Post on 28-Feb-2021
Written by: Larry E. Collins
Geometry:A Complete Course
(with Trigonometry)
Module A – Instructor's Guidewith Detailed Solutions for
Progress Tests
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
© 2014 VideoTextInteractive Geometry: A Complete Course 3
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’
© 2014 VideoTextInteractive Geometry: A Complete Course8
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.
NameUnit I, Part A, Lesson 3, Quiz Form A—Continued—
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. _______________________________
© 2014 VideoTextInteractive Geometry: A Complete Course 17
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} { }
⊆
⊂
NameUnit I, Part A, Lesson 5, Quiz Form A—Continued—
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.
© 2014 VideoTextInteractive Geometry: A Complete Course18
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.
© 2014 VideoTextInteractive Geometry: A Complete Course20
⊆
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.
These are the elements that are common to both sets
These are the elements that are common to both 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.
© 2014 VideoTextInteractive Geometry: A Complete Course 29
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.
© 2014 VideoTextInteractive Geometry: A Complete Course 31
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” ________ ________
base height Area Perimeter
7. ________ 7’ 63 sq. ft. ________
© 2014 VideoTextInteractive Geometry: A Complete Course34
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’
NameUnit I, Part C, Lesson 1, Quiz Form A—Continued—
Refer to the rectangle at the right to complete exercises 8 and 9.
base height Area Perimeter
8. y units ________ (y2 - 5y) sq. units ________
base height Area Perimeter
9. ________ 10 units ________ 56 units
© 2014 VideoTextInteractive Geometry: A Complete Course 35
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
© 2014 VideoTextInteractive Geometry: A Complete Course 41
Unit I - The Structure of GeometryPart C - MeasurementLesson 2 - Parallelograms
1. Find the area and perimeter of the given parallelogram.
Area:____________
Perimeter:____________
2. Find the area and perimeter of the given parallelogram.
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
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—
© 2014 VideoTextInteractive Geometry: A Complete Course42
Name
3. Find the area and perimeter of the given rhombus.
Area:____________
Perimeter:____________
4. Find the perimeter and the height of this parallelogram.
Area:____________
Perimeter:____________
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
⋅
Perimeter Sum of lengths of the sides
P
== 166 13 16 13
29 29
58
+ + += += yards
Area Base Height
A a
a square cm
= ⋅= ⋅=
5 3 5
17 5
.
.
Perimeter Sum of lengths of the sides
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:____________
Perimeter:____________
base:____________
6. Find the area and perimeter of the given parallelogram.
Area:____________
Perimeter:___________
NameUnit I, Part C, Lesson 2, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course 43
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
⋅ ⋅ ⋅
Perimeter Sum of lengths of the sides
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
© 2014 VideoTextInteractive Geometry: A Complete Course 45
Quiz Form B
Unit I - The Structure of GeometryPart C - MeasurementLesson 2 - Parallelograms
1. Find the height of the given parallelogram.
Area:____________
height:____________
2. Find the area and perimeter of the given parallelogram.
Area:____________
Perimeter:____________
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
Perimeter Sum of lengths of the sides
P
== 122 8 12 8
20 20
40
+ + += += inches
7 inches
sq. inches
40 inches
48 2
© 2014 VideoTextInteractive Geometry: A Complete Course46
Unit I, Part C, Lesson 2, Quiz Form B—Continued—
Name
3. Find the area and perimeter of the given rhombus.
Area:____________
Perimeter:____________
4. Find the perimeter of the given parallelogram.
Area:____________
Perimeter:____________
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
Perimeter Sum of lengths of the sides
P
== 577 48 57 48
210
+ + += cm
Area base height
A
square me
= ⋅= ⋅=
36 34
1224 tters
Perimeter Sum of lengths of the sides
P
== 366 36 36 36
144
+ + += meters
1224 sq. meters
144 meters
210 cm
Name
Class Date Score
Quiz Form A
© 2014 VideoTextInteractive Geometry: A Complete Course 49
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:____________
Perimeter:___________
2. Area:____________
Perimeter:___________
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
Perimeter Sum of lengths of the sides
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
Unit I, Part C, Lesson 3, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course50
Name
3. Area:____________
Perimeter:___________
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
NameUnit I, Part C, Lesson 3, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course 51
5. Find the area and perimeter of the given triangle.
Area:____________
Perimeter:____________
6. Find the area and perimeter of the shaded square in the given figure.
Area:____________
Perimeter:____________
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 Sum of lengths of the sides
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
© 2014 VideoTextInteractive Geometry: A Complete Course 53
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
Perimeter Sum of Lengths of the Sides
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
Perimeter Sum of Lengths of the Sides
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:____________
Perimeter:___________
2. Area:____________
Perimeter:___________
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
Unit I, Part C, Lesson 3, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course54
Name
3. Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the Sides
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—
© 2014 VideoTextInteractive Geometry: A Complete Course 55
5. Find the area and perimeter of the given figure. Area:____________
Perimeter:____________
6. Find the area and perimeter of the given figure. Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the Sides
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
© 2014 VideoTextInteractive Geometry: A Complete Course 57
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:____________
Perimeter:___________
2. Area:____________
Perimeter:___________
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
Area height sum of the bases
A
= ⋅ ⋅
= ⋅ ⋅ +(
1
21
24 11 9))
=⋅
=⋅ ⋅
= square inches
4 20
22 2 20
240
Perimeter sum of lengths of the sides
P
== 100 6 7 7 6 3
30
+ + +=
. .
P inches
Perimeter sum of lengths of the sides
P
== 4 ++ + +=
11 3 9
27P inches
40 sq. inches
27 inches
51 sq. inches
30 inches
Unit I, Part C, Lesson 4, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course58
3. Area:____________
Perimeter:___________
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
Area height sum of the bases
A
= ⋅ ⋅
= ⋅ ⋅ +
1
21
27 10 18(( )
=⋅
=⋅ ⋅
= square inche
7 28
27 2 14
298 ss
Area height sum of the bases
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
NameUnit I, Part C, Lesson 4, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course 59
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:____________
Area height sum of the bases
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
Area height sum of the bases
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
Unit I, Part C, Lesson 4, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course62
Name
3. Area:____________
Perimeter:___________
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
Area height sum of the bases
A
= ⋅ ⋅
= ⋅ ⋅ +(
1
21
25 15 7))
=⋅
=⋅ ⋅
= square feet
5 22
25 2 11
255
Area height sum of the bases
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
NameUnit I, Part C, Lesson 4, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course64
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
Unit I, Part C, Lesson 5, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course66
3. Area:____________
Perimeter:___________
4. Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the sides
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
Area measure of a side Apothem
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
NameUnit I, Part C, Lesson 5, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course 67
5. Area:____________
Perimeter:___________
6. Find the area and perimeter of a regular decagon with apothem 6.8cm and side length 4.4cm.
Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
5 16
80
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
10 4 4
44
( . )
Area measure of a side Apothem
number of sides
= ⋅ ⋅1
2
s a n
or
a Perimeter
A
= ⋅ ⋅ ⋅
= ⋅ ⋅
= ⋅
1
2
1
21
2111 80
11 2 40
2440
⋅
=⋅ ⋅
= square cm
Area measure of a side Apothem
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
NameUnit I, Part C, Lesson 5, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course68
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
Perimeter Sum of Lengths of the sides
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
© 2014 VideoTextInteractive Geometry: A Complete Course 69
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:____________
Perimeter:___________
2. Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
6 20
120
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
4 18
72
Area measure of a side Apothem
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
Area measure of a side Apothem
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
NameUnit I, Part C, Lesson 5, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course70
3. Area:____________
Perimeter:___________
4. Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
3 16
48
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
8 15
120
Area measure of a side Apothem
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
Area measure of a side Apothem
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
NameUnit I, Part C, Lesson 5, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course 71
5. Area:____________
Perimeter:___________
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
Perimeter Sum of Lengths of the sides
P N
== uumber of sides
length of each side
P n s
⋅
= ⋅P
cm
= ⋅=
5 10
50
Area measure of a side Apothem
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
Area measure of a side Apothem
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
© 2014 VideoTextInteractive Geometry: A Complete Course 73
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
Unit I, Part C, Lesson 6, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course74
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
NameUnit I, Part C, Lesson 6, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course 75
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 _____________
NameUnit I, Part C, Lesson 6, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course76
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
NameUnit I, Part C, Lesson 6, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course80
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
NameUnit I, Part C, Lesson 7, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course84
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
Unit I, Part C, Lesson 7, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course88
Name
2. Lateral Area = ________
Total Area = ________
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
NameUnit I, Part C, Lesson 7, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course90
4. Lateral Area = ________
Total Area = ________
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
•
•
Unit I, Part C, Lesson 8, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course94
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.
2. Lateral Area = ________
Total Area = ________
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
Unit I, Part C, Lesson 8, Quiz Form A—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course96
Name
Find the lateral area, total area, and volume of the right pyramid shown below.
4. Lateral Area = ________
Total Area = ________
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.
Use Pythagorean Theorem to find h.
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
© 2014 VideoTextInteractive Geometry: A Complete Course 99
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.
1. Lateral Area = ________
Total Area = ________
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.
Use Pythagorean Theorem to find h.
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
Unit I, Part C, Lesson 8, Quiz Form B—Continued—
© 2014 VideoTextInteractive Geometry: A Complete Course102
Name
Find the lateral area, total area, and volume of the right circular cone shown below.
4. Lateral Area = ________
Total Area = ________
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
Unit I, Part C, Lesson 9, Quiz Form A—Continued—
Name
3. If the volume of a sphere is 12 cubic units, find the radius and the surface area.
radius = ________
Surface Area = ________
© 2014 VideoTextInteractive Geometry: A Complete Course 107
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
© 2014 VideoTextInteractive Geometry: A Complete Course 109
Unit I, Part C, Lesson 9, Quiz Form A—Continued—
Name
5. The volume of a sphere is cubic meters. Find the radius and surface area.
radius = ________
Surface Area = ________
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 � � � � � � � � � � � � � � � � � � � � � � � � � � � �
Surface� Area� of � a� sphere = 4 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
© 2014 VideoTextInteractive Geometry: A Complete Course 111
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
Surface� Area� of � a� sphere = 4 r
�
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
Unit I, Part C, Lesson 9, Quiz Form B—Continued—
Name
3. The volume of a sphere is 36 mm3. Find the radius and the surface area.
radius = ________
Surface Area = ________
© 2014 VideoTextInteractive Geometry: A Complete Course
π
113
Surface� Area� of � a� sphere = 4 r
�
2⋅ ⋅π= 4 3� mm 3� mm
�
⋅ ⋅ ⋅π= 4 3 3⋅ ⋅ ⋅π mm
� =
2
336 sq� mmπ
Volume� of � sphere =4
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
Unit I, Part C, Lesson 9, Quiz Form B—Continued—
Name
© 2014 VideoTextInteractive Geometry: A Complete Course
5. The radius of a sphere is inches. Find the surface area and volume.
Surface Area = ________
Volume = ________
8 2
115
Surface� Area� of � a� sphere = 4 r
�
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
Volume� of � sphere =4
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—
© 2014 VideoTextInteractive Geometry: A Complete Course118
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—
© 2014 VideoTextInteractive Geometry: A Complete Course120
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—
© 2014 VideoTextInteractive Geometry: A Complete Course 123
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
© 2014 VideoTextInteractive Geometry: A Complete Course 125
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
© 2014 VideoTextInteractive Geometry: A Complete Course126
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—
© 2014 VideoTextInteractive Geometry: A Complete Course128
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.
© 2014 VideoTextInteractive Geometry: A Complete Course136
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
Conjunction: Conjunction:
Disjunction: Disjunction:
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—
© 2014 VideoTextInteractive Geometry: A Complete Course 137
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
2) Negated True ––> False
Scientists are not uneducated or geology involves the study of the earth. This is a true statement.
1) True or True ––> True
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.
1) True and True ––> True
2) ~True and ~True ––> False
~p ~q
~p ~q
© 2014 VideoTextInteractive Geometry: A Complete Course140
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.
Conjunction: Conjunction:
Disjunction: Disjunction:
Negation of q: Negation of q:
10. both p and q are false. 11. p is false and q is true.
Conjunction: Conjunction:
Disjunction: Disjunction:
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—
© 2014 VideoTextInteractive Geometry: A Complete Course 141
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.
1) True or True ––> True
It is not the case that algebra is a branch of mathematics or geometry is not worthless. This is a false statement.
1) True or True ––> True
2) Negated True ––> False
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.
1) True and True ––> True
2) Negated True ––> False
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
© 2014 VideoTextInteractive Geometry: A Complete Course 143
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