Geometry: A Complete Course - VideoTextGeometry, we must understand that there are different...

Written by: Larry E. Collins

Geometry:A Complete Course

(with Trigonometry)

Module A – Instructor's Guidewith Detailed Solutions for

Progress Tests

RobbinsCreative

Errata March 2015

Geometry: A Complete Course (with Trigonometry)Module A -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 1-59676-093-11 2 3 4 5 6 7 8 9 10 - RPInc - 18 17 16 15 14

Letter from the author . . .I want to personally thank you for your interest in the VideoTextInteractive Geometry course. I alsowant to applaud you for the research you have done which led you to add this program to youreducational resources. Congratulations. You have chosen a delivery system that is allowing studentsnationwide to enjoy remarkable success, in a variety of settings. I'm sure you will continue to besatisfied with your choice, as you see your students develop a strong understanding of Geometryconcepts, and an increased confidence in their own mathematical ability.

Please be sure to carefully review the program overview on pages iv and v, as well as the scope andsequence rationale on pages vi and vii. I wrote them to help you, as the instructor, to understand thedynamic quality of the instructional process in this course, and to afford you the most efficient use of thematerials. In fact, if your students do not involve themselves personally, and actively, in the learningprocess, as we suggest - pausing the lessons frequently to answer questions before the narrator does,and engaging in analysis - their understanding of the concepts will not be as deep as it should be, andthose concepts will not be remembered as long.

Further, as you now officially begin Unit I, you must be aware that the focus of these lessons is to makesure your student is ready to engage in a formal study of Geometry. That means we must do somere-teaching. Most students, by this time, have memorized some elements of Geometry, including thenames for basic geometric shapes and the formulas for perimeter, area, and volume. They do not,however, have any notion as to the “why” of these terms and formulas. So, in Unit I, we must explore,analytically, the development of these concepts. Further, to really be successful in the study ofGeometry, we must understand that there are different Geometries, and a decision must be made as towhich Geometry we will study. Finally, to be completely ready, we must engage in a thoroughoverview of inductive and deductive reasoning, as well as a comprehensive introduction to logic,and the nature of formal proof.

That means, of course, that your student will encounter both familiar and unfamiliar material in Unit I. However, because this is not simply a review of geometric ideas your student has alreadystudied, it is essential that you give the proper attention to the “re-learning” of these concepts, without areliance on the formulas and rules the student may have memorized in the past. So, we will explore, indetail, the reasons “why” these formulas and rules work the way they do. And that will be the tonethroughout this course. To understand concepts, and retain knowledge, we simply must find thereasoning behind the processes we use.

Finally, I strongly encourage you to do every exercise in the lessons in this unit. Most of them areshort-answer exercises and I want to be sure you understand every aspect of the concepts. You will needthat understanding as you go more deeply into the study of Plane Geometry.

Again, we at VideoTextInteractive are very happy that you have taken this step, and we look forward toworking with you and your student, as well as hearing your comments and suggestions. We especiallywant to know about the successes you have enjoyed as a result of using the program. Of course, asalways, we are available to help you on the help-line, should you need some assistance.

We wish you well!

Thomas E. Clark, Author

Table of ContentsInstructional Aids

Program Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .iiiScope and Sequence Rationale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .v

Detailed Solutions for Progress TestsUnit I - The Structure of GeometryPart A - What is Geometry?

LESSON 1 - Origin and StructureLESSON 2 - More on Things

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .1Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .3

LESSON 3 - More on OperationsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .9

LESSON 4 - More on RelationsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .13Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .15

LESSON 5 - More on GroupingsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .17Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .19

Part B - The Scope of Our GeometryLESSON 1 - Undefined Terms

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .21Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .23

LESSON 2 - Simple Closed Plane CurvesLESSON 3 - Polygons

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .27

LESSON 4 - SolidsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .29Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .31

Part C - MeasurementLESSON 1 - Rectangles

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .33Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .37

LESSON 2 - ParallelogramsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .41Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .45

LESSON 3 - TrianglesQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .49Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .53

LESSON 4 - TrapezoidsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .57Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .61

Module A - Table of Contents i

combined

combined

LESSON 5 - Regular PolygonsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

LESSON 6 - CirclesQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

LESSON 7 - PrismsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

LESSON 8 - PyramidsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

LESSON 9 - SpheresQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

Part D - Inductive ReasoningLESSON 1 - General Nature

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

LESSON 2 - Applications in MathematicsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Part E - Deductive ReasoningLESSON 1 - General Nature

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

LESSON 2 - Applications in MathematicsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

Part F - LogicLESSON 1 - Simple Statements

Quiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

LESSON 2 - ConditionalsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

LESSON 3 - Negations of ConditionalsQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

LESSON 4 - FallaciesQuiz A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155Quiz B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

Unit I Test - Form A . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159

Unit I Test - Form B . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181

ii Module A - Table of Contents

Program OverviewThe VideoTextInteractive Geometry program addresses two of the most important aspects ofmathematics instruction. First, the inquiry-based video format contributes to the engaging ofstudents more personally in the concept development process. Through the frequent use of the pausebutton, 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 thelesson every time a question is asked, and trying to answer it correctly before continuing. Of course, thestudent may answer incorrectly, but the narrator will be sure to give the right answer when the playbutton is pressed to resume the lesson. Right or wrong, however, the student is regularly engaging inanalytical and critical thinking, and that is a healthy exercise, in and of itself. Second, eachincremental concept is explored in detail, using no shortcuts, tricks, rules, or formulas, and no stepin the process is ignored. As such, the logic and the continuity of the development assure students thatthey understand completely. Subsequently, learning is more efficient, and all of the required concepts(topics) of the subject can be covered with mastery. Of course, the benefits of these efforts can be seeneven more clearly in 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 computergenerated graphics, and animation, students are never shortchanged when it comes to the insightnecessary to 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 studentsanything 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

iii

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 someexplanation, either verbal or in writing, to you as the instructor. In fact, at this stage, you might evenconsider grading your student only on the completion of the work, not on its accuracy. Remember,this is the first time the student has tried to demonstrate understanding of a concept, and he or she maystill need some fine-tuning. So, because this is part of the initial learning process, the focus should beon a careful analysis of the logic behind the work, not just the answers. Finally, it is time to assessyour student's mastery of the concept behind the work. Just be sure you are not testing on the sameday the exercises were completed. Short-term memory can trick you into thinking that you “have it”,when, in fact, you are just remembering what you did moments before. A more accurate evaluation canbe made on the next day, before moving on to the next lesson. Further, the quizzes and tests in theprogram often utilize open-response questions which will require your student to state, in writing,his or her understanding of the concept. This often reveals much more about a student'sunderstanding than just checking to see if an answer on a test is correct. Remember too, that there aretwo versions of every quiz and test, allowing you to retest, if necessary, in order to make sure that yourstudent has mastered the 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 greateropportunity than usual to grow mathematically, at a personal level, and develop confidence in theirability. That can have a tremendous impact on a student's future pursuits, especially in an age whereapplications of mathematics are so important.

iv

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 concernedprimarily with rates of change, Geometry is a comprehensive exploration of “shapes” (as sets of points),the measurements associated with those shapes, and the relationships that can be established betweenthose shapes. As such, no treatment of Geometry should ever investigate those relationships onlyindividually, or in isolation. This is especially noticeable with traditional textbooks, which generally usea format which addresses them in different “chapters”. In the VideoText Interactive Geometry course,concepts are discussed from a “Unit” perspective, pursuing and connecting, in an exhaustive way,all of the outcomes associated with various possibilities for a specific relationship. Of course, asmuch as is possible, 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 thebenefits of visualization, in life in general, and in Geometry in particular. So, in the VideoTextInteractive Geometry program, computer-generated graphics are used extensively, along withanimation 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 andconviction that both inductive and deductive reasoning can bring to the table. In other words, it isalways desirable, and helpful, for students to “experiment”, inductively, with a geometric relationship, inan effort to come to some general conclusion. Once that general conclusion has been arrived at,however, it is even more convincing if the student is able to “prove”, deductively, that the conclusionabsolutely must follow, logically, from the given information. No, formal proof is not often asked for ineveryday life. On the other hand, the exercise of developing that kind of thinking is invaluable, not onlyin some specific job-related activities, but, more generally, in the daily problem-solving situations thatconfront us. The VideoText Interactive Geometry program is formatted in such a way that formal proofis 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

v

Name

Class Date Score

Quiz Form A

Unit I - The Structure of GeometryPart A - What is Geometry?Lesson 1 - OriginLesson 2 - Structure

1. In our study of Algebra, the symbols used to name numbers were examples of the “things” of mathematics or the objects around which our study revolves. Name two of the new “things” we have discussed in Lesson 2.

2. Tell what part of mathematical speech each of the following is.

a) _________________________________________

b) • (as in 3 • 4) _________________________________________

c) _________________________________________

d) _________________________________________

e) _________________________________________

f) _________________________________________

g) _________________________________________

h) _________________________________________

3. Name the line shown in three ways.

© 2014 VideoTextInteractive Geometry: A Complete Course 1

[ ] G

5 87.

e

7

4

m

X

WY

Z

Q ml

P

M

l

N

l

Z Q

Line Plane

line m line XY XW

(also WY)

(also Point or Space)

Grouping Symbol

Operation Symbol

Relation Symbol

Number Symbol

Number Symbol (or placeholder)

Operation Symbol

Number Symbol

Relation Symbol

>

Name

Class Date Score

Quiz Form B

Unit I - The Structure of GeometryPart A - What is Geometry?Lesson 1 - OriginLesson 2 - Structure

1. In our study of Algebra, the symbols used to name numbers were examples of the “things” of mathematics, or the objects around which our study revolves. How many new things did we discuss in Lesson 2? _______Name them.

2. Tell what part of mathematical speech each of the following is:

a) _________________________________________

b) _________________________________________

c) _________________________________________

d) _________________________________________

e) _________________________________________

f) _________________________________________

g) _________________________________________

h) _________________________________________

3. Name the plane shown in two ways


4

p

||

17

+

{} G

e

>t

P

Z

RP

Q

Point Line Plane Space

Plane Z Plane PQR; Plane PQR;

4

Operation Symbol

Number Symbol

Relation Symbol

Number Symbol

Operation Symbol

Grouping Symbol

Number Symbol (or placeholder)

Relation Symbol

3. Draw the image of the given rectangle after a rotation of 90o clockwise around the center of rotation Q.

1) Draw a segment from vertex A to the center of rotation point Q

2) Measure a 90 o Angle clockwise at point Q. Draw the angle.

3) Use a ruler to locate A’ on the ray of the angle forming angle AQA’.

The measure of segment QA’ will equal the measure of segment QA.

4) Repeat steps 2 through 4 for each vertex. Connect the vertices to form rectangle A’B’C’D’.

© 2014 VideoTextInteractive Geometry: A Complete Course6

NameUnit I, Part A, Lesson 3, Quiz Form A—Continued—

Q

A

B

C

D

90°

Q

A

B

C

D A’D’

C’B’


6. Dilate triangle ABC so that each side of the image is twice as long as in the original triangle.

Draw rays OA, OB, and OC. Measure line segments OA, OB, and OC. Then draw line segments OA’, OB’, and OC’ to lengths 2 times the lengths of line segments OA, OB, and OC, respectively. Triangle A’B’C’ is the dilation image of triangle ABC.


Q Q

A B

C

A’ B’

C

A B

C

A’ B’

C’

Y

X

W

Z

Y’

Z’

W’

X’

Y

X

W

Z

Y’

Z’

W’

O

A

B

C

O

A

B

C

A’

B’

C’

B’C’

Name

Class Date Score

Quiz Form A

Unit I - The Structure of GeometryPart A - What is Geometry?Lesson 5 - More on Groupings

1. Write D = {1, 3, 5, ...} using set-builder notation. _______________________________________

2. Write using the roster method. _________________________

Rewrite the statements in exercises 3 and 4 using set notation. Use the roster method if possible.

3. The set made up of even counting numbers less than ten is an improper subset of the set made up of even counting numbers less than ten. ________________________________

4. The set whose only element is 0 is not a subset of the empty set. _______________________________


B I= = + ∈ ≥{ , , }x x y y y3 1 0

D = {x | x is an odd natural number}

or D = {x | x is an odd counting number}

B = {1, 4, 7, 10, 13, 16...}

{2, 4, 6, 8} {2, 4, 6, 8}

{0} { }

⊆

⊂


Consider these sets for questions 5 through 10.

A = {a, e, i, o, u} B = {c, m, n, r, t}

C = {m, i, n, t} D = {e, i}

5. B C = _______________________________________________________________

6. B C = _______________________________________________________________

7. A D = _______________________________________________________________

8. D B = _______________________________________________________________

9. Is D A? _______________________________________________________________

10. Does A = B? _______________________________________________________________

11. Of 73 men surveyed, 54 would rather ride a golf cart when they play golf, and 20 others would prefer to walk when they play golf. However, of those two groups, 4 men also said they would be comfortable with either. How many of the 73 men surveyed would rather do neither? In other words, how many wouldprefer not even to play golf? ____________ Use a Venn Diagram to show your work.


U

⊂⊂

{c,m, n r, t} These are the elements in one, or the other, or both of the sets.

{m, n t} These are the elements that are common to both sets.

{e, i} These are the elements that are common to both sets.

{ } These are the elements that are common to both sets.

Yes. All of the elements in set D are also in set A

No. Sets A and B do not contain exactly the same elements.

Ride Walk

3

50 164

U

U

U

U

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

Consider these sets for questions 5 through 10.

A = {a, b, c, d, e} B = {a, b, c, d, e, f, g}

C = {m, i, n, t} D = {d, e}

5. A B = _______________________________________________________________

6. A C = _______________________________________________________________

7. A D = _______________________________________________________________

8. C D = _______________________________________________________________

9. Is D {d, e}? _______________________________________________________________

10. Does A = C? _______________________________________________________________

11. Of 68 people surveyed, 33 most often drive to work, 57 usually take the bus to work, and 27 do both equally as often. How many of these surveyed did neither? _____________________ Use a Venn Diagram to show your work.


⊆

U

Bus

5

6 3027

5

a b c d e∩ { , , , , }

a b c d e m i n t∪ { , , , , , , , , }

d e∩ { , }

{ }

Yes, all of the elements in set D are also in the set {d,e}

No, sets A and C do not contain exactly the same elements.

These are the elements that are common to both sets

These are the elements in one, or the other, or both of the sets.



Drive

U

U

U

U

Name

Class Date Score

Quiz Form A

Unit I - The Structure of GeometryPart B - The Scope of Our GeometryLesson 4 - SolidsComplete each sentence in exercises 1 through 5 with the appropriate geometric term(s).

1. The three basic three dimensional shapes in all the world are_________________, __________________, and _________________.

2. A prism has _______________________ for sides.

3. A pyramid has _______________________ for sides.

4. Cones and cylinders have _______________________ for bases.

5. A sphere is a surface which is everywhere the same _______________________from a fixed point.

Identify the solids in exercises 6 through 11 as prisms, cylinders, pyramids, cones, or spheres. Notice the shapeof the base when naming a prism or pyramid and be as specific as possible. NOTE: For these exercises only,you may assume that lines are parallel if they look parallel.

6. 7. 8.

9. 10. 11.


Prisms Pyramids

Spheres

Parallelograms

Triangles

Circles

Distance

Triangular PrismCylinder

(or circular prism)Square Prism

(or parallelogram prism)

Trapezoidal Pyramid Triangular PyramidCone

(or circular pyramid)

Name

Class Date Score

Quiz Form B

Unit I - The Structure of GeometryPart B - The Scope of Our GeometryLesson 4 - SolidsComplete each sentence in exercises 1 through 5 with the appropriate geometric term(s).

1. A prism is named after the shape of its _________________.

2. A cone has a circular_______________________.

3. The sides of a prism are _______________________ .

4. The sides of a pyramid are _______________________.

5. A cylinder has two_______________________bases.

Identify the solids in exercises 6 through 11 as prisms, cylinders, pyramids, cones, or spheres. Notice the shapeof the base when naming a prism or pyramid and be as specific as possible. NOTE: For these exercises only,you may assume that lines are parallel if they look parallel.

6. 7. 8.

9. 10. 11.


Rectangular Pyramid Cone (Circular Pyramid) Triangular Prism

Elliptical Cylinder (or Elliptical Prism) Triangular Pyramid

Rectangular Pyramid (top view)

base

base

parallelograms

triangles

circular

Refer to the rectangle at the right to complete exercises 6 and 7.

base height Area Perimeter

6. 13” 9” ________ ________


7. ________ 7’ 63 sq. ft. ________


NameUnit I, Part C, Lesson 1, Quiz Form A—Continued—

h

b

Perimeter = 2 base+ 2 height⋅ ⋅

63 = b 7

63 = 7b

17

63 =17

7 b

637

or� 9 = 1 b

9 = 1 b

9' = b

⋅

⋅ ⋅ ⋅

⋅

⋅aase

= 2(9)� � � � + 2(7)

P = 18� � � � � � � +114

P = 32'

Area = base height= 13" 9"= 117 square inc

⋅( )( )

hhes

Perimeter = 2 base + 2 height⋅ ⋅

Perimeter = 2 13" + 2 9"Perimeter = 26" +18"

⋅ ⋅

PPerimeter = 44 inches

Area = base height⋅

117 sq. in. 44 in.

9’ 32’


Refer to the rectangle at the right to complete exercises 8 and 9.


8. y units ________ (y2 - 5y) sq. units ________


9. ________ 10 units ________ 56 units


h

b

Perimeter = 2 base+ 2 height⋅ ⋅

Perimeter = 2 base + 2 height⋅ ⋅

56 = 2b+ 2(10)56 = 2b+ 20

56 - 20 = 2b+ 20 - 2036 = 2b+ 0

12

⋅336 = 12

2b

1 2 182 1

= 12

21

b

18 = 1 b18 units = b

⋅

⋅ ⋅⋅

⋅ ⋅

⋅

Area = base heighty - 5y = y hy - 5y = yh

1y

(y

2

2

⋅

⋅

22

2

- 5y) = 1y

(yh)

1y

y - 5y1

= 1y

yh1

1 y(y - 5)y

⋅ ⋅

⋅⋅11

= 1 y hy 1

y - 5 = h

⋅ ⋅⋅

Area = base heightArea = 18 10Area = 180 square uni

⋅⋅

tts

Perimeter = 2(y)� � � � + 2(y - 5)

Perimeter = 2y� � � � � � Ä Ä +2y - (2)(5)

Perimeter = 2y� � � � � � � + 2y - 10

Perimetter = (4y - 10)� units

(y - 5) units (4y - 10) units

18 units 180 sq. units

Name

Class Date Score

Quiz Form A


Unit I - The Structure of GeometryPart C - MeasurementLesson 2 - Parallelograms

1. Find the area and perimeter of the given parallelogram.

Area:____________

Perimeter:____________


Area:____________


2ft.

8“6.4” 6‘

60o

13’

3 3

6x h yd

16 yd

13 yd

5 cm

3.5 cm 4 14

6 13

5 58

1ft

8“

12“

45o

4 2

mm

mm

mm

‘

“

b

112

“

6‘

60o

13’

3 3

h yd

16 yd

13 yd

5

‘

Area base height

A

= ⋅

= ⋅

= ( )( )( )13 3 3

13 3 3

' '

square feet= 39 3

Perimeter Sum of lengths of the sides

P

== 133 6 13 6

19 19

38

' ' ' '

' '

+ + += += feeet

Area base height

ft in in

A a

= ⋅= ⋅ == ⋅

2 2 12 24

24 6 4" . ""

.

.

a square inches

A a inches

=

=

153 6

153 6 2

Perimeter sum of lengths of the sides

P

== 244 8 24 8

32 32

64

" " " "

" "

"

+ + += +=

153.6 sq. inches

64 inches

38 feet

39 3 square� feet

Unit I, Part C, Lesson 2, Quiz Form A—Continued—


Name

3. Find the area and perimeter of the given rhombus.

Area:____________


4. Find the perimeter and the height of this parallelogram.

Area:____________


height:____________

6‘

60o

13’

3 3

h yd

16 yd

13 yd

5

4 14

6 13

5 58

8“

12“

45o

4 2

mm

mm

mm

‘

“

6x h yd

16 yd

13 yd

5 cm

3.5 cm 4 14

6 13

5 58

1ft

8“

12“

45o

4 2

34m 36m44cm 48cm

b

3x2 1

2

7 14

1 38

mm

mm

mm

‘

“

b

12

“

176 sq. yds.

Area = base height

176 = 16 h

116

176 =1

1616 h

176

⋅⋅

⋅ ⋅ ⋅

116= 1 h

11 yards = h

⋅


P

== 166 13 16 13

29 29

58

+ + += += yards

Area Base Height

A a

a square cm

= ⋅= ⋅=

5 3 5

17 5

.

.


P

== 5 ++ + +=

5 5 5

20cm

17.5 sq. cm.

20 cm.

11 yds.

58 yds.

5. Find the perimeter and the base of the given rhombus.

Area:____________


base:____________


Area:____________

Perimeter:___________



2ft.

8“6.4” 6‘

60o

13’

3 3

6x h yd

16 yd

13 yd

5 cm

3.5 cm 4 14

6 13

5 58

1ft

8“

12“

45o

4 2

34m 36m44cm 48cm

b

3x2 1

2

7 1

1 38

mm

mm

mm

‘

“

b

112

“ 6‘

60o

13’

3 3

h yd

16 yd

13 yd

5

4 14

6 13

5 58

8“

12“

45o

4 2

44cm 48cm

b

mm

mm

mm

‘

“

48x2 sq. units

Area = Base Height

48x = b 6x

16x

48x1

= b6x1

16

2

2

⋅

⋅

⋅ ⋅ ⋅xx

6 8 x x6x

= b

8x units = b

⋅ ⋅ ⋅


P

== 8xx x x x

P x units

+ + +=

8 8 8

32

Area = base height

A 61

34

1

419

3

17

4

⋅

⋅

⋅

19 17

3 4

A323

12square mm� or� 26

11

12mm2

⋅⋅

Perimeter = Sum of lengths of the sides

P = 6113

+ 558

+613

+ 558

=193

+458

+193

+458

=193

88

+458

33

+193

88

+458

33

⋅ ⋅ ⋅ ⋅

=152 +135 +152+135

24

P =57424

mm� or � 2322224

mm� or � 231112

mm

32x units

8x units

2611

122mm

m12

2311

12mm

Name

Class Date Score


Quiz Form B

Unit I - The Structure of GeometryPart C - MeasurementLesson 2 - Parallelograms

1. Find the height of the given parallelogram.

Area:____________

height:____________


Area:____________


5 cm

3.5 cm 4 14

6 13

5 58

1ft

8“

12“

45o

4 2

34m 36m44cm 48cm

b

3x2 1

2

7 14

1 38

mm

mm

mm

“

4

6‘

60o

13’

3 3

h yd

16 yd

13 yd

5

4 14

6 13

5 58

8“

12“

45o

4 2

44cm 48cm

b

2 1 1 3

mm

mm

mm

‘

“

84 square inches

Area base height

h

h

= ⋅= ⋅

⋅ = ⋅ ⋅

⋅

84 12

1

12

84

1

1

12

12

17 122

127

=

=

h

h

Area base height

A a

a

= ⋅

= ⋅

= ( ) ⋅( ) ⋅

12 4 2

12 4 2

" "

(( )=a square inches48 2


P

== 122 8 12 8

20 20

40

+ + += += inches

7 inches

sq. inches

40 inches

48 2


Unit I, Part C, Lesson 2, Quiz Form B—Continued—

Name

3. Find the area and perimeter of the given rhombus.

Area:____________


4. Find the perimeter of the given parallelogram.

Area:____________


h yd

1

13 yd

5

4 14

6 13

5 58

8“

12“

45o

4 2

44cm 48cm

b

2 12

7 14

1 38

mm

mm

mm

‘

“

1ft

8“

12“

45o

4 2

34m 36m44cm 48cm

b

3x2 1

2

7 14

1 38

“

2508 sq. cm.

Area base height

b

b

= ⋅= ⋅

⋅ = ⋅ ⋅

2508 44

1

44

2508

1

44

1

1

4441 44 57

4457

⋅ ⋅=

=

b

cm b


P

== 577 48 57 48

210

+ + += cm

Area base height

A

square me

= ⋅= ⋅=

36 34

1224 tters


P

== 366 36 36 36

144

+ + += meters

1224 sq. meters

144 meters

210 cm

Name

Class Date Score

Quiz Form A


Unit I - The Structure of GeometryPart C - MeasurementLesson 3 - Triangles

Find the area and perimeter of the given triangles in exercises 1 through 3. [Note: You may first have to use thePythagorean Theorem (a2 + b2 = c2) to find some missing parts.]

1. Area:____________


2. Area:____________


17

810

6

15 4“

3“6“

9“5

2 6

9

5 13 yds

6 34 yds

4 5

4

5

45

4

5

15” 21”

12”

16”

9‘

3‘

4‘ 9 cm

6 cm

10.5 cm

8.4 cm

6.3 cm

6m

3m

4m

yds

4“

3“6“

9“5

2 6

5 13 yds

4 5

4

5

45

4

5

15” 21”

12”

16”

10.5 cm

8

Area base height

A

= ⋅ ⋅

= ⋅ +( ) ⋅

=⋅

1

21

215 6 8

1 211 8

21 21 2 4

221 4

84

⋅

=⋅ ⋅ ⋅

= ⋅= square uniits


P

== 100 17 15 6

48

+ + +=

( )

P units

Perimeter = Sum of Lengths of the Sides

Perimeter = 6 ++ 9 + 5

Perimeter = 20� inches

a b c

c

c

c

c

2 2 2

2 2 2

2

2

3 4

9 16

25

5

+ =

+ =

+ =

==

Pythagorean Theorem

Area base height

A

= ⋅ ⋅

= ⋅ ⋅

=⋅ ⋅

1

21

26 4

1 6 4

2

square inches

=⋅ ⋅ ⋅

=

1 2 3 4

212

84 sq. units

48 units

12 sq. inches

20 inches



Name

3. Area:____________


4. Find the area of a triangle with base (2x + 3) units and height (4x - 2) units.

Area:____________

5

2 6

15” 21”

12”

16”

9

10.5 cm

8.4 cm

6.3 cm

6

Perimeter Sum of Lengths of the Sides

P

=

Pythagorean TheoremArea base height

A

= ⋅ ⋅

= ⋅ ⋅

=⋅

1

21

28 5

8 5

2

==⋅ ⋅

= ⋅=

2 4 5

24 5

20 square units

a b c

x

2 2 2

2

+ =

+2 5

4 25

29

29

2 2 2

2

2

+ =

+ =

=

=

x

x

x

x

6 5

36 25

61

61

2 2 2

2

2

+ =

+ =

=

=

y

y

y

y

Perimeter units= + +( )8 29 61

Area base height

A x x

A

= ⋅ ⋅

= ⋅ +( ) ⋅ −( )

1

21

22 3 4 2

x x

Ax x x x

=⋅ +( ) −( )

=⋅ + ⋅ + ⋅ − +

1 2 3 4 2

22 4 3 4 2 2 3( ) ⋅⋅ −

=+ + +

=+ −

− −

( )2

28 12 4 6

28 8 6

2

2

2

x x x

Ax x

Ax x

A x x square units

=+ −( )

= + −( )

2 4 4 3

2

4 4 3

2

2

20 sq. units

units= + +( )8 29 61

x x square units

=

= + −( )

2

4 4 3

2

2



5. Find the area and perimeter of the given triangle.

Area:____________


6. Find the area and perimeter of the shaded square in the given figure.

Area:____________


17

810

6

15 4“

3“6“

9“5

2 6

9

5 13 yds

6 34 yds

4 5

4

5

45

4

5

15” 21”

12”

16”

9‘

3‘

4‘ 9 cm

6 cm

10.5 cm

8.4 cm

6.3 cm

6m

3m

4m

yds

4“

3“6“

9“5

2 6

4 5

4

5

45

4

5

15” 21”

12”

16”

9 cm

6 cm

10.5 cm

8.4 cm

6.3 cm

6

Area base height

A a

a

= ⋅ ⋅

= ⋅ ⋅

= ⋅

1

21

26

3

45

1

31

2

277

4

16

327 16

2 4 33 9 4 2 2

2 4 3

⋅

=⋅

⋅ ⋅

=⋅ ⋅ ⋅ ⋅

⋅ ⋅

A a

a

a

a square yards

= ⋅=

9 2

18


Perimeter

=

= 633

45

1

39

27

4

16

39

27 3

4

+ +

= + +

=⋅

Perimeter

Perimeter⋅⋅

+⋅⋅

+⋅⋅

= + +

3

16 4

3 4

9 12

1 1281

12

64

12

108Perimeter

11281 64 108

12253

12

Perimeter

Perimeter

=+ +

= or � 2211

12yards

Area of Larger Square: Pythagorean Theorem helps us find c.

c) Area of Shaded Square d) Perimeter is the sum of the lengths ofthe sides.

Area

square units

= +( ) ⋅ +( )= ⋅=

5 4 5 4

9 9

81

a b c

c

c

c

c

2 2 2

2 2 2

2

2

4 5

16 25

41

41

+ =

+ =

+ =

=

=

Area

square units

= ⋅

= ⋅

==

41 41

41 41

1681

41

Perimeter

units

= + + +

= + + +( )=

41 41 41 41

1 1 1 1 41

4 41

18 sq. yards

41 sq. units

4 41� units

211

12yards

Name

Class Date Score


Quiz Form B

Unit I - The Structure of GeometryPart C - MeasurementLesson 3 - Triangles

5

2 6

5

4

5

4

15” 21”

12”

16”

9

cm10.5 cm

8.4 cm

6.3 cm

6

Area base height

A

= ⋅ ⋅

= ⋅ ⋅

=⋅ ⋅

1

21

212 15

1 12 155

21 2 6 15

26 15

90 square inch

=⋅ ⋅ ⋅

= ⋅= ees


P

== 211 16 12

21 28

49

+ += += inches

Area =1

2base height

Area =1

29 3

Area =1 9 3

2 1

⋅ ⋅

⋅ ⋅

⋅ ⋅⋅ ⋅⋅1

Area =27

2or� 13

1

2square feet


P

=

Pythagorean Theorem

a b c

x

2 2 2

2

+ =

+

3 4

9 16

25

25

5

2 2 2

2

2

+ =

+ =

=

==

x

x

x

x

x

3 5

9 25

34

34

2 2 2

2

2

+ =

+ =

=

=

y

y

y

y

Two Missing Pieces:

feet

P feet

= + +( )= +( )

9 5 34

14 34

90 sq. inches

49 inches

131

2sq.� feet

14 34+( ) feet

Find the area and perimeter of the given triangles in exercises 1 through 3. Note: You may first have to use thePythagorean Theorem (a2 + b2 = c2) to find some missing parts.

1. Area:____________


2. Area:____________


17

810

6

15 4“

3“6“

9“5

2 6

9

5 13 yds

6 34 yds

4 5

4

5

45

4

5

15” 21”

12”

16”

9‘

3‘

4‘ 9 cm

6 cm

10.5 cm

8.4 cm

6.3 cm

6m

3m

4m

yds



Name

3. Area:____________


4. Find the area of a triangle with base (2x - 4) units

and height (x - 2) units Area:____________

4 5

4

5

45

4

5

15” 21”

12”

16”

9

9 cm

6 cm

10.5 cm

8.4 cm

6.3 cm

6


P

=

Pythagorean TheoremArea base height

A

= ⋅ ⋅

= ⋅ ⋅

=⋅ ⋅

1

21

26 9

1 6 9

2

square cm

=⋅ ⋅ ⋅

=

1 2 3 9

227

a b c

c

c

c

c

c

2 2 2

2 2 2

2

2

2

6 9

36 81

117

117

3 13

3

+ =

+ =

+ =

=

=

⋅ =

113 = c

cm

= + +

= +

6 9 3 13

15 3 13( )

Area base height

A a x x

A a

= ⋅ ⋅

= −( ) −( )

=

1

21

22 4 2

1

222 8 8

4 4

2

2

x x

A a x x square units

− +( )= − +( )

27 sq. cm

cm

=

+15 3 13( )

4 4

2

2x x square units

−

− +( )

NameUnit I, Part C, Lesson 3, Quiz Form B—Continued—


5. Find the area and perimeter of the given figure. Area:____________


6. Find the area and perimeter of the given figure. Area:____________


4

5

15” 21”

12”

16”

9

10.5 cm

8.4 cm

6.3 cm

6

17

810

6

15 4“

3“6“

9“5

2 6

9

5 13 yds

6 34 yds

4 5

4

5

45

4

5

15” 21”

12”

16”

9‘

3‘

4‘ 9 cm

6 cm

10.5 cm

8.4 cm

6.3 cm

6m

3m

4m

yds

Area base height

A

= ⋅ ⋅

= ⋅ ⋅ ( )

=

1

21

28 4 6 3( . ) .

88 4 6 3

22 4 2 6 3

226 46

. .

( . ) ( . )

.

( )( )

=⋅ ⋅

= cm22


P

== 8.. . .

.

4 6 3 10 5

25 2

+ += cm

Pythagorean Theorem

a) find c. b) Find x:

c) Perimeter is: d) Area is half the base times the height

a b c

c

c

c

c

2 2 2

2 2 2

2

3 4

9 16

25

5

+ =

+ =

+ =

==

a b x

x

x

x

x

2 2 2

2 2 2

2

2

6 5

36 25

61

61

+ =

+ =

+ =

=

=

Perimeter

Perimeter meters

= + + +

= +

6 3 4 61

13 61( )

Area base height base height

Area

= ⋅ ⋅ + ⋅ ⋅

= ⋅

1

2

1

21

24 ⋅⋅ + ⋅ ⋅

=⋅ ⋅⋅ ⋅

+⋅ ⋅ ⋅

⋅ ⋅=

31

25 6

1 4 3

2 1 1

1 5 2 3

2 1 1Area

Area 66 15

21

+=Area square meters

26.46 cm2

25.2 cm

21 sq. meters

13 61+( )meters

Name

Class Date Score

Quiz Form A


Unit I - The Structure of GeometryPart C - MeasurementLesson 4 - Trapezoids

Find the area and perimeter of each of the given trapezoids in exercises 1 through 3. Assume all measures arein inches.

1. Area:____________


2. Area:____________


6.3

7

6

10

6.7 9

4

11

3

18

10

87 9

x

x +7

x + 412

9

8

10

3.5

10

4.3

14

7

15

66.3 5

x

4

2x = 14

x

8m

x + 9

9

4

11

3

18

10

87 9

12

9

8

10

3.5

10

4.3

14

7

Area height sum of the bases

A

= ⋅ ⋅

= ⋅ ⋅ +(

1

21

26 10 7))

=⋅

=⋅ ⋅

= square inches

6 17

22 3 17

251


A

= ⋅ ⋅

= ⋅ ⋅ +(

1

21

24 11 9))

=⋅

=⋅ ⋅

= square inches

4 20

22 2 20

240


P

== 100 6 7 7 6 3

30

+ + +=

. .

P inches


P

== 4 ++ + +=

11 3 9

27P inches

40 sq. inches

27 inches

51 sq. inches

30 inches



3. Area:____________


4. The area of a trapezoid is 100 square centimeters. The sum of the

lengths of the bases is 50 centimeters. Find the height. height:____________

Name

18

10

87 9

9

3.5

10

4.3

14

7 x

8m

x + 9


A

= ⋅ ⋅

= ⋅ ⋅ +

1

21

27 10 18(( )

=⋅

=⋅ ⋅

= square inche

7 28

27 2 14

298 ss


h

= ⋅ ⋅

= ⋅ ⋅ ( )

1

2

1001

250

2

1⋅⋅ = ⋅ ⋅ ⋅

= ⋅ ⋅

⋅ = ⋅ ⋅

1002

1

1

250

200 1 50

1

50

200

150

1

50

h

h

h

550 4

50

50

504

⋅= ⋅

=

h

cm h

Perimeter sum of lengths of the sides== + + +=

8 10 9 18

445 inches

98 sq. inches

45 inches

4 cm



5. The area of a trapezoid is 420m2. The height is 12m. One base is 20m. Find the length b2 of the other base.

other base:____________

6. Application: The area of a trapezoid is 66 square units. The length of its longer base is 4 units longer than the length of its shorter base, and its height is 7 units longer than the length of its shorter base. Find the length of each base and the height of the trapezoid. (draw a diagram and label the necessary parts)

length of short base:____________

length of long base:____________

height:____________


b

= ⋅ ⋅

= ⋅ ⋅ +

1

2

4201

212 20 2(( )

= ⋅ ⋅ +

=⋅ ⋅

⋅⋅ +( )

4201

2

12

120

4201 2 6

2 120

42

2

2

( )b

b

00 6 20 6

420 120 6

420 120 120 120 6

2

2

2

= ⋅ + ⋅= + ⋅

− = − +

b

b

b

3300 0 6

1

6

300

1

1

66

1 6 50

6 1

6

650

2

2

2

= +

⋅ = ⋅ ⋅

⋅ ⋅⋅

= ⋅

b

b

b

m == b2

6.3

7

6

10

6.7 9

4

11

3

18

10

87 9

x

x +7

x + 412

9

8

10

3.5

10

4.3

14

7

15

66.3 5

x

4

2x = 14

x

8m

x + 9


x x

= ⋅ ⋅

= ⋅ + ⋅ ( )

1

2

661

27( ) ++ +( )

= ⋅ +( ) ⋅ +( )

⋅ = ⋅ ⋅ +

x

x x

x

4

661

27 2 4

2 662

1

1

27(( ) +( )

= +( ) +( )= ⋅ + ⋅ + ⋅ +

2 4

132 7 2 4

132 2 7 2 4 7

x

x x

x x x x ⋅⋅

= + +

− = + + −

=

4

132 2 18 28

132 132 2 18 28 132

0 2

2

2

x x

x x

x22

2

18 104

0 2 9 52

0 2 4 13

+ −

= + −( )= −( ) +( )

x

x x

x x

0 2

0 4

0 4 4 4

4 0

4

0 13

0 13

=

= −+ = − +

= +=

= +− =

( )false

x

x

x

x

x

xx

x

x can t be negative

+ −

= +

=

−

−

13 13

13 0

13 ( ' )

x base

x height

x base

=+ =+ =

4

7 11

4 8

( )

( )

( )

50 meters

4 units

8 units

11 units



Name

3. Area:____________


4. The longer base of a trapezoid is twice the length of the shorter base. The longer base measures 14 inches and the height is 4 inches. Find the area. (Draw a diagram and label the appropriate parts.)

Area:____________

x + 412

9

4.3

1

7

15

66.3 5

x

4

2x = 14

x

8m

x + 9


A

= ⋅ ⋅

= ⋅ ⋅ +(

1

21

25 15 7))

=⋅

=⋅ ⋅

= square feet

5 22

25 2 11

255


A

= ⋅ ⋅

= ⋅ ⋅ +(

1

21

24 14 7))

=⋅

=⋅ ⋅

= square inches

4 21

22 2 21

242

Perimeter sum of the lengths of the sides

P

=

feet

= + + +=

15 6 7 6 3

34 3

.

.

55 sq. feet

34.3 feet

42 sq. inches

.7 9

4

11

3

18

10

87 9

12

9

8

10

3.5

10

4.3

14

7 x

4

2x = 14

x

8m

x + 9

Longer � base� is� twice� the� shorter� base

14 = 2xx

7 =� x



6. Application: One base of a trapezoid is 9m longer than the other. The height is 8m. The area is 88m2.Find the lengths of the bases. (Draw a diagram and label the necessary parts)

Shorter base:____________

Longer base:____________

9

3.5

10

4.3

14

7 x

8m

x + 9

Area =1

2height sum of the bases

88 =1

28 x + x + 9

⋅ ⋅

⋅ ⋅ ( )

⋅ ⋅ ( )

⋅⋅

⋅ ( )

88 =1

28 2x + 9

88 =1

2

2 4

12x + 9

88 = 4 2x + 99

88 = 4 2x + 4 9

88 = 8x + 36

-36 +88 = 8x + 36 + 36

52 = 8

-

( )⋅ ⋅

xx

1

8

52

1=

1

88x

4 13

4 2=

8

8x

13

2or� 6

1

2= x (shor

⋅ ⋅

⋅⋅

⋅

tter � base)

61

2+ 9 = x + 9

151

2= x + 9 (longer � base)( )

13

2or� 6

1

2

151

2



3. Area:____________


4. Area:____________


Name

12

2 3

12

12

12

ar

12

12

12

121

6

20

310

1

33

18.115

16

2.4

2

412

2 3

6

3 3

11

16

12

12

12

ar

12

12

12

12

12

1212

12

12

1212

12

12

12

12

12

12

1212

12

12

1212

12

6 6

20

310

18 16

8 33

18.115

1

Perimeter Sum of Lengths of the sides

P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

3 12

36


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

6 6

36

Area measure of a side Apothem

number of sides

= ⋅ ⋅1

2

s a n

or

a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

2

1

21

222 3 36

3 36

36 3

( ) ⋅

= ⋅

= square cm


the number of si

= ⋅ ⋅1

2ddes

A s a n

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

=

1

21

21

2⋅⋅ ( ) ⋅

=⋅ ⋅ ⋅

=

3 3 36

3 3 2 18

2

54 3

A

square cm

36 cm

36 3

(

square cm

36 cm

54 3

A

square cm



5. Area:____________


6. Find the area and perimeter of a regular decagon with apothem 6.8cm and side length 4.4cm.

Area:____________


412

6

3

11

16

12

12

12

ar

12

12

1

12

121

12

12

12

12

12

1212

12

12

1212

12

6 6

20

310

18 16

8 33

10

6 9

18.115

16

16

ra


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

5 16

80


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

10 4 4

44

( . )


number of sides

= ⋅ ⋅1

2

s a n

or

a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

2

1

21

2111 80

11 2 40

2440

⋅

=⋅ ⋅

= square cm


number of sides

= ⋅ ⋅1

2

s a n

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

21

21

26.88 44

2 3 4 44

2149 6

( ) ⋅

=⋅( ) ⋅

= square cm

.

.

80 cm

440 square cm

44 cm

149.6 square cm



7. Application: Find the indicated measures in the given regular hexagon. a = ________

r = ________

Area = _______

Perimeter = ________

12

2 3

12

12

12

ar

12

12

12

121

6

20

310

1

33

18.115

16

1


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

units

= ⋅=

6 12

72

412

2 3

11

16

12

12

12

ar

12

12

12

12

12

1212

12

12

1212

12

6 6

20

310

1 16

8 33

18.115

1

16

ra

Hint #1: Complete the inside of the hexagon with line segments drawn from the center to all 6 vertices. The smalltriangles appear to be (and actually are) equilateral. Therefore, r = 12 units

Hint #2: To find “a”, we can reason this way. Each triangle is an equilateral triangle and therefore an isosceles triangle. The Apothem, “a”, is an altitude in the triangle and will bisect the base of 12. So, our hexagon could look like this: Observe what the apothem does to the side of length 12.

Pythagorean Theorem.

Area measure of a side Apothem number of sides= ⋅ ⋅1

2

s a n

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

=

1

21

2

2.4

2

412

2 3

6

3 3

11

16

12

12

12

ar

12

12

12

12

12

1212

12

12

1212

12

12

12

12

12

12

1212

12

12

1212

12

6 6

20

310

18 16

8 33

10

6.9

18.115

16

16

ra

= ⋅

1

1

26 3 ⋅⋅

=⋅ ⋅ ⋅

=

72

2 3 3 72

2

216 3

A

square units

a b c

a

a

a

2 2 2

2 2 2

2

2

6 12

36 144

36 36 144 36

+ =

+ =

+ =

+ + = +− −

aa

a

a

a units

2 108

108

36 3

6 3

=

=

= ⋅

=

6 3 units

12

216 3 square � units

72 units

Name

Class Date Score


Quiz Form B

Unit I - The Structure of GeometryPart C - MeasurementLesson 5 - Regular PolygonsFind the perimeter and area of each regular polygon in exercises 1 through 6. Assume all dimensions are in centimeters.

1. Area:____________


2. Area:____________


12

12

12

12

ar

20

310

1 18.115

1

2.4

2

412

2 3

6

3 3

11

16

12

12

12

ar

12

12

12

12

12

1212

12

12

1212

12

12

12

12

12

12

1212

12

12

1212

12

6 6

20

310

18 16

8 33

10

6.9

18.115

16

16

ra


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

6 20

120


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

4 18

72


number of sides

= ⋅ ⋅1

2

s a n

or

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

2

1

21

2110 3 120

2 5 3 120

2

600 3

⋅

=⋅ ⋅

= square cm


number of sides

= ⋅ ⋅1

2

s a n

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅ ⋅

1

21

21

29 772

9 2 36

2324

A

square cm

=⋅ ⋅

=

120 cm

600 3 square cm

72cm

324 square cm

Note: a = 18/2 or 9



3. Area:____________


4. Area:____________


12

12

12

12

1212

12

6 6

310

16

8 33

18.115

16

16

ra

12

2 3

12

12

12

ar

20

310

18.115

1


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

3 16

48


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

8 15

120


number of sides

= ⋅ ⋅1

2

s a n

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

21

2

1

2

8 3

3348

8 3 2 3 8

2 3

64 3

⋅

=⋅ ⋅ ⋅⋅

= square cm


number of sides

= ⋅ ⋅1

2

AArea s a n

Area a Perimeter

Area

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

21

21

218..

.

.

1 120

18 1 2 60

218 1 60

( ) ⋅

= ( ) ⋅ ⋅

= ( ) ⋅square cm= 1086

120 cm

1086 square cm

48 cm

64 3 square cm



5. Area:____________


6. A regular dodecagon has a side of length 2 in. and an approximate area of 44.78 in2.Find the length of the apothem to the nearest tenth of an inch.

apothem:____________

18 16

8 33

10

6.9

18.115

16

16

ra


P N

== uumber of sides

length of each side

P n s

⋅

= ⋅P

cm

= ⋅=

5 10

50


number of sides

= ⋅ ⋅1

2

s a n

A a Perimeter

A

= ⋅ ⋅ ⋅

= ⋅ ⋅

= ⋅

1

21

21

26.99 50

6 9 2 25

26 9 25

17

( ) ⋅

=( ) ⋅ ⋅

= ( ) ⋅=

.

.

22 5. square cm


number of sides

= ⋅ ⋅1

2

444 781

2

44 781

22 12

44 782 12

2

.

.

.

= ⋅ ⋅ ⋅

= ⋅ ⋅ ⋅

= ⋅ ⋅

s a n

a

a

444 78 12

1

1244 78 12

1

123 731 1

3 731

.

.

.

.

= ⋅

⋅ = ⋅ ⋅

=

a

a

a

innches a=

50 cm

172.5 square cm

3.7 inches

Name

Class Date Score

Quiz Form A


Unit I - The Structure of GeometryPart C - MeasurementLesson 6 - Circles

1. If the radius of a circle is 16.2 inches, the diameter of the circle is ______________.

2. If the area of a circle is 15.7 sq. cm., the radius of the circle is _______. (use 3.14 to approximate .)π

32 4. inches

Diameter D = 2 radius r

Diameter = 2 16.2

Dia

( ) ⋅ ( )⋅( )

mmeter = 32.4 � � inches

Area r

r

π ⋅

⋅

⋅ ⋅

2

215 7 3 14

1

3 14

15 7

1

1

3 14

3

. ( . )

.

.

.

.114

13 14 5

3 14

3 14

3 14

5

5

2

2

2

⋅

( ) ⋅( ) ⋅

r

r

r

cm r

.

.

.

.

5 cm

5

8

8

8

8

8



Name

radius diameter Circumference Area3. ________ 13” ________ ________

radius diameter Circumference Area4. 7.5 cm ________ ________ ________

Find the missing radius or diameter, as indicated in exercises 3 through 6. Then find the circumference andarea of the circle. Note: Approximate your answers to the nearest hundredth.

Diameter = 2 radius� (r)⋅⋅

⋅ ⋅ ⋅

13 = 2 r

1

2

13

1=

1

2

2

1r

13

22or� 6

1

2= r

= ⋅( )=

2 7 5

1

.

55

Area r

Area inches inche

= ⋅

( )

π 2

3 1413

2

13

2. ss

Area

Area

( ) ⋅ ( ) ⋅ ( )( ) ⋅3 14 6 5 6 5

3 14

. . .

. 442 25

132 665

.

.

( )Area !

Area r

A

π ⋅

( )( )( ) ⋅

2

23 14 7 5

3 14 56 25

. .

. .(( )A 176 625.

Circumference� � � � � = 2 rπ

CircumferenceB� 2 3.14! (( )

( )( )

13

2

CircumferenceB� 3.14 13

Circu

!

mmferenceB� 40.82!

Circumference D

Circumference

Cir

=( )( )π3 14 15.

ccumference ! 47 10.

Diameter D radius r

D

( ) = ⋅ ( )=

2

6.50 m. 40.82 in. 132.67 sq”

15 cm 47.10 cm 176.625 cm2

8

8

8

8

8

8

5

8

8

8

8

8



radius diameter Circumference Area5. ________ 3.5 m ________ ________

radius diameter Circumference Area6. 6 ft. ________ ________ ________

312

2

12

72

12

2

1 72 2

1 22 1 1

74

1

= ⋅

⋅ = ⋅ ⋅

⋅⋅

=⋅ ⋅⋅ ⋅

r

r

r

or ..75 = r

= ⋅=

2 6

12

Area r

A

= ⋅

= ( ) ⋅ ( )= ( )

π 2

23 14 1 75

3 14 3 06

. .

. . 225

9 61625

( )= .

Area r

Area

Area

Area

= ⋅

( )( )( ) ⋅

π 2

23 14 6

3 14 36

! .

.

1113 04.

Circumference r

C

=( )(

2

2 3 14 1 75

π. . ))

( )( )C 6 28 1 75

1

. .

00 99.

Circumference r

C e ft

=( )( )

2

2 3 14 6

π.

e

e

6 28 6

37 68

.

.

( )( )

1.75 m 10.99 m 9.61625 m2

Diameter D radius r

D

( ) = ⋅ ( )=

2

Diameter D radius r

D

( ) = ⋅ ( )=

2

12 ft. 37.68 ft. 113.04 ft2

8

8

8

8

8

8

8

8

8

In exercises 7 and 8 find the approximation, correct to the nearest hundredth, of the circumference of a circlewith the given radius. Use 3.142 to approximate .

7. radius = cm Circumference 8 ________ 8. radius = 8.6 in. Circumference 8 ________

9. Find the area of the shaded region. Use 3.142 for , and approximate the answer to the nearest tenth.

Area 8 _____________



3

5

5

5 2

π

π

73

Area r r

Area

= −( )( )( ) − ( )( )(π π2 2

3 142 5 5 3 142 3 3! . . ))( ) − ( )( ) −( )

Area

Area

! 3 142 25 3 142 9

3 142 25 9

. .

.

AArea

Area

Area

! 3 142 16

50 272

50 27

.

.

.

( )( )

14.66 cm. 54.04 in.

Area of shaded region = area of large circle - area of small circle.

50.3 units2

Circumference = 2πr

2 3.142 73

cm

6.2

( )

884 73

cm

14.66 cm

( )

Circumference� � = 2 r

Circumference 2 3.142 8.

π( ) 66 in

Circumference 6.284 8.6 � in

Circumf

( )( )( )!

eerence 54.04 � in!

8

8

8

8

8

8

8

8

8

8

8



In exercises 7 and 8 find the approximation, correct to the nearest hundredth, of the circumference of a circlewith the given radius. Use 3.142 for .

7. r = 2.1 km Circumference = ________ 8. r = cm. Circumference = ________

9. Find the area of the shaded region. Use 3.142 for . Approximate the answer to the nearest tenth

Area = _____________

5

5 2

7

5

π

π

Circumference r

C

== ( )(

2

2 3 142 2 1

π. . ))

= ( )( )=

6 284 2 1

1

. .

33 1964

13 20

.

. km=

Circumference r

Circumference

=

( )

2

2 3 1427

5

π

.

( )( )Circumference

Circumferen

! 2 3 142 1 4. .

cce

Circumference

Circumf

! 6 284 1 4

8 7976

. .

.

( )( )

eerence cm! 8 80.

Area of shaded region Area of circle Area of squar= − ee

Area r s

Area

Area

= −

= ⋅ ⋅ − ( )( )= ⋅ ⋅ −

π

π

π

2 2

5 5 5 2 5 2

5 5 55 2 5 2

25 5 5 2 2

25 25 2

( )( )= ⋅ − ⋅ ⋅ ⋅( )= − ⋅( )

Area

Area

π

πAArea

Area

Area

A

= −( ) −

−

25 50

25 3 142 50

78 55 50

π.

.

rrea square unitsB 28 55.

13 20

.

. km= 8.80 cm

28.6 square units

8

8

8

8

8

8

8

8



4. Lateral Area = ________

Total Area = ________

Volume = ________

10ft

19ft

1

8m

6m12m

8m

6m

4 in 4 in

3 in

5 in

2.23 in

3“

12“

3 2“2

6“

4“3“

3“

4“

9“

5“

4“

2 3“

10 ft

4 ft

8

6cm

6 ft

12 ft

90o

Lateral Area of a Prism (L.A.) = Perimeter of the Base Multiplied by the height of the prism

Total Area of a Prism (T.A.) = The sum of the lateral area plus the area of the bases

T.A.= L.A.+1

2b h+

1

2b h

Area = 55 +1

24 2.23+

1

2

⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅⋅ ⋅

⋅ ⋅

4 2.23

Area = 55 + 2 2.23+ 2 2.23

Area = 55 + 4.46 + 4..46

Area = 63.92� in2

L.A.= P h

L.A.= 4 + 4 + 3 5

L.A.= 11 5

L.A.= 55 in2

⋅( ) ⋅

⋅

Volume of Prism� (V)� =� Area of one base

� multiplieed by� the�

� height � of � the� prism

V =1

2b⋅ ⋅⋅ ⋅

⋅ ⋅ ⋅

⋅ ⋅

h 5

V =1

24 2.23 5

V = 2 2.23 5

= 4.46 ⋅⋅ 5

= 22.3� in3

55in2

63.92in2

22.3in3



Name



Volume = ________

3 in

5

2.23 in

3“

12“

3 2“2

3“

4“

9“

5“

4“

2 3“

10 ft

4 ft

8cm

6cm

6 ft

12 ft

90o

144 in2

(144 +� 16 3 )� in2

72 3 � in3

Lateral Area of a Prism (L.A.) = Perimeter of the Base Multiplied by the height of the prism.

Perimeter = Sum of the lengths of the sides.

Perimeter of Base = 3 + 4 + 5 + 4= 16 in.

Total Area of a Prism (T.A.)= The sum of the L.A. and the area of the bases

L.A.= P h

� = 16 � 9

� = 144 � in2

⋅⋅

Area� of � one� base =12

h (b + b )

�

1 2⋅ ⋅

=12

2 31

(3 5)

�

⋅ ⋅ ⋅

= 3 8� sq.� in.� or � 8 3 � in2⋅

Area� of � the � two� bases = 8 3 � in + 8 3 � in

�

2 2

= 16 3 � in

�

2

Total � area �= 144 +16 3 � in

�

2

= (144 +16 3 � )in2

Volume of a Prism (V) = Area of the Base x height

Volume =� B h = 8 3 � in 9� in

Volume =� 72 3 � in � or � 7

2

3

⋅ ⋅

22 3 � cubic� inches





Volume = ________

10 ft

4

8cm

6cm

6

(48 + 24 )cm2π

(33 48)� cm2π +

36 � cm3π

Lateral Area of a Cylinder (L.A.) = Circumference of the Basemultiplied by height

Diameter = 2 • Radiusd = 2 • r

Circumference = • d

(*full cylinder) Since this figure has a semi-circle for a base, itis of a cylinder. The Lateral Area will be of the Lateral Areaof the full cylinder plus the Area of the rectangular side createdby cutting the full cylinder in half.

Total Area of a Cylinder (T.A.)= The sum of the L.A. and the area of the bases

L.A.= Å d h

� = 6 cm 8� cm

� =

ππ

⋅ ⋅⋅ ⋅

πππ⋅

∗48� cm

� = 48 � cm � �

2

2

Area� of � one� base = radius radius

�

π ⋅ ⋅

� = r

�

2π ⋅= 3� cm 3� cm

�

π ⋅ ⋅� = 9� cm

� *

2π ⋅= 9 � cm

�

2π== 2 r

� 6 cm�= 2 r

�

⋅⋅

12

6 cm�=12

2 r

�

⋅ ⋅ ⋅

3 cm� = r

Area� of � one� base� of �12

cylinder =1∴22

9 cm =9

Area� of � two� bases = 292

2⋅

⋅

π

π cm

� =21

92

2

⋅ ⋅π ccm

= 9 Å cm

2

2π

Total � Area = 24 � cm + 48� cm 9 cm

�

2 2 2π π+= (24 + 48 + 9 )� cm

�

2π π� = (24 + 9 48)� cm

� = (33 4

2π ππ

++ 88)� cm2

Volume of a Cylinder (V) = Area of the Base x height

Volume� of � a� full � cylinder =� 9 � cm � � 8� cm = 722π ππ cm

� V

3

oolume� of � a� half � cylinder =12

72 � cm

�

3π

= 36 � cm or�3π 336 � cubic� centimetersπ

π

1

2

1

2

12

Lateral � Area� of � full � cylinder � =�12

48 Å⋅ ⋅ π ccm 24 � cm

Area� of � Rectangle = w = 8� cm 6 � cm

2 2=

⋅ ⋅

π

= 48� cm

Lateral � Area� of �12

cylinder = 24 Å cm

2

∴ π 22 2 2+ 48� cm = (48 + 24 )� cmπ

∴ ⋅Area� of � one� base� of �12

cylinder =12

9 cm =92π22

cm2π

•

•



Name

Find the lateral area, total area, and volume of the right circular cone shown below. Note: Do not approximate , but round the decimal numbers in your answers to the nearest hundredths.



Volume = ________

12.7mm

6.3mm

8cm

8cm

10cm

1

16 in

7 in

10”

6“6

20cm16cm

8

mm

Lateral Area of a Cone (L.A.) = times the circumference of the base times the slant of the height.

C = • d

d = 2 • r

Use Pythagorean Theorem to find slant height.

L.A.= Å1

2c

=1

2d

=

l

l

⋅ ⋅

⋅ ⋅ ⋅π

1

22 r

� =1 2 6.3� mm 14.17

l⋅ ⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅

π

π 77 mm

2 1 1 1⋅ ⋅ ⋅ ⋅ 1= 89.32� mm

� = 89.3

2π ⋅22 sq� mmπ

1

2

π

Total Area of a Cone (T.A.) = the sum of the Base Area and the Lateral Area.

T.A.= Å B.A.� +� L.A.

� = (Area� of � a� circle))+ L.A.�

� = r � +� 89.32Å � sq.� mm

�

2π ⋅

= (6.3� mm) � +� 89.32 � mm

� = 39.69

2 2π ππ

⋅⋅ mm � +� 89.32 � mm

� = (39.69 � +� 89.32 )

2 2ππ π � mm

� = 129.01 � mm

2

2π

Volume of a Cone (V) = times the Area of the Base times the height of the cone.

Use Pythagorean Theorem to find h.

V = Å1

3B h

� =1

3(Area� of � a� circl

⋅ ⋅

⋅ ee) h

� =1

3r h

� =1

3(6.

2

⋅

⋅ ⋅

⋅ ⋅

π

π 33mm) (6.3mm) (12.7mm)

� =1 504.06

⋅ ⋅

⋅ ⋅π 33

3 1 1

� =504.063

3

168.0

⋅ ⋅

≈

π

22 mm3π

1

3

89.32π sq mm

129.01 mm2π

168.02 mm3π

8

8

8

8

8

8

8

8

8 168.021

π

8

8

8



Name

Find the lateral area, total area, and volume of the right pyramid shown below.



Volume = ________

5cm

3cm

12.7mm

6.3mm

8cm

8cm

10cm

10’8‘

16 in

7 in

10”

6“6’

12’

20cm16cm

8 mm

4 mm

12

14

Lateral Area of Pyramid (L.A.) = times the perimeter of the Base multiplied by the slant height.

Use Pythagorean Theorem to find the length of each side of thesquare.

L.A.= Å1

2P

=1

2(12 +12 +12 +12) 10

�

⋅ ⋅

⋅ ⋅

"

=1

248� ft � 10� ft

� =1

2

2 2

⋅ ⋅

⋅ ⋅ 44 10

1 1ft

� = 240� ft

2

2

⋅⋅

1

2Total Area of Pyramid (T.A.) = the sum of the Base Area and

the Lateral Area.

T.A.= B.A.� +� L.A.

� =144 � ft + 240� ft

�

2 2

� = 384� ft 2

Volume of a Pyramid (V) = times the Area of the Base times the height of the pyramid.


V = Å1

3B h

� =1

3144 � ft 8� ft

�

2

⋅ ⋅

⋅ ⋅

=1

3

3 48� ft 8� ft

1 1ft

� = 38

23⋅

⋅ ⋅⋅

44 ft 3

1

3

1

2

240 ft2

384 ft2

384 ft3

Name

Class Date Score


Quiz Form B

Unit I - The Structure of GeometryPart C - MeasurementLesson 8 - Pyramids

Find the lateral area, total area, and volume of the right square pyramid shown below.



Volume = ________

8cm

8

10’8‘

16 in

7 in

10”

6“6’

12’

20cm16cm

8 mm

4 mm

12

14

Lateral Area of Pyramid (L.A.) = times the perimeter of the Base multiplied by the slant height.

Use Pythagorean Theorem to find slant height.

L.A.=� �1

2P

=1

2(6 + 6 + 6 + 6) 153

�

⋅ ⋅

⋅ ⋅

"

=1

2(24 � ft)� 153 ft

� =1

2

2 1

⋅ ⋅

⋅⋅ 22 153

1 1ft 2⋅

⋅

⋅

⋅

= 12 153 ft

= 12 9 17 ft

= 12 3 17 ft

= 3

2

2

2

66 17 ft 2

1

2Total Area of Pyramid (T.A.) = the sum of the Base Area and

the Lateral Area.

T.A.= B.A.� +� L.A.

� = 36 � ft + 36 17 � ft

�

2 2

= 36(1� + 17 )� ft 2

Volume of a Pyramid (V) = times the Area of the Base times the height of the pyramid.


V = Å1

3B h

� =1

336 � ft 12� ft

�

2

⋅ ⋅

⋅ ⋅

=1

3

3 12� 12�

1 1ft

� = 144� ft

3⋅ ⋅ ⋅⋅

33

1

3

36 17 � ft 2

36(1�+ 17 )� ft 2

144 � ft 3



Name

Find the lateral area, total area, and volume of the right circular cone shown below.



Volume = ________

10”

6“6’

12’

20cm16cm

8 mm

4 mm

12

14

Lateral Area of a Cone (L.A.) = times the circumference of the base times the slant height of the cone.

C = • d

d = 2 • r

Use Pythagorean Theorem to find the slant height.

L.A.= Å1

2c

=1

2d

=

⋅ ⋅

⋅ ⋅ ⋅

"

π

1

22r

� =1

22

17

4mm

17 5

⋅ ⋅ ⋅

⋅ ⋅ ⋅ ⋅

π

π44

mm

� =1 2 17 17 5

2 1 1 4 4mm

�

2⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅

π

=289 5

16mm

� =289 5

16mm

2

2

π

π

⋅

1

2

π

Total Area of a Cone (T.A.) = the sum of the Base Area and the Lateral Area.

T.A.= B.A.� +� L.A.

� = (Area� of � a� circle)+ L.A.�

� = r � +�289 5

16mm

� =

2 2π π

π

⋅

⋅⋅ ⋅

⋅

17

4mm

17

4mm�+� �

289 5

16mm

� =1

17

4

2π

π⋅⋅17

4mm +� �

289 5

16mm

� =289

16mm +� �

28

2 2

2

π

π 99 5

16mm

� =289 289 5

16mm

� =289

2

2

π

π π+

ππ(1+ 5)

16mm2

Volume of a Cone (V) = times the Area of the Base times the height of the cone.

V = Å1

3B h

� =1

3(Area� of � a� circl

⋅ ⋅

⋅ ee) h

� =1

3r h

� =1

3

17

4

2

⋅

⋅ ⋅

⋅

π

π mmm17

4mm

17

2mm

� =1 17 17 17

3 1

⋅ ⋅

⋅ ⋅ ⋅ ⋅⋅

π⋅⋅ ⋅ ⋅4 4 2

mm

� =4913

96mm

3

3π

1

3

289π 516

mm2

289π 1+ 5

16mm2( )

4913π96

mm3


Name

3. If the volume of a sphere is 12 cubic units, find the radius and the surface area.

radius = ________

Surface Area = ________


π

Surface� Area� of � a� sphere = 4 r

�

2⋅ ⋅π

= 4 9 9

�

3 3⋅ ⋅ ⋅π

= 4 9

�

23⋅ ⋅π

= 4 81

�

3⋅ ⋅π

= 4 2⋅ ⋅π 77 3

�

3 3⋅

== 4 3 3

�

3⋅ ⋅ ⋅π

= 12 3 � sq.� units3π ⋅

Volume� of � sphere =4

3r

12

3⋅ ⋅π

ππ π

π π

� =4

3r

3 12 � � � = 34

3r

3

3

⋅

⋅ ⋅ ⋅

36 � � � = 4 r

�1

4

36

1

3π π

ππ⋅ =

1

4

4

1r

1

4

4 9

1= 1 r

�

3

3

ππ

ππ

⋅ ⋅

⋅⋅

⋅

9 = r

� 9 units

3

3 == r

12 3 � sq.� units3π ⋅

9 units3



Name

5. The volume of a sphere is cubic meters. Find the radius and surface area.

radius = ________


π9

16

Volume� of � sphere� � =4

3r 3⋅ ⋅π

9

16� =

4

3r

48�

9

16 1

3⋅ ⋅ ⋅

⋅ ⋅

π π

ππ

=48 4

3 1r

3 16�

9

16 1� =

3 1

3

ππ

ππ

⋅ ⋅ ⋅

⋅⋅ ⋅

⋅ 66 4

3 1r

27 = 64r

3

ππ

⋅ ⋅ ⋅

33

31

64

27

1=

1

64

64

1r⋅ ⋅ ⋅

27

64= r 3

27

64� = r

27

64

3

3

33= r

3

4meter� � =� � r � � � � � � � � � � � � � � � � � � � � � � � � � � � �


=4

1 1�

3

4

2⋅ ⋅

⋅ ⋅

ππ 3

4

=4

1 1�

3

4

3

4

=1

9

4

=9

4square

⋅

⋅ ⋅ ⋅

⋅

π

π

π meters

34

meter

94

square metersπ

Name

Class Date Score

Unit I - The Structure of GeometryPart C - MeasurementLesson 9 - Spheres


Quiz Form B

Find the surface area and volume of the sphere illustrated below. Round all answers to the nearest tenth. Use 3.14 as an approximation for .

1. Surface Area = ________

Volume = ________

π

6cm

92

0.76cm

5 in

ft

Volume� of � sphere� � =4

3r 3⋅ ⋅π

=4

35 in 5� in 5� in

�

⋅ ⋅ ⋅ ⋅π

=4 3.14 5 5 5

3

⋅ ⋅ ⋅ ⋅iin

=1570

3i

3

nn

= 523.3� � c

3

uubic� inches


�

2⋅ ⋅πTotal � Area� � � = 4 r r

�

⋅ ⋅ ⋅π= 4 3.14 5� in 5� in

�

⋅ ⋅ ⋅� = 4 3.14⋅ ⋅ 55 5� in

�

2⋅= 314� in � or � 314.0� sq� inches2

314.0� sq� inches2

523.3� � c

3

uubic� inches


Name

3. The volume of a sphere is 36 mm3. Find the radius and the surface area.

radius = ________


© 2014 VideoTextInteractive Geometry: A Complete Course

π

113


�

2⋅ ⋅π= 4 3� mm 3� mm

�

⋅ ⋅ ⋅π= 4 3 3⋅ ⋅ ⋅π mm

� =

2

336 sq� mmπ


3r

36

3⋅ ⋅π

ππ π

ππ

ππ

=4

3r

3

4

36

1=

3

4

4

3r

3

3

⋅

⋅ ⋅ ⋅

3

4

4 9

1= 1 r

�

3

ππ

⋅⋅

⋅

227 = r

� 27 � =� � r � �

�

3

3

3 mm�=� � r � �

3 mm� =� � r � �

336 sq� mmπ


Name

© 2014 VideoTextInteractive Geometry: A Complete Course

5. The radius of a sphere is inches. Find the surface area and volume.


Volume = ________

8 2

115


�

2⋅ ⋅π

Total � Area� � = 4 8 2 � in 8 2 � in

�

⋅ ⋅ ⋅π

= 4 8 8 2 �⋅ ⋅ ⋅ ⋅ ⋅π 22 in

�

2

== 4 64 2� in

�

2⋅ ⋅ ⋅π= 512 � square� inπ


3r 3⋅ ⋅π

=4

38 2 � in 8 2 � in 8 2 � in

�

⋅ ⋅ ⋅ ⋅π

=4 8 2 � 8 2 8 2

3 1 1 1 1� i

⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅⋅ ⋅ ⋅ ⋅

πnn

=4 8� 8 8 � 2

3

⋅ ⋅ ⋅ ⋅ ⋅π ⋅⋅ ⋅

⋅ ⋅

� 2 � 2

3in

� =4

3

π 5512 2 � 2

3in

� =40

3⋅ ⋅

996 2

3cubic� in

π ⋅

512 � square� inπ

40

3

996 2

3cubic� in

π ⋅

NameUnit I, Part D, Lesson 1, Quiz Form A—Continued—


For each group listed in exercises 4 through 6, read the accompanying scenario illustrating a good use ofinductive reasoning. Then write a scenario of your own. Check with another person for its validity.

4. Football Players

The opponent in the Cougar’s next game throws a pass on first down 8 out of 10 times according to statistics from the first five games. The Cougars expect they will need to be prepared to use theirpass defense the majority of the time on first down.

5. Employees

Employees of the Discount Mart Variety Store have a meeting every Friday morning one hour before the store opens. Their supervisor has been ten to fifteen minutes late to the meeting for the last 7 weeks. There has been a noticeable increase in the number of employees who are late since the meeting never seems to start on time anyway.

6. Police Officers

The intersection of 5th Street and Cumberland Avenue has been the scene of nine accidents in the last four weeks. Over the last three months the number of speeding citations issued on Cumberland Avenue has increased by 5% over the previous three month period. The Police Department has requested that a study of the daily traffic patterns be conducted to determine a remedy for the dangerous situation atthis intersection.

Answers will vary.

Answers will vary.

Answers will vary.

NameUnit I, Part D, Lesson 1, Quiz Form B—Continued—


For each group listed, in exercises 4 through 6, read the accompanying scenario illustrating a good use ofinductive reasoning. Then write a scenario of your own. Check with another person for its validity

4. Manufacturers

Each year, auto makers introduce new colors for their cars. One way auto makers choose colors for new cars is to find out which colors sold well in the past. Trends which are observed over a long period of time, say five years, help automakers to decide what color of automobile to produce, in an attempt to sell more cars.

5. Thieves

The owner of a small photography shop is observed leaving his business to go to the bank at approximately the same time every day. A thief would use such an observation to plan a confrontation and possible robbery. For the businessman, his responsibility is to avoid following the same routine everyday.

6. Explorers

Throughout history, there have been many explorers from Columbus to Lewis and Clark to Space Shuttle Astronauts. Early explorers had only limited knowledge about conditions they would encounter. More modern explorers can use technology to help them prepare for their travels. However, at all levels, explorers were required to make observations and record patterns of activity which might effect their success. For example, how did Columbus acquaint himself with the prevailing winds needed to push him across the ocean? How did Lewis and Clark know when the best time was to move? When is the best “window” for launching a space shuttle? Planning had to be based on observations such as weather patterns and moon phases, and using that information to make assumptions to be acted upon, to successfully complete the mission.

Answers will vary.

Answers will vary.

Answers will vary.

NameUnit I, Part D, Lesson 2, Quiz Form A—Continued—


4. Use inductive reasoning to find the next two terms of the sequence given below. Describe how you found these terms.

A, B, D, G, K, ____, _____

Look for a pattern and predict the next two numbers in each sequence in exercises 5 through 8. Write a sentence describing how you found these numbers.

5. 1, 10, 100, 1000, __________, __________

6. 180, 360, 540, 720, __________, __________

7. __________, __________

8. 2, 20, 10, 100, 50, __________, __________

1

6

1

3

1

2

2

3, , ,

Assign numbers 1 to 26 to the letters of the alphabet and numbers 1 to 7 to the position of the letter in the pattern A, B,C, G, K, P, V. The sum of the position of the letter in the alphabet and the positions of the letter in the pattern gives thenumerical position in the alphabet of the next letter in the pattern.

Example: A –––> 1 1 1+1 =2

B –––> 2 2 2+2 =4

D –––> 4 3 4+3 =7

G –––> 7 4 7+4 =11

K –––> 11 5 11+5 =16

P –––> 16 6 16+6 =22

V –––> 22

|̂Position of letter in alphabet + Position of letter in pattern = Position of next letter in the alphabet.

10,000 100,000Each term is multiplied by 10 to get the next term.

900 1080Each term is found by adding 180 to the previous term.

500 250The pattern is formed by multiplying the first term by 10 to get the

second term. Then divide the second term by 2 to get third term.

Repeat this process– multiply by 10, divide by 2.

Each new term is found by adding to the previous term.

Alternate solution: Each fraction divided by the next fraction in the

sequence, gives the following fraction. This would give anwers of and .

1

65

6

6

6or 1

P V

3

4

8

9

Name

Class Date Score


Quiz Form B

Unit I - The Structure of GeometryPart D - Inductive ReasoningLesson 2 - Applications in Mathematics

1. In each of the past 8 months, Joe has received the rent payment for his rental house on the fourth or fifth day of the month. He makes a conjecture that he will receive next month’s rent on the fourth or fifth of the month. Is this a good example of inductive reasoning?

Explain your answer.

2. Do you think there is a connection between inductive reasoning and the stock market investments made by investors in our country?

Explain your answer.

3. Anthony noted that 5 = 12 + 22 and 13 = 22 + 32. He concluded that every prime number may be expressed as the sum of the squares of two positive integers. Was he correct? You might want to test some cases.

Explain your answer2 1 1 1 1

3 1 1 1 1 1 2 1 4

5 1

2 2

2 2 2 2

2

= + = +

= + = + + = +

= +

Yes

or No

22 1 4 5

7 1 2 1 4 1 3 1 9

11 1 3

2

2 2 2 2

2

= + =

= + = + + = +

= +

Yes

or No22 2 2

2 2

2

1 9 2 3 4 9

13 2 3 4 9 13

17 1

= + + = +

= + = + =

= +

or No

Yes

44 1 16 17

19 2 4 4 16 20

23 3 4 9

2

2 2

2 2

= + =

= + = + =

= + = +

Yes

No

116 25

29 2 5 4 25 292 2

=

= + = + =

No

Yes

Yes, this is a good example of inductive reasoning. His tenants probably get paid on the first

of each month and do not mail any bill payments until their pay check is securely in the bank. Their pattern has been

established and will probably continue.

Yes, there is a connection. We hear news reports mentioning “economic indicators” which

are factors occurring within our economic system that have been related to good times for growth or bad times for

growth. A stock broker will be aware of these indicators in deciding when and in what investments to place money.

In the first 10 prime

numbers, there are 5 which cannot be expressed as the

sum of squares of two positive integers.

Yes

Yes

No


NameUnit I, Part D, Lesson 2, Quiz Form B—Continued—

4. Write a general formula for the sum of any number (n) of consecutive odd integers by examining the following cases and using inductive reasoning.

integers number of integers sum of integers

1 1 1

1,3 2 4

1,3,5 3 9

1,3,5,7 4 16

1,3,5,7,9 5 25

For “n” consecutive odd integers, the sum will be ____________________________________________

Look for a pattern and predict the next two numbers in each sequence in exercises 5 through 8. Write a sentence describing how you found these numbers.

5. 0, 10, 21, 33, 46, 60, __________, __________

6. 1, 3, 4, 7, 11, 18, __________, __________

7. 3, ––12, 48, ––192, 768, __________, __________

8. , __________, __________

1

29

2

310

5

611 1, , , , , ,

Let n = the number of integers added.

Sum can be found by squaring “n” or Sum of “n” odd integers is n2.

n = 2� � � � � � 2

n = 3� � � � � � 3

n = 4 � � � � � � 4

n = 5

2

2

2

=

=

=

4

9

16

� 5 2 = 25

Start at 0. The pattern of numbers is found by adding

consecutive integers to each number beginning with 10.

0, 0+10=10, 10+11=21, 21+12=33

75 91

The pattern is formed by adding two adjacent numbers to get

the next number, assuming that you start with 1 and 3.

29 47

Each new number is found by multiplying the previous

number by ––4.

––3072 12,288

Every number in an odd position in the pattern is found by

adding to the previous number in an odd position. Every

number in an even position in the pattern is found by adding

1 to the previous number in an even position.

121

6

11

6

Unit I, Part E, Lesson 1, Quiz Form A—Continued—


3. General Statement - No one living in Oklahoma has a house on the beach.

Specific Statement - Jeremy has a beachfront house.

Conclusion -

4. General Statement - The new fluoride toothpaste, Dent-Sure, prevents cavities.

Specific Statement - Carl had no cavities at his dental check-up today.

Conclusion -

.

.

Name

We might want to conclude that Jeremy does not live in Oklahoma. However, this is not a valid

conclusion. The condition of the general statement is that someone must live in Oklahoma. The specific statement does not

satisfy that condition. It states the Jeremy has a beachfront house. Okay, but where does he have his house? For the

reasoning to be valid, the specific statement must state whether or not Jeremy lives in Oklahoma.

We cannot arrive at a conclusion here. The condition of the general statement is that someone uses

Dent-Sure toothpaste. The specific statement does not satisfy this condition. It states Carl had no cavities. We don’t know

what he does to prevent them.


NameUnit I, Part F, Lesson 1, Quiz Form A—Continued—

7. Suppose p stands for “Triangle DEF is a right triangle” and q stands for “Triangle DEF is scalene”. Use these two statements to form, and state in words, a conjunction, a disjunction, and a negation of q.

Conjunction:

Disjunction:

Negation of q:

Given two statements as indicated in exercises 8 through 11, indicate whether the conjunction, disjunction, and negation of p are true or false.

8. both p and q are false 9. p is false and q is true

Conjunction: Conjunction:

Disjunction: Disjunction:

Negation of p: Negation of p:

10. p is true and q is false 11. both p and q are true



Negation of p: Negation of p:

nDEF is a right triangle and nDEF is scalene.

nDEF is a right triangle or nDEF is scalene.

nDEF is not scalene.

false and false false

false or false false

not false true

false and true false

false or true true

not false true

true and false false

true or false true

not true false

true and true true

true or true true

not true false

NameUnit I, Part F, Lesson 1, Quiz Form A—Continued—


Suppose p stands for “Scientists are not uneducated” (a true statement) and q stands for “Geology involves the study of the earth” (a true statement). Write, in words, each of the statements in exercises 12 through 16.Then decide the truth of each compound statement.

12.

13.

14.

15.

16.

~ ( )p q∧

~ ( )p q∨

( )p q∨

∧

∨

∧

∨

∨

It is false that scientists are not uneducated and geology involves the study of the earth. This is a false statement.

1) True and True ––> True

2) Negated True ––> False

It is not the case that scientists are not uneducated, or it is not the case that geology involves the study of the earth OR

scientists are uneducated or geology does not involve the study of the earth. This is a false statement.

1) False or False ––> False

It is false that scientists are not uneducated or geology involves the study of the earth. This is a false statement.

1) True or True ––> True


Scientists are not uneducated or geology involves the study of the earth. This is a true statement.


It is false that scientists are not uneducated and it is false that geology involves the study of the earth. Or, we could write it

another way. Scientists are uneducated and geology does not involve the study of the earth. This is a false statement.


2) ~True and ~True ––> False

~p ~q

~p ~q


NameUnit I, Part F, Lesson 1, Quiz Form B—Continued—

7. Suppose p stands for “Triangle ABC is right isosceles” and q stands for “Triangle ABC is a right triangle”. Use these two statements to form a conjunction, a disjunction, and a negation of p.

Conjunction:

Disjunction:

Negation of p:

Given two statements as indicated in exercises 8 through 11, indicate whether the conjunction, disjunction, and negation of q are true or false.

8. both p and q are true. 9. p is true and q is false.



Negation of q: Negation of q:

10. both p and q are false. 11. p is false and q is true.



Negation of q: Negation of q:

True and True True

True or True True

not True False

True and False False

True or False True

not False True

False and False False

False or False False

not False True

False and True False

False or True True

not True False

nABC is a right isosceles and nABC is a right triangle.

nABC is a right isosceles or nABC is a right triangle.

nABC is not right isosceles.

NameUnit I, Part F, Lesson 1, Quiz Form B—Continued—


Suppose p stands for “Algebra is a branch of mathematics” (a true statement) and q stands for “Geometry is not worthless” (a true statement). Write, in words, each of the statements in exercises 12 through 16. Thendecide the truth of each compound statement.

12.

13.

14.

15.

16.

p q∨

~ ( )p q∨

~ ~p q∧

~ ( )p q∧

~ ~p q∨

∧

∨

∨

∧

Algebra is a branch of mathematics or geometry is not worthless. This is a true statement.


It is not the case that algebra is a branch of mathematics or geometry is not worthless. This is a false statement.



It is not the case that algebra is a branch of mathematics and it is not the case that geometry is not worthless. We could

say this another way by saying algebra is not a branch of mathematics and geometry is worthless. This is a false

statement. 1) ~True and ~True

False and False ––> False

It is not the case that algebra is a branch of mathematics and geometry is not worthless. This is a false statement.



It is not the case that algebra is a branch of mathematics or it is not the case that geometry is not worthless. This is a

false statement. 1) ~True or ~True

False or False ––> False

∨

Name

Class Date Score

Quiz Form A


Unit I - The Structure of GeometryPart F - Deductive ReasoningLesson 2 - Conditionals

In each of exercises 1 and 2 below, there are two premises which lead to a valid conclusion. In each case, state the conclusion.

1. Eating too much candy makes one overweight. Paul eats too much candy.

2. All triangles contain three angles. Polygon ABC is a triangle.

3. Complete this sentence: “In a valid syllogism, the predicate of the ____________________ is the same as the predicate of the_______________________ premise.”

State the hypothesis and conclusion in each statement in exercises 4 and 5.

4. If I pass Algebra, then I will take Geometry.

Hypothesis:

Conclusion:

5. Two straight lines intersect, only if vertical angles formed are equal.

Hypothesis:

Conclusion:

Paul is overweight.

Polygon ABC contains three angles.

I pass Algebra.

I will take Geometry.

conclusion

major

(Alternate answer, considering what “only if” means)

Hypothesis:Vertical angles formed are equal

Conclusion: Two straight lines intersect

Two straight lines intersect

Vertical angles formed are equal

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