property of every square is a propertyof every rectangle. _________

g) Every property of every trapezoid is a propertyof every parallelogram. _________

h) Every property of a parallelogram is a property of every rhombus. _________

True

False

False

False

False

False

False

True


Name

Class Date Score

Quiz Form A

Unit V - Other PolygonsPart A - Properties of PolygonsLesson 2 - Parallelograms

Use parallelogram ABCD to the right forproblems 1 – 6.

1. Name two pairs of congruent sides. ________________________________

2. Name two pairs of congruent angles. ________________________________

3. Name pairs of congruent segments that arenot sides of the parallelogram. ________________________________

4. Name two pairs of supplementary angles. ________________________________

5. If m�CDB = 40, find m�ABD. ________________________________

6. If m�ADC = 95, find m�ABC and m�BAD. ________________________________

Use parallelogram ABCD shown to the right to complete each statement in problems 7 – 11.

7. If AB = 3x and CD = x + 10, then AB = __________

8. If AD = 3x + 15 and BC = 21, 10. If m�BAD = 100O,then AD = __________ then m�DCE = __________

9. If AD = and BC = 2x – 12, 11. If m�ADC = 85O and m�ABD = 40O

then BC = __________ then m�DBC = __________

AB � CD; AD � CB

�DAB � �BCD; �ADC � �CBA

DE � BE; AE � EC�DAB and �ADC (or �ADC and �BCD; or �BCD and �CBA; or �CBA and �DAB)

m �ABD = 40

m �ABC = 95 and m �BAD = 85

15

4 45O

21 80O

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

AB= CD3x = x+102x = 10x = 5

AB= 3xAB= 3 5AB= 15

⋅

AD = BC3x+15 = 213x = 6x = 2

AD = 3x+15= 3 2+15

AD = 21⋅

AD = BCx2 = 2x -12x = 4x - 24

-3x = -24x = 8

BC = 2x -12= 2 8 -12= 16 -12

BC = 4

⋅

x2

180 -100 = 80

85 - 40 = 45


Name

Class Date Score

Quiz Form A

Unit V - Other PolygonsPart A - Properties of PolygonsLesson 3 - Special Parallelograms (Rectangle, Rhombus, Square)

1. State three properties relating the diagonals of a square.

________________________ ________________________ ________________________

2. A parallelogram with perpendicular diagonals is a ________________________

or a ________________________.

3. Every rectangle is also a ________________________.

4. Are the diagonals of a rhombus congruent? ________________________

5. Are the diagonals of a rectangle perpendicular to each other? ________________________

6. Are the diagonals of a square congruent? ________________________

7. Name the properties of a rhombus that are not properties of every parallelogram.

________________________ ________________________ ________________________

8. Name two kinds of parallelograms that are equilateral and two that are equiangular.

________________________________________________

________________________________________________

9. Name the properties of a square that are not properties of every rhombus.

________________________________________________________

congruent perpendicular bisect each other

rhombus

square

parallelogram

Not necessarily

Not necessarily

Yes

rhombus and square are equilateral

square and rectangle are equiangular

sides are congruent diagonals are perpendicular

A square has all angles congruent; A square has congruent diagonals

diagonals bisect interior angles


Name

Class Date Score

Quiz Form A

Unit V - Other PolygonsPart A - Properties of PolygonsLesson 4 - TrapezoidsLesson 5 -Kites

1. In trapezoid ABCD, AB || DC, m�B = 8x – 15 and m�C = 15x – 12. Find m�B. m�B = ____________

2. PQRS is a kite. Find SR and QR SR = ____________QR = ____________

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

57O

21

8.1

m�B + m�C = 1808x – 15 + 15x – 12 = 18023x – 27 = 18023x = 207x = 9

m�B = 8x – 15m�B = (8)(9) – 15m�B = 72 – 15m�B = 57

Name

Class Date Score

Quiz Form A

Unit V - Other PolygonsPart A - Properties of PolygonsLesson 6 -Midsegments

Use the figure to the right for problems 1 and 2.

1. Point H is the midpoint of GJ. GH = ____________

Point L is the midpoint of GK. HJ = _____________

If GJ = 12, and HL = 9, Find GH, HJ, and KJ. KJ = _____________

Problems 1 and 2

2. Point H is the midpoint of GJ.

Point L is the midpoint of GK. m�K = ____________

If m�GLH is 21 degrees and KJ = 14 ,

find m�K and HL. HL = ____________

3. Using the figure to the right, find DE, BC, m�A, m�B, and m�C. DE = ____________

BC = ____________m�A = ____________m�B = ____________m�C = ____________

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4


D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

GH = 12 GJ

GH = 12 12

GH = 6

⋅

⋅

HJ = GHHJ = 6

KJ = 2 HLKJ = 2 9KJ = 18

⋅⋅

m K = m GLHm K = 21∠ ∠∠

HL = 12 KJ

HL = 12 14 12

HL = 12292

HL = 294 = 7 14

⋅

⋅

⋅

DE = 12 FG

DE = 12 6

DE = 3

⋅

⋅

BC = 2 FGBC = 2 6BC = 12

⋅⋅

m A= 180 - 90 - 40m A= 50

m B= 90

m C = 40

∠∠

∠

∠

GH = 12 GJ

GH = 12 12

GH = 6

⋅

⋅

HJ = GHHJ = 6

KJ = 2 HLKJ = 2 9KJ = 18

⋅⋅

m K = m GLHm K = 21∠ ∠∠

HL = 12 KJ

HL = 12 14 12

HL = 12292

HL = 294 = 7 14

⋅

⋅

⋅

DE = 12 FG

DE = 12 6

DE = 3

⋅

⋅

BC = 2 FGBC = 2 6BC = 12

⋅⋅

m A= 180 - 90 - 40m A= 50

m B= 90

m C = 40

∠∠

∠

∠

GH = 12 GJ

GH = 12 12

GH = 6

⋅

⋅

HJ = GHHJ = 6

KJ = 2 HLKJ = 2 9KJ = 18

⋅⋅

m K = m GLHm K = 21∠ ∠∠

HL = 12 KJ

HL = 12 14 12

HL = 12292

HL = 294 = 7 14

⋅

⋅

⋅

DE = 12 FG

DE = 12 6

DE = 3

⋅

⋅

BC = 2 FGBC = 2 6BC = 12

⋅⋅

m A= 180 - 90 - 40m A= 50

m B= 90

m C = 40

∠∠

∠

∠

12

6

6

18

21O

7

3

12

50O

90O

40O

14



4. In the figure to the right, W, T, and S are midpoints

of the sides of triangle DEF. If WT = 5, ST = 8, and

SW = 7, What is the perimeter of �DEF?

Permimeter of �DEF = ____________

5. Which of the following named quadrilaterals are parallelograms?

a) b) c)

__________________ __________________ ____________________________________ __________________ ____________________________________ __________________ __________________

6. In the figure to the right, ABCD is a trapezoid with median MN as shown.a) If BC = 10t and MN = 15t, find AD. AD = ____________

b) If AD = 35x and MN = 28x, find BC. BC = ____________

c) If AD = and BC = , find MN. MN = ____________

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

ABCD is not a parallelogram

Since point A and D are not

midpoints.

XYZW is a parallelogram GHIJ is a parallelogram

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

MN = BC+AD2

15t = 10t +AD2

30t = 10t +AD20t = AD

MN = BC + AD2

28x = BC +35x2

56x = BC +35x21x = BC

MN = BC + AD2

MN = 9 3 +5 62

21x

9 3 +5 62

MN = BC+AD2

15t = 10t +AD2

30t = 10t +AD20t = AD

MN = BC + AD2

28x = BC +35x2

56x = BC +35x21x = BC

MN = BC + AD2

MN = 9 3 +5 62

21x

9 3 +5 62

MN = BC+AD2

15t = 10t +AD2

30t = 10t +AD20t = AD

MN = BC + AD2

28x = BC +35x2

56x = BC +35x21x = BC

MN = BC + AD2

MN = 9 3 +5 62

21x

9 3 +5 629 3 5 6

40

20t

21x

MN = BC+AD2

15t = 10t +AD2

30t = 10t +AD20t = AD

MN = BC + AD2

28x = BC +35x2

56x = BC +35x21x = BC

MN = BC + AD2

MN = 9 3 +5 62

21x

9 3 +5 62

DE = 2 • WT EF = 2 • SW DF = 2 • ST

DE = 2 • 5 EF = 2 • 7 DF = 2 • 8

DE = 10 EF = 14 DF = 16


NameUnit V, Part A, Lessons 6, Quiz Form B—Continued—

4. In the figure to the right, point D is the midpoint of AC, and point E is the midpoint of BC.AD = x + 5, DC = 2y + 6, DE = 2x – 5, and AB = y + 8. Find DE and AB.

AB = ____________

DE = ____________

5. Which of the following named quadrilaterals are parallelograms?

a) b) c)

__________________ __________________ __________________

6. In the figure to the right, ABCD is a trapezoid with median MN as shown.

a) If BC = 2x + 5 and MN = 10x – 1.2, find AD. b) If BC = and AD = , find MN.AD = ____________ MN = ____________

c) If BC = 6.7 and AD = 14.4, find MN. MN = ____________

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

WXYZ is not a parallelogram MNPQ is a parallelogram CDEF is not a parallelogram

D C

A B D

A B

E

C

A

B C

D

E

Q

P

S

T

R

V

D

A B

C

F E

A (0,0)

B (3,4) C (9,4)

D (6,0)

y

x

y

x A (3, -2) B (8, -2)

C (8, -5) D (3,-4)

R

S T

U

X y

(x)

24 o

o 95 o (y) o

110 o

(2x) o 64 o

45 o

(2x) o

(3y) o

X Y

Z W

R T

16

18

(x) o (y) o 3x 2y

x-y 12

x+y 8

D C

A B

E

R

W

Q

T

S D C

A B

D

A B

C

E A

D

B

C E

27

D C

A B

S

R

P

Q

21 8.1 B

E

C

D A

R N

M Q T P

S (-3,5)

R (2,0)

U (8,2)

T (9,9)

D C

B A

M

N

P O

Q

B

E

C

D A

R N

M Q T P G

J K

L H

A

C B

F

D E

G

8 6

10

5 4

5 4

40 O

F

E D

W T

S

A 5 3 3 D 4

4 C 4

3 B

3

Z 2 2 4 W

4

3 X 3 5

Y 5

G 4 4

3

H

3

2 I 2

2 J

2

B C

N M

D A

V

Z

X W

Y

B

20

C

D

20 A

10 E

C

B A

D E

W 7 7 4

X 4

3 Z 3

5 Y 4

3 3

3

3

3 3

3

3

M

N Q

P

3 C 3 5 B

M

C

A

N

D

D

6

2 E 2 4

F 4

MN = BC+AD2

10x -1.2 = 2x+5+AD2

20x - 2.4 = 2x+5+AD18x --7.4 = AD

MN = BC + AD2

MN = 3 2 +7 22

MN = 10 22

MN = 2 5 22

MN = 5 2

⋅

MN = BC + AD2

MN = 6.7 +14.42

MN = 21.12

MN = 2 10.552

MN =

⋅

110.55

AD = DCx +5 = 2y +6x - 2y = 1

-8x +2y = -36-7x = -35

x = 5

x +5 = 2y +65 +5 = 2y +6

10 = 2y +64 = 2y2 = y

DE = 2x - 5DE = 2 5 - 5DE = 10 - 5DE = 5

AB = y +8AB = 2 +8AB = 1

( )

00

AB = 2 DEy +8 = 2 2x - 5y +8 = 4x - 10

-4x + y = -182(-4x +

( )( )

yy = -18)-8x +2y = -36

MN = BC+AD2

10x -1.2 = 2x+5+AD2

20x - 2.4 = 2x+5+AD18x --7.4 = AD

MN = BC+AD2

MN = 3 2 +7 22

MN = 10 22

MN = 2 5 22

MN = 5 2

⋅

MN = BC + AD2

MN = 6.7 +14.42

MN = 21.12

MN = 2 10.552

MN =

⋅

110.55

AD = DCx +5 = 2y +6x - 2y = 1

-8x +2y = -36-7x = -35

x = 5

x +5 = 2y +65 +5 = 2y +6

10 = 2y +64 = 2y2 = y

DE = 2x - 5DE = 2 5 - 5DE = 10 - 5DE = 5

AB= y+8AB= 2+8AB= 1

( )

00

AB = 2 DEy+8 = 2 2x - 5y+8 = 4x -10

-4x+y = -182(-4x+

( )( )

yy = -18)-8x+2y = -36

MN = BC+AD2

10x -1.2 = 2x+5+AD2

20x - 2.4 = 2x+5+AD18x --7.4 = AD

MN = BC + AD2

MN = 3 2 +7 22

MN = 10 22

MN = 2 5 22

MN = 5 2

⋅

MN = BC + AD2

MN = 6.7 +14.42

MN = 21.12

MN = 2 10.552

MN =

⋅

110.55

AD = DCx +5 = 2y +6x - 2y = 1

-8x +2y = -36-7x = -35

x = 5

x +5 = 2y +65 +5 = 2y +6

10 = 2y +64 = 2y2 = y

DE = 2x - 5DE = 2 5 - 5DE = 10 - 5DE = 5

AB = y +8AB = 2 +8AB = 1

( )

00

AB = 2 DEy +8 = 2 2x - 5y +8 = 4x - 10

-4x + y = -182(-4x +

( )( )

yy = -18)-8x +2y = -36 5

10

18x – 7.4

MN = BC+AD2

10x -1.2 = 2x+5+AD2

20x - 2.4 = 2x+5+AD18x --7.4 = AD

MN = BC+AD2

MN = 3 2 +7 22

MN = 10 22

MN = 2 5 22

MN = 5 2

⋅

MN = BC + AD2

MN = 6.7 +14.42

MN = 21.12

MN = 2 10.552

MN =

⋅

110.55

AD = DCx +5 = 2y +6x - 2y = 1

-8x +2y = -36-7x = -35

x = 5

x +5 = 2y +65 +5 = 2y +6

10 = 2y +64 = 2y2 = y

DE = 2x - 5DE = 2 5 - 5DE = 10 - 5DE = 5

AB= y+8AB= 2+8AB= 1

( )

00

AB = 2 DEy+8 = 2 2x - 5y+8 = 4x -10

-4x+y = -182(-4x+

( )( )

yy = -18)-8x+2y = -36

MN = BC+AD2

10x -1.2 = 2x+5+AD2

20x - 2.4 = 2x+5+AD18x --7.4 = AD

MN = BC + AD2

MN = 3 2 +7 22

MN = 10 22

MN = 2 5 22

MN = 5 2

⋅

MN = BC + AD2

MN = 6.7 +14.42

MN = 21.12

MN = 2 10.552

MN =

⋅

110.55

AD = DCx +5 = 2y +6x - 2y = 1

-8x +2y = -36-7x = -35

x = 5

x +5 = 2y +65 +5 = 2y +6

10 = 2y +64 = 2y2 = y

DE = 2x - 5DE = 2 5 - 5DE = 10 - 5DE = 5

AB = y +8AB = 2 +8AB = 1

( )

00

AB = 2 DEy +8 = 2 2x - 5y +8 = 4x - 10

-4x + y = -182(-4x +

( )( )

yy = -18)-8x +2y = -36

10.55

In Problems 1 - 3, find the number of sides of a polygon if the sum of the measure of its angles is:

1. 8640O sides = ______ 2. 1440O sides = ______ 3. 1800O sides = _______

In Problems 4 - 6, if the measure of each interior angle of a regular polygon is the given measure,how many sides does the polygon have?

4. 162O sides = _______ 5. 150O sides = _______ 6. 108O sides = _______

In Problems 7 - 9, find the sum of the measures of the interior angles of a polygon with the given number of sides.

7. 11 sides sum = _______ 8. 9 sides sum = _______9. 102 sides sum = _______

In Problems 10 - 12, find the measure of each exterior angle of a regular polygon with the givennumber of sides.

10. 3 angle = _______ 11. 5 angle = _______ 12. x angle = _______

37

Name

Class Date Score

Quiz Form A

Unit V - Other PolygonsPart A - Properties of PolygonsLesson 7 -General Polygons

50 10 12

20 12 5

1620O 1260O 18,000O

120O 72O

n - 2 180 = 8640n - 2 = 48n = 50

( ) ⋅ n - 2 180 = 1440n - 2 = 8n = 10

( ) ⋅ n - 2 180 = 1800n - 2 = 10n = 12

( ) ⋅

n - 2 180n = 162

n - 2 180 = 162n180n - 360 = 162n

-3

( ) ⋅

( ) ⋅

660 = -18n20 = n

n - 2 180n = 150

n - 2 180 = 150n180n - 360 = 150n

-3

( ) ⋅

( ) ⋅

660 = -30n12 = n

n - 2 180n = 108

n - 2 180 = 108n180n - 360 = 108n

-3

( ) ⋅

( ) ⋅

660 = -72n5 = n

n - 2 18011- 2 180

9 1801620

( ) ⋅( ) ⋅

⋅

n - 2 1809 - 2 180

7 1801260

( ) ⋅( ) ⋅

⋅

n - 2 180102 - 2 180

100 18018,000

( ) ⋅( ) ⋅

⋅

360n

3603

120

360n

3605

72

360n

360x

360x

o

© 2014 VideoTextInteractive Geometry: A Complete Course


Name

Class Date Score

Quiz Form A

Unit V - Other PolygonsPart B - Areas of PolygonsLesson 1 -Postulate 14 - AreaLesson 2 - Triangles

For problems 1 – 6, find the area of the given polygon using the appropriate Postulate, Theorem, or Corollary from lessons 1 and 2.

1. A = _______ 2. A = _______

3. A = _______ 4. A = _______

5. A = _______ 6. A = _______

24 units2 45 units2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

a +b = ca +5 = 12a +25=144a = 119a = 119

2 2 2

2 2 2

2

2

A = 12 b h

A = 12 12 4

A = 12 42 = 2 6 4

2 = 24 units2

⋅ ⋅

⋅ ⋅

⋅ ⋅ ⋅

A = b hA = 9 5A = 45 units2

⋅⋅

A = 12

A = 12 4 12

A = 4 122 = 2 2 12

2 = 24 units

⋅ ⋅

⋅ ⋅

⋅ ⋅ ⋅

1 2

22

5 119 units2

A = b hA = 119 5A = 5 119 units2

⋅

⋅

3 2 = x 23 = x

A = b hA = 3 3A = 9 units2

⋅⋅

6 = x 262

= x

3 2 22 2

= 3 2 22 = x

3 2 = x

⋅ ⋅⋅

⋅ ⋅

A = 12

A = 12 3 2 3 2

A = 3 3 22 = 9 units2

⋅ ⋅

⋅ ⋅

⋅ ⋅

1 2

24 units2

9 units2 9 units2


Name

Class Date Score

Unit V - Other PolygonsPart B - Areas of PolygonsLesson 1 -Postulate 14 - AreaLesson 2 - Triangles

For problems 1 – 6, find the area of the given polygon using the appropriate Postulate, Theorem, or Corollary from lesson 1 and 2.

1. A = _______ 2. A = _______

3. A = _______ 4. A = _______

5. A = _______ 6. A = _______

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

36 3 units2352 units2

x 32 where x = 12

12 32 = 2 6 3

2 = 6 3⋅ ⋅

A = 12

A = 12 5 7

A = 352 units2

⋅ ⋅

⋅ ⋅

1 2

A = 12

b h

A = 12

6 55

A = 6 552

= 2 3 552

= 3 55units2

⋅ ⋅

⋅ ⋅

⋅ ⋅

3 55units2

25 3 units2812 units2

A = s s or sA = 8A = 64 units

2

2

2

⋅

Height = 10 32 = 5 3

A = 12 b h

A = 12 9 2 9 2

2

A = 9 2 9 22 2

A = 81 22 2 = 81

2 u

⋅ ⋅

⋅ ⋅

⋅⋅

⋅⋅

nnits2

x2 where x = 12

122 = 6

A = b hA = 6 3 6A = 36 3 units2

⋅

⋅

x +3 = 8x +9 = 64x = 55x = 55

2 2 2

2

2

x 32

Area = 12 b h

= 12 10 5 3

= 10 5 32 = 2 5 5 3

2= 25 3 uni

⋅ ⋅

⋅ ⋅

⋅ ⋅ ⋅

tts2

x 2 = 9Altitude of triangle is

x where

x = 92

x = 9 22 2

= 9 22⋅

64 units2

Quiz Form B

Base: Side opposite 60O angle Height: Side opposite 30O angle

Height is the long leg of the righttriangle with short leg equal to 3and hypotenuse equal to 8.

Base of triangle is 10. Height of triangle is the long leg of a right triangle with a hypotenuse of 10. So, the height is where x = 10.


Name

Class Date Score

Unit V - Other PolygonsPart B - Areas of PolygonsLesson 3 -ParallelogramsLesson 4 - Trapezoids


1. A = _______ 2. A = _______

(Parallelogram) (Trapezoid)

3. A = _______ 4. A = _______

(Trapezoid)

(Parallelogram)

5. A = _______ 6. A = _______

(Parallelogram)

(Trapezoid)

Quiz Form A

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

96 units2

Height is x where x 2 = 8

x = 82

= 8 22 2

= 8 22

x = 2⋅

⋅⋅

⋅ ( )⋅ ⋅ ( )

⋅

4 22 = 4 2

A = 12 h b +b

A = 12 4 2 15 +3

A = 4 2 182

1 2

== 2 2 2 182 = 36 2 units2⋅ ⋅

A = b hA = 12 8A = 96 units2

⋅⋅

1332 units2

A = 12

h b +b

A = 12

7 11+8

A = 1 7 192

= 1332

un

1 2⋅ ⋅ ( )⋅ ⋅ ( )⋅ ⋅ iits2

A = 12

h b +b

A = 12

4 9 +5

A = 1 4 142

= 2 2 142

1 2⋅ ⋅ ( )⋅ ⋅ ( )⋅ ⋅ ⋅ ⋅ == 28 units2

40 3 units2

Height is x 32

where x = 10

Height is 10 32

= 2 5⋅ 332

= 5 3

A = b h

A = 8 5 3

A = 40 3 units2

⋅⋅

36 2 units2

Height is x2

where x = 5

Height is 52

A = b h

A =

⋅

66 52

A = 6 52

= 2 3 52

= 15 units2

⋅

⋅ ⋅ ⋅

28 units2

15 units2

egion.

1532 units2 102+9 3

40 units2

The area of the rectangle is b h or 13 ⋅ ⋅110 or 130 units

Area of each trapezoid is

2

112 h b +b or 12 5 13+5

= 5 182 = 5 2 9

2

1 2( ) ⋅ ⋅ ( )⋅ ⋅ ⋅ == 45 units

Area of shaded triangles is the

2

aarea of the rectangle minus the areaof eachh trapezoid. 130 - 45 - 45 = 40 units2

A = 12 h b +b = 1

2 5 15 +12 = 5 272 = 135

2 units1 22( ) ⋅ ( ) ⋅

A = s s = 3 3 = 9 units2⋅ ⋅

Total area of the polygonal region is 1352 +99 = 135

2 + 182 = 153

2 units2

3 33 3

A = 12 h b +b

A = 12 3 3 12+6

A = 3 3 182 = 3 3 2 9

2 = 2

1 2( )

⋅ ( )⋅ ⋅ ⋅ 77 3 units2

6 3 units2

Height of trapezoid and parallelogram is 3 3..(Use 30 ,60 ,90 right triangle propertieO O O ss)

Area of shaded triangle is area of trapezzoid minus area of parallelogram.12 h b +b1⋅ ⋅ 22 - b h = 1

2 3 3 9 +5 - 5 3 3 = 3 3 142 -15 3 = 42 3

2 - 1( ) ⋅ ⋅ ⋅ ( ) ⋅ ⋅ 55 31

= 42 32 - 15 3 2

2 = 42 32 - 30 3

2 = 12 32 = 6 3 units2⋅

A = b hA = 6 17 - 3 3

A = 102 - 18 3

⋅

⋅ ( )

So, total area equals 27 3 +102 -18 3 = 102 +9 3


NameUnit V, Part B, Lessons 3&4, Quiz Form A—Continued—

For problems 7 and 8, find the area of each polygonal region.

7. A = _______ 8. A = _______

For Problems 9 and 10, find the area of the shaded region.

9. A = _______

10. A = _______

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

Divide the region into a trapezoid (with bases equal to 10and 6 and height equal to ), and a parallelogram (withbase equal to 6 and height equal to 17 - )

Divide the region into a trapezoid (with one base equal to15, a second base equal to 12 and an altitude equal to 5)and a 3 x 3 square.

A = b hA = 12 7

A = 121

152

A = 6 2 152 = 90 units

12

2

⋅⋅

⋅

⋅ ⋅

15 1194 units2

a +b = c52 +b = 6

254 +b = 36

b = 1444 - 25

4

2 2 2

22 2

2

2

== 1194

b = 1194 = 119

4= 119

2

A = 12 h b +b = 1

21192 10 +5 = 119 15

2 2 = 15 11 2⋅ ⋅ ( ) ⋅ ⋅ ( ) ⋅

⋅119

4 units2

a +b = c3 +h = 79 +h = 49h = 40h = 40 = 4 10 = 2 10

A =

2 2 2

2 2 2

2

2

⋅

bb h

A = 8 2 10

A = 16 10 units2

⋅⋅

28 units2

A = 12

h b +b

A = 12

4 8 +6

A = 4 142

= 2 2 142

= 2

1 2⋅ ⋅ ( )⋅ ⋅ ( )⋅ ⋅ ⋅ 88 units2

16 10 units2

A = b h or s s

s = 72

= 7 22 2

= 7 22

A = 7 22

7 22 = 7 7

⋅ ⋅

⋅

⋅ ⋅ ⋅⋅⋅

22 2 = 49

2 units2

492 units2

A = 12 h b +b

h = 4 (Use 45 ,45 ,90 right tr

1 2

O O O

⋅ ⋅ ( )

iiangle properties)

A = 12 4 12 +8

A = 4 202 = 2 2

⋅ ⋅ ( )⋅ ⋅ ⋅⋅20

2 = 40 units2

90 units2


Name

Class Date Score

Unit V - Other PolygonsPart B - Areas of PolygonsLesson 3 -ParallelogramsLesson 4 - Trapezoids


1. A = _______ 2. A = _______

(Trapezoid) (Parallelogram)

3. A = _______ 4. A = _______


5. A = _______ 6. A = _______


Quiz Form B

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

40 units2

Height of a trapezoid is the altitudeof the isosceles triangle. The altitudebisects the base of the triangle.

92 1+ 2 units2( )

6 3 units2

392 units2

Area = b h= 3 2 = 6= 4 1 = 4= 5 1 = 5= 6 1 = 6= 7 1 = 7= 8 1 =

⋅⋅⋅⋅⋅⋅⋅ 88

= 9 1 = 9Total Area = 45 units2

⋅

Area = 12 b h where h = 2, and b = 2 3.

(Use 30 ,60O

⋅ ⋅OO O,90 right triangle properties)

Total Area of the polygonal regions is 3 12

b h⋅ ⋅ ⋅

⋅ ⋅ ⋅

.

31

12

2 31

21

= 6 3 units2

For the triangle on the left

A = 12 3 3

A = 92

⋅ ⋅

Area of the shaded region is the area of thee triangle minus

the area of the trapezoid.. Height of the trapezoid is 62 or 3.

= 12 b⋅ ⋅⋅⋅ ( )

⋅ ⋅ ⋅ ( )h - 1

2 h b +b

= 12

81

91 - 1

231 8 +3

(Use 30

1 2

O )seitreporp elgnairt thgir 09,06,,

= 12

2 4O O

⋅ ⋅11

91 - 1

231

112

= 36 - 332 = 72

2 - 332 = 39

2 units2

⋅ ⋅ ⋅

For the triangle on the right

A = 12 3 3 2

(Use

⋅ ⋅

45 ,45 ,90 right triangle properties)

A = 3

O O O

⋅⋅3 22 = 9 2

2

Total Area = 92 + 9 2

2

= 9 +9 22

or9 1+ 2

2or 9

2 1+ 2 un

( )( ) iits2


NameUnit V, Part B, Lessons 3&4, Quiz Form B—Continued—

For Problems 7 and 8, find the area of each polygonal region.

7. A = _______ 8. A = _______

For Problems 9 and 10, find the area of the shaded region.

9. A = _______

10. A = _______

45 units2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

(working from the top down.)

4 4

42 2o30 o30

5.13

o60

x x

o60o30 o30

5.13

10.23

360n = 360

3 = 3 1203 = 120o⋅ 360

n = 3608 = 8 45

8 = 45o⋅

360n = 360

12 = 12 3012 = 30o⋅ 360

n = 36010 = 10 36

10 = 36o⋅

3

5. p = n 5p = 3 4p = 12

2 = x 32

4 = x 343

= x

4 33 = x

⋅⋅

Apothem = 12 x

= 12

4 33

= 2 2 32 3

= 2 33

A = 12 a p

A = 12

⋅

⋅ ⋅⋅

⋅ ⋅

⋅⋅⋅

⋅ ⋅⋅

2 33

121

A = 2 3 3 42 3

A = 4 3 units2

2 33 4 3 units2

6. A = 12

a p

12 = 12

31

p

12 = 32

p

23

121

= 243

= p

⋅ ⋅

⋅ ⋅

⋅

⋅

24 33

= 3 8 33

= 8 3 = p

p= n s

8 3 = 6 s

8 36

= s

2 4

⋅ ⋅

⋅

⋅

⋅ 332 3

= s

4 33

= s

⋅

10.23

7. p = n s20.4 = 6 s20.4

6= s

2 10.22 3

= s

10.23

= s

5.1

⋅⋅

⋅⋅

33= x

210.2 = 3x10.2

3= x

Apothem = x 32

=

10.23

31

2=

10.2 33

12

12

21

=

2 5.1⋅ ⋅ ⋅

⋅

⋅ ⋅⋅⋅

⋅

⋅ ⋅

32 3

22

= 5.1 33

= 3 1.7 33

= 1.7 3 units

A = 12

a p

A = 112

1.7 3 20.4

A =1.7 3 10.2 2

2 A = 17.3

⋅ ( ) ⋅ ( )

( ) ⋅ ( ) ⋅

44 3 units2

17.34 3 units21.7 3


Name

Class Date Score

Unit V - Other PolygonsPart B - Areas of PolygonsLesson 5 -Regular Polygons

For problems 1-4, find the degree measure of each central angle of each regular polygon with thegiven number of sides.

1. 3 degree measure = _______________ 2. 8 degree measure = _______________


For problems 5-7, complete the chart for each regular polygon described.

n s P a A

5. 3 4 ________ ________ ________

6. 6 ________ ________ 6 3 units2

7. 6 ________ 20.4 ________ ________

Quiz Form A

120O 45O

30O 36O

12

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

2 12


NameUnit V, Part B, Lesson 5, Quiz Form A—Continued—

8. Find the area of an equilateral triangle inscribed in a circle, with a radius of units. Area = ______________

9. Find the area of a square with an apothem of 8 inches and a side of length 16 inches. Area = ______________

10. Find the area of a regular hexagon with an apothem of meters and a side of length 22 meters. Area = ______________

A = 12 s a n or A = s = 16 = 256 inches

A = 12

2 2 2⋅ ⋅ ⋅

⋅116 8 4

A = 16 8 2 22

A = 16 16A = 256 inches2

⋅ ⋅

⋅ ⋅ ⋅

⋅

256 inches2

A = 12 s a n

A = 12 22 11 3 6

A = 726 3 = 1257.47 meters2

⋅ ⋅ ⋅

1257.47 meters 2

Apothem = 12 4 3

= 4 32 = 2 2 3

2 = 2 3

s = 2 2 3 3s = 4 3 3s = 4

⋅

⋅

⋅ ⋅

⋅⋅⋅3

s = 12

A = 12 s a n

A = 12 12 2 3 3

A = 12 2 3 32 = 36 3 units2

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅

36 3 units2

11 3

4 3

•

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

360n = 360

9 = 9 409 = 40o⋅ 360

n = 36018 = 18 20

18 = 20o⋅

360n = 360

6 = 6 606 = 60o⋅ 360

n = 3605 = 5 72

5 = 72o⋅

5. p = n s12 = 4 s124 = s

3 44 = s

3 = s

Apothem = 32

⋅⋅

⋅

A = 12 a p

A = 12

32

121

A = 3 2 2 32 2 = 9 units2

⋅ ⋅

⋅ ⋅

⋅ ⋅ ⋅⋅

32

6. p = n sp = 6 8p = 48

4 = 12

x

8 = x

Apothem = x 32

= 8 32

= 2 4

⋅⋅

⋅ 332

= 4 3

4 3 96 3 units2

7. p = n sp = 8s

A = 12

a p

A = 12

a1

8s1

A = a 2 4s2

= 4as

⋅

⋅ ⋅

⋅ ⋅

⋅ ⋅ units2


Name

Class Date Score

Unit V - Other PolygonsPart B - Areas of PolygonsLesson 5 -Regular Polygons

For problems 1-4, find the degree measure of each central angle of each regular polygon with thegiven number of sides.



For problems 5-7, complete the chard for each regular polygon described.

n s P a A

5. 4 ________ 12 ________ ________

6. 6 8 ________ ________ ________

7. 8 s _________ a ________

Quiz Form B

40O 20O

60O 72O

3 9 units2

48

4as units28s

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

A= 12 s a n

A = 12131

8.91

51

A = 13 8.9 52

A = 578.

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅

552

A = 289.25 feet 2

289.25 feet 2

186 cm2

27 3 units2

A = 12 s a n

A = 12 6 1

5 7 12 8

A = 1 31 3 5 2 2 22 5

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅⋅

⋅ ⋅ ⋅2

A = 31 3 2 = 31 6A = 186 cm2

A = 12 s a n

A = 12 6 3 3 3

A = 6 3 3 32 = 2 3 3 3 3

2A = 2

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅

77 3 units2


NameUnit V, Part B, Lesson 5, Quiz Form B—Continued—

x

12

10 4 5

9

9 9

6

6

4

4

4

4

4

4

8

7 7

2 3 9

9 4

5

4

12 5 12 3 2

45 6

9

7

7

7

7

O

12

20

8

15 5

9 16

20

3 12

10

17

5

7 8

8

8

6 60 O

30 O

10 45

9

O

5

5

5

5

8

10

o

o

60

o 60

o 60

o 60

o 45

o 30 o 30 o 30

2

2

2

2 11

9 9

9 9

8

12

9

12 3

3

8

12

11

7

8

9

4

5

15

3

8

45

6

17

4

7

7

8

8

3

8

3

3

3

3

4 4 8

3

Apothem = 3

4 4

4 2 2

3

6

6

6

6 10

13

13 9

5

6

6 6 6

9 3

8

6

5 10

12

7

5

5

5

7

4 4

= 3

5

10 6

o 30 5

6

4 2

8

12

3 2

8

9

1 2

1 1 1 1

1 1 1 1

1 1 1 1

o 30 o 30 o 60

3 2

6 6 3

6 o 30 6

2 5.1 3

o 60

x x

o 60 o 30 o 30

5.1 3

10.2 3

4 3 2 3

3

3 2

8. Find the area of a regular pentagon with apothem 8.9 feet and a side of length 13 feet. Area = ______________

9. Find the area of a regular octagon with apothem 71/2 centimeters and a side of length 61/5 centimeters. Area = ______________

10. Find the area of an equilateral triangle inscribed in a circlewith radius of 6 units. Area = ______________


NameUnit V, Part C, Lessons 1&2, Quiz Form A—Continued—

9. �ABC is a right triangle with AB = 12 and BC = 5. BD is a median of the triangle. What is the area of �ABD? ____________

10.Two similar triangles have areas of 81 square inches and 36 square inches. Find the length of a side of the larger triangle if a corresponding side of the smaller triangle is 6.

Side = ____________

11.Make a complete schedule for a round robin tournament with 6 teams.

Week 1 ________________

Week 2 ________________

Week 3 ________________

Week 4 ________________

Week 5 ________________

Area ABC is

Area ABC is AB BC

12 b h

12A

⋅ ⋅

( )( )

rrea ABC is 12 12 5

Area ABC is 30 units2

⋅

So, Area ABD = 12 0

Area ABD = 15 units2

⋅ 3

36 = 6

81 = 99 = x

15 units2

9

1–23–54–6

2–31–45–6

1–34–52–6

1–52–46–3

3–42–51–6

C

A

1

2

34

56

7

8E

D

B

Y

W 5

U4

Z

X

A

B A

D45

8 C

8

B8

8

135CD

P

M

10

8

O

Q

1

2

3

45

6

7

N

T

R 5

V9 8

U

S

A B

C H

J I

A

B C

D16

KD

O O

M

N

QP

4

91

2

34

56

1

2

34

5

6

1

2

34

56

1

2

34

56

7

7

7

7

10

10

8

8 x10 10

60O

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

C

A

1

2

34

56

7

8E

D

B

Y

W 5

U4

Z

X

A

B A

D45

8 C

8

B8

8

135CD

P

M

10

8

O

Q

1

2

3

45

6

7

N

T

R 5

V9 8

U

S

A B

C H

J I

A

B C

D16

KD

O O

M

N

QP

4

91

2

34

56

1

2

34

5

6

1

2

34

56

1

2

34

56

7

7

7

7

10

10

8

8 x10 10

60O

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

C

A

1

2

34

56

7

8E

D

B

Y

W 5

U4

Z

X

A

B A

D45

8 C

8

B8

8

135CD

P

M

10

8

O

Q

1

2

3

45

6

7

N

T

R 5

V9 8

U

S

A B

C H

J I

A

B C

D16

KD

O O

M

N

QP

4

91

2

34

56

1

2

34

5

6

1

2

34

56

1

2

34

56

7

7

7

7

10

10

8

8 x10 10

60O

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

C

A

1

2

34

56

7

8E

D

B

Y

W 5

U4

Z

X

A

B A

D45

8 C

8

B8

8

135CD

P

M

10

8

O

Q

1

2

3

45

6

7

N

T

R 5

V9 8

U

S

A B

C H

J I

A

B C

D16

KD

O O

M

N

QP

4

91

2

34

56

1

2

34

5

6

1

2

34

56

1

2

34

56

7

7

7

7

10

10

8

8 x10 10

60O

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

C

A

1

2

34

56

7

8E

D

B

Y

W 5

U4

Z

X

A

B A

D45

8 C

8

B8

8

135CD

P

M

10

8

O

Q

1

2

3

45

6

7

N

T

R 5

V9 8

U

S

A B

C H

J I

A

B C

D16

KD

O O

M

N

QP

4

91

2

34

56

1

2

34

5

6

1

2

34

56

1

2

34

56

7

7

7

7

10

10

8

8 x10 10

60O

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

C

A

1

2

34

56

7

8E

D

B

Y

W 5

U4

Z

X

A

B A

D45

8 C

8

B8

8

135CD

P

M

10

8

O

Q

1

2

3

45

6

7

N

T

R 5

V9 8

U

S

A B

C H

J I

A

B C

D16

KD

O O

M

N

QP

4

91

2

34

56

1

2

34

5

6

1

2

34

56

1

2

34

56

7

7

7

7

10

10

8

8 x10 10

60O

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1

2

3

45

6

7

1–2, 3–5, 4–6

2–3, 1–4, 5–6

3–4, 2–5, 1–6

1–3, 4–5, 2–6

1–5, 2–4, 6–3


NameUnit V, Unit Test Form A—Continued—

3. Find the length of the sides of parallelogram ABCD if the perimeter of the parallelogram is 110cm. and the measure of two consecutive sides is 3x – 2 and 2x + 12 respectively. AB = __________

BC = __________CD = __________DA = __________

4. RSTU is a parallelogram. RV = 8, and UV = 5.Find RT and US. Give a reason to justify your answers.

RT = ________ US = _________________________________________________________________________________________________________

5. ABCD is a parallelogram. m�A = 37, find m�B, m�C, and m�D. Give a reason to justify your answers.

m�B = ________ m�C = _______m�D = ________________________________________________________________________________________________________

(A-2)

(A-2)

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

Perimeter = AB + BC + CD + DA

AB = CD and BC = DA (Theorem 41 - If a quadrilateral is a parallelogram, then both pairs of opposite sides are congruent.)

110cm = 3x - 2 + 2x+12 + 3x - 2 + 2x+12110cm = 3x

( ) ( ) ( ) ( )++2x+3x+2x+ 2+ 2+12+12

110cm = 10x+2090cm = 10x9

- -

ccm = x

So, 3x - 2 = 3 9 - 2 = 27 - 2 = 25and, 2x +12 = 2 9

( )( )++12 = 18 +12 = 30

RT = RV +VTRV = VT

RT = 8 +8RT = 16

US = UV +VSUV = VS

US = 5 +5US = 10

37 +m B = 180m B = 143

m C = 37m D = 143

∠∠

∠∠

(A-2)

30

25

30

25

(1 possibility)

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

RT bisects SU and SU bisects RT (Theorem 43 -

If a quadrilateral is a parallelogram, then its

diagonals bisect each other.

m�A + m�B = 180 (Theorem 42 - If a

quadrilateral is a parallelogram, then any pair

of supplementary angles are supplementary.

16 10

143O 37O

143O

Corollary 42a - If a quadrilateral is a parallelogram, then opposite angles are congruent.



6. MNOP is a rectangle as shown. Find MO and NP. Give a reason to justify your answers.

MO = ________ NP = _________________________________________________________________________________________________________

7. Quadrilateral FGHI is a rhombus as shown. Find FG, GH, and HI. Give a reason to justify your answers.

FG = ________ GH = ________HI = ________________________________________________________________________________________________________

8. Quadrilateral STUV is a rhombus as shown. Find m�1, m�2, and m�T if m�V = 50. Give a reason to justify your answers.

m�1 = ________ m�2 = _______m�T = ________________________________________________________________________________________________________

(A-3)

(A-3)

(A-3)R

V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

MO = NP (Theorem 49 - If a quadrilateral is a

rectangle, then its diagonals are congruent.)

A rhombus is a quadrilateral with four

congruent sides. FG = GH = HI = IF, so all

sides equal 6 units.

a2 + =

+ = ( )+ = ( )= ( )

b cMOMO

MO

2 2

2 2 2

2

2

11 5121 251461466

146

=

=

MO

NP

146 146

If m V = 50, then m T = 50

m V +m S = 18050 +m S = 180

∠ ∠

∠ ∠∠

mm S = 130

m 1 = m 2and

m 1+m 2 = m Sm 1+m 2 = 130m 1

∠

∠ ∠

∠ ∠ ∠∠ ∠∠ ++m 1 = 130

2 m 1 = 130m 1 = 65 = m 2

∠⋅ ∠

∠ ∠

6 6

6

A rhombus is a parallelogram, so opposite angles are

congruent consecutive angles are supplementary, and

diagonals bisect opposite interior angles.

65O 65O

50O

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5



9. Quadrilateral MNPQ is a rhombus. Find NR if PQ = 8and MR = 4. Give a reason(s) to justify your answers.

NR = ________________________________________________________________________________________________________________________________________________________________________________________________________

10. ABCD is an isosceles trapezoid. If m�D = 60, find m�A m�B, and m�C. Give a reason(s) to justify your answer.

m�A = ________ m�B =________m�C = ________________________________________________________________________________________________________________________________________________________________________

11. WXYZ is an isosceles trapezoid. If WZ = 12 and WY = 16, find XY and XZ. Give a reason(s) to justify your answer.

XY = ________ XZ = ________

________________________________________________________________________________________________

(A-4)

(A-4)

(A-3)R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

A rhombus is a parallelogram, so the diagonals

bisect each other (Theorem 43) So, if MR = 4,

then PR = 4. Diagonals of a rhombus are

perpendicular.(Theorem 51) This means QPR is

a right triangle. NP is equal to PQ since a

rhombus has four congruent sides.

m�D = m�C since each pair of base angles of

an isosceles trapezoid are congruent (Theorem

53) AB || DC by the definition of a trapezoid.

�A and �D are supplementary, and �B and

�C are supplementary (Theorem 17)

NRNR

NR

( ) + ( ) = ( )( ) + ( ) = ( )( ) + =

2 2 2

2 2 2

24 816 64

PR NP

NNR( ) =

= = ⋅ = ⋅ =

2 4848 16 3 16 3 4 3NR

4 3

m C = 60 = m Dm D +m A = 180m A = 120 (180 - m D)m B = 1

∠ ∠

∠ ∠

∠ ∠

∠ 220

120O 120O

60O

If WY = 16, then XZ = 16 (Corollary 53a)

If WZ = 12, then XY = 12 (Definiton of Isosceles

Trapezoid)

12 16

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5



16. a) Find the area of the kite shown to the right Area = __________

b) If m�EFG = 38O, Find m�HFG m�HFG = ________

c) Find m�EIF. m�EIF = __________

17. Points P and Q are midpoints of the sides of �DEF, shown to the right. Complete each of the followinga) FE = 18; PQ = __________

b) FE = 2x2 – 7x + 10; PQ = x2 – 9; FE = _________; PQ = _________.

c) PQ = x + 3; FE = 1/3x + 16; PQ = __________

d) PQ = 18; FE = __________

a) b) c) d)

(A-5)(C-1)

(A-6)

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

80 units2

19O

90O

9

14 7

9

36

a b cEIEI

EI

2 2 2

2 2 2

2

2

12 13144 169

+ =

( ) + ( ) = ( )( ) + =

( ) ==

=

+ =

=

+ =

=

255

5 510

EI

EI IG EGEI IG

EGEG

a b cEI IH HE

IH

2 2 2

2 2 2

2 2 25 41

25

+ =

( ) + ( ) = ( )

( ) + ( ) = ( )++ ( ) =

( ) =

=

+ =

+ =

=

IHIH

IH

FI IH FHFHFH

2

2

41164

12 416

Area of EFGH = 12 EG FH

= 12 10 16

= 2 5 162 = 8

( ) ⋅ ( )

⋅ ⋅

⋅ ⋅ 00 units2

m HFG = 12

m EFG Corollary 54a

= 12

38

= 2 192

= 19

∠ ⋅ ∠

⋅

⋅

m EIF = 90 Corollary 54c∠

PQ = 12 FE

PQ = 12 18

PQ = 9

⋅

⋅

PQ = 12 FE

2 PQ = FE2 x - 9 = 2x -7x +102x -18 = 2x -

2 2

2 2

⋅

⋅

( )77x +10

-18 = -7x +10-28 = -7x

4 = x

PQ = 12 FE

x +3 = 12

13 x +16

2x +6 = 13 x +16

6x +18 =

⋅

xx +165x = 30x = 6

PQ = x +3PQ = 6 +3 = 9

PQ = 12 FE

18 = 12 FE

36 = FE

⋅

⋅

FE x xFEFE

= − += ( ) − ( ) += − + =

2 7 102 4 7 4 1032 18 10 14

2

2

PQ xPQ

= −=( ) − = − =

2

2

94 9 16 9 7

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5



18. Find the sum of the measures of the interior angles of a 12-sided polygon. Sum = ________

19. The sum of the measures of the interior angles of a polygon is 1980O. How many sides does the polygon have? ________________

20. Find the measure of each angle of a regular 15-gon. ________________

21. The measure of an exterior angle of a regular polygon is 18O.How many sides does the polygon have? ________________

(A-7)

(A-7)

(A-7)

(A-7)

13

n - 2 18012 - 2 18010 1801800

( ) ⋅( ) ⋅

⋅

n - 2 180 = 1980180n - 360 = 1980

180n = 2340

n = 23401

( ) ⋅

880 = 180 13180 = 13⋅

n - 2 180n

15 - 2 18015

13 5 3 125 3 = 13 12 = 15

( ) ⋅

( ) ⋅

⋅ ⋅ ⋅⋅

⋅ 66

360n = 18

360 = 18n20 = n

156O

20

1800O



22. Find the area of each of the following labeled polygonal regions using the appropriate postulate, theorem, or corollary. (Note: figures which appear to be regular are regular)

a) Area = _________ b) Area = _________

c) Area = _________ d) Area = _________

e) Area = _________ f) Area = _________

g) Area = _________ h) Area = _________

(B-2) (B-4)

(B-3) (B-3)

(B-1) (B-5)

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

Area of Triangle = 12 b h

= 12 7 5

= 352 units2

⋅ ⋅

⋅ ⋅

Area of Trapezoid = 12 h b +b

= 12 5 9 +5

= 1

1 2⋅ ⋅ ( )⋅ ⋅ ( )

⋅⋅⋅⋅⋅⋅ 5 142 = 1 5 2 7

2 = 35 units2

Area of Rhombus (Parallelogram) = b h⋅ Area of Parallelogram = b h = 11 5 = 55 uni

⋅⋅ tts2

Area of Rectangle = b h or l w

= 8 4 = 32 u

⋅ ⋅⋅ nnits2

A = 12

s a n

A = 12

A =2

units2

⋅ ⋅ ⋅

⋅ ⋅ ⋅5 3 5

75

Where x = 42 2 +2 2

4 2

A = 12 s a n

A = 12 4 2 2 2 4

A = 4 2 2 2 42 = 4 2 2 4

A = 3

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅ ⋅

22 units2

352 units2

35 units2

55 units28 3 units2

32 units2 25 3 units2

32 units2752 units2

h = long leg of a 30 - 60 - 90

= x 32 = 4 3

2 2 3=

A b h= ⋅⋅= 4 2 3

= 8 3 units2

12

A = 53

5 33

A = 12 s a n

A = 12

A = uni

⋅ ⋅ ⋅

⋅ ⋅ ⋅

= ⋅ ⋅ ⋅⋅

10 5 3 3

2 5 5 3 32 3

25 3

A

tts2

(B-5) (B-5)

(Triangle)(Trapezoid)

(Rhombus)(Paralleloram)

(Rectangle) (Regular Triangle)

(Regular Pentagon)

(Square)

Draw line segments from the center of the triangle to the vertices, considering the apothem to be A. We now have two, 30-60-90 right triangles. The long leg of either is • 10 or 5. So, or .



i) Area = _________ j) Area = _________

k) Area = _________ l) Area = _________

m) Area �BCD = _________

(Area �ABC = 42; with median BD)

23. In the figures below, �ABC ~ �DEF; Area �ABC = 15 units. Find the area of �DFE = _______

(B-5) (B-1)

(B-1)(B-2)

(C-1)

(C-1)

(C-1)

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

R V

U

A B

C D

5

5

9 4

11 8

4 10 3

4 5

6 o

o

60 6

6

2 1 1

2 8

10

30 12

5 5

6

7

11

T

S A

D C

B M N F G

H I

6

11

5

O P

S T

U V

1 2 M N

P

R

Q

W X

Y

R

S H

E

I F P

D

Q

E F G

12

13 41

V

T U

Z

5 5

A

D C

B A

B C A C

B D

D F 5 2

E

A

D C

B 10

x 4

x x

x 5 5

o 30

o 60

o 45

o 45 o 90 o 45

o 90 o 90

2

2 2

6

10

1

2

3 4

5

2

6

x x o 60

o 30 o 30

x 3

2 x

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

1

2

3 4

5

6.5 4.5

A = 12 s a n

A = 12 6 3 3 6

A = 2 3 3 3 62

A = 54 3 units

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅

22

A = 12 4 + 2 1 + 2 1A = 48 +2+2A = 52 units2

⋅( ) ⋅( ) ⋅( )

A = b h + 12 b h

A = 8 5 + 12 5 3 5

A = 4

⋅( ) ⋅

⋅( ) ⋅ ⋅

00 + 25 32

= 802

+ 25 32

A = 80 +25 32

units2

h = 102

= 5

b = 10 32

= 5 3

A = 12

d d

A = 12

9 13

A = 1172

units

1 2

2

⋅ ⋅

⋅ ⋅

21 D is the midpoint, so BD is a median

Area ABCArea DFE

15Area DFE

2

= ( )( )

= (

ACDF

2

2

))( )

=

⋅

2

2515

Area DFE4

25

Area DFE = 15 254 = 375

4 == 93 34 units2

54 3 units2 52 units2

1172 units280 +25 3

2 units2

21units2

93 34 units2

(Regular Hexagon)

(Trapezoid) (Rhombus)


Unit V, Unit Test Form B Name

Class Date Score

Unit V - Other Polygons

1. Find the area of trapezoid LMNO shown to the right. Area = ____________

2. Points A, J, H, and K are midpoints of the sidesof quadrilateral MNPQ shown to the right. AJHK is a ____________What type of quadrilateral is AJHK?

a) Square b) Rectanglec) Parallelogram d) Trapezoid

3. The sum of the measures of the interior angles of a polygon is 26,640O.Find the number of sides of the polygon. Sides = ______________

4. Find the area of parallelogram RTXK Area = ______________as shown, with KX = 24; KQ = 9

(B-4)

(A-6)

(A-7)

(B-3)

(c) Parallelogram

M J N

R T

X K Q 30 O

H

P K

Q

A

P Q B

C

E

H

G

F

E P

D

3

8

C

B

A

D D E

A

R S

K J

F N L

G

H M

L M

N

E

H G

F 10y – 4

9y – 9

7y + 5 O P

D

A

E

C B

P Q 3

6 M

Q

N

R O P

5

4 2 2

R S T

J K

L M (5x + 18)

(7x – 20)

O

O 5 9

5

7 7

Q

S

R T P

L M 9

9 9 9

11

N K

?

Q 9

x

60 O

45 O 45 O

30 O

x 2

27 O

11

L M 9

11

N 27 O

11

L M 8

32 4

16

8

128 N O P 5

13 20

18

A

D

24

24

x

2

6

4 x

C

B P Q

R S T 6

h

9 A

D (3,0)

C (1,–3)

D (–1,–6)

A (–4,–1)

B (–2,2)

B D

C

A 3x – 7

4x + 3

3y + 5

5y + 8

B

C D

6

2

4

A B

C

12

Altitude is a +b = ch +9 = 11h +81 = 121

h = 40

2 2 2

2 2 2

2

2

hh = 40 = 4 10 = 2 10⋅

36 10 units2

Area of Trapezoid LMND = 12 h b +b

= 12 2 10

1 2⋅ ⋅ ( )⋅ ⋅ 227 +9

= 2 10 362 = 36 10 units2

( )⋅

n - 2 180 = 26,640180n - 360 = 26,640

180n = 27,000

n

( ) ⋅

== 27,000180

= 90 2 15090 2

= 150⋅ ⋅⋅

Altitude is = or3

Area = b h

= 24 3 3

= 72 3 units2

⋅⋅

72 3 units2

9 3 3

M J N

R T

X K Q 30 O

H

P K

Q

A

P Q B

C

E

H

G

F

E P

D

3

8

C

B

A

D D E

A

R S

K J

F N L

G

H M

L M

N

E

H G

F 10y – 4

9y – 9

7y + 5 O P

D

A

E

C B

P Q 3

6 M

Q

N

R O P

5

4 2 2

R S T

J K

L M (5x + 18)

(7x – 20)

O

O 5 9

5

7 7

Q

S

R T P

L M 9

9 9 9

11

N K

?

Q 9

x

60 O

45 O 45 O

30 O

x 2

27 O

11

L M 9

11

N 27 O

11

L M 8

32 4

16

8

128 N O P 5

13 20

18

A

D

24

24

x

2

6

4 x

C

B P Q

R S T 6

h

9 A

D (3,0)

C (1,–3)

D (–1,–6)

A (–4,–1)

B (–2,2)

B D

C

A 3x – 7

4x + 3

3y + 5

5y + 8

B

C D

6

2

4

A B

C

12

M J N

R T

X K Q 30 O

H

P K

Q

A

P Q B

C

E

H

G

F

E P

D

3

8

C

B

A

D D E

A

R S

K J

F N L

G

H M

L M

N

E

H G

F 10y – 4

9y – 9

7y + 5 O P

D

A

E

C B

P Q 3

6 M

Q

N

R O P

5

4 2 2

R S T

J K

L M (5x + 18)

(7x – 20)

O

O 5 9

5

7 7

Q

S

R T P

L M 9

9 9 9

11

N K

?

Q 9

x

60 O

45 O 45 O

30 O

x 2

27 O

11

L M 9

11

N 27 O

11

L M 8

32 4

16

8

128 N O P 5

13 20

18

A

D

24

24

x

2

6

4 x

C

B P Q

R S T 6

h

9 A

D (3,0)

C (1,–3)

D (–1,–6)

A (–4,–1)

B (–2,2)

B D

C

A 3x – 7

4x + 3

3y + 5

5y + 8

B

C D

6

2

4

A B

C

12

150

M J N

R T

X K Q 30 O

H

P K

Q

A

P Q B

C

E

H

G

F

E P

D

3

8

C

B

A

D D E

A

R S

K J

F N L

G

H M

L M

N

E

H G

F 10y – 4

9y – 9

7y + 5 O P

D

A

E

C B

P Q 3

6 M

Q

N

R O P

5

4 2 2

R S T

J K

L M (5x + 18)

(7x – 20)

O

O 5 9

5

7 7

Q

S

R T P

L M 9

9 9 9

11

N K

?

Q 9

x

60 O

45 O 45 O

30 O

x 2

27 O

11

L M 9

11

N 27 O

11

L M 8

32 4

16

8

128 N O P 5

13 20

18

A

D

24

24

x

2

6

4 x

C

B P Q

R S T 6

h

9 A

D (3,0)

C (1,–3)

D (–1,–6)

A (–4,–1)

B (–2,2)

B D

C

A 3x – 7

4x + 3

3y + 5

5y + 8

B

C D

6

2

4

A B

C

12

21. If ABCD is a parallelogram named in standard notation, which of the following must always be true? ____________

a) �C � �D b) �A � �C c) m�B + m�D = 180

d) AB || BC e) AC � BD d) All of these

22. Find the area of the regular pentagon shown to the right. Area = ____________(Note: P is the center of the pentagon)

23. If four angles of a pentagon have measures of 105O, 75O, 145O, and 130O, then the measure of the fifth angle is? ____________

a) 95O b) 80O c) 100O d) 85O e) 145O

24. The area of the parallelogram shown to the right is _______________

a) b)

c) d)


NameUnitV, Unit Test Form B—Continued—

(A-2)

(B-5)

(A-7)

(A-2)

(b) �A = �C

60 units2

M J N

R T

X K Q 30 O

H

P K

Q

A

P Q B

C

E

H

G

F

E P

D

3

8

C

B

A

D D E

A

R S

K J

F N L

G

H M

L M

N

E

H G

F 10y – 4

9y – 9

7y + 5 O P

D

A

E

C B

P Q 3

6 M

Q

N

R O P

5

4 2 2

R S T

J K

L M (5x + 18)

(7x – 20)

O

O 5 9

5

7 7

Q

S

R T P

L M 9

9 9 9

11

N K

?

Q 9

x

60 O

45 O 45 O

30 O

x 2

27 O

11

L M 9

11

N 27 O

11

L M 8

32 4

16

8

128 N O P 5

13 20

18

A

D

24

24

x

2

6

4 x

C

B P Q

R S T 6

h

9 A

D (3,0)

C (1,–3)

D (–1,–6)

A (–4,–1)

B (–2,2)

B D

C

A 3x – 7

4x + 3

3y + 5

5y + 8

B

C D

6

2

4

A B

C

12

M J N

R T

X K Q 30 O

H

P K

Q

A

P Q B

C

E

H

G

F

E P

D

3

8

C

B

A

D D E

A

R S

K J

F N L

G

H M

L M

N

E

H G

F 10y – 4

9y – 9

7y + 5 O P

D

A

E

C B

P Q 3

6 M

Q

N

R O P

5

4 2 2

R S T

J K

L M (5x + 18)

(7x – 20)

O

O 5 9

5

7 7

Q

S

R T P

L M 9

9 9 9

11

N K

?

Q 9

x

60 O

45 O 45 O

30 O

x 2

27 O

11

L M 9

11

N 27 O

11

L M 8

32 4

16

8

128 N O P 5

13 20

18

A

D

24

24

x

2

6

4 x

C

B P Q

R S T 6

h

9 A

D (3,0)

C (1,–3)

D (–1,–6)

A (–4,–1)

B (–2,2)

B D

C

A 3x – 7

4x + 3

3y + 5

5y + 8

B

C D

6

2

4

A B

C

12

85O

A = 12 s a n

A = 12 8 3 5

A = 2 4 3 52 = 60 units2

⋅ ⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅ ⋅

n - 2 180 - 105 +75 +145 +1305 - 2 180 - 445

3 18

( ) ⋅ ( )( ) ⋅

⋅ 00 - 455540 - 455

85O

30 3

45

15 32

90 3

(d) 90 3units2

∠ ≅ ∠∠S Q

m S = 60In this 30 - 60 - 90 triangle, h is 6 3 So, Area of PQRS = b h

= 6 +9 6 3= 15 6 3= 90 3 u

⋅( ) ⋅ ( )

⋅nnits2


NameUnit VI, Part B, Lessons 1,2&3, Quiz Form A—Continued—

2. Use the figure to the right to complete the following statements. In the figure, JT is tangent to �Q at point T.

a) If QT = 6 and JQ = 10, then JT = ____________________

b) If QT = 8 and JT = 15, then JQ = ____________________

c) If m�JQT = 60 and QT = 6, then JQ = ____________________

d) If JK = 9 and KQ = 8, then JT = ____________________

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

Since QT JT by Corollary 68a, �QTJ is a right triangle. a +b = cQT + JT = JQ6 + JT = 1036 + JT

2 2 2

2 2 2

2 2 2

( ) ( ) ( )( )( ))( )

2

2

= 100JT = 64JT = 8

a +b = cQT + JT = JQ

8 +15 = JQ64+225

2 2 2

2 2 2

2 2 2

( ) ( ) ( )( )

== JQ289 = JQ17 = JQ

2

2

( )( )

QT = 12 JQ

6 = 12 JQ

12= JQ

a2 + =

( ) + ( ) = ( )+ ( ) = +( )

b cQT JT JQ

JT JK KQ

2 2

2 2 2

2 2 28644 9 864 1764 289

2 2

2 2

2

+ ( ) = +( )+ ( ) = ( )+ ( ) =

JTJTJTJT(( ) =

=

2 22515JT

8

17

12

15

Since QT JT by Corollary 68a, �QTJ is a right triangle.

Since QT JT, �QTJ is a right triangle. If m�JQT = 60,�QTJ is a 30-60-90 triangle, and QT is the short leg ofthe triangle opposite the 30 degree angle. QT = 1/2 JQ in a 30-60-90 triangle.

Since QT JT, �QTJ is a right triangle.


NameUnit VI, Part B, Lessons 4&5, Quiz Form A—Continued—

5. 6. m�1 = ________

m�1 = ________

7. m�1 = ________ 8. y = _________m�2 = ________ x = _________

9. x = ________ 10. m�1 = ______y = ________ x = _________

y = _________

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

Q

A

C

B D

Q T

J

K

Q

D C

A

X

B

Q V X

W U

Q

M

N

O P

T

Q

H

I

G

Q

W

R T

Q J

K

N M

Q

A

C

B Q B

C D

A T

J

Q 60

Q

A

C

B D

Q

A

C

B

D Q

A

B C

D

Q

U

V

W

X

Q E F

D

Q

Z Y

X

Q

A

B

C

D Q

A D

X

B

C

Q A C

B

D

Q

N P R

M X

126

1

1

1 1

24

1

1

2

2

140

Q

70

24

30 35 Q Q

E

D

C

C

A

x x x

30

A

B F x

y

x

x 70

50

18

1

1

30

40 B

D

C

A A Q 95

174

96

x

B

C

1

1

1 120

Q

Q

E

B A

F

C D y

Q A

C

B 110

16 28

66

16

80

124

120 260

200

Q Q

B

Q

1

Q

Q

A B x

C D

20 20 Q

x

y

210

20

x

x

84

96

Q Q B

A

E

y

C

D 112

F

Q Q

Q Q

Q

x

88

24

y

mBXC = 360 - mBC= 360 - 95= 265

m 1 = 12 mBXC - mBC

m 1 = 12 265 - 95

= 12 170

= 2 8

∠ ( )

∠ ( )

⋅

⋅ 552

= 85

m 1 = 12 174 - 96

= 12 78

= 2 392

= 39

∠ ( )

⋅

⋅

m 1 = 12 70 - 24

= 12 46

= 2 232

= 23

∠ ( )

⋅

⋅

m = 12 30 +70

= 12 100

= 2 502

= 50

∠ ( )

⋅

⋅

2

m AFE = 12

x + y

88 = 12

x + y

176 = x + y

m ACE = 12

y - x

∠ ( )

( )

∠ ( )

224 = 12

y - x

48 = y - x

( )

176 = y + x48 = y - x

224 = 2y112 = y

48 - y = -x-48 + y = x

-48 +112 = x64 = x

m CED = 180 - 90 +30 = 180 -120= 60

∠ ( )

∠

∠ ( )( )

1 = 180 - 60= 120

m 1 = 12 mBC +mAF

120 = 12 160 + x

1120 = 80 + 12

40 = 12 x

80 = x

xy = 180 - x= 180 - 80= 100

85

39

35

35

23

50

64

112

120

80

100


NameUnit VI, Part C, Lessons 1,2&3, Quiz Form A—Continued—

5. Find EQ in �Q, EQ = _________ 6. Find mCB in �Q, mCB = _________if CD = 10 and DQ = 9. if mCD = 96O

7. Find DC in �Q. DC = _________ 8. Find BD and AC in �Q. BD = _______AC = ________

9. Find CB in �Q, given CB = _________ 10. Find CE in �Q. CE = _______that CD = 16, AQ = 9 and EQ = 5

Q

A DE

4 B

C

Q

A

DE

B

C43x

8-x

Q

AC

DB

7

E

3

Q

A C

B

D

E 2

9

x8

Q

B DE

C

A

Q E

C

B

D

A

Q

A

D

E4x

x4

B

C

Q

AB

CD

6Q

BD

E

C

A

Q

A

C

D

B

6

E3

4x2

Q

C

A

B

DE

x

9 3

2

Q

CA

D

BE

34

12

Q

B

D

C

A

QB

D

C

AQ

B

DE

120

C

A

Q

B

DE

C A

Q

AB

CD

4Q

AC

D

B

6

Ex-54

x QEC

B

D

A

Q

AC

D

BEx

1

4x-3

E

Q

A DE

4 B

C

Q

A

DE

B

C43x

8-x

Q

AC

DB

7

E

3

Q

A C

B

D

E 2

9

x8

Q

B DE

C

A

Q E

C

B

D

A

Q

A

D

E4x

x4

B

C

Q

AB

CD

6Q

BD

E

C

A

Q

A

C

D

B

6

E3

4x2

Q

C

A

B

DE

x

9 3

2

Q

CA

D

BE

34

12

Q

B

D

C

A

QB

D

C

AQ

B

DE

120

C

A

Q

B

DE

C A

Q

AB

CD

4Q

AC

D

B

6

Ex-54

x QEC

B

D

A

Q

AC

D

BEx

1

4x-3

E

Q

A DE

4 B

C

Q

A

DE

B

C43x

8-x

Q

AC

DB

7

E

3

Q

A C

B

D

E 2

9

x8

Q

B DE

C

A

Q E

C

B

D

A

Q

A

D

E4x

x4

B

C

Q

AB

CD

6Q

BD

E

C

A

Q

A

C

D

B

6

E3

4x2

Q

C

A

B

DE

x

9 3

2

Q

CA

D

BE

34

12

Q

B

D

C

A

QB

D

C

AQ

B

DE

120

C

A

Q

B

DE

C A

Q

AB

CD

4Q

AC

D

B

6

Ex-54

x QEC

B

D

A

Q

AC

D

BEx

1

4x-3

E

Q

A DE

4 B

C

Q

A

DE

B

C43x

8-x

Q

AC

DB

7

E

3

Q

A C

B

D

E 2

9

x8

Q

B DE

C

A

Q E

C

B

D

A

Q

A

D

E4x

x4

B

C

Q

AB

CD

6Q

BD

E

C

A

Q

A

C

D

B

6

E3

4x2

Q

C

A

B

DE

x

9 3

2

Q

CA

D

BE

34

12

Q

B

D

C

A

QB

D

C

AQ

B

DE

120

C

A

Q

B

DE

C A

Q

AB

CD

4Q

AC

D

B

6

Ex-54

x QEC

B

D

A

Q

AC

D

BEx

1

4x-3

E

Q

A DE

4 B

C

Q

A

DE

B

C43x

8-x

Q

AC

DB

7

E

3

Q

A C

B

D

E 2

9

x8

Q

B DE

C

A

Q E

C

B

D

A

Q

A

D

E4x

x4

B

C

Q

AB

CD

6Q

BD

E

C

A

Q

A

C

D

B

6

E3

4x2

Q

C

A

B

DE

x

9 3

2

Q

CA

D

BE

34

12

Q

B

D

C

A

QB

D

C

AQ

B

DE

120

C

A

Q

B

DE

C A

Q

AB

CD

4Q

AC

D

B

6

Ex-54

x QEC

B

D

A

Q

AC

D

BEx

1

4x-3

E

Q

A DE

4 B

C

Q

A

DE

B

C43x

8-x

Q

AC

DB

7

E

3

Q

A C

B

D

E 2

9

x8

Q

B DE

C

A

Q E

C

B

D

A

Q

A

D

E4x

x4

B

C

Q

AB

CD

6Q

BD

E

C

A

Q

A

C

D

B

6

E3

4x2

Q

C

A

B

DE

x

9 3

2

Q

CA

D

BE

34

12

Q

B

D

C

A

QB

D

C

AQ

B

DE

120

C

A

Q

B

DE

C A

Q

AB

CD

4Q

AC

D

B

6

Ex-54

x QEC

B

D

A

Q

AC

D

BEx

1

4x-3

E

48O

1212

8

DE = 5 from Theorem 73and AB CD, So AB bisects CD

2 14

8 3

4 5

DQ DE EQEQ

EQ

( ) = ( ) + ( )( ) = ( ) + ( )

= + ( )

2 2 2

2 2 2

2

9 581 2555656

4 142

2= ( )± =

⋅ =

EQEQ EQ cannot be negativeEQ

( )

114 = EQ

AE BE = DE EC12 4 = x x48 = x2

⋅ ⋅⋅ ⋅

± =48 xx cannot be nega( ttivexx

so DC

)

,

16 34 38 3

⋅ =

=

=

AQ QC = BQ QD (call=

⋅ ⋅⋅ ⋅

⋅ = ⋅

the radius rr r rr rr

r

" ")666

6r

r =

BD BQ DQBDBD

= += +=6 612

AC AQ CQACAC

= += +=6 612

AE BE = DE EC=

⋅ ⋅

⋅ ⋅

=

=

± =

6 4 324 388

2

2

2

xxxxx cannot be( nnegativexx

)4 22 2⋅ =

=

CE x=

= ( )=

2

22 28

CB CE BECBCB

( ) = ( ) + ( )( ) = ( ) + −( )( ) = +

2 2 2

2 2 2

2

8 9 564 4464 168080

2

2

2

( )( ) = +

( ) =

= ±

CBCBCBCB cannot be ne( ggativeCBCB

)= ⋅

=

16 54 5

((

(mCB = 48O from Corollary 73aand AB CD, So AB bisects CD

(

Theorem 73 - AB bisects CD, So CE = 8

(

(AC and BD are diameters)


NameUnit VI, Part C, Lessons 4,5,6&7, Quiz Form A—Continued—

3. Find x in �Q. x = _________ 4. Find mAD in �Q. mAD = _________

5. Find CD in �Q. CD = _________ 6. Find AC and AE in �Q. AC = _________AE = _________

7. Find AB and BC in �Q. AB = _________ 8. AB and AC are m�BAE = _________BC = _________ tangents to �Q, and

m�BAC = 42O.Find m�BAE.

QC

E

D

B

A5

xx

5

Q

C

DA

B

8

8

x

Q

B

A

C

12

x -x2

Q

C

D

A

B

94

80

Q

A

B

C

D12

Q

A

B

CE

D

x+2 x+6

x+0 x

Q

A

B

D

C

1.2

0.4x

0.5x Q E

B

C

DA

12

12

Q

54C

E

D

B x2

A

Q

A B

CD

80

110

Q

AB

C

D

Q

E

B

C

A

D

Q

AB

CD

Q

A

B

C

E

D3x

2

0.5

x

QC

D

A

B

20

10

x

Q

E

B

C

A

D

Q

A

B

CE

D

3

5

4

x Q

A

B

C

D

Q

A

BC

D

QC

D

A

B

9

8

x

Q

E

B C

D

A

1010

85

x

O

Q

A

B

C

E

D

8

54

x

Q

0.8x

CE

D

B

0.40.5

A

x

135

QC

E

D

B

A5

xx

5

Q

C

DA

B

8

8

x

Q

B

A

C

12

x -x2

Q

C

D

A

B

94

80

Q

A

B

C

D12

Q

A

B

CE

D

x+2 x+6

x+0 x

Q

A

B

D

C

1.2

0.4x

0.5x Q E

B

C

DA

12

12

Q

54C

E

D

B x2

A

Q

A B

CD

80

110

Q

AB

C

D

Q

E

B

C

A

D

Q

AB

CD

Q

A

B

C

E

D3x

2

0.5

x

QC

D

A

B

20

10

x

Q

E

B

C

A

D

Q

A

B

CE

D

3

5

4

x Q

A

B

C

D

Q

A

BC

D

QC

D

A

B

9

8

x

Q

E

B C

D

A

1010

85

x

O

Q

A

B

C

E

D

8

54

x

Q

0.8x

CE

D

B

0.40.5

A

x

135

QC

E

D

B

A5

xx

5

Q

C

DA

B

8

8

x

Q

B

A

C

12

x -x2

Q

C

D

A

B

94

80

Q

A

B

C

D12

Q

A

B

CE

D

x+2 x+6

x+0 x

Q

A

B

D

C

1.2

0.4x

0.5x Q E

B

C

DA

12

12

Q

54C

E

D

B x2

A

Q

A B

CD

80

110

Q

AB

C

D

Q

E

B

C

A

D

Q

AB

CD

Q

A

B

C

E

D3x

2

0.5

x

QC

D

A

B

20

10

x

Q

E

B

C

A

D

Q

A

B

CE

D

3

5

4

x Q

A

B

C

D

Q

A

BC

D

QC

D

A

B

9

8

x

Q

E

B C

D

A

1010

85

x

O

Q

A

B

C

E

D

8

54

x

Q

0.8x

CE

D

B

0.40.5

A

x

135

QC

E

D

B

A5

xx

5

Q

C

DA

B

8

8

x

Q

B

A

C

12

x -x2

Q

C

D

A

B

94

80

Q

A

B

C

D12

Q

A

B

CE

D

x+2 x+6

x+0 x

Q

A

B

D

C

1.2

0.4x

0.5x Q E

B

C

DA

12

12

Q

54C

E

D

B x2

A

Q

A B

CD

80

110

Q

AB

C

D

Q

E

B

C

A

D

Q

AB

CD

Q

A

B

C

E

D3x

2

0.5

x

QC

D

A

B

20

10

x

Q

E

B

C

A

D

Q

A

B

CE

D

3

5

4

x Q

A

B

C

D

Q

A

BC

D

QC

D

A

B

9

8

x

Q

E

B C

D

A

1010

85

x

O

Q

A

B

C

E

D

8

54

x

Q

0.8x

CE

D

B

0.40.5

A

x

135

QC

E

D

B

A5

xx

5

Q

C

DA

B

8

8

x

Q

B

A

C

12

x -x2

Q

C

D

A

B

94

80

Q

A

B

C

D12

Q

A

B

CE

D

x+2 x+6

x+0 x

Q

A

B

D

C

1.2

0.4x

0.5x Q E

B

C

DA

12

12

Q

54C

E

D

B x2

A

Q

A B

CD

80

110

Q

AB

C

D

Q

E

B

C

A

D

Q

AB

CD

Q

A

B

C

E

D3x

2

0.5

x

QC

D

A

B

20

10

x

Q

E

B

C

A

D

Q

A

B

CE

D

3

5

4

x Q

A

B

C

D

Q

A

BC

D

QC

D

A

B

9

8

x

Q

E

B C

D

A

1010

85

x

O

Q

A

B

C

E

D

8

54

x

Q

0.8x

CE

D

B

0.40.5

A

x

135

QC

E

D

B

A5

xx

5

Q

C

DA

B

8

8

x

Q

B

A

C

12

x -x2

Q

C

D

A

B

94

80

Q

A

B

C

D12

Q

A

B

CE

D

x+2 x+6

x+0 x

Q

A

B

D

C

1.2

0.4x

0.5x Q E

B

C

DA

12

12

Q

54C

E

D

B x2

A

Q

A B

CD

80

110

Q

AB

C

D

Q

E

B

C

A

D

Q

AB

CD

Q

A

B

C

E

D3x

2

0.5

x

QC

D

A

B

20

10

x

Q

E

B

C

A

D

Q

A

B

CE

D

3

5

4

x Q

A

B

C

D

Q

A

BC

D

QC

D

A

B

9

8

x

Q

E

B C

D

A

1010

85

x

O

Q

A

B

C

E

D

8

54

x

Q

0.8x

CE

D

B

0.40.5

A

x

135

( (4 or –3 86

AB ACx xx xx x

x or x

=

= −

= − −= −( ) +( )= − =

120 120 4 30 4 0

2

2

++= − ==

34 3

4 or –3x or x

x

mAD mAB mBC mCDmADmAD

= − − −

= − − −

360360 94 86 94

= 86

CD ABCD ABCD

≅== 12

AB AC AD AEx x x x x x

⋅ = ⋅

+ +( ) = +( ) +( ) + +( ) 6 1 1 2 +( ) = +( ) +( )+ = + +

= +

x x x xx x x x

x x

2 6 1 2 32 6 2 5 3

6 5 32 2

xx = 3AC x xACAC

= + += ( ) + ( ) +=

63 3 6

12

AE x xAEAEAE

= +( ) + +( )= +( ) + +( )= +=

1 23 1 3 2

4 59

ACAD

ADAB

xxx

=

=

( )( ) = ( )

0 91 2

1 20 4

1 2 1 2 0 9 0

..

..

. . . .441 44 0 361 440 36

44

2

2

2

xx

xxx x cannot

( )=

=

=± =

. ...

( bbe negativexx

)42==

AB xABAB

= ( ) ⋅= ( )=

0 40 4 2

0 8

.

..

BC xBCBC

= ( ) ⋅= ( )=

0 50 5 2

1

.

.

m BAE m BAC∠ = ⋅ ∠

= ⋅

=

1212 4221

12 12

9

0.8

1AE bisects �BAC byCorollary 80a

21O

If we draw radii QA, QB, QC, andQD, we can prove triangles congruent.We can then use Theorem 73 and prove AB � CD.


Name

Class Date Score

Unit VI - CirclesPart D - Circle ConcurrencyLesson 1 - Theorem 83 - “If you have a triangle, then that triangle is cyclic.”

Lesson 2 - Theorem 84 - “If the opposite angles of a quadrilateral aresupplementary, then the quadrilateral is cyclic.”

1. Quadrilateral ABCD is cyclic. Find x and y. x = _________y = _________

2. Quadrilateral (Kite) ABCD is cyclic. Find mAB. mAB = _________

Quiz Form A

A

B

Q

D

A

B

C 100

100

C

D

110

75

y

x

QX

Y

W

X

Z A B

Q

C W

Z

Q

B

A

D

C

160

Q

X Y

W

Y

A

Q

X

Z

C B Z

Q

A B

C

D

E F

G 36

84

Q

DA

B

C

134

A

B

Q

D

A

B

C 100

100

C

D

110

75

y

x

QX

Y

W

X

Z A B

Q

C W

Z

Q

B

A

D

C

160

Q

X Y

W

Y

A

Q

X

Z

C B Z

Q

A B

C

D

E F

G 36

84

Q

DA

B

C

134

�ABC and �ADC are supplementary. (Corollary 67b)

m ABC m ADCy

y

m BAD mBD Th

∠ + ∠ =+ =

=

∠ = ⋅

18090 180

90

12 ( eeorem

mBD

mBD or mBCD

67

75 12

150

)

( )

= ⋅

=

mAB mBCD mDAx

xx

+ + =+ + =

+ ==

360150 110 360

260 3601100

mAB mBC ABC is a semicirclemAB

+ =

+ =

180134 18

( )00

46mAB =

100

90

46

Quadrilateral ABCD is a kite, so CD � CB,BC � DC (Theorem 81), and mBC = 134.

(

( ( (

(


NameUnit VI, Part D, Lessons 1&2, Quiz Form A—Continued—

3. Given: Quadrilateral XYWZ is cyclic. ZY is a diameter of �Q.XY � WZ

Prove: �XYZ � �WZY

4. The angle bisectors of the angles of �XYZ meet at point Q. QX = 75 and QC = 20. Find QB. Explain your answer.

QB = _______________

Complete the following statements by choosing “sometimes”, “always”, or “never”.

5. Rectangles are ____________________ cyclic quadrilaterals.

6. Irregular quadrilaterals are ____________________ cyclic.

7. Regular polygons are ____________________ cyclic.

8. A kite is ____________________ a cyclic quadrilateral.

9. Opposite angles of a cyclic quadrilateral ____________________ add up to 180 degrees.

10. Isosceles trapezoids are ____________________ cyclic quadrilaterals.

STATEMENT REASON

1. Quadrilateral XYZW is cyclic 1. Given2. ZY is a diameter of �Q 2. Given3. XY � WZ 3. Given4. �ZXY is a right angle 4. Corollary 67a - If you have an angle inscribed in a semicircle,

then that angle must be a right angle.5. �ZXY is a right triangle 5. Definition of Right Triangle6. �YWZ is a right angle 6. Corollary 67a 7. �YWZ is a right triangle 7. Definition of Right Triangle8. XY � WZ 8. Theorem 82 - If two minor arcs are congruent, then their chords

are congruent.9. YZ � ZY 9. Reflexive Property for Congruent Line Segments10. �ZXY � �YWZ 10. Postulate Corollary 13c - If the hypotenuse and one leg of one

right triangle are congruent to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent. (HL)

11. �XYZ � �WZY 11. C.P.C.T.C.

( (

A

B

Q

D

A

B

C 100

100

C

D

110

75

y

x

QX

Y

W

X

Z A B

Q

C W

Z

Q

B

A

D

C

160

Q

X Y

W

Y

A

Q

X

Z

C B Z

Q

A B

C

D

E F

G 36

84

Q

DA

B

C

134

A

B

Q

D

A

B

C 100

100

C

D

110

75

y

x

QX

Y

W

X

Z A B

Q

C W

Z

Q

B

A

D

C

160

Q

X Y

W

Y

A

Q

X

Z

C B Z

Q

A B

C

D

E F

G 36

84

Q

DA

B

C

134

The bisectors of the angles of a triangle are concurrent at a point which is equidistant

from the sides of the triangle. (Corollary 83b) Therefore, QC = QB. So, QB = 20.

20

always

sometimes

always

sometimes

always

always

( (


NameUnitVI, Unit Test Form A—Continued—

Determine whether each of the following is always, sometimes, or never true.

________ 13. Congruent chords of different circles intercept congruent arcs.

________ 14. An angle inscribed in a semicircle is a right angle.

________ 15. Two circles are congruent if their radii are congruent.

________ 16. Two externally tangent circles have only two common tangents.

________ 17. A radius is a segment that joins two points on a circle.

________ 18. A polygon inscribed in a circle is a regular polygon.

________ 19. A secant is a line that lies in the plane of a circle, and contains a chord of the circle.

________ 20. The opposite angles of an inscribed quadrilateral are supplementary.

________ 21. If point X is on AB, then mAX + mXB = mAXB.

________ 22. The common tangent segments of two circles of unequal radii are congruent.

________ 23. Tangent segments from an external point to two different circles are congruent.

________ 24. Cyclic quadrilaterals are congruent.

________ 25. If two circles are internally tangent, then the circles have three common tangents.

sometimes

always

always

never

never

sometimes

always

always

always

always

sometimes

sometimes

never

( ( ( (

(C-7)

(B-2)

(A-1)

(A-3)

(A-1)

(A-3)

(A-1)

(B-2)

(Postulate 8 - p198)

(A-3)

(A-3)

(D-2)

(A-3)



Use the given figure to answer problems 26 to 35.(Note: AB is tangent to Q at point B)

26. If mDF = 96, find m�DEF. 27. If mCD = 62 and m�EGF = 110, find mEF.

m�DEF = ________ mEF = ________

28. If mDF = 96 and mCE = 40, find m�FAD. 29. If mBFD = 170 and mBC = 110, find m�BAD.

m�FAD = ________ m�BAD = ________

30. Find m�ABQ. 31. If m�ADE = 26, find mCE.

m�ABQ = ________ mCE = ________

48 158

28 30

90 52

m DEF mDF Theorem∠ = ⋅

= ⋅

= ⋅ ⋅

=

12 6712 96

1 2 482

48

( )

m EGF mCD mEF Theorem

mE

∠ = ⋅ +( )= ⋅ +

12 69

110 12 62

( )

FF

mEFmEF

( )= +

=

220 62158

m FAD mDF mCE Theorem∠ = ⋅ −( )= ⋅ −( )

=

12 7012 96 40

( )

112 56

1 2 282

28

⋅

= ⋅ ⋅

=

m BAD mBFD mBC Theorem∠ = ⋅ −( )= ⋅ −

12 7112 170 110

( )

(( )

= ⋅

= ⋅ ⋅

=

12 60

1 2 302

30

m ABQ Corollary a∠ = 90 68( )

m ADE mCE Theorem

mCE

mCE

∠ = ⋅

= ⋅

=

12 67

26 12

52

( )

( ( (

( ( ( ((

(

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

(B-2) (B-4)

(B-4) (B-5)

(B-3) (B-4)

(



32. If m�ADE = 26, find m�AFC. 33. �DCF � ________.

m�AFC = ________

34. If m�FAB = 18 and mBE = 80, find mBF. 35. If m�BQF = 90, find mBF.

mBF = ________ mBF = ________

For problems 36 to 41, find the value of x, or the indicated angle.

36. x = ________ 37. x = ________

26

�FED

116 90

20 6

∠ ≅ ∠

∠ = ⋅

= ⋅

ADE AFC Corollary c

m ADE mCE

m

( )6712

26 12 CCE

mCE52 =

∠FED Corollary( )67

m AFC mCE∠ = ⋅

= ⋅

= ⋅ ⋅

=

1212 52

1 2 262

26

m FAB mBF mBE Theorem

mBF

∠ = ⋅ −( )= ⋅ −

12 71

18 12 8

( )

00

36 80116

( )= −

=

mBFmBF

mBF m BQF= ∠

mBF = 90

AC AB AE AD Theoremx

( )( ) = ( )( )+( )( ) = +( )

772 2 4 7 4(( )

+ = += + −==

4 2 16 282 16 28 42 40

20

xxxx

AE EB CE ED Theoremx

( )( ) = ( )( )( )( ) = ( )( )

764 12 8

488 86==

xx

The measure of a central angle is the same as themeasure of its intercepted arc.

( ( (

((

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

(B-2) (B-2)

(B-5) (A-2)

(C-4) (C-3)

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5



38. Find m�C m�C = ________ 39. x = ________

40. Find m�D m�D = ________ 41. x = ________

42. Find QE QE = ________ 43. Find BD. BD = ________

95

m A m C Corollary bm Cm C

∠ + ∠ = ( )+ ∠ =

∠ =

180 6785 180

95

ACAD

ADAB Theorem

xx

x xx

= ( )+ =

⋅ = ( ) +( )=

784 6

44 4 642 ⋅⋅

= ± ⋅

= ⋅

=

104 10

4 102 10

xx cannot be negative

xx

( )

m D m B

m D

∠ + ∠ =

∠ + ==

180

100 18080

QF FC QCQFQF

( ) + ( ) = ( )( ) + ( ) = ( )( ) + =

2 2 2

2 2 2

2

4 516 25

QQFQF

QF cannot be negativeQFQF

( ) =

= ±

==

2 99

93

( )

If DC then FCTheorem

= =8 473,

( )

2 10

80 17

3 10

A trapezoid inscribed in a circleis cyclic. (Corollary 67b)

x = 17; AB � AC (Theorem 80)

BD = 10, since AD = CD = BD(Corollary 83a)

so, QE = 3. Congruent Chords in the same circle are equidistant from the center of the circle.(See Unit VI, Part C, Lesson 7, Exercise 3)

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

(B-2) (C-4)

(B-2) (C-6)

(C-1) (D-1)

Given: l1 and l3 are perpendicularbisectors of the sides of �ABC. AC = 20.

Given: AB � CD, DC = 8, the radius of �Q is 5.


NameUnitVI, Unit Test Form B—Continued—

Use the figure to the right and the given information to answer problems 26 to 35.

26. Find m�ABH m�ABH = ______ 27. Find m�ABF . m�ABF = ______

28. Find m�ACF m�ACF = ______ 29. Find m�DQH m�DQH = ______

30. Find m�BQE m�BQE = ______ 31. Find m�CFB m�CFB = ______

90 14

6 140

52 8

m ABH Corollary a∠ = 90 68( )

m HBG mHG Theorem

mHG

mHG

m

∠ = ⋅

= ⋅

=

12 67

76 12

152

( )

BBG mHG= −= −=

180180 15228

m ACF mBE mBD Theorem∠ = ⋅ −( )= ⋅ −( )

=

12 7112 52 40

( )

112 12

1 2 62

6

⋅

= ⋅ ⋅

=

m DQH mDHmDH mBD

m DQH

∠ =

= −= −=

∠ =

180180 40140140

m BQE mBEmBE mEG mBG

∠ =

= += +=24 2852

m CFB mBD mEG Theorem∠ = −( )= −( )

=

12 7012 40 2412

( )

⋅⋅

= ⋅ ⋅

=

16

1 2 82

8

m ABF mBGTheorem

∠ = ⋅

= ⋅

= ⋅ ⋅

=

12

6812 28

1 2 142

14

( )

mBE mBG mEG

mBD m BQD

= += +=

= ∠=

28 2452

40

(The measure of a centralangle is equal to the measureof its intercepted arc.)

Q

F

B

E A

CG

D

Q

ED A

B

C

x

2

BC

E

D

A

8

12x

4

47

Q Q

A

B

CD

85

Q

D

A

B

C6

4

x

B

CD

A100

Q

Q

H

C

D IE

GA

F

B

B

A

C

17

xQ

AE B

A

B

CD

l1

l3

l2

C

F

D

Q

M

L

P

RT U

N

5 432

Q

T

RC

B

A

D

E

6

x

7

68 17

Q

Qx

6

40

y

C

Q

B

Q

CB

A

DE

xx

45

3AxD

A

BD

CA G

D

FC

E

B

10

8

10

Q

5

(B-3) (B-2)

(B-5) (A-2)

(A-2) (A-4)

The measure of a centralangle is equal to the measureof its intercepted arc.

BH is a diameter of �Q and CAis tangent to �Q at point B.

mEG = 24m�HBG = 76m�BQD = 40

(

geometry: a complete course - videotextthese course notes are replications of the essential content...

Documents