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Wavelet Theory. An Elementary Approach with Applications
A self–contained, elementary introduction to wavelet theory and applications
Exploring the growing relevance of wavelets in the field of mathematics, Wavelet Theory: An Elementary Approach with Applications provides an introduction to the topic, detailing the fundamental concepts and presenting its major impacts in the world beyond academia. Drawing on concepts from calculus and linear algebra, this book helps readers sharpen their mathematical proof writing and reading skills through interesting, real–world applications.
The book begins with a brief introduction to the fundamentals of complex numbers and the space of square–integrable functions. Next, Fourier series and the Fourier transform are presented as tools for understanding wavelet analysis and the study of wavelets in the transform domain. Subsequent chapters provide a comprehensive treatment of various types of wavelets and their related concepts, such as Haar spaces, multiresolution analysis, Daubechies wavelets, and biorthogonal wavelets. In addition, the authors include two chapters that carefully detail the transition from wavelet theory to the discrete wavelet transformations. To illustrate
the relevance of wavelet theory in the digital age, the book includes two in–depth sections on current applications: the FBI Wavelet Scalar Quantization Standard and image segmentation.
In order to facilitate mastery of the content, the book features more than 400 exercises that range from theoretical to computational in nature and are structured in a multi–part format in order to assist readers with the correct proof or solution. These problems provide an opportunity for readers to further investigate various applications of wavelets. All problems are compatible with software packages and computer labs that are available on the book's related Web site, allowing readers to perform various imaging/audio tasks, explore computer wavelet transformations and their inverses, and visualize the applications discussed throughout the book.
Requiring only a prerequisite knowledge of linear algebra and calculus, Wavelet Theory is an excellent book for courses in mathematics, engineering, and physics at the upper–undergraduate level. It is also a valuable resource for mathematicians, engineers, and scientists who wish to learn about wavelet theory on an elementary level.
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"The book, putting emphasize on an analytic facet of wavelets, can be seen as complementary . to the previous Patrick J. Van Fleet's book, DiscreteWavelet Transformations: An Elementary . Approach with Applications, focused on their algebraic properties." (Zentralblatt MATH, 2011)
"Requiring only a prerequisite knowledge of calculus and linear algebra, Wavelet theory is an excellent book for courses in mathematics, engineering, and physics at the upper–undergraduate level. It is also a valuable resource for mathematicians, engineers, and scientists who wish to learn about wavelet theory on an elementary level." (Mathematical Reviews,Wavelet Theory. An Elementary Approach with Applications
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Provide students with a college-prep Algebra II course that will allow them to easily progress onto even more difficult mathematical challenges. Saxon Algebra 2, 4th Edition prepares students for calculus through explicit embedded geometry instruction. Trigonometry concepts, statistics, and applications for other subjects such as physics and chemistry are also included. Incremental lessons include a Warm Up activity; New Concepts section that introduces new concepts through examples with sidebar hints and notes; and Lesson Practice questions with lesson reference numbers underneath the question number. Real-world applications and continual practice & review provide the time needed to master each concept, helping students to build confidence in their mathematical abilities. Distributed,...
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Provide students with a college-prep math course that will give them the foundation they need to successfully move into higher levels of math. Saxon Algebra 1, 4th Edition covers all of the traditional first-year algebra topics while helping students build higher-order thinking skills, real-world application skills, reasoning, and an understanding of interconnecting math strands. Saxon Algebra 1 focuses on algebraic thinking through multiple representations, including verbal, numeric, symbolic, and graphical, while graphing calculator labs model mathematical situations. Incremental lessons include a Warm Up activity; New Concepts section that introduces new concepts through examples with sidebar hints and notes; and Lesson Practice questions with lesson reference numbers underneath the...
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Template By Froo! Algebras Golden Rule Math POSTER Algebra's Golden Rule New 12 x 18 Poster on High Quality Paper FREE SHIPPING !!! (U.S. only)This new poster is a PosterEnvy Exclusive! It's a great way to help your students learn the golden rule and it's great for review. It makes a great addition to any classroom. And it makes a great gift! All of our posters are printed on quality of the heavy 80lb satin cover paper - they are durable and can stand up to all kinds of abuse and they won't p
Repetition, drills, and application exercises ensure mastery of computational skills with Lifepac Math: Pre-Algebra and Pre-Geometry 2. Students will progress to higher-level cognitive reasoning and analysis as their problem solving ability increases. Lifepac math programs use mastery-based learning along with spiraling review to encourage long-term student success. Perfect for students who flourish in a self-paced, individualized learning format, each consumable Lifepac combines lessons, exercises, projects, reviews and tests. This set includes 10 Lifepac Workbooks and the Teacher's Guide. Grade 8. 2014 Edition. Subjects covered include: The Real Number System Modeling Problems in Integers Modeling Problems with Rational Numbers Proportional reasoning More with Functions Measurement...
LessKiss My Math: Showing Pre-Algebra Who's Boss Children Young Adults FEATURES DESCRIPTION Read Danica McKellar's posts on the Penguin Blog.The New York Times bestseller-from the Dancing With The Stars contestant and popular author of Math Doesn't Suck, Hot X: Algebra Exposed , and Girls Get Curves -teaches girls how to kick pre-algebra butt In her New York Times bestselling books, actress and math genius Danica McKellar shatters the math nerd stereotype and gives girls the tools to ace middle | 677.169 | 1 |
Math
Following are course descriptions. Check the Annual Schedule to plan your schedule.
MATHEMATICS 007
Self-Paced Mathematics Review (2)
On-campus, computer-delivered. Intended for students who want to improve their math placement, or who want a review before beginning a math course. Students work with an instructor to design and complete an individual plan of study. Students work independently to review mathematics topics. Mandatory P/NC Grading.
MATHEMATICS 060
Basic Math (5)
Addition, subtraction, multiplication, and division of whole numbers, common fractions, and decimals. Also includes an introduction to percentages, ratio and proportion, estimation and narrative problems. Student option grading.
MATHEMATICS 070
Introduction to Algebra (5)
An introduction to fundamental algebraic skills needed for Intermediate Algebra I. Course includes a review of arithmetic and an introduction to graphing, simplifying algebriac expressions, exponents, and solving linear equations. Prerequisites: MATH 060 (2.0 or better), or a score of 35 or higher on the Pre-Algebra COMPASS test. Student option grading.
MATHEMATICS 091
Intermediate Algebra, Individualized - Module 1 (1)
On-campus computer-based instruction. Simplifying and evaluating rational expressions, solving equations involving rational expressions, applications. This is the 1st92
Intermediate Algebra, Individualized - Module 2 (1)
On-campus computer-based instruction. Functions, linear models and graphs, solving linear inequalities, solving absolute value equations and inequalities. The 291 is required. Mandatory P/NC grading.
MATHEMATICS 093
Intermediate Algebra, Individualized - Module 3 (1)
On-campus computer-based instruction. Using rational exponents, simplifying radical expressions, and solving radical equations. This is the 392 is required. Mandatory P/NC grading.
MATHEMATICS 094
Intermediate Algebra, Individualized - Module 4 (1)
On-campus computer-based instruction. Solving quadratic equations by completing the square & the quadratic formula, graphing quadratic functions, applications. The 4 btter), or a score of 51 or higher on the Algebra COMPASS test, or a score of 28 or higher on the College Algebra COMPASS test. Previous credit for or concurrent enrollment in MATH 093 is required. Mandatory P/NC grading.
MATHEMATICS 095
Intermediate Algebra, Individualized - Module 5 (1)
On-campus computer-based instruction. Evaluating exponential and logarithmic expressions, solving these types of equations. This is the 594 is required. Mandatory P/NC grading.
MATHEMATICS 096
Intermediate Algebra, Individualized - Modules (5)
On-campus computer-based sequence of 1-credit self-paced modules. MATH 091-095. Rational, radical, and quadratic expressions and equations. Introduction to functions: linear, quadratic, exponential, and logarithmic. MATH 081-085 and MATH 091-095 together are equivalent to MATH 080 and MATH 099. Must be completed within one year. Prerequisite: MATH 080 (2.0 or better),99
Intermediate Algebra II (5)
Simplifying and evaluating linear, quadratic, polynomial, radical, rational, exponential, and logarithmic expressions. Solving these same types of equations and inequalities with graphs and applications to real world modeling and investigating functions. Prerequisite: MATH 098 (2.0 or better), or a score of 51 or higher on the Algebra COMPASS test, or a score of 28 or higher on the College Algebra COMPASS test. Student option grading.
MATHEMATICS & 107
Math in Society (5)
Practical applications of mathematics as they arise in everyday life. Includes finance math, probability & statistics, and a selection of other topics. Designed for students who are not preparing for calculus. Student option grading.
MATHEMATICS 111
Elements of Pre-Calculus (5)
Algebra topics including mathematical modeling, graphing & problem solving w/ polynomial, rational, exponential & logarithmic functions. Applications. Topics from mathematics of finance. Intended for students in business, social sciences & some biological sciences 141
Precalculus I (5)
The elementary functions and their graphs, with applications to mathematical modeling. Examples include linear, quadratic, polynomial, rational, exponential and logarithmic functions, composite functions, inverse functions and transformation of graphs 146H
Introduction to Stats AH (5)
Analysis of data through graphical and numerical methods, linear regression, the Normal distribution, data collection, elementary probability, confidence intervals and hypothesis testing. Emphasis on applications, AND placement in ENGL 099 or EAP 099. Student option grading.
MATHEMATICS & 148
Business Calculus (5)
Differential and Integral Calculus of elementary functions with an emphasis on business and social science applications. Designed for students who want a brief course in Calculus. (No credit given to those who have completed MATH& 151.) Prerequisite: MATH 111 preferred (2.0 or better) or MATH& 141 (2.0 or better), or a score of 70 or higher on the College Algebra COMPASS test. Student option grading.
MATHEMATICS & 151
Calculus I (5)
Definition, interpretation and applications of the derivative. Derivatives of algebraic and transcendental functions. Prerequisite: MATH& 142 (2.0 or better), or a score of 70 or higher on both the College Algebra and Trigonometry COMPASS Tests. Student option grading.
MATHEMATICS & 171
Math for Elem Ed I (5)
Fundamental concepts of numbers and operations related to topics taught at the K-8 level. Topics include problem solving, algebraic thinking, numberation, and arithmetic with rational numbers. Recommended for future elementary teachers. Prerequisite: MATH 098 (2.0 or better), MATH 099 (2.0 or better), or MATH 095 (2.0 or better), or a score of 69 or higher on the Algebra COMPASS test or a score of 35 or higher on the College Algebra COMPASS test, AND placement into ENGL 099 or EAP 099. Student option grading.
MATHEMATICS 297
Individual Project in Mathematics (1)
Individual project in a specific area of mathematics. By arrangement with instructor. Prerequisite: Instructor permission, based on evaluation of student's educational and work experience. Student option grading.
MATHEMATICS 298
Individual Project in Mathematics (2)
Individual project in a specific area of mathematics. By arrangement with instructor. Prerequisite: Instructor permission, based on evaluation of student's educational and work experience. Student option grading.
MATHEMATICS 299
Individual Project in Mathematics (3)
Individual project in a specific area of mathematics, by arrangement with instructor. Prerequisite: Instructor permission based on evaluation of student's educational and work experience. Student option grading. | 677.169 | 1 |
THEORY OF NUMBERS
THEORY OF NUMBERS
2013 Fall Term
3 Units
Mathematics 417
A study of the properties of integers, representation of integers in a given base, properties of primes, arithmetic functions, module arithmetic. Diophantine equations and quadratic residues. Consideration is also given to some famous problems in number theory.
Other Requirements: PREREQ: MATH 280 OR MATH 415 OR CONSENT OF INSTRUCTOR
Class Schedule
There are no sections offered for this course and term that meet your criteria. | 677.169 | 1 |
Students may choose to work at their own pace across all three subject areas, or to select individual content areas. Pretests will determine any learning deficits, which can then be mastered through self-paced learning modules. Not forgetting the importance of the human touch, this course is overseen by a trio of reading, writing, and mathematics professors who will be available to assist and encourage students along their journey to college readiness.
Want to be an engineer or scientist? Lack mathematical confidence? Learn to think mathematically and explore essential concepts. This course is aimed at those who aspire to study science or engineering foundation courses at university level. It draws upon the experience of staff from the Mathematics Education Centre at Loughborough University - a centre that has specialised for many years in mathematics teaching and mathematics support for science and engineering students who find the transition to university mathematics particularly challenging.
Explore and solve encrypted maths puzzles, in which numbers are replaced by letters or symbols, with this free online course. On this free online course, we'll look at three types of maths puzzle, exploring their history, meeting the puzzlists who created them, and studying and practising problem-solving techniques.
This class presents the fundamental probability and statistical concepts used in elementary data analysis. It will be taught at an introductory level for students with junior or senior college-level mathematical training including a working knowledge of calculus. A small amount of linear algebra and programming are useful for the class, but not required.
The purpose of this course is to review the material covered in the Fundamentals of Engineering (FE) exam to enable the student to pass it. It will be presented in modules corresponding to the FE topics, particularly those in Civil and Mechanical Engineering | 677.169 | 1 |
West Palm Beach Microsoft ExcelMy I have been an ESL student myself and that helped me understand what teaching techniques work best.
Reginald H.Hongjun C.
...Precalculus includes algebra, coordinate geometry, trigonometry, and most of all, functions---the general concept as well as specific functions. Students come to this course familiar with basic algebra and geometry. A precalculus course builds on all of their previous mathematical knowledge and experience to understand calculus that is essential to succeed in a college. | 677.169 | 1 |
0201308606umerical Methods with MATLAB : Implementations and Applications
This thorough, modern exposition of classic numerical methods using MATLAB briefly develops the fundamental theory of each method. Rather than providing a detailed numerical analysis, the behavior of the methods is exposed by carefully designed numerical experiments. The methods are then exercised on several nontrivial example problems from engineering practice. This structured, concise, and efficient book contains a large number of examples of two basic types One type of example demonstrates a principle or numerical method in the simplest possible terms. Another type of example demonstrates how a particular method can be used to solve a more complex practical problem. The material in each chapter is organized as a progression from the simple to the complex. Contains an extensive reference to using MATLAB. This includes interactive (command line) use of MATLAB, MATLAB programming, plotting, file input and output. For a practical and rigorous introduction to the fundamentals of numerical computation | 677.169 | 1 |
I'll never forget the first time I installed Mathematica in college. I was excited by the demos, and wanted to see how much it could help me take my calculus knowledge further — and take the drudgery out of math. Turns out, it was far more complicated to use than I ever anticipated, even more so than my trusty TI-89.
Couldn't CAS — computer algebra systems — be a bit less complex and more accessible to everyone who doesn't have time to take a whole class on using them? Computers were designed originally to solve complex math, but normal calculators, spreadsheets, and CAS systems have remained too basic on the one end and too complex on the other to change the way most of us feel about math.
It's more than understandable that we'd tend to be skeptical when a new app claims to make math simpler for everything from engineering to basic budgets at the same time — but that's exactly what Calca claims. It's a markdown text editor fused with a CAS; can it possibly be the answer to the frustrations of math? | 677.169 | 1 |
This review is from: The Calculus Diaries: How Math Can Help You Lose Weight, Win in Vegas, and Survive a Zombie Apocalypse (Paperback)
The Calculus Diaries: How Math Can Help You Lose Weight, Win in Vegas, and Survive a Zombie Apocalypse0143117378Jennifer OuellettePenguin BooksThe Calculus Diaries: How Math Can Help You Lose Weight, Win in Vegas, and Survive a Zombie ApocalypseBooks
should have been rejected by an editorMartyMOctober | 677.169 | 1 |
The typical introductory real analysis text starts with an analysis of the real number system and uses this to develop the...
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The typical introductory real analysis text starts with an analysis of the real number system and uses this to develop the definition of a limit, which is then used as a foundation for the definitions encountered thereafter. While this is certainly a reasonable approach from a logical point of view, it is not how the subject evolved, nor is it necessarily the best way to introduce students to the rigorous but highly non-intuitive definitions and proofs found in analysis.This book proposes that an effective way to motivate these definitions is to tell one of the stories (there are many) of the historical development of the subject, from its intuitive beginnings to modern rigor. The definitions and techniques are motivated by the actual difficulties encountered by the intuitive approach and are presented in their historical context. However, this is not a history of analysis book. It is an introductory analysis textbook, presented through the lens of history. As such, it does not simply insert historical snippets to supplement the material. The history is an integral part of the topic, and students are asked to solve problems that occur as they arise in their historical context.This book covers the major topics typically addressed in an introductory undergraduate course in real analysis in their historical order. Written with the student in mind, the book provides guidance for transforming an intuitive understanding into rigorous mathematical arguments. For example, in addition to more traditional problems, major theorems are often stated and a proof is outlined. The student is then asked to fill in the missing details as a homework We Got from There to Here: A Story of Real Analysis to your Bookmark Collection or Course ePortfolio
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Elementary Differential Equations with Boundary Value Problems is written for students in science, engineering, and...
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Elementary Differential Equations with Boundary Value Problems is written for students in science, engineering, and mathematics who have completed calculus through partial differentiation.An elementary text should be written so the student can read it with comprehension without too much pain. I have tried to put myself in the student's place, and have chosen to err on the side of too much detail rather than not enough. An elementary text can't be better than its exercises. This text includes 1695 numbered exercises, many with several parts. They range in difficulty from routine to very challenging. An elementary text should be written in an informal but mathematically accurate way, illustrated by appropriate graphics. I have tried to formulate mathematical concepts succinctly in language that students can understand. I have minimized the number of explicitly stated theorems and definitions, preferring to deal with concepts in a more conversational way, copiously illustrated by 250 completely worked out examples. Where appropriate, concepts and results are depicted in 144 figures.Although I believe that the computer is an immensely valuable tool for learning, doing, and writing mathematics, the selection and treatment of topics in this text reflects my pedagogical orientation along traditional lines. However, I have incorporated what I believe to be the best use of modern technology, so you can select the level of technology that you want to include in your course. The text includes 336 exercises – identified by the symbols C and C/G – that call for graphics or computation and graphics. There are also 73 laboratory exercises – identified by L – that require extensive use of technology. In addition, several sections include informal advice on the use of technology. If you prefer not to emphasize technology, simply ignore these exercises and the advice Elementary Differential Equations with Boundary Value Problems to your Bookmark Collection or Course ePortfolio
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This is a "first course" in the sense that it presumes no previous course in probability. The mathematical...
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This is a "first course" in the sense that it presumes no previous course in probability. The mathematical prerequisites are ordinary calculus and the elements of matrix algebra. A few standard series and integrals are used, and double integrals are evaluated as iterated integrals. The reader who can evaluate simple integrals can learn quickly from the examples how to deal with the iterated integrals used in the theory of expectation and conditional expectation. Appendix B provides a convenient compendium of mathematical facts used frequently in this work. And the symbolic toolbox, implementing MAPLE, may be used to evaluate integrals, if desired.In addition to an introduction to the essential features of basic probability in terms of a precise mathematical model, the work describes and employs user defined MATLAB procedures and functions (which we refer to as m-programs, or simply programs) to solve many important problems in basic probability. This should make the work useful as a stand-alone exposition as well as a supplement to any of several current textbooks.Most of the programs developed here were written in earlier versions of MATLAB, but have been revised slightly to make them quite compatible with MATLAB 7. In a few cases, alternate implementations are available in the Statistics Toolbox, but are implemented here directly from the basic MATLAB program, so that students need only that program (and the symbolic mathematics toolbox, if they desire its aid in evaluating integrals).Since machine methods require precise formulation of problems in appropriate mathematical form, it is necessary to provide some supplementary analytical material, principally the so-called minterm analysis. This material is not only important for computational purposes, but is also useful in displaying some of the structure of the relationships among Applied Probability to your Bookmark Collection or Course ePortfolio
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Linear Algebra: A First Course presents an introduction to the fascinating subject of linear algebra. As the title suggests,...
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Linear Algebra: A First Course presents an introduction to the fascinating subject of linear algebra. As the title suggests, this text is designed as a first course in linear algebra for students who have a reasonable understanding of basic algebra. Major topics of linear algebra are presented in detail, with proofs of important theorems provided. Connections to additional topics covered in advanced courses are introduced, in an effort to assist those students who are interested in continuing on in linear algebra. Each chapter begins with a list of desired outcomes which a student should be able to achieve upon completing the chapter. Throughout the text, examples and diagrams are given to reinforce ideas and provide guidance on how to approach various problems. Suggested exercises are given at the end of each section, and students are encouraged to work through a selection of these exercises.A brief review of complex numbers is given, which can serve as an introduction to anyone unfamiliar with the topic.Linear algebra is a wonderful and interesting subject, which should not be limited to a challenge of correct arithmetic. The use of a computer algebra system can be a great help in long and difficult computations. Some of the standard computations of linear algebra are easily done by the computer, including finding the reduced row-echelon form. While the use of a computer system is encouraged, it is not meant to be done without the student having an understanding of the computations Linear Algebra to your Bookmark Collection or Course ePortfolio
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This lesson uses WebImage, a Web-based customized version of ImageJ, to investigate the relationship between the...
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This lesson uses WebImage, a Web-based customized version of ImageJ, to investigate the relationship between the circumference and diameter of a circle. Images of circular objects are provided, and, by measuring diameter and circumference, students are able to obtain an approximate value of Pi. They also explore the history behind Pi and how to find itsice of Pi to your Bookmark Collection or Course ePortfolio
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This module comes from Visionlearning, an educational resource funded by the National Science Foundation. This particular...
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This module comes from Visionlearning, an educational resource funded by the National Science Foundation. This particular module introduces the history of wave theories, basic descriptions of waves and wave motion, and the concepts of wave speed and frequency. The module, available in Spanish, also includes illustrations, embedded definitions of key terms, additional links, and questions & quizzes Waves and Wave Motion: Describing Waves to your Bookmark Collection or Course ePortfolio
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This online exercise lets students practice vector addition. They choose the precision of the test by selecting a target...
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This online exercise lets students practice vector addition. They choose the precision of the test by selecting a target size, then estimate the sum of the two vectors by dragging and dropping a third arrow. Points are awarded; a higher degree of precision scores more Vectors to your Bookmark Collection or Course ePortfolio
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Think Stats is an introduction to Probability and Statistics for Python programmers.Think Stats emphasizes simple techniques...
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Think Stats is an introduction to Probability and Statistics for Python programmers.Think Stats emphasizes simple techniques you can use to explore real data sets and answer interesting questions. The book presents a case study using data from the National Institutes of Health. Readers are encouraged to work on a project with real datasets.If you have basic skills in Python, you can use them to learn concepts in probability and statistics. Think Stats is based on a Python library for probability distributions (PMFs and CDFs). Many of the exercises use short programs to run experiments and help readers develop Stats: Probability and Statistics for Programmers to your Bookmark Collection or Course ePortfolio
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Created by Lang Moore for the Connected Curriculum Project, the purpose of this module is to study solutions of...
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Created by Lang Moore for the Connected Curriculum Project, the purpose of this module is to study solutions of initial/boundary value problems for the one-dimensional wave equation. This is one of a much larger set of learning modules hosted by Duke One-Dimensional Wave Equation to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material The One-Dimensional Wave Equation
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This book provides an introduction to quadratic forms, building from basics to the most recent results. Professor Kitaoka is well known for his work in this area, and in this book he covers many aspects of the subject, including lattice theory, Siegel's formula, and some results involving tensor products of positive definite quadratic forms. The reader should have a knowledge of algebraic number fields, making this book ideal for graduate students and researchers wishing for an insight into quadratic forms.
Reviews:
' … ideal for graduate students and researchers wishing for an insight into quadratic forms.'
Extrait de L'Enseignement Mathématique | 677.169 | 1 |
A First Course in Abstract Algebra
Browse related Subjects ...
Read More level courses at schools with strong math programs. *New exercises have been written, while past exercises have been revised and modifed. *Classical approach to abstract algebra. *Focus on applications of abstract algebra. *Classic text for the high end of the market - known and loved in this discipline. It is geared towards high level courses at schools with strong maths programs *Accessible pedagogy includes historical notes written by Victor Katz (author of another AWL book The History of Mathematics), an authority in this area *By opening with a study of group theory, this text provides students with an easy transition to axiomatic mathematics
Read Less
New. Considered a classic by many, A First Course in Abstract Algebra is an in-depth introduction to abstract algebra. Focused on groups, rings and fields, this text gives students a firm foundation for more specialized work by emphasizing an understanding of the nature of algebraic structures | 677.169 | 1 |
9780824719555 experienced teachers and recognized experts in electrical engineering, Handbook of Electrical Engineering Calculations identifies and solves the seminal problems with numerical techniques for the principal branches of the field -- electric power, electromagnetic fields, signal analysis, communication systems, control systems, and computer engineering. It covers electric power engineering, electromagnetics, algorithms used in signal analysis, communication systems, algorithms used in control systems, and computer engineering. Illustrated with detailed equations, helpful drawings, and easy-to-understand tables, the book serves as a practical, on-the-job | 677.169 | 1 |
Plan Algebra
The student will demonstrate an understanding of algebra by solving linear equations with one variable. Also understanding expressions, equations, inequalities, linear and quadratic functions can be represented in multiple ways.
Lesson Plan Algebra
1.
Jill Fleming
Date/Period:
Subject: Math
Class: 10th grade
Topic: Algebra 1, Expressions, Equations, Inequalities, Linear and Quadratic
Time: Between 3 to 4 weeks based on students' process within this topic
Learning Objectives
The student will demonstrate an understanding of algebra by solving linear equations with one
variable. Also understanding expressions, equations, inequalities, linear and quadratic functions
can be represented in multiple ways.
Computer software and hardware, Presentation software, Webcam, Video equipment, personal
device (i.e. cell phone, tablet), iPod touch or iPads.
Resources/Materials
Activities
There are two activities for this assignment the first is to learn about algebra. The second will be
to describe the contributions made by the Babylonians, Egyptians, Scipione del Ferro, René
Descartes, Muhammad ibn Musa al-Khwarizmi and Arthur Cayley to algebra.
Description of class
30 students-18 females, 12 males all students in this class are African American and from middle
to lower class families. There are three students with exceptionalities; however, the students have
been mainstreamed into classes from a young age. The students learner styles is a combination of
visual and auditory thus the selection of how student's work with be presented.
Procedures
Analyze algebra problems to identify the given information, create a strategy for solving each
problem. Apply strategies to solve routine and non-routine problems and identify connections
between algebra and the real world.
To prepare for this lesson I will introduce students to algebra by providing the origin of the
subject, mathematical terms, formulas and equations.
2.
Students will then work together in groups of five members to create in a newspaper or magazine
style format what they have learned in mathematical format to include the contributions of
individuals that I mentioned previously.
Technology inclusion
Students will work in a group with a maximum of 5 members. This lesson is to begin with an
introduction to lessons involving expressions, equations, inequalities, linear and quadratic
algebra this should take at least one week..
Although this is a group assignment student are still responsible for their own portion of their
work and feedback to me on what is working or where assistance is needed.
Listed below are websites that you can access to aid in working on mathematical problems, the
lite versions of these apps are free:
Free Graphing Calculator is a powerful, flexible graphing calculator this is an app for apple
products
Hands-On Equations Lite- Level 1 is a fun introduction to algebra in an entertaining way it is an
app on android and apple products
MathAlly Graphing Calculator can be used as with your own handwriting to graph, Symbolic,
and Scientific material android app
MathPad is an educational and productive note taking app
MathPlanet
MyScript Calculator student can enter operations naturally using your handwriting this is an app
on android and apple products
YourTeacher Math Help this is also an app for apple products
Listed below are websites that you can access to aid in researching history of algebra:
Brief history of Algebra
Encyclopedia Brittanica
History
Khan Academy
3.
As a class we will use chat format or notebook for communication through the lessons. The
following website for instant chat is an email address is not required
but computer is required. For the chat I will give you a time when I am available and provide a 4
digit code. If you prefer having more communication you can also use MySchoolnotebook you
can sign up with an existing facebook account or if you do not have one just register through
their site the link is Once you are signed into this site
please enable publishing and provide me with your information so that I can add you and we will
exchange information. Google + is another
option so which ever you prefer just send the information to my school email provided in the
syllabus so I know how you are connecting.
During the second or third week students can use any and all technology devices and/or software
to create their presentations the minimum is to use at least two forms of technology iPod touch,
iPads/Tablet, webcam from personal computer or Smartphone. This assignment is to be
presented to the class in any of the following ways as newspaper, magazine, blog, brochure or
video format, however, if another format is selected please make sure it works for the group and
submit information about the site to me.
The recommendation for different sites to display finished material are Calameo, Toontastic,
Zinepal, Youtube, Pinterest, Appafolio or Vcasmo.
Each student will use a chat, email and other social networking sites to communicate with
instructor and other students.
Evaluation/Assessment
Teacher observation
Teacher-made tests and quizzes
Interaction and submission in chat and material through MySchoolnotebook or Google+
Rubric for the presentation of completed work on media sites.
Procedures
Possible issues with software and hardware is that students may not have access at home to edit
projects.
In-school access to computers could be limited based on other classes needing use of the devices.
If students do not have their own device they can use loaner equipment from school.
Risk Analysis
Possible conflict in completing group tasks.
Student lacking motivation or creativity when completing assignments. | 677.169 | 1 |
a visual, graphical approach that emphasizes connections among concepts, this text helps students make the most of their study time. The authors show how different mathematical ideas are tied together through their zeros, solutions, and x-intercepts theme; side-by-side algebraic and graphical solutions; calculator screens; and examples and exercises. By continually reinforcing the connections among various mathematical concepts as well as different solution methods, the authors lead students to the ultimate goal of mastery and success in class. | 677.169 | 1 |
The famous mathematical cat Penrose takes us on a trip though puzzleland, while sharing fascinating and challenging puzzles in this uniquely designed mini book. Each page's puzzle is introduced by our star Penrose or one of his quirky friends. Readers are treated to page after page of Penrose antics and problems, while the over 100 mind teasers... more...... more...
Written in an engaging style that renders complex concepts accessible to a wide readership, and with more than 450 exercises, this textbook introduces the basics of real analysis as well as more advanced topics such as measure theory and Lebesgue integration. more...
L.E.J. Brouwer: Collected Works, Volume 1: Philosophy and Foundations of Mathematics focuses on the principles, operations, and approaches promoted by Brouwer in studying the philosophy and foundations of mathematics. The publication first ponders on the construction of mathematics. Topics include arithmetic of integers, negative numbers, measurable... more...
In this monograph, which is an extensive study of Hilbertian approximation, the emphasis is placed on spline functions theory. The origin of the book was an effort to show that spline theory parallels Hilbertian Kernel theory, not only for splines derived from minimization of a quadratic functional but more generally for splines considered as piecewise... more... | 677.169 | 1 |
Find a Corte Madera AlIt needs a strong foundation but the concepts carry through the course that is basically differentiation and integration. After learning the basic skills, application becomes very important. But the depth of understanding in the course by a student leads to a better prepared thinker on a higher level | 677.169 | 1 |
This module will allow you to apply the information you have learned through ten situations with complex information, each with three calculation problems. You will move through the steps to solve the problems interactively, with feedback helping you along the way. The challenge for you here is to sort through complex information and know what you must use to solve the problems.
COMMON PATHWAYS
These are combinations of information which can give you certain information: | 677.169 | 1 |
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Sixth-form students will be able to take a new qualification in practical maths alongside their other A levels to make them better prepared for university.
A panel of mathematicians has published outline plans for a course covering areas such as statistics, probability, advanced calculation and modelling to develop problem-solving skills.
These will be used by awarding bodies to design new qualifications in core maths that should be ready for schools to start teaching from autumn 2015. They are intended to be studied in addition to a student's A-level choices. | 677.169 | 1 |
97804714520rief Calculus: An Applied Approach
This accessible introduction to Calculus is designed to demonstrate how calculus applies to various fields of study. The text is packed with real data and real-life applications to business, economics, social and life sciences.
Applications using real data enhances student motivation. Many of these applications include source lines, to show how mathematics is used in the real world.
NEW! Conceptual problems ask students to put the concepts and results into their own words. These problems are marked with an icon to make them easier to assign.
More opportunities for the use of graphing calculator, including screen shots and instructions, and the use of icons that clearly identify each opportunity for the use of spreadsheets or graphing calculator.
Work problems appear throughout the text, giving the student the chance to immediately reinforce the concept or skill they have just learned.
Chapter Reviews contain a variety of features to help synthesize the ideas of the chapter, including: Objectives Check, Important Terms and Concepts, True-False Items, Fill in the Blanks and Review Exercises | 677.169 | 1 |
It is very important that you learn to pose and solve equations in practical problems. Ernest Mach, a famous scientist of the nineteenth century, said that algebra is characterized by a lightening of mind, because the solution of a problem, after building the equation, you can "forget" all the practical situation to focus on the mathematical expression; everything that is not necessary to solve the problem no longer interfere with your mind. Another famous scientist, Isaac Newton, wrote that the language of algebra is the equation. To see a problem concerning abstract relations of numbers or amounts, simply translate the problem of colloquial language to the algebraic language. Here I leave the application for first order equations developed in 2016 Maple.
Once you've run the command, move the slider from side to side. Neat, isn't it?
With this single line of code, you have built an interactive application that shows the graph of x to the power of various exponent powers.
The Explore command is an application builder. More specifically, the Explore command can programmatically generate interactive content in Maple worksheets.
Programmatically generated content is inserted into a Maple worksheet by executing Maple commands. For example, when you run the Explore command on an expression, it inserts a collection of input and output controllers, called Embedded Components, into your Maple worksheet. In the preceding example, the Explore command inserts a table containing:
a Slider component, which corresponds to the value for the exponent k
a Plot component, which shows the graph of x raised to the power for k
Together these components form an interactive application that can be used to visualize the effect of changing parameter values.
Explore can be viewed as an easy application creator that generates simple applications with input and output components. Recently added packages for programmatic content generation broaden Maple's application authoring abilities to form a full development framework for creating customized interactive content in a Maple worksheet. The DocumentTools package contains many of these new tools. Components and Layout are two sub-packages that generate XML using function calls that represents GUI elements, such as embedded components, tables, input, or output. For example, the DocumentTools:-Components:-Plot command creates a new Plot component. These key pieces of functionality provide all of the building blocks needed to create customizable interfaces inside of the Maple worksheet. For me, this new functionality has completely altered my approach to building Maple worksheets and made it much easier to create new applications that can explore hundreds of data sets, visualize mathematical functions, and more.
I would go so far as to say that the ability to programmatically generate content is one of the most important new sources of functionality over the past few years, and is something that has the potential to significantly alter the way in which we all use Maple. Programmatic content generation allows you to create applications with hundreds of interactive components in a very short period of time when compared to building them manually using embedded components. As an illustration of this, I will show you how I easily created a table with over 180 embedded components—and the logic to control them.
Building an interface for exploring data sets:
In my previous blog post on working with data sets in Maple, I demonstrated a simple customized interface for exploring country data sets. That post only hinted at the much bigger story of how the Maple programming language was used to author the application. What follows is the method that I used, and a couple of lessons that I learned along the way.
When I started building an application to explore the country data sets, I began with an approach that I had used to build several MathApps in the past. I started with a blank Maple worksheet and manually added embedded components for controlling input and output. This included checkbox components for each of the world's countries, drop down boxes for available data sets, and a couple of control buttons for retrieving data to complete my application.
This manual, piece-by-piece method seemed like the most direct approach, but building my application by hand proved time-consuming, given that I needed to create 180 checkboxes to house all available countries with data. What I really needed was a quicker, more scriptable way to build my interface.
So jumping right into it, you can view the code that I wrote to create the country data application here:PECCode.txt
Note that you can download a copy of the associated Maple worksheet at the bottom of this page.
I won't go into too much detail on how to write this code, but the first thing to note is the length of the code; in fewer than 70 lines, this code generates an interface with all of the required underlying code to drive interaction for 180+ checkboxes, 2 buttons and a plot. In fact, if you open up the application, you'll see that every check box has several lines of code behind it. If you tried to do this by hand, the amount of effort would be multiplied several times over.
This is really the key benefit to the world of programmatic content generation. You can easily build and rebuild any kind of interactive application that you need using the Maple programming language. The possibilities are endless.
Some tips and tricks:
There are a few pitfalls to be aware of when you learn to create content with Maple code. One of the first lessons I learned was that it is always important to consider embedded component name collision and name resolution.
For those that have experimented with embedded components, you may have noticed that Maple's GUI gives unique names to components that are copied (or added) in a Maple worksheet. For example, the first TextArea component that you add to a worksheet usually has the default name TextArea0. If you next add another TextArea, this new TextArea gets the name TextArea1, so as to not collide with the first component. Similar behaviour can be observed with any other component and even within some component properties such as 'group' name.
Many of the options for commands in the DocumentTools sub-packages can have "action code", or code that is run when the component is interacted with. When building action code for a generated component, the action code is specified using a long string that encapsulates all of the code. Due to this code being provided as a long string, one trick that I quickly picked up is that it is important to separate out the names for any components into sub-strings inside of a longer cat statement.
For example, here is a line that is contained within a longer cat statement in the preceding code:
Doing so ensures that when the components are created, the names are not hard-coded to always just look for a given name. This means that the GUI can scrape through the code and update any newly generated components with a new name when needed. This is important if "ComboBox_0" already exists so that the GUI can instead create "ComboBox_1".
Another challenge for coding applications is adding a component state. One of the most common problems encountered with running any interactive content in Maple is that if state is not persistent, errors can occur when, for example, a play button is clicked but the required procedures have not been run. This is a very challenging problem, which often require solutions like the use of auto-executing start-up code or more involved component programming. Some features in Maple 2016 have started working to address this, but state is still something that usually needs to be considered on an application by application basis.
In my example, I needed to save the state of a table containing country names so that the interface retains the information for check box state (checked or unchecked) after restart. That is, if I saved the application with two countries selected, I wanted to ensure that when I opened the file again those same two countries would still be selected, both in the interface as well as in the table that is used to generate the plot. Now accomplishing this was a more interesting task: my hack was to insert a DataTable component, which stored my table as an entry of a 1x1 Matrix rtable. Since the rtable that underlies a DataTable is loaded into memory on Maple load, this gave me a way to ensure that the checked country table was loaded on open.
Here, for example, is the initial creation of this table:
"if not eval( :-_SelectedCountries )::Matrix then\n",
" :-_SelectedCountries := Matrix(1,1,[table([])]):\n",
"end if;\n",
For more details, just look for the term: ":-_SelectedCountries" in the preceding code.
I could easily devote separate posts to discussing in detail each of these two quick tips. Similarly, there's much more that can be discussed with respect to authoring an interface using programmatic tools from the DocumentTools packages, but I found the best way to learn more about a command is to try it out yourself. Once you do, you'll find that there are an endless number of combinations for the kinds of interfaces that can be quickly built using programmatic content generation. Several commands in Maple have already started down the path of inserting customized content for their output (see DataSets:-InsertSearchBox and AudioTools:-Preview as a couple of examples) and I can only see this trend growing.
Finally, I would like to say that getting started with programmatic content generation was intimidating at first, but with a little bit of experimentation, it was a rewarding experience that has changed the way in which I work in Maple. In many cases, I now view output as something that can be customized for any command. More often than not, I turn to commands like 'Explore' to create an interface to see how sweeping through parameters effects my results, and any time I want to perform a more complex analysis or visualization for my data, I write code to create interfaces that can more easily be customized and re-used for other applications.
If you are interested in learning more about this topic, some good examples to get started with are the examples page for programmatic content generation as well as the help pages for the DocumentTools:-Components and DocumentTools:-Layout sub-packages.
To download a copy of the worksheet used in this post, click here (note that the code can be found in the start-up code of this worksheet): CountryDataPEC.mw To create the datasets interface, simply run the CountrySelection(); command.
General description of the method of solving underdetermined systems of equations. As a particular application of the idea proposed a universal method of calculation for all kinds of linkage (lever) mechanisms. With the description and examples. The method can be used for powerful CAD linkages.
Some examples of a much larger number calculated by the proposed method. Examples gathered here not to look for them on the forum and opportunity to demonstrate the method. Among the examples, I think, there are very complicated.
However, the animations do not work properly. Instead of the animations, I get series of sets of plots. Each such set consists of four plots: one with a green edge, one with a blue edge, one with blue edges, and one with points (nodes).
I guess, it happens because of a new version of maple: the application was created in 2008 for Maple 12, and I use Maple 2015.
Could anybody, please, tell me what exactly should I change in the code so that it will work properly?
Valery Ochkov and Volodymyr Voloshchuk have developed a series of thermal engineering applications in Maple 2016. The applications explore steam turbine power generation and refrigeration cycles, and use the ThermophysicalData package for fluid properties.
Their work can be found at the following locations on the Application Center.
"Nonlinear dynamics shape the world around us, from the harmonious movements of celestial bod- ies, via the swirling motions in fluid flows, to the complicated biochemistry in the living cell. Mathematically these beautiful phenomena are modeled by nonlinear dynamical systems, mainly in the form of ordinary differential equations (ODEs), partial differential equations (PDEs) and delay differential equations (DDEs). The presence of nonlinearities severely complicates the mathe- matical analysis of these dynamical systems, and the difficulties are even greater for PDEs and DDEs, which are naturally defined on infinite-dimensional function spaces. With the availability of powerful computers and sophisticated software, numerical simulations have quickly become the primary tool to study the models. However, while the pace of progress increases, one may ask: just how reliable are our computations? Even for finite-dimensional ODEs, this question naturally arises if the system under study is chaotic, as small differences in initial conditions (such as those due to rounding errors in numerical computations) yield wildly diverging outcomes. These issues have motivated the development of the field of rigorous numerics in dynamics"
I am learning to use maple for my notes preparation for the subject Finite Element Analysis. It is interesting to know that how often we blame maple or computer for the silly mistakes we made in our commands and expect the exact answers. I have used a small file and find it easy to analyse my mistakes fatser. If we make a small mistake in a big file, it not only gives us problem finding our mistakes, it leads to more mistakes in other parts as well. A command working in one document need not necessarily work the same way in other document.
I have made my first document and people will come with suggestions to make appropriate modifications in the various sections to improve my knowledge on maple as well as the subject.
Here the potential of maple 2015 to the quantitative study of the decomposition of a vector table is shown in two dimensions. Application for the exclusive use of engineering students, which was implemented with embedded components.
We find recent applications of the components applied to the linear momentum, circular equations applied to engineering. Just simply replace the vector or scalar fields to thereby reasoning and use the right button.
Here we have an application to understand how algebraic expressions, calculating degrees relative abosulutos polynomial operations and introduction to work.Here we have an application to understand how algebraic expressions, calculating degrees relative abosulutos polynomial operations and introduction to work. | 677.169 | 1 |
This book is a course on arithmetic designed for college students. It covers whole numbers, fractions, decimals, percents, ratios and proportions, measurement, and integers. Geometry and statistics... More > are integrated throughout the text rather than covered in independent sections.
This book is also available for online reading at:
A collection of handouts/worksheets is available at:
This text was created by the Monterey Institute for Technology and Education (MITE) under a Creative Commons Attribution License and was remixed by David Lippman.< Less
This is a free, open textbook covering a two-quarter pre-calculus sequence including trigonometry. The first portion of the book is an investigation of functions, exploring the graphical behavior of,... More > interpretation of, and solutions to problems involving linear, polynomial, rational, exponential, and logarithmic functions. An emphasis is placed on modeling and interpretation, as well as the important characteristics needed in calculus.<Beginning and Intermediate Algebra was designed to reduce textbook costs to students while not reducing the quality of materials. This text includes many detailed examples for each section along with... More > several problems for students to practice and master concepts. Complete answers are included for students to check work and receive immediate feedback on their progress.
Topics covered include: pre-algebra review, solving linear equations, graphing linear equations, inequalities, systems of linear equations, polynomials, factoring, rational expressions and equations, radicals, quadratics, and functions including exponential, logarithmic and trigonometric.
Each lesson also includes a World View Note which describes how the lesson fits into math history and into the world, including China, Russia, Central America, Persia, Ancient Babylon (present day Iraq) and more!< Less
Modeling, Functions, and Graphs includes the material found in a typical algebra course, along with introductions to curve-fitting and display of data. The order and presentation of topics are... More > organized around families of functions and their applications.
Appendix A includes review material on polynomial products and factoring, laws of exponents, and facts from geometry. Appendix B provides a note on the structure of the real number system and an introduction to complex numbersA traditional coverage of Beginning and Intermediate Algebra, covering linear equations, graphing, inequalities, systems of equations, polynomials, factoring, rationals, radicals, quadratics, and... More > functions. A free download of the text, and a student solutions manual and workbook are available at Less
This book is designed to provide ample practice material for students preparing for the Foundation Based Education (FBE) exam in mathematics. The book progresses smoothly by providing a small step by... More > step method for working out problem and then providing numerous drills to ensure retention of the skills. Higher order thinking skills are re-enforced by the numerous word problems scattered throughout. The book then provides sample exams that closely approximate the actual FBE Exit Exams in order to provide the students with the feel of the actual exam.< Less
The Fundamentals of Teaching English as a Foreign Language Course Book is used for TEFL classes with International TEFL Academy that begin in 2015.
TEFL is the acronym for Teaching English as a... More > Foreign Language, or simply, English language instruction for non-native speakers. Also known as Teaching English to Speakers of Other Languages (TESOL) or English Language Teaching (ELT), the field of TEFL/TESOL is one of the fastest growing educational fields in the world, presenting thousands of excellent professional opportunities in all corners of the globe to native and fluent English speakers who achieve their TEFL/TESOL certification | 677.169 | 1 |
Synopses & Reviews
Publisher Comments
This text prepares readers for fluency with analytic ideas, such as real and complex analysis, partial and ordinary differential equations, numerical analysis, fluid mechanics, and differential geometry. This book is designed to challenge advanced readers while encouraging and helping readers with weaker skills. Offering readability, practicality and flexibility, Wade presents fundamental theorems and ideas from a practical viewpoint, showing readers the motivation behind the mathematics and enabling them to construct their own proofs.
ONE-DIMENSIONAL THEORY; The Real Number System; Sequences in R; Continuity on R; Differentiability on R; Integrability on R; Infinite Series of Real Numbers; Infinite Series of Functions; MULTIDIMENSIONAL THEORY; Euclidean Spaces; Convergence in Rn; Metric Spaces; Differentiability on Rn; Integration on Rn; Fundamental Theorems of Vector Calculus; Fourier Series
For all readers interested in analysis.
About the Author
William Wade received his PhD in harmonic analysis from the University of California—Riverside. He has been a professor of the Department of Mathematics at the University of Tennessee for more than forty years. During that time, he has received multiple awards including two Fulbright Scholarships, the Chancellor's Award for Research and Creative Achievements, the Dean's Award for Extraordinary Service, and the National Alumni Association Outstanding Teaching Award.
Wade's research interests include problems of uniqueness, growth and dyadic harmonic analysis, on which he has published numerous papers, two books and given multiple presentations on three continents. His current publication, An Introduction to Analysis,is now in its fourth edition.
In his spare time, Wade loves to travel and take photographs to document his trips. He is also musically inclined, and enjoys playing classical music, mainly baroque on the trumpet, recorder, and piano. | 677.169 | 1 |
3DMath Explorer Desciption:
3DMath Explorer is a computer program that pilots 2D and 3D graphs of mathematical functions and curves in unlimited graphing space. It has many useful feature such as 3D curve ploting in real time, perspective drawing, graph scaling (zooming), active graph rotation, fogging effect, cubic draw, unlimited space ploting, four view plot screen, auto rotate animation, drawing many curves in the same screen,working with many graph screen est.
Review 3DMath Explorer
Math Mechanixs is an easy to use scientific and engineering FREE math software program. (FREE registration is required after 60 days). The typical tool for solving mathematical problem has been the calculator. Unfortunately, a calculator can be very...
Mark Jacobs Graph Plotter is an application that allows you to easily build a graph of multiple mathematical functions.You can insert up to 10 functions and the program automatically plots the resulting chart, which you can save in a separate file.
Graph polynomials and view prime numbers on a ulam spiral graphing plot. Only integer numbers. Based on an old GPL version of the JEP equation parser. See website for additional examples/live applets Seems to be closely related to Cellular automaton.
RISAFloor designs floor systems and works hand in hand with RISA-3D and RISAFoundation to provide a more complete structural engineering software solution for building design. RISAFloor will manage loads, design beams and columns, create quality CAD...
Math 3rd Grade Place Values was created as an accessible and easy-to-use software utility that allows kids to test their math knowledge.Math 3rd Grade Place Values is very useful if you want to test and improve the mathematics knowledge of your...
Star Graphs is a simple and small piece of software that allows its users to learn more about graphs.Star Graphs allows you to click on the "+" button to increase the number of vertices. The "-" button can be used to decreases the number of vertices....
Degree Sequences of Graphs was designed as a simple software that allows the users to draw a graph and view the degree sequence. The corresponding degree graph will be shown in the text field available below the graph.Degree Sequences of Graphs is a...
Connected Components of a Graph was specially developed as an accessible and handy software that lets you learn more about graphs.Connected Components of a Graph is a software that lets you draw a graph in the left pane and view the connected | 677.169 | 1 |
9780072965339Beginning and Intermediate Algebra
Miller/O'Neill's "Beginning and Intermediate Algebra" is an insightful textbook written by instructors who have first-hand experience with students of developmental mathematics. Through specially designed exercise sets, student-friendly writing, carefully organized page-layout, and helpful hints and tips, "Beginning and Intermediate Algebra" engages students in their study of mathematics and paves the pathway for success | 677.169 | 1 |
Product details
ISBN-13: 9780792334989
ISBN: 0792334981
Publication Date: 1995
Publisher: Springer London, Limited
AUTHOR
Hazewinkel, Michiel
SUMMARY
The Encyclopaedia of Mathematics is the most up-to-date, authoritative and comprehensive English-language work of reference in mathematics which exists today. With over 7,000 articles from 'A-integral' to 'Zygmund Class of Functions', supplemented with a wealth of complementary information, and an index volume providing thorough cross-referencing of entries of related interest, the Encyclopaedia of Mathematics offers an immediate source of reference to mathematical definitions, concepts, explanations, surveys, examples, terminology and methods. The depth and breadth of content and the straightforward, careful presentation of the information, with the emphasis on accessibility, makes the Encyclopaedia of Mathematics an immensely useful tool for all mathematicians and other scientists who use, or are confronted by, mathematics in their work. The Enclyclopaedia of Mathematics provides, without doubt, a reference source of mathematical knowledge which is unsurpassed in value and usefulness. It can be highly recommended for use in libraries of universities, research institutes, colleges and even schools.Hazewinkel, Michiel is the author of 'Encyclopaedia of Mathematics/Unabridged', published 1995 under ISBN 9780792334989 and ISBN 0792334981 | 677.169 | 1 |
theory of elliptic curves involves a blend of algebra, geometry, analysis, and number theory. This book stresses this interplay as it develops the basic theory, providing an opportunity for readers to appreciate the unity of modern mathematics. The book's accessibility, the informal writing style, and a wealth of exercises make it an ideal introduction for those interested in learning about Diophantine equations and arithmetic geometry.
Editorial Reviews
Review
From the reviews:
"The authors' goal has been to write a textbook in a technically difficult field which is accessible to the average undergraduate mathematics major, and it seems that they have succeeded admirably..."--MATHEMATICAL REVIEWS
"This is a very leisurely introduction to the theory of elliptic curves, concentrating on an algebraic and number-theoretic viewpoint. It is pitched at an undergraduate level and simplifies the work by proving the main theorems with additional hypotheses or by only proving special cases. … The examples really pull together the material and make it clear. … a great book for a first introduction to the subject of elliptic curves. … very clearly written and you will understand a lot when you are done." (Allen Stenger, The Mathematical Association of America, August, 2008)
Top Customer Reviews
This wonderful book is an excellent introduction to elliptic curves over the rational numbers. It is self-contained and easily accessible, but still takes the reader quite far, thus giving an undergraduate reader some exciting glimpes of deeper mathematics. This book is ideally suited as a text book for an undergraduate course (I have myself enjoyed it as a course), but is written in a lively style that also makes it fun to read on one's own. It covers such topics as the Nagell-Lutz Theorem, Mordell's Theorem over rational numbers, elliptic curves over finite fields and reduction modulo p, Thue's Theorem and diophantine approximation, and even an introduction to complex multiplication. An appendix provides the reader with a basic background on projective geometry. This book is a must for any student wanting to see beyond the ordinary coursework, and at the same time provides a natural stepping stone to a more advanced treatment of the subject, such as "The Arithmetic of Elliptic Curves", also by Silverman, which has become pretty much the standard text on the subject.
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The authors do a fantastic job of introducing elliptic curves for individuals and students interested in this area. Because of the importance of elliptic curves to cryptography, in integrable models in statistical mechanics, in superstring theory in physics, in mirror symmetry in algebraic geometry, in mechanics in the solution of the spinning top, and even in financial engineering, this book will be useful in building intuition about these interesting objects. Be careful in reading this book though...the theory of elliptic curves is beautiful and addicting, and you will want no doubt to read more about them after finishing it. There are two other books by Silverman that will alleviate the monkey on your back for more knowledge about elliptic curves. Happy reading......
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The virtue of this book is its leisurely style; and the subject is very attractive as well. The idea that lies behind this whole theory is the way in which one imposes an arithmetic on the points of an elliptic curve. There is much to be said about this: the whole thing becomes a group, the rational points form a finitely generated group, etc. That's chapters 1-3. In chapters 4-5 we finally get to some number-theoretical applications. First applications to factorisation (and thus to cryptography). Then chapter 5 is on "integer points on cubic curves", which sounds nice and classical, but things quickly turn quite messy. And finally there is chapter 6, on complex multiplication and Kronecker's Jugendtraum, for those who are interested in this extension field business that everybody seems to love. One could certainly read chapter 1-3 as a very elementary and easy-going introduction to the basics of elliptic curve arithmetic. I think the applications are treated better elsewhere, in Koblitz's books for instance.
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MATH 1036 / 1037 HELP SITE MAIN PAGE
Welcome to Nipissing
University's MATH
1036/1037 Help Site. This webpage contains tutorials, tests, and sample
problems that will help students to better understand the concepts of
the first year calculus course. This site is meant to aid students in
their understanding of calculus. Although this site covers most
concepts of the course, many have been left out and will be covered in
the lectures or in the textbook. Since this website is not a substitute
for these resources, it is strongly recommended that students continue
to go to lectures and read the chapters in their textbooks. The various
sections of the help site are summarized below.
TESTS
The tests are divided into 3 main groups, online
preliminary tests, online general tests and advanced tests. The online
tests can be found on the the ilrn website. For more information about
ilrn, visit the tests section at the link below. The preliminary tests
section covers general mathematical concepts at the high school level.
The online general tests section covers the calculus concepts from the
MATH 1036 course. The general tests are composed of questions similar
to
those that will appear on assignments and midterm exams. The advanced
tests section can be found on the Math 1036/1037 Help Site. This
section contains more challenging prolems that will further prepare
students for midterm and final examinations.
All of the tests are constructed to give students
feedback after completion. This will help students to realize which
sections they may be a little weaker on.
For more information about the tests, visit the
Tests Home page at the link below.
TUTORIALS
Each tutorial contains definitions, formulas,
graphs and examples to aid in the understanding of the given subject.
The information contained in these tutorials is written in simpler
terms than textbooks may express in order to help the students having
trouble to grasp the more complicated concepts.
The lessons covered in this section are divided
into two groups. The first group contains tutorials on general
mathematical concepts, such as trigonometry, logarithms or
inequalities. These tutorials are intended to give students the
background information that they will require to understand the
concepts covered in the Math 1036/1037 course. The second group
contains tutorials on calculus related concepts such as limits, curve
sketching and integration.
For more information about the tutorials section,
visit the Tutorials Home page at the link below.
SAMPLE PROBLEMS
The sample problems section contains questions
for the students to attempt to solve. The solutions for these problems
will be posted after each specific concept is covered in the course.
This will allow the students to compare their solutions to the full
correct solutions, helping them to prepare for tests and exams.
For more information about the sample problems
section, visit the Sample Problems Home page at the link below.
COMMON MISTAKES
Have you ever thought you had answered a problem
on a test or assignment perfectly, only to have it returned full of red
ink because you used an incorrect formula? The common mistakes section
is devoted to listing some of the "infamous" formulas that may seem
logical to students, but are completely meaningless to their
professors. This section will ensure that students do not make these
mistakes again.
For more information about the common mistakes
section, visit the Common Mistakes Home page at the link below.
STUDY TIPS
The study tips page is one of the most useful
sections on the Math 1036/1037 Help Site. It provides help for students
in areas that are not usually covered in textbooks or lectures. This
page contains helpful information on topics such as reading the
textbook or preparing for exams. The tips in this section can help
students to get the best grades possible in the first year calculus
course.
For more information about the study tips section,
visit the Study Tips Home page at the link below.
CALCULUS APPLICATIONS
Do you constantly wonder why you have to take calculus? Why is it
important and when will you ever use it in your life? The calculus applications page
is designed to give you some answers to your questions. It outlines what the basic
calculus techniques can be used for and why calculus is so important.
For more information about the applications section, visit the
Calculus Applications Home page at the link below.
MATH HUMOUR
Everyone enjoys a good laugh now and then. Take a break from
your studies and check out the humour section. Every joke contains something
essential about mathematics and the mathematical way of thinking.
For more information about the humour section, visit the
Math Humour Home page at the link below.
NOTE: This website has
been designed for and tested for functionality on Internet Explorer
6.0, Netscape Navigator 7.2, Netscape Browser 8.0,
and
Mozilla Firefox 1.0. Other browsers may not display information
correctly, although future versions of the abovementioned browsers
should function properly. | 677.169 | 1 |
Jackson Heights PrealgebraNancy O.Sharon F.
...We learn terms, polynomials, equations, and algebraic structures. In algebra, the concept of variables is introduced. Variables represent numbers and generalizations can be made using the rules of operations, i.e., addition, subtraction, multiplication, and division. | 677.169 | 1 |
$13.43Should be required for grade school students! I use it in my home school
The Saxon series is wonderfully progressive in pace; not too fast or slow, and presented well enough where young minds can grasp the material and gain a foothold by doing every problem in each section, and working step-by-step as they progress through the book. I find the examples and explanations to be better than most. High on content, low on fluff. This should be taught in grade school!
I am a firm believer that it is up to me (not public school) as a parent to require a high level of performance from my children. I use this book in my home school arsenal to teach my youngest children (in grade school) the most important subject matter in academics- higher math and science! Physics, Chemistry, Engineering principles, coding, etc... My daughter will not wind up "doing hair" at the local mall for a living! Saxon math/science books are one weapon that I use to keep their minds both hungry and taxed.Read full review
This book is a great for homeschooling families. I love the fact that it covers everything a student needs to learn in Algebra. The teachers book that is sold separately is awesome too. Actually show the teacher how the equation is done, by breaking down the steps. If your a homeschooling parent, you should consider this math book for your children.
I have used Saxon math from 54 through Algebra 1. I love how Saxon has plenty of equations for the new material as well as the old. The only draw back I've had using Saxon's Algebra 1 is for the student that has difficulty understanding algebra. For that student I found teaching one chapter in two sessions helps, but I have discovered another publisher was better for him.
my son had his Algebra 1 book on his basement bedroom floor one night the sump pump quit and we had a slight flood damaging his book. since it was an older book and the teachers book & tests were older we needed a maching book. looking on ebay for a replacement book was the awnser I replaced the book at under half the cost of a new book.He was able to finish his school work with his replaced book. I will be looking for more homeschool books on ebay. | 677.169 | 1 |
Links
Drop-In Mentoring
What it is In the Drop-In Tutoring center, specially-trained tutors are available to answer your questions about Mathematics concepts. Menotrs interact with students on a question-by-question basis, and students can expect multiple, brief interactions with available mentors. Drop-In Mentoring is a great place to study alone or in groups, and is especially good for completing homework assignments.
Courses mentored in Drop-In Mentoring:
Math
021, 101, 102, 103, 115, 120, 123, 124, 225, 281
Why it works Getting your homework done and understanding your homework problems are two different things. In Drop-In Tutoring, tutors help you get oriented toward the right answer, but won't hand it to you. They are much more focused on you understanding each part of the problem along the way. In many cases when you study, you only need a little boost to get on the right path – that is exactly what Drop-In Tutoring is here to provide.
How it works Drop-In Mentoring is free and available to all students. Simply come to the Math Assistance Center (JS 154) and a mentor will get you signed-in and will explain the Drop-In Tutoring layout and how to request assistance from a mentor. Have a seat in one of the spaces and feel free to study alone or with other students. Our question-and-answer system allows you to request a tutor's assistance when necessary. Questions are answered on a first-come, first-served basis. In general, only one question will be answered per interaction, and tutors try to keep each interaction to 3 to 5 minutes. When you are done studying, simply check out at the receptionist desk as you leave.
If you're looking for a more personalized, individualized experience you might want to try Scheduled Mentoring, which allows you to meet one-on-one with a mentor for a one hour session. We may also be able to assist you and your friends or members of a student organization - who are all enrolled in the same course - arrange for tutors in a small group setting. Contact us for more information: MAC@bhsu.edu.
Scheduled Mentoring
What it is In one hour sessions, you get to partner with a specially trained, academically qualified mentor. Together, you and your mentor tackle difficult homework problems and review course concepts.
Why it works Our mentors are trained to provide you assistance, but they do not do the work for you. This means that when you return to your dorm – or to your classroom for a test – you're better prepared to do the work on your own.
How it works Simply complete the online form selecting a time that works with your schedule. Read the mentoring agreement, and submit the form to schedule the session – that's all there is to it!
If you're looking for a group setting where you can work on homework problems with other students, briefly assisted by mentors, you might want to try Drop-In Mentoring. We may also be able to assist you and your friends or members of a student organization - who are all enrolled in the same course - arrange for mentors in a small group setting. Contact us for more information: MAC@bhsu.edu. | 677.169 | 1 |
Syllabi and notes for math courses from arithmetic review to beginning calculus. Download MathHelp a tutorial program with problem sets for Mac or PC, or learn how to use MathHelp to create your own tutorials. Brief Mathematics Articles present concepts such as unit conversion, pi, and trigonometry, with suggestions for where to find more information on the Web. | 677.169 | 1 |
Matrix groups are a beautiful subject and are central to many fields in mathematics and physics. They touch upon an enormous spectrum within the mathematical arena. This textbook brings them into the undergraduate curriculum. It is excellent for a one-semester course for students familiar with linear and abstract algebra and prepares them for a graduate course on Lie groups.
Matrix Groups for Undergraduates is concrete and example-driven, with geometric motivation and rigorous proofs. The story begins and ends with the rotations of a globe. In between, the author combines rigor and intuition to describe basic objects of Lie theory: Lie algebras, matrix exponentiation, Lie brackets, and maximal tori. The volume is suitable for graduate students and researchers interested in group theory.
Readership
Undergraduates and beginning graduate students interested in group theory.
Reviews
"this is an excellent, well-written textbook which is strongly recommended to a wide audience of readers interested in mathematics and its applications. The book is suitable for a one-semester undergraduate lecture course in matrix groups, and would also be useful supplementary reading for more general group theory courses." | 677.169 | 1 |
Details about Basic College Mathematics:
Pat McKeague's passion and dedication to teaching mathematics and his ongoing participation in mathematical organizations provides the most current and reliable textbook series for both instructors and students. When writing a textbook, Pat McKeague's main goal is to write a textbook that is user-friendly. Students develop a thorough understanding of the concepts essential to their success in mathematics with his attention to detail, exceptional writing style, and organization of mathematical concepts. BASIC COLLEGE MATHEMATICS: A TEXT/WORKBOOK, Third Edition offers a unique and effortless way to teach your course, whether it is a traditional lecture course or a self-paced format. In a lecture-course format, each section can be taught in 45-to-50-minute class sessions, affording instructors a straightforward way to prepare and teach their course. In a self-paced format, Pat's proven EPAS approach (Example, Practice Problem, Answer and Solution) moves students through each new concept with ease and assists students in breaking up their problem-solving into manageable steps. The Third Edition of BASIC COLLEGE MATHEMATICS has new features that will further enhance your students' learning, including boxed features entitled Improving Your Quantitative Literacy, Getting Ready for Chapter Problems, Section Objectives, and Enhanced and Expanded Review Problems. These features are designed so your can students to practice and reinforce conceptual learning. Furthermore, Enhanced WebAssign for Developmental Math, is an assignable assessment and algorithmic homework system that consists of videos, tutorials and active examples keyed to problem level. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.
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Rent Basic College Mathematics 3rd edition today, or search our site for other textbooks by Charles P. McKeague | 677.169 | 1 |
Additional Mathematics is a UK qualification pilot scheme in its final year of implementation for a GCSE level qualification in mathematics which is applied to a range of problems set out in a different format to the standard Mathematics GCSE. This has been formed because of the standard secondary schools in England offering two GCSE qualifications in Science and English but only one in Mathematics and as Mathematics is also a core subject it needs to be viewed on the same level as the other two core subjects (Science and English.)
In Singapore, Additional Mathematics is an optional subject offered to pupils in Secondary School who have an aptitude in Mathematics and are in Normal (Academic) stream or Express stream. The syllabus covered is slightly more in depth as compared to Mathematics. Additional Mathematics is also a prerequisite for students who are intending to offer H2 Mathematics and H2 Further Mathematics at A-level if they choose to enter a Junior college after Secondary School. Students without Additional Mathematics at the 'O' level will usually offer H1 Mathematics instead.
Paper 1 (Duration 2 Hours) A variety of compulsory questions are being set, with a mark weighting from 1 Mark to maximum mark of 7. The total weighting of the paper is 80 marks and constitutes to 44% of the O Level Additional Mathematics grade.
Paper 2 (Duration 2 Hours 30 Minutes) A variety of compulsory questions are set. One of the questions would be a plotting question where one is required to plot a given graph on to a piece of graph paper. This can be either tested in Paper 1 or Paper 2. The total weighting of the paper is 100 marks and constitutes to 56% of the O Level Additional Mathematics grade.
In Northern Ireland, Additional Mathematics was offered as a GCSE subject by the local examination board, CCEA. There were two examination papers: one which tested topics in Pure Mathematics and one which tested topics in Mechanics and Statistics. It was discontinued in 2014 and replaced with GCSE Further Mathematics.
Starting from 2012, Edexcel and AQA have started a new course, now that Additional Maths has been abolished, which is an IGCSE in Further Maths. Edexcel and AQA both offer completely different courses, Edexcel including the calculation of solids formed through integration, AQA not touching on integration. AQA's syllabus mainly offers further algebra, with the factor theorem and more complex algebra such as algebraic fractions as well as differentiation up to and including the calculation of normals to a curve. AQA's syllabus also includes a lot of matrices work, which is an AS Further Mathematics topic. AQA's syllabus is much more famous than Edexcels, mainly for its controversial decision to award an A* with Distinction(A^), a grade higher than the maximum possible grade in any Level 2 qualification; it is known colloquially as a Super A* or A**.
In Malaysia, Additional Mathematics is offered as an elective to upper secondary students studying within the public education system. This subject is included in the Sijil Pelajaran Malaysia examination. Science stream students are required to apply Additional Mathematics as one of the subjects in the Sijil Pelajaran Malaysia examination. Additional Mathematics is an optional subject for students who are from arts or commerce streams.
In Mauritius, Additional Mathematics is offered in secondary school as an optional subject in the Arts Streams and it is a compulsory subject in the Science, Technical and Economics Stream. This subject is included in the University of Cambridge International Examinations. Topics that are covered in the Additional Mathematics syllabus include functions, quadratic equations, differentiation and integration (calculus). It is more commonly known as Add Maths . Many students found it difficult as it involves a lot of practice to know the subject itself. For at least the past three years, the JOHN KENNEDY COLLEGE of Mauritius had produced the best students ranked 1st . | 677.169 | 1 |
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About the Book
The resource math teachers have been waiting for is finally here!
Volume Two of the Van de Walle Professional Mathematics Series provides practical guidance along with proven strategies for practicing teachers of grades 3 through 5. In addition to many of the popular topics and features from John Van de Walle's market-leading textbook, "Elementary and Middle School Mathematics," this volume offers brand-new material specifically written for these grades. The expanded grade-specific coverage and unique page design allow readers to quickly and easily locate information to implement in the classroom. Nearly 200 grade-appropriate activities are included. The student-centered, problem-based approach will help students develop real understanding and confidence in mathematics, making this series indispensable for teachers! Big Ideas provide clear and succinct explanations of the most critical concepts in 3-5 mathematics. Problem-based activities in Chapters 2-12 provide numerous engaging tasks to help students develop understanding. Assessment Notes illustrate how assessment can be an integral part of instruction and suggest practical assessment strategies. Expanded Lessons elaborate on one activity in each chapter, providing examples for creating step-by-step lesson plans for classroom implementation. A Companion Website ( provides access to more than 50 reproducible blackline masters to utilize in the classroom. The are provided in the appendix for teachers' reference. About the Authors
John Van de Walle is Professor Emeritus at Virginia Commonwealth University. He is a co-author of "Scott Foresman-Addison Wesley Mathematics," a K-to-6 textbook series, and the author of "Elementary and Middle School Mathematics: Teaching Developmentally," the best-selling text and resource book on which this professional series is based.
LouAnn Lovin is a former classroom teacher and is currently an assistant professor in mathematics education at James Madison University, where she teaches mathematics methods and mathematics content courses for Pre-K-8 prospective teachers and is involved in the mathematical professional development of teachers in grades 4-8.
Collect all three volumes in the Van de Walle Professional Mathematics Series! Each volume provides in-depth coverage at specific grade levels. Learn more about the series at | 677.169 | 1 |
wide range of courses have an intake that requires a basic, easy introduction to the key maths topics for engineering - Basic Engineering Mathematics is designed to fulfil that need.
Unlike most engineering maths texts, this book does not assume a firm grasp of GCSE maths, yet unlike low-level general maths texts the content is tailored for the needs of engineers. The result is a unique text written for engineering students, but which takes a starting point below GCSE level. The textbook is therefore ideal for students of a wide range of abilities, and especially for those who find the theoretical side of mathematics difficult.
John Bird's approach is based on numerous worked examples, supported by 525 worked problems and followed by 925 further problems. The content has been designed to match current level 2 courses, including Intermediate GNVQ and the new specifications for BTEC First.
Level 3 students who struggle with their maths will also find this book particularly useful. With this in mind, all topics within the compulsory units of the AVCE (Applied Mathematics for Engineering) and the new specifications for BTEC National (Mathematics for Technicians) are covered.
Lecturers' support materials: Throughout the book Assignments are provided that are ideal for use as tests or homework. These are the only problems where answers are not provided in the book. Full worked solutions are available to lecturers only as a free download from the Newnes website:
* Unique in being written for engineering students but taking a starting point below GCSE level * Coverage fully matched to the requirements of the core units of the new BTEC First and BTEC National specifications * Ideal for a wide range of Level 2 courses including City & Guilds certificates and EMTA/EAL NVQs
Editorial Reviews
Review
"The book is excellent for revision practises you have a full understanding of how to apply mathematical skills, it's good for learning transposition and simultaneous equations, It's worth the 4 STARS." - Amazon review of the 2nd edition by a U.K. reader (Mar 2001)
--This text refers to an out of print or unavailable edition of this title.
Book Description
A truly introductory mathematics course written for engineering students
--This text refers to an out of print or unavailable edition of this title. | 677.169 | 1 |
Lesson content has been expanded and includes more video instruction for each lesson. In addition to the video examples each lesson includes interactive practice problems, challenge questions, and worksheets.
Description
Everything you get from a traditional test prep book PLUS one-on-one instruction covering all the math on the ASVAB.
YourTeacher – ASVAB Math Prep Course
******
"It's like a private math classroom, but you are the only student."
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"I just love YourTeacher and the way you explain things. I felt like I was in a classroom instead of just looking at examples."
Diane
"I was using the program to improve my ASVAB score to qualify for Marine Corps Officer's Candidate School. I improved my score from a 60 something to a 77. Many points above the necessary 70. I really enjoyed the program, and if you diligently use it you will accomplish what needs to be done."
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Need more than practice problems to get ready for the ASVAB?
YourTeacher's ASVAB Math Prep Course provides everything you would get in a traditional test prep review book (i.e. written practice problems, reading material, etc.) PLUS the one-on-one instruction you need to truly learn the math covered on the ASVAB.
Our lessons include:
-Multiple video example problems
(similar to how a teacher starts class at the board by explaining the examples from the textbook)
-Interactive practice problems with built-in support
(similar to how a teacher assigns practice and walks around the class providing help)
-A Challenge Problem
(similar to how a teacher assigns a higher level problem which students must work on their own to prove mastery)
-Extra problem worksheets
(similar to how a teacher assigns additional problems for homework)
-Review notes
(similar to how a teacher provides summary handouts or refers you to your textbookMaze Trap
In Maze Trap you are a blue ball (the hero) inside a maze and your goal is to trap the enemies,
the black balls (aka monsters), so they have nowhere to move, i.e. no empty spaces to move to.
To achieve this goal you have to push around the green maze building blocks in order to build
walls to first block and then trap your enemies. This is not an… | 677.169 | 1 |
An innovative course that offers students an exciting new perspective on mathematics,Modeling With Mathematicsexplores how mathematics can help explore problems real people encounter in their jobs and lives. Mathematical modeling and a data-driven approach to exploring functions helps students deepen their mathematical skills and maturity. Modeling With Mathematics: A Bridge To Algebra IIhas been designed for students who have completed Algebra I or Algebra I and Geometry but need review practice and motivation to succeed in Algebra II. In addition the course gives students a look ahead to many Algebra II topics. Modeling With Mathematics: A Bridge To Algebra II list serv whfreeman. com/bridgelistserv. pdf As a service to instructors usingModeling With Mathematics: A Bridge To Algebra II, a listserv has been designed as a forum to share ideas, ask questions and learn new ways to enhance the learning experience for their students.
With the emphasis the Common Core State Standards (CCSS) places on modeling, Modeling With Mathematics: A Bridge to Algebra II (Bridge 2e) addresses these modeling requirements while helping prepare students for success in Algebra II. Intended for students who have taken Algebra I and Geometry but who are not yet ready for Algebra II, this program helps solidify their understanding by providing a different kind of learning experience. With Bridge 2e students model real-world applications with a functions approach netting a deeper grasp of the important concepts necessary for success in Algebra II and on the forthcoming Common Core assessments | 677.169 | 1 |
Tewksbury AlgebraThorough understanding of the theoretical underpinnings of this powerful tool can be left to the math majors. Those who ask for help in a calculus course are most often taking it as a requirement for a technical field. Here, the practical application of derivatives and integrals are what is important
Ellen C. | 677.169 | 1 |
* Endorsed by Edexcel for the new Two-Tier exam* Clear learning objectives and...* Endorsed by Edexcel for the new Two-Tier exam* Clear learning objectives and summaries for each chapter* 'Internet Challenges' ensure regular and integrated use of ICT* Revision of key concepts at the start of the Foundation course* Plenty of...more
EdexcelGCSEMathematics The right formula for success! Edexcel's own teaching...Edexcel GCSE Mathematics The right formula for success! Edexcel's own teaching resources for the two-tier GCSE Maths specifications.more
Targeted at the B to A grade range in the Foundation tierGCSE, this title off...Targeted at the B to A D to B grade range in the Foundation tierGCSE, this title off...Targeted at the D to B grade range in the Foundation tier GCSE, this title offers: a review test focusing on prior topics for reinforcement; two sets of questions that relate to individual lessons in the units providing practice; a synoptic homework to...more
Books/Subjects/Science & Nature/Mathematics/Education/Teaching Aids Tailored t...Books/Subjects/Science & Nature/Mathematics/Education/Teaching Aids Tailored to both the specification and the tier, this Student Book delivers exactly what students and teachers need to cover the unit in exactly the right depth.more
Edexcel offers a suite of maths resources designed to give students more chanc...Edexcel offers a suite of maths resources designed to give students more chances to succeed in the Linear and Modular Edexcel GCSE Mathematics specifications.more
his revision guideis matched to the new EdexcelGCSEMathematics linear specif...his revision guideis matched to the new Edexcel GCSE Mathematics linear specification (2540). It provides full coverage of everything that may be tested in the two terminal exam papers. Just as importantly, it contains no superfluous material which...more
Providing material for consolidation, homework and independent study, this Pra...Providing material for consolidation, homework and independent study, this Practice Book is tailored to both the specification and the tier to deliver focused support.more
Providing material for consolidation, homework and independent study, this Pra...Providing material for consolidation, homework and independent study, this Practice Book is tailored to both the specification and the tier to deliver focused highy targeted support for those working at the Foundation level.more
Edexcel's own suite of resources for the two-tierGCSEMathematics specificati...Edexcel's own suite of resources for the two-tier GCSE Mathematics specification - designed to give your students the best chance of success.more
Targeted at the E to C grade range in the Foundation tierGCSE, this title off...Targeted at the E to C G to E grade range in the Foundation tierGCSE, this title off...Targeted at the G to E grade range in the Foundation tier GCSE, this title offers: a review test focusing on prior topics for reinforcement; two sets of questions that relate to individual lessons in the units providing practice; a synoptic homework to...more | 677.169 | 1 |
you may want to try DOTMATH. I have a website at
that can give you an introduction to what
dotmath is and why it is needed. There is information there that can
help you get in contact with me if you still need help with this.
dotmathman. | 677.169 | 1 |
"...look at the language of numbers through common situations, such as playing games or cooking. Put your decision-making skills to the test by deciding whether buying or leasing a new car is right for you, and predict how much money you can save for your retirement by using an interest calculator."
MathWorld is a comprehensive and interactive mathematics encyclopedia ``intended for students, educators, math enthusiasts, and researchers. The ``site is continuously updated to include new material and incorporate new discoveries.
QuickMath is an automated service for answering common math problems over``the internet. Think of it as an online calculator that solves equations``and does all sorts of algebra and calculus problems - instantly and``automatically. | 677.169 | 1 |
Developmental Mathematics
The Developmental Mathematics Program assists students in developing and applying mathematics skills and concepts and developing critical thinking skills related to mathematics. The program provides a solid foundation for success in college-level mathematics | 677.169 | 1 |
Chapter 12: Probability
Introduction
In Math 7, the learning content is divided into Concepts. Each Concept is complete and whole providing focused learning on an indicated objective. Theme-based Concepts provide students with experiences that integrate the content of each Concept. Students are given opportunities to practice the skills of each Concept through real-world situations, examples, guided practice and independent practice sections. There are also video links provided to give students an audio/visual way of connecting with the content. In this Chapter, students will engage with Concepts on writing and calculating theoretical probability, experimental probability, disjoint events, complementary events, overlapping events, tree diagrams, sample spaces, the Counting Principle, factorials, permutations, combinations, tree diagrams, independent and dependent events.
Chapter Summary
Summary
Having completed this chapter, students are now finished with this book. Each Concept has provided students with an opportunity to engage and practice skills in many Concepts including writing and calculating theoretical probability, experimental probability, disjoint events, complementary events, overlapping events, tree diagrams, sample spaces, the Counting Principle, factorials, permutations, combinations, tree diagrams, independent and dependent events. | 677.169 | 1 |
At a Glance - In the Real World
Believe it or not, there are lots of jobs where you'll need to know algebra—including some exciting ones involving explosives. Especially jobs where you're hired to blow up inequalities.
Check out the US Military site and search for algebra; it appears in quite a few job descriptions. You know, like You + Algebra = America.
Algebra is useful because it provides a convenient way to solve lots of different types of problems. If we can turn a problem into an equation or inequality, then we can solve the problem by solving the equation or inequality. We don't need to think about the original problem until it comes time to actually write down the answer, when we need to make sure we're answering the right question.
It's like the bus that takes you from Point A to Point B, and you only need to wake up five minutes before you get there to make sure you're getting off at the right stop. Point B, in case you've forgotten. | 677.169 | 1 |
Product Description
Beginning Algebra: Expanded Edition is the first book in the Life of Fred High School Mathematics Series, and is designed for students in 9th grade. This expanded edition of Algebra replaces the both the earlier Life of Fred Beginning Algebra and Fred's Home Companion Algebra books; it also contains all problems completely worked out. The sold-separately Zillions of Practice Problems for Beginning Algebra book is an optional add-on for students who want more practice opportunities.
Twelve chapters with multiple sub-lessons are included. Each lesson ends with a "Your Turn to Play" segment with a small number of thought-provoking questions. Answers are provided on the next page for students to go over themselves after attempting to solve the problems. Chapters conclude with three problem sets, each of which is named after a city. This full-year curriculum will provide students with a thorough knowledge of freshman algebra I.
Life of Fred is a unique, complete (not supplemental), everywhere544 pages, indexed; hardcover, non-consumable textbook with Smyth-sewn binding. Students write their answers on separate paper. Answers are also included in the text and written directly to the student.
Several years ago, I had two kids who absolutely hated math. The drill and kill technique of the many math curriculums we tried nearly destroyed a love of mathematical learning along with any confidence they had in their abilities to learn. At the advice of another homeschooling friend, we tried Life of Fred. We backed up and started at the Intermediate Series (Kidneys, Liver, & Mineshaft), and I sat back and watched in awe as my children slowly began gaining confidence and skills under Fred's teachings.
I now have an up-and-coming 7th and 9th grader who are already working their way through LOF Algebra, way ahead of where we would have been had we stuck with traditional math programs. We just got their yearly testing scores back, and both children scored upper high school level or beyond in ALL areas of math.
My family is proof that LOF works. No, it's not easy, and my children sometimes get stuck. But I very rarely need to step in and help them. LOF taught my children not only that they can "do" math, but they have also gained the confidence and knowledge to teach and learn for themselves. And that's the best education a person can have!
We have not yet reached Life of Fred: Beginning Algebra, but expect to get there sometime next fall. After reading the negative review written on Oct 14, 2014, I wanted to provide my perspective. Author Stanley Schmidt teaches a process for laying out word problems, and if you miss the earlier books, you will not understand what he wants you to do. He explains everything thoroughly -- not just the "how" but the "why" (which is what my son loves).
If you think your child is ready for algebra and has been using a regular curriculum or if your child is struggling in regular algebra, I would take him or her back to Fred's 3-book Pre-Algebra series. If it's easy, proceed at double-pace and feel good about yourself. But I promise your child will learn new things and also learn Mr. Schmidt's process, which my son independently applies to all sorts of science and everyday life problems. If you think your child is ready for Pre-Algebra or is struggling in Pre-Algebra now, go back and do Fred Fractions and then Decimals & Percents. They really set the foundation of Mr. Schmidt's process and will make the subsequent books fully understandable. Again, proceed at double-pace if chapters seem easy. But going "back" will be worth it in the long run. I think most of us would rather our child understand -- and even enjoy -- math than keep on a math schedule that might not fit.
But truly, if you are willing to go back, your child will soon be so interested in the math and science she is learning that she'll soon be ahead of her peers without even realizing it. Fred Math has transformed our experience. My son went from a child who understood math but hated it and felt frustrated by the lack of explanations to a child who loves math and can apply it to life easily and happily. He has done four books this year and will continue through the summer (at a reduced pace) just because it's fun.
If you do not understand algebra already, this is not the book for you. There is not much help in it to understand how to solve algebra equations. It can be entertaining with this silly backstory about Fred, but when it comes to teaching the subject, it is sorely lacking. | 677.169 | 1 |
Description
Computational Geometry is an area that provides solutions to geometric problems which arise in applications including Geographic Information Systems, Robotics and Computer Graphics. This Handbook provides an overview of key concepts and results in Computational Geometry. It may serve as a reference and study guide to the field. Not only the most advanced methods or solutions are described, but also many alternate ways of looking at problems and how to solve them.
Details
About the editors
Reviews
@from:H.-D. Hecker
@qu:The Handbook gives an excellent overview of the central topics of both classic and new lines of computational geometry.......The authors of the chapters are without exception excellent specialists in the topic discussed. This Handbook discusses fundamentals and the most interesting applications.
@source:Mathematical Reviews | 677.169 | 1 |
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Differential calculus
How would you like to follow in the footsteps of Euclid and Archimedes? Would you like to be able to determine precisely how fast Usain Bolt is accelerating exactly two seconds after the starting gun? Differential calculus deals with the study of the rates at which quantities change. It is one of the two principal areas of calculus (integration being the other). | 677.169 | 1 |
Glimpses of Soliton Theory is a textbook (published by the American Mathematical Society in 2010) that aims to introduce the algebro-geometric structure of soliton equations to undergraduate math majors.
Solitons are solutions to certain very special differential equations that have applications in science and engineering. Aside from these practical applications, however, soliton theory is also amazing in the way that it ties together seemingly unrelated branches of mathematics. Unlike most nonlinear differential equations, soliton equations can be solved explicitly using algebraic methods and the set of all of the solutions has a rich geometric structure. This textbook allows undergraduate students to appreciate this "gem" of mathematics with only courses in calculus and linear algebra as prerequisites. (In particular, this book does not require prior experience with physics or differential equations.)
Because of its interdisciplinary nature - combining aspects of algebra, geometry, analysis, and applied mathematics - this book would make an ideal textbook for a "capstone class" in mathematics. Moreover, carefully constructed examples and carefully selected topics also make it ideal for a reading course for a student wishing to learn this material independently.
Mathematica Notebook: One of the ways the book is able to address topics generally inaccessible to undergraduates is through the use of mathematical software. For instance, rather than introducing elliptic functions abstractly (which would require advanced experience with complex analysis), the computer manipulates these functions for the reader much as students first learning about trigonometric functions benefit from being able to graph and compute the values of the sine function on their calculator. In addition, although the reader will learn how to multiply differential operators "by hand", some homework exercises require the reader to compute products involving Lax operators that would be tedious to do without computer assistance.
For the reader's convenience, I am making the following Mathematica Notebook available for download. It contains all of the commands used in the textbook (except those which the student is expected to write as part of a homework exercise), organized by chapter.
(Note: If your browser opens the file as a text file in the browser window rather than saving it to your disk, simply use the "file" menu on your computer, select "save" and save it to the location of your choice with the suffix ".nb" in the filename.)
Errata:
Page 39: The summation notation for the Navier-Stokes equations on page 39 is humorously incorrect. Obviously, the sum is supposed to be from j=1 to j=4; anyone who has used TeX can guess what went wrong. (Apparently, this was noticed by Andrew Hone, though it was Anton Dzhamay (University of Northern Colorado) who actually told me about it.)
Page 43: The equation in question 7b should say
(ut-uxxx)(1+ux2)=0.
Thanks to Bob Riehemann (Thomas More College) for catching the error and bringing it to my attention.
Page 96: Hideshi Yamane (Kwansei Gakuin University) notes that the n appearing as a limit on the summation in the denominator of the first displayed equation at the top of the page is not the same as the n appearing elsewhere in Definition 5.1. So, it would be better if both occurrences of n in this paragraph were replaced with something else, like m'.
Page 97: Anton Dzhamay also pointed out that the τ-function in Example 5.2 was mistypeset, although you can easily figure out that it was supposed to say
Page 129: Iain Findlay points out that the 3rd line of last paragraph, should say "kernel of L" rather than "kernel of V".
Page 217: Thanks to Paul Rigge (University of Michigan) for pointing out that Question 7 mistakenly refers to "the τ-function in part (c)" when it should say "the τ-function in part (b)".
Page 236: The second line should begin with Φk not Φ2.
(The first subscript on the second line should be a k not a 2.)
Page 269: Iain Findlay also noted that the middle parenthetical term and the final parenthetical term in Equation 2.1 should be (bc + ad).
Page 282: Paul Rigge also noticed a problem in Project VII which refers to "Equation (7.48)" in the paper by Ascher and McLachlan. The relevant equation appears at the top of page 86 in that paper, but is not given an explicit number or name. (A partial explanation for this mysterious error involves noting that the same equation appears as Equation (7.48) in the textbook by Drazin and Johnson.)
(If you notice any other errors I should list here, please let me know.)
Official AMS Website:
Each book in the Student Mathematical Library series gets a webpage on the American Mathematical Society's server. The page for this book is | 677.169 | 1 |
PreCalculus Algebra 2 Domain and Range Match Up Activity *Task Cards*
This task card activity is a sorting and matching activity for Algebra 2 or Pre-Calculus students. There are 10 piece-wise function cards showing the graph, an equation card, a
This is a great mini-unit on Graphing Trigonometric Functions for Algebra 2 or Pre-Calculus students. The mini-unit includes 16 pages of worksheets on graphing trigonometric functions, a sort and match activity, and a general review of
AP Calculus AB, AP Calculus BC, College Calculus
As we approach the AP Calculus AB Exam, many students need a topical approach to the types of questions they will be asked to complete on the AP Calculus AB Exam.
In this packet, you will find the
PreCalculus Algebra 2: Logarithms Stations Activity
This Algebra 2 Activity is great for an end-unit review of logarithmic and exponential functions. Students work through 10 stations in groups of 3 or 4 on timed intervals to review the unit in
PI DAY ACTIVITIES FOR SECONDARY STUDENTS
Have a little fun celebrating PI DAY. Whether you spend an entire class period or just assign one of the activities for a homework assignment, there is a variety of engaging fun inside.
The product
Here's a task card activity to get your students in Algebra 2 and Pre-Calculus thinking about Polynomial Functions. There are 10 graph cards, 10 description cards, 10 equation cards, and 10 zeros cards. Students work cooperatively to match each
This is an activity with 32 cards meant for the beginning unit in Calculus AB, BC, or Calculus Honors. There are 8 graph cards that have matching Equation Cards, Limit Cards, and Description Cards to create a unique set. There is also a student
PreCalculus Solving Trigonometric Equations Task Cards QR Codes
Here's a fun way to use smartphone apps and keep your PreCalculus or Trigonometry students engaged in solving trig equations. There are 12 task cards of varying difficulty, including
This is a cooperative matching activity for students in Algebra 2 and above. All the graphs are basic sine or cosine curves with amplitude and period changes only (no phase shifts). It's truly a great beginning sort and match for your students.
Algebra 2: Radical Functions Stations Activity
This Algebra 2 Activity covers all the topics in the unit on Radical Functions and Rational Exponents. There are eight stations for students to work through in groups of 3 or 4 students. Time should be
Your Algebra 2 students will practice solving quadratic equations with both REAL and COMPLEX SOLUTIONS by various methods, including square root, factoring, and quadratic formula.
The 12 task cards include integral and rational solutions, real and
If you like group activities, here's a fun RELATED RATES MATCH UP. Eight task cards contain a rate problem to be solved. Calculus students match the question card with the correct equation card and then the proper derivative card. For an extension
AP Calculus AB: Full Year Curriculum Early Transcendentals
This is everything you need to teach a Full year of AP Calculus AB with Early Transcendentals. You can purchase the products all together instead of separately. All units are complete and
This unit is an introduction to trigonometry for courses of Trigonometry and PreCalculus levels.
The unit includes over 30 pages of extra practice handouts beginning with angles, radians, conversions,
complements, supplements, trigonometric
Algebra: Graphing Linear Functions Basics
This packet is a comprehensive set of extra practice for your Algebra I students. The packet contains group activities, calculator discovery activities that help your students explain, through writing,
Here is my COMPLETE ALGEBRA 2 CURRICULUM for students enrolled in ALGEBRA 2 HONORS. There are 81 lessons. Each of the lessons and activities are rigorous and engaging. An extra unit has been added to cover the beginning concepts for Trigonometry
Algebra 2 students work in groups of 3 or 4 to solve linear equations. The 12 task cards include multi-step problems, distributive property, variables on both sides, and fractional equations. The activity can be completed in a single class
PreCalculus Trigonometric Graphs Match Up Activity *Task Cards*
Don't you love task cards? I know my PreCalculus students do. Here is a great activity for practice with transformations of trigonometric functions. Students work in groups of 3-4 to
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TEACHING EXPERIENCE
I have taught grades 8 through 12 for over 23 years in the Central Florida area. I consider it an honor to share strategies with colleagues through local and national math conferences.
MY TEACHING STYLE
My friends say that I am way too energetic.I love hands-on strategies that engage students to learn cooperatively. There is always something new happening in my classroom.
HONORS/AWARDS/SHINING TEACHER MOMENT
1998-Teacher of the Year at Thomas Jefferson Junior High, Merritt Island, FL.; 2001-Teacher of the Year at George Jenkins High School, Lakeland, FL.; 2001-AIChE Mathematics Teacher of the Year,Polk County, FL.; National Board Certified Teacher, Adolescent Young Adult Mathematics,2001.2009-2010 Math Teacher of the Year, Orange County Public Schools
I teach AP Calculus AB, Pre-Calculus, and Algebra 2 Honors in the Metro-Orlando area. I want to blog about ways to engage, encourage, and empower teachers at if I could only find the time. I am an active teacher-trainer when opportunities arise. I love to share how Foldables(R) can change student engagement within your classroom. | 677.169 | 1 |
Binding is in very good condition. Has clean, unmarked pages. Ships fast from Amazon's warehouse. THe upper right corner of the cover is embossed by the former owner. Other than that the book looks great - the pages are in like new elementary text introduces basic quantum mechanics to undergraduates with no background in mathematics beyond algebra. Containing more than 100 problems, it provides an easy way to learn part of the quantum language and apply it to problems. Emphasizing the matrices representing physical quantities, it describes states simply by mean values of physical quantities or by probabilities for possible values. This approach requires using the algebra of matrices and complex numbers together with probabilities and mean values, a technique introduced at the outset and used repeatedly. Students discover the essential simplicity of quantum mechanics by focusing on basics and working only with key elements of the mathematical structure--an original point of view that offers a refreshing alternative for students new to quantum mechanics Publisher
This simple text makes basic quantum mechanics accessible with a minimum of mathematics. The focus is on the matrices representing physical quanitities. States are described simply by mean values of physical quantities or by probabilities for possible values. This approach reveals the essential simplicity of quantum mechanics by focusing on basics and working only with key elements of mathematical structure. Introduces all mathematics involved with using algebra of matrices, complex numbers, probabilities, and mean values. Offers over l00 problems.
--This text refers to an out of print or unavailable edition of this title.
Top Customer Reviews
This review is written from the point of view of a philospher, poorly trained in mathematics, but still wanting to get to the meat of quantum mechanics from a methematical point of view. Wow. In this book I found what I thought I never would. It describes the mathematical world of quantum physics using the majestic simplicity of matrices and the algebra of complex numbers. As the author states in the preface, no calculus or trigonometry is required. While the math isn't downright simple, neither is beyond the grasp of someone who is bright, but hasn't taken claculus or even precalc. For those who want to journey past this book another excellent intro level quantum mechanics text that introduces wave mechanics and does assume a knowledge of basic calculus is "Fundamentals of Quantum Mechanics" by J.E. House. Both are excellent!
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There are few books which explain quantum mechanics with such grace and simplicity. Starting with the basics the author sets out to explain the ideas and mathematics behind qunatum mechanics. The author also provides the historical references leading to the birth of quantum mechanics. The layout and presentation of the material is pure mathematical poetry. Whilst the material would never make light bedtime reading, I would seriously recommend this book for both phyisicists and electronic engineering at the undergraduate and graduate level. The book has been a great source of information for my own research into the mysteries of quantum mechanics.
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I have quite a few books on Quantum Mechanics. This book does what the others do not. The first half is about simple math. Understanding that QP - PQ = ih/2pi is the matrix form of an equation and the QP - PQ is not zero because the matrices do not commute is critical. This is basic stuff that a lot of books just skip. The second half uses the math to explain some of the features of Quantum Mechanics. For me I needed the detailed first half even though the math was not too hard. Now I can read my other books with a new understanding and finally I am starting to understand Quantum Mechanics.
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What an interesting book, but not for beginners from chapter 7 and on! The first 6 chapters are a walk in the park; a simple review of complex numbers, matrix algebra, and probability. Then the author sneaks up on his trusting student and delivers a knock-out punch! You better know eigen vectors/values for this chapter to make sense. The uninitiated has been left in the dust!
The subject gets interesting in the second half (chapters 8 to 27). It takes prior knowledge of quantum mechanics to appreciate! The book deserves 5 stars for a wonderful display of demonstration and historical references. It was a joy to read.
I was hoping for some elegantly simple explanation of a rather intense subject. I never found it. Instead, I got a fresh perspective and better appreciate the history of the subject. There is nothing simple about quantum mechanics, other than it's all about solving the eigen value problem . . . but keep that to yourself, it's a little secret!
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It's true that this book requires absolutely no calculus. Or linear algebra for that matter. This book doesn't even assume you've ever seen a complex number or a matrix before. All that is necessary is introduced in the first few chapters.
However, as this book progresses it slowly reveals itself for what it truly is: a first book on the operator formalism in quantum mechanics, where commutation relations for observable quantities are promoted to central importance.
While I'm certain that students with only a very modest background in physics and mathematics will be able to get something out of this book at least in the early chapters, the last third of this book is more suitable for fairly advanced students of quantum mechanics looking to make their way from state vectors to operators as required by quantum field theory. To such students I would recommend already having The Principles of Quantum Mechanics (International Series of Monographs on Physics) under your belt.
This is ultimately a challenging book masquerading as an elementary one.
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I was attracted to this book by its title, which communicates a theme now rarely encountered. Way back when, the opening lecture in elementary quantum mechanics explained that the theory came in two distinct but equivalent forms: wave mechanics and matrix mechanics. Thereafter nothing more was said about matrix mechanics. Yet, this is where it all began and the story of matrix mechanics is itself a lesson in how theoretical physics advances. Thus, this book is as much a brief history as it is a manual of instruction.
Though billed as a simple introduction to quantum mechanics for the mathematically unprepared, I don't see it as such. Yes, the mathematical development is simple enough, the writing is crystal clear, the problems illuminate the text and exercise the beginner in the subject matter, yet were I to pick up this book and read it without already knowing much about quantum mechanics, I think I would walk away understanding little. The outcome would be better if read in conjunction with lectures based on the text. And this is the underlying scenario for clearly it is a course book. Still, I am not sure that I would "get it" even then.
But coming at this book with a background in physics I found it a good read. My conclusion is that this book is most accessible to people like me with some prior training in physics. Its accessibility to others will depend on the strength of their technical background or on their ability to reason, their fundamental curiosity and their determination to learn. | 677.169 | 1 |
Glencoe Precalculus 2011is a comprehensive program that prepares students to be successful in college or AP Calculus programs. Features of this program include: Chapter 0, Graphing Technology Labs, leveled exercise sets, H.O.T. (Higher-Order Thinking) Problems, and Preparation for AP Calculus lessons within every chapter.Glencoe Precalculusalso includes a complete technology suite that contains an online student edition, online teacher edition, Interactive Classroom, Advance Tracker, andExamView Assessment Suite. | 677.169 | 1 |
Tussy and Gustafson's fully integrated learning process is designed to expand students' reasoning abilities and teach them how to read, write, and think mathematically. In this text, students get a thorough review of arithmetic and geometry along with all the topics covered in a standard elementary algebra course. The authors build the strong mathematical foundation necessary to give students confidence to apply their newly acquired skills in further mathematics courses, at home, or on the job. | 677.169 | 1 |
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Can you find the right steps to solve each challenge? StepUp is an addicting puzzle game that requires you to rethink what you know about solving equations. Simply choose a step to get closer to solving for x, but be careful, the path to the ...
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Prepare to achieve the SAT scores you need to get into your top choice schools. Get ready for the test day with this most comprehensive app on SAT Math: Algebra & Functions. Features include: HIGHEST QUALITY and QUANTITY Over 420 questions and 42 ...
By , Michael Robinson Chavez The Washington Post Men don't wear clip-on ties. That sartorial directive from algebra and geometry teacher Shaka Green was just one of the many lessons being absorbed by the initial class of 100 or so students ...
For the past seven summers, hundreds of students have been gathering at churches across the state of California to study algebra. The students are taking part in a unique program sponsored by California State University through a partnership with | 677.169 | 1 |
Algebra resources
Overview of the rules of partial differentiation and methods of optimization of functions in Economics and Business Studies. This leaflet has been contributed to the mathcentre Community Project by Morgiane Richard (University of Aberdeen) and reviewed by Anthony Cronin (University College Dublin).
Overview of the properties of the functions e and ln and their applications in Economics. This leaflet has been contributed to the mathcentre Community Project by Morgiane Richard (University of Aberdeen) and reviewed by Shazia Ahmed (University of Glasgow) and Anthony Cronin (University College Dublin).
An interactive version of the refresher booklet on Algebra including links to other resources for further explanation. It includes revision, exercises and solutions on fractions, indices, removing brackets, factorisation, algebraic frations, surds, transpostion of formulae, solving quadratic equations and some polynomial equations, and partial fractions. An interactive version and a welsh language version are available.
Factorising can be thought of as a reversal of the process of removing brackets. When we factorize an expression, it is written as a product of two or more terms, and these will normally involve brackets.
In many business applications, two quantities are related linearly. This means a graph of their relationship forms a straight line. This leaflet discusses one form of the mathematical equation which describes linear relationships.
This leaflet provides information on symbols and notation commonly used in mathematics. It shows the meaning of a symbol and, where necessary, an example and an indication of how the symbol would be said. For further information from mathcentre resources, a search phrase is given. This Quick Reference leaflet is contributed to the mathcentre Community Project by Janette Matthews and reviewed by Tony Croft, University of Loughborough.
Fractions involving symbols occur frequently. It is necessary to be able to simplify these and rewrite them in different, but equivalent forms. In this leaflet, we revise how these processes are carried out.
On occasions you will come across two or more unknown quantities, and two or more equations relating to them. These are called simultaneous equations and when asked to solve them yo umust find values of the unknowns which satisfy all the given equations at the same time. On this leaflet we will illustrate one way in which this can be done.
Equations always involve one or more unknown quantities which we try to find when we solve the equation. The simplest equations to deal with are linear equations. On this leaflet we describe how these are solved.
Mathematics provides a very rich language for the communication of concepts and ideas, and a set of powerful tools for the solution of problems. In order to use this language, it is essential to appreciate how symbols are used to represent quantities, and to understand the conventions which have been developed to manipulate them.
The letter e is used in many mathematical calculations to stand for a particular number known as the exponential constant. This leaflet provides information about this important constant, and the related exponential function.
When a number is to be multiplied by itself, a power or index can be used to write this compactly. In this leaflet, we remind you of how this is done, and state a number of rules, or laws, which can be used to simplify expressions involving indices.
There are a number of rules known as the laws of logarithms. These allow expressions involving logarithms to be rewritten in a variety of different ways. The laws apply to logarithms of any base, but the same base must be used throughout a calculation.
It is often useful to be able write a quadratic expression in an alternative form - that is as a complete square plus or minus a number. The process for doing this is called completing the square. This booklet explains how this process is carried out.
This booklet explains what is meant by a cubic equation and discusses the nature of the roots of cubic equations.
It explains a process called synthetic division which can be used to locate further roots when one root is known.
The graphical solution of cubic equations is also described.
This is a complete workbook covering the removal of brackets
from expressions. It contains lots of examples and exercises.
It can be used as a free-standing resource, or can be read in conjunction with mathtutor - the companion on-disk resource.
This is a complete workbook on Indices covering definitions, rules and lots of examples and exercises.
It can be used as a free-standing resource, or can be read in conjunction with mathtutor - the companion on-disk resource.
This is a complete workbook introducing the solution of a single linear equation in one variable. It contains plenty of examples and exercises.
It can be used as a free-standing resource or in conjunction with the mathtutor DVD.
This booklet explains what is meant by a logarithm. It states and illustrates the laws of llogarithms. It explains the standard bases 10 and e.
Finally it shows how logarithms can be used to solve certain types of equations.
This is a complete workbook introducing the solution of a pair of simultaneous linear equations. It contains plenty of examples and exercises.
It can be used as a free-standing resource or in conjunction with the mathtutor DVD.
Formulae are used to relate physical quantities to each other. They provide rules so that if we know the values of certain quantities we can calculate the values of others. This booklet discusses several formulae.
Diagnostic test for completing the square. (1 of 2).
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Diagnostic test for completing the square. (2 of 2).
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Diagnostic test for factorising quadratics. (1 of 2).
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Diagnostic test for factorising quadratics. (2 of 2).
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Diagnostic test for solving inequalities.
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Diagnostic test for mathematical language
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Diagnostic test for simplifying fractions.
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Diagnostic test for simultaneous equations.
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Diagnostic test for substituting formulae.
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Diagnostic test for transposing formulae.
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Online exercise on completing the square.
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Online exercise on completing the square using maxima and minima.
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Online exercise on factorising quadratics.
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Online exercise on factorising quadratics (2).
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Online exercise on simultaneous equations.
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In this unit we consider how quadratic expressions can be written in an equivalent form using the technique known as completing the square. This technique has applications in a number of areas, but we will see an example of its use in solving a quadratic equation.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Completing the square is an algebraic technique which has several applications. These include the solution of quadratic equations. In this unit we use it to find the maximum or minimum values of quadratic functions.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In this unit we see how to expand an expression containing brackets. By this we mean to rewrite the expression in an equivalent form without any brackets in. Fluency with this sort of algebraic manipulation is an essential skill which is vital for further study.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
An essential skill in many applications is the ability to factorise quadratic expressions. In this unit you will see that this can be thought of as reversing the process used to 'remove' or 'multiply-out' brackets from an expression.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
Logarithms appear in all sorts of calculations in engineering and science, business and economics. Before the days of calculators they were used to assist in the process of multiplication by replacing the operation of multiplication by addition. Similarly, they enabled the operation of division to be replaced by subtraction. They remain important in other ways, one of which is that they provide the underlying theory of the logarithm function. This has applications in many fields, for example, the decibel scale in acoustics.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
This introductory section provides useful background material on the importance of symbols in mathematical work. It describes conventions used by mathematicians, engineers, and scientists.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
After viewing this tutorial, you should be able to explain the meaning of the terms 'proper fraction' and 'improper fraction', and express an algebraic fraction as the sum of its partial fractions. (Mathtutor Video Tutorial) algebraic fraction as the sum of its partial fractions.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In order to simplify certain sorts of algebraic fraction we need a process known as polynomial division. This unit describes this process.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
A knowledge of powers, or indices as they are often called, is essential for an understanding of most algebraic processes. In this section you will learn about powers and rules for manipulating them through a number of worked examples.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In this unit we give examples of simple linear equations and show you how these can be solved. In any equation there is an unknown quantity, x say, that we are trying to find. In a linear equation this unknown quantity will appear only as a multiple of x, and not as a function of x such as x2, x3, sin x and so on. Linear equations occur so frequently in the solution of other problems that a thorough understanding of them is essential.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
This video explains how algebraic fractions can be simplified by cancelling common factors. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
The purpose of this section is to look at the solution of simultaneous linear equations. We will see that solving a pair of simultaneous equations is equivalent to finding the location of the point of intersection of two straight lines.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
All cubic equations have either one real root, or three real roots. In this video we explore why this is so. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
This video explains linear and quadratic inequalities and how they can be solved algebraically and graphically. It includes information on inequalities in which the modulus symbol is used. (Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
This unit is about the solution of quadratic equations. These take the form ax2+bx+c = 0. We will look at four methods: solution by factorisation, solution by completing the square, solution using a formula, and solution using graphs.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
In mathematics, engineering and science, formulae are used to relate physical quantities to each other. They provide rules so that if we know the values of certain quantities; we can calculate the values of others. In this video we discuss several formulae and illustrate how they are used.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd.
It is often useful to rearrange, or transpose, a formula in order to write it in a different, but equivalent form. This unit explains the procedure for doing this.
(Mathtutor Video Tutorial)
This resource is released under a Creative Commons license Attribution-Non-Commercial-No Derivative Works and the copyright is held by Skillbank Solutions Ltd. | 677.169 | 1 |
Elementary Technical Mathematics: Basic Select VersionELEMENTARY TECHNICAL MATHEMATICS helps you develop the math skills so essential to your success on the job! Ewen and Nelson show you how technical mathematics is used in such careers as industrial and construction trades, electronics, agriculture, allied health, CAD/drafting, HVAC, welding, auto diesel mechanic, aviation, and others. The authors include plenty of examples and visuals to assist you with problem solving, as well as an introduction to basic algebra and easy-to-follow instructions for using a scientific calculator. Each chapter opens with useful information about a specific technical career. | 677.169 | 1 |
Undergraduates
Welcome!
Mathematics is the language of the sciences, a cultural phenomenon with a rich historical tradition, and a model of abstract reasoning. With a vibrant community of over 750 declared majors and minors, Mathematics is also one of the more popular subjects to study at Michigan.
Undergraduate students fill Mathematics classes with approximately 13,000 enrollments each year. While many of these students enroll in Courses for Freshmen, approximately two thirds enroll in one or more of the Department's diverse collection of advanced courses.
We invite you to explore the many opportunities Michigan Mathematics affords. While commonly used links may be found below, the main exploration tool is the navigation bar on the left. If you have any additional questions about the undergraduate program, please contact us at math-undergrad-office@umich.edu. | 677.169 | 1 |
Sales Handle A no-nonsense practical guide to trigonometry, providing concise summaries, clear model examples, and plenty of practice, making this workbook the ideal complement to class study or self-study, preparation for exams or a brush-up on rusty skills. About the Book Established as a successful practical workbook series with over 30 titles... more...
Tough Test Questions? Missed Lectures? Not Enough Time? Fortunately, there's Schaum's. This all-in-one-package includes more than 600 fully solved problems, examples, and practice exercises to sharpen your problem-solving skills. Plus, you will have access to 20 detailed videos featuring Math instructors who explain how to solve the most commonly... more...
Trigonometry has always been an underappreciated branch of mathematics. It has a reputation as a dry and difficult subject, a glorified form of geometry complicated by tedious computation. In this book, Eli Maor draws on his remarkable talents as a guide to the world of numbers to dispel that view. Rejecting the usual arid descriptions of sine, cosine,... more...
This encyclopedia contains trigonometric identity proofs for some three hundred identities. The book is presented in the form of mathematical games for the reader's enjoyment and includes a concordance of trigonometric identities, enabling easy reference. Trig or Treat is a must-have for:. • every student of trigonometry, to find the proofs... more...
This volume offers a concise, highly focused review of what high school and beginning college undergraduates need to know to successfully solve the trigonometry problems they will encounter on exams. Rigorously tested examples and coherent, to-the-point explanations are presented in an accessible form and will provide valuable assistance in conquering... more...
Boost Your grades with this illustrated quick-study guide. You will use it from high school all the way to graduate school and beyond. Clear and concise explanations. Difficult concepts are explained in simple terms. Illustrated with graphs and diagrams. Search for the words or phrases. Access the guide anytime, anywhere - at home, on the train, in... more...
500 Ways to Achieve Your Best Grades We want you to succeed on your college algebra and trigonometry midterm and final exams. That's why we've selected these 500 questions to help you study more effectively, use your preparation time wisely, and get your best grades. These questions and answers are similar to the ones you'll find on a typical college... more...
This book on symmetric geometric patterns of Islamic art has educational, aesthetic, cultural and practical purposes. Its central purpose is to bring to the attention of the world in general, and the people of Islamic culture in particular, the potential of the art for providing a unified experience of science and art in the context of mathematical | 677.169 | 1 |
Advantage™ is a Unit Aware™ calculator that lets you work with feet-inch-fraction dimensional values. Unit Aware™ means that you can effortlessly compute with units (not just convert). Advantage also includes options to make the calculator work the way you want it to such as: algebraic and RPN entry modes, customizable buttons for unit entry, display precision options, and much more. Our calculators have a clean interface that's easy to read and easy to use and are available for both Windows and Windows Mobile™ Pocket PC.
KEYPADS AND ENTRY LOGIC - Algebraic and RPN entry modes (user selectable) - Allows entry of feet-inches-fractions using the decimal key. For example, to enter 4' 3-21/32", you would press [4] [.] [3] [.] [2] [1] - Allows entry of any denominator (after entering numerator, just press [.] to enter a new denominator) - Dedicated buttons for picking the units that you use most often (you can assign the units that you want to these buttons)
DISPLAY OPTIONS - Choice of display precision, including fixed decimal places or significant figures - Option to reduce inch fractions to the least common denominator or leave them unreduced (e.g., 1/2" vs 16/32") - Option to show or not show a zero before inch fractions (e.g., 5' 0-3/4" vs 5' 3/4")
ADDITIONAL FEATURES - "Paperless tape" that keeps your last 20 entries (you can also recall tape values to the main display) - Memory with 5 locations - all support store, recall, add (M+), and subtract (M-) - Advanced math functioBasic Advantage Combo Match at Super SharewareEducation / Mathematics Popular Software
Student Grade Calculator - Student Grade Calculator for Excel makes easy work of grading students. It quickly weights the students grades on a per-assignment basis.Student Grade Calculator for Excel makes easy work of grading students. It quickly weights the students grades...
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Construction Advantage for Windows - Construction Advantage is a valuable tool for construction professionals.Construction Advantage is a valuable tool for construction professionals. It develop many functions such memory with 10 locations ( all support store, recall, add (M+), and | 677.169 | 1 |
Elementary Methods in Number Theory begins with "a first course in number theory" for students with no previous knowledge of the subject. The main topics are divisibility, prime numbers, and congruences. There is also an introduction to Fourier analysis on finite abelian groups, and a discussion on the abc conjecture and its consequences in elementary... more...
In this book the author treats four fundamental and apparently simple problems. They are: the number of primes below a given limit, the ap proximate number of primes, the recognition of prime numbers and the factorization of large numbers. A chapter on the details of the distribution of the primes is included as well as a short description of a recent... more... | 677.169 | 1 |
High School: Algebra I & II
This subject resource guide is designed to help high school teachers of algebra find information that will be useful to prepare lesson plans for the classroom.
Catalog Search Strategies
Suggested below are a few search terms to help focus your research. When using these search terms, it is best to do a Subject Search to better target items that will be relevant to your search.
The following list of terms is provided to help you begin a catalog search:Algebra
Algebra exercises
Algebra Functions
Algebra Graphing methods
Algebra Outlines
Algebra Problems
Algebra Study and teaching
Algebra Study guides
Algebra syllabi
Algebra Word problems
Books
Below you will find library items of interest to teachers of Algebra I & II. You will find books of interest to students (as well as teachers) of these subjects on the Mathematics Grades 9-12 Book List.
Exemplary Practices for Secondary Math Teachers by Alfred S. Posamentier, Daniel Jaye, and Stephen Krulik, 2007. 231 pages. 372.1 P855-1 While this book is directed at the new high school math teacher the information it provides is useful to math teachers at all points in their careers. It includes chapters on effectively setting yourself up for the year, designing lessons, reasoning and problem solving, and assessment as well as other things. Also included are extensive references to other materials you will find interesting and useful.
The Colossal Book Of Short Puzzles And Problems: Combinatorics, Probability, Algebra, Geometry, Topology, Chess, Logic, Cryptarithms, Wordplay, Physics And Other Topics Of Recreational Mathematics by Martin Gardner, 2006. 494 pages. 510.76 G227-17 This book compiles puzzles from renowned puzzle guru Martin Gardner (think Scientific American) and arranges them by topic and level of difficulty. An excellent source of different problems you can challenge your students with.
Algebra Unplugged by Kenn Amdahl and Jim Loats, 1995. 258 pages. 512 A497 While this book is aimed at the student of Algebra rather than the teacher it provides an interesting narrative explanation of the subject that can provide you with ideas for ways to approach specific topics as you prepare your own lessons.
100+ Ideas for Teaching Mathematics by Mike Ollerton, 2007. 140 pages. 372.51 O49-2 2007 This slim book provides you with short, 1-page ideas designed to help you explore highly specific problems with your students. The problems are organized by subject.
Discipline Survival Guide for the Secondary Teacher by Julia G. Thompson, 2011. 348 pages. 371.5 T473 2011 This book provides you with many practical ideas and strategies for setting up and managing your high school classroom.
Teaching Outside the Box: How To Grab Your Students By Their Brains by LouAnne Johnson, 2011. 297 pages. 371.3 J67-1 2011 This book is chock full of very pragmatic and useful ideas for teachers who are teaching in crowded schools with students who range from engaged to unmotivated.
DVDs
Algebra for Students, 200. 23 minutes/disk. DVD 512 A3945 This is a 10 volume DVD series on the topic of Algebra. Topics covered include: analyzing inequalities, functions & relations, polynomials, and variables, expressions & equations. It explores the topics it presents thoroughly, making use of real-world, student friendly examples. This is a tool for you to use with students, either as a supplement in class or for those who need another look at a topic. You can accompany the DVD with online teacher guides and worksheets.
Databases
Research Library (Proquest). This database includes full text coverage of a number of journals from the fields of Mathematics (~49) and Education (~192). Use this database to keep up with your professional development, research specific questions, or browse through topical articles. Noted journals you will be able to find here include: Journal for Research in Mathematics Education, Mathematics and Computer Education, The Mathematics Teacher and Mathematics Teaching; American Teacher and Harvard Educational Review.
General OneFile (Gale). This is another general interest database that includes full text coverage of a number of journals from the fields of Mathematics and Education. Noted journals you will be able to find here include: Mathematics Magazine, Teaching Children Mathematics, The Mathematical Intelligencer and The American Journal of Mathematics; American Journal of Education, The Education Innovator, Equity & Excellence in Education and T H E Journal (Technological Horizons In Education).
LearningExpressLibrary. This online tool provides you with access to practice tests, courses, eBooks, and exercises to help you in a wide range of subjects, including Algebra. This is a good source to direct students to for additional help.
Live Homework Help. This online tool provides one-on-one tutoring with an expert, live tutor to k-12 students between the hours of 3pm-10pm. It also includes a Homework Resources section (available 24 hours a day) with lessons, worksheets, study guides, and videos. Algebra is among the subjects they are more than capable of helping a student with and the chat interface includes a white board component to make sure that that Algebra assistance is effective. This is a good source to direct students to for additional help and is a source of lesson planning resources for you.
Websites
Below you will find websites of interest to you as a teacher of algebra. In the section titled "For You" you will find those sites aimed particularly at teachers. In the section titled "For You & Your Students" you will find those sites aimed for use with or by your students.
For You: ERIC – Education Resources Information Center Here you will find an enormous collection of articles on the subject of education, many of them available to you full text. A standard search interface is provided with limiting options to refine your results. So, for example, a keyword search on algebra instruction that is limited to full text results yields more than 1300 articles.
NCTM Illuminations Here you will find the National Council of Teachers of Mathematics resources for teaching math. Included are activities, lessons, the standards, and web links.
ElibM – The Electronic Library of Mathematics This is a repository of online journals, article collections, monographs, and other electronic resources in the field of mathematics.
DOAJ – Directory of Open Access Journals Here you will find free, full text, quality controlled scientific and scholarly journals, covering all subjects and many languages. Included are 197 mathematics journals and 44 statistics journals.
For You and Your Students: Kahn Academy This site is a source of online video tutorials on many subjects. It was originally started as a math site so its collection of math tutorials is, and remains, excellent. The link provided will start you on a couple of collections of algebra tutorials. By the way, the site also offers exercises and if you set up an account and have your students set up accounts naming you as coach then you will be able to access a suite of tools you can use to track your students' progress.
Shodor Interactivate Here you will find an excellent collection of online materials for math and science education. Each activity provides information for the learner as well as for the instructor to maximize its effectiveness. Click on the algebra link to limit the item list to those that are algebra specific and explore.
Wolfram MathWorld This site is an online mathematics dictionary & encyclopedia. Your students can use it to look up specific terms or to browse articles on various math topics.
Wolfram Alpha This is a computational knowledge engine. It is a fun site for you and your students to play with and it covers many areas. In the area of algebra this engine will do things like perform calculations, graph equations, find the solution(s) of an equation, etc.
Desmos Graphing Calculator This is an excellent online graphing calculator. You can use it for basic arithmetic calculations but it really shines as a graphing tool. Well worth playing with.
Follow LAPL
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Product Description
Algebra Readiness is designed for the middle school learner and provides a smooth transition from Course 1 (Grade 6) and Course 2 (Grade 7) Math.
The Prentice-Hall mathematics series is designed to help students develop a deeper understanding of math through an emphasis on thinking, reasoning, and problem-solving. A mix of print and digital materials helps engage students with visual and dynamic activities alongside textbook instruction. This Algebra Readiness transition course covers algebraic expressions and integers, solving one-step equations and inequalities, area and volume, and linear functions.
In the "Getting Ready to Learn" portion of the textbook lesson, "Check your readiness" exercises help students see where they might need to review before the lesson. "Check skills you'll need" list out the skills used in the lesson, and new vocabulary is listed before it's introduced. Sidebar helps tell students where to go for help in the textbook if they need to review, or note when an online tutor video is available.
The lesson itself includes "quick check" problems for students to see if they understand the concept just introduced; vocabulary sidebars and features that help focus on the language of math; and multiple types of practice activities that feature new material, integrate older material, and provide challenges. A homework video tutor for every lesson is provided online. Designed to especially help students prepare for high-stakes tests like the SAT and ACT, as well as standardized tests, test-taking strategies are included in each chapter. Skills handbook, Spanish/English glossary, instant check answers, and selected answers are included in the student textbook.
The workbook provides complete daily support for the lesson, and includes a daily notetaking guide, guided problem solving exercises, and additional practice for every lesson. Vocabulary and study skills are emphasized. The Daily Notebooking Guide is designed to be used while working through the text; each section corresponds to a section in the text, with objectives clearly laid out. Perforated, newsprint-like pages, softcover.
The Teacher Center CD-ROM includes all the tools parents will need to successfully teach this course. This CD-ROM includes the TeacherExpress CD which includes a lesson planner, teacher's edition, and teaching resources, as well as the MindPoint QuizShow CD and the Presentation Express that has lesson PowerPoint Presentations.
This Middle-school kit includes:
Parent Guide Pamphlet for Homeschoolers
Algebra Readiness Textbook
Algebra Readiness Workbook
Teacher's Center CD-ROM
System Requirements for Teacher's Center CD-ROM:
Windows 2000, XP
4X CD-ROM Drive
100 MB hard drive Space
128 MB RAM (256 MB RAM or higher recommended)
Mac Users, please note: the Teacher's Resource DVD-ROM will only work on PowerPC & G4 Macs, and is not supported for new versions. | 677.169 | 1 |
...
Show More concepts first, so you'll ease into the basics. You'll gradually master functions, graphs of functions, logarithms, exponents, and more. As you progress, you'll also conquer topics such as absolute value, nonlinear inequalities, inverses, trigonometric functions, and conic sections. Clear, detailed examples make it easy to understand the material, and end-of-chapter quizzes and a final exam help reinforce key ideas. It's a no-brainer! You'll learn about: Linear questions Functions Polynomial division The rational zero theorem Logarithms Matrix arithmetic Basic trigonometry Simple enough for a beginner but challenging enough for an advanced student, Precalculus Demystified, Second Edition, Second Edition, helps you master this essential | 677.169 | 1 |
Euclid's Elements (Geometry) to your Bookmark Collection or Course ePortfolio
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This site "provides an eclectic mix of sound, science, and Incan history in order to raise students' interest in Euclidean...
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This site "provides an eclectic mix of sound, science, and Incan history in order to raise students' interest in Euclidean geometry. Visitors will find geometry problems, proofs, quizzes, puzzles, quotations, visual displays, 'scientific speculation', and moreometry Step by Step from the Land of the Incas to your Bookmark Collection or Course ePortfolio
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Lines, segments, and rays are among the most fundamental objects in geometry, as well as Euler and Simpson lines. These nine...
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Lines, segments, and rays are among the most fundamental objects in geometry, as well as Euler and Simpson lines. These nine JavaSketchpad applets can help you to understand and investigate some of the different lines that come up in geometry. Each line is defined, and its important properties are illustrated with interactive Geometry Dictionary: Lines in Geometry to your Bookmark Collection or Course ePortfolio
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This site contains several hundred articles concerned with mathematics and physics. General topics include Number Theory,...
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This site contains several hundred articles concerned with mathematics and physics. General topics include Number Theory, Combinatorics, Geometry, Algebra, Calculus & Differential Equations, Probability & Statistics, Set Theory & Foundations, Reflections on Relativity, History, and Physics. The articles under each general heading are highly varied, many are quite advanced, and there is no apparent organizational scheme. For example, under Calculus & Differential Equations there is a proof that pi is irrational, a examination of the Limit Paradox, a discussion of Ptolemy's Orbit, and an historical review of the cycloid among many other articles. Visitors can browse by topics or search by keyword. (Anyone with information on the identity of the site author please contact the MERLOT submitterPages to your Bookmark Collection or Course ePortfolio
Select this link to close drop down of your Bookmark Collection or Course ePortfolio for material MathPages
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This site contains a java applet that finds medians and area bisectors of a triangle. There is also a related discussion of...
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This site contains a java applet that finds medians and area bisectors of a triangle. There is also a related discussion of deltoids. Links on this page to other math topics include: an animated proof of the Theorem of Pythagoras and a page of formulas for the area of a triangleians and Area Bisectors of a Triangle to your Bookmark Collection or Course ePortfolio
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You can alter the geometric construction dynamically in order to test and prove (or disproved) conjectures and gain...
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You can alter the geometric construction dynamically in order to test and prove (or disproved) conjectures and gain mathematical insight that is less readily available with static drawings by hand. Requires Java Plug-in 1.3 or higher. Please be patient while the applet loads on your computer. If you are using a dial-up connection, it may take a few minutes but is well worth the waitge & d'Alembert Three Circles Theorem I with Dynamic Geometry to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material Theorem of Pythagoras Triangle Given 3 Sides to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material Triangle Given 3 Sides Whats My Angle? to your Bookmark Collection or Course ePortfolio
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This book is written for people who wish to learn MATLAB for the first time. The book is really designed for beginners and students. In addition, the book is suitable for students and researchers in various disciplines ranging from engineers and scientists to biologists and environmental scientists. One of the objectives of writing this book is to introduce MATLAB and its powerful and simple computational abilities to students in high schools In addition, the MATLAB Symbolic Math Toolbox is emphasized in this book. There are also over 230 exercises at the ends of chapters for students to practice. Detailed solutions to all the exercises are provided in the second half wanted this book as a complete n00b of programming. It was helpful, but i had to refer to other sources in order to understand some of the maths that it assumes you know. Also it isnt very good at explaining matrices at the start - again too much assumed knowledge. If you know a bit about programming/maths, this is probably perfect, but it suggests it is for use by high school students/teachers learning this stuff for the first time, and for that I would say it's not particularly useful. I like the exercises and worked examples.
Indeed, Matlab for beginners. Easily accessible, very understandable, clear examples. You can't do anything really professional with Matlab after reading the book, but at least you know enough to give it a start. Don't think the book will solve all your problems, but if you never used Matlab before, it can help you swift through the first steps. | 677.169 | 1 |
Miami Shores, FL PrecalculusNadeem A.
...More formally, discrete mathematics has been characterized as the branch of mathematics dealing with countable sets (sets that have the same cardinality as subsets of the integers, including rational numbers but not real numbers). However, there is no exact, universally agreed, definition of the | 677.169 | 1 |
About this item
Comments:ALTERNATE EDITION: Instructor Edition: Same as student edition but has free copy markings. Almost new condition. SKU:9780495553953-2-0-12 book bridges the gap between traditional and reform approaches to algebra encouraging users to see mathematics in context. It presents fewer topics in greater depth, prioritizing data analysis as a foundation for mathematical modeling, and emphasizing the verbal, numerical, graphical and symbolic representations of mathematical concepts as well as connecting mathematics to real life situations drawn from the users' majors.Lothar Redlin, Saleem Watson, Phyllis Panman James Stewart is the author of 'College Algebra: Concepts and Contexts (Includes Webassign)' with ISBN 9780495387893 and ISBN 04953878 | 677.169 | 1 |
Details about Basic Technical Mathematics with Calculus:
This is the eBook of the printed book and may not include any media, website access codes, or print supplements that may come packaged with the bound book. This tried-and-true text from Allyn Washington preserves the author's highly regarded approach to technical math, while enhancing the integration of technology. Appropriate for a two- to three- semester course, BASIC TECHNICAL MATHEMATICS WITH CALCULUS shows how algebra, trigonometry, and basic calculus are used on the job. It addresses a vast number of technical and pre-engineering fields, including computer design, electronics, solar energy, lasers fiber optics, and the environment. Known for its exceptional problem sets and applied material, the book offers practice exercises, writing exercises, word problems, and practice tests. This edition features more technical applications, over 2300 new exercises, and additional graphing calculator screens.
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Rent Basic Technical Mathematics with Calculus 10th edition today, or search our site for other textbooks by Allyn J. Washington. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Pearson. | 677.169 | 1 |
Many topics in the history of mathematics can be used both as a motication and as an introduction to concepts that seem formidable to the uninitiated. The task of conveying these ideas becaomse a lot easier (and much more successful) using Mathematica. In this paper we present the way we taught two classical topics of mathematics using Mathematica's functions and packages to enhance the student's understanding. | 677.169 | 1 |
Numerical analysis provides the theoretical foundation for the numerical algorithms we rely on to solve a multitude of computational problems in science. Based on a successful course at Oxford University, this book covers a wide range of such problems ranging from the approximation of functions and integrals to the approximate solution of algebraic, transcendental, differential and integral equations. Throughout the book, particular attention is paid to the essential qualities of a numerical algorithm - stability, accuracy, reliability and efficiency. The authors go further than simply providing recipes for solving computational problems. They carefully analyse the reasons why methods might fail to give accurate answers, or why one method might return an answer in seconds while another would take billions of years. This book is ideal as a text for students in the second year of a university mathematics course. It combines practicality regarding applications with consistently high standards of rigour. less | 677.169 | 1 |
Homework Assignments
Demonstrate
and practice integration by parts including double application.
Read p. 525-529
examples and box on p. 530.
Hand in Wed. page 531 # 11, 15-27
odd, 30, 31,33 ,35. I will grade #11,27 and 30 and spot check the rest.
Remember that this is for YOU!!!
Tuesday 5/14/13
2C
7.4
Arc Length/Surface Area
Demonstrate
and practice the formulas for arc length of a rectifiable curve and for
surface area of a solid of revolution.
Read/do examples 4,5,7
of Section 7.4. Hand in Friday #10,40. Remember that this is for YOU!!!
BRING TEXT TO CLASS for a
homework grade.
Wednesday 5/15/13
1D
AP History, Euro
Met Trip
8.2 hand-in due.
Textbook check for a
homework grade.
8.8
Improper Integrals
Define
and demonstrate all forms of improper integrals.
Read your notes. Hand in Tuesday P.
585 #18, 27, then 8. DO IN THAT ORDER. They are in increasing order of
"sophistication."
Thursday 5/16/13
1E
Great Adventure
7.5, 8.6
Work; use of integration
tables revisited.
Derive
the formula for work when force is variable. Review the use of integral
tables.
Carefully read p.
487-492. Hand-in
Wednesday
p. 493 #9,10,23.
Friday 5/17/13
Special
7.4 hand-in due.
Catch-up
Day
Complete
all objectives for the week.
Catch-up
Monday 5/20/13
1F
Calculus-Food Event
Student
presentations to illustrate calculus concepts.
Catch-up
Tuesday 5/21/13
2A
8.8 hand-in due.
Calculus-Food Event
Student
presentations to illustrate calculus concepts.
Catch-up
Wednesday 5/22/13
1B
AP English Language
7.5 hand-in due.
Textbook collection
Film: The Proof
(Andew Wiles work on Fermat's
Last Theorem)
Inspire graduating students with the true
story of contemporary mathematician Andrew Wiles.
None
Thursday
5/23/13
1C
(cont.)
(cont.)
None
Friday 5/24/13
SPECIAL D
Period 3,4 Exams
AP English Lit
No
exam. Underclassmen report to Room 302 as usual.
8.3
Trig Integrals
Work
all examples in 8.3 pages 534-39.
Read
p. 534-539 working all examples.
No
school Monday: Memorial Day
Final
Exam scheduled for Wednesday, June 5 for Period 4, 9:30. Non-seniors report to
Room 302 for a final test for Quarter 4. Topics to be announced which will include a selection of
problems related to the work assigned above.
********************************************
AP Calculus
Week 33-34
Date
Lesson
Objectives
Homework
Monday 4/29/13
1F
Give students all remaining exam materials.
Multiple choice practice.
Provide students
with more practice on the subtleties of answering multiple choice questions.
Look at materials given and determine your picks for more
free response work. Work on multiple choice materials as your time allows.
Tuesday 4/30/13
Special A
Obtain a list of student preferences for free response
review.
Multiple choice practice.
Provide students
with more practice on the subtleties of answering multiple choice questions.
Work on multiple choice materials as your time allows. Be
prepared with questions from the multiple choice problems of the recent released
exam.
Wednesday 5/1/13
12:30 dismissal B
Released AP exam
Answer questions on the released AP exam and other
multiple choice questions.
Work on multiple choice materials as your time allows.
Questions from any multiple choice materials.
Thursday 5/2/13
1C
Extra help 3p.m.
Multiple choice questions.
Answer questions on the released AP exam and other
multiple choice questions.
Be sure you have read p. 417 (Newton's Law
of Cooling) and p. 427-28 (logistic growth) before the test.
Tuesday 3/12/13
2C
AP Problem due
7.2Volume of
Solids
Disks and Washers
Use party favors to demonstrate
volumes of solids of revolution. Use y=x2[0,3] to develop the volume formula. Begin volume problems using
the same function and interval and all combinations of vertical and
horizontal axes of rotation required for the AP exam include washers.
Answer questions related to the test. Introduce formulas for integrals involving inverse trig functions and do examples. Provide opportunities to decide what to do in an integral rather than working out the entire problem.
Reinforce the formulas introduced for differentiation and integration of an exponential function. Use these in applications including max/min, area, tangent lines, differential equations and Newton's Method.
Derive differentiation and integration formulas to parallel those in 5.1,3,4 but using bases other than e. Practice the mechanics of using these formulas. Compare derivatives of exponent expressions including those with both a variable base and a variable exponent.
FOOD DRIVE DONATION.
Page 366 #1-35odd as needed; #37-47odd,51-71odd
Read 5.5. Do page 366 #78,91,93(d is important!),101-106(TTT).
Thursday 2/14/13
D Day
Valentine's Day
Extra help 3 p.m.
"The Romance of the Dot and the Line"
Complete 5.5
Begin 5.6 Inverse Trig Derivatives (time permitting)
Wish students Happy Valentine's Day with an award winning short that combines mathematics, art and language. Answer some questions related to the quiz. Discuss p. 368 #93 as "classic AP" if time allows. Review the restrictions related to inverse trig functions and using right triangles with variable sides with inverse trig functions.
FOOD DRIVE DONATION.
Study for short quiz on differentiation and integration mechanics from 5.1,2,3,5.
Friday 2/15/13
E Day
Quiz: Mechanics of Derivatives/Integrals 5.1-5-5 (partial period)
Continue 5.6
Determine students' ability to do basic differentiations and integrations related to logarithmic and exponential functions. Introduce the derivatives of the inverse trig functions without proof andprovide examples . (Proofs in next class) Do p. 368 #93 if not discussed on Thursday. | 677.169 | 1 |
concise, insightful introduction to the field of numerical linear algebra. The clarity and eloquence of the presentation make it popular with teachers and students alike. The text aims to expand the reader's view of the field and to present standard material in a novel way. All of the most important topics in the field are covered with a fresh perspective, including iterative methods for systems of equations and eigenvalue problems and the underlying principles of conditioning and stability. Presentation is in the form of 40 lectures, which each focus on one or two central ideas. The unity between topics is emphasized throughout, with no risk of getting lost in details and technicalities. The book breaks with tradition by beginning with the QR factorization - an important and fresh idea for students, and the thread that connects most of the algorithms of numerical linear algebra.
Audience: Written on the graduate or advanced undergraduate level, this book can be used widely for teaching. Professors looking for an elegant presentation of the topic will find it an excellent teaching tool for a one-semester graduate or advanced undergraduate course. A major contribution to the applied mathematics literature, most researchers in the field will consider it a necessary addition to their personal collections.
$58.56I have used Numerical Linear Algebra in my introductory graduate course and I have found it to be almost the perfect text to introduce mathematics graduate students to the subject. I like the choice of topics and the format: a sequence of lectures. Each chapter (or lecture) carefully builds upon the material presented in previous chapters, providing new concepts in a very clear manner. Exercises at the end of each chapter reinforce the concepts, and in some cases introduce new ones. …The emphasis is on the mathematics behind the algorithms, in the understanding of why the algorithms work. …The text is sprinkled with examples and explanations, which keep the student focused. --Daniel B. Szyld, Department of Mathematics, Temple University.
Just exactly what I might have expected--an absorbing look at the familiar topics through the eyes of a master expositor. I have been reading it and learning a lot. --Paul Saylor, University of Illinois at Urbana-Champaign
This is a beautifully written book which carefully brings to the reader the important issues connected with the computational issues in matrix computations. The authors show a broad knowledge of this vital area and make wonderful connections to a variety of problems of current interest. The book is like a delicate soufflé --- tasteful and very light. --Gene Golub, Stanford University.
Book Description
This is a concise, insightful introduction to the field of numerical linear algebra. The authors' clear, inviting style and evident love of the field, along with their eloquent presentation of the most fundamental ideas in numerical linear algebra, make it popular with teachers and students alike.
Top Customer Reviews
I used this text for a two-semester graduate sequence in numerical linear algebra (NLA) while I was a graduate student in the Mathematics Department at The University of Kentucky. If you do not have a substantial background in linear algebra and numerical analysis, which I did not when I first used this book, the material covered and the presentation can seem to be quite daunting. But while the presentation is very thorough, it is not unnecessarily so. After I had used this text for about three months, I grew accustomed to the very detailed nature of the writing and grateful for the sheer level of information contained in a meer 419 pages. Many introductury numerical analysis books include several chapters covering the commonly used algorithms in NLA but usually not in complete detail. While this format is friendlier to use for an overview of the "basics," in the real world, the standard ways of solving numerical systems such as GEPP, SVD, QR, Cholesky decompostions, Gauss-Siedel iterations, and other methods do not always work in a nice cookbook-like fashion. When one of these standard methods that engineers and research scientists use to solve "standard" problems fails, and it will sometimes, this book will give you a good starting point to figure out what went wrong and what alternate methods can be used to solve a linear system that is not as easy as it first appeared to be. If you are learning NLA, you are probably doing so because you either want to or have to apply it in your professional life, by which I mean your job or the job that you hope to get. In my current position, I develop and design statistical and deterministic simulators for human genetics research.Read more ›
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This book should be required reading for anyone interested in computational numerics, especially those who are starting in the field. The authors concentrate on the few fundamental topics that underlie and unite the subject. The presentation, while rigorous, is simple, clear and friendly. The authors follow a logical thread and eliminate unnecessary and disorienting aspects that plague other books on the subject. It is easy to pick up the book, read several chapters at a stretch without looking up, and come away with new insights. Unquestionably the most valuable book I have read to date on the subject.
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This book on Linear Algebra is excellent. In particular chapters seven through thirty (as far as I have read) are great for self-directed study. However, I found chapters one through six ( through Projectors) a bit terse. Therefore I would highly recommend this book for self-study ONLY IF you already have a good idea of the concept of basic linear algebra including matrix norms, the singular value decomposition, and projectors, and also the correct way to perform a proof...and by a "good idea" I mean you already know how to use these ideas in a practical way. Otherwise, you should only use this book if you have a truly good instructor to guide you through the early material. I started out taking a class using this book four years ago from a poor instructor, and I and the entire class, as far as I could tell from casual conversation, were completely lost. I dropped the class and retook it just recently with an excellent instructor. Her help and insight made a world of difference. It will also help to have a copy of "Matrix Computations" by Golub and Van Loan for reference, especially when you get to the later chapters and eigenproblems.
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I can't speak to the entire book, as I've only made significant use of the section of matrix solvers. Having said that, his explanation of Krylov methods was the most clear and well organized I've ever seen. His book is the first I've seen that so nicely ties together all such methods. It's true that his book is probably not going to be enough if you are planning to focus on this as your research topic. But for those of us who simply need to apply the field to their research, it is the best book I've found, and he goes out of his way to be helpful to the practitioner, a rare thing in a math book. (For example, he has a wonderful flowchart in Chapter 6 providing a rough guideline for selecting a linear system solver based on the properties of one's problem.)
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The chapters on numerical stability of algorithms and conditioning of numerical problems are excellent. While the focus is of course linear algebra, these principles can be readily extended to all computational mathematics. If you regularly use computational methods and have not yet studied elementary error analysis, this book may revolutionize how you perceive numerical problems.
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Face it, most math textbooks are awful: boring to read, not much insight, little more than a compendium of definitions, theorems, proofs, and examples. Trefethen and Bau is an exception to that rule. Indeed, the field of numerical linear algebra is unusual in having available several top-notch textbooks: Golub and Van Loan, Stewart's two volumes, Saad's books on iterative methods, Demmel's introduction, Watkins' undergraduate level treatment, and T&B. All of these are excellent (and any student in numerical analysis should delve into all of them) but to my tastes T&B and Stewart are the standouts for insight and simply being fun to read.
If you're a student using T&B in a course, to use it effectively you need to understand that T&B is a book to be read carefully for understanding; it's not a typical textbook suited only for "mining" for examples and solutions to homework problems. My students have sometimes complained -- accurately -- that T&B is short on details and worked examples, and many of the proofs are just sketches. But that's a feature, not a bug: you can learn much by filling in the missing steps. This is book for reading, so take the time to read it, to think about what you've read, and to fill in the gaps; it's worth it. If you need some worked examples, Watkins has them in great detail and would be a good supplement to T&B (though see the caveat below).
The only minor gripe I have about T&B is that the order of topics (QR before LU before Cholesky) is unusual, which makes it a little awkward to coordinate with other books such as Watkins which do Cholesky before LU before QR.
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Calculator use opens door to high school algebra
01/26/1998
CORVALLIS - Another study on the use of calculators in mathematics education has concluded that some of the more powerful instruments - which actually are more like a hand-held computer but still only cost around $150 - can be of significant help in learning high school algebra.
Progress in teaching this level of mathematics, which is often a major hurdle for average-ability high school students, could be a significant step forward in math education and a blessing for any freshman who's ever struggled with "y equals three x squared minus nine divided by six."
And use of such calculators in solving algebraic equations doesn't mean the students won't eventually learn how to do it themselves, educators say. It just means that at first they concentrate more on the concepts and less on the procedures.
"Based on this and other studies similar to it, there's no doubt in my mind that using calculators is a superior way to teach algebra," said Barbara Edwards, an assistant professor of mathematics at Oregon State University. "And that's important. Right now there's a widespread and valid concern that the traditional approach to math education isn't working.
"Math is often taught in a way that makes it seem boring and difficult," she said. "Our approach helps students understand how things work and how math can be applied to the real world."
The recent study was done by Edwards and other collaborators in a high school freshmen math class composed mostly of minority students in a Washington, D.C., classroom. It was part of an overall curriculum there that emphasized math concepts, not just skills.
Inexpensive, hand-held calculators have been used for some time now - and with success - in teaching arithmetic at elementary school levels, Edwards said. Other studies have shown the value of more sophisticated instruments in teaching high school and college calculus. But in between lies algebra, which for many students is the highest level of math expertise they attain, and the toughest.
Calculators, or many software programs available for personal computers, can easily solve the quadratic equations used in beginning high school algebra and even far more difficult equations that might challenge a college student, Edwards said.
"In our use of these calculators with freshmen high school students, we found they removed the grunge work and let the students concentrate on the ideas, setting up the equations," Edwards said. "That's really what math is all about, it 's not doing the things a calculator can do quicker and easier."
In one class exercise, for instance, students were asked to develop an optimal solution for running a high school talent show. Many variables, such as advertising, numbers of participants, concession sales, available audience and ticket prices came into play.
"If you charge too little for admission, you won't cover costs or make a profit," Edwards said. "Charge too much and no one will come. That's an algebraic problem and the teacher asked the students to create the best solutions, using their computers."
In the process, she said, even some students who previously had shown little interest or aptitude in algebra got excited, involved and did surprisingly well on the exercise.
Edwards said that as more research is done in this area and more schools increase their use of calculators and computers, such approaches will become far more common. Studies continue to show that using calculators really doesn't interfere with a student eventually learning more basic skills, whether it's addition tables or how to solve an algebraic equation.
"I know these approaches work," Edwards said. "I have a son who was never really drilled on his multiplication tables. He's now a professional mathematician." | 677.169 | 1 |
Details about Introduction to Geometric Probability:
The purpose of this book is to present the three basic ideas of geometrical probability, also known as integral geometry, in their natural framework. In this way, the relationship between the subject and enumerative combinatorics is more transparent, and the analogies can be more productively understood. The first of the three ideas is invariant measures on polyconvex sets. The authors then prove the fundamental lemma of integral geometry, namely the kinematic formula. Finally the analogues between invariant measures and finite partially ordered sets are investigated, yielding insights into Hecke algebras, Schubert varieties and the quantum world, as viewed by mathematicians. Geometers and combinatorialists will find this a most stimulating and fruitful story.
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Rent Introduction to Geometric Probability 1st edition today, or search our site for other textbooks by Daniel A. Klain. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Cambridge University Press.
Need help ASAP? We have you covered with 24/7 instant online tutoring. Connect with one of our Geometry tutors now. | 677.169 | 1 |
Details about Elementary Statistics: A Step By Step Approach:
ELEMENTARY STATISTICS: A STEP BY STEP APPROACH is for general beginning statistics courses with a basic algebra prerequisite. The book is non-theoretical, explaining concepts intuitively and teaching problem solving through worked examples and step-by-step instructions. This edition places more emphasis on conceptual understanding and understanding results. This edition also features increased emphasis on Excel, MINITAB, and the TI-83 Plus and TI-84 Plus graphing calculators; computing technologies commonly used in such courses.
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Rent Elementary Statistics: A Step By Step Approach 7th edition today, or search our site for other textbooks by Allan G. Bluman. Every textbook comes with a 21-day "Any Reason" guarantee. Published by McGraw-Hill Science/Engineering/Math. | 677.169 | 1 |
Careers for a Math MajorMathematics is usually divided into the broad areas of theoretical (pure) mathematics and applied mathematics, but there is a fairamount of overlap between these categories.
4.
Careers for a Math MajorTheoretical mathematicians are dedicated to advancing mathematical knowledge by developing new principles and recognizing previously unknown relationships between existing principles of mathematics.
5.
Careers for a Math Major Although theoretical mathematicians focus on increasing basic knowledge withoutnecessarily considering its practical use, it is a fact that the knowledge they obtain has played a crucial role in many scientific and engineering achievements.
6.
Careers for a Math MajorOn the other hand, applied mathematiciansare focused on solving practical problems inbusiness, government, engineering, and the physical, life, and social sciences by using mathematical theories and techniques likemathematical modeling and computational methods.
7.
Careers for a Math Major Applied mathematicians start with a real- world problem, consider the separate elements of the problem, and then reduce the elements to mathematical variables so that they can analyze relationships amongthe variables, and solve complex problems by developing mathematical models.
8.
Careers for a Math MajorTo find out more about careers for Mathmajors and the possible jobs available tothose with a degree in Math, please visit | 677.169 | 1 |
Calculus I -- Fall 2016
From Intelligent Perception
MTH 229 - Calculus with Analytic Geometry I (CT). An introduction to analytic geometry. Limits, derivatives, and integrals of the elementary functions of one variable, including the transcendental functions. (PR: MTH ACT of 27 or above, or MTH 130 and 122, or MTH 127 and 122, or MTH 132) This course meets a Core I/Critical Thinking requirement.
Contents
1 Lectures
2 Projects
The student will be given a short, one or two sentences, description of a problem that uses no calculus terminology. Example: "How should you throw a ball from the top of a 100 story building so that it hits the ground at 100 feet per second?" The student's first task is to recast the problem in the language of the appropriate mathematics. Then he is to explore the problem numerically and graphically with a spreadsheet and make an educated guess what the answer might be. Finally, the problem is solved analytically with the calculus tools and the answer is confirmed. All the steps, the explanations of the methods, the data, the illustrations, and the computations are presented in writing.
3 Description
List of topics
Brief review of basic concepts of algebra
Number systems. Distance formula. Slope of a line. Standard equations of lines.
The limit of a function at a point. One-sided limits. Continuity and the intermediate value theorem. Infinite limits. Limits at infinity. Applications of limits to engineering and science.
Differentiation and applications
Definition of the derivative at a point and on an interval. Slope of a tangent line. Derivatives of polynomials. Derivatives of trigonometric functions. Derivatives of exponential and logarithmic functions. Rules for differentiation. Mean value theorem. Implicit differentiation. Maxima and minima. Critical points and intervals of increase and decrease. Concavity and inflection points. Newton's Method. Differentials and linear approximation. Applications of derivatives to engineering and science.
Integration and applications
Area as an integral. Antiderivatives. Riemann sums. Definite integrals as limits of Riemann sums. The Fundamental Theorem of Calculus. The substitution method for integrals. Applications of integrals to engineering and science.
Learner outcomes
1. Students will be able to evaluate limits, derivatives, and integrals symbolically.
2. Students will be able to approximate limits, derivatives, and definite integrals from tabular and graphical data.
3. Students will be familiar with the definitions of limits, derivatives, and integrals; be able to apply these definitions to test properties of these concepts; and be able to produce verbal arguments and examples showing that basic properties hold or do not hold.
4. Students will be able to apply the techniques of calculus to answer questions about the analytic geometry of functions, including vertical and horizontal asymptotes, tangent lines, local extrema, and global extrema.
5. Students will be able to verbally explain the meaning of limits, derivatives, and integrals in their own words, both in general terms and in the context of specific problems.
6. Students will be able to select or construct an appropriate function to model an applied situation for which calculus is applicable, based on a verbal description of the situation.
7. Students will be able to apply techniques of calculus to solve applied problems from fields such as engineering and the sciences.
8. Students will be able to interpret symbolic and numerical results in real-world terms, and analyze the validity of their results in a real-world setting.
Course goals
To give students an understanding of the fundamental concepts of calculus and an appreciation of its many applications.
To develop critical thinking skills by asking students to convert real-world problems into forms suitable for calculus, and interpret the results of calculus in real-world terms.
To provide students with a deeper understanding of the mathematics that is used in their science and engineering courses.
To develop facility in using graphing calculators and computers to solve mathematics problems.
To satisfy program requirements.
4 Tutoring
The math tutoring lab will be open this semester during the following hours: | 677.169 | 1 |
,...
Show More, understandable language that is mathematically accurate. Dugopolski also includes a double cross-referencing system between the examples and exercise sets, so no matter where the students start, they will see the connection between the two. Finally, the author finds it important to not only provide quality but also a wide variety and quantity of exercises and applications | 677.169 | 1 |
Dave's Math Tables is an excellent mathematical resource. The mathematical reference tables include General Math, Algebra,...
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Dave's Math Tables is an excellent mathematical resource. The mathematical reference tables include General Math, Algebra, Geometry, Odds and Ends, Trigonometry, Calculus, Statistics, and Advanced Topics. In addition, the site features an interactive area for posting and answering mathematical questions and a list of related Internet resources. This site is also available in Spanish, and an English-Spanish math dictionary is provided Dave's Math Tables to your Bookmark Collection or Course ePortfolio
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This subject provides an introduction to the mechanics of materials and structures. You will be introduced to and become...
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This subject provides an introduction to the mechanics of materials and structures. You will be introduced to and become familiar with all relevant physical properties and fundamental laws governing the behavior of materials and structures and you will learn how to solve a variety of problems of interest to civil and environmental engineers. While there will be a chance for you to put your mathematical skills obtained in 18.01, 18.02, and eventually 18.03 to use in this subject, the emphasis is on the physical understanding of why a material or structure behaves the way it does in the engineering design of materials and structures050 Engineering Mechanics I to your Bookmark Collection or Course ePortfolio
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The course material emphasizes mathematical models for predicting distribution and fate of effluents discharged into lakes,...
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The course material emphasizes mathematical models for predicting distribution and fate of effluents discharged into lakes, reservoirs, rivers, estuaries, and oceans. It also focuses on formulation and structure of models as well as analytical and simple numerical solution techniques. Also discussed are the role of element cycles, such as oxygen, nitrogen, and phosphorus, as water quality indicators; offshore outfalls and diffusion; salinity intrusion in estuaries; and thermal stratification, eutrophication, and sedimentation processes in lakes and reservoirs. This course is a core requirement for the Environmental MEng77 Water Quality Control to your Bookmark Collection or Course ePortfolio
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This course introduces programming languages and techniques used by physical scientists: FORTRAN, C, C++, MATLAB®, and...
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This course introduces programming languages and techniques used by physical scientists: FORTRAN, C, C++, MATLAB®, and Mathematica. Emphasis is placed on program design, algorithm development and verification, and comparative advantages and disadvantages of 12.010 Computational Methods of Scientific Programming (MIT) to your Bookmark Collection or Course ePortfolio
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The subject introduces the principles of ocean surface waves and their interactions with ships, offshore platforms and...
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The subject introduces the principles of ocean surface waves and their interactions with ships, offshore platforms and advanced marine vehicles. Surface wave theory is developed for linear and nonlinear deterministic and random waves excited by the environment, ships, or floating structures. Following the development of the physics and mathematics of surface waves, several applications from the field of naval architecture and offshore engineering are addressed. They include the ship Kelvin wave pattern and wave resistance, the interaction of surface waves with floating bodies, the seakeeping of ships high-speed vessels and offshore platforms, the evaluation of the drift forces and other nonlinear wave effects responsible for the slow-drift responses of compliant offshore platforms and their mooring systems designed for hydrocarbon recovery from large water depths. This course was originally offered in Course 13 (Department of Ocean Engineering) as 13.022. In 2005, ocean engineering subjects became part of Course 2 (Department of Mechanical Engineering), and this course was renumbered 2.2424 Ocean Wave Interaction with Ships and Offshore Energy Systems (13.022) to your Bookmark Collection or Course ePortfolio
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This course provides an integrated introduction to electrical engineering and computer science, taught using substantial...
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This course provides an integrated introduction to electrical engineering and computer science, taught using substantial laboratory experiments with mobile robots. Our primary goal is for you to learn to appreciate and use the fundamental design principles of modularity and abstraction in a variety of contexts from electrical engineering and computer science. Our second goal is to show you that making mathematical models of real systems can help in the design and analysis of those systems. Finally, we have the more typical goals of teaching exciting and important basic material from electrical engineering and computer science, including modern software engineering, linear systems analysis, electronic circuits, and decision-making01SC Introduction to Electrical Engineering and Computer Science I (MIT) to your Bookmark Collection or Course ePortfolio
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Select this link to open drop down to add material 6.01SC Introduction to Electrical Engineering and Computer Science I441 Information Theory (MIT) to your Bookmark Collection or Course ePortfolio
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This is a free textbook offered by BookBoon.'A good knowledge of Fluid mechanics is essential for Chemical, Mechanical and...
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This is a free textbook offered by BookBoon.'A good knowledge of Fluid mechanics is essential for Chemical, Mechanical and Civil engineers. As a result it is taught at a very early stage in degree courses on those disciplines.A First Course in Fluid Mechanics covers the basics of the engineering fluid mechanics without delving into deeper more mathematical concepts.Building from most basic concepts such as physical properties of fluids, it covers the topics in fluid statics and dynamics. Hydrostatic pressure, buoyancy and forces on immersed bodies are discussed under fluid statics.Under fluid dynamics, Bernoulli's principle is introduced. Furthermore, the nature of fluid flows is discussed in engineering context. Laminar and turbulent flows in pipes are explained in detail.Finally hydraulic design is discussed paying attention to pump capacity calculations.This textbook is levelled at first year undergraduate students Fluid Mechanics for Engineers to your Bookmark Collection or Course ePortfolio
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This is a free textbook offered by BookBoon.'A First Course on Aerodynamics is designed to introduce the basics of...
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This is a free textbook offered by BookBoon.'A First Course on Aerodynamics is designed to introduce the basics of aerodynamics to the unfamiliar reader. This text avoids lengthy and complicated derivations, focusing on primary equations and fundamental concepts. Detailed figures and explanations of important mathematical equations are interspersed throughout the text. This e-book can be downloaded free of charge.The field of aerodynamics studies the motion of air around an object, such as an aircraft. After introducing fundamental concepts such as fluid flow, Thin Airfoil Theory, and Finite Wing Theory, A First Course on Aerodynamics presents the fundamentals of three key topics: Inviscid Compressible Flow, Viscous Flow, and wind tunnels. Important subtopics include one dimensional flow, quasi one dimensional flow, oblique shocks and expansion waves, boundary layer, and low speed and supersonic speed wind tunnels.Following each chapter are multiple choice questions, designed to help the reader put theoretical concepts into practice and identify equations that are vital to the continued study of aerodynamics. A list of references for further reading is included at the end of the text on Aerodynamics to your Bookmark Collection or Course ePortfolio
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Calculator Edge is an outstanding free online website for Engineers/Engineering students worldwide (translation in 12 major...
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Calculator Edge is an outstanding free online website for Engineers/Engineering students worldwide (translation in 12 major Languages) to help them solve complex equations and formula calculations with a collection of more than few hundred calculators aimed at solving complex equations and formulas found in the filed of Electrical, Mechanical, Electronics, Civil, Mettalurgy, Oil and Gas, Optical, Plastics, Ceramics if Calculator Edge to your Bookmark Collection or Course ePortfolio
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Product Description
The Developmental Mathematics workbook series covers basic mathematics through early algebra. Workbooks aren't grade-specific, but rather focus on individual skills, making it an ideal curriculum for self-paced learners at any ability level. This is a self-teaching curriculum that was specifically designed for students to read, learn, and complete themselves, cultivating independent learning skills.
Lessons begin with an explanation of the concept and example problems that are solved step-by-step. A number of practice problems are provided on the following "Applications" pages.
We've used all levels up to 11 of Developmental Mathematics after trying many other math curriculum. This style of learning is what works for my daughter who is challenged by math to a great degree. We'll likely use it for our 6 year old son. I highly recommend it for those who've "tried it all"; it's inexpensive compared to others and you're not out alot of money if it doesn't work for your child, as you can buy one level at a time. ChristianBook is one of only a few places you can find Developmental Mathematics and these prices are quite fair. We've ordered many things from ChristianBook and will continue to shop here; a very reliable store | 677.169 | 1 |
Details about Mathematics for ESL Learners:
A practical, hands-on book to help students and immigrants adjust to the life and applications of mathematics in day-to-day living in the U.S.
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Rent Mathematics for ESL Learners 1st edition today, or search our site for other textbooks by Janet C. Arrowood | 677.169 | 1 |
Description:
The primary purpose of the book is to provide a convenient source of reference to those people who are appearing for engineering entrance examinations One has to refer a good number of books to understand the basics of graphs but this book will surely reduce the number of books that each of them needs to perform his job. This book has been designed for convenience.
In JEE Maths syllabus, graphs are not explicitly mentioned but For last 15-20 years questions based on graphs are being asked in IIT JEE question papers
We have tried our level best to present the graphs in easier way and good number of problems based on basic graphs and inequalities are discussed and solved in the book. All types of transformations are also coveredEllipse-Coordinate geometry
Ellipse Coordinate geometry ebook grade 11,12, engneering, high school students
Definition, Auxiliary circle & Eccentric Angle, Parametric form of the Ellipse, Position of a Point w.r.t. Ellipse, Intersection of a Line and an Ellipse, Number of tangents from a point, Equation of the Pair of Tangents Drawn From the Point (X1, Y1), Properties of…
Adventure Church
Welcome to the official Adventure Church Android app. The app compliments many things that Adventure Church has to offer:
- Sermon notes have never been this convenient! With a single button you will access the current sermon and from there you can take notes on any note or verse and then email or save the notes when you are done
- Facebook, In…
Grade-2-EVS-Part-1
Air Around Us
Air Can Move Things
Animals that Help us
Light & Shadow
This workbook contains worksheets on Air Around Us, Air Can Move Things, Animals that Help us and Light & Shadow for Grade 2 students. There are total 18 worksheets with 170+ questions.
Pattern of questions : Multiple Choice Questions, Fill in the blanks, True and f…
NORTHGO COLLEGE
Grade-2-EVS-Part-2
Bones & Muscles
Food for Health
The Water Cycle
This workbook contains worksheets on Bones & Muscles, Food for Health and The Water Cycle for Grade 2 students. There are total 15 worksheets with 130+ questions.
Pattern of questions : Multiple Choice Questions, Fill in the blanks, True and false…
This book is from LP's Kids series. In… | 677.169 | 1 |
Courses in the Math Emporium
Developmental Math Courses (Residential)
Math 100: Fundamentals of Mathematics
A review of basic arithmetic and elementary algebra. Open to all students but required of students with low scores on Liberty University placement tests and inadequate preparation in mathematics. A grade of C or better is required in order to go on to a higher-numbered mathematics course. This course may not be used in meeting General Education requirements in mathematics. (3 credit hours)
Math 110: Intermediate Algebra
Review of exponents, polynomials, factoring, roots and radicals, graphing, rational expressions, equations and inequalities, systems of linear equations and problem solving. This course may not be used to meet the General Education requirement. (3 credit hours)
General Studies Math Courses (Residential)
MATH 115: Mathematics for Liberal Arts
Prerequisite: MATH 110, minimum grade of "C". A survey course for liberal arts majors including a review of algebra and an introduction to logic, probability and statistics, mathematical structure, problem solving, number theory, geometry and consumer applications.
Math 116: Logic and Social Reasoning
Prerequisite: Math 110, minimum grade of "C". A survey course for liberal arts majors including an introduction to logic and various financial math applications, applications of mathematics to elections, measuring political power, effective ways of sharing good and services, and apportionment of votes.
Online Course Offerings
The Distance Learning mathematics courses all use the MyMathLab computer learning system. All work (homework, quizzes, and tests) are done in the MyMathLab system. The textbook and the MyMathLab tutorials and other resources are available to help the students to learn the material and each class is proctored by a mathematics teacher who can give assistance as needed. Only students in residential MATH 100, 110, 115, 116, 121, 201 and BUSI 230 courses may use the Math Emporium.
MATH 100, 110, 115, 116, 121, 201 and BUSI 230 are currently offered through Liberty University Online. Residential students may not take MATH 100 or MATH 110 online. | 677.169 | 1 |
Are you allowed to use calculators? At the university I went to, the math department didn't allow any computation helpers (calculators, slide rules, conversion matrices) even for the most advanced math courses.
In the end they will lay their freedom at our feet and say to us, 'Make us your slaves, but feed us.'
Are you allowed to use calculators? At the university I went to, the math department didn't allow any computation helpers (calculators, slide rules, conversion matrices) even for the most advanced math courses.
Are you allowed to use calculators? At the university I went to, the math department didn't allow any computation helpers (calculators, slide rules, conversion matrices) even for the most advanced math courses.
That's messed up, my university is like Serivor's, they just don't allow graphing calculators54 | 677.169 | 1 |
Find a Bedford, TXIt is a foundation course for both higher math and science education. It is also very useful in several quantitative business and economics courses. As a math major in college and a management science PhD, I can attest to the importance of mastering basic Algebra. | 677.169 | 1 |
Mathematica
for Students Version 4.1 Japanese Edition
Mathematica for Students Japanese Edition provides language resources exclusively for Japanese-speaking
Mathematica users. Mathematica Japanese Edition includes a full Japanese user interface and online documentation for
Mathematica, as well as translations of two of the most well-known and definitive guides to
Mathematica: Getting Started with Mathematica and The Mathematica
Book, 4th ed., by Stephen Wolfram, the system's creator.
The entire Mathematica interface–including menus, palettes, dialog boxes, error and warning messages, and over a thousand pages of online
help–has been translated by technical professionals
with years of experience with the Mathematica system.
With Mathematica for Students Japanese Edition, Japanese-speaking
students can work faster and more comfortably in their own language.
Mathematica
for Students Version 4.1 Japanese Edition
was released in April 2001. This
popular student version of Mathematica is available for Microsoft Windows
and Macintosh platforms. | 677.169 | 1 |
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Comment: Very nice clean study reading copy with no underlining marks or notes. Very mild shelf ware on top corner edge.Tight binding and Among the topics covered in this wide-ranging text are: mathematics before Euclid, Euclid's Elements, non-Euclidean geometry, algebraic structure, formal axiomatics, the real numbers system, sets, logic and philosophy and more. The emphasis on axiomatic procedures provides important background for studying and applying more advanced topics, while the inclusion of the historical roots of both algebra and geometry provides essential information for prospective teachers of school mathematics. The readable style and sets of challenging exercises from the popular earlier editions have been continued and extended in the present edition, making this a very welcome and useful version of a classic treatment of the foundations of mathematics. "A truly satisfying book." — Dr. Bruce E. Meserve, Professor Emeritus, University of Vermont.
Editorial Reviews
From the Back Cover
Top Customer Reviews
Howard Eves presents this five-star story of mathematics as two intertwined threads: one describes the growing content of mathematics and the other the changing nature of mathematics. In exploring these two elements, Eves has created a great book for the layman. I find myself returning to his book again and again. My few semesters of calculus, differential equations, and other applied math failed to formally introduce me to abstract algebras, non-Euclidian geometries, projective geometry, symbolic logic, and mathematical philosophy. I generally considered algebra and geometry to be singular nouns. Howard Eves corrected my grammar. "Foundations and Fundamental Concepts" is not a traditional history of mathematics, but an investigation of the philosophical context in which new developments emerged. Eves paints a clear picture of the critical ideas and turning points in mathematics and he does so without requiring substantial mathematics by the reader. Calculus is not required. The first two chapters, titled "Mathematics Before Euclid" and "Euclid's Elements", consider the origin of mathematics and the remarkable development of the Greek axiomatic method that dominated mathematics for nearly 2000 years. In chapter three Eves introduces non-Euclidian geometry. Mathematics is transformed from an empirical method focused on describing our real, three-dimensional world to a creative endeavor that manufactures new, abstract geometries. This discussion of geometries, as opposed to geometry, continues in chapter four.Read more ›
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There are several books available on the history of mathematics. Some give an account on the development of a certain area, others focus on a group of persons and some do hardly more than story telling. I was looking for one that tells the story of the development of the main ideas and the understanding of what mathematics and science in general is (or what people thought it is and should be). Howard Eves' book is the first book I bought that gives me the answers I was looking for. Starting with pre-Euclidean fragments, going on with Euclid, Aristotle and the Pythagoreans, straight to non-Euclidean geometry it focuses on the axiomatic method of geometry. What pleased me most here is that the author really takes each epoch for serious. He quotes longer (and well chosen) passages from Euclid, Aristotle and Proclus to demonstrate their approaches. Each chapter ends with a Problems section. I was surprised to see how much these problems reveal of the epoch, its problems and thinking. The book goes on with chapters on Hilbert's Grundlagen, Algebraic Structure etc, always showing not only the substance of these periods but also the shift in the way of thinking and the development towards rigor. The last chapter is titled Logic and Philosophy. Eves divides "contemporary" philosophies of mathematics into three schools: logistic (Russel/Whitehead), intuitionist (Brouwer) and the formalist (Hilbert). The book ends with some interesting appendices on specific problems like the first propositions of Euclid, nonstandard analysis and even Gödel's incompleteness theorem. Bibliography, solutions to selected problems and an index are carefully prepared to round up an excellent book. Should you buy this book ? Yes.Read more ›
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Though originally published in 1958, Howard Eves' book was a completely new find for me. Fortunately this classic text has found extended life through Dover Publications, which is making many great older volumes available for newer generations. I am not a mathematician by vocation or training and I am usually only interested in more philosophically focused books concerning logic or meta-logical issues. But I found this book extremely enlightening, showing the interrelations of (what had previously been to my mind) unrelated historical streams of thought. In the following I will give a brief summary and point out some of, what I consider, the highlights of Eves' volume.
In the first chapter Eves gives a brief but good historical overview of mathematics in ancient civilizations. He deals with the early Egyptians, Babylonians, and of course the Greeks. This approach naturally segues into an emphasis upon Euclid and his monumental Elements. Eves pays particular attention to Euclid's methodology, the material axiomatic, discussing its origin and ensuing problems.
Other texts that I have read on the subject of mathematical logic tend to give quite a bit of time to Euclid's fifth (or parallel) postulate. Not until reading Eves' book have I understood why though. Euclid's fifth postulate has the appearance of being quite different from the first four; any non-mathematician can perceive this fact from a mere browsing of the first several postulates. Euclid needed this fifth statement for his geometry; and since he could never prove it as a theorem, he made it a postulate in his system. Eves notes that a good deal of mathematical history is devoted to this same exact project that Euclid failed to accomplish.Read more ›
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Number Systems
1
Module 1
Algebra
Since the dawn of civilization, man tried to have a count of his belongings–goods, stones, animals, trees, etc. Whenever the animals were taken out of the enclosure for grazing, for each animal taken out, a scratch was put on the ground/stone. Thus, the number of animals taken out for grazing were equal to the number of scratches. (This is analogous to tally marks used now a days). On return, for every one animal returning to the enclosure a scratch was erased. In this way, even without knowing the counting, they used to save their belongings. Slowly the civilizations advanced and first came the counting numbers (natural numbers). You will be happy and proud to know that the present system of numeration, including numerals 0, 1, 2, 3, ..., 9 and place value system were the discoveries of our ancient Indians to the world. From India, these numbers reached the reign of Arabian king Al-mansur whose wise man Al-khowarizmi, translated the works of Indian scholars and mathematicians. From Arabia, the numerals reached the western world. Therefore, these are called Hindu-Arabic Numerals. You know that Algebra is generalised form of Arithmetic in which variables are used for numbers. Aryabhatt (476 AD) and Brahmgupta (578 AD) were the first Indian Mathematicians who used variables for numbers and called them "Yavat-Tawat". They illustrated the sum, difference, product and division of expressions using variables and even found their squares, cubes, square-roots, cube-roots. Aryabhatt and Brahmgupta worked on solving linear, quadratic and indeterminate equations also. They called the method of solving indeterminate equations as "Chakrawal" and gave "Avyakat Ganit" to algebra. Bhaskaracharya and Mahaviracharya also contributed a lot to this, especially ratio and proportion and extended the works of previous Mathematicians on equations and indices and surds. The name "Beejganit" was given to Algebra by Bhaskaracharya. The credit of calling this as Algebra goes to Al-khowarizmi, the wise man of Al-Mansur. In this module, we shall study about number system, polynomials, factorisation of algebraic expressions, simplificaiton of rational expressions, solving linear and quadratic equations, indices and surds, Arithmetic and Geometric Progressions.
2
Mathematics
1
Number Systems
1.1 INTRODUCTION One of the greatest inventions in the history of civilization is the creation of NUMBERS. You can imagine the state of confusion, if you did not know about natural numbers or counting numbers. The introduction of those numbers enabled us to answer the question 'How many'? You may recall your familiarity with the concepts of natural numbers, whole numbers, integers, fractions and rational numbers. These helped us to count, to know about the number zero representing nothingness, parts of a whole, describe opposites like profit and loss, rise and fall, going towards east and west etc. But one could have easily got along without the introduction of these numbers except the natural numbers. However, the working out of problems became easier with the introduction of numbers upto rational numbers. All these numbers mentioned above are rational numbers. Recall that a natural number is a rational number, a whole number is a rational number, a fraction is also a rational number and so also is an integer. At this stage, it is pertinent to ask ourselves a question. Are all numbers rational numbers ? Is it possible to solve all mathematical and life problems with the help of rational numbers? The answer to this question is an emphatic 'NO'. For example, given a unit of length, we are not in a position to exactly measure all distances if we have with us rational numbers only. Likewise, it is not possible to find out the square root of an arbitrary natural number as a rational number, as for example, you know that 2 is not a rational number. To overcome such difficulties, it is essential to extend to the system of real numbers. In this lesson, you will be able to recall all that you may be knowing about rational numbers but also extend the same to the system of real numbers. This is a big leap forward in the study of Mathematics. 1.2 OBJECTIVES After studying this lesson, the learner will be able to :
z
illustrate the extension of system of numbers from natural numbers to real (rational and irrational) numbers. identify different types of numbers. express an integer as a rational number
z z
Number Systems
3
z
express a rational number as a terminating or non terminating but recurring decimal and vice-versa. find a rational number between two given numbers. represent a rational number on the number line cite examples of irrational numbers represent 2 , 3 , 5 on the number line
z z z z z z z
find an irrational number between two given numbers. round off a rational or irrational number to a given number of decimal places. perform the four fundamental operations of Arithmetic i.e., addition, subtraction, multiplication and division on real numbers.
1.3 EXPECTED BACKGROUND KNOWLEDGE
z z z
Concept of Natural numbers, whole numbers, fractions, integers and rational numbers. Prime and composite numbers and co-prime numbers. HCF and LCM of two or more natural numbers.
1.4 RATIONAL NUMBERS You may recall your familiarity with rational numbers. For example 1 , 2 , 3 , 1 , 4 , 4 , 0 , −9 , 22 , ... 2 5 7 4 4 are all rational numbers. 1, 2, 3, 4, ... are counting numbers or Natural Numbers. With the help of these you are able to answer the question 'How many' ? If we introduce another number 'zero' into the system of natural numbers, we get the system of Whole Numbers as 0, 1, 2, 3, 4, ... Notice that 0 is not a natural number but it is a whole number. All natural numbers are whole numbers as well. By introducing into the system of whole numbers the opposite (negatives) of natural numbers, we get the numbers ... –4, –3, –2, –1, 0, 1, 2, 3, 4, 5, ... called integers. Similarly, the numbers 1 , 2 , 6 , 5 , ... 2 3 4 7
4
Mathematics
are called fractions. You may have a look at some numbers like −2 , 5 , 6 , ... 3 −8 −21 These numbers do not fall into the categories mentioned above. If we include these numbers as well in our system of numbers, we get what is known as the system of 'Rational Numbers'. Thus,
A number of the form numbers.
p where p and q are integers and q ≠ 0 is called a rational q
Here p is called the numerator and q the denominator of the rational number. You may notice that all natural numbers, whole numbers, fractions, integers are rational numbers but a rational number may not be a natural number, a whole number, a fraction or 11 an integer. For example, the number is not a natural number. It is not a whole number. −7 11 It is neither a fraction nor an integer. But certainly is rational number. Can you justify −7 11 p this statement ? Surely, you may see that is a number of the form where p = 11, q −7 q = –7 and q ≠ 0. A rational number whose numerator and denominator are both positive integers or both negative integers is a positive rational number. Similarly, if in a rational number, one out of its numerator and the denominator is positive and the other is negative is a negative rational number. Thus, for examples 2 , −7 are positive rational numbers 3 −11 and −2 , 11 are negative rational numbers. 7 −7
0 is the zero rational number. 1
14.1 Rational Number in its Lowest Terms Consider the rational number
24 . Here its numerator and denominator are 24 and 36. But these 36 two natural numbers have a Highest Common Factor, namely 12 and so the number could as well as be written as 2 × 12 2 or simply 3 × 12 3 −12 2 is said to be rational number in the lowest terms. Similarly, the rational number , could 48 3 easily be rewritten as −1 × 12 −1 = 4 × 12 4
The numerator and denominator of a rational number in lowest terms are co-prime numbers. Note : The number
12 −12 is as well written as − . 48 48
1.5 DECIMAL REPRESENTATION OF A RATIONAL NUMBER You may be familiar with the process of representing a rational number in the form of a decimal. We illustrate this process with the help of a few examples.
6
Mathematics
Example 1.1 : Represent the numbers
1 3 ,− , in the decimal form. 4 5
Solution : By actual division, we know that (i)
1 = 0.25 4
4) 1.0 (0.25 8 20 20 0 5) 3.0 (0.6 30 0
(ii) −
3 = – 0.6 5
In the cases above, we see that in the process of repeated division by the denominator, we end up after a finite number of steps, when we get the remainder as zero. But this is not always so. To illustrate this we consider another example. Example 1.2 : Express the following rational numbers (i) −
2 3
(ii)
6 7
in the decimal form. Solution : (i) −
2 = – 0.6666 ..... 3
Please note that in this case, the process of repeated division never comes to an end. In fact at every step, we get the remainder 2. (ii)
6 = 0.857142 857142 ..... 7
In this example, we find that the remainder at each step keeps changing till we arrive at a stage that the remainder repeats and then on further division the quotient will start repeating. In examples, 1.1 and 1.2 above, we see that either the remainder will be zero after a finite number of steps or it will start repeating after a finite number of steps. In the first case we say that the decimal is terminating and in the other we say that the decimal is non-terminating but repeating. In fact we have the following important statement.
A rational number is either a terminating decimal or a non-terminating but recurring (repeating) decimal. In the following examples, we try to represent a decimal (terminating or non-terminating but repeating decimal) in the form
The examples help us to assign at a result which can be stated in the following form.
8
Mathematics
A terminating decimal or a non-terminating but recurring decimal is a rational number. Note : Recurring decimals such as 0.66..., 0.234234..., 3.142857142857... are also written as 0.6 , 0. 234 , 3.142857 . In fact a digit or a group of digits which repeat are put below a bar to indicate that these repeat again and again. CHECK YOUR PROGRESS 1.2 1. Represent the following rational numbers in the decimal form : (i)
21 40
(ii)
16 25
(iii)
13 8
(iv)
15 6
(v)
91 35
2. Represent the following rational numbers in the decimal form : (i)
5 7
1.6 RATIONAL NUMBERS BETWEEN ANY TWO RATIONAL NUMBERS Given two rational numbers, can you find a rational number between them ? To answer this question, we consider the following examples : Example 1.5 : Find a rational number between two rational numbers 1 3 + Solution : Consider the number 2 4 2
1 3 and . 2 4
Number Systems
9
2+3 5 You may simplify this to get 4 i.e., 8 2 Surely, 5>1 3 5 5 1 3 5 and > . You may verify this by computing − and − which come 8 2 4 8 8 2 4 8 5 is a rational number such that 8
out to be both positive numbers, Thus,
1 5 3 < < 2 8 4
Think for a while, what we have done. We added the two numbers and divided their sum by 2. The resulting number is between the two given numbers. It will be greater than the smaller of the two numbers and less than the larger number. Let us now consider the following question : "How many rational numbers lie between two given rational numbers ?" Can you guess the answer ? In fact you can find as many rational numbers between two given numbers as you like. Can you think of as to how you may find many rational numbers lying between two rational numbers ? Example 1.6 : Find a rational number between 0.12 and 0.13. Solution : Consider the number
012 . + 013 . 2
= 0.125 It is a number greater than 0.12 and less than 0.13. Can you find some other rational numbers between 0.12 and 0.13 ? How many such like numbers you can find ? These are some of interesting aspects of the study of rational numbers. CHECK YOUR PROGRESS 1.3 1. Find a rational number between the following rational numbers (i) 7 8 and 8 7 (ii) 2 and 3 3 1 (iii) − , − 4 3
1.7 THE NUMBER LINE Rational numbers can be represented on a line which is known as the number line in the following manner. Consider a line as shown in Fig 1.1. We fix a point O on its line. We choose a convenient unit of length and mark points on the line on both sides of O at fixed distances equal to the unit of length chosen. These points represent the number 0, 1, 2, 3, 4, .... –1, –2, –3, –4, ..... 7 on the line. We as shown in the Fig 1.1. Suppose now we have to represent the number 4 divide the distance between 1 and 2 into four equal parts and mark the point after three equal parts as shown in the figure.
Fig. 1.1
In this way we can represent any positive rational numbers by a point on the number line to the right of the point O. Similarly if we were to represent the number −
1 on the number line, we divide the line segment 2
between –1 and 0 into two equal parts and mark the point −
1 as shown in the Fig 1.1. In 2 this manner you may see that any given rational number can be represented by a point on the number line. You may also note that the number 0 is represented by the point O on the number line.
Example 1.7 : Represent the number 1.2 on the number line. Solution :
Fig. 1.2
Consider the number line with points on it marked as –2, –1, 0, 1, 2 as shown in Fig. 1.2 above. Divide the line segment between 1 and 2 into 10 equal parts. Mark a point P as 1.2, as is shown
Number Systems
11
in the Fig, 1.2, 2 steps ahead of the point representing the number 1. P is the point representing the number 1.2 on the number line. You may note that any rational number whether in the form be represented by a point on the number line. CHECK YOUR PROGRESS 1.4 1. Represent the following numbers by points on the number line (i)
3 2
p or in the decimal form can q
(ii) −
1 3
(iii) 1.5
(iv) – 1.3
2. Find the numbers corresponding to the points O, P, Q and R on the number line as shown in the Fig. 1.3.
Fig. 1.3
1.8 IRRATIONAL NUMBERS In section 1.5, you have seen that when a rational number is represented as a decimal, then either this decimal is terminating or it is a non-terminating but repeating decimal. The question that arises is this. Are there decimals which are neither terminating nor non-terminating but repeating decimals ? The answer to this question is 'Yes'. Consider for example the decimal 0.101001000100001 ... This decimal has been written in such a manner that it has a definite pattern and so you can keep on writing it indefinitely. But this pattern is such that no block of digits repeats again and again. It is an example of a non-terminating and non-repeating decimal. Similarly, you may consider the decimal 0.123456789 10 11 12 13 ... Can you write the next nine digits in the decimal ? Note that, all that has been done is that we have written successively all natural numbers in the ascending order. The next six digits will be 14 15 16. Note once again you can continue this process endlessly. Thus, there is a definite pattern to write the digits but no block of digits is repeating again and again. This is another example of a non-terminating and non-repeating decimal.
12
Mathematics
The two examples we have discussed above indicate that there are decimals which are not rational numbers. Why ? We conclude that if we have to possess in our fold all decimals, namely. (i) terminating decimals (ii) non-terminating but repeating decimals (iii) non-terminating and non-repeating decimals then the system of rational numbers is not adequate and so we must extend this system to include numbers which are not rational i.e., irrational numbers.
p Note that the name rational is derived from the word 'ratio', as a number q , q ≠ 0 is the ratio
of two integers p and q. Also number which is not a ratio of integers p and q, q ≠ 0 is named as an irrational number. 1.8.1 Inadequacy of rational numbers You may recall that we would not have been able to answer the question 'how many' ?, if we did not know counting numbers or natural numbers. Let us now try to examine what we shall not be able to do if we have in our possession the rational numbers only. We put to ourselves the following question. Can we measure the length of any given line segment in terms of a prescribed unit of length, with the help of rational numbers ? The answer to this question is an emphatic 'No', as is clear from the following example. Consider a square ABCD of each side of unit length. The diagonal AC of this square has its length 2 units. You also know that 2 is not a rational number as there is no rational number whose square is 2.
Fig. 1.4
We conclude that we cannot exactly measure the length of the line segment AC in terms of rational numbers, if the unit of length given to us is AB. It is this inadequacy which necessitates the extension of the system of rational numbers to a system which includes in it rational as well as irrational numbers. There is yet another point which necessitates the extension of rational numbers. We discuss the same in the following paragraph. Recall that you have seen in section 1.7, that to each rational number these corresponds a unique point on the number line. Let us now consider the converse of it. Given a point on the number line, will it always correspond to a rational number ? The answer to this question is also a 'NO', as is clear from the following example.
1. there is a point namely P. In fact.5
Construct a square on the line segment. The diagonal of the square has length as centre and radius OA. again an irrational number. the
point P on the number line corresponds to the number 2 which is irrational.
p 1 2 3 you can get points on the line corresponding to number like 2 2 .6
Similarly. there are infinitely many points on the number line which do not correspond to rational numbers. 3 3 .
Fig. q 7 . etc. on the number line which does not correspond to a rational number. − 4 5 .
Fig.Number Systems
13
On the number line mark the points corresponding to the number 0 and 1. 1.
. The diagonal of the rectangle will be OQ = 3 . Thus. How many such points are those on the line which do not correspond to rational numbers ? In fact there are many such points and their number is infinite i. 1. With O as centre and OQ as radius draw an arc cutting the number line in the point R. This point P
5 . if P corresponds to the number
2 as in Fig 1. 5 etc. Then the point R corresponds to the irrational number
3. Thus.e.7
Once you have the points on the number line corresponding to irrational number
2 .5. draw an arc cutting the number line in P. by constructing the rectangle with sides 2 and 1 as shown in Fig 1. draw an arc cutting the number line in P. as shown in Fig 1. Then OP =
2 .
Fig.6. 3 . With O 2 . we construct a rectangle with one
side OP and the other side 1.7 and with O as centre and diagonal OA = corresponds to
5 .
2 is
. All these points correspond to irrational numbers. We may therefore say that the rational line is not complete whereas the real line is complete.8 : Find an irrational number between the numbers 1 and 2. This discussion leads us to the following conclusion.. Write down the next three digits in the following numbers : (i) 0..9 IRRATIONAL NUMBER BETWEEN TWO GIVEN NUMBERS Recall that you had found rational numbers between two given rational numbers. 2. It is the square root of the product of the two numbers 1 and 2.14
Mathematics
p where q is any rational numbers. Represent the following numbers on the number line : (i)
1 2 3
(ii) 1 + 2
(iii) − 3
(iv) − 2 + 3
1. We have thus extended the system of rational numbers to include in it all irrational numbers as well.
The number line consisting of points corresponding to rational numbers has gaps on it. Example 1. This number line is called the real line and the one corresponding to rational numbers is known as the rational line. In other words it is not complete.35 .e..
1 × 2 i. You may refer to Fig 1.246 ..6912 . (iv) 0. The system of numbers consisting of all rational and irrational numbers is called the system of real numbers or the Real Number System. Solution : Consider the number 1 × 2 . But the number line consisting of points corresponding to all rational and irrational numbers is without any gaps and therefore is complete. (ii) 1.11011110 . In the following examples we illustrate how do we find irrational numbers between any two numbers... (iii) 3.
CHECK YOUR PROGRESS 1. So an irrational number lying between the number 1 and 2.5 and see that the point P corresponding to 1 and 2 and so
2 lies between the points
2 is such that 2 is greater than 1 and less than 2.5 1. This system is called the Real Number System....
178.6 1. 2
∴ The required number is
CHECK YOUR PROGRESS 1. 1. Solution : We look up the 4th place after the decimal point.
3+ 2 . it is less than 3. We illustrate this process with the help of the following examples.178473. In this case. 2
Also. Example 1. you can verify than 3 is greater than the number
3+ 2 . it is 4 which is less than 5. So The approximate value of 2.
3− 2 is a positive real number 2
3+ 2 is greater than 2
∴
2.9 : Find an irrational number lying between the number
3+ 2 2
2 and 3.Number Systems
15
Example 1.
. upto three places of decimal is 2.10 : Express the number 2.
Solution : Consider the number
We have As ∴
3+ 2 3+ 2 − 2 2 3− 2 − 2 = = 2 2 2
2 is less than 2.10 ROUNDING OFF NUMBERS TO A GIVEN NUMBER OF DECIMAL PLACES Quite often it is convenient to write the approximate value of a real number to a specified number of decimal places.178473 approximately by rounding it off to three places of decimal. How many irrational numbers are there between 1 and 2 ? Give three examples of irrational numbers between these two numbers. Find an irrational number between the following pairs of numbers : (i) 2 and 3 (iii) (ii)
3 and 2 2 and 8
2 and
3
(iv)
2.
in the form q
(i) 0.. we observe the next digit in the decimal part of number.... So.e.. we observe that.34567 . Write down the approximate value of the numbers.11 : Find the approximate value of 2.. (iii) 1. (iii) 0.. With the help of real numbers..14159 .3125 (ii) 0..1111 .142857 LET US SUM UP
z z
(ii) 7.7 1. we add 1 to the preceding digit to get the required number. there are no gaps on it.1785 Thus. when represented as a decimal is either terminating or non-terminating but repeating decimal.9999 .. TERMINAL EXERCISE
z z z
z z
p 1. (ii) If the digit is 5 or more than 5. correct upto 4 places of decimal. we can measure exactly any length in terms of a unit of length.. correct upto 4 places of decimals (i) 0. Solution : The fifth place of decimal (one after the fourth) is 7.. ∴ The required approximate value of the number is 2.16
Mathematics
Example 1. there lie an infinite number of rational as well as irrational numbers. An irrational number is a non-terminating and non-repeating decimal. (iv) 12. we add to 4 the fourth place 1. (iv) 3.
CHECK YOUR PROGRESS 1.
. There are gaps on the rational line.7777 ..325444 .... because 7 is greater than 5.01101 . Express the decimal. we ignore it and give the answer. A rational number. The real line is complete i. to round off a number to some decimal places.178473. (iv) 3.. Between any two given numbers.
The Real Number System consists of rational and irrational numbers. (i) If the digit is less than 5.
seven is being multiplied by itself 4 times.1 INTRODUCTION You have learnt.
2. In 5 × 5 × 5. Isn't it difficult to write 15 × 15 × . simplify expressions involving exponents using laws of exponents. 2. five is being multiplied by itself three times. We shall explain its meaning. express a natural number as a product of powers of prime numbers uniquely.. In 7 × 7 × 7 × 7.2 OBJECTIVES After studying this lesson. identify the base and the exponent of a number in the exponential notation. 20 times? In this lesson we shall try to overcome this difficulty by the introduction of Exponential notation. We shall learn the application of laws of exponents.. In the end of the lesson we shall also learn how to express very large and very small numbers using scientific notation. Think of the problem when 11 is multiplied by itself 10 times or 15 is multiplied by itself 20 times. explain the meaning of a0. state the laws of exponents.20
Mathematics
2
Indices (Exponents)
2. We shall express real numbers as product of powers of prime numbers.
p q
z z z
express very large and very small numbers using scientific notation.3 EXPECTED BACKGROUND KNOWLEDGE
z z z
Multiplication of numbers Division of numbers Prime Numbers
. the learner will be able to :
z z z z
write a repeated multiplication in exponential notation and vice-versa. You can easily write the product 5 × 5 × 5 = 125 or 7 × 7 × 7 × 7 = 2401. a–m and a . state and derive laws of exponents. how to multiply two or more real numbers.
. 6 is the base and 5 is the exponent. in (iii) multiply 4 by itself 6 times. 11 times as 811 and (–3) × (–3) × (–3) × . the number 5 is called the base and 2 is called the exponent or index. We write 5 × 5 as 52. The notation for writing the product of a number by itself several times is called the exponential notation. Similarly 34 is read as 'three raised to the power 4. yes. 52 is read as 'five raised to the power 2' or 'second power of five' In 52. Again in 46.. and 6 × 6 × 6 × 6 × 6 as 65. Similarly (–2) × (–2) × (–2) × (–2) = (–2)4 where –2 is the base and 4 is the exponent. and in (iv) multiply 6 by itself 5 times. You can now write 8 × 8 × 8 × . or 'fourth power of three'.. 4 × 4 × 4 × 4 × 4 × 4 as 46.
2. Write the base and exponent of (–3)7 Base : Exponent :
. In 811. 3 × 3 × 3 × 3 as 34.Indices (Exponents)
21
z z
Rational Numbers Four fundamental operations on rational numbers. 4 is the base and 6 is the exponent and in 65. 7 times as (–3)7. the base is 8 and exponent is 11. in (ii) multiply 3 by itself 4 times.4 EXPONENTIAL NOTATION Let us consider the following products : (i) 5 × 5 (ii) 3 × 3 × 3 × 3 (iii) 4 × 4 × 4 × 4 × 4 × 4 (iv) 6 × 6 × 6 × 6 × 6 In (i) we have to multiply 5 by itself 2 times. Can you identify base and exponent in 811 ? Of course. 3 is the base and 4 is the exponent..
e. Similarly. We write it as 3–2 and read it as '3 raised to the power 32
b−5g
is
1
4
.7 NEGATIVE INTEGERS AS EXPONENTS In the previous sections.
1 is written as a–m and is read as 'a raised to the power (–m)'. We know that : The reciprocal of 3 is 1 . we have been taking non-negative integers as exponents. am 1 = a–m am
. We write it as (–5)–4 and read it as '–5 raised to the power
b g
1 .
4 7 and − are 1 and 1 respectively. We write them as 5 9 45 −7 9 −1 −1 4 7 7 4 and − and read them as ' raised to the power (–1)' and ' − raised to the 9 5 9 5 power (–1)' respectively. We write it as (–4)–1 and read it as '–4 raised to the power −4
The reciprocal of –4 is (–1)'. the reciprocals of
FH IK
FH IK
Similarly reciprocal of 32 is (–2)'. From the above we get.. then the reciprocal of am i. The reciprocal of (–5)4 is (–4)'. We write it as 7 5 18 7 18
FH IK
−5
and read it as '
7 raised to the 18
If a is any non-zero rational number and m is any positive integer. Thus. The reciprocal of power (–5)'.32
Mathematics
(e)
FH 8 I 3K
0
=0
4 0
(f) 8I = F H 25 K
4
FH − 4 I 5K
2
16 = − 25
(g)
8 I ×F 7 I FH 25 K H 19 K
2. Now we shall try to assign meaning to negative integers as exponents. 3 1 . We write it as 3–1 and read it as 3 raised to the power (–1). 7I FH 18 K
5
1 .
e.. Let us consider an example to illustrate this point..
Thus.. If a is a positive rational number.. then
a
Now
p q
=
q
ap
a × a × a × ... 4 3 means cube root of 42. Let us multiply a
1 1 q. We state that a q is the qth root of a and is written as
q
a . p is an integer and q is a natural number. We know that am × an = am+n for all integral values of m and n. a q means the qth root of ap. We observe that if the exponent is a rational number.36
Mathematics
2..9 MEANING OF a . how should we define a q ... q factors = a
∴
p
p q
p q
p q
p p p q + q + q . qth power of a q = a..
2
. Thus. 3 ×3 ×3 ×3 ×3
1 5 1 5 1 5 1 5 1 5
=
3
1+1+1+ 1+ 1 5 5 5 5 5 1 5
=3
5 5
= 31 = 3
or
3
1 5
is the fifth root of 3 which is written as
3
=
5
3
Now we define a rational power of a.. × a q = aq = a aq q q = q times
1 1
q
In other words.
1 + 1 + 1 + . its numerator denotes the index and the denominator denotes the root.
1 1 1
i. q times p q
=a
pq q
=a
p
a
= q ap . q times a q × a q × a q × . If a is a positive real number. and q is a natural number. 1
p q
q times...
where A is terminating decimal lying between 1 and 10 i. A × 10n is in scientific notation.e.0 = 1. 10000 1 = 3.7 = 1. Let us write 0.7 × 10 = 1.1 × 10–5.7 × 101 1.38
Mathematics
2. multiplied by an integral power of 10.7 × 101 170 = 1. (ii) 170 17 = 1.7 ×
1 = 1.7 × 100 0. 105
(iii) 1700
(iv) 17000 ?
0.7 to get (i) 17 Clearly.17 = 1. Sometimes it is also referred to as writing a number in standard form.7 × 102 1700 = 1. Now we shall see how to write a number less than 1 in the scientific notation. A number is said to be in scientific notation. in (i) (ii) (iii) (iv) Similarly.7 ×
Similarly.7 × 10–1 10 1 = 1.7 × 103 17000 = 1.0017 = 1.7 × 10 5 1700000 = 1.7 × 104 170000 = 1.017 = 1.7 ×
0. In symbols. 1 < A < 10 and n is an integer. 17.1 ×
.7 × 10–2 100 1 = 1.7 × 106 and so on.00017 in the scientific notation. when it is expressed as a number between 1 and 10.00017 = 1.
0.7 × 10–3 1000 1 = 1.7 × 10–4 and so on. Scientists and Engineers often find it convenient to write numbers in this form.7 × 1 = 1.000031 = 3.10 SCIENTIFIC NOTATION Let us consider the following problem : Can you calculate what powers of 10 should be multiplied to 1.7 ×
0.
57 (b) 143000 (d) 0. Express each of the following numbers in scientific notation : (a) 0.43000 × 105 = 1. Write it in scientific notation.6 1. 4. Express it in scientific notation. Example 2.00000567 (b) 0. Express it in the scientific notation.00031 = 3.18 : It is said that the distance of earth from the sun is 149000000 km. Solution : The distance of earth from the sun = 149000000 km = 1.49 × 108 km CHECK YOUR PROGRESS 2. The diameter of the earth is said to be 13000 km.1 × 10–4 Example 2.
.43 × 105 (c) 0. (b) The exponent of 10 is a positive integer equal to the number of places.7 × 10–1 (d) 0. we observe that a number written in scientific notation has the following characteristics : (a) There is only one digit to the left of decimal point. the decimal point has been moved when the given number is greater than or equal to 10 (c) The exponent of 10 is zero when the given number is greater than or equal to 1 but less than 10 (d) The power of 10 is a negative integer equal to the number of places the decimal point has been moved when the given number is less than 1.Indices (Exponents)
39
From the above. Express each of the following numbers in scientific notation (a) 15 lakh (b) 72 × 106 (b) 3730000 (d) 317 × 104
Solution : (a) Seven thousand = 7000 = 7.00079 (c) 0.0 × 103
2. It is said that distance of moon from the earth is 380000 km.00031 (b) 143000 = 1.11 (d) 33 × 10–5
3.17 : Write each of the following in scientific notation : (a) Seven thousand (b) 0.57 = 5.
In the last lesson we gave meaning to the number a q as the qth root of a.1 INTRODUCTION In the last lesson. In this lesson we shall call q a or n x a radical. you have learnt the laws of exponents for positive integral indices and negative integral indices. We shall find the rationalising factor of a radical and rationalise the denominator of a radical and also simplify expressions involving radicals. We shall also discuss the laws of radicals. In this lesson we shall study about a special type of numbers a which are irrational numbers. We shall find the simplest (lowest) form of a radical.
. q or n as index and a or x as the radicand.44
Mathematics
3
Radicals (Surds)
2.2 OBJECTIVES After studying this lesson. You have also learnt the meaning of numbers of the type a . the learner will be able to :
z z z z z z z z z z z
p q
p q
1
identify radicals from a given set of irrational numbers identify index and radicand of a surd state the laws of radicals (surds) express a given surd in the simplest form classify similar and non-similar surds reduce surds of different orders to those of the same order perform the four fundamental operations on surds arrange the given surds in ascending/descending order of magnitude find a rationalising factor of a given surd rationalise the denominator of a given surd simplify expressions involving surds. We shall also name this as surd. 3.
3 . In 7 ..Radicals (Surds)
45
3. 5 . in the 3 5 .
4
32 and
4
50 are all surds. π are not surds. When the order is not mentioned it is taken as 2. Similarly 5 + 3 .5 PURE AND MIXED SURDS A surd which has unity as its rational factor. 3. For example. A surd having rational factor other than unity along with irrational factor is called a mixed surd.
Again 2 + 2 . is not a surd because it is the square not of an irrational number. 3 3 5 and 4 7 are mixed surds. where it is not possible to find exactly the nth root of x. Similarly.
4
4 For example 2 3 .. as the radicands are not rational numbers. then if a is irrational. A surd is defined as a positive irrational number of the type n x . but
3
8 is not a surd as its value 2 is rational. that surd is an irrational number in which the radicand is a positive rational number. If n is a positive integer and a be real number. type of the radicand should be kept unchanged.4 SURD You have already studied in lesson 1 that numbers of the type 2 . then also n a is not a surd. Thus a number
n
x is a surd if and only if
(a) it is an irrational number (b) it is a root of positive rational number In the surd n x . 5 It may be noted that in the cases of conversion. We may repeat. order relation in numbers Laws of exponents Meaning of a0.
6
8. other factor being irrational is called a pure surd. where x is a positive rational number.
27 . order of the surd is 3 and 5 is the radicand.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z
Four fundamental operations on numbers Prime numbers. the order is 2 and the radicand is 7. a–m and a .
. the symbol is called the radical sign. Again if n a is rational. . Now we shall study irrational numbers of a particular type called radicals or surds. though it is an irrational number. 5 112 and 3 50 are pure surds. For example
n
a is not a surd.
p q
3. Definition. The index n is called the order of the surd and x the radicand. are irrational numbers. Note.
6 ORDER OF SURDS In the surd 7 3 2 . 3 is order of the surd and 2 is radicand. ∴
3
128 is a surd.
∴ (d)
3
81 is not a surd
3
128 =
4×4×4×2
= 4 3 2 .
2 × 5×5 = 5 2
50 =
∴
50 is an irrational number
Hence
50 is a surd. (i)
21
(ii)
3
18
(iii) 3 125
(iv) 2 7 141
(v) 5 5 1125
Solution : Pure surds (i) and (ii). which is an irrational number. which is a rational number. (iv) and (v)
. 25 is not a surd.2 : Identify index and radicand of each of the following surds : (a) (c)
5 6
17 123
(b) (d)
82
11
517
Solution : (a) Here index is 5 and radicand is 17 (b) Here index is 2 and radicand is 82 (c) Here index is 6 and radicand is 123 (d) Here index is 11 and radicand is 517. Example 3. mixed surds (iii).
Example 3.1 : State which of the following are surds and which are not : (a) (c)
4
25 81
(b) (d)
3
50 128
Solution : ∴ (b)
25 = 5 which is a rational number.46
Mathematics
3. When there is no coefficient in a surd.
(c)
4
81 =
4
4
3 × 3 × 3 × 3 = 3.3 : Identify the pure and mixed surds from the following. 7 is called coefficient of the surd. Example 3. it is assumed that coefficient is 1.
we proceed as follows :
The order of the surds are 2 and 3. LCM of 2 and 3 is 6 ∴ We shall change both the surds to surds of order 6.12 MULTIPLICATION AND DIVISION IN SURDS In the last section. we change them to the surds of the same order. Similarly two surds can be multiplied or divided if they are of the same order. For example
3 × 2 = 3× 2 = 6 12 = 6 2
Q 3 and 2 both are surds of order 2
and
12 ÷ 2 =
Q 12 and 2 are surds of order 2
But if we have to multiply
3 by
3
2 .10 : (a) Multiply 5 3 16 and 11 3 40 (b) Multiply 5 3 16 and 5 4 3 Solution : (a) 5 3 16 × 11 3 40 = 5 × 11 × 2 3 2 × 2 3 5 = 55 × 2 × 2 × 3 2 × 3 5 = 220 3 2 × 5 = 220 3 10
. before multiplying or dividing. ∴ and ∴
3
3 =
6 6
33 = 6 27 22 = 6 4
2 =
3×3 2 =
=
6 6
6
27 × 6 4 27 × 4 =
27 = 6 27 4 4
6
108
and
3
3 = 2
6
Example 3.54
Mathematics
3. we can perform the operation directly. Thus. However if the given surds are of the same order. you have seen that operation of addition and subtraction can be performed when the surds are similar surds. You have also studied that order of a surd can be changed by multiplying or dividing the index of the surd and index of the radicand by the same positive number.
Arrange in descending order
3
2. Their sum and product are always rational 2. In such cases each surd is called a rationalising factor the other.
3
4
3. we search a factor which when multiplied by the given surd gives us a rational number. Thus (i) (ii) (iii) and (iv)
3
3 is a rationalising factor of 3 is a rationalising factor of
3
3 9 and vice versa
11 7 5
11 4 5
is a rationalising factor of
4
and vice versa
4
7 is a rationalising factor of
7 3 and vice versa. 31 2 = 3 (b) 32 3 . The process of converting the surds to rational numbers is called rationalisation. For example. Rationalisation is usually used to rationalise the denominator of a rational expression involving irrational surds. 71 4 = 7 We observe that on multiplying the two surds we get the result as a rational number.Radicals (Surds)
57
6.
6
3. Which is smaller :
5
10 or
4
9 ?
7. 31 3 = 3 (c) 57 11 .14 RATIONALISATION OF SURDS Consider the following products (a) 31 2 .
.
Thus we see here. Arrange in ascending order
3
2. how to multiply a surd by another surd in such a way that the product is a rational number. The quantities x − y and x + y are called conjugate surds.
4
3. 54 11 = 5 (d) 7 3 4 .
3
4
8. the rationalising factor of
x is
x and rationalising factor of
3 + 2 is
3− 2
Note : 1. Thus to rationalise.
b > 0)
n
a
n
n
=a
n
a
b = n ab
a na = n b b
z
n
Operations on surds
x1 n × y1 n = xy x1 n x = 1 n y y
b g
1n
FG IJ HK
1n
dx i
dx i
m
z z
1n 1m
= x1 mn = x1 m
d i
1n
m 1n
= xm n
x a = mn x an or x a
d i
1m
= xa
m
= x an mn = x an
d i
1 mn
Surds are similar if they have the same irrational factor. The order of
n
x is n
Laws of radicals (a > 0. the radical has the smallest possible index with no fraction under the radical sign and no factors of the form an under the radical sign Similar surds can be added and subtracted Orders of surds can be changed by multiplying index of the surd and index of the radicand but the same number Surds of the same order are multiplied and divided. we change the surds to surds of the same order. Then they can be compared by their radicands
z z
z z
z z
x + y is called rationalising factor of x − y and vice versa.62
Mathematics
z z z z
A surd having factor other than unity along with irrational factor is called a mixed surd The order of the surd is the number that indicates the root.
If the product of two surds is rational. In the simplest form of a surd.
. then each is called rationalising factor of the other. To compare surds.
2 OBJECTIVES After studying this lesson. p represents an unknown quantity. z. Variables are generally denoted by alphabets x. you will have to pay Rs 10 × 2 or Rs 20 three pens. p. you will have to pay Rs 10 × p or Rs 10p. c. Thus. then for one pen. In this lesson. The branch of mathematics which deals with variables and four fundamental operations on them is called Algebra.66
Mathematics
4
Algebraic Expressions and Polynomials
4. you will have to pay Rs 10 two pens. Here p can take different values from one onwards. the learner will be to able to :
z z z z z
identify a variable cite examples of algebraic expressions classify algebraic expressions into monomials. takes different values which you assign to it and behaves like that (those) numbers (s). y. b.1 INTRODUCTION If you go to the market to purchase pens for your class and suppose each pen is available for Rs 10. binomials and trinomials perform four fundamental operations on algebraic expressions evaluate an algebraic expression for given values of the variables
. q etc. called a variable. Therefore. a variable p or x. you will have to pay Rs 10 × 3 or Rs 30 In general for p pens. a. you will learn about some basic concepts of algebra. algebraic expressions and polynomials and four fundamental operations on them. 4.
15 . 15
3x . x. y and are real numbers 3x . 17
You can see that p. z . –14 etc. y and z respectively. 4a . 8
z
21y 3 4 .
Thus. − 14 . x. 2 .
3 . 2
− 4 .
. subtraction. − 3z .4 IDENTIFYING A VARIABLE Consider numbers of the type
4 . 3z + 6 . say x state the degree of a polynomial perform four fundamental operations on polynomials
4.Algebraic Expressions and Polynomials
67
z z z z z
understand and identify a polynomial as a special case of an algebraic expression write examples of polynomials in one and two variables define a polynomial in one variable. Can you pick up variables from the following ? 13 . . a and b are variables and all others are constants. − 2 15 8 z respectively and therefore do not have a fixed value like 4. as their values do not change. etc. 2x + y 4 The expressions of the above type. − 14 . y – 3 . which involve variables and constants connected by operations of addition.3 EXPECTED BACKGROUND KNOWLEDGE
z
Different number systems and four fundamental operations on them
4.5 ALGEBRAIC EXPRESSIONS Consider expressions of the form 3 . An algebraic expression is a number or a combination of numbers including variables joined by the four fundamental operations. 7 b . 4. multiplication and division are called Algebraic Expressions.
21y . − 14 . z contain unknown x.
2 . A variable is a number which can have different values whereas a constant has a fixed value. x + 4 . x . y. y and z are called variables. z. p 4 4 y . You know that 4 . Their values will depend on x.
Identify variables and constants in the following : 2x. x and 3y. namely. (ii) 2x+y 3
(iv) x2 + 2xy + y2
. Binomial : An algebraic expression which has two terms is called a binomial. x – 3y + 3 has three terms. in the following algebraic expressions : (i) 3x2 – 4y 5 (iii) 3x + 4 y − z 2 x 8 (v) 7 − y 4. namely 3x. namely. x – 3y has two terms. For example. there are two terms 4x and 3y and the coefficient of x in 4x is 4 and coefficient of y in 3y is 3. Coefficient : The constant in a term is called the coefficient of the variable. Similarly 4x2 + 3 has two terms 4x2 and 3 and x has only one term. Monomial : An algebraic expression which has only one term is called a monomial. For example. 3 3 3 3
CHECK YOUR PROGRESS 4. 3
Thus a term contains either a variable or a constant or both variables(s) and constant connected by the operation of multiplication. 2x – 3y + z is a trinomial.6 TYPES OF ALGEBRAIC EXPRESSIONS 1. 41.1 1. Trinomial : An algebraic expression which has three terms is called a trinomial. the coefficient of x2 in x is because x = 1 × x 2 . 3x has only one term. Thus. For example 4x – 3y is a binomial : 3. each of which is called a term. 4y. 3t. 3y and 3. x. 4x is a monomial. For example. in 4x + 3y. Write the terms and the coefficients of the variables in the terms. 2.68
Mathematics
Term : When one or more of the symbols + or – occur in an algebraic expression.
2 2 1 Similarly. 17 2. they separate the algebraic expression into parts.
2x ÷ 2x = 1
x+1 ∴ 2x + 3) 2x2 + 5x + 3( 2x2 + 3x – – ––––––––– 2x + 3
Step 4 : Multiply the divisor by the second term of the quotient and subtract the result from the resultant dividend of Step 3.e.74
Mathematics
Example 4.8.
x+1 2x + 3) 2x2 + 5x + 3( 2x2 + 3x – – 2x + 3 2x + 3 – – 0
Step 5 : Repeat this process i. Step 3 and 4 till you get the remainder zero or the highest exponent of the remainder is less than that of the highest exponent in the divisor in the above example. Remainder = zero.
(2x + 3)x = 2x2 + 3x x 2 2x + 3) 2x + 5x + 3( 2x2 + 3x – – 2x + 3
Step 3 : Divide the first term of the resultant dividend obtained in Step 2 by the first term of the divisor and write the result as the second term of the quotient. Quotient = x + 1
. Divide 2x2 + 5x + 3 by 2x + 3 Step 1 : Divide the first term of the dividend by the first term of the divisor and write the result as the first term of the quotient. 2x + 3) 2x2 + 5x + 3 2x2 ÷ 2x = x 2x + 3) 2x2 x + 5x + 3
( (
Step 2 : Multiply the divisor by the first term of the quotient and subtract the result from the dividend.
. 3x − y .8 1. a1. 3 3 x z
4. .10 POLYNOMIALS Polynomial : An algebraic expression in which the variable(s) does (do) not occur in the denominator and the exponents of the variable (or variables) are whole numbers and the co-efficients of different terms are real numbers is called a polynomial. + an–1x + an where a0.11 GENERAL FORM OF A POLYNOMIAL IN ONE VARIABLE An expression of the form a0xn + a1xn–1 + a 2 x n− 2 + . a2. . y = 1 .
CHECK YOUR PROGRESS 4.. 3x3 – y3z. and x − 2 y 2 are all polynomials whereas x 3 − x is not a 3 3
polynomial.Algebraic Expressions and Polynomials
77
x2 4. x −9 4
. an are real constants. The degree of a constant is taken as zero. y = 1
5.. 5 x + y .. 4 − 3y. 3xyz – x3 – y3 + z3 at x = 2 .. y + 3xy − 11 at x = 2 . the degree of the polynomial 2x3 – 9x + 3 is 3. 3x 3 + 4 y 2 + 2 . z = –3. 4. x is a variable and n is a non-negative integer is called a polynomial in the variable x.12 DEGREE OF A POLYNOMIAL : The degree of a polynomial in one variable is the greatest exponent of the variable occurring in the various terms of the polynomial. Which of the following algebraic expressions are polynomials ?
1 2 x 3 + 1 5x 2 − y 2 . Also
3
x + y is not a polynomial.
3 1 a −b For example 5 . 4. For example.. Similarly the degree of is 1 and the degree of 3 is zero.
. + an. In case of division of polynomial.Algebraic Expressions and Polynomials
81
CHECK YOUR PROGRESS 4. The process of substituting a numerical value of the variable in the algebraic expression is called evaluating the algebraic expression for that value of the variable. Subtract 12x3 – 3x2 – 11x +13 from 5x3 + 7x2 + 2x – 4 5.. Find the remainder and quotient for each of the following : (i) Divide x4 –1 by x – 1 (ii) Divide x3 – 3x2 + 5x – 8 by x – 2 LET US SUM UP
z z
An unknown quantity. x is a variable and n is a non-negative integer is called a polynomial in x of degree n in standard form. The terms which are not like are called unlike terms. a1. where a0. Find the product of
FG 2 x H3
2
5 + x − 3 and (3x2 + 4x + 1) 4
IJ K
7... Subtract 2x3 + 7x – 5x2 + 2 from 5x + 7 – 3x2 + 5x3 4. the degree of remainder is less than that of divisor. . is called a variable. Four fundamental operations on polynomials yield another polynomial. The terms of an expression having the same variable with same exponent are called like terms.10 1.. which can have various values. a binomial and with three terms is called a trinomial. The degree of a constant is zero. Add
7 3 7 3 2 2 2 3 x + x − 3x + and x3 + x2 − 3x + 5 5 3 5 3 5
3. An expression of the type a0xn + a1xn–1 + . Add 2x3 + 7x2 – 5x + 7 and –2x 2 + 7x3 – 3x – 7 2. An algebraic expression with one term is called a monomial. an are real numbers. The degree of a polynomial in one variable is the greatest exponent of the variable occurring in the various terms of the polynomial. Find the product of (x2 + 1) and (x3 – 1) 6.
z
z
z z
z
z
z z
. with two terms. An expression which has one or more terms involving variable joined by + or – sign is called an algebraic expression. Four fundamental operations on algebraic expressions yield another algebraic expression.
Try to recall how you would find the square of a number.e. Simply you have to multiply 103 with itself. Now. Method 1 : Direct Multiplication 103 × 103 309 000 103 10609 or 1032 = 10609 ∴ or ∴ The answer is the same in both cases. Notice carefully the expression in the box in the second method. This can easily be generalised to the square of any binomial (or an algebraic expression with two terms).Special Products and Factorisation
85
5
Special Products and Factorisation
5. how would you find the cube of a number.. say. This is the most important step in this calculation. Is this the only method ? No! Here is another method. i. algebraic expressions and polynomials in a previous lesson. 993 ? 1032 = 1002 + 2 × 100 × 3 + 32 1032 = 10000 + 600 + 9 1032 = 10609 Method 2 : Breaking up 103 103 = 100 + 3 1032 = (100 + 3)2 = (100 + 3) (100 + 3) = 100 × 100 + 100 × 3 + 3 × 100 + 3 × 3
.1 INTRODUCTION You have already studied different number systems. say 1032. 1032 = (100 + 3)2 = square of the first term + two times the product of the first and the second terms + square of the second term.
99 = 100 – 1 ∴ 993 = (100 – 1)3 Thus.86
Mathematics
Direct multiplication will surely make the calculations lengthy and difficult.4 SQUARE OF A BINOMIAL (a) Consider a binomial say. In this lesson.2 OBJECTIVES After studying this lesson.. the learner will be able to :
z z z z z z z z
write formulae for special products calculate squares and cubes of numbers using special products factorise a given algebraic expression and a polynomial in one variable factorise a given quadratic polynomial by splitting the middle term determine the HCF and LCM of two or more polynomials cite examples of rational expressions in one or two variables express a given rational expression in the simplest form perform four fundamental operations on rational expressions. you can easily calculate 993 if you know the formula for the cube of a binomial.
5..x + x.y = x2 + 2xy + y2 .3 EXPECTED BACKGROUND KNOWLEDGE
z z z z
Knowledge of four fundamental operations on numbers and algebraic expressions GCD and LCM of numbers Polynomials and operations on them Area of squares and rectangles. x + y Let us find (x + y)2 (x + y)2 = (x + y) (x + y) = x (x + y) + y(x + y) = x.y + y.(i)
.x + y. you will study these as well as many other useful results which will make calculations simpler and also help you to understand polynomials better. but if you write.
5. 5.
Two methods are given below: Method 1 : Direct Multiplication (x – y)2 = (x – y) (x – y) = x(x – y) – y(x – y) = x. in such a way that AP = AQ = 'x' (say) and PB = QD = 'y' (say) [See Fig.1]
Fig. intersecting 'BC' at S.. 5.2
[Using the formulae for the areas of a square and a rectangle] ∴ (x + y)2 = x2 + 2xy + y2
(b) Consider the square of the binomial (x – y) There are many ways of finding this product. Method Step 1 : Take a square piece of cardboard ABCD and mark points 'P' and 'Q' on sides AB and AD. 'l' and 'm' divide ABCD into two squares and two rectangles.Special Products and Factorisation
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The square of a binomial is equal to the sum of the square of the first term. 5. (–y) = x2 – xy – xy + y2 ∴ (x– y)2 = x2 – 2xy + y2 .x + x (–y) – y.. or (x + y)2 = x2 + 2xy + y2 .2) ∴ Area of ABCD = x2 + xy + y2 + xy
Fig.(i)
You can easily make a model and verify this formula geometrically. (See Fig.(ii)
. two times the product of both the terms and the square of the second term. Step 3 : Through 'Q' draw a line 'm' parallel to 'AB'. 5.. 5..x –y.1
Step 2 : Through 'P' draw a line 'l' parallel to AD. respectively. intersecting 'DC' at 'R'.
. It is the largest number which divides all the given numbers.11. the next common factor is 3 and it occurs at least once in both the numbers ∴ The H. 24 = 23 × 3 and 36 = 22 × 32 The first common factor is 2 and it occurs at least twice in the factorisation of the two numbers and so it must occur twice in the H. the H.C. x2 – x – 56 8. 5.C.F.F. x2 – 2x – 8 6. of 24 and 36 = 22 × 3 = 12 Similarly. 4x2 + 12x + 5 7.F.F. 3x2 – 2x – 5 10. you can find the H.F.F. of Polynomials We already know the meaning of the term 'H. 6x2 + 7x – 10 5. x2 + 4x – 60 9. of two or more polynomials is the polynomial of the highest degree which divides all the given polynomials completely. We used the method of prime factorisation. of 12 and 18 is 6 The H.F.C.' of any given numbers.C.C.C. Let us recall how to find the H.F. of two numbers.1 H.C. say 24 and 36.Special Products and Factorisation
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= 2x2 – 4x – 7x + 14 = 2x(x – 2) – 7(x – 2) = (2x – 7) (x – 2) ∴ 2x2 – 11x + 14 = (2x –7)(x – 2) CHECK YOUR PROGRESS 5.11 APPLICATIONS OF FACTORISATON Factorisation helps us not only in expressing a given polynomial in terms of simpler polynomials. Let us take some examples and illustrate. x2 + 9x + 20 3. For instance. x2 + x – 20 4. of two or more polynomials.C. but also enables us to generate new polynomials using the given polynomials.8 Factorise each of the following polynomials by splitting the middle terms : 1. 9x2 – 12x + 4 2. Similarly. 15x2 – 14x + 3
5.
C. (2) Check that the H. Now 2x – 3 is the next common factor and it occurs only once in the first polynomial. The next common factor is 2x + 5 and occurs at least once.F. of 4x2y and x3y2 is x2y
Example 5. Example 5.F. of x2 – 4 and (x + 2)2 is x + 2. The second polynomial can be written as (3x – 1)3 (2x + 5)(2x – 1) The first common factor is 3x – 1 and occurs at least three times.F.16 : Find the H. ∴ H.C.F. occur in the first polynomial and so does not occur in the H. of (3x –1)4 (2x + 5)2 and (3x –1)3 (2x + 5) (2x – 1) = (3x – 1)3(2x + 5) Note : (1) 2x – 1 does not.F.C. It must occur two times in the H. ∴ H.F. x2 – 4 = x2 – 22 = (x +2) (x – 2) The only common factor is x + 2 and it occurs at least once ∴ H. ∴ The H.
.15 : Find the H.F.104
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Example 5. of the given polynomials = (x – 2)2 (2x – 3)
Let us take some more examples.F.C. divides both the polynomials completely.14 : Find the H. of each of the following pairs of polynomials : (i) (3x – 1)4 (2x + 5)2 and (2x – 1) (2x + 5) (3x –1)3 (ii) x2 – 4 and (x + 2)2 Solution : (i) (3x – 1)4 (2x + 5)2 is the first polynomial.C. of (x – 2)3(2x – 3) and (x – 2)2(2x – 3)3 Solution : The first common factor is x – 2 and it occurs at least twice in the given polynomials.C.C.C.F.C.C.F. of 4x2y and x3y2 Solution : 'x' occurs as a factor at least twice in the two monomials and 'y' occurs at least once. (ii) x2 – 4 and (x + 2)2 Now.
M.
. 3 occurs twice and 5 occurs once.M.9 Find the H.C.C. at the most ∴ L.C.Special Products and Factorisation
105
CHECK YOUR PROGRESS 5.M.11. you can easily find the L. of (x – 2)3 (2x – 3) and (x – 2)2 (2x – 3)3 = (x – 2)3 (2x – 3)3 (ii) (3x – 1)4(2x + 5)2 and (2x – 1)(2x + 5)(3x – 1)3 The L. say 36 and 40. of two or more polynomials is defined as the polynomial of the least degree which is a multiple of all the given polynomials. of two numbers. The L.M. using each prime factor the maximum number of times it occurs in the factorisation of any of the numbers. ∴ L.C.M. of Polynomials Recall how we found the L. of the given polynomials = (3x – 1)4(2x + 5)2(2x – 1). We write the prime factorisation of the numbers and find the product of all the different prime factors of the numbers.M. Let us take some examples to illustrate.M.C.M. Similarly.17 : Find the L.C. 36 = 2 × 2 × 3 × 3 and 40 = 2 × 2 × 2 × 5
2 occurs three times.C. of 36 and 40 is 2 × 2 × 2 × 3 × 3 × 5 which is 360.F. (i) (x – 2)3 (2x – 3) and (x – 2)2 (2x – 3)3 x – 2 occurs 3 times at the most and 2x – 3 also occurs 3 times at the most.2 The L.C.C. of each of the following pairs of polynomials : (i) (x + 1)3 and (x + 1)2 (x – 1) (iii) (x + 2)3 and x3 + 8 (ii) x2 + 4x + 4 and x + 2 (iv) (x + 1)2 (x + 5)3 and x2 + 10x + 25
(v) (2x – 5)2 (x + 4)3 and (2x – 5)3(x – 4) 5. Example 5. of polynomials. of each of the following pairs of polynomials : (i) (x – 2)3(2x – 3) and (x – 2)2(2x – 3)3 (ii) (3x – 1)4(2x + 5)2 and (2x – 1)(2x + 5)(3x – 1)3 (iii) x2 – 4 and (x + 2)2 Solution.
C. = (x – 2)2(2x – 3)(x – 2)3(2x – 3)3 = (x – 2)5(2x – 3)4 which is the same as the product of the given polynomials.C.M.C.M. of the given polynomials is (x – 2)2 (2x – 3) and the L.C.M. of the given polynomials is (x – 2)3(2x – 3)3.10 Find the L.F.F.C.F. and their L.12 THE RELATIONSHIP BETWEEN H.C. of 6 and 4 ? What is their L.F.C.F. and the L.18 : Is the result given above true for each of the following pairs of polynomials ? (i) (x – 2)3 (2x – 3) and (x – 2)2(2x – 3)3 (ii) x2 – 1 and x3 – 1 (iii) (x – 1)3 and (x + 1)2 Solution : (i) (x – 2)3 (2x – 3) and (x – 2)2 (2x – 3)3 The product of (x – 2)3 (2x – 3) and (x – 2)2(2x – 3)3 = (x – 2)3 (2x – 3) (x – 2)2(2x – 3)3 = (x – 2)3 + 2 (3x – 3)1 + 3 = (x – 2)5(2x – 3)4 The H. is 12.C.C.C.M.M.M. of 6 and 4 is 2. Example 5.C.C. AND L. × L.C. what is 2 × 12 ? Is it not the product 6 × 4 ? Yes! The product of the H. Hence H. of each of the following pairs of polynomials : (i) (x + 1)3 and (x + 1)2 (x – 1) (iii) (x + 2)3 and x3 + 8 (iv) (x + 1)2( x + 5)3 and x2 + 10x + 25 (v) (2x – 5)2(x + 4)3 and (2x – 5)3 (x – 4) 5.M.106
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CHECK YOUR PROGRESS 5. The same result is true even for polynomials. (ii) x2 – 1 and x3 – 1 x2 – 1 = (x + 1)(x – 1) (ii) x2 + 4x + 4 and x + 2
. Let us verify this result with the help of some examples. is the same as the product of the numbers. What is the H.F. Now.? You know very well that the H.
For all other values of x. whereas and (ii) and (iii) are equations in two variables. (ii) and (iii) can be true for some specific value of the variables involved. (ii) is true for x = 10. it is true. The same thing can be said for (iii) also. some are true. y = 2. (i) is false. Thus. because they involve two statements involving variables and constants connected by a sign of equality. Now consider the statements : (i) x + 10 = 22 (ii) x + y + 2 = 14 (iii) 2x – y + 4 = 20 Statements (i). if x equals 12. (ii) and (iii) are all called equations. involve two variables x and y. (iv) and (v) are true statements whereas (ii) and (vi) are false statements.1 INTRODUCTION Consider the following statements (i) 6 + 5 = 11 (iii) 8 × 7 = 56 (v) 16 ÷ 4 = 4 (ii) 7 + 6 = 18 (iv) 6 – 17 = –11 (vi) 18 × 7 = 116
Of the above statements. (i) is an equation in one variable. (i) involves one variables x. (i). Further. (iii). Similarly. Can you say which of them are true and which are false ? Of course. your reply will be that (i). whereas (ii) and (iii).122
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6
Linear Equations
6. Can you see that (ii) is true for some other pairs of values of x and y. We are able to call them true or false after finding their value. In (i). whereas the others are false.
. All the above statements (i) to (vi) are called numerical statements.
write a linear equation in one variable and also give its solution.4. 6. Linear polynomials in one or two variables. translate a word problem into a linear equation in one or two variables. it is said to be an equation in one variable or one unknown. solve real life problems involving linear equations in one or two variables. Knowledge of exponents. Four fundamental operations on numbers. We shall also solve word problems involving one or two variables.Linear Equations
123
In this lesson.
. Plotting of points on a graph. 6. You will learn about them later. find the solution of a system of two linear equations graphically. the learner will be able to :
z z z z z z z z z z
identify a linear equation from a given collection of equations.2 OBJECTIVES After studying this lesson. We shall also solve system of two linear equations graphically and algebraically.4 EQUATIONS IN ONE VARIABLE If one equation has only one variable. write a linear equation in two variables draw the graph of a linear equation in two variables. check-the consistency or otherwise of systems of linear equations solve system of equations algebraically. Four fundamental operations on algebraic expressions.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z z z
Algebraic expressions and polynomials in one or two variables. cite examples of linear equations.
6.
Each of these is an equation in one unknown.
6. namely x. There are equations having more than one unknown.1 Degree of an Equation Consider the following equations (i) 3x – 4 = 8 (iii) x2 – 16 = 0 (ii) (x – 4) – 8 = 2(x + 3) (iv) x3 – 3x = x. we shall study about linear equations in one and two variables and discuss about methods of finding their solutions – algebraic as well as graphic.
6. For example. Thus. is called a linear equation. Let us take some example to identify linear equations. By solving an equation. x = 2 is a solution of the equation x + 3 = 5 because on substituting 2 for x we get. Thus. In equation (iii). the degree of equation (iii) is 2 and of the equation (iv) is 3. the greatest exponent of x is 2 and in equation (iv) it is 3.6 SOLUTION OF A LINEAR EQUATION The value (or values) of the variable (or variables) which make the equation a true statement is (are) called solution(s) of the equation. Example 6. the degree of equations (i) and (ii) is one whereas. the degree of a linear equation is one. An equation in which the greatest exponent of the variable after simplification is one.1 : Which of the following equations are linear equations ? (i) 5x – 3 = 5 (ii) 8m – 3 = 12 (ii) 4x + 4 = 4 (iv) 2(x + 3) – 1 = x2
Solution : (i) .124
Mathematics
You will observe that the greatest exponent of the variable in (i) and (ii) is one.
. 6. Equation (ii) can be simplified as follows : x – 4 – 8 = 2x + 6 or or or x – 2x – 12 – 6 = 0 –x –18 = 0 x + 18 = 0
The greatest exponent of the variable in an equation in one unknown is called the degree of the equation. (ii) and (iii) are equation of degree one and hence linear equations whereas equation (iv) is of degree 2 and hence is not a linear equation.5 GENERAL FORM OF A LINEAR EQUATION A linear equation in one unknown can be written as : ax + b = 0 where 'a' and 'b' are real numbers and a ≠ 0. we mean finding the solution (or solutions) of the equation.
Example 6. and 4. Solving it for x. Example 6. To do so. Then. this is true
∴ The required number is 2. we come across many problems which can be solved using equations. Find its side. Perimeter = 4x
∴ The equation becomes 4x = 64
or or
x = 64 ÷ 4 x = 16
∴ The side of the square is 16 cm.3 : Which number when added to 6 gives 8 ? Solution : Let x be the given number 6 added to x mean x + 6 According to the question x + 6=8 This is a linear equation in x.5 : Renu is 20 years younger to her mother. Translate the given statement into an equation 3.t etc. by putting it in the original problem..
. her mother will be twice as old as Renu will be then. 2. follow the steps given below : 1.7 WORD PROBLEMS In our day-to-day life. Represent the unknown by an alphabet say x.4 : The perimeter of a square is 64 cm. Yes. Her mother's present age = (x + 20) years After 10 years.z.
Example 6.e.y. we get x = 2 i.128
Mathematics
6. Let us now take some examples. 2 added to 6 gives us 8. After 10 years.p. Solve the equation.. Solution : Let the side of the square be x cm. Check the value found.n. How old is Renu now ? Solution : Let Renu's present age be x years.
Renu's mother will be 2 times as old as Renu after 10 years. Solution : Let the number be x Then.2 1.. CHECK YOUR PROGRESS 6.6 : Twice a certain number increased by 10 equals 32. twice the number = 2x
∴ The equation is
2x + 10 = 32 or or or or 2x = 32 – 10 2x = 22 x = 22 ÷ 2 x = 11
Thus. 3
. the required number is 11. the 2 fraction becomes . A man is 20 year older than his son.e. Find their present ages. Example 6. Renu's age = 10 + 10 = 20 years After 10 years. Find the fraction. Renu's mother's age = 30 + 10 = 40 years and 40 is twice of 20 Thus. After 10 years his age becomes twice the age of his son. The denominator of a fraction is 2 more than its numerator. Renu's present age is 10 years.Linear Equations
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Renu's age will be = (x + 10) years and her mother's age will be[(x + 20) + 10] years According to the problem. If 1 is added to each. 2. Find the number.
∴
(x + 20) + 10 = 2(x + 10) x + 20 + 10 = 2x + 20 x + 30 = 2x + 20 x – 2x = 20 – 30 x = 10
or or or or
i. Renu is presently 10 years old Check : Renu's present age = 10 years Renu's mother's age = 30 years After 10 years.
e.e.8 A LINEAR EQUATION IN TWO VARIABLES Recall that ax + b = 0 is the general form of a linear equation in one variable. a pair of numbers taken in a fixed order. 3. The sum of three consecutive natural numbers is 42. and its graph is the diagrammatic form of this representation. 4 added to 5 times a number equals 2 subtracted from 7 times the number.130
Mathematics
3. Let us now understand how to represent a point in a plane. 5. 6. Find the number. A linear equation in two unknowns is an equation which after simplification contains two unknowns. The angles of a triangle are such that sum of two angles equals the third and the ratio between the acute angles is 2 : 3. a point on a plane represents an ordered pair of numbers i. Given below are some linear equations in two unknowns : 1. For that we go through the following steps :
. each one of them in a separate term and having the exponent one. 6.. x – y =
5. (1) Representation of a Point in a plane In order to draw the graph of any equation. on a sheet of paper). Similarly. your must first know how to plot points in a plane (i. 2x = 5y – 7 2. 4. Find the numbers. Find the angles of the triangle.
3 3 x = 5− y 2 2 2 x + 3y = 6 π 2
4. This plane is called the coordinate plane or cartesian plane.9 THE GRAPH OF A LINEAR EQUATION IN TWO UNKNOWNS An equation in two unknowns represents a relationship between the unknowns. b and c are real numbers with a ≠ 0 and b ≠ 0. 2x – 1 = (y – 3) + (2y – x) A general linear equation in two unknowns is written as ax + by + c = 0 where a. You already know that a point on the number line represents a real number.
Fig.3).4
Along the x-axis.3
Fig. These two coordinate axes divide the plane into four parts. positive integers are marked to the right of O. 3).1). 6. They are called the x-axis and y-axis respectively. 6. 6. the two axes can be taken as two perpendicular number lines.
Fig. and O is called the origin (See Fig. (See Fig. Along the y-axis. positive integers are marked above O. 6. each of which is called a quadrant (See Fig. Thus. 6.2) Step 3. 6. To represent on ordered pair say (2. and negative integers below O. we proceed as follows :
. The integers are marked along both the axes with O representing as zero on both the axes.2
Step 2. and negative integers to the left of O. 6.1
Fig.Linear Equations
131
Step 1 : Two perpendicular lines XOX′ and YOY′ are drawn intersecting each other at O.
6. 5) on the coordinate plane. 'l' and 'm' intersect each other at point P.7
Similarly.132
Mathematics
Step 4.5
Fig. Step 5. 6. P represents the point (2. 6. The first number in the ordered pair is –3. you can represent any ordered pair (x. Let us take another example. and so we will take 3 units along the x-axis to the left of O. Let us first draw the x-axis and the y-axis and mark points on them representing integers as in Fig. Similarly take 5 units along the y-axis above O and draw lines 'l' and 'm' parallel to the y-axis and the x-axis. The point A represents (–3. y) by a point in the cartesian plane. respectively. Mark a point B on mark 3 on y-axis and draw a line 'm' through B parallel to the x-axis (See Fig.5). 3) because P is at a distance of 2 units measured along the x-axis and 3 units measured along the y-axis. Note that : 1. Mark a point A on mark 2 on x-axis and draw a line 'l' through A parallel to the y-axis (See Fig. Solution.
.7. 6.7 : Represent the point (–3. Example 6. 6. 6.6. The first number in the ordered pair tells us the number of units to be taken along the x-axis and is thus called the x-coordinate or the abscissa of the point.
Fig. 5) as in Fig. 6.6
Fig.4).
you can easily find the point in the plane which corresponds to this ordered pair. 2).
Similarly. given an ordered pair of numbers. 5) and (–5.Linear Equations
133
2. a point on the y-axis will have the x-coordinate equal to zero. then P lies in : (i) the first quadrant if both 'x' and 'y' are positive real numbers. Thus. (iv) the fourth quadrant if 'x' is positive and 'y' is negative. (–3. (ii) the second quadrant if 'x' is negative and 'y' is positive. The second number in the ordered pair tells us the number of units to be taken along the y-axis and is thus called the y-coordinate or the ordinate of the point. Example 6.
CHECK YOUR PROGRESS 6. If (x. –2). (3. 0). and will thus have the coordinates (0. "What happens to 'P' if y = 0 ?" Obviously 'P' is then 'x' units along the x-axis and zero units along the y-axis and thus lies on the x-axis. draw a line 'l' parallel to the y-axis and mark A as the point of
. 0). O has the coordinates (0. 0) and (0. y) is a point 'P' in the cartesian plane. –2). 2). –4). (iii) the third quadrant if 'x' and 'y' both are negative. 0). 3. (–2.
∴ Any point on the x-axis has the coordinates (a. (3.8 : 'P' is a point in a plane. Let us take an example to illustrate this. Therefore. can we associate an ordered pair of numbers with it ? Yes.10 FINDING ORDERED PAIR CORRESPONDING TO A POINT IN THE PLANE Given a point in a plane. 4) 6. b). –5) (ii) (–5.3 Represent the following ordered pairs as points in the cartesian plane : (i) (5. we can find its coordinates with respect to a given pair of axes. –4). Find its coordinates with respect to the coordinate axes XOX′ and YOY′. (2. (3. Solution : Step 1 : Through 'P'. (0.
Fig. along the y-axis. the ordered pair associated with 'P' is (a. GRAPH OF A LINEAR EQUATION Let us draw the graphs of some linear equations. If OA = a units.134
Mathematics
intersection of 'l' with the x-axis (See Fig. 6.
∴ The y-coordinate of 'P' is 'b'. b). Step 2 : Through 'P' draw a line 'm' parallel to the x-axis and mark B as the point of intersection of 'm' with the y-axis (See Fig. i.11. 6. CHECK YOUR PROGRESS 6.9
If B is 'b' units from O.10. (ii) If a point is not on the graph of an equation.8).
Hence the coordinates of 'P' are (a. its coordinates do not satisfy the equation. 6. Here are some examples of graphs of linear equations. 6. 0). b). its coordinates satisfy the equation..10
.
Fig. then B has the coordinates (0.e. 6. Q.9). R and S in Fig.8
Fig. Remember that : (i) If a point lies on the graph of an equation.4 Find the coordinates of the points P. b). the point A has the coordinates (a. 6. 6.
–2).136
Mathematics
Equation (ii) 2x – 5y – 10 = 0 or 2x = 5y + 10 or x =
5 y+5 2
Table x y 10 2 5 0
This is the x-form of the equation. (2. some of the ordered pairs satisfying it are (2. (5. we an draw the graph of a linear equation by plotting two ordered pairs only because two points determine a line uniquely. But what happens if out of 'a' and 'b' one of them is zero ? Let us take an example. 2) and (2. (5. In fact..y = 2. Example 6. 3).13).. . The graph of a linear equation in two unknowns is a straight line which is neither the x-axis. 2). say (2. –2). –1). Plotting the third pair helps us to verify the correctness of the graph. nor the y-axis and is not parallel to either of the axis. 6. 2). 2. Three of its ordered pairs are (10. Hence.10 : Draw the graphs of each of the following equations : (i) x = 2 (ii) 2y + 5 = 0 Table x y 2 0 2 2 Solution : (i) The equation x = 2 can be written as x + 0. This is because for a linear equation in two unknowns. 0) and (0. 0). –1)
2 –1
. Q and R to get the line 'm' (as shown in Fig. 'm' is the required line which is the graph of the linear equation 2x – 5y – 10 = 0
Fig. 6. 0). Join the points P. 0) and (0.13
Notes 1. (2. We plot three ordered pairs. (2. 2). (2. we get the three points P.
0 –2
Plotting the ordered pairs (10. ax + by + c = 0. neither 'a' nor 'b' is zero. Q and R.
5 –2.14
y = –2. 'l' is the required graph of x = 2.5 –2. B (2. CHECK YOUR PROGRESS 6. –1) on the cartesian plane.Linear Equations
137
We get three points A (2. 6. N and R. 2. –2.5
y –2. or is a line parallel to either of the axes. –2.5
∴ Any point whose y-coordinate is –2.5)
Plotting these ordered pairs. Three of the ordered pairs satisfying the equation are (–1. Join them to get the line 'l' as in Fig. Note that 'l' is parallel to the y-axis. we get the three points M. The x-axis is the graph of the equation y = 0. 3. 6.14. Join them to get the line 'p' (See Fig. The graph of a linear equation in one unknown is either the x-axis or the y-axis. Thus.15
.5 1.5). The y-axis is the graph of the equation x = 0. –2. (0. 6. (ii) 2y + 5 = 0 This equation can be written as 0. Draw the graphs of each of the following equations : (i) 2x – 3y = –1 (iv) 2x + 3 = 0 (vii) x = –5 (ii) 3x + y = 4 (v) x + y = 0 (viii) 2x – 3y = 8 (iii) y = 5 (vi) 3x + 3y = 6 (ix) x – y = 0
Fig. Notes : 1.5) and (1. Draw the graphs of x = 0 and of y = 0.15) p is the required graph of the equation 2y + 5 = 0 This is a line parallel to the x-axis. 0) and C (2. 6. 2).5 is a solution of this equation. 2.x + 2y + 5 = 0 and the y-form of this equation is y= or
−5 2
Table x –1 0 1
Fig.
6.12 THE SYSTEM OF LINEAR EQUATIONS IN TWO UNKNOWNS You know that ax + by + c = 0 (where a. 4) (iii) (8.16). –2) Also. or x + y = –2 y = –x – 2 x y x y Table 1 x – 2y = 7 1 –3 7 0 3 –2
Table 2 x + y = –2 0 –2 –2 0 1 –3
Three of ordered pairs satisfying it are (0. x – 2y = 7 x = 7 + 2y Three of ordered pairs satisfying it are (1. Example 6. b. 0) and (1. a'. 0). 2) (ii) (5. 0) and (3. –4) (ii) (0. –2).
6. The graphs of these equations are straight lines. b ≠ 0 and a. b. 16) 4. GRAPHICAL SOLUTION OF A SYSTEM OF LINEAR EQUATIONS In order to solve a system of linear equations in two unknowns graphically.138
Mathematics
3. (7. b' and c' are all real numbers and a. (–2. Which of the following ordered pairs make the equation 2x + y = 10 a true statement? (i) (0. 8) (iv) (0.13. a' and b' are non-zero. (i) (7. 10) (iii) (8. 6. A system of linear equation in two unknowns is given as : ax + by + c = 0 a'x + b'y + c' = 0 where a. b. 8) (iv) (–3. c. c are real numbers) is a linear equation in two unknowns (variables). –3). State which of the following ordered pairs are not the solutions of the equation 4x = 3y + 8.
.11 : Solve the following system of linear equations graphically : x – 2y = 7 x + y = –2 Solution : You have already learnt how to draw the graph of a linear equation in two unknowns. Let us now learn to solve a system of linear equations in two unknowns. –3)
By plotting these ordered pairs we get a pair of straight lines as the graphs of the given equations (See Fig. we draw the graphs of the equations in the same cartesian plane and find the point(s) of intersection of their graphs. Let us take some examples.
x = 1 and y = –3 is the solution of the given system of equations. Example 6. 8) and (3. how many points are common to both the lines 'l' and 'm' ? From the graph.
Now. 4). 'P'. (0. 4) and B (0.Linear Equations
139
Fig. 6. 0) and (3. Also. namely. we get line 'l' as the graph of 4x + 3y = 24. P (3. Thus. 2). 4) x y 6 0 0 8 3 4
Plotting these ordered pairs. 0). whose coordinates are (1. 8) and on joining them. or 3y – 2x = 6 y= 6 + 2x 3 x y Table 2 3y – 2x = 6 Three of ordered pairs satisfying it are (0. we get three points A (6.12 : Solve graphically the following system of linear equations : 4x + 3y = 24 3y – 2x = 6 Solution : 4x + 3y = 24 or y= 24 − 4 x 3 Table 1 4x + 3y = 24 Three of ordered pairs satisfying it are (6. –3). how many ordered pairs are the solutions of both the equations ? In other words. 0).16. it is clear that there is only one such point. (–3. 0 2 –3 0 3 4
.
2x = 3
Table y – 2x = 3 x 0 1 2 or y = 2x + 3 y 3 5 7 ∴ On plotting the ordered pairs (0. 2). (i) y– 2x = 3 4x = 10 solution. 7). The only point common to both the lines 'l' and 'm' is point 'P' with coordinates (3.
Fig. 0) and P (3. Draw the graph of : y– 2y – and find their common Solution. and on joining them. 3) (1.13. In other words x = 3 and y = 4 is the only solution of the given system of linear equations Example 6. we get the line 'm' as the graph of 3y – 2x = 6 (See Fig. we get the straight line 'p' as the graph of y – 2x = 3 (ii) 2y – 4x = 10 Table or y = 2x + 5 2y – 4x = 10 x 0 1 2 y 5 7 9
. 6.17
To get the common solution of both the equations.17). 4). 5) and (2.140
Mathematics
Plotting these ordered pairs in the same cartesian plane we get three points M(0. 4). we look for the points that are common to both the lines 'l' and 'm'. if any. 6. N(–3.
we get the line 'l' as the graph of y – 2x = 3. (See Fig. Hence. 7) and on plotting these we get three points P. 9). (1. 3). 2y – 4x = 6
. Similarly.
Fig.18
It is very clear from the graph that the straight lines 'p' and 'q' are parallel.Linear Equations
141
∴ On plotting the ordered pairs (0. 5) (1. the given system of equations has no solution. and hence do not have any point common. 7) and (2. Example 6. 6.18). if any Table 1 y – 2x = 3 Solution : or y – 2x = 3 y = 2x + 3 x y 0 3 1 5 2 7
Three of its solutions are the ordered pairs (0.14 : Draw the graphs of : y – 2x = 3 2y – 4x = 6 and obtain the common solutions. 6. Thus. we get. the straight line 'q' as the graph of 2y – 4x = 10 in the same plane. 5) and (2. there is no solution common to both the equations. Q and R. On joining them.
Solve graphically each of the following system of equations :
3. 5y – x = 14 and y – 2x = 1 respectively. This means that all the points of the line 'l' are also on the line 'm' and all the points of line 'm' are also on the line 'l'. Thus. it is clear that the lines 'l' and 'm' coincide or are the same. Find graphically the vertices of a triangle whose sides have the equations 2y – x = 8. Thus all points of the lines 'l' and 'm' are common.19.
From Fig. 6. CHECK YOUR PROGRESS 6.19). Draw the graphs of each of the following systems of equations : (i) 2x – y – 3 = 0 4x – 2y – 10 = 0 (i) 2x – y = 5 x + 3y = 6 (ii) x + y = 7 2x + 6y = 14 (ii) x + y = 8 2x – y = 1
2. 6.142
Mathematics
or y = 2x + 3. 6.6 1. (Hint : Draw the graphs of the three equations using the same cartesian plane and scale.
Fig. these two equations have an infinite number of common solutions. and so the same ordered pairs satisfy the second equation also (See Fig.
.19. 1 cm = 1 unit on both axes and find the points of intersection of two lines at a time).
are coincident straight lines. and hence is an inconsistent system. This is also called a consistent system of equations. then the system has one and only one solution. Equations 1. The system in Example 6. (iii) are one and the same. the system has a unique solution and is said to be a consistent or compatible system.12 had a unique solution. Example 6. These three examples help you to conclude that : If the graphs of the pair of equations ax + by + c = 0 and a'x + b'y + c' = 0: (i) intersect each other. then the system has no solution.14 of this lesson. whereas the system in Example 6.Linear Equations
143
6.14 and put the result in the form of a table.14 CONSISTENCY AND INCONSISTENCY OF A SYSTEM OF EQUATIONS You now know.e. There are other algebraic methods to do so. Example 6. the system of equations had no solution. (ii) are parallel straight lines. let us study the examples solved in Section 6. Comparison of y-forms Let us compare the y-forms of the equation in Examples 6.e.14 y – 2x = 3 2y – 4x = 6 y= y= 24 − 4 x 3 2x + 6 3 No solution Inconsistent Unique solution Consistent y-forms Solutions Type of System
y = 2x + 3 y = 2x + 5 y = 2x + 3 y = 2x + 3
Infinite number of solutions
Dependent
.12. To understand some of these methods. Example 6.13 y – 2x = 3 2y – 4x = 10 3. Notes : Any system of two linear equations in two unknowns has to be one of the types of systems described above.13 and 6. In Example 6.12 4x + 3y = 24 3y – 2x = 6 2. i. how to solve a system of equations graphically. Solving the system graphically is not necessary if you want to determine whether the system of equations is consistent or not.11 and 6..13. i. the system has infinite number of solutions and thus is a dependent system. Method I.6.14 had infinite solutions.
and a ′ 7 b′ 2 c′ a ≠ b . the system of equations has a unique solution. find whether the system is consistent. The coefficients of 'x' and the constant terms are the same. the two y-forms are identical. The coefficients of x are different (even if the constant terms are the same). the system of equations has no solution. (b) Method 2 : The Ratio Method Here a = 5. The coefficients of x are the same but the constant terms are different. b′ = –4 and c' = –2
∴
a = 5 b =−1 c =8 .144
Mathematics
From the above table you can easily conclude that if in the y-forms of the equations : 1.. Method 2.15 : In each of the following. 2. the system of equations is consistent. The Ratio Method This is another method to determine whether a system of equations is consistent or not.
. i. the system of equations is consistent. b = 2 and c = –16 and a′ = 7.e. inconsistent or dependent : (i) 5x + 2y = 16 (ii) 5x + 2y = 16 (iii) 5x + 2y = 15 7x – 4y = 2
6 3x + y = 2 5 15 x + 3y = 24 2
Solution : (i) (a) Method 1 : Using y-forms The y-forms of the two equations are : 5 y = − x+8 2 and y= 7 x− 1 4 2
Since the coefficient of 'x' are different in the two y-forms. a ′ b′
Since. 3. A system of linear equations ax + by + c = 0 and a'x + b'y + c' = 0 is : (i) consistent if a ≠ b a ′ b′ a = b ≠ c a ′ b′ c′
(ii) inconsistent if (iii) dependent if
a = b = c a ′ b′ c′
Let us take some examples and illustrate : Example 6. the system has infinite number solutions.
Substance the value of 'x' in one of the equations and get the value of 'y'
3x + 15 = 8x – 10 5x = 25 x=5 2y – x = 5
or or or
2y – 5 = 5 2y = 10 y=5
Thus.H.1 Elimination by Comparison Let us understand this method with the help of an example.S.16 : Solve : 2y – x = 5 4x – 3y = 5 Solution : Method 1. Equate the right hand side (R. Example 6. x = 5 and y =5 is the solution of the system of equations : 2y – x =5 and 4x – 3y = 5
. Write the y-forms (or x-forms) of the given equations. Solve for equation for 'x' or or or 4. Some algebraic methods are used to solve the system of equations and these are usually called elimination methods. Find the values of 'k' for which the system of equations given below has infinite number of solutions : 3x + y = 0 x – ky = 2 6.15. Solution of the Example y-forms are : 2y – x = 5 or y =
x 5 + 2 2 4x 5 − 3 3
4x – 3y = 5 or y =
x 5 4x 5 + = − 2 2 3 3
2.) of the y-forms 3. Some of these methods are discussed below.15 ALGEBRAIC METHODS OF SOLVING A SYSTEM OF EQUATIONS It is not always possible to find the exact solutions of a given system of equations graphically.146
Mathematics
4. 6.
Find the value of the first unknown by substituting the value of the second unknown found in either of the two given equations.. we get x – 2y = 7 or x = 7 + 2y 3(7 + 2y) + y = 35
2.
Solving for 'x' we get
Thus..(2)
Expressing 'y' in terms of 'x' from equation (1) we get y= 2 x − 14 3 3
Substituting the value of 'y' in terms of 'x' in (2).(1) .
Example 6.
Solving for 'y'. Solve the resulting equation for the second unknown.
∴ x = 11 and y = 2 is the solution of the given system of equations. Express one unknown in terms of the other unknown form either of the equations.150
Mathematics
Method 1.. we get
3. Substitute the value of the first unknown in the other equation.
Solution of the Example Expressing 'x' in terms of 'y' from (1).20 : Solve by elimination by substitution the following system of equations : 3y = 2x – 14 and x – y = 10 Solution : The system of equations is 3y = 2x – 14 and x – y = 10 . we get 3(7 + 2y) + y = 35 or or or or 21 + 6y + y = 35 21 + 7y = 35 7y = 14 y=2 x – 2 × 2 = 7 from (1) or or x–4=7 x = 11. we get x − 2 x − 14 3 3 or
FH
IK = 10
x − 2 x + 14 = 10 3 3
.
Substituting for 'x' in (2)..
4. x = 11 and y = 2.
"The perimeter of a rectangle is 52 cm. For solving these problems. you will have to (i) first express the statements given in the problem algebraically. substitute (iii) x – y = 0. [Hint : Find 'x' and 'y' by solving the given equations and substitute their values in 'A' and 'B'] 5.1 Expressing the Statements Algebraically Example 6. Solution : This is the problem discussed in the beginning of the lesson. Find the solution for each of the following systems of equations and check your answers : (i) 3 − 3 =1 u v 1 + 1 =1 u v 6. then (ii) find their common solution. Let us take some examples to understand these two steps. First simplify the second equation] 6. If 2x + y = 23 and 4x – y = 19.Linear Equations
153
4. 6 (i) 4 p + q = 15 6p − 8 = 14 q [Hint. find the value of A = 3x – 2y and B = 5y + 2x. The breadth of the rectangle is 2 cm more than one third its length. Solve each of the following system of equations and check your answers. Let 'x' denote the length and 'y' denote the breadth of the rectangle (ii) 8 − 9 =1 u v 10 + 6 = 1 u v (iii) 2u + 3v = 6uv 5u – 2v = 3uv
.9 1 = y in (i)] q
y (ii) x + = 4 3 4
3x – y = 23
11 2 x+y = 1
b g
[Hint. 6.17 WORD PROBLEMS You already know that there are many problems in daily life which can easily be solved by using the methods you have learnt in this lesson.17.22 : Express the following statements algebraically : 1.
154
Mathematics
Statement (i) The perimeter is 52 cm (ii) The breadth is 2 cm more than one-third the length
Equation 2 (x + y) = 52 x y = 2+ 3
Thus. the system of equations representing the statements in the problem is : 2(x + y) = 52 y = 2+
x 3
or
x + y = 26 –x + 3y = 6
2. If the digits are reversed. Then : Statement (i) The larger number is '3' times the smaller number (ii) The sum is 8 more than twice the smaller number y = 3x x + y = 2x + 8 3. the new number is 10u + t Statement (i) The sum of the digits is 11 (ii) The new number is 20 less than twice the given number
∴ The given statements can be expressed as : u + t = 11
Equation y = 3x x + y = 8 + 2x
Thus. When the digits are reversed. Let 'x' represent the smaller number and 'y' represent the larger number. Their sum is 8 more than twice the smaller number. the system of equations representing the statement given is
Equation u + t = 11 10u + t = 2(10t + u) – 20
and
10u + t = 2(10t + u) – 20. Solution.
. 'The larger of two numbers is three times the smaller number. the new number is 20 less than twice the original number. Let 't' represent the digit in the ten's place and 'u' represent the digit in the unit's place of the number. the given number is 10 t + u. Then. Solution. The sum of the digits of a two-digit number is 11.
we have x = 4y – 9 Substituting in (2).(2)
Substituting y = 7 in (2).24 : The length of a rectangle is 5 cm less than twice its breadth. Solutions : Let the present age of Atul be 'x' years and Parul's age be 'y' years Then. Five years hence (from now) Atul's age will be two times Parul's age then.2 Solving the Word Problems Let us now find solutions to word-problems by taking some examples..[Ages 5 years from now] . we get 2(2y – 5 + y) = 110 or 2 (3y – 5) = 110
.. we get x – 2 × 7 = 5 Hence. Three times its breadth exceeds twice the length by 3 cm....17. the conditions in the problem can be expressed as follows : x – 3 = 4(y – 3) and or and From (1). Example 6.(2)
Substituting the value of 'x' in terms of 'y' from (1) in equation (2).. 6. Solution : Let the length of the rectangle be 'x cm' and the breadth be 'y' cm.(1) .. find the area of the rectangle.[Ages 3 years ago] ..23 : Three years back (ago) Atul's age was four times Parul's age then. Examples 6. (ii) The sum of two numbers is 64 and the larger number is three times the smaller number.Linear Equations
155
CHECK YOUR PROGRESS 6. Then... If the perimeter is 110 cm..(1) .12 Express each of the following statements as a system of equations : (i) The perimeter of a rectangle is 72 cm. Atul's present age is 19 years and Parul's present age is 7 years.. or 2y = 14 x + 5 = 2(y + 5) x – 4y = –9 x – 2y = 5 . we get 4y – 9 – 2y = 5 or or y=7 x = 19.. the statements in the problems can be expressed as : x = 2y – 5 and 2(x + y) = 110 . Find their present ages.
Solve each of the following problems by writing each of them as a system of equations : (a) The sum of two numbers exceeds three times the smaller by 3.
Thus. Find the numbers.y cm2
Also. 2. the area of the rectangle = x. it is equal to the area of the plot given).13 1. Five years ago 'A. was twice as old as 'B' was then. perimeter = 2(35 + 20) = 2 × 55 = 110 cm. If the length is increased by 2m and the breadth is reduced by 1m. Hence. (b) Two numbers are in the ratio 4 : 7. (c) 2 pens and 5 note-books cost Rs 16 and 3 pens and 4 note books cost Rs 17. the area of the plot remains unchanged (i.
∴ The first condition is verified.. If three times the larger number is added to two times the smaller.156
Mathematics
or ∴
6y – 10 = 110 y = 20
or
6y = 120
Substituting the value of 'y' in (1).e. we get. the second condition is also verified. The number obtained on reversing the digits is one more than twice the given number. (e) The perimeter of a rectangular plot of land is 32 m. (b) 'A' is five years older than 'B'. the sum is 59. Express each of the following statements as a system of equations (a) The sum of two numbers is 12 and their difference is 2.
. The difference of the numbers is 5. x = 2 × 20 – 5 or x = 35 = 35 × 20 cm2 = 700 cm2 Check : Here. CHECK YOUR PROGRESS 6. Find their present ages. (d) The sum of the digits of a two-digit number is 10. length = 35 cm and breadth = 20 cm 2 × breadth – 5 = 2 × 20 –5 = 40 – 5 = 35 = length.
b are real numbers is the general form of a linear equation in one variable.
z z z
z
z
z
To solve a word problem. a ≠ 0 and a. c are real numbers is called the general form of a linear equation in two variables. ax + b = 0. (e) 400 people attended a school fete. To solve a word problem. find the length and breadth of the rectangle. Find the number of people who gave Rs 100 each. we translate it into linear equations and solve them. inconsistent or dependent : (a) 2x + 3y = 1 y–x=2 (c) x – y = 1 3y – 3x = 6 (b) 3x + y = 4 x + 3y = 4 (d)
x + 3y = 2 2 x + 6y = 4
. TERMINAL EXERCISE
1. it is translated to algebraic statement(s) and then solved. The weight of the son is one-sixth that of the father. LET US SUM UP
z z
An equation in one variable of degree one is called a linear equation in one variable. b. (d) The sum of the weights of a father and his son is 105 kg. To draw the graph of a linear equation. we find three ordered pairs and plot them. The algebraic methods of solving a system of linear equations are (i) elimination by substitution (ii) elimination by equating the co-efficients. Draw the graphs of each of the following system of equations and state whether they have (i) a unique solution (a) x + y = 2 x–y=4 (d) x + 2y = 2 2x + 4y = 4 (ii) infinite number of solutions (iii) (b) 2x + 3y = 5 3x – 2y = 1 (e) x + y = 0 2x – 3y = 0 no solution
(c) x + 2y = 2 2x + 4y = 6
2. a ≠ 0. find whether the system is consistent. The equation ax + by + c = 0. If the difference between the length and the breadth is 4 cm. Solution of a linear equation is also called its root. Without solving the system of equations. The line joining the points is the graph of the linear equation. Find their respective weights. Some people gave Rs 100 as a donation and some gave Rs 50. The total collection was Rs 34000.Linear Equations
157
(c) The perimeter of a rectangle is 20 cm. b ≠ 0 and a.
Find the values of 'k' for which the system of equations
2−1=2 . The sum of the present ages of 'A' and 'B' is 85 years.158
Mathematics
3. Also find out its area. If 18 is added to the number. Solve each of the following system of equations : a b (i) x + y = 2 a −b =0 [a. Also. The perimeter of a rectangular is 48 cm. 'A' was twice as old as 'B' was then. 5. y ≠ 0] x y 1 (ii) 3x + y = 4 2x − 1 = 1 . 8. What is the fraction ? 2 5 times the breadth. x. 9. If its length is the length of the rectangle. its digits are reversed. y ≠ 0 y (iii) 1+1 =4 u v (b) x – 4y = –3 x+y=2 (d) x + y = 4 x – 4y = 4
4. u. (b) infinite number of solutions. Find their present ages. the cost of a table is Rs 5 less than twice the cost of a chair. 7. it becomes is increased by 5. Five years ago. Find the number. find out 2
10. If the numerator of a fraction is decreased by one. but if the denominator 3
1 .
. the fraction becomes 2 . The sum of the digits of a two-digit number is 10. Find the total cost of 10 chairs and 3 tables. Solve each of the following system of equations : (a) 3x + y = 2 x – 2y = 1 (c) 5x−y =1 2 4x + y = 5 4x + y = 5 x + ky = 20 has (a) a unique solution. b are non-zero constants. v ≠ 0 u v 6. Six chairs and two tables cost Rs 240.
The above type of problems can be solved by a systematic method. form a quadratic equation with given roots. it becomes difficult to answer the problem by a guess. It is again difficult to answer the problem by making a guess. similarly if the product is 56 again we can find the number as 7 and 8. b and c. If we put different conditions on the length and breadth. For example in the first case if the numbers are x and x + 1.162
Mathematics
7
Quadratic Equations
7. Again in 2nd case. write quadratic equations in standard form. and in this lesson we shall deal with solutions of such equations. (ii) Similarly suppose we are to find the dimensions of a rectangle with area 168 square metres and the length exceeds breadth by 2 m. 7. For example (i) Suppose we are to find two consecutive natural numbers whose product is 12. solve a quadratic equation by (i) factorisation and (ii) using the quadratic formula. Such type of equation is called a quadratic equation.2 OBJECTIVES After studying this lesson. we have to solve the equation x (x + 1) = 552 or x2 + x – 552 = 0.1 INTRODUCTION Many times in life we come across problems whose solution we find by hit and trial method with the help of available information. the length will be x + 2 and we will have to solve the equation x (x + 2) = 168 or x2 + 2x – 168 = 0. but if the product is 552.
. we shall again get an equation of the type ax2 + bx + c = 0 but with different values of a. we can easily guess that the numbers are 3 and 4. the learner will be able to :
z z z z
identify a quadratic equation from a given collection of equations. when we take breadth as x.
where a. Thus. you will find that for x = Thus. consider a quadratic polynomial p(x) = 2x2 – 5x + 2. 2
The values of the variables for which a polynomial p(x) vanishes are known as zeroes of the polynomial p(x). In particular.
7. Can you find some more values of x for which p(x) becomes zero ? After some more trials. 3x2 – 5x + 7 = 0 is a quadratic equation in x and 2y2 + 3y – 7 = 0 is a quadratic equation in y. 2 3 a 2 + 5 = 0 . using quadratic equations. b and c are real numbers. x ≠ 0. is called a quadratic equation in x. Can you find a value of x for which p(x) = 0 ? By hit and trial method you can find that for x = 2. Example 7. You know that for every real value of x. where a ≠ 0 and a. For example p(x) = –1 for x = 1 and p(x) = 5 for x = 3.Quadratic Equations
163
z z
translate a word problem into a quadratic equation.
7. For example. 3 x2 – 5x = 0.e. p(x) has a real value. x2 = 0 are some other examples of quadratic
equations. b and c are real number and a ≠ 0. An equation of the form ax2 + bx + c = 0. b and c are real numbers and a ≠ 0 be a quadratic polynomial. x = 2 and x = 2 and 1 are the numbers for which p(x) = 0. p(x) becomes zero. 2
1 are called the zeroes of p(x).3 EXPECTED BACKGROUND KNOWLEDGE
z z z
identifying and solving Linear equations in one or two variables finding square root of natural numbers factorisation of polynomials. ax2 + bx + c = 0 is called a quadratic equation. p(x) = 0.
.4 QUADRATIC EQUATIONS Recall that p(x) = ax2 + bx + c is the general form of a quadratic polynomial where a. (i) (x + 3) (x – 2) = 5 (ii) 2x2 + 3x = 2x(x – 7) (iii) 3x2 – 5x = 5(x2 – x + 3) 1 (iv) x − x = 0. then p(x) = 0 i. Quadratic Equation : Let p(x) = ax2 + bx + c. solve word problems. 2 1 .1 : Determine which of the following are quadratic equations in x.
(ii) If D = 0 then the quadratic equation have two equal roots. The age of father is 10 times the age of his son. LET US SUM UP
z
An equation of the form ax2 + bx + c = 0. A two digit number is such that the product of the digits is 12. A quadratic equation whose roots are α. (i) If D > 0 then the quadratic equation have two real and unequal roots. It is usually denoted by D. x2 – 8x + 15 = 0 4. find the age of father. (iii) If D < 0 then the quadratic equation have no real roots. 6x2 + x – 15 = 0
. 7. Find two consecutive odd positive integers whose squares have the sum 290. If one side of a right triangle exceeds the other by 7 cm and the hypotenuse is 13 cm. 2x2 – 5x = 0 3. Find the number . Roots of the quadratic equation ax2 + bx + c = 0. 8. a ≠ 0 are given by
z
− b ± b 2 − 4ac 2a
z
b2 – 4ac is called discriminant of the quadratic equation. When 9 is added to the number. a ≠ 0 and a. Determine the number. the digits interchange their places. (x – 8) (x+ 4) = 13 2. If the product of their ages is 160. The sum of a number and its reciprocal is
50 . β is given by x2 – (α + β)x + αβ = 0 The value (s) of the variable which satisfy an equation is (are) called a root(s) or solution(s) of the equation.178
Mathematics
3. TERMINAL EXERCISE
z
z
Solve the following equations by factorisation method : 1. b and c are real numbers. Find the sides. 9x2 + 15 x – 14 = 0 5. If the length of a rectangle exceeds the breadth by 4 cm and its area is 525 cm2. 5. find its dimensions. 7
4. 6. is called a quadratic equation in x.
(ii). (iii) and (iv) ? A little careful study of the above patterns shows that the next figures in (i).Number Patterns
181
8
Number Patterns
8. They make the learners motivated to form such new patterns. This becomes a topic of interest and knowledge to predict the next figure in a pattern.1 INTRODUCTION In our day-to-day life.
and
respectively.1
Can you predict the next figures in (i).
Think about these and try to find the reasons for those. pictures. we see patterns of geometric figures on clothes. Consider the following patterns : (i)
(ii)
(iii)
(iv)
Fig. posters etc.
. 8.
. (ii). (iii) and (iv) are
.
.4 RECOGNITION OF NUMBER PATTERNS Suppose you want to purchase a handkerchief whose cost is Rs 5. 6. 4.. 2 2 (iii) 10. In this lesson. Number patterns play an important role in the field of Mathematics. . 8. It is an interesting study to find whether some specific names have been given to some of the above number patterns and the methods of finding some next terms of the given patterns.P. 8. . 11111. If you want to purchase two handkerchiefs..2 OBJECTIVES After studying this lesson. 4... 16. (vi) 1.. . 8. 2 16 128 1 1 1 .
z z
8. you will study about number patterns called Arithmetic Progressions and Geometric Progressions. –2.. . 1 1 1 . . then you have to pay Rs 10. 2 3 4
(vii) 1. 2.182
Mathematics
Likewise number patterns are also faced by learners in their study. ... 1 .3 EXPECTED BACKGROUND KNOWLEDGE
z z z z
Idea of numbers Idea of the number line Knowledge of various number systems Four fundamental operations on numbers
8. 32. 1111. 2 .. Let us study the following number patterns : (i) 2. 7. determine nth term of an Arithmetic progression (A.. .. the learner will be able to :
z
recognise number patterns and identify those which are Arithmetic or Geometric progressions. (v) 4. 3. . (iv) 2.P) find the sum to 'n' terms of an A. 10.
. 1 1 (ii) 1. . 11. . 4. 111. .. You will also study methods of finding general term and sum to 'n' terms of an Arithmetic Progression.. 1.
three. . can be obtained by subtracting a constant number 2 and 4 respectively from the preceding term.. 15. 1 and 5 respectively. 6. .. the respective costs of one. –4. 10. 2. 8.. In number patterns (a). (b) 4. But in the number patterns (f).Number Patterns
183
Therefore. .. 0. 9. you will find that each successive number. if the number of handkerchiefs is increased by one successively. (c) 3. 12. other than the first term.. 3. these numbers form a number pattern. . (d) 13. other than the first number.. . 1. Therefore. Can you recognize the relationship between any consecutive numbers of the above pattern ? If you observe the pattern carefully. Some examples of numbers patterns are given below : (a) 1.. 1. can be obtained by adding a constant number 1. other than the first term... 4
(h) 5. ... each term.. 4.5 ARITHMETIC PROGRESSION You are all aware that salaries of employees are often calculated on the basis of their basic salary. the respective cost (in Rs) would increase by 5 every time. 2.. Each number of the number pattern is called a term. 14 . .. plus fixed increments (increases) for each year of service and all other usual allowances. –3. 4. handkerchiefs would be 5. 3 and 4 respectively to the preceding term.. (c) and (e). can be obtained by multiplying the preceding term by a constant number 2. 8. is obtained by adding a constant number to the preceding number. 5. 6 . 4
. Suppose a person begins to work for a firm in the scale of Rs 4500-125-7000. 125. 4 4 4 4 (f) 1.e. 625. 1 1 1 1 (e) 2 .. two. . In the number patterns (b) and (d). 10 ... 4. each successive term. . other than the first term. –7. (g) 16... each successive term.. 1 . . 16. 25. i. –2. 2. (g) and (h). 9.
. 4. . 4. 2. 25. 8. The constant number is called common difference. The first term of this pattern is 4500 and each successive term.2
. . 16. Probably the simplest arithmetic progression is of natural numbers : 1. 7.P)... 6. 19.. 4625.. 9. as in the A.. 14 1 . 3. 4. 6. 8. 6. A progression is said to be an Arithmetic Progression (abbreviated as A. except the first. An arithmetic progression can be represented on the number line by a series of points placed the same distance apart.P 10. 17. other than the first term.
Fig. 10 1 . –5. . –2... 4 4 4 4 –7.. The multiplication tables are all familiar A.... Rational numbers can serve as the common difference. 4875. Such a number pattern is called an Arithmetic Progression (abbreviated as A.. 8.184
Mathematics
Then in successive years his basic salary (in Rs) will be 4500. –2. if the difference of each term. 17. 15. from its preceding term is always the same.. 5000 and he will get usual allowances on the basic salary. 4. 20. 21.2. .. An arithmetic progression can start from any number. . For example. 5. 10. the progression 13. 8.. 1.P : 1 1 9. 12. 6 1 . 2 2 and negative numbers.. 2. 11 . 2 1 . 14. .P). 3. We can say that the terms of the pattern progress arithmetically. 3.. as in the A. 8. can be obtained by adding a constant number (here increament in Rs) 125 to the preceding number. is shown in Fig.P's. 11. 4750. . 14. . positive or negative for example : 5. 16 . ... 12. 8..
. . . 10 1 . . a + 3d .. (c) 13. 19. ? (a) 3.. (d) 21.. 12. 6. 22. 18.P : 3. .. t3.. 13 1 .P are usually denoted by t1.P in the following manner :
In general. 14. 15. 12. It is clear from the above A.. 5. .. ... a + d .P.. 13. the common difference may be negative in which case the terms of the progression would decrease. t2. . 2 2
The first term of this A.P is 15. 7. . 24. 17. 6...P that its first term is 3 and the common difference is 2 (usually denoted by 'd') You can rewrite the above A. the standard form of an Arithmetic Progression would be a ..1 : Which of the following progressions are A. 9.. . Observe the following A. For example : 15. 9. 11.Number Patterns
185
The successive terms of an A. 1 1 1 1 The common difference = 13 – 15 = 12 – 13 = 10 – 12 = –1 2 2 2 2 Let us take some examples to illustrate : Example 8.
Solution : (a) The given progression is 3. It should be noted that. 9.. The common difference is usually denoted by 'd'. if the first term is denoted by 'a' and the common difference by 'd'.
.. 12. (b) 9. 9. a + 2d ..
. 18. as far as the required term by adding 2 successively to the first term.. Here the first term is 9 and the common difference is = 14 – 9 = 19 – 14 = 24 – 19 = 5
∴ The given progression is an A.6 THE GENERAL TERM (OR nTH TERM) OF AN ARITHMETIC PROGRESSION If we need to find a particular term of an A. We can find an easier method if we look at the progression more closely. Here the first term is 21 and the common difference is 17 – 21 = 13 – 17 = 9 – 13 = – 4
∴ The given progression is an A..
. 13. 22.. But its common difference is not the same as 15 – 13 ≠ 18 – 15 ≠ 22 – 18
∴ The given progression is not an A. we can. 10. . 24. of course. . The first term (a) for an A. The successive terms can be formed by adding 2 to the first term. 27. 14... .
(c) The given progression is 13. .
CHECK YOUR PROGRESSION 8. 12.P.P. But this method would become rather laborious if you had to find. 29. Here the first term is 13. 18..P. as 3.P is 3 and the common difference (d) = 2. 12... (b) 7. 9.
(b) The given progression is 9. 15. . 24.P.P's ? (a) 6.P.186
Mathematics
Here the first term is 3 and the common difference = 6 – 3 = 9 – 6 = 12 – 9 = 3
∴ The given progression is an A. 14.. . .1 1. –3. say 102nd term of the progression.. for example 12th term whose first term is 3 and common difference is 2... (d) 31. 19. –7. 5.P.. 9. 17.. . Which of the following progression are A.
(d) The given progression is 21. –5.. 25. build up the A. (c) –1. 7.
8.
.. tn – 1. .P's : (i) 6. (ii) 12. Example 8. whose first term is 'a' and common difference is 'd'.. 14.
∴
a = 5 and d = 8 – 5 = 3 tn = 5 + (n – 1). –15...188
Mathematics
If you see the above pattern. 10.. 8.. a + 2d. a + d. (ii) –7. 11. 18. the formula for the nth term or tn would be a + (n – 1)d Hence.P. 6. –19. –15. the first term (a) is 12 and the common difference (d) = 9 – 12 = –3.
(ii) The given A... 14.. is 12. 8. is given by tn = a + (n – 1)d Here..P..P is 5. we can rewrite the standard form of an A...P is given by tn Here.. .2 : Identify the first term and the common difference of each of the following A.
∴
=a + (n – 1)d a = –7 and d = –11 – (–7) = –11 + 7 = –4 tn = –7 + (n – 1).P's and find the 100th term of each of the following : (i) 5. 18.. –11. a + (n – 1)d where tn–1 denotes the (n – 1)th term of the A.. 14. 6.3 = 5 + 3n – 3 = 3n + 2
∴
100th term = t100 = 3 × 100 + 2 = 302. the first term (a) is 6 and the common difference (d) = 10 – 6 = 4 (ii) The given A.P as a. . 11. We know that the nth term of an A. ... 10. – 19. .P.3 : Write the expressions for nth terms of the following A. Solution : (i) The given A. . Solution : (i) The given A.P is –7. –11. . Here. 14. nth term of an A.(–4)
. 9. Here.P is 6.. Example 8. 9.. ...
. Find the first term and the common difference. 5 1 .P be 'a' and 'd' respectively. It's nth term is given by tn = a + (n – 1)d
∴
2nd term = t2 = a + (2 – 1)d = a + d 6th term = t6 = a + (6 – 1)d = a + 5d
By the given condition a + 5d = 0 and a + d=4 .P is 6 . ..7 : Which term of the A.(i) . Solution : Let the first term and common difference of the A. we get a + 5(–1) = 0 or a=5 or d = –1
∴ The first term of the A.(ii)
Subtracting (ii) from (i).P.. 1 .
Example 8.. is 6 ? 4 4 2 4 4 Solution : The first term (a) = Common difference (d) = 1 Let the nth term be 6 4 or 1 + n −1 1 25 = 4 4 4 1 4
1−1= 1 2 4 4
b g
or 1 + (n – 1) = 25 or n = 25
1 Hence.P is zero and the 2nd term is 4... 1. is 5 and the common difference is –1..190
Mathematics
Example 8.. 4
. 25th term of the above A. 3 .P 1 . we have 4d = –4 Putting d = –1 in (i).6 : The 6th term of an A.
(i) t3 = a + 2d
. that of 2nd brother is 30 years and of the eldest brother is 45 years.(ii) .(ii)
Putting a = 30 in (ii).(i)
3a – 3d = a + d 2a = 4d a = 2d (a – d) + a + (a + d) = 90 3a = 90 a = 30 30 = 2d . t4 = a + 3d t7 = a + 6d and By the given condition a + 3d = 3a or and or or 2a – 3d = 0 a + 6d = 2(a + 2d) + 1 a – 2d + 1 = 0 a = 2d – 1 .9 : The fourth term of an A.. Find the ages of all. The age of the youngest brother is onethird the age of the eldest. a + d By the given condition a – d = or or or and or or
1 (a + d) 3
..... we get or d = 15
Putting the values of a and d in (i).P is equal to three times its first term and 7th term exceeds twice the third term by 1. Solution : Let the first term and common difference of the A. Find the first term and common difference. we get a – d = 15 a = 30 and a + d = 45
∴ The required age of the youngest brother is 15 years. be 'a' and 'd' respectively.P...
Example 8. a .8 : The ages of three brothers are in A. Solution : Let the ages (in years) of three brothers be a – d .P.Number Patterns
191
Example 8.. if the sum of their ages is 90 years.
P is equal to n times its nth term.192
Mathematics
Putting the value of a in (i) we get. d = +11.P. Solution : Let the three terms of the A.. 2(2d – 1) – 3d = 0 or 4d – 2 – 3d = 0 or d=2 2a = 6 or
∴
Substituting d = 2 in (i) . Solution : Let the first term and common difference of the A. 7. a + d By the given condition (a – d) + a + (a + d) = 21 or or and or or or or or or or 3a = 21 a = 7 (a – d)2 + a2 + (a + d)2 = 389 a2 – 2ad + d2 + a2 + a2 + 2ad + d2 = 389 3a2 + 2d2 = 389 3 × 72 + 2d2 = 389 147 + 2d2 = 389 2d2 = 242 d2 = 121 d = ± 11 [using (i)] . the three numbers are 18.P be 'a' and 'd' respectively
∴
tm = a + (m – 1)d tn = a + (n – 1)d
By the given condition m × tm = n × tn
.
Example 8. 18 When a = 7. –4. d = –11. the three numbers are –4.11 : If m times the mth term of an A. prove that its (m + n)th term is zero. 7. are 3 and 2 respectively.P whose sum is 21 and sum of their squares is 389.we get a=3
The required first term and common difference of the A. a.. Example 8.(i)
When a = 7.P be a – d.10 : Find three numbers in A.
It is shown by following figure :
Step 0 : ........ In this case it is 3 : 1 as 9 = 27 = 81 = 243 = 3 3 9 27 81 1 It is usually denoted by 'r' Such a progression is called a Geometric progression (abbreviated as G....... 27......... .. 243.........
.(i)
If you observe each term of the above pattern/progression............... Consider that he/she conveys this message to two persons and both of them pass on this message to two other persons each and so on..
Step 1 : . 81.........) Let us take another example to consolidate : Consider that a person reads from the newspaper that the Government is stressing on more security for the railway passengers.......................... If we consider that the starting prize in a game of 'Triple Your Money' is 3... then repeated tripling (multiplication by 3) would form the progression of possible prizes as 3.P.... 9.....
Step 2 : .....
Step 3 : ............... .198
Mathematics
be formed by repeatedly multiplying or dividing its preceding term by a non-zero constant............ you will find that there is a common ratio between two consecutive terms.....
.. (ii) 5. 12.(ii)
The ratio of any two successive terms of a Geometric Progression is called its common ratio.. 27. Both can be rewritten as as follows :
In general if 'a' be the first term and 'r' be the common ratio. A progression is said to be a Geometric Progression (abbreviated as G.. 16.. 8.. 8.P's : (i) 3.). if the ratio of any two successive terms is always the same. . You can verify that common ratio in (ii) is always the same as 2 = 4 = 8 = 16 = 2 1 2 4 8 . The following progressions are all examples of G. .P.P's in (i) and (ii) 3. 20. and 1... 2. 10. This type of progression are called a Geometric Progression.. 1. 40. Now observe the G. . you will get the following progression. 192. 2. the standard form or general form of a Geometric progressions would be a .. 9. ar2 .. 48. . ar3 . If you calculate the number of persons informed at each step.. 16. . .
. ar . 81.. 4. 4.Number Patterns
199
Generally rumours spreads in this manner..
. . 15.). LET US SUM UP
z
A progression is said to be an Arithmetic Progression (abbreviated as A.P will decrease.. The successive terms of an A. (d) 45. except the first. .P.P. Identify Geometric Progressions from the following progressions : (a) 4. ... tn where 'a' and 'd' are the first term and common difference respectively. 5.. Find the common ratio of each of the progressions given in Question Number 2... .P is given by a . Here the first term = –3 Common ratio =
+15 −75 = −5 . 20. 500. = −3 +15
CHECK YOUR PROGRESSION 8. 17. .. The common difference in usually denoted by 'd'. 25. 15. (b) 24.. 4.. from its preceding term is always the same.. 8. . .P's : (a) 16. the terms of an A... . 3 9 27 (c) 125... –75.
2. 6..4 1. 1. a + 2d . In that case. t2. The standard form of an A. 20.Number Patterns
201
(b) The given progression is –3. are usually denoted by t1. 5. .. 12.. .. The general term (nth term) of an A.
z z z
z
The common difference may be negative also.P is given by tn = a + (n – 1)d.. 100. a + d . Show that the following progressions are in G.
z
. 14. 1 1 1 (b) 1. tn–1 . . 1. . 2... if the difference of each term. (c) 11. 3. t3.
z
The ratio of any two successive terms of a G. 3 . 8.. TERMINAL EXERCISE
z
1.P. ar2 . . 2
b g
z
A progression is said to be a Geometric Progression (abbreviation as G... 8 . .'s ? (a) 13 . 19 . .. 2 2 22 . ar . 20 . .P is called its common ratio and is usually denoted by 'r'. 2 . is 56 ?
. : (a) 5...P whose first term is 'a' and common difference is 'd' is given by Sn = = n 2a + n − 1 d 2
m b gr
n a+l ... 1. 2. 3 3
3.. . 1. (c) 3 .. .. 11.P whose nth term is given by (a) n (c) 2n +1 3 (b) 3 – 5n n (d) – + 1 2
4.P. 52. 2. Show that the following progressions are in A. Which of the following progressions are A... –5. 7 7 5. The standard form of a Geometric Progression is given by a .. .. where l is the last term... Which term of the following A. 4.P's : (a) –1. .. . 1 1 1 (d) 1.. 2 3 4
(c)
(e) 22. . . where 'a' and 'r' are the first term and common ratio respectively. 5 ..) if the ratio of any two successive terms is always the same. 7 7 (b) –1.. Find the general term (or nth term) of the following A. 13 .202
Mathematics
z
The sum of the first n terms of an A. 2 2 (b) (d) 15 . . . –3. 82.P. –1. 1 . ..
9. 25... 4 16 (c) 16.P are 8 and 14 respectively.P. 125 25 5
.. 28. − . as 1625 ? 6. 11. 11. –186. . 49. 10.. 12. 6.. . The sum of three consecutive terms of an A. is –14 ? 5.. is 506 ? (c) –190.... –182.P is twice its 8th term.. Find the sum of the following A...Number Patterns
203
(b) 1. How many terms are needed to make the sum of the A. 7.. 9. If its 6th term is –8.. 16 terms (c) 1 + 9 + 17 + .000 in 10 years. 8. . The 14th term of an A.P's for given number of terms : (a) 7 + 3 – 1 – 5 .. 20 terms
(b) 15 + 13 + 11 + . 20 terms
7. find the first term and the common difference.. Find the terms. 1 1 (b) 1. .. A man saved Rs 20. Show that the following progressions are in G.. If 5th and 9th terms of an A....P.. Find the sum of all even numbers upto 1001. . If the saving of each year is Rs 100 more than the saving in the preceding year. 31. 1 . find the first term and common difference. .P is 36 and their product is 1428. find his saving in the first year. . 1 . . : (a) 1. 12. (d) 1 ..
. For that they levy certain taxes on citizens. Sometimes when we are not able to return loaned money on time. called interest. Sometimes we have to buy articles on instalments because of non-availability of adequate funds.e. the financer starts charging interest on interest also. Due to greater supply of goods or sub-standard goods they are to be sold on loss. Due to this the students are taught to calculate interest when they buy articles on instalment plan. The formulae of compound interest is also used in finding increase or decrease in prices of things.) worked on fractions and ratio and proportion.. In 900 A. Brahmgupta. Taking the lead from this. in their nest or caves. The latent meaning of this is to save something for difficult times. there is a description of levying taxes. This encourages citizens to save and keep the money safe.D. Banks and other financial institutions keep the money of their customers and on the expiry of the period pay extra money. This is why calculation of interest on deposits in banks is included for teaching. This is also taught under "Appreciation and Depreciation" of value. The learners are. Bakshali Manuscript was discovered which had a number of problems on Arithmetic. Due to this the study of compound interest has been included in this module. introduced to percentage and profit and loss. Sridharacharya).
. The Government provides a number of facilities to the citizens. Yodoksu (370 B.C. the students have been told about the importance and need of savings in this module. keep your expenditure less than your income. In the reigns of Ashoka and Chandragupta.208
Mathematics
Module 2
Commercial Mathematics
It is a common saying by elders "spend within your limits" i. To keep your savings safe is another tough task. Two of these taxes are sales tax and income tax to which the learners are introduced in this module. Mahavira. There is a description of many mathematicians working on practice and proportion (like Aryabhatt. Financial transactions about buying and selling are generally done for profit. therefore. Many Indian mathematicians have worked on the topic of commercial Mathematics. You must have seen birds and animals saving eatables for rainy season. which is called compound interest. in addition to the money deposited.
9.Ratio and Proportion
209
9
Ratio and Proportion
9. direct and inverse proportion (variation) and use these concepts to solve real life problems.
9. we talk of the ratio of two numbers. Another way is to see one number is how many times the other number.3 EXPECTED BACKGROUND KNOWLEDGE
z z z
Four Fundamental operations on numbers Comparison by division Knowledge of different units of measures and their conversions. we say that we have formed a ratio of the two quantities. partnership. we compare the numbers by difference and in the second we compare the numbers by division. In this lesson we shall study to write a ratio in the simplest form. work and wages etc.
9. time and work. In the first case. the learner will be able to :
z z z z
write a ratio in the simplest form determine whether given four numbers are in proportion illustrate the concept of direct and inverse proportion (variation) solve real life problems pertaining to time and distance. work and wages and partnership. introduce the concept of proportion. pertaining to time and distance.4 CONCEPT OF RATIO When we compare the two quantities by division.2 OBJECTIVES After studying this lesson. In the second case. time and work.1 INTRODUCTION You are aware of the concept of comparison of numbers in two different ways.
. One way is to see which number is larger and which is smaller.
Here. For example. the order of two terms in a ratio is very important. Example 9.e.
. Remarks : From Example 1 given above.. we do not compare 5 boys and 6 cows or 10 kg and 500 m or 5 toys and 3 fruits.2 : A labourer earns Rs 3200 a month and spends Rs 2500. is 90 : 37.1 : In a class. we compare (3 × 30) days and 37 days. thus. b
In the ratio 90 : 40. This is also written as a . to compare 3 months and 37 days. the ratio of any number 'a' and 'b' is written as a : b. there are 28 girls and 16 boys. we can say that the ratio of the weight of a bag of wheat to the weight of a bag of rice is 90 : 40 (read as 90 is to 40) The symbol ':' is used to denote a ratio. Find the ratio of his (i) expenditure to income (ii) savings to income (iii) savings to expenditure Solution : Here Income = Rs 3200. Remark : We cannot compare two quantities if they are not of the same kind i. it is necessary to express the two quantities in the same unit. It may be noted that in the above examples ratio is not 3 : 37. 90 and 40 are called terms of the ratio. Thus. Hence. the ratio a : b is different from the ratio b : a. Find (i) the ratio of the number of boys to that of girls (ii) the ratio of the number of girls to that of boys. The ratio. Solution : (i) The required ratio is 16 : 28.210
Mathematics
Consider an example of a bag of wheat with weight 90 kg and a bag of rice with weight 40 kg. Expenditure = Rs 2500
∴
Savings = Income – Expenditure = Rs 3200 – Rs 2500 = Rs 700
(i) Ratio of expenditure to income = 2500 : 3200 (ii) Ratio of savings to income = 700 : 3200 (iii) Ratio of savings to expenditure = 700 : 2500. The first term (90) is called the antecedent and the second term (40) is called the consequent. (ii) The required ratio is 28 : 16. Example 9. To compare two quantities as a ratio.
Hence. because 10 is a common factor of its two terms. Solution : (i) Since the given quantities are in different units. Hence. Divide each of the terms by the H. How to obtain the simplest form of a given ratio ? For a given ratio.4. getting 1000 ml.C. obtained. The simplest form of the ratio 20 : 50 is 2 : 5. 9. or 4 : 10 or 2 : 5. getting 7000 m. find the H. i. other than 1. the required ratio is 7000 : 750. the required ratio is 250 : 1000.F. the required ratio is 5 : 7.F.2 Ratio in the Simplest Form A ratio a : b is said to be in the simplest form if its two terms do not have a common factor.Ratio and Proportion
211
Example 9. therefore first convert 7 km into m. Hence. of its two terms a and b. For example.1 Equal Ratios A ratio remains unchanged when both of its terms are multiplied or divided by the some nonzero number. Note : A ratio can be expressed in several ways. The ratio 20 : 50 can be written as 10 : 25. 9. the ratio 20 : 50 is not in the simplest form. HCF of the two terms is 1. The ratio formed of the two new numbers obtained is the simplest form of the given ratio. say a : b. (iii) The two quantities are not in same units.C.4. the two terms 2 and 5 do not have a common factor other than 1. etc. we convert 1 week into days. We. we therefore convert 1 litre into ml. as seen above. Hence.
.e. (ii) The two quantities are not in the same units. getting 7 days. 20 : 50 is equal to 10 : 25 or or 4 : 10 2:5
Note that in the ratio 2 : 5.3 : Find the ratio of (i) 5 days to 1 week (ii) 7 km to 750 m (iii) 250 ml to 1 l.
is (v) The ratio 5 : 4 is of the ratio 100 : 80. Narmita earned Rs 84000 and paid Rs 1200 as income tax. 49 of them are men and the rest are women.C. Find the ratio (in the simplest form) of : (i) the number of men to that of women (ii) the number of women to the total number of workers
. Fill in the blanks to make each of the following a true statement : (i) In the ratio 3 : 7.
∴
6 : 24 =
6 : 24 =1:4 6 6
Thus. 5.50.
2. Find the ratio (in the simplest form) of : (i) 45 cm to 5 m (iii) 90 paise to Rs 3 (v) 1 hour to 15 seconds (ii) 200 g to 5 kg (iv) 35 minutes to 45 seconds (vi) Rs 38 to Rs 9. . CHECK YOUR PROGRESS 9.32 : 1. In a year. in the simplest form. Find the ratio (in the simplest form) of her (i) income to income tax (ii) income tax to income. Express each of the following ratios in the simplest form : (i) 39 : 65 (iv) 120 : 144 (ii) 200 : 350 (v) 0. the simplest form of the ratio is 3 :8. of 6 and 24 is 6.Ratio and Proportion
213
∴
45 : 120 =
45 : 120 =3:8 15 15
Thus.
(iii) The simplest form of the ratio 15 : 20 is
(iv) The ratio of 2 months to 2 days. So.1 1.
4. the simplest form of the ratio is 1 :4.2 (iii) 172 : 528 (vi) 4860 : 8370
3. (iii) The ratio of 6 hours to 1 day is the same as the ratio for 6 hours to 24 hours (1 day = 24 hours). the antecedent is (ii) The ratio 3 : 5 is and consequent is
from the ratio 5 : 3. Total number of workers in a factory is 196.F. . the ratio is 6 : 24 The H.
Solution : We have Sum of the terms of the ratio = 3 + 5 + 7 = 15
∴
3 × 1050I FH 15 K F5 I B's share = Rs H 15 × 1050K A's share = Rs C's share = Rs 7 × 1050I FH 15 K
= Rs 210 = Rs 350 = Rs 490. l + m.
and
. Example 9. the two parts of 144 with given ratio are 63 and 81. we follow the procedure given below : Step 1 : Find the sum of the two terms of the ratio. A rectangular sheet of paper is 30 cm long and 21 cm wide. Solution : We have Sum of the two terms of the ratio = 7 + 9 = 16
∴
First part = second part =
7 × 144 = 63 16 9 × 144 = 81 16
and
Thus. Step 2 : Use the following formula to obtain the two parts of the given number x. Find the ratio (in the simplest form) of its (i) width to length (ii) length to perimeter. First part = Second part = l ×x l+m m ×x l+m
Remark : The same procedure is followed even if the given number is to be divided in more than two parts in the given ratio.5 DIVISION OF A NUMBER IN THE GIVEN RATIO To divide a given number x in the given ratio l : m.e.6 : Divide 144 in two parts in the ratio 7 : 9. 9. i.214
Mathematics
6. Example 9. C in the ratio 3 : 5 : 7.7 : Divide Rs 1050 among A. B.
If the perimeter is 36 cm. Thus. On expressing these ratios in the simplest form.2 1. PROPORTION An equality of two ratios constitutes a proportion. The ratio between two quantities is 2 : 7. [Hint : Sum of the three angles of a triangle is 180°] 5. then find the number of teachers.e. (i) Divide 15 in the ratio 2 : 3. 2. c. (ii) Divide 184 in the ratio 3 : 5 (iii) Divide 7780 in the ratio 7 : 8 : 5. d (in order) are said to be in proportion. If the total number of persons in the camp is 159. b. Divide : (i) Rs 140 in the ratio 2 : 5 (ii) Rs 154 in the ratio 3 : 4 (iii) 9 cm 8 mm in the ratio 2 : 5 (iv) 10 cm 5 mm in the ratio 1 : 4.
. 3. Find the angles. if the ratio of the first two is equal to ratio of the last two i.Ratio and Proportion
215
CHECK YOUR PROGRESS 9. the ratio of the number of clerks to the number of officers is 17 : 2. Consider two ratios 8 : 14 and 20 : 35. the ratio of the number of teachers to the number of students is 3 : 50. we find that 8 : 14 = 4 : 7 and Therefore.8 kg. 9. find the number of clerks in the bank. 6. The sides of a triangle are in the ratio 2 : 3 : 4. In a camp.6. we define proportion as follows : "Four numbers a. In a bank. a : b = c : d". If the total number of persons working in the bank is 57. how much is the first ? 4. 7. 20 : 35 = 4 : 7 8 : 14 = 20 : 35 8 : 14 = 20 : 35 is a proportion. Similarly. If the second quantity is 9. 20 : 70 = 2 : 7 is a proportion. The angles of a triangle are in the ratio 3 : 5 : 7. then find its sides. Thus.
30. 5.8 above. 10. 60 are in proportion. Example 9. a : b=c : d ⇒ a=c b d or ad = bc. 5 Solution : (i) Here. 25. 60 in proportion ? Solution : Ratio of first two terms is 5 : 10. we find that the order of four numbers for a proportion is important. which is equal to 1 : 2
∴
5 : 10 = 30 : 60
Hence. (ii) Here. then we write a:b::c :d which is read as "a is to be as c is to d" or "a to b as c to d". second. c.
. b. 5.9 : Are. 15. 10. In other words a : b : : c : d if and only if ad = bc. Example 9. Aliter Here. b. The product of extremes = 3 × 25 = 75 The product of means = 5 × 15 = 75 Since the two products are equal. the second and third terms are called means. 5. 25 (ii) 3. c and d are the first. 30. In a proportion a : b : : c : d. the four numbers are not in proportion.
Note : In a proportion. which is equal to 1 : 2 Ratio of last two terms is 30 : 60.8 : Which of the following four numbers constitute a proportion ? (i) 3.216
Mathematics
When four numbers a. d (in order) are in proportion. third and fourth terms of the proportion. the product of extremes is equal to the product of means. then given numbers are in proportion. The first and fourth terms are called extremes. Product of extremes = 5 × 60 = 300 Product of means = 10 × 30 = 300 Since the two products are equal. Remark : From Example 9. The product of extremes = 3 × 5 = 15 The product of means = 15 × 25 = 375 Since the two products are not equal. the four numbers are in proportion. 15. a.
We know that for a proportion. then find the other mean. 8. ⇒ ⇒ 5 × 30 = 10 × x x=
5 × 30 = 15 10
Thus.Ratio and Proportion
217
Example 9. In this case. Example 9. If one of means is 24.0 : 4.10 : The first. if a. 8. c are in proportion. 8. Thus. the four numbers 4. we say that a. 10 and 30 respectively. we say that 4. Solution : Let the other mean be x. x.12 : Are 4. product of means = product of extremes ⇒ ⇒ ⇒ 24 × x = 12 × 28 x=
12 × 28 24
x = 14
Thus. 16 are in proportion. b. Which of the following statements are true ? (i) 4 : 6 = 8 : 12 (iii) 16 : 24 : : 30 : 20 (ii) 4 : 28 = 7 : 1 (iv) 6. 16 are in continued proportion. Remark : In Example 5 above. 30 are in proportion.6
. 8. 8. c are in continued proportion. Solution : Let the third term be x so that 5. Find the third therm. b. b2 = ac and b is called the mean proportional of a and c. Example 9. 16 in proportion ? Solution : Here.11 : The two extremes of a proportion are 12 and 28.5 : : 4.3 1. CHECK YOUR PROGRESS 9. Product of means = 8 × 8 = 64 Product of extremes = 4 × 16 = 64 Since the two products are equal. second and fourth terms of a proportion are 5. the third term is 15.8 : 3. In such a case. 10. the means are the same. b. the other mean is 14.
12. you see that as the number of pens purchased increases. 27. more money you have to pay. you will have to pay only Rs 8. find the value of x. 12 are in continued proportion. Here you see that more the number of pens purchased. where k is a constant. less the number of pens purchased. lesser the number of days. 10 (iv) 40. We say that the number of persons employed to complete a job and the number of days taken by them are in inverse proportion (variation). 9. x. the number of days required to complete the job decreases. if you buy 4 pens. 10. 16. 7.7. where k is a constant of b
. the price you have to pay also increases. 50 are in proportion. 30. Thus. If 3. called the constant of variation or proportionality. We also say that a varies directly as b and write a = kb. x. we say that the number of pens purchased and the price paid are in direct proportion. 15 (ii) 12. 6. Determine which of the following numbers are in proportion : (i) 18. 9.7 DIRECT PROPORTION (VARIATION) If you buy 6 pens for Rs 12. if x : 6 : : 5 : 3. 4. Thus. Find the mean proportional between 5 and 125. Find the value of x. two quantities a and b are said to be in direct proportion if increase/decrease in one quantity results in the corresponding increase/decrease in the other. less money you have pay. k or ab = k. Again. we see that lesser the number of persons employed. if a = proportionality. In general. x. 18 (iii) 2. 18.1 Inverse Proportion (Variation) If two persons. then 4 persons employed to do the same job will take 12 days to complete it. We may note that as the number of persons employed increases. 35 and 42. To indicate the above situation. take 24 days to complete it. Likewise. 5. 6 persons will take only 8 days to complete the job.218
Mathematics
2. employed to construct a wall. 60. We say that a varies inversely (indirectly) to b. 45
3. you will have to pay Rs 16 for 8 pens and Rs 24 for 12 pens. 14. find x. more the number of days Or more the number of persons employed. 3. In general. Set up all possible proportions from the numbers 15. two quantities a and b are said to be in inverse proportion if increase/decrease in one quantity results in corresponding decrease/increase in the other quantity. If 18.
Example 9. How many pages can he type in 7 days? Solution : Let y denote the number of pages that the typist would type in 7 days Since the number of pages vary directly as the number of days.Ratio and Proportion
219
9. the cost of 8 books is Rs 112 Example 9. we get the following : Quantity 5 8 5 : 8 = 70 : x ⇒ ⇒
5 70 = 8 x
Cost (in rupees) 70 x
Since the cost varies directly as quantity. we get 5 : 7 = 70 : y ⇒ ⇒ 70 5 = y 7 y=
70 × 7 = 98 5
Number of days 5 7
Thus. Equating the two ratios we get
x=
8 × 70 = 112 5
Thus. how many men should be employed to build 120 m long wall of the same height in the same time.8 APPLICATIONS OF DIRECT PROPORTION We will take some examples to illustrate the application of direct proportion (variation) in daily life situations.15 : If 15 men can build a 100 m long wall in some days. we get the following : Number of pages 70 y Equating the two ratios. Example 9.14 : A typist types 70 pages in 5 days. Since the number of men varies directly as the length of the wall. the typist will type 98 pages in 7 days. how much will 8 books cost ? Solution : Let Rs x denote the cost of 8 books.13 : If 5 books cost Rs 70. we get the following :
. Solution : Let n denote the number of persons to be employed to build a wall 120 m long.
220
Mathematics
Number of men 15 n Equating the two ratios. 60
Thus.17 : If a deposit of Rs 3000 carries an interest of Rs 900 in some time.16 : A jeep travels 50 km in 1 hour. Example 9. we get 100 : 120 = 15 : n ⇒ ⇒
100 15 = 120 n
Length of wall (in metres) 100 120
n=
15 × 120 = 18 100
Thus. the jeep would travel 10 km in 12 minutes. 18 men will build 120 m long wall in the same time. what interest would a deposit of Rs 5000 earn in the same time ? Solution : Let I denote the interest that Rs 5000 would earn. Since the distance travelled and time taken vary directly we get the following : Time (in minutes) 60 12 Equating the two ratios we get 60 : 12 = 50 : d ⇒ ⇒
60 50 = 12 d
Distance (in km) 50 d
d=
50 × 12 = 10. Since the interest earned and the amount deposited vary directly. we get the following : Principal (in Rs) 3000 5000 Interest (in Rs) 900 I
. How much distance would it travel in 12 minutes ? Solution : Let d denote the distance (in km) travelled by the jeep in 12 minutes. Example 9.
In partnership. If 7 pens cost Rs 98.e. 2. we get 3000 : 5000 = 900 : I ⇒ ⇒
3000 900 = 5000 I
I=
5000 × 900 = 1500 3000
Thus. how many erasers can be bought for Rs 50 ?
. These individuals are called partners. all or some partners invest money in business for same or different periods of time. and that of B is Rs (35000 – 14000) i. 21000. The following examples illustrate the above concept.9 PARTNERSHIP When two or more individuals enter into business jointly we say that they enter into a partnership.18 : A and B start a business by investing Rs 40000 and Rs 60000 respectively. we get 40000 : 60000 = x : (35000 – x) ⇒ ⇒ ⇒ ⇒ ⇒ ⇒
40000 x = 60000 35000 − x 2 x = 3 35000 − x
2(35000 – x) = 3x 70000 – 2x = 3x 5x = 70000 x = 14000
Thus A's share is Rs 14000. 9. If the cost of 12 erasers is Rs 30. Then B's share is Rs (35000 – x). equating the two ratios.4 1.Ratio and Proportion
221
Equating the two ratios. Since share of profit varies directly as the investment. the deposit of Rs 5000 would earn an interest of Rs 1500 in the given time. Solution : Let A's share be Rs x. The profits or losses of a partnership business are shared among the partners in the ratio of their investments. find the share of each in the profit. Rs. CHECK YOUR PROGRESS 9. If at the end of the year they earn a profit of Rs 35000. Example 9. find the cost of 35 pens.
If 5 litres of petrol cost Rs 140. find the thickness of a pile of 30 similar cardboards ? 16. A and B start. A and B start a business with Rs 70000 and Rs 80000 respectively. Rs 18000 paid for the rent from the profit of Rs 50700. They earn Rs 50700 as the annual profit. In how many days would 15 men dig the same trench? 14. A man earns Rs 1750 per week. In how many days would he prepare 27 such chairs ? 5.222
Mathematics
3. 19. should be appointed. 4 taps of equal capacity can fill 6 tubs in a given time. How many tubs can be filled by 10 taps in the same time ? 9. A carpenter prepares 18 chairs in 4 days. they earned a profit of Rs 40500.e. A car travels 160 km in 4 hours. A woman can pack 270 bundles in 3 days. Find the sum invested by each. how many meters of cloth can be woven by 240 men in a day ? 8. How much will it earn in 5 years ? 17. a business with Rs 60000 and Rs 75000 respectively. [Hint. If the thickness of a pile of 20 cardboards is 50 mm. At the end of the year. Subtract Rs (1500 × 12) i. find the share of each after paying the rent. If the interest on Rs 3000 is Rs 150 for a certain period. If 30 men can weave 40 metres of cloth in a day. How long will it take to travel 400 km? 11. 20 farmers plough a field in 8 days. How much will he earn in 4 days ? 13. one gets Rs 35000 and the other gets Rs 18000 as profit. 15. How many bundles can she pack in 7 days? 7. If a family of 5 members consumes 50 kg of cereals in a month.12 lakhs in a joint business. After a year. How many teachers. 3 teachers are appointed. How many days will 16 farmers take to plough the field. Two partners together invested Rs 2. what quantity of cereals would be consumed by family of 6 members in a month ? 10. How many trousers can he stitch in 10 days ? 6. A deposit earns Rs 120 in 24 years. The balance of profit will be shared in the ratio 70000 : 80000] 20. A tailor stitches 10 trousers in 4 days. when the rate of interest remains the same ? 18. Find the share of each partner in the profit.
. For every 50 students. what will be the interest on Rs 4500 for the same period. If Rs 1500 per month out of the profit is paid as rent. if a school has 1250 students ? 12. how much will 8 litres of petrol cost ? 4. 10 men dig a trench in 9 days.
we get the following : Speed 55 40 The ratios of like terms are 55 : 40 and 4 : t. the car will cover the same distance in 5.5 40 Thus.5 hours. How many days will 6 girls take to do the some work ? Solution : Let x be the number of days in which 6 girls can do the work. to cover the same distance. Equating the ratio with the inverse of the other. How much time will it take to cover the same distance at an average speed of 40 km per hour ? Solution : Let t (in hours) denote the time that the car will take at the speed of 40 km per hour. we get the following : Girls 4 6 The ratios of like terms are 4 : 6 and 3 : x Equating ratio with the inverse of the other.20 : A car takes 4 hours to cover some distance at an average speed of 55 km per hour. Example 9. Since the number of girls varies inversely as the number of days.10 INVERSE PROPORTION We will now take some examples to illustrate the application of inverse proportion (variation) in daily life situations. 6 girls will complete the work in 2 days. t=
. we get 4 : 6=x : 3 ⇒ ⇒ 4 x = 6 3 x= 4×3 =2 6 Days 3 x
Thus. Example 9. travelling at a speed of 40 km per hour. Since speed varies inversely as time.Ratio and Proportion
223
9.19 : 4 girls can do a work 3 days. we get 55 : 40 = t : 4 or or
t 55 = 40 4
Time 4 t
55 × 4 = 5.
a is the first term (antecedent) and b the second term (consequent).5 km per hour ? 7. how much time will it take to cover the same distance ? 4.F. b. A man deposited Rs 5000 in a bank for one year to get a fixed amount of interest.C. b. c are said to be in continued proportion. After 5 days. 120 men had provisions for 200 days. b.5 km per hour. If 60 more men join them. of its two terms is 1.5 1. how long will the provisions last ? 3. If 3 men can build a wall in 15 days. If a : b = b : c. d are in proportion if and only if ad = bc. The order terms in a ratio is important a : b is not the same as b : a
z z z z
A ratio is said to be in the simplest form if the H. 6 taps of equal capacity can fill a reservoir in 15 minutes. 400 men have provision for 23 days. In a ratio a : b. how long will 5 men take to build the same wall? 2. Two quantities are said to be in inverse proportion if an increase/decrease in one results in a corresponding decrease/increases in the other. How many taps can fill it in 10 minutes ? 6. c are in continued proportion.5 hours. then a. How long will the remaining provision last ? LET US SUM UP
z z z
Ratio is a comparison of numbers or quantities of the same kind. If a. A journey takes 21 minutes if I walk at 4. a and d are called the extremes and b and c are called the means. To cover some distance. c. Two quantities are said to be in direct proportion if the increase/decrease in one results in a corresponding increase/decrease in the other. If the same interest is to be earned in 8 months at the same rate of interest. then b2 = ac and b is called mean proportion of a and c. what sum head to be deposited ? 5. A proportion has four terms. a train.224
Mathematics
CHECK YOUR PROGRESS 9. In a proportion a : b : : c : d. Four numbers a. A statement indicating equality of two ratios is called a proportion. running at a speed of 80 km per hour. If it runs at a speed of 100 km per hour.
z z z
z
z
. How long will it take if I walk at 3. takes 12. 30 men left.
If the second quantity is 9. Are 5. 5 km (iii) 16 mm. Find the ratio of the following in the simplest form : (ii) 50 minutes. Write each of the following ratios in the simplest form : (i) 64 : 160 (i) 450 m. In how much time will it gain 2 minutes 6 seconds ? 12. A and B started a business by investing Rs 50000 and Rs 75000 respectively. 5.5 : 9 2.12 kg.
. 6 kg 400 g. what will be the sales tax on the purchase worth Rs 1500 ? (ii) 216 : 384 (iii) 1. Find the share of each out of a profit of Rs 15000. How many cups of sugar are needed for 150 persons ? 9. 45 in proportion ? 8. how many boxes will be required to hold 105 dozen ? 10. A 12 m length of an angle iron weighs 24. How many litres of diesel are needed for 600 km ? 13. Divide (i) Rs 154 in the ratio 5 : 9 (ii) Rs 9292 in the ratio 1 : 3.Ratio and Proportion
225
TERMINAL EXERCISE 1. The ratio between two quantities is 2 : 5. A is a working partner and gets Rs 300 per month for the same. 1 hour 5 minutes (iv) 5 kg 600 g. A mini bus uses 35 litres of diesel for a journey of 420 km.5 kg. Determine the value of p if 20 : p : : 25 : 450. If 12 boxes are required to hold 60 dozen apples. find the first quantity. 81. 9. 14. 6. What will a piece of length 8 m of the same weigh ? 11. Find the share of each in an annual profit of Rs 13500. 15. A and B entered into a partnership investing Rs 35000 and Rs 42000 respectively. A watch gains 42 seconds in 3 days 8 hours. If the sales tax on a purchase worth Rs 600 is Rs 42. 3 cm 3. A dish for 250 persons uses 15 cups of sugar. 4. Find the angles. 7. The angles of a triangle are in the ratio 3 : 7 : 8.
you see a banner in the market. you come across situations in which the word 'percent' is made use of very frequently.2 OBJECTIVES After studying the lesson. and the rate of discount
z z z z
. no.5 percent" There are many such situations in different walks of life where the concept of percentage finds its use. instalments etc. more than three) find a single discount equivalent to a given discount series calculate the discount and the selling price of an article. sales tax.1 INTRODUCTION In every day life. 10. the learner will be able to :
z z z z z
write a fraction and a decimal as a percent and vice-versa calculate specified percent of a given number or a quantity solve problems based on percentage solve problems on profit and loss calculate simple interest and amount when a given sum of money is interested for a specified time period on a given rate of interest. given marked price of the article. discount.5 percent to 6. "Sale upto 50 percent off" You read news in the newspaper "Votes turnout in the poll was over 65 percent" "Banks have lowered the rate of interest on fixed deposits from 7. For example. In this lesson we shall study percent as a fraction or a decimal and its application in solving problems of profit and loss. state the need for given discount define discount and discount series (successive discounts.228
Mathematics
10
Percentage and its Applications
10.
10. This means that she has secured 80 marks out of 100 or 40 marks out of 50. when we say a man has spent 20% of his income on food. given the marked price of the article and the rate of sales tax or commission solve inverse problems pertaining to sales tax solve inverse problems on commission determine the amount of each instalment when goods are purchased under investment plan (case of equal instalments only) determine the rate of interest when equal instalments are given
z z
z
10. for example percent. For instances 3 out of 4 equal parts. A ratio whose second term is 100 is also called a percent. we wish to compare two fractions 3 4 and . fractions and decimals.
Fig. The symbol '%' is used for the term percent.3 EXPECTED BACKGROUND KNOWLEDGE
z
Four fundamental operations on whole numbers. it means that out of every hundred rupees of his income. 4 5
Since these are fractions having different denominators we need to convert them into fractions with common denominator. 33 out of 100 small squares are shaded.
. This means 33 of the larger square is shaded. he has spent Rs 20 on food.'
10.1. A fraction denotes part of a whole.Percentage and its Applications
229
z z
solve inverse problems pertaining to discount calculate the sales tax on commission and the selling price of an article. In Fig.1
When we say Anita has secured 80% marks in mathematics. Suppose. 10. Similarly. 100 5 is read as five 100 3 means 4
7 17 means 7 out of ten equal parts and means 17 out 10 100
The word 'percent' is abbreviated form of the Latin word "per centum" which means "per hundred" or "hundredths". A fraction whose denominator is 100 is read as a percent.4 PERCENT You have learnt a lot about fractions. Similarly. of 100 equal parts.
A class of a school had 45% girls.02 (h) 1. Kavita read 84 pages of 100-page book. What percent of the class were boys ? 10.8 (b) 14% (f) 25% (b) 40% (c) 0. What was her percentage of marks ? 6. 100
To write a fraction as a percent. Write each of the following decimals as percent : (a) 0. x% = Conversely. What is the percentage of cotton in the suit piece ? 8.56 (e) 0.03 (f) 0. What percent of the book did she read ? 9.75 (g) 0. What percent of the shoes were there on normal price ?
. Aruna obtained 19 marks in a test of 25 marks. A suit piece consists of cotton and rayon fibre in which cotton is 3 out of 8 parts.1 1. Gurmeet got half the answers correct. Write each of the following percents as fraction : 4.
x . Convert each of the following fractions into percent : (a) 3 4 (b) 1 5 (c) 3 10 (d) 4 25
5. One-fourth of the shoes in a shop were on sale. simplify it and suffix the % sign. For example
1 1 = 4 × 100 % = 25% 4 2
FH IK 3 F3 I = H 2 × 100K % = 150%
3 3 = 5 × 100 % = 60% 5
FH
IK
CHECK YOUR PROGRESS 10.Percentage and its Applications
231
In general. we multiply the fraction by 100.04 (c) 3% (g) 400% (c) 70% (d) 0.4 (d) 115% (h) 350% (d) 85%
2. Write each of the following percents as decimal :
3. What percent of their answers were correct ? 7.97 (a) 75% (d) 2% (a) 35% (b) 0.
Y spent Rs 500 out of Rs 800 every week.1 : A family spends 35% of its monthly budget of Rs 7500 on food. Example 10.70 135% of 80 = 1. 60 candidates appeared in an examination and 45 of them passed. There are 20 eggs in a fridge and 6 of them are brown.35 × 500 = 175 Number of shrubs = 20% of 500 = 0. 16.35 × 80 = 108 Example 10. there are 500 plants of which 35% are trees. What percent of eggs are brown ? 14.45 × 90 = 40. what per cent of students of the class do not wear glasses ? 13. What is the percentage of students that failed ? 10. The rest are creepers. In a class of 40 students. Compute their spending as percentages and state who spends higher percentage. we first change the percent to a fraction or a decimal and then multiply with the number.7 CALCULATION OF PERCENT OF A QUANTITY To determine the percent of a number or quantity.00 or Rs 2625.50 100
60% of 120 = 0. and 13 just qualified. what percent of the letters are E's ? 12. What percent of candidate passed ? 15.232
Mathematics
11. Mr.18 × 215 = 38. X spends Rs 310 out of Rs 500 and Mr. 15 secured second division. shrubs.50 45% of 90 =
45 × 90 = 40.25 × 500 = 125 Since the remaining plants are creepers.2 : In a garden. herbs and creepers. How much do they spend on food ? Solution : Expenditure on food = 35% of Rs 7500 = 0. Solution : Number of trees = 35% of 500 = 0. For example.35 × 7500) = Rs 2625. 10 secured first division.60 × 120 = 72.00 18% of 215 = 0.35 × Rs 7500 = Rs (0. Find out the number of trees.20 × 500 = 100 Number of herbs = 25% of 500 = 0. 20% are shrubs and 25% are herbs. Number of Creepers = 500 – (175 + 100 + 125) = 500 – 400 = 100
. If three fourths of students of a class wear glasses. or 45% of 90 = 0. In the word PERCENTAGE.
7 : If 80 is increased to 125.25%. 60
Example 10. of boys in the school = 1240 – 434 = 806
Aliter Since 35% of the students in the school are girls. Example 10. (100% – 35%) i.Percentage and its Applications
233
Example 10. then Reduction = 60 – 45 = 15 Reduction percent = 15 × 100 % = 25%. then find a. Solution : We have 27% of a = 54 27 ⇒ 100 × a = 54 ⇒ a= 54 × 100 = 200 27
Thus.5 : If 27% of 'a' is 54.4 : What percent of 240 is 96 ? Solution : Percent = 96 × 100 % 240
= 40% Example 10. 65% of the students in the school are boys. What is the reduction percent ? Solution : Let 45 is less than 60 by x%. what is the increase percent ? Solution : Increase = 125 – 80 = 45 Increase percent = 45 × 100 % = 56.6 : 60 is reduced to 45. 80
.35 × 1240 = 434
∴ No.e.3 : 35% of students in a school are girls. Solution : Number of girls in the school = 35% of 1240 = 0.65 × 1240 = 806
Example 10.
∴
Number of boys = 65% of 1240 = 0.. If the total number of students is 1240. find the number of boys in the school. the value of a is 200.
Thus. Find : (a) 15% of 440 (c) 47% of Rs 1200 (b) 16% of 1250 (d) 39% of 1700 metres. Solution : Since Ramesh secures 178 marks and fails by 22 marks. Example 10.. So. 56% of the students are boys. By the given condition. If the number of girls is 6 less than the number of boys. Find the maximum marks. Example 10.10 : Raman has to secure 40% marks for passing. 12 of the number of students = 6 100
∴
Number of students =
6 × 100 = 50 12
Thus. they collected Rs 29000. Let x be maximum marks.8 : A voluntary organisation was collecting money for a relief camp. but they exceeded their target by 45%.45 × 20000) = Rs 20000 + Rs (9000. there are 12% less girls than boys.234
Mathematics
Example 10. how many students are there in the class ? Solution : Given that 44% of the students are girls. the maximum marks are 500. there are 50 students in the class. How much money did they collect ? Solution : The money collected = 20000 + 45% of 20000 = Rs 20000 + Rs (0. Their target was Rs 20000.
.2 1. CHECK YOUR PROGRESS 10.00) = Rs 20000 + Rs 9000 = Rs 29000 Hence.e. He gets 178 marks and fails by 22 marks. the pass marks are 178 + 22 = 200. (100 – 44)% i. then 40% of x = 200 or or 40 × x = 200 100 x=
200 × 100 = 500 40 Thus.9 : 44% of the students of a class are girls.
On a particular day 12. If 70000 people cast their votes. In the following. A candidate secured 40% of the votes polled and was defeated by 900 votes. What amount does he save per month ? 6. If a piece of this alloy weighs 150 kg . we discuss applications of the concept of percentage in different fields. find the original price per kg. During a general election. It takes me 45 minutes to go to school and I spend 80% of the time travelling by bus. A rise of 25% in the price of sugar compels a person to buy 1.5% of them were absent.
.5 kg of sugar less for Rs 240. What percent of 160 is 64 ? 11. 15% for his personal expenses. 14. What is the net increase or decrease percent ? 17. If he has Rs 11704 with him now. how much zinc it contains ? 5. At a school. By what percent did the cost increase ? 10. 16. then find the C's income. A man donated 5% of his monthly income to a charity and deposited 12% of the rest in a bank. Naresh earns Rs 15400 per month. If A's income is Rs 4050. Find the increased price as well as the original price per kg of the sugar. An alloy is a combination of zinc and copper with 30% zinc and 70% copper. A number is first increased by 10% and then decreased by 10%. What is the reduced price per kg of tea ? Also. 25% for expenditure on his children and the rest he saves. 15. what was his monthly income ? 10. 25% voters did not cast their votes. 40% of the students come on foot to the school. A's income is 25% more than B's and B's income is 8% more than C's. If 120 is reduced to 96. How many are boys ? How many are girls ? 4. A reduction of 10% in the price of tea enables a dealer to purchase 25 kg more tea for Rs 22500. There are 600 students in the school. He keeps 50% for household expenses. 13. How many boys were absent on that day ? 7.8 APPLICATIONS OF PERCENTAGE We come across a number of situations in our day to day life wherein we use the concept of percent. Find the total number of voters. How long does the bus journey last ? 8. The cost of a saree was Rs 450. There are 32 boys in a class. 70% of the population voted. what is the population of the town ? 9.Percentage and its Applications
235
2. Its cost has increased to Rs 495. In an election. How many students come on foot to the school ? 3. There are 36 children in a class 25% of them are boys. what is the percentage reduction ? 12.
Thus, the women should sell 21 oranges for Rs 112. CHECK YOUR PROGRESS 10.3 1. A shopkeeper buys an article for Rs 320 and sells it for Rs 240. Find his gain or loss percent. 2. A dealer buys a wrist watch for Rs 450 and spends Rs 30 on its to repairs. If he sells the same for Rs 600, find the profit per cent. 3. A dealer sold two machines at Rs 2400 each. On selling one machine he gained 20% and on selling the other he lost 20%. Find the dealer's gain or loss percent. 4. A sells an article costing Rs 1000 to B and earns a profit of 6%. B, in turn sells it to C at a loss of 5%. At what price did C purchase the article ? 5. By selling 90 ball pens for Rs 160, a person loses 20%. How many ball pens should he sell for Rs 96, so as to have a gain of 20% ? 6. If the selling price of 20 articles is equal to the cost price of 23 articles, find the loss or gain percent. 7. A watch was sold at a profit of 12%. Had it been sold for Rs 33 more, the profit would have been 14%. Find the cost price of the watch. 8. By selling a book for Rs 258, a publisher gains 20%. For how much should he sell it to gain 30% ? 9. A vendor bought bananas at 6 for 5 rupees and sold them at 4 for 3 rupees. Find his gain or loss percent. 10.8.2 Simple Interest All the transactions that take place around us involve money. Sometimes, a person has to borrow some money as a loan from his friends, relatives, bank etc. He promises to return it after a specified time period. So, he has to give back not only the money he borrows but also some extra money to the lender for using his money. The money borrowed is called the principal, usually denoted by P. The extra money paid is called interest, usually denoted by I.
Percentage and its Applications
241
The sum of principal and the interest is called the amount usually denoted by A. A=P + I Interest is calculated on the principal Interest is mostly expressed as a rate percent per year (per annum). Interest depends on how much money (P) has been borrowed and the duration of the time (T) for which it is borrowed. Interest is calculated according to an agreement, which specifies a certain percent of the principal for each year's use, called the rate of interest. I=P × R × T Interest as calculated above is called simple interest. Example 10.17 : A man borrowed Rs 50000 from a finance company for buying a motor bicycle for a period of 2 years. If the finance company charged simple interest at the rate of 15% per annum, how much interest was paid by the man to the finance company. Solution : Here, Principal (P) = Rs 50000 Time (T) = 2 years Rate (R%) = 15% = 0.15 We have I=P × R × T = Rs (50000 × 0.15 × 2) = Rs 15000 Hence, the man paid Rs 15000 as the interest to the finance company. Example 10.18 : A certain sum of money was deposited for 5 years. Simple interest at the rate of 12% was paid. Calculated the sum deposited if the simple interest received by the depositor is Rs 1200. Solution : Let the sum deposited be Rs P. Given that I = Rs 1200; T = 5 years; R = 12% We have I = P × R × T or P= I R×T 1200 = Rs 2000 012 . ×5
= 8 years 4 months Thus, in 8 years 4 months, Rs 8000 will amount to Rs 12000 at the rate of interest 6% per annum. CHECK YOUR PROGRESS 10.4 1. Ramesh borrowed Rs 7000 from his friend at 8% per annum simple interest. He returned the money after 2 years. How much did he pay back altogether ? 2. Jaya deposited Rs 15600 in a bank. The bank pays interest at 8% per annum. Find the interest she will receive at the end of 3 years. 3. Subnam lent Rs 25000 to her friend. She gave Rs 10000 at 10% per annum and the remaining at 12% per annum. How much interest did she receive in 2 years ? 4. Nalini borrowed Rs 5000 from her friend at 8% per annum. She returned the money after 6 months. How much amount did she pay to her friend ?
Percentage and its Applications
243
5. Find the interest received by Anil if he deposits Rs 16000 for 8 months at the rate of 9% per annum. Also, find the amount. 6. Shalini deposited Rs 14500 in a finance company for 3 years and received Rs 4785 as simple interest. What was the rate of interest per annum ? 7. In how many years will Rs 8000 amount to Rs 16000, if simple interest is earned at the rate of 12% per annum ? 8. In how much time will simple interest be annum ? 9. In which case, is interest earned more : (a) Rs 5000 deposited for 5 years at 4% per annum (b) Rs 4000 deposited for 6 years at 5% per annum ? 10. At what rate of interest will simple interest be half the principal in 5 years ? 10.9 DISCOUNT You must have seen all around advertisements of the following types, especially during the festival seasons. SALE Discount upto 50% A discount is a reduction in the marked (or list) price. "25% discount" means a reduction of 25% in the marked price of an article. For instance, if the marked price of an article is Rs 100, it is sold for Rs 75, i.e., Rs 25 less than the marked price. Note. Discount is always calculated on Marked Price. Marked Price (or list price) : The marked price (M.P) of an article is the price at which it is listed for sale. Discount : The discount is the reduction from the marked price of the article. Net Selling price : In case of discount sale, the price of the article obtained by subtracting discount from the list price is called the Net selling Price. Let us consider the following example to illustrate. Example 10.21 : A shirt with marked price Rs 165 is sold at a discount of 10%. Find its net selling price. Solution : Here, Marked Price (M.P.) of the shirt = Rs 165, Discount = 10%
∴
= 20 Hence, the discount being offered is 20%. 10.9.1 Discount Series Sometimes a manufacturer, offers a discount of 10% on discounted price in addition to a previous discount of 20%, because suddenly he gets a supply of cloth at a very low price. He may allow another discount of 5% on the discounted price, to some of his customers for prompt payments. In other words, he allows a discount series. In a discount series, the first figure denotes the discount on the list price, the second denotes the discount on the discounted price and so on. If a shirt is marked for Rs 120 and a discount series 20%, 10% and 5% is offered, then computation for calculating net selling price is as under :
Example 10.24 : An old scooter is sold at three successive discounts of 10%, 5% and 2%. If the marked price of the scooter is Rs 18000, find the selling price of the scooter. Solution : Here, list price = Rs 18000 First discount of 10% = Rs 18000 × 10 = Rs 1800 100
110 × Rs 1100 100 = Rs 1210 Thus, the selling price of the table is Rs 1210. CHECK YOUR PROGRESS 10.5 1. A coat is marked at Rs 1200. Find its selling price if a discount of 15% is offered. 2. A man pays Rs 2100 for a machine listed at Rs 2800. Find the rate of discount offered. 3. An article listed at Rs 2650 is sold at a discount of 10%. Due to festival season, the shopkeeper allows a further discount of 5%. Find the selling price of article. 4. Find a single discount equivalent to a discount series given in each of the following discount series : (a) 25%, 20% and 10% (b) 20%, 15% and 10% (c) 20%, 10% and 5%. 5. Which of the following discount series is better for a customer ? 20%, 10% and 5% OR 10%, 5% and 20%.
6. The list price of a table fan is Rs 840 and it is available to a retailer at 25% discount. For how much should the retailer sell it to earn a profit of 15% ? 7. The marked price of a TV set is Rs 25000. A discount series of 20%, 10%, 5% is allowed on it. How much money does one have to pay for the TV set ? 8. If a shopkeeper marks his goods 50% more than their cost price and allows a discount of 40%, find his gain or loss per cent. 9. The list price of a watch is Rs 320. After two successive discounts it is sold for Rs 244.80. If the first discount is 10%, what is the rate of second discount ? 10. A retailer buys shirts from a manufacturer at the rate of Rs 75 per shirt and marked them at Rs 100 each. He allows some discount and gets a profit of 30% on the cost price. What percentage discount does he allow to his customers ?
248
Mathematics
10.9.3 Sales Tax Government levies some taxes to have earning called revenue. One such tax which is levied on the sale of goods is called sales tax. The rates of sales tax are different for different commodities. Some essential commodities are exempted from sales tax. This tax is charged on the net selling price of commodities and its rate is expressed as a percentage. For example, if an article is sold for Rs 750 and the rate of sales tax is 8%, then Sales tax = Rs 750 × 8 = Rs 60 100 Price inclusive of sales tax = Rs (750 + 60) = Rs 810 The customer will have to pay Rs 810. Example 10.27 : The marked price of a pair of shoes is Rs 320. If the rate of sales tax is 4%, calculate the amount to be paid by a customer for the purchase of the shoes. Solution : Marked price of shoes = Rs 320 Rate of sales tax = 4% ∴ Sales tax = 4% of Rs 320 = 4 × Rs 320 100
= Rs 12.80 Thus, the customer has to pay (Rs 320 + Rs 12.80) = Rs 332.80 for purchasing the shoes. Examples 10.28 : Anita purchased a shirt for Rs 594 including sales tax. If the rate of sales tax is 8%, find the list price of the shirt. Solution : Let the list price of the shirt be Rs P Then P + 8% of P = 594 or or or 108% of P = 594
108 P = 594 100
P= 594 = 550 108 .
Thus, the list price of the shirt is Rs 550. Example 10.29 : Hari Om bought a radio set for Rs 1870, after getting 15% discount on the list price and then 10% sales tax on the reduced price. Find the list price of the radio set. Solution : Let the list price or the radio set be Rs P. Thus, selling price of the radio after discount= Rs P – 15% of Rs P = 85% of Rs P
Percentage and its Applications
249
= Rs
85 P 100
85 I FH 100 K P
Sales tax = 10% of Rs =
10 × 85 P 100 100 85 P 1000
= Rs
The net price to be paid for the purchase of the radio set is Rs 85 P + 85 PI FH 100 935 1000 K = Rs 1000 P
Thus, the list price of the radio set is Rs 2000. Example 10.30 : The list price of a washing machine is Rs 9000. The dealer allows a discount of 5% on the cash payment. How much money will a customer pay to the dealer in cash, if the rate of sales tax is 10% ? Solution : Here, list price = Rs 9000 and discount = 5%
∴ Cash price of the washing machine
= Rs 9000 − 5 × 9000 100 = Rs (9000 – 450) = Rs 8550
∴
FH
IK
Sales tax = 10% of Rs 8550 = 10 × Rs 8550 100
= Rs 855 Hence, the customer has to pay Rs 8550 + Rs 855 = Rs 9405 for the purchase of the washing machine. Example 10.31 : The list price of the air-conditioner is Rs 25630. The rate of sales tax is 10% The customer requests the dealer to allow a discount to such an extent that the price of the
1. The marked price of a sewing machine is Rs 3500. If the sales tax on sewing machine is charged at the rate of 6%, find how much a customer has to pay for purchasing the machine. 2. Amita purchases a pair of socks whose list price is Rs 44. The shopkeeper charges sales tax at the rate of 5%. Find how much money Amita has to pay for purchasing the socks. 3. Mrs. Mohini purchased a saree for Rs 1100 including sales tax. If the list price of the saree is Rs 1000, find the rate of sales tax charged. 4. A refrigerator is available for Rs 13915 including sales tax. If the rate of sales tax is 10%, find the selling price of the refrigerator. 5. Radhika purchased a car with a marked price of Rs 2.1 lakhs at a discount of 5%. If the sales tax is charged at the rate of 12%, find the amount Radhika had to pay for purchasing the car. 6. Dayakant bought a set of cosmetic items for Rs 345 including 15% sales tax and a purse for Rs 110 including 10% sales tax. What percent is the sales tax charged on the whole transaction ? [Hint. C.P. of cosmetic items = Rs (345 ÷ 1.15) ; and C.P. of purse = Rs (110 ÷ 1.1)] 7. Kamal wants to buy a suitcase whose list price is Rs 504. The rate of sales tax is 5%. He requests the shopkeeper to reduce the list price to such an extent that he has to pay Rs 504 only. Calculate the discount given in the price of the suit-case. 10.10 COMMISSION Manufacturers of goods, farmers and owners of properties, frequently uses the services of a middle man to find a buyer in order to sell their goods or properties. The middle man is called an agent, who gets some money for the services rendered by him. This money paid is called commission. In general, commission is expressed in terms of percentage.
This extra amount is actually the interest charged on the amount of money which the customer owes to the seller at different times of instalments paid. His commission was 1. colour TV etc. Solution : Cash payment price = Rs 2600 Cash down payment = Rs 1000
. Find the rate of interest charged under the instalment plan. Instalments : It is the amount which is paid by the customer at regular intervals towards the remaining part of the selling price of the article. which are needed by him and his family. it is a part of the selling price.25%. Cash Price : It is the amount for which the article can be purchased on full payment i. scooters etc. A commission merchant charged Rs 4212 as commission at rate of 3% for selling rice at Rs 450 per bag. The remaining payment is made in easy monthly. video cameras. The remaining part of the cost is paid on subsequent dates.1 What is Instalment Buying ? Instalment purchase scheme. refrigerator. but makes a partial payment in the beginning and takes away the article for use. how much did he sell altogether ? 10. enables a person to buy costly goods like colour TV sets. His commission is then raised to 12%. Cash down Payment : It is the partial payment made by the customer at the time of signing the agreement and taking away the article for use. fridge. Example 10.e. It may be noted that in the instalment plan only part payment of the total cost is paid by the customer at the time of purchase. on convenient terms of payment. An auctioneer sold a property worth Rs 17. We shall study about instalment purchase scheme.35 : Bimla buys a sewing machine. the customer does not make full payment of the cost of the article at the time of purchase. In the following. the selling price of the article. which is available for Rs 2600 cash payment or under an instalment plan for Rs 1000 cash down payment and 3 monthly instalment of Rs 550 each.. 10. How much did the auctioneer earn as commission ? 3. In this scheme. quarterly or half yearly instalments. If his total earning (commission) were Rs 8000.252
Mathematics
2.8 lakhs. and therefore the seller charges some extra amount for deferred payments. we solve a few examples to illustrate the process. A commission merchant sells a certain amount of goods at a commission of 10% in the first two weeks of a month.11. Such articles are available on easy instalments. In fact. and he sells an equal amount of goods in the remaining part of the month. How many bags of rice did he sell ? 4. as per the agreement signed between the customer and the shopkeeper.11 INSTALMENT BUYING With the cost of articles going up day by day it has become difficult for the common man to buy some articles like scooter.
simplify it and suffix the % sign. A T. set is available for 21000 cash or for Rs 4000 cash down payment and 6 equal monthly instalments of Rs 3000 each. LET US SUM UP
z z z z
Percent means 'per hundred'. 3.a. we drop the % sign and divide the number by 100. we multiply the fraction by 100. Calculate the rate of interest charged under the instalment plan. If the rate of interest charged is 22 % per annum.254
Mathematics
Thus. Percents can be written as fractions as well as decimals and vice-versa. To write a percent as a fraction.
. A scooter is available for Rs 30000 cash or for Rs 15000 cash down payment and 4 equal 1 monthly instalments. find the 9 amount of each instalment. 3 find the amount of each instalment. To write a fraction as a percent.
∴
(70000 – 10P) ×
1 × 30 = 5P – 14000 12 100 1 = 5P – 14000 40
or or or or or
(70000 – 10P) ×
70000 – 10P = 40(5P – 14000) 70000 – 10P = 200P – 560000 210 P = 560000 + 70000 P= 630000 210
= 3000
∴
Amount of each instalment = Rs 3000. 2. 4. Anil purchased a type writer priced at Rs 6800 cash payment under the instalment plan by making a cash down payment of Rs 2000 and 5 monthly instalments of Rs 1000 each. he had to pay interest on Rs (70000 – 10P) for 1 month at the rate of 30% p. Find the rate of interest charged under the instalment plan. CHECK YOUR PROGRESS 10. If the rate of interest charged under the instalment plan is 33 %.8
1.V. A microwave oven is available for Rs 9600 cash or for 4000 cash down payment and 2 3 equal monthly instalment.
– S. When the selling price is more than the cost price of the goods. P.3% (c) 11. (Marked price – discount).P. = Gain × 100 C. Sales tax is charged on the sale price of goods. P. – C. gives the price.P. 100
z z
z
The simple interest (S. Write each of the following as fraction : (a) 0.P. TERMINAL EXERCISE
z z z z
z
1. there is a profit (or gain) When the selling price is less than the cost price of the goods.4 (d) 0. Two or more successive discounts are said to form a discount series.3% (d) 113%
.3%
3. Loss = S.P.13% (b) 1.I.I.Percentage and its Applications
255
z
To determine the specific percent of a number or quantity. .25 (c) 1. A discount series can be reduced to a single discount. is calculated. Commission is paid to an agent for his services in arranging the sale or purchase of goods from some one else.P. we change the percent to a fraction or a decimal and then multiply. 100 + Gain% × C.07
2. . = Loss × 100 C. Discount is always calculated on the marked price of the goods.P.) on a principal (P) at the rate of R% for a time T years. An Instalment plan enables a person to buy costlier goods. which a customer has to pay while buying an article. using the formula S. there is a loss Profit (Gain) = S. P. Write each of the following as a per cent : (a) 7 20 (b) 0. P. Write each of the following as a decimal : (a) 63% (b) 13% (c) 3% (d) 0. 100 − Loss% × C. Gain% = Further S. = P × R × T
z z z
Discount is a reduction in the list price of goods. Loss = C. 100 .
Find the price of one bag at which it was sold ? 13. Find the rate of interest. What number increased by 10% of itself is 352 ? 9.
. At what rate per annum. What percent of 700 is 294 ? 6.a. Find each of the following : (a) 37% of 400 5. Ahmad purchased a bicycle by making a cash down payment of Rs 400 and 3 monthly instalments of Rs 275 each. The bicycle was also available on cash payment of Rs 1200. For what amount did he buy the scooter ? 12. was interest paid to him ? 14.5% of 800
15. By what percent is 60 more than 45 ? 7. Find the amount of each instalment. If the rate of interest charged in the instalment plan is 18% p. Find the number whose 15% is 270 ? 8. Rita purchased a washing machine for Rs 4000 cash down payment and 4 equal monthly instalments.74 ? 10. Simple interest on a sum of money is 1 rd of the sum itself and the number of years is 3 thrice the rate percent. Arun at the beginning of a year had a bank balance of its 17500 and at the end of the year he had a balance of Rs 21350. The washing machine was also available for Rs 15000 cash payment. (b) 3. Shalim deposited Rs 14000 in a bank for 2 years and received Rs 4200 as simple interest.256
Mathematics
4. Find the rate of interest per annum charged under the instalment plan. A man loses 25% by selling a scooter for Rs 8400. By what percent did his balance increase ? 11. What number decreased by 7% of itself is 16. 16. A commission merchant charged Rs 700 as commission for selling 100 bags of cotton at 5%.
calculate compound interest. forecast the population of a town when rate of growth of population is given for a specific period of time.4 THE CONCEPT OF COMPOUND INTEREST Recall that interest is the extra money paid by the user to the bank for the use of money. The money which has been borrowed is called the principal (P). you will learn to calculate compound interest.
11. a vehicle. the user will have to pay more interest for the second period of time. This type of interest is called compound interest. Recall that interest is the extra money paid by the user to the bank or money lender for the use of his money for some specified time period. the time interval for which the
.1 INTRODUCTION You have already studied about simple interest in your earlier classes.Compound Interest
259
11
Compound Interest
11. the user will pay interest on the amount borrowed as well as on the interest accrued for the first time period. calculate amount of a sum invested for a given time. a building etc.3 EXPECTED BACKGROUND KNOWLEDGE
z z
Knowledge of percentage Knowledge of simple interest.2 OBJECTIVES After studying this lesson. If the money is retained by the user for the next time period also. calculate the depreciated value of a machine. 11.
z
11. the learner will be able to :
z z z z z
illustrate the concept of compound interest. Thus. find the difference between simple interest and compound interest. In this lesson.
The interest charged in this way is called the compound interest. Therefore. The interest is calculated for the next time period on the new principal. then it becomes a part of principal for the second year. then it becomes a part of the principal and added to the principal for the next time period. The time period after which the interest is added to the principal for the next time period is called the conversion period. This is expressed as percentage of the principal. the interest due after every year is added to the principal and then the interest is calculated for the next year on this new principal. half yearly. the interest is called simple interest and is given by S.I = PRT But if this interest is due (not paid) after the decided time period. This way of calculating interest is called compound interest.
.260
Mathematics
money has been taken is called the time duration (T) and the extra money paid by the user to the bank upto the time the principal amount is returned is called the rate of interest (R). The total interest to be paid at the end of two years = Rs (200 + 220) = Rs 420. three months or a month and the interest is said to be compounded annually. 10 Interest for one year = Rs 2000 × 100 × 1 = Rs 200 If this interest is not paid at the end of one year. six month. When the interest is calculated on the principal for the entire time period of loan. quarterly or monthly respectively. Let us go through the following example : Suppose you borrow Rs 2000 from a bank at the rate of interest 10% per annum. the principal for the second year becomes Rs (2000 + 200) = Rs 2200 Now interest for second year 10 = Rs 2200 × 100 × 1 = Rs 220
∴ Amount payable at the end of two years
FH
IK
FH
IK
= Rs (2200 + 220) = Rs 2420. Thus for calculating the compound interest. The conversion period may be one year.
50) = Rs 4630. Calculate the compound interest on Rs 8000 for 3 years at 5% per annum when the interest is compounded annually.50 4000
i. Calculate the compound interest on Rs 390625 for 2 years at 8% per annum when the interest is compounded half yearly. 3. on Rs 8000.11 : Find the rate at which Rs 4000 will give Rs 630. 5.50 as compound interest in 9 months. 7.50 and n = 3 quarters Let the rate of interest = R Now
∴
A = P(1 + R)n 4630. the interest being compounded annually. Find the difference between simple interest and compound interest for 2 years at 10% per annum. A = Rs (4000 + 630. Calculate the compound interest on Rs 62500 for 1 year at 8% per annum when the interest is compounded quarterly. 2. The difference between simple interest and compound interest for a certain sum of money at 8% per annum for 1½ years when the interest is compounded half-yearly is Rs 228.50 = 4000 (1 + R)3 (1 + R)3 = 4630.a. 4. How much money will become Rs 194481 after 2 years at 10% per annum when the interest is compounded semi-annually ? 6.
9261 = 21 = 8000 20
∴
FH IK
3
1+R= R=
21 20 1 = 5 = 5% per quarter. Solution : Here P = Rs 4000.
. when the interest is compounded semi-annually. 20 100
⇒
∴ Rate of interest = 20% p. Find the sum. Find the sum of money which will amount to Rs 27783 in 3 years at 5% per annum.e.
CHECK YOUR PROGRESS 11.1 1.Compound Interest
267
Example 11. interest being compounded quarterly.
cars etc. 10. we observe about growth of population. Find the sum and the rate of interest per annum.. it would have earned Rs 881 more than in the previous case. 9..268
Mathematics
8. If the rate of growth/depreciation for period is denoted by r%. If the interest were compounded half yearly. then Vn is given by Vn = V0 1 +
Vn =
FG r IJ FG1 + r IJ FG1 + r IJ . (For growth) H 100K H 100K H 100K F r IJ FG1 − r IJ FG1 − r IJ . (For depreciation) V G1 − H 100K H 100K H 100K
1 2 3 0 1 2 3
We solve some examples to illustrate the above concepts.. r = 4% and n = 3
∴
Vn = 281250 1 + 4 100
FH
IK
3
26 26 26 = 281250 × × × 25 25 25 = 316368 Hence the population of the town after 3 years = 316368.. interest being compounded half-yearly ? 11. Find the sum.
If the rate of growth/depreciation varies for each conversion period. after use.. A sum of money is invested at compound interest for 2 years at 10% per annum when the interest is compounded yearly... A sum of money becomes Rs 18522 in three years and Rs 19448.. What will be its population after 3 years if the rate of growth of population is 4% per year ? Solution : Here V0 = 281250. and depreciation of articles like machinery. plants.5 RATE OF GROWTH AND DEPRECIATION In our day to day life. then Vn is given by : r Vn = V0 1 + 100 r Vn = V0 1 − 100
FH FH
IK IK
n
in case of growth and
n
in case of depreciation.
.12 : The population of a town is 281250. V0 is the value of article in the beginning and Vn is the value after 'n' conversion periods.10 in 4 years at the same rate of interest when the interest is compounded annually. Examples 11. viruses etc. At what rate of interest per annum would the compound interest on Rs 12500 be Rs 9100 in 1½ years.
was 3. 5. CHECK YOUR PROGRESS 11. A tree gains its height at the rate of 2% of what it was in the beginning of the month. The value of property increases at the rate of 20% of what it was in the beginning of the year.00 A. then r Vn = V0 1 + 100 Vn
0
z z
FH IK r I = VF H1 − 100 K
n
in case of growth and
n
in case of depreciation.00 A.5 × 108. Rate = r%. The population of a town is 50000.270
Mathematics
Example 11. the rate of growth of the population was 5%.M. the value of property will be almost double ? LET US SUM UP
z
If Principal = P. find the virus count at 11. 3. The population of a city is 2.
. In the third year. 4. on the same day.2256 × 108
FH
IK
2
Hence virus count at 11. In the second year. R = 4% and n = 2
∴
4 Vn = 3. find its height at the end of the April of the same year (Give your answer correct to 2 decimal places). Find the population of the town after 3 years.16 : The virus of a culture decreases at the rate of 4% per hour due to a medicine. the population growth rate was noticed as 4%.2 1.5 × 108 1 − 100 = 3. 2.7 × 108.5 m in the beginning of January 2002. If the rate of depreciation is 15% for the first year and 10% for the subsequent years. Solution : Here V0 = 3. is 3. Time = n conversion periods then Amount = Principal + Compound Interest = P (1 + R)n C.M. It its height was 1. In how many years. If the virus count in the culture at 9.I. If V0 is the value of article in the beginning. find its population after 3 years. Vn is the value after 'n' conversion periods and r be the rate of growth/depreciation per period. find its cost after 3 years.2256 × 108. The cost of a car was Rs 215000 in January 2001.M. the population decreased by 10%. If the rate of growth of population is 5% of what it was in the beginning of the year. because of some epidemics. In the first year.00 A. = P [(1 + R)n – 1] Compound Interest is greater than simple interest except for the first conversion period.5 × 108.
. If the cost of scooter is 15625. if the present population is 16000 ? 6. The price of a scooter depreciates at the rate of 20% per annum of its value at the beginning of each year. During second year.. the population decreases by 10%. interest being compounded half yearly. = V G H 100K H 100K H 100K
2 3 0 1 2 3
TERMINAL EXERCISE 1. Find the population at the end of 2 years. 5.. for growth H K H 100K H 100K F1 − r IJ FG1 − r IJ FG1 − r IJ . The population of a town is 20000. 3. when the interest is compounded quarterly. The population of a town increases at the rate of 5% per annum. Find the rate of interest per annum if Rs 31250 amounts to Rs 35152 in 1½ years. for depreciation. find the time period.Compound Interest
271
z
If the rate of growth/depreciation varies for each conversion period. A sum of Rs 1000 amounts to Rs 1331 after some time. 4. what will be the population of the town after two years. Find the sum of money which will amount to Rs 26460 in six months at 20% per annum. what will be its value at the end of three years ? 7. If the rate of interest is 10% per annum compounded annually.. It increases by 10% during first year.. 2. then
r1 Vn = V0 1 + 100
and
Vn
FG IJ FG1 + r IJ FG1 + r IJ . when the interest is being compounded annually on Rs 2500. Find the difference between simple and compound interests for 2 years at 10% per annum.
4.
. 5. banks also help the people in various kinds of financial transactions. 8.1 INTRODUCTION All of us need money to meet our day-to-day expenses. 10. For the convenience of customers. Issuing/Encashing travellers cheques in Local/Foreign currency. 6. Exchange of Foreign currency etc. Giving loans to the borrowers on interest. 2. 3. Keeping money of the depositors and pay interest on deposits. Transferring money from one place to another. Collection of taxes – income tax. Receiving payments – Telephone bills. We earn money. We would like to keep the money in safe custody. save some money. Providing All Time Money (ATM) facility to the customers. Besides this. sales tax. school fees etc. the banks offer different types of accounts (deposits) some of which are : (i) Savings Bank Account (ii) Current Account (iii) Fixed deposit Account. Issuing credit cards. water bills. Buying and selling security bonds. Electricity bills. 7. house tax etc.Banking
273
12
Banking
12. Banks are the institutions where we keep the money in the form of deposits and those who need money can borrow as loans on payment of interest with certain conditions. (iv) Recurring deposit Account. Some of the functions of a bank are : 1. 9. spend money and also.
12. S.
12. for which money was invested are given i. The general format of a page of the savings bank passbook is as given below :
.. calculate interest/maturity value of fixed/term deposit given the rate of interest compounded yearly. (ii) Compound interest when P. deposits and withdrawals can be made and the record is maintained by the bank in a 'Pass-Book' given to the Account holder (i. n is number of conversion periods. = A – P where A = P(1 + R)n. the person who holds the account).I.e. In this account. The bank pays interest for the money that an account holder keeps in the account. Some of these are : (a) Savings bank account (b) Fixed or term deposit account (c) Current account 12. Anybody can open this account with a minimum sum of Rs 1000 with cheque book facility. n are given i. C. a pass book is issued to the account holder.e. It contains date wise entries of deposits withdrawals and the interest earned. The rate of interest changes from time to time. calculate interest for a saving bank account.4. On opening a savings bank account. It bears savings bank account number also. the learner will be able to :
z z z
state different types of accounts.274
Mathematics
12. The prevailing rate of interest is 4% per annum compounded half yearly. given the rate of interest. = PRT. rate of interest (R) and time (T).I.e.4 TYPES OF ACCOUNTS You can open different types of accounts in a bank depending upon your needs. R.1 Savings Bank Account This is the most popular account offered by the banks.2 OBJECTIVES After studying this lesson. semi-annually or quarterly (not more than four conversion periods).3 EXPECTED BACKGROUND KNOWLEDGE (i) Knowledge of simple interest when principal (P). This account encourages people to develop the habit of saving.
12.
. (ii) The interest is credited to the account every six months. (i) The bank pays interest for the month on the minimum closing balance from the 10th day of the month to the last day of the month. (iii) The present rate of interest is 4% per annum compounded semi-annually.
Fi.2 Sample of Cheque
Note : In a savings bank account. (iv) Usually not more than 25 withdrawals are allowed in a quarter from Savings Bank account.1 Sample of Withdrawal slip
Fig.Banking
275
Date — — —
Particulars — — —
Amount withdrawn Rs P — — —
Amount Deposited Rs P — — —
Balance — — —
Money can be withdrawn from the bank through a withdrawal slip (sample shown) or by a cheque (sample shown below).g 12.
1/12 year.e.00 5000. Solution : Balance on 5th March. Find the principal for which he will earn interest for March 2003. 2003.00 — 20000. (ii) Add all the minimum balances for each month as per step (i) to obtain principal for one month.
. (iii) Calculate the simple interest on this sum for one month using the formula Interest = P × R ×
1 12
where P is Principal.00 10000.1 : Sidharth opened a savings bank account in State Bank of India on 5th March 2003 with a deposit of Rs 5000. He deposited Rs 1500 on 10th March 2003 and withdrew Rs 3000 on 29th March.4. no interest is payable for that month.00 20000.2 : Anisha's savings bank account pass-book has the following entries : Date 2002 Jan 6 Feb 11 April 5 April 27 May 3 June 15 By Cash By Cash By Cheque To Cheque By Cash To Cheque — — — 18000.00 30000. Let us illustrate the above with some examples. (iv) If account is opened after the 10th day of month.00 17000. the principal for which interest is earned for the month of March 2003 is Rs 3500.276
Mathematics
12. 2003 = Rs 3500 Here minimum balance between 10th and 31st March.00 Particulars Amount withdrawn Amount Deposited Balance Rs P Rs P Rs P
Find the sum on which Anisha will earn interest from January to June 2002. (v) No interest is paid for the month in which the account is closed. 2003 = Rs 3500. R is rate of interest per annum and time 1 month i. Example 12. 2003 = Rs 5000 Balance on 10th March. Example 12.00 30000. Therefore.2 Computation of Interest (i) Write down the minimum balance between the closing balance on 10th to the last day of the month.00 — 20000.00 10000.00 — 13000. 2003 = Rs 6500 Balance on 29th March.00 35000.
10 April 3 June 5 June 25 B/F To Cash By Cheque To Cheque By Cash By Cheque By Cheque — Rs 7000.00 20000.00 10000.00 20000. find the interest earned by Ritu at the end of June on her savings bank account.00 15000. 11 Feb.00
.00 — — — — — 20000. Feb.00 30000. 9 Jan. the sum on which Anisha will earn interest for one month = Rs 127000. Solution : Details of qualifying amount for the calculation of interest are as below: Month Jan.00 30000.00 5000.00 — 5000.00 20000.00 15000.00 25000. Principal for January = Rs 20000 20000 30000 17000 30000 10000
Principal for February = Rs Principal for March Principal for April Principal for May Principal for June Total = Rs = Rs = Rs = Rs
Rs 127000
Thus.00 105000.00 Particulars Amount withdrawn Amount Deposited Balance Rs P Rs P Rs P
If the rate of interest is 4% per annum.00 12000.00 35000. March April May June Total Rs Rs Rs Rs Rs Rs Rs Amount 5000. Example 12.00 — Rs 10000. 1 Jan.00 5000.00 15000.3 : Ritus's pass book has the following entries : Date 2002 Jan.Banking
277
Solution : The qualifying amount for the interest is the minimum balance between the 10th and the last day of the month.
6 : Salim opened a savings bank account with a bank on 9th January 2002 with a cash deposit of Rs 10000.00 2000. By cash By Cash — — — — 3000.00 15000.00 11250.00 2000.2002 By Interest — 250. His pass book has the following entries :
.280
Mathematics
Example 12.00 2000.00 10000.00 14000.7.00 13000.1 1.00 — — 6000.00 — 10000. If the bank pays interest at the rate of 4% per annum.00 16000.00 17000. Write all entries of the passbook. Subsequently he deposited Rs 2000 on the 8th day of every month. calculate the interest upto the last day of 30th June 2002 and make the entry in the passbook alongwith the balance. Solution : The entries in the passbook are as below :
Date Particulars Amount withdrawn Rs 2002 9th Jan.00 12000.00 P Amount Deposited Rs P Balance Rs P
8th March By Cash 8th April By Cash
25th April To Cash 8th May 8th June By Cash By Cash
28th June To Cash
Details of qualifying amount for the calculation of interest are : Month Amount (in Rs) January February 10000 12000 March 14000 April 13000 May 15000 June 11000 Total 75000
∴ Interest upto 30th June. Krishna Murthy opened a savings bank account in State Bank of India on 7th July 2002.00
CHECK YOUR PROGRESS 12. 8th Feb.00 — 2000.00 2000. He withdrew Rs 3000 on 25th April and Rs 6000 on 28th June 2002.00 11000. 2002 = Rs
75000 × 1 × 4 = Rs 250 100 × 12
The interest entry is made on 1–7–2002 as below : 1.
00 — 6000. 2002 To Cheque December 19.00 10000.00 20000.00 5000. 5.00 Particulars Amount withdrawn Amount Deposited Rs P Rs P Balance Rs P
If the rate of interest is 6% per annum.00 — — — 9000. She withdraws Rs 4000 on 3rd April and Rs 12000 on 10th June 2002. 2002.
. 2000 By Cash
The account is closed on 2nd January 2003. write all the entries. Find the amount received if the rate of interest is 4% per annum.00
November 25.00 Balance Rs P 2000. including interest.00 10000.00 — 4000. 2002 July 11. 6. If the bank pays interest at the rate of 5% per annum. payable at the end of June and December. 10 Sept. calculate the interest entry on 1st January 2003 in the passbook alongwith the balance.00 1500.00 10000. Subsequently she deposited Rs 6000 on the 6th day of every month. find the interest earned if the account is closed on : (i) 30th June. 2002 (ii) 3rd July.00 — 15000. Madhu's Savings Bank Account passbook has the following entries :
Date 2002 July 1 July 9 Sept.00 1000. Kavita opens a savings bank account with a bank on 8th January.00 — 4000.282
Mathematics
If the rate of interest is 4% per annum. 2002 Particulars B/F By Cheque By Cheque Amount withdrawn Amount Deposited Rs P Rs P — — — 15000. 2002 August 9.00 11000.00 4500. 2002 with a cash deposit of Rs 10000.00 9500.00 — — 8000. which are made upto 1st July 2002.00 20000.00 2000. A page from the pass book of savings bank account is given below :
Date July 1. 14 December 5 December 10 December 23 B/F By Cheque To Cheque By Cash By Cash By Cash To Cheque — — 9000.00 5000. 4.
For this account. government organisations etc.80
Example 12. (iii) For one year and more. (ii) For six months and above but less than 1 year. 5% 6% 6. find the amount she receives at the end of 60 days. Rather.4. they charge some money as service charges. Here the depositor agrees to keep the money with the bank for a fixed time. Hence the banks offer higher rates of interest on such deposits depending upon the period of deposits. Solution : Here P = Rs 2920. banks allow the current account holder to get money over and above their deposits called overdraft. The rate of interest per annum on term deposits is as below : (i) For 46 days and above but less than 179 days.4 Fixed Deposit Account Suppose you have some money which is not required for some time. the bank can use this money more freely than the money kept in the savings bank account. If the bank pays interest at 6% per annum. The total amount receivable after the expiry of the time is called maturity value. there is no limit on number of withdrawals or on the amount of withdrawals but the banks do not pay interest. the minimum balance for individuals account is Rs 5000 while for big concerns. Depending upon the goodwill of the individual/company.80
FG 2920 × 6 × 60IJ H 100 × 365 K
∴
Amount Received by Anju = Rs (2920 + 28. 12..5%
For senior citizens. it is Rs 10000. The scheme suitable for depositing such money is the Fixed Deposit or Term deposit.5% is given on deposits.Banking
283
12.80) = Rs 2948. Big business concerns. have to do a number of transactions every day.
.7 : Anju deposited Rs 2920 in a fixed deposit scheme in a bank for 60 days. R = 6% per annum and T =
60 year 365
∴
Interest = PRT = Rs = Rs 28. find the maturity value of the money deposited by him. For them banks offer a different type of account called current account. Obviously. sometimes. If the rate of interest is 8% per annum compounded half yearly.8 : Joginder makes a fixed deposit of Rs 31250 in a bank for 1½ years.4. and additional interest of 0.3 Current Account In a saving bank account. Here. the account holder is allowed to have a limited number of withdrawals in a half year. companies. Example 12.
How much money should Shanta deposit in a fixed deposit account in a bank so as to enable her to receive a sum of Rs 9261 after 1½ years. Find the amount at the end of 9 months if Rs 50000 is deposited and the interest 8% per annum is compounded quarterly. 3. n = 4 We know that A = P(1 + R)n
∴
456976 = P 1 + 4 100 P = Rs
FH
IK
4
= P 26 25
FH IK
4
456976 × 25 × 25 × 25 × 25 26 × 26 × 26 × 26
= Rs 390625 Hence amount to be deposited = Rs 390625.Banking
285
Example 12. 4. A = Rs 456976.e. 2. R = 4% per half year and T = 2 year = 4 half years i. Some of these are : (i) Saving bank account (ii) Fixed or term deposit account (iii) Current account
. It the rate of interest is 10% per annum and the interest is compounded half yearly. Pankaj deposits Rs 75000 in a fixed deposit account for 3 years. find the maturity value of the money deposited by him. Charu makes a fixed deposit of Rs 8000 in a bank for 1½ years. CHECK YOUR PROGRESS 12. the rate of interest being 10% per annum compounded half yearly ? 5. find the maturity value of the money deposited by her.11 : How much money should Nirmal deposit in a fixed deposit account in a bank so that she gets Rs 456976 after two years. Solution : Here P = ?. the rate of interest being 8% per annum compounded half yearly. the rate of interest being 8% per annum compounded quarterly ? LET US SUM UP
z
There are different types of accounts in a bank. If the rate of interest is 10% per annum compounded annually.2 1. How much money should Kamal deposit in a fixed deposit account in a bank so as to enable him to receive a sum of Rs 1061208 after nine months.
z
Step for computing interest are : (i) Write down the minimum balance between the closing balance on 10th to the last day of the month. (v) No interest is paid for the month in which the account is closed. (iii) Calculate the simple interest on this sum for one month using the formula Interest = P × R ×
1 12
where P is Principal. Dass's Saving bank Account in a particular year is given below : Date Particulars Amount withdrawn Amount Deposited Rs P Rs P By Cash By Cheque To Cheque By Cash By Cash — — 20000. (ii) Add all the minimum balances for each month as per step (i) to obtain principal for one month.00 — — 15000. 9 Feb.00 3000. 1/12 year. 10 April 25 June 7 June 11
If the rate of interest is 4% per annum. (ii) the interest is credited to the account every six months.
. R is rate of interest per annum and time 1 month i.00 8000.00 5000.00
Jan.00 23000. no interest is payable for that month. TERMINAL EXERCISE 1.00 12000.00 — 2000. A page for the pass book of Mr.00 Balance Rs P 15000. (iv) If account is opened after the 10th day of month. Dass at the end of June on his savings bank account. find the interest earned by Mr.286
Mathematics
z
In a saving bank account : (i) the bank pays interest for the month on the minimum closing balance from the 10th day of the month to the last day of the month.e.00 17000.
00 Particulars Amount withdrawn Rs P Amount Deposited Rs P Balance Rs P
The account is closed on 10th January. Sujata has a savings bank account in a bank. Vandana makes a fixed deposit of Rs 62500 in a bank for 1½ years.00 — — 2500. Smith makes a fixed deposit of Rs 10000 in a bank for a year. 20 Dec. 27 B.00 — 9600. find the maturity value of money deposited by her. Her passbook has the following entries :
Date 2002 July 1 July 9 August 10 October 19 November 2 Dec. How much money should Sarla deposit in a fixed deposit account in a bank so that she gets Rs 194481 after two years. If the rate of interest is 8% per annum compounded half yearly.00 4000.Banking
287
2.00 12600. If the rate of interest is 8% per annum compounded quarterly. 5.F. By Cheque By Cash To Cheque By Cash To Cash By Cheque — — — 6000. Find the amount received if the rate of interest is 4% per annum.00 — 9200.00 — 10600.00 3400. 2003. the rate of interest being 10% per annum compounded half-yearly ?
. find the maturity value of the money deposited by her.00 3000.00 9000.00 14000. 4. 3.00 2500.00 5000.
Since then efforts have been made to give it a perfect logical shape.Lines and Angles
1
Module 3
Geometry
Geometry is a branch of Mathematics which deals with the study of different types of figures and their properties. is a difficult task. Euclid a Greek mathematician collected all available knowledge in geometry till his times (330 B.). Egyptians and Babylonians discovered many formulae for finding the areas of different rectilinear figures and used them practically.
. quadrilaterals and circles along with their properties. In this module on geometry. arranged it in a systematic way and gave it a logical approach based on deductive reasoning.C. who profounded and proved. To study geometry in its full and complete logical form. we shall study about lines. verifying and stating properties of figures and proving logically only a few important properties known as theorems. As such we shall be studying geometry in a very informal way. As such geometry originated in ancient times when man started measuring land for making his home and boundaries for his fields. angles. what is now known as Pythagoras theorem are contributions to geometry worth mentioning from the glorius past of India. triangles. Sulbasutras used during vedic period. and the work of the great mathematician Baudhayan. The Indian Mathematicians also had contributed a lot towards development of knowledge of geometry as is evident from the civilizations of Harappa and Mohenjodaro. defining terms with suitable examples. Geometry means measurement of earth.
1
. angle. 13. rays and angles. plane. you have studied about a point. illustrate and verify properties of parallel lines prove that the sum of angles of a triangle is 180° explain the concept of locus and find the locus of a point under certain conditions. 13. Let us quickly recall these concepts. a plane and an angle. LINE AND ANGLE In earlier classes. intersecting lines. plane.2 OBJECTIVES After studying this lesson. we get a fine dot.
Fig.2
Mathematics
13
Lines and Angles
13. the learner will be able to :
z
illustrate the concepts of point. its corner that of a point and the edges meeting at a corner give an idea of an angle. line. parallel lines. intersecting lines and pair of angles made by them. Its edges give an idea of a line. which is called a point. It gives an idea of a plane.4 POINT.3 EXPECTED BACKGROUND KNOWLEDGE We assume that the learner is familiar with the geometrical concepts and figures such as :
z z
point. Now move your hand on the top of your table. a line. Point : If we press the tip of a pen or pencil on a piece of paper. parallel lines
13.
z z z
13.1 INTRODUCTION Observe the top of your desk or table. line.
6
. (See Fig. A line is named using any two points on it. This gives us a straight line or simply called a line.4)
Fig.3 Plane If we move our palm on the top of a table. Join them with the help of a ruler or a scale and extend it on both sides.
Fig.2 Ray If we mark a point X and draw a line. starting from it extending infinitely in one direction only.5
X is called the initial point of the ray XY. (See Fig.13. 13.4
13. 13.1 Line Now mark two points A and B. then we get a ray XY. 13. viz. 13. m etc.4.3
The part of the line between two points A and B is called a line segment and will be named AB.
Fig.
Fig. 13. Observe that a line segment is the shortest path between two points A and B. B. 13.2
In geometry a line is extended infinitely on both sides and is marked with arrows to give this idea.4. we get an idea of a plane.3)
Fig.4. 13.Lines and Angles
3
A point is used to show the location and is represented by capital letters A. AB or by a single small letter l. 13. C etc.
at some distance from A.
. floor of a room also gives the idea of part of a plane. we talk of collinear points only when their number is three or more. For example points P. Take another point B. only one passes through B. We conclude that one and only one line can be drawn passing through two given points.4
Mathematics
Similarly. If a line can not be drawn passing through all three points (or more points) then they are said to be non-collinear. B and C in the Fig.9.9
We observe that a line may or may not pass through the three given points. Since two points always lie on a line. We can again draw an infinite number of lines passing through B. For example. Plane also extends infinitely lengthwise and breadthwise. Mark a point A on a sheet of paper.
Fig.9 are collinear points. 13. 13.7
In fact we can draw an infinite number of lines through a point. then these points are said to be collinear. points A. 13. Now we take three points in plane. 13.
Fig. Thus. If a line can pass through three or more points.
Fig. Q and R in the Fig. How many lines can you draw passing though this point ? As many as you wish. 13.8
Out of these lines how many pass through both the points A and B ? Out of all the lines passing through A. only one line passes through both the points A and B.
or or (ii) They may intersect in one point only as in Fig.4 Angle Mark a point O and draw two rays OA and OB starting from O. [In such a case they are called intersecting lines] or (ii) no points in common as in Fig. Now observe three (or more) distinct lines in plane.4.Lines and Angles
5
Let us now take two distinct lines AB and CD in a plane. In such a case they are called concurrent lines.10
How many points can they have in common ? We observe that these lines can have. In such a case they are called parallel lines. 13.10 (a) and (b). The figure we get is called an angle. 13.
Fig. an angle is a figure consisting of two rays starting from a common point.11(b).
.11
What are the possibilities ? (i) They may interest in more than one point as in Fig.11(c). Thus. either (i) one point in common as in Fig. 13.10(c). (iii) They may be non intersecting lines parallel to each other as in Fig.11(a) and 13. 13.
13. 13.
Fig. 13. 13.11 (d).
. [see Fig.13
Two lines or rays making a right angle with each other are called perpendicular lines.6
Mathematics
Fig. An angle less than 90° is called an acute angle. 13. If we take any point O and draw two rays starting from it in opposite directions then the measure of this angle is taken to be 180° degrees. 13.11] An angle is measured in degrees.14(a). ∠XOY is an obtuse angle in Fig. written as 180°. 13. 13. 13. In Fig.
Fig. we can say OA is perpendicular to OB or vice-versa.
Fig.12
This measure divided into 180 equal parts is called one degree (1°). An angle of 90° is called a right angle.14(b). For example ∠POQ is an acute angle in Fig. For example. An angle greater than 90° but less than 180° is called an obtuse angle. and is written as ∠AOB or ∠BOA or ∠O.13. 13. 13.11
This angle may be named as angle AOB or angle BOA or simply angle O. for example ∠BOA or ∠BOC. Angle obtained by two opposite rays is called a straight angle.
(a) Fig. Each pair has a common vertex O and a common side OA in between OB and OC.
.14
(b)
13. A pair of angles.16
(b)
Observe the angles in each pair in Fig.5
PAIRS OF ANGLES
Fig. is called a pair of complementary angles.15. 13. whose sum is 90°.16[(a) and (b)]. Such a pair of angles is called a 'pair of adjacent angles'. Each angle is called the complement of the other. They add up to make a total of 90°. 13. 13. 13. 13.Lines and Angles
7
(a) Fig.15
Observe the two angles ∠1 and ∠2 in each of the figures in Fig.
8
Mathematics
(a) Fig. Note that they also make a pair of supplementary angles. The pair of angles so formed as in Fig. Draw two intersecting lines AB and CD. Each angle is called the supplement of the other. 13. 13.
Fig. Draw a line AB. whatever be the position of the ray CD. 13.
Fig. is called a pair of supplementary angles. From a point C on it draw a ray CD making two angles ∠X and ∠Y. 13.18 is called a linear pair of angles. A pair of angles whose sum is 180°. These add up to make a total of 180°. 13. We conclude If a ray stands on a line then the sum of the two adjacent angles so formed is 180°.19
. intersecting each other at O.17[(a) and (b)]. we will always find the sum to be 180°.17
(b)
Again observe the angles in each pair in Fig.18
If we measure ∠X and ∠Y and add.
13. You will always find that ∠AOC = ∠DOB.
Fig. Measure them. keeping the other in position and observe that the pairs of vertically opposite angles thus formed are always equal.22
.
Fig. Attach two strips with a nail or a pin as shown in the figure. 13. 13. you will again find ∠AOD = ∠BOC We conclude : If two lines intersect each other. eight angles are formed. These make a pair of vertically opposite angles.21
When a transversal intersects two lines. For example line l in Fig.20
Rotate one of the strips. ∠AOD and ∠BOC is another pair of vertically opposite angles. Activity for you. A line which intersects two or more lines at distinct points is called a transversal. 13.
Fig. On measuring.21 is a transversal. the pairs of vertically opposite angles are equal.Lines and Angles
9
∠AOC and ∠DOB are angles opposite to each other.
. 13. lines m and n are not parallel.
that is. which we study in the following. we shall always find that ∠1 = ∠5. You may also verify the truth of these results by drawing a pair of parallel lines (using parallel edges of your scale) and a transversal and measuring angles in each of these pairs. Some of the useful pairs are as follows : (a) ∠1 and ∠5 is a pair of corresponding angles. ∠3 = ∠6 and ∠4 = ∠5 ∠3 + ∠5 = 180° and ∠4 + ∠6 = 180°
that is. Hence we conclude : When a transversal intersects two parallel lines. 13. there may not exist any relation between the angles of any of the above pairs.22 above. angles in each pair of alternate angle are equal. when lines are parallel. In Fig. Also. ∠3 = ∠7 and ∠4 = ∠8
that is. then (i) each pair of corresponding angles. whatever be the position of parallel lines or the transversal
Fig. ∠3 and ∠7 and ∠4 and ∠8 are other pairs of corresponding angles.23
If we measure the angles. are equal (ii) each pair of alternate angles are equal (iii) each pair of interior angles on the same side of the transversal are supplementary. However. (b) ∠3 and ∠6 is a pair of alternate angles. When a transversal intersects two parallel lines. (c) ∠3 and ∠5 is a pair of interior angles on the same side of the transversal. ∠4 and ∠6 is another pair of interior angles. Also ∠2 = ∠6. ∠4 and ∠5 is another pair of alternate angles. eight angles are formed. there are some very useful relations in these pairs. ∠2 and ∠6. as such. sum of angles in each pair of interior angles is 180°. angles in each pair of corresponding angles are equal.10
Mathematics
These angles in pairs are very important in the study of properties of parallel lines.
as shown in Fig. 13.24. we shall find that they do not intersect each other.
Fig.24
At C and D. 13. Example 13.1 : Choose the correct answer out of the alternative options in the following multiple choice questions. 13. say 50°. Hence we conclude that When a transversal intersects two lines in such a way that (i) any pair of corresponding angles are equal or or (ii) any pair of alternate angles are equal (iii) any pair of interior angles on the same side of transversal are supplementary
then the two lines are parallel. To verify the truth of the first converse. they are parallel. we construct two angles ACF and CDH equal to each other. that is.Lines and Angles
11
Converse of each of these results is also true.25
. In a similar way. On producing EF and GH on either side. we draw a line AB and mark two points C and D on it. we can verify the truth of the other two converses.
Fig.
namely three angles (i) ∠ABC or ∠B (ii) ∠ACB or ∠C (iii) ∠CAB or ∠A and three sides : (iv) AB (v) BC (vi) CA It is named as ∆ ABC or ∆ BAC or ∆ CBA and read as triangle ABC or triangle BAC or triangle CBA. [∆ABC in Fig.34(ii)].33
It is a closed figure formed by three line segments having six elements.34
(iii)
(i) Equilateral triangle : A triangle in which all the three sides are equal is called an equilateral triangle. 13.Lines and Angles
15
13. ITS TYPES AND PROPERTIES Triangle is the simplest of all the closed figures formed in a plane by three line segments. (a) On the basis of sides
(i)
(ii) Fig.34(i)] (ii) Isosceles triangle : A triangle in which two sides are equal is called an isosceles triangle. 13. 13. is called a scalene triangle [∆LMN in Fig. 13. 13.1 Types of Triangles Triangles can be classified into different types in two ways.35
(iii)
. (iii) Scalene triangle : A triangle in which all sides are of different lengths.
Fig. 13. 13. [∆DEF in Fig.34(iii)] (b) On the basis of angles :
(i)
(ii) Fig.6 TRIANGLE.6.
13. 13.36 (a).35(ii)] (iii) Acute angled triangle : A triangle in which all the three angles are acute is called an acute angled triangle or acute triangle [∆XYZ in Fig. We will prove this result in a logical way naming it as a theorem.36 (b). ∠Q = 40°.
What do you observe ? Sum of the angles of triangle in each case in 180°.
Fig. ∠A = 80°.
Fig. 13.16
Mathematics
(i) Obtuse angled triangle : A triangle in which one of the angles is an obtuse angle is called an obtuse angled triangle or simply obtuse triangle [∆PQR is Fig. Theorem : The sum of the three angles of a triangle is 180°. [∆UVW in Fig 13.37
Given : A triangle ABC To prove : ∠A + ∠B + ∠C = 180° Construction : Through A draw a line DE parallel to BC. 13.2 Angle Sum Property of a Triangle We draw two triangles and measure their angles. ∠ R = 110° ∠P + ∠Q + ∠R = 30° + 40° + 110° = 180° In Fig. ∠B = 40° and ∠C = 60° ∴ ∴ ∠A + ∠B + ∠C = 80° + 40° + 60° = 180° ∠P = 30°. 13.35(iii)] Now we shall study some important properties of angles of a triangle.
. 13. 13.35(i)] (ii) Right angled triangle : A triangle in which one of the angles is a right angle is called a right angled triangle or right triangle.6.36
In Fig.
13. We measure these angles.38. ∠5 and ∠6.39
In Fig.
Fig. there are two interior opposite angles. Interior opposite angles are the angles of the triangle not forming a linear pair with the given exterior angle.3 Exterior Angle of a Triangle Let us produce a side BC of the ABC to a point D.. 13.Lines and Angles
17
Proof : Since DE is parallel to BC and AB is a transversal. namely ∠1. (Angles making a straight angle) (pair of alternate angles) (pair of alternate angles) . ∠2. observe that there are six exterior angles of the ∆ABC. For example ∠A and ∠B are the two interior opposite angles corresponding to the exterior angle ACD of ∆ABC. 13. 13. Thus.
.. ∠ACD so obtained is called an exterior angle of the ∆ABC. (1)
Now adding ∠A to both sides of (1)
Fig.38
In Fig. ∠3.6.39. ∴ Similarly ∴ ∠B = ∠DAB ∠C = ∠EAC ∠B + ∠C = ∠DAB + ∠EAC ∠A + ∠B + ∠C = ∠A + ∠DAB + ∠EAC = 180° 13. Corresponding to an exterior angle of a triangle. The angle formed by a side of the triangle produced and another side of the triangle is called an exterior angle of the triangle. ∠4.
Examples 13. Thus. 2
.3 : Choose the correct answer out of the given alternatives in the following multiple choice questions : (i) Which of the following can be the angles of a triangle ? (a) 65°.41. (c)
(iii) In a triangle. Ans. 13. (d)
Fig. 45° and 80° (c) 60°. one angle is twice the other and the third angle is 60°. 60° and 60°. Then the largest angle is (a) 60° (c) 100° Example 13. (b)
Fig. we may conclude : An exterior angle of a triangle is equal to the sum of the two interior opposite angles. 13. 13. 60° and 59° (b) 90°. 13.40
(ii) In Fig. bisectors of ∠PQR and ∠PRQ intersect each other at O.18
Mathematics
∠A = 60° ∠B = 50° and ∠ACD = 110° We observe that ∠ACD = ∠A + ∠B.41
In Fig.40 ∠A is equal to (a) 30° (c) 45° (b) 35° (d) 75° Ans. 30° and 61° (d) 60°. This observation is true in general. Prove that ∠QOR = 90° + 1 ∠P.4 : (b) 80° (d) 120° Ans.
2 2 CHECK YOUR PROGRESS 13. In Fig.42
In Fig. 13.2 1. Choose the correct answer out of given alternatives in the following multiple choice questions: (i) A triangle can have (a) Two right angles (b) Two obtuse angles
(c) At the most two acute angles (d) All three acute angles (ii) In a right triangle. one exterior angle is 120°.43 ABCD is a trapezium such that AB || DC. 3.
. Prove that the sum of the four angles of a quadrilateral is 360°. 13. The smallest angle of the triangles is (a) 20° (c) 40° (iii) (b) 30° (d) 60°
Fig. Find ∠D and ∠C and verify that sum of the for angles is 360°.42.Lines and Angles
19
Solution :
1 1 ∠QOR = 180°− 2 ∠PQR + 2 ∠PRQ 1 = 180°− ∠PQR + ∠PRQ 2
FH
IK
b 1 = 180°− b180°−∠P g 2
g
1 1 = 180°− 90°+ ∠P = 90°+ ∠P . find the three angles. CD is parallel to BA. 13. 4. ∠ACB is equal to (a) 55° (c) 65° (b) 60° (d) 70°
2. The angles of a triangle are in the ratio 2 : 3 : 5.
Find the angles of the triangle.20
Mathematics
Fig. 13. 13. before being caught or touching the ground. ABC is a triangle such that ∠ABC = ∠ACB. In Fig. when a player hits the ball. 13.44
The path described is called Locus.44
13. 13.
Fig. also a point P between them equidistant from both the lines . Prove that if one angle of a triangle is equal to the sum of the other two angles. 13.
Fig. For example : (1) Given two parallel lines l and m. it describes a path.7 LOCUS During the game of cricket.
Fig. A figure in geometry is a result of the path traced by a point (or a very small particle) moving under certain conditions. 6. then it is a right triangle.45
If the particle moves so that it is equidistant from both the lines.44.43
5. what will be its path?
.
Fig. 13. 13. 13.49
.46
The path traced by P will be a line parallel to both the lines and exactly in the middle of them as in Fig. 13. 13. Strike it hard with a pencil or a stick so that it leaves the table with a certain speed and observe its path after it leaves the table.48
The path of the moving point P will be a circle as shown in Fig. what will be its path.
Fig.
Fig.48. (3) Place a small piece of chalk stick or a pebble on top of a table. (2) Given a fixed point O and a point P at a fixed distance d. 13.Lines and Angles
21
Fig.47
If the point P moves in a plane so that it is always at a constant distance d from the fixed point O.46.
it can easily be verified that every point on PM is equidistant from A and B.1 Locus of a point equidistant from two given points. every point of which satisfies the given condition(s). Joint AB.49. we may conclude the following : The locus of a point equidistant from two given points is the perpendicular bisector of the line segment joining the two points. Activity for you : Mark two points A and B on a sheet of paper and join them.50
We have to find the locus of a point P such that PA = PB.
.51
Thus. 13. Fold the paper along mid-point of AB so that A coincides with B.
Fig.2 Locus of a point equidistant from two lines intersecting at O Let AB and CD be two given lines intersecting at O. Make a crease along the line of fold. Also ∠AMP = ∠BMP = 90° That is. This is the locus of the point equidistant from the given points A and B.7. 13. 13.22
Mathematics
The path traced by the pebble will be a curve (part of what is known as a parabola) as shown in Fig. locus of a point moving under certain conditions is the path or the geometrical figure. Mark mid point of AB as M. PM is the perpendicular bisector of AB. This crease is a straight line.7. Using a pair of divider or a scale. Let A and B the two given points. 13. Join PM and extend it on both sides. It can be easily checked that every point on it is equidistant from A and B.
Fig. Mark another point P using compasses such that PA = PB. 13. Thus.
13.54
In a similar way find the other bisector by folding again and getting crease 2. on a sheet of paper.Lines and Angles
23
Fig. Draw bisectors of ∠BOP and ∠BOC. that is. 13. 13.53
If we take any point P on any bisector l or m. Any point on this crease 2 is also equidistant from both the lines.
. Take a point P on this crease which is the bisector of ∠BOD and check using a set square that PL = PM
Fig. we will find perpendicular distances PL and PM of P from the lines AB and CD are equal. Fold the paper through O so that AO falls on CO and OD falls on OB and mark the crease along the fold. we may conclude : The locus of a point equidistant from two intersecting lines is the pair of lines. PL = PM
Thus. Activity for you : Draw two lines AB and CD intersecting at O.
Fig.52
We have to find the locus of a point P which is equidistant from both AB and CD. bisecting the angles formed by the given line.
show in a diagram the locus of the point P. Find the locus of the centre of a circle passing through three given points A.5 : Find the locus of the centre of a circle passing through two given points. Show in a diagram the possible locations of the post.24
Mathematics
Example 13. 13. Find the locus of a point which is always at a distance 5 cm from a given line AB. 4. 2. Solution : Let the two given points be A and B. Representing the villages by points A and B and the well by point P. 3.56
CHECK YOU PROGRESS 13. There are two villages certain distance apart.
.55
Point O must be equidistant from both the points A and B. As we have already learnt.
Fig. Two straight roads AB and CD are intersecting at a point O. 3.
Fig. the locus of the point O will be the perpendicular bisector of AB. We have to find the position or positions of centre O of a circle passing through A and B. A well is to be dug so that it is equidistant from the two villages such that its distance from each village is not more than the distance between the two villages.3 1. An observation post is to be constructed at a distance of 1 km from O and equidistant from the roads AB and CD. B and C which are non-collinear.
then they are called collinear points. whose sum is 90° is called a pair of complementary angles. Two rays starting from a common point form an angle. 13. TERMINAL EXERCISE
1.
Fig. A pair of angles. If three or more lines intersect in one point only then they are called cocurrent lines.Lines and Angles
25
LET US SUM UP
z
A line extends to infinity on both sides and a line segment is only a part of it between two points. 13. (ii) alternate angles are equal. B and C. if x = 42.57. A pair of angles whose sum is 180° is called a pair of supplementary angles. Two distinct lines in a plane may either be intersecting or parallel.57
2. An exterior angle of a triangle is equal to the sum of the two interior opposite angles. Find ∠1 and ∠2. 13. If a line passes through three or more points. When a transversal intersects two parallel lines.
z z z z z z z z z
z z
The sum of the angles of a triangle is 180°. then determine (a) y (b) ∠AOD
Fig. then (i) corresponding angles in a pair are equal.
. If a ray stands on a line then the sum of the two adjacent angles. In Fig.58
In the above figure p. q and r are parallel lines intersected by a transversal l at A. If two lines intersect each other the pairs of vertically opposite angles are equal. so formed is 180°. (iii) interior angles on the same side of the thransversal are supplementary.
13. What type of triangle is it ? 4. Find the third angle. 13. 13.61 ABC is a triangle in which bisectors of ∠B and ∠C meet at O. 6.60
In Fig.60.
Fig.
5.26
Mathematics
3.
Fig. sides ABC of the triangle ABC have been produced as shown. 13. Show that ∠BOC = 125°.59
In Fig. The sum of two angles of a triangle is equal to its third angle. 13.
Fig. Show that the sum of the exterior angles so formed is 360°.61
In Fig. 13. Find the angles of the triangle.59 sides of ∆ABC have been produced as shown.
.
Fig.
. in ∆PQR. AD is perpendicular to BC and AE is bisector of ∠BAC. 10. ∠C. 8. 12. ∠A.62
In Fig.
Fig. 9.62 above. Prove that the sum of the (interior) angles of a pentagon is 540°.
Fig. 13.64
In Fig.63 in ∆ABC. ∠F. 13. PT is bisector of ∠P and PR is produced to S show that ∠PQR + ∠PRS = 2∠PTR. 13.64 above. 13. find the sum of the angles. ∠B and ∠E. ∠D.13.Lines and Angles
27
7.63
In Fig. Find ∠DAE.
Congruence of Triangles
29
14
Congruence of Triangles
14. the learner will be able to :
z z z z z
verify and explain whether two given figures are congruent or not. prove that angles opposite to equal sides of a triangle are equal. In this lesson you will study congruence of two triangles. 14. prove that if two sides of triangle are unequal.1 INTRODUCTION You might have observed that leaves of different trees have different shapes. solve problems based on the above results.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z z
Recognition of plane geometric figures Equality of lines and angles Types of angles Angle sum property of a triangle Paper cutting and folding. The geometrical figures which have same shape and same size are called congruent and the property is called congruence. state and verify inequalities in a triangle. then the longer side has the greater angle opposite to it.
z z
14. but leaves of the same tree have almost the same shape. prove that sides opposite to equal angles of a triangle are equal.
. state the criteria for congruency of two triangles and apply them in solving problems.2 OBJECTIVES After studying this lesson. some relations between their sides and angles in details. Although they may differ in size.
14. 14.2
(iii) Which have same shape and same size as shown in Figures 14. (i) Which have different shapes and sizes as shown in Fig.1
(ii) Which have same shapes but different sizes as shown in Fig. 14.2
Fig. (a) two' one rupee coins
Fig. 14.4 and 14.30
Mathematics
14.1
Fig.5.3. 14.4 CONCEPT OF CONGRUENCE In our daily life you observe various figures or objects.4
.3
(b) two postage stamps or post cards
Fig. 14. These figures or objects can be categorised in terms of its shape and size in the following manner. 14.
1 ACTIVITY FOR YOU Take a sheet of paper.Congruence of Triangles
31
(c) two photo prints of same size from the same negative.4. These appear to be one. Two figures. when they are of equal length.6
(2) Two squares are congruent if their sides are equal. 14. In other words. 14. on the upper part of the sheet. these are congruent figures. For example : (1) Two line segments are congruent.
Fig. Now draw a figure of a leaf or a flower or any object which you like. two figures will be congruent.
Fig. The figure you drew and its carbon copy are of the same shape and same size. You will get a carbon copy of it.5
We will deal with the figures which have same shapes and same sizes.2 CRITERIA FOR CONGRUENCE OF SOME FIGURES Congruent figures when placed one over the other. 14. on the sheet below. exactly coincide with one another or cover each other. which have the same shape and same size are called congruent figures and this property is called congruence. 14. if parts of one figure are equal to the corresponding parts of the other. Thus.
Fig. Observe a butterfly folding its two wings.4. 14.7
. fold it in the middle and keep a carbon (paper) between the two folds.
32
Mathematics
(3) Two circles are congruent. 14. 14. if all the sides and all the angles of one are equal to the corresponding sides and angles of other.5 CONGRUENCE OF TRIANGLES Triangle is a basic rectilinear figure in geometry.8
14. Two triangles are congruent. if their radii are equal. having minimum number of sides. ∠Q = ∠Y and ∠R = ∠Z Thus we can say ∆PQR is congruent to ∆XYZ and we write ∆PQR ≅ ∆XYZ Relation of congruence between two triangles is always written with corresponding or matching parts in proper order. implying their circumferences are also equal.
Fig. PR = XZ.9
PQ = XY. Q corresponds to Y and R corresponds to Z. Here ∆PQR ≅ ∆XYZ
also means P corresponds to X. For example. QR = YZ ∠P = ∠X. As such congruence of triangles plays a very important role in proving many useful results.
. 14.9
Fig. in triangles PQR and XYZ in Fig. Hence this needs a detailed study.
∠Q = ∠B and PQ = AB. 14. (elements) are equal.11). QP = YX. we need to know that all the six parts of one triangle are equal to the corresponding six parts of the other triangle. We shall now learn that it is possible to prove the congruence of two triangles. 14. Consider a triangle ABC in Fig. we will observe that one covers the other exactly. we may say they are congruent. 14.11
If we trace or cut out triangle ABC and place it over triangle PQR. ∠Q = ∠Y.10
Construct another triangle PQR such that QR = BC.
Fig. ∆RPQ ≅ ∆ZXY but NOT as Or NOT as ∆PQR ≅ ∆YZX.6 CRITERIA FOR CONGRUENCE OF TRIANGLES In order to prove. even if we are able to know the equality of three of their corresponding parts. ∆PQR ≅ ∆ZXY.Congruence of Triangles
33
This congruence may also be written as ∆QRP ≅ ∆YZX which means. and observe that
. Thus. ∠R = ∠Z and ∠P = ∠X. we can also measure the remaining parts.10
Fig. (See Fig. RP = ZX. whether two triangles are congruent or not. R corresponds to Z and P corresponds to X. Q corresponds to Y. This congruence may also be written as
14. namely QR = YZ. 14. It also means corresponding parts. Alternatively.
13)
Fig. It should be noted here that in constructing ∆PQR congruent to ∆ABC we used only two pairs of sides PQ = AB. we observe that ∠P = ∠A.13
By superimposition or by measuring the remaining corresponding parts. QR = BC and the included angle between them ∠Q = ∠B. which again means that equality of the three corresponding parts (two angles and the included side) of two triangles results in congruent triangles. 14. 14. then the two triangles are congruent. (See Fig.
Fig. This criterion is referred to as ASA or AAS. This criterion is referred to as SAS (side angle side). QR = BC. Again.12.
. We also know that the sum of the three angles of a triangle is 180°. as such if two angles of one triangle are equal to the corresponding angles of another triangle. Thus we have Criterion 1 : If any two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of the other triangle. consider ∆ABC in Fig. then the third angles will also be equal. ∠Q = ∠B and ∠R = ∠C. 14. the two triangles are congruent. This means that equality of these three corresponding parts results in congruent triangles. Thus instead of included side we may have any pair of corresponding sides equal.34
Mathematics
AC = PR. ∠A = ∠P and ∠C = ∠R Showing that ∆PQR ≅ ∆ABC. PQ = AB and PR = AC establishing that ∆PQR ≅ ∆ABC. Thus we have Criterion 2 : If any two angles and a side of one triangle are equal to corresponding angles and the side of the other triangle. 14.12
Construct another ∆PQR such that.
side. we find that.
. ∠P = ∠P′ = ∠A.14
Now take three thin sticks equal in lengths to sides AB. 14.15
By measuring the corresponding angles. This is referred to as SSS (side. results in congruent triangles. 14.14)
Fig. then the two right triangles are congruent.1 ACTIVITY FOR YOU In order to explore another criterion we again take a triangle ABC (See Fig.15).6. we can establish one more criterion which will be applicable for two right triangles only. Similarly. then the two triangles are congruent. ∠Q = ∠Q′ = ∠B and ∠R = ∠R′ = ∠C.
Fig. Thus we have Criterion 3 : If the three sides of one triangle are equal to the three corresponding sides of the other triangle.Congruence of Triangles
35
14. 14. side). establishing that ∆PQR ≅ ∆P′Q′R ≅ ∆ABC which means that equality of the three corresponding sides of two triangles. 14. BC and CA of ∆ABC. Place them in any order to form ∆PQR or ∆P′Q′R′ near the ∆ABC (Fig. Criterion 4 : If the hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.
36
Mathematics
This criterion is referred to as R.16. (b) Example 14.2 : Two rectilinear figures are congruent if they have only (a) all corresponding sides equal (b) all corresponding angles equal (c) the same area (d) all corresponding angles and all corresponding sides equal. (Right Hypotenuse side). Example 14.S. 14. (d) Example 14.1 : In which of the following criteria. Using these criteria we can easily prove. (a) All corresponding sides are equal (b) All corresponding angles are equal (c) Two corresponding sides and their included angles are equal (d) All corresponding angles and any pair of corresponding sides are equal.H. Show that AX = AY. PX and QY are perpendicular to PQ and PX = QY. two given triangles are NOT congruent. whether two triangles are congruent and establish the equality of remaining corresponding parts. Ans. knowing three corresponding parts only. Ans.16
Solution : In ∆PAX and ∆QAY ∠XPA = ∠YQA (Each is 90°)
.
Fig.3 : In Fig 14.
Fig.Congruence of Triangles
39
3. (c) Two angles and a side of one are equal to two angles and the corresponding side of the other (d) One angle and two sides of one are equal to one angle and two sides of the other.
Fig. If O is the mid point of BC. ∠B = ∠C and AB = AC. Two triangles are congruent.. show that it is also the mid point of AD.22
. In Fig. 14. 5. along with equality of two corresponding angles. AB is parallel to CD. In order that two given triangles are congruent. if (a) All three corresponding angles are equal (b) Two angles and a side of one are equal to two angles and a side of the other. 14.21.22.21
6. we must know the equality of (a) No corresponding side (b) Minimum one corresponding side (c) Minimum two corresponding sides (d) All the three corresponding sides 4.14. In Fig. 14. Hence show that CD = BE. Prove that ∆ABE ≅ ∆ACD.
14.24
14. In ∆ABC (Fig. BE is ⊥ AC and AD = BE. Theorem : The angles opposite to equal sides of a triangle are equal. From Fig. show that the triangles are congruent and make pairs of equal angles.23
8.24. Proof : In ∆ABD and ∆ACD. AD is ⊥ BC. Prove that AE = BD.40
Mathematics
7.
Fig. 14. Given : A triangle ABC in which AB = AC.25
.
Fig. To prove : ∠B = ∠C.23). 14. Construction : Draw bisector of ∠BAC meeting BC at D. 14. we shall now prove some important theorems. AB = AC ∠BAD = ∠CAD and AD = AD (Given) (By construction) (Common)
Fig. 14.7 ANGLES OPPOSITE TO EQUAL SIDES OF A TRIANGLE AND VICE VERSA Using the criteria for congruence of triangles.
In Fig.31
2.
Fig. 14.34
. Prove that ∠BCD is a right angle.
Fig.32).Congruence of Triangles
43
CHECK YOUR PROGRESS 14. Prove that ∆ABC is a isosceles triangle. ∆ABC is an isosceles triangle such that AB = AC. 14. If the line l in Fig.
Fig.33
4. 14. Side BA is produced to a point D such that AB = AD.32
3. Prove that ∠PQS = ∠PRS.31.33 is parallel to the base BC of the isosceles ∆ABC find the angles.
Fig. 14.2 1. 14. PQ = PR and SQ = SR. 14. if the altitude AD bisects the base BC (Fig. 14.
36. Prove that ∆ABC is an isosceles triangle. 14.38).38
. D is the mid point of BC and perpendiculars DF and DE to sides AB and AC are equal in length.36
7. P is any point in the interior of the triangle such that ∠ABP = ∠ACP.35.
Fig.37
8. 14. Prove that ∠PQS = ∠PRS.35
6. 14. 14. 14. PQ = PR. QS and RT are the angle bisector of ∠Q = ∠R respectively Prove that QS = RT. 14.
Fig.
Fig. ∆PQR and ∆SQR are isosceles triangles on the same base QR (Fig. Prove that AP bisects ∠BAC. 14.44
Mathematics
5. In ∆ABC.
Fig. In Fig. AB = AC (Fig. In Fig.37). 14.
Again compare ∠C and ∠A and measure sides AB and BC opposite to these angles.39.1 Theorem If two sides of a triangle are unequal. ∠ACB > ∠ABC
Fig. To prove.Congruence of Triangles
45
14. i. It is clear that ∠C is greater than ∠B. You will find that these angles are not equal and ∠C is greater than ∠B. 14.8 INEQUALITIES IN A TRIANGLE We have learnt the relationship between sides and angles of a triangle when they are equal. (Fig. 14. AD = AC ∴ But ∠ACD = ∠ADC ∠ADC > ∠ABC (Exterior angle is greater than opposite interior angles) Again ∴ ∠ACB > ∠ACD(Point D lies in the interior of the ∠ACB). A triangle ABC in which AB > AC. when they are unequal.41
. Now compare sides AB and AC opposite to these angles by measuring them. Mark a point D on the side AB such that AD = AC and join DC. We observe that ∠C > ∠A and AB > BC.41) compare ∠C and ∠B.39
In Fig. Let us examine.
Fig. In ∆ABC.40
(Angles opposite equal sides)
What can we say about the converse of this theorem. This can be proved easily. then the longer side has the greater angle opposite to it. 14. as follows.e. We shall now study some relations among sides and angles of a triangle. We observe that AB is longer than AC. Measure ∠B and ∠C. 14. triangle ABC has side AB longer than the side AC. Given. Proof : In ∆ADC. 14. If you repeat this experiment. side opposite to greater angle is longer.
Fig. 14.8. you will always find that this observation is true. ∠ACB > ∠ABC Construction.
we observe that (i) AB + BC > CA (ii) BC + CA > AB and (ii) CA + AB > BC.. and CA + AB separately and compare each sum of a pair with the third side. we shall now study whether the three sides of a triangle are related in some way.
. Measure any pair of angles in a triangle.46
Mathematics
Comparing ∠A and ∠B. Compare them and then compare the sides opposite to them by measurement. 14. the sum of the three angles of a triangle is 180°. which we state as a property. side opposite to greater angle is longer. ACTIVITY FOR YOU Fix three nails P.e. BC + CA. In a triangle. Now find the sum of different pairs AB + BC. Observe that in a triangle if one angle is right or an obtuse angle.e.42
Measure its three sides AB. You can also verify this property by drawing any type of triangle. i. the greater angle has longer side opposite to it. we observe a similar result. Draw a triangle ABC. a right triangle or an obtuse triangle. ∠A > ∠B and BC > AC. BC and CA. You will find the above result always true. Q and R on a wooden board or any surface.
Fig. You have already learnt the relationship among the three angles of a triangle i. then the side opposite to that angle is the longest. Thus we conclude that Sum of any two sides of a triangle is greater than the third side.
49
LET US SUM UP
z z
Figures which have the same shape and same size are called congruent figures. If two sides of a triangle are unequal. In Fig. if AB = AD then prove that BC > CD. Sides opposite to equal angles of a triangle are equal.
Fig. AB is parallel to CD. 14. 14. Sum of any two sides of a triangle is greater than the third side. when placed one over the other completely cover each other.Congruence of Triangles
49
6. then the longer side has the greater angle opposite to it.
z z
. 14. the greater angle has the longer side opposite to it. 14. In a triangle. (iii) SSS (ii) ASA or AAS (iv) RHS
z
z z z
Angles opposite to equal sides of a triangle are equal. In Fig. These corresponding parts must satisfy one of the four criteria (i) SAS.48. To prove that two triangles are congruent we need to know the equality of only three corresponding parts. [Hint : ∠ADB = ∠ABD]. If ∠A > ∠B then prove that BC > AD.48
7.
Fig. Congruent figures.49. All parts of one figure are equal to the corresponding parts of the other figure.
51. Prove that ∠C > ∠A and ∠D > ∠B.53.52
5. 14.50) 2.50
Fig.51
3.
Fig.50
Mathematics
TERMINAL EXERCISE 1.52. 14. Prove that CA = BD (Fig. If AC is parallel to DB then prove that O is also the mid point of CD. In a right triangle. if the median AD is perpendicular to the base BC then prove that the triangle is an isosceles triangle. Prove that the hypotenuse is twice the side opposite to the angle of 30°. then prove that the two triangles are congruent. 14. In Fig.
Fig. In Fig. 7. 14.
Fig. If ∠B = 60°. [Hint : Join AC and BD]. 4. one of the acute angles is 30°. Prove that ∆ABC ≅ ∆PQR. ∆ABC and ∆CDE are such that BC = CE and AB = DE. In Fig. 14. In a ∆ABC. 14. Line segments AB and CD intersect each other at O such that O is the midpoint of AB. two sides AB and BC and the altitude AD of ∆ABC are respectively equal to the sides PQ and QR and the altitudes PS. 6.53
. 14. 14. ∠ACE = 30° and ∠D = 90°. AB is the longest side and DC is the shortest side of a quadrilateral ABCD. Two lines AB and CD bisect each other at O.
54
9. Prove that BD = DC.Congruence of Triangles
51
8. 14.
Fig. 14. [Hint : Show that ∆DBC ≅ ∆ECB)
Fig.55
. Prove that the medians bisecting the equal sides of an isosceles triangle are also equal. ABC is an isosceles triangle in which AB = AC and AD is the altitude from A to the base BC.
15.Concurrent Lines
53
15
Concurrent Lines
15. 15. we shall study the concurrency property of these lines.
(a)
(b) Fig. which can be drawn in a triangle. i.1
(c)
. medians.1(a)] or intersecting [See Fig. right bisectors of sides.
z
15. the learner will be able to :
z
distinguish between an angle bisector and perpendicular bisector of a side. angle bisectors and altitudes. altitudes and medians of a triangle. state and apply the concurrency property of angle bisectors. 15. an altitude and a median of a triangle. In this lesson.2 OBJECTIVES After studying this lesson. such as :
z
Two lines in a plane can either be parallel [See Fig. You have also studied about triangles and some special lines. which are quite useful. in the lesson on lines and angles. perpendicular bisectors of sides.1(b) and (c)].e. 15.1 INTRODUCTION You have already learnt about concurrent lines.3 EXPECTED BACKGROUND KNOWLEDGE Properties of intersecting lines.
2(b).3
(c)
15.2(d)]
(a)
(b) Fig. or (ii) intersect each other in exactly one point [Fig. 15. 15. 15. intersect in no point [See Fig 15.2(a).1 ANGLE BISECTORS OF A TRIANGLE In triangle ABC.54
Mathematics
z
Three lines in a plane may : (i) be parallel to each other..4
.4. 15. or (iii) intersect each other in two points [Fig. 15.2
(c)
(d)
15. i. the line AD bisects ∠A of the triangle.2(c). 15. 15.
(a)
(b) Fig.15.3). (See Fig.4 CONCURRENT LINES Three or more lines in a plane which intersect each other in exactly one point or which pass through the same point are called concurrent lines and the common point is called the point of concurrency (See Fig.4)
Fig.e. or (iv) intersect each other at the most in three points [Fig.
Can you reason out. (See Fig. 15. We may take any type of triangle— acute.5).7)
Fig. 15. why the name incentre for this point ? Recall that the locus of a point equidistant from two intersecting lines is the pair of angle bisectors of the angles formed by the lines.Concurrent Lines
55
A line which bisects an angle of a triangle is called an angle bisector of the triangle.5
Fig. How many angle bisectors can a triangle have ? Since a triangle has three angles.7
Thus we conclude the following : Angle bisectors of a triangle pass through the same point. The point of concurrency I is called the 'Incentre' of the triangle. and draw its angle bisectors. Let us draw second angle bisector BE of ∠B (See Fig 15. 15. AD is one of the three angle bisectors of ∆ABC. Also I is a point on angle bisector of ∠ABC. that is they are concurrent. In other words they are concurrent and the point of concurrency is I. (see Fig.8). we will always find that the three angle bisectors of a triangle are concurrent. we can draw three angle bisectors in it. Since I is a point on bisector of ∠BAC. We observe that this angle bisector of the triangle also passes through I. 15. 15.6
The two angle bisectors of the ∆ABC intersect each other as I. it must be equidistant from them. right or obtuse triangle.
Fig. it must
. Let us draw the third angle bisector CF of ∠C (See Fig. 15.6).
15. Since a triangle has three sides.10
Fig. we have IL = IM = IN (Fig. In other words.
Fig. DP is one of the three perpendicular bisector of ∆ABC (Fig.9).11). 15. then we observe that it also passes through the point O (Fig. 15. called 'Incircle' of the triangle. line DP bisects side BC at right angle. I being the centre of the incircle is called the Incentre and IL the radius of the incircle is called the inradius of the triangle. Thus this point of concurrency I is at the same distance from the three sides of the triangle. we can draw a circle touching all the three sides. we can say that the three perpendicular bisectors of the sides are concurrent at O. Taking I as the centre and IL as the radius. 15. 15.8).9
Fig. We draw the second perpendicular bisector EQ.11
. 15.10).8
Thus. 15. Now if we also draw the third perpendicular bisector FR. so we can draw three perpendicular bisectors in a triangle. of the triangle. intersecting DP at O (Fig. 15. A line which bisects a side of a triangle at right angle is called the perpendicular bisector of the side. 15.56
Mathematics
also be equidistant from them.
Fig. Note : That incentre always lies in the interior of the triangle.2 Perpendicular bisectors of the sides of a triangle ABC is a triangle.4.
15. we have AO = BO = CO.12
(b)
Thus we conclude that : The three perpendicular bisectors of the sides of a triangle pass through the same point.
(a) Fig. Thus.13
The point O also lies on the perpendicular bisector of AC.13). why the name circumcentre for this point ? Recall that the locus of a point equidistant from two given points is the perpendicular bisector of the line joining the two points. so it must be equidistant from both A and C. they are concurrent. that is. that is.Concurrent Lines
57
We may repeat this experiment with any type of triangle. so it must be equidistant from both the point B and C. AO = CO.
Fig.
. The point of concurrency O is called the 'circumcentre' of the triangle Can you reason out. 15. 15. so that BO = CO (Fig. but we will always find that the three perpendicular bisectors of the sides of a triangle pass through the same point. Since O lies on the perpendicular bisector of BC.
AL is one of these altitudes. 15. This shows that the three altitudes of the triangle pass through the same point.14
Perpendicular drawn from a vertex of a triangle to the opposite side is called its altitude. 15. 15. We also draw the third altitude CN and observe that it also passes through the point H (Fig. How many altitudes can be drawn in a triangle ? There are three vertices in a triangle.4. so we can draw three of its altitudes. 15.3 ALTITUDES OF A TRIANGLE In ∆ABC.16
. we can draw a circle passing through the three vertices. 15.11) 2. called 'circumcircle' of the triangle.16).14)
Fig. the line AL is the perpendicular drawn from vertex A to the opposite side BC. 15.58
Mathematics
If we take O as the centre and AO as the radius. Note that the circumcentre will be 1. on the hypotenuse for a right triangle [(Fig.12(b)]. which intersects the first altitude at a point H (see Fig. in the interior of the triangle for an acute triangle (Fig. O being the centre of this circle is called the circumcentre and AO the radius of the circumcircle is called circumradius of the triangle. B and C of the triangle.
Fig.15). in the exterior of the triangle for an obtuse triangle [(Fig. 1515
Fig. 15. 15. A. (Fig. 15.12(a)] 3. Now we draw the second altitude BM.
that is.19)
(a) Fig.4 MEDIANS OF A TRIANGLE In ∆ABC. Again observe that the orthocentre will be 1. The point of concurrency is called the 'Orthocentre' of the triangle. 15.4. in the interior of the triangle for an acute triangle (Fig.16) 2.19
(b)
. in the exterior of the triangle for an obtuse triangle (Fig.18) 15.18
Thus we conclude that : In a triangle. at the vertex containing the right angle for a right triangle (Fig.Concurrent Lines
59
We may take any type of triangle and draw its three altitudes.
Fig. they are concurrent. 15. the three altitudes pass through the same point.17
Fig. 15. We always find that the three altitudes of a triangle are concurrent. 15. 15.17) 3. 15. 15. AD joins the vertex A to the mid point D of the opposite side BC (Fig.
On measurement.20
(c)
Here in each of the triangles ABC given above (Fig.20 (a). Try to balance the triangle by placing the tip of a pointed stick or a needle of compasses below the point G or at G. 15.21
. 15. the point of concurrency G divides each of the medians in the ratio 2 : 1 Thus we conclude that : Medians of a triangle pass through the same point. we always find that the three medians pass through the same point [Fig.60
Mathematics
A line joining a vertex to the mid point of the opposite side of a triangle is called its median. (c)]
(a)
(b) Fig. BG = 2GE and CG = 2GF that is. which divides each of the medians in the ratio 2 : 1.
Fig. Clearly.21). If the position of G is correctly marked then the weight of the triangle will balance at G (Fig. 15. (b). The point of concurrency G is called the 'centroid' of the triangle. 15. In each triangle we measure the parts into which G divides each median. If we draw all the three medians in any triangle. ACTIVITY FOR YOU Cut out a triangle from a piece of cardboard.20) the three medians AD. 15. AD is one of the medians. BE and CF are concurrent at G. we observe that AG = 2GD. three medians can be drawn in a triangle. Draw its three medians and mark the centroid G of the triangle.
Example 15.3 : Find the circumradius of circumcircle and inradius of incircle of an equilateral triangle of side a. AD is also the angle bisector of ∠A. why the point of concurrency of the medians of a triangle is called its centroid. the angle bisector of ∠A is also a perpendicular bisector of BC.22
⇒ AD is perpendicular bisector of side BC
[Q ∠ADB = 90° ⇒ AD is an altitude also] Example 15. altitudes and medians of the ∆ABC. Solution : We draw perpendicular from the vertex A to the side BC.1 : In an isosceles triangle. since AB = BC and BC = AC ∴ BE and CF. Solution : In ∆ABD and ∆ACD AB = AC ∠BAD = ∠CAD AD = AD ∴ ∴ ∆ABD ≅ ∆ACD BD = CD (Given) [Q AD is bisector of ∠A]
⇒ AD is also a median ⇒ Also ∠ADB = ∠ADC = 90°
Fig. perpendicular bisector of side BC and a median joining vertex to the midpoint of BC
. an altitude and a median of the triangle. an altitude and a median of the ∆ABC (Refer Example 1 above) Similarly.2 : In an equilateral triangle. show that the three angle bisectors are also the three perpendicular bisectors of sides. three altitudes and the three medians of the triangle. It is the point where the weight of the triangle is centred or it is the point through which the weight of the triangle acts. 15. Solution : Since AB = AC ∴ AD. show that the bisector of the angle formed by the equal sides is also a perpendicular bisector. We consider some examples using these concepts.Concurrent Lines
61
Can you reason out. are also perpendicular bisectors.23
Example 15. 15.
Fig. angle bisectors of ∠B and ∠C respectively.
15.8 cm. an angle bisector. If AG is 4.
.62
Mathematics
Fig. 3 2 6
CHECK YOUR PROGRESS 15. In an equilateral triangle show that the incentre. 15.25
2.1 1.24
∴
AD =
3 a as BC = a. find BE. In an equilateral ∆ABC (Fig. G is the centroid of the triangle. 3. In the given figure if BF = FC. 15. ∠BAE = ∠CAE and ∠ADE = ∠GFC = 90° then name a median.26). 2 2 3 3 × a = a 3 2 3
⇒
AG = circumradius in this case =
and
GD = inradius in this case =
1 3 3 × a = a. the circumcentre. and altitude and a perpendicular bisector of the triangle.
Fig. the orthocentre and the centroid are the same point.
26
4. In a triangle (i) Angle bisectors are concurrent and the point of concurrency is called incentre.Concurrent Lines
63
Fig. (ii) Perpendicular bisectors of the sides are concurrent and the point of concurrency is called circumcentre. 15. then show that A is the orthocentre of the ∆HBC. the point equidistant from vertices of a triangle is called its (a) centroid (c) circumcentre (b) incentre (d) orthocentre
(ii) In the plane of a triangle. (iii) Altitudes are concurrent and the point of concurrency is called orthocentre. If H is the orthocentre of ∆ABC. A line which bisects a side of a triangle at right angle is called a perpendicular bisector of the triangle.
z z
z
z
z
. which divides each of the medians in the ratio 2 : 1. A line drawn perpendicular from a vertex of a triangle to its opposite side is called an altitude of the triangle. A line which joins a vertex of a triangle to the mid-point of the opposite side is called a median. 5. A line which bisects an angle of a triangle is called an angle bisector of the triangle. (iv) Medians are concurrent and the point of concurrency is called centroid. the point equidistant from the sides of the triangle is called its (a) entroid (c) circumcentre LET US SUM UP
z
(b) incentre (d) orthocentre
Three or more lines in a plane which intersect each other in exactly one point are called concurrent lines. Choose the correct answers out of the given alternatives in the following questions : (i) In a plane.
the incentre. ABC is an isosceles triangle such that AB = AC = 17 cm and base BC = 16 cm. 4. Draw an equilateral triangle. Show that the centroid. Also draw the circumcircle of the triangle. all lie on AD.
. ABC is an equilateral triangle of side 12 cm. 15. the circumcentre and the orthocentre.27 D.28
3. If G be its centroid. Draw a triangle ABC and find its circumcentre. 3. Draw the circumcircle and the incircle for an equilateral triangle of side 5 cm. E and F are the mid points of the sides of ∆ABC. 2. 15. Show that BE + CF >
3 BC. ACTIVITIES FOR YOU : 1. In the given Fig. Draw its incircle and circumcircle.64
Mathematics
TERMINAL EXERCISE 1. ABC is an isosceles triangle such that AB = AC and D is the midpoint of BC.27
2. 15. If G is the centroid of ∆ABC. 2
Fig. Find its incentre and circumcentre.
Fig. find AG. find AG.
you will find many objects bounded by four lines. rectangles. the floor of your room are all examples of a closed figure bounded by four line segments. 16.
z z
z
z
z z
z
. such a figure is called a quadrilateral. verify that triangles on the same or equal bases and between the same parallels are equal in area and its converse. window door. prove that parallelograms on equal (or same) bases and between the same parallels are equal in area. verify that the line drawn through the mid-point of a side of a triangle parallel to another side bisects the third side. slice of bread.1 INTRODUCTION If you look around. verify properties of different types of quadrilaterals. verify that a diagonal of a parallelogram divides it into two triangles of equal area. enclosing a part of the plane. the learner will be able to :
z
describe various types of quadrilaterals viz. verify that if there are three or more parallel lines and the intercepts made by them on a transversal are equal. Thus. trapeziums. a quadrilateral is that geometrical figure which has four sides. we shall study about terms and concepts related to quadrilateral with their properties. A book. verify that in a triangle the line segment joining the mid-points of any two sides is parallel to the third side and is half of it. rhombuses and squares. some parts of window-grill. parallelograms.2 OBJECTIVES After studying this lesson. the corresponding intercepts on any other transversal are also equal. In this lesson.66
Mathematics
16
Quadrilaterals
16. The word quadrilateral has its origin from the two words "quadric" meaning four and "lateral" meaning sides.
angles denoted by 1. C and D. ∠2. then the closed figure made up of four line segments is called a quadrilateral with vertices A. 16. (iii) AB and BC . ABCD. BC and CD are two pairs of consecutive sides or adjacent sides.4 QUADRILATERAL Recall that if A. B. both the quadrilaterals can be named as quad. Can you name the other pairs of consecutive angles ? (v) AC and BD are the two diagonals. ABCD. In Fig. Angles denoted by 5. Four fundamental operations on numbers.2. B. CD and DA do not intersect except at their end points. 16. ∠3 and ∠4
.Quadrilaterals
67
16. 6. Drawing parallel and perpendicular lines. 3 and 4 are the interior angles or the angles of the quad. C and D is generally denoted by quad. 2. 16.1 (i) and (ii). BC and AD are two pairs of opposite sides. ABCD.1
(ii)
(i) AB and DC .
(i)
Fig.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z
Drawing line-segments and angles of given measure. A quadrilateral with vertices A.
16. B. 7 and 8 are the exterior angles of the quad. ∠B and ∠D are two pairs of opposite angles. Drawing circles/arcs of given radius. (ii) ∠A and ∠C . ABCD. BC. In Fig. C and D are four points in a plane such that no three of them are collinear and the line segment AB. In quadrilateral ABCD. ∠B and ∠C are two pairs of consecutive angles or adjacent angles. Measure ∠1. Can you name the other pairs of consecutive sides ? (iv) ∠A and ∠B .
. sum of exterior angels of a quadrilateral is also 360°.3
Trapezium
Kite
Rhombus
Let us describe them one by one. You also know how to name them. In Fig.5. 16.3 below : Quadrilateral Parallelogram Rectangle Square
Fig. 16.68
Mathematics
(i)
Fig.2
(ii)
What is the sum of these angles ? You will find that ∠1 + ∠2 + ∠3 + ∠4 = 360°. A family tree of quadrilaterals is given in Fig. However.e. sum of interior angles of a quadrilateral equals 360°. Also what is the sum of exterior angles of the quadrilateral ABCD ? You will again find that ∠5 + ∠6 + 7 + ∠8 = 360° i. 16. 16.1 Trapezium A quadrilateral which has one pair of opposite sides parallel is called a trapezium. we will now study different types of quadrilaterals in a systematic way. 16.5 TYPES OF QUADRILATERALS You are familiar with quadrilaterals and their different shapes. 16.e.
. i.4 [(i) and (ii)] ABCD and PQRS are trapeziums with AB || DC and PQ || SR respectively.
In Fig.6
.5 [(i) and (ii)] ABCD and PQRS are parallelograms with AB||DC and AD||BC.3 Rhombus A rhombus is a parallelogram in which any pair of adjacent sides is equal. 16.5
(ii)
16. 16.5.
Fig. 16.5.6 ABCD is a rhombus.4
(ii)
16.Quadrilaterals
69
(i)
Fig.
(i)
Fig. These are denoted by ||gm ABCD and ||gm PQRS. 16. is called a parallelogram. 16.2 Parallelogram A quadrilateral which has both pairs of opposite sides parallel. In Fig.
ABCD is a square in which AB||DC.5.8. each pair of adjacent sides is equal. a parallelogram having all sides equal and each angle equal to a right angle is called a square.5.
Fig.8
In Fig. 16. Let us take some examples to illustrate. AD||BC and ∠A = ∠B = ∠C = ∠D = 90°.7
16.e..
Fig. with a pair of adjacent sides equal. 16.70
Mathematics
You may note that ABCD is a parallelogram with AB = BC = CD = DA i. AD||BC.7. 16. 16.4 Rectangle A parallelogram one of whose angles is a right angle is called a rectangle. and AB = BC = CD = DA and ∠A = ∠B = ∠C = ∠D = 90°. ABCD is a rectangle in which AB||DC. In other words. In Fig. 16.
.5 Square A square is a rectangle.
PQR is a triangle.1 : In Fig.9.2 : The three angles of a quadrilateral are 100°. the measure of fourth angle is 140°.9
Solution : Quadrilateral STRQ is a trapezium. 16. S and T are two points on the sides PQ and PR such that ST||QR.
(i)
(ii)
(iii)
(iv)
Fig. Name the type of quadrilateral STRQ so formed. 50° and 70°. CHECK YOUR PROGRESS 16. Find the measure of the fourth angle. Then 100° + 50° + 70° + x° = 360° 220° + x° = 360° x = 140 Hence. Let the fourth angle be x.
Fig.10
(v)
(vi)
. Name each of the following quadrilaterals. because ST||QR. 16.1 1. Example 16. 16. Solution : We know that the sum of the angles of a quadrilateral is 360°.Quadrilaterals
71
Example 16.
The angles of a quadrilateral are in the ratio 5 : 7 : 7 : 11.72
Mathematics
2. (vi) A square is a parallelogram. all its angles are equal. (iii) A rectangle is a parallelogram.1 Properties of a Parallelogram We have learnt that a parallelogram is a quadrilateral with both pairs of opposite sides parallel. 3.11
(ii)
. States which of the following statements are correct ? (i) Sum of interior angles of a quadrilateral is 360°. If a pair of opposite angles of a quadrilateral are supplementary. (vii) A parallelogram is a rhombus. (viii) A trapezium is a parallelogram. 16. 4. Find the measure of each angle. PROPERTIES OF DIFFERENT TYPES OF QUADRILATERALS 16. Find the measure of each angle. Draw a pair of parallel lines l and m as shown in Fig.11.6.
(i)
Fig. 5. angles and diagonals of a parallelogram. what can you say about the other pair of angles ? 16. (ii) All rectangles are squares. (ix) A trapezium is a rectangle. 16.6. (iv) A square is a rhombus. Draw another pair of parallel lines p and q such that they intersect l and m. (x) A parallelogram is a trapezium. You observe that a parallelogram ABCD is formed. In a quadrilateral. They intersect each other at O. (v) A rhombus is a parallelogram. Now let us establish some relationship between sides. Join AC and BD.
What do you find ? You will find that OA = OC and OB = OD
Draw another parallelogram and repeat the activity you will find that (i) The opposite sides of a parallelogram are equal.12. Now place ∆ADC on ∆ABC in such a way that the vertex D falls on the vertex B and the side CD falls along the side AB. 16. In other words ∆ABC ≅ ∆ADC. What do you find ? You will find that AB = DC and BC = AD. BC. Thus. (iii) The diagonals of a parallelogram bisect each other. the parallelogram has been divided into two parts and each part is a triangle. ∠BCD. Also measure ∠ABC. Draw its diagonal AC as shown in Fig 16. Now cut this parallelogram along the diagonal AC. Also AB = CD and BC = AD and ∠B = ∠D. What do you find ? You will find that ∠DAB = ∠DCB and ∠ABC = ∠CDA Again.Quadrilaterals
73
Now measure the sides AB. ∆ABC and ∆ADC. you get two triangles.12
(ii)
In other words. Where does the point C fall ? Where does the point A fall ? You will observe that ∆ADC will coincide with ∆ABC. OB and OD. ∠CDA and ∠DAB. Draw any parallelogram ABCD on it.
. CD and DA. (ii) The opposite angles of a parallelogram are equal.
(i)
Fig. Cut the parallelogram ABCD from the cardboard. The above said properties of a parallelogram can also be verified by Cardboard model which is as follows : Let us take a card board. OC. Measure OA.
You may also verify the following properties which are the converse of the properties of a parallelogram proved earlier. Draw its diagonals PR and QS which intersect each other at O as shown in Fig. 16.14.e.
. (ii) A quadrilateral is a parallelogram if its opposite angles are equal. ∆POS and ∆ROQ. Where does the point S fall ? Where does the side OS fall ? Is ∆ROS ≅ ∆POQ ? Yes.. (i) A quadrilateral is a parallelogram if its opposite sides are equal. In Fig. what do you observe ? We find that RO = PO and OS = OQ You may also verify this property by taking another pair of triangles i. We know that a rhombus is a parallelogram in which a pair of adjacent sides are equal. 16. Now cut the parallelogram PQRS. the diagonals of a parallelogram bisect each other. Now you can prove the third property of the parallelogram. ABCD is a rhombus. 16. you will always get the same results as verified earlier.e. (iii) A quadrilateral is a parallelogram if its diagonals bisect each other. proving the above two properties of the parallelogram. 16.13.
Fig.2 Properties of a Rhombus In the previous section we have defined a rhombus.74
Mathematics
You may repeat this activity by taking some other parallelograms. You will again arrive at the same result. Now place ∆ROS and ∆POQ in such a way that the vertex R coincides with the vertex P and RO coincides with the side PO. Draw any parallelogram PQRS on it. So. it is.6. i. Again take a thin cardboard.13
Also cut ∆POQ and ∆ROS. thus.
(i) All sides of a rhombus are equal (ii) The opposite angles of a rhombus are equal (iii) The diagonals of a rhombus bisect each other at right angles. Thus. the diagonals of a rhombus bisect each other at right angles. we have the following properties of a rhombus. You may repeat this experiment by taking different rhombuses.. What is the measures of these angles ? You will find that each of them equals 90° Also ∠AOB = ∠COD and ∠BOC = ∠DOA ∠AOB = ∠COD = ∠BOC = ∠DOA = 90° Thus.. (Each pair is a vertically opposite angles)
. i. therefore all the properties of a parallelogram are also true for rhombus. (i) Opposite sides are equal. i.14
Thus.e.e.e. the diagonals of a rhombus bisect each other. you will find in each case. i.Quadrilaterals
75
Fig.e. all the sides of a rhombus are equal. ∠A = ∠C and ∠B = ∠D (iii) Diagonals bisect each other i. AB = DC and AD = BC (ii) Opposite angles are equal. Since every rhombus is a parallelogram.. Measure ∠AOD and ∠BOC. ABCD is a parallelogram with AB = BC. AO = OC and DO = OB Since adjacent sides of a rhombus are equal and by the property of a parallelogram AB = BC = CD = DA Thus. 16.
Also measure AO. Also measure AO.76
Mathematics
16. BO and DO. with a pair of adjacent sides equal. it is You will also find that AO = OC and BO = DO. Join AC and BD in each case. OC.6. ∠BCD and ∠ADC. Join AC and BD as shown in the Fig.15
Measure ∠ΒAD.e. OC and BO.15
Fig. Label them again by ABCD. can you conclude from definition of a square that a square is rectangle and possesses all the properties of a rectangle ? Yes it is. Do you find that AC = BD ? Yes. 16. Thus. we can conclude that ∠A = ∠B = ∠C = ∠D = 90° i. we have the following properties of a rectangle : (i) The opposite sides of a rectangle are equal (ii) Each angle of a rectangle is a right-angle. OD for each rectangle.6. Can you say whether a rectangle possesses all the properties of a parallelogram or not ? Yes it is.4 Properties of a Square You know that a square is a rectangle. Thus. 16. Draw some more rectangles of different dimensions. Let us now study some more properties of a square.3 Properties of a Rectangle We know that a rectangle is a parallelogram one of whose angles is a right angle. (iv) The diagonals of a rectangle bisect each other. (iii) The diagonals of a rectangle are equal. each angle of a rectangle measures 90°. what do you find ? What are the measures of these angles ? The measure of each angle is 90°. Let them intersect each other at O. In each case you will find that The diagonals of a rectangle are equal and they bisect each other.. Now.
. Now measure the diagonals AC and BD. Let us study some more properties of a rectangle Draw a parallelogram ABCD in which ∠B = 90°. 16.
∠BOC. we conclude that the diagonals AC and BD of a square bisect each other at right angles. Now measure ∠AOB.
Fig. Thus. ∠COD and ∠AOD. we have the following properties of a square (i) All the sides of a square are equal. (ii) Each of the angles measure 90°. therefore a square ABCD is a rhombus also. AD = BC (ii) ∠A = ∠B = ∠C = ∠D = 90° (iii) AC = BD and AO = OC. 16. What do you find ? Does each angle measure 90° ? Yes Thus.16
Since ABCD is a rectangle.16. therefore we have (i) AB = DC.Quadrilaterals
77
Draw a square ABCD as shown in Fig. (iii) The diagonals of a square are equal. (iv) The diagonals of a square bisect each other at right angles. ∠AOB = ∠BOC = ∠COD = ∠AOD = 90° Thus. BO = OD But in a square we have AB = AD ∴ By property (i) we have AB = AD = CD = BC. Let us study some examples to illustrate :
. You may also observe that since a square is a parallelogram also with AB = AD. 16.
∠A = 62°.7 MID POINT THEOREM Draw any triangle ABC. DE =
1 BC 2
Fig. Find the value of x. 2. Find the measure of ∠CAB.80
Mathematics
CHECK YOUR PROGRESS 16. You will always find that DE = Thus.22. 16. it is. ABCD is a parallelogram in which ∠DAB = 70° and ∠CBD = 55°. find the measure of ∠OPS. 3. AC is one diagonal of a square ABCD.21
Again. In a parallelogram ABCD. If ∠ROQ = 40°. the diagonals of a rectangle PQRS intersect each other at O. we conclude that
1 BC and DE || BC. they are equal. Mark them as D and E respectively. 16. In a parallelogram ABCD. the lines are parallel. Find ∠CDB and ∠ADB.22 Fig. 6. 16. 5. Find the mid points of the side AB and AC. Find the measures of the other angles. measure ∠ADE and ∠ABC. 16. Join DE. Find the measure of ∠ACD. 7. as shown in Fig.21. ∴ DE || BC
You may repeat this experiment with another two or three triangles and naming each of them as triangle ABC and the mid points as D and E of sides AB and AC. Are these angles equal ?
Yes.2 1. 16. In Fig. ABCD is a rhombus in which ∠ABC = 58°. Measure BC and DE. Find all the angles of the parallelogram. you know that these angles make a pair of corresponding angles. 4. What relation do you find between the length of BC and DE ? Of course. ∠A = (2x + 10)° and ∠C = (3x – 20)°. You know that when a pair of corresponding angles are equal. 2
. The sum of the two opposite angles of a parallelogram is 150°.
You may repeat with different triangles and by naming each of them as PQR and taking each time L as the midpoint RQ and drawing a line LM || PQ.e. We can also verify the converse of the above stated result. Show that F is the mid-point of BC. 16. 16. ABCD is a trapezium in which AD and BC are its non-parallel sides and E is the mid-point of AD. DE || BC and D is the mid point of AB ∴ E is also the mid point of AC i.Quadrilaterals
81
In a triangle the line-segment joining the mid points of any two sides is parallel to the third side and is half of it. Measure PM and MR. EF||AB. we conclude that
Fig.25.24
= 4 cm Hence AE = 4 cm.25
. Solution : In ∆ABC. Solution : Since EG||AB and E is the mid-point of AD ∴ G is the mid point of DB In ∆DBC. If AC = 8 cm. they are equal.24. 16. GF||DC and G is the mid-point of DB. ∴ F is the mid-point of BC. AE =
1 AC 2
=
FG 1 × 8IJ H2 K
cm
[Q AC = 8 cm]
Fig. From L. D is the mid-point of the side AB of ∆ABC and DE || BC. Example 16. 16. Example 16.8 : In Fig. and mark it as L. PR at M. Are they equal ? Yes.23
"The line drawn through the mid point of one side of a triangle parallel to the another side bisects the third side. Let us consider some examples. Thus. 16. Draw any ∆PQR. you will find in each case that RM = MP.
Fig. which intersects. find AE. Find the mid point of side RQ. draw a line LX || PQ.7 : In Fig.
DG||EF.31. 16.
Fig. 16. In Fig.28. 16. In Fig. find DE. 16. 16.28
3.
Fig.30
5. M is the mid-point of AB and MN||BC. Show that ∆AMN is also an isosceles. AD is a median of a ∆ABC and E is the mid-point of AD. In Fig. meets AC at G.30.
. AB||CD||QR.29
4. A and C divide the side PQ of ∆PQR into three equal parts. find AF. If AC = 9 cm.Quadrilaterals
83
2. BE is produced to meet AC at F. 16. In Fig. Prove that B and D also divide PR into three equal parts.
Fig. D and E are the mid-points of the sides AB and AC respectively of a ∆ABC. 16.29. ABC is a isosceles triangle in which AB = AC. If BC = 10 cm.
16.33
(ii)
.32.8 INTERCEPT THEOREM Recall that a line which intersects two or more lines is called a transversal. 16. The segment cut off from the transversal by a pair of lines is called an intercept.31
16.84
Mathematics
Fig. Thus.33 (ii).32
The intercepts made by parallel lines on a transversal have some special properties which we shall study here. 16.
Fig.
(i)
Fig. in Fig. 16. there will be two intercepts AB and BC as shown in Fig. If there are three parallel lines and they are intersected by a transversal. Let l and m be two parallel lines and XY is an intercept made on the transversal "n". 16. XY is an intercept made by line l and m on transversal n.
On a page of your note-book. q. Repeat this experiment by taking another set of two or more equidistant parallel lines and measure their intercepts as done earlier. r and s as shown in Fig. A.
Fig. they are equal. 16. These transversal make different intercepts. Z respectively such that XY = YZ. Thus.35. measure LM. Y. Measure the intercept AB. Solution. Given that XY = YZ ∴ and AB = BC LM = MN (Intercept theorem)
Thus. M. the other pairs of equal intercepts are AB = BC and LM = MN. Are they equal ? Yes. You will find in each case that the intercepts made are equal. 16.Quadrilaterals
85
Now let us learn an important property of intercepts made on the transversals by the parallel lines. MN and NX. Let us illustrate it by some examples : Example 16. B. draw any two transversals l and m intersecting the parallel lines p.35
. we conclude the following : If there are three or more parallel lines and the intercepts made by them on a transversal are equal. p || q || r.34
Also. Name the other pairs of equal intercepts. N.
Fig.34. The transversals l. BC and CD. 16. m and n cut them at L. they are. C and X. the corresponding intercepts made on any other transversal are also equal.10 : In Fig 16. Do you find that they are also equal ? Yes.
(ii)
Fig. AB = 3.2 cm. XP and BZ. Draw DP ⊥ DC and QC ⊥ DC. are said to be on the same or equal bases and between the same parallels. 16. Let us prove it logically.. LM = MZ = 3 cm. Now.
16.(i)
.39
16.4 cm.
Fig.Quadrilaterals
87
3. Area ∆ADC = Area of ∆ACB = As ∴
1 DC × PD 2 1 AB × QC 2
. Join its diagonal AC. we conclude the following : A diagonal of a parallelogram divides it into two triangles of equal area. therefore PD = QC.40
AB = DC and PD = QC
Area (∆ADC) = Area (∆ACB)
Thus.5 cm and YZ = 3. 16. Because AB || DC. find XY.9 THE DIAGONAL OF A PARALLELOGRAM AND DIVISION OF AREA Draw a parallelogram ABCD..10 PARALLELOGRAMS AND TRIANGLES BETWEEN THE SAME PARALLELS Two parallelograms or triangles. having equal or same bases and having their other vertices on a line parallel to their bases. We will prove an important theorem on parallelogram and their area Theorm : Parallelograms on the same base (or equal bases) and between the same parallels are equal in area. In Fig..39. Given that l || m || n and PQ = 3.. Consider the two triangles ADC and ACB in which the parallelogram ABCD has been divided by the diagonal AC.
. 16.
16. on the same base and between the same parallels.11 TRIANGLES ON THE SAME OR EQUAL BASES HAVING EQUAL AREAS HAVE THEIR CORRESPONDING ALTITUDES EQUAL Recall that the area of triangle =
1 (Base) × Altitude 2
. 16..(iii)
Fig.41
(Opposite sides of a parallelogram) (Opposite sides of a parallelogram) (Corresponding Angles)
Area (||gm ABCD) = Area(∆ABP) + Area(Trap.88
Mathematics
Given : Parallelograms ABCD and PBCQ stand on the same base BC and between the same parallels BC and AQ.41. BCDP)
From (i). Result : Triangles... we have and ∴ ∴ Now.. To prove : Area (ABCD) = Area (BCQP) Proof : Consider two triangles ABP and DCQ.(i) . we get Area (||gm ABCD) = Area (||gm BCQP) Note : ||gm and Trap. stands for parallelogram and trapezium respectively. Thus.. ∴ and ∴ Area (∆BCQ) = Area (∆PBQ) Area(∆ABC) = Area (∆CAD) Area (∆ABC) = Area (∆BCQ) Parallelogram on the same base (or equal bases) and between the same parallels are equal in area.. 16. Join the diagonals BQ and AC of the two parallelograms BCQP and ABCD respectively. BCDP) Area (||gm BCQP) = Area (∆CQD) + Area(Trap. AB = DC BP = CQ ∠1 = ∠2 ∆ABP ≅ ∆DCQ Area (∆ABP) = Area (∆DCQ) . We know that a diagonal of a ||gm divides it in two triangles of equal area.are equal in area Consider Fig. (ii) and (iii).(ii) . we also conclude the following : Triangles on the same base (or equal bases) and between the same parallels are equal in area.
z In a parallelogram : (i) opposite sides and angles are equal.Quadrilaterals
91
3. z Triangles on equal bases having equal areas have their corresponding altitudes equal. 16. enclosing some area of the plane.
Fig. find the altitude of ||gm BCFE. Which of the following are trapeziums ?
(i)
Fig. z A parallelogram is a rectangle if its one angle is 90°. The area of ∆ACD in Fig 16. z The diagonals of a rhombus bisect each other at right angle.46 is 8 cm2.
A quadrilateral is a parallelogram if both pair of sides are parallel.47
(ii)
(iii)
. z The diagonals of a square intersect at right angles. (ii) diagonals bisect each other. z A parallelogram is a rhombus if its adjacent sides are equal. z The triangles on the same base (or equal bases) and between the same parallels are equal in area. z The diagonal of a parallelogram divides it into two triangles of equal area. TERMINAL EXERCISE 1.46
LET US SUM UP
z z z z
A quadrilateral is a four sided closed figure. The sum of the interior or exterior angles of a quadrilateral is each equal to 360°. A quadrilateral is a trapezium if its one pair of opposite sides is parallel. If EF = 4 cm. z The diagonals of a rectangle are equal. z Parallelogram on the same base (or equal bases) and between the same parallels are equal in area. z A rectangle is a square if its adjacent sides are equal. 16.
92
Mathematics
2. In Fig. 16.48, PQ || FG || DE || BC. Name all the trapeziums in the figure.
Fig. 16.48
3. In Fig. 16.49, ABCD is a parallelogram with an area of 48 cm2. Find the area of (i) shaded region (ii) unshaded region.
Fig. 16.49
4. Fill in the blanks in each of the following to make them true statements : (i) A quadrilateral is a trapezium if ... (ii) A quadrilateral is a parallelogram if ... (iii) A rectangle is a square if ... (iv) The diagonals of a quadrilateral bisect each other at right angle. If none of the angles of the quadrilateral is a right angle, it is a ... (v) The sum of the exterior angles of a quadrilateral is ... 5. If the angles of a quadrilateral are (x – 20)°, (x + 20)°, (x – 15)° and (x + 15)°, find x and the angles of the quadrilateral. 6. The sum of the opposite angles of a parallelograms is 180°. What special type of a parallelogram is it ?
Quadrilaterals
93
7. The area of a ∆ABD in Fig. 16.50 is 24 cm2. If DE = 6 cm, and AB || CD, BD || CE, AE || BC, find
Fig. 16.50
(i) Altitude of the parallelogram BCED. (ii) Area of the parallelogram BCED 8. In Fig. 16.51, the area of parallelogram ABCD is 40 cm2. If EF = 8 cm, find the altitude of ∆DCE.
17
Similarity of Triangle
17.1 INTRODUCTION Looking around you will see many objects which are of the same shape but of same or different sizes. For examples, leaves of a tree have almost the same shape but same or different sizes. Similarly, photographs of different sizes developed from the same negative are of same shape but different sizes, the miniature model of a building and the building itself are of same shape but different sizes. All those objects which have the same shape but different sizes are called similar objects. Let us examine the similarity of plane figures : (i) Two line-segments of the same length are congruent but of different lengths are similar.
(ii) Two circles of the same radius are congurent but circles of different radii are similar.
(iii) Two equilateral triangles of different sides are similar.
96
Mathematics
(iv) Two squares of different sides are similar.
In this lesson, we shall study about the concept of similarity, especially similarity of triangles and the conditions thereof. We shall also study about various results related to them. 17.2 OBJECTIVES After studying this lesson, the learner will be able to :
z z z z z z
identify similar figures distinguish between congurent and similar plane figures cite the criteria for similarity of triangles viz. AAA, SSS and SAS. verify and use unstarred results given in the curriculum based on similarity experimentally prove the Baudhayan/Pythagoras Theorem apply these results in verifying experimentally (or proving logically) problems based on similar triangles.
17.3 EXPECTED BACKGROUND KNOWLEDGE Knowledge of
z z z z z
plane figures like triangles, quadrilaterals, circles, rectangles, squares, etc. criteria of congruency of triangles finding squares and square-roots of numbers ratio and proportion internal and external bisectors of angles of a triangle.
17.4 SIMILAR PLANE FIGURES
Fig. 17.2
Similarity of Triangle
97
In Fig. 17.2, the two pentagon seem to be of the same shape. We can see that ∠A = ∠A′, ∠B = ∠B′, ∠C = ∠C′, ∠D = ∠D′ and ∠E = ∠E′ and AB = BC = CD = DE = EA . We say that the two pentagons are similar. Thus we say A ' B' B' C' C' D' D' E' E' A ' that Any two polygons, with corresponding angles equal and corresponding sides proportional, are similar Thus, two polygons are similar, if they satisfy the following two conditions : (i) Corresponding angles are equal (ii) The corresponding sides are proportional. Even if one of the conditions does not hold, the polygons are not similar as in the case of a rectangle and square given in Fig. 17.3. Here all the corresponding angles are equal but the corresponding sides are not proportional.
Fig. 17.3
17.5 SIMILARITY OF TRIANGLES Triangles are special type of polygons and therefore the conditions of similarity of polygons also hold for triangles. Thus, Two triangles are similar if (i) their corresponding angles are equal, and (ii) their corresponding sides are proportional
Fig. 17.4
We say that ∆ABC is similar to ∆DEF and denote it by writing ∆ABC ~ ∆DEF
98
Mathematics
The symbol '~' stands for the phrase " is similar to" If ∆ABC ~ ∆DEF, then by definition ∠A = ∠D, ∠B = ∠E, ∠D = ∠F and 17.5.1 AAA criterion for similarity We shall show that if either of the above two conditions is satisfied then the other automatically holds in the case of triangles. Let us perform the following experiment. Construct two ∆'s ABC and PQR in which ∠P = ∠A, ∠Q = ∠B and ∠R = ∠C as shown in Fig. 17.5. AB = BC = CA DE EF FD
Fig. 17.5
Measure the sides AB, BC and CA of ∆ABC and also measure the sides PQ, QR and RP of ∆PQR.
AB BC CA Now find the ratio PQ , QR and . RP
What do you find ? You will find that all the three ratios are equal and therefore the triangles are similar. Try this with different triangles with equal corresponding angles. You will find the same result. Thus, we can say that If in two triangles, the corresponding angles are equal the triangles are similar. This is called AAA similarity criterion. 17.5.2 SSS criterion for similarity. Let us now perform the following experiment : Draw a triangle ABC with AB = 3 cm, BC = 4.5 cm and CA = 3.5 cm
Similarity of Triangle
99
(i) Fig. 17.6
(ii)
Draw another ∆PQR as shown in Fig. 17.6 (ii)
AB BC AC We can see that PQ = QR = PR
i.e., the sides of the two triangles are proportional Now measure ∠A, ∠B and ∠C of ∆ABC and ∠P, ∠Q and ∠R of ∆PQR. You will find that ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R. Repeat the experiment with another two triangles having corresponding sides proportional, you will find that the corresponding angles are equal and so the triangle are similar. Thus, we can say that If the corresponding sides of two triangles are proportional the triangles are similar. 17.5.3 SAS Criterian for Similarity Let us conduct the following experiment. Take a line AB = 3 cm and at A construct an angle of 60°. Cut off AC = 4.5 cm. Join BC
Thus, we conclude that If one angle of a triangle is equal to one angle of the other triangle and the sides containing these angles are proportional, the triangles are similar. Thus, we have three important criteria for the similarity of triangles. They are given below: (i) If in two triangles, the corresponding angles are equal, the triangles are similar. (ii) If the corresponding sides of two triangles are proportional, the triangles are similar. (iii) If one angle of a triangle is equal to one angle of the other triangle and the sides containing these angles are proportional, the triangle are similar. Example 17.1 : In Fig. 17.8 are given two triangles ABC and PQR
17.6 BASIC PROPORTIONALITY THEOREM We state below the Basic Proportionality Theorem : If a line is drawn parallel to one side of a triangle, the other two sides of the triangle are divided proportionally. Thus, in Fig. 17.13, DE || BC, According to the above result AD = AE DB EC We can easily verify this by measuring AD, DB, AE and EC. You will find that AD = AE DB EC We state the converse of the above result as follows : If a line divides any two sides of a triangle in the same ratio, the line is parallel to third side of the triangle. Thus, in Fig. 17.13, if DE divides sides AB and AC of ∆ABC such that DE || BC. We can verify this by measuring ∠ADE and ∠ABC and finding that ∠ADE = ∠ABC These being alternate angles, the lines DE and BC are parallel. We can verify the above two results by taking different triangles. Let us solve some examples based on these. Example 17.3 : In Fig. 17.14, DE || BC. If AD = 3 cm, DB = 5 cm and AE = 6 cm, find AC. Solution : DE || BC (Given). Let EC = x ∴ ∴ ⇒ ⇒ ∴ ∴ AD AE = DB EC 3 6 = 5 x 3x = 30 x = 10 EC = 10 cm AC = AE + EC = 16 cm.
Fig. 17.14 Fig. 17.13
2. In Fig. 17.17 [(i), (ii) and (iii)], find whether DE is parallel to BC or not ? Give reasons for your answer.
Fig. 17.17
17.7 BISECTOR OF AN ANGLE OF A TRIANGLE We now state an important result as given below : The internal bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle
5 cm. we may verify the result. The bisector AD of ∠A intersects the opposite side BC in D such that BD = 4.104
Mathematics
Thus. AB and AC and finding the ratios. Example 17. Find the lengths of the line-segments into which the smallest side is divided by the bisector of the angle opposite to it. ∴ ⇒ BD 36 = 3 = DC 48 4 4BD = 3DC or BD = 36 DC = 3 DC 48 4
Fig. we have BD AB = DC AC (Q AD is the internal bisector of ∠A of ∆ABC) or ⇒ 4.e. 17.5 6 = x 8 6x = 4. 17. Example 17. the length of line-segment CD = 6 cm. 17. Solution : The smallest side is of length 28 cm and the sides forming the angle. DC. then BD AB = DC AC We can easily verify this by measuring BD.6 : The sides of a triangle are 28 cm.5 × 8 x=6 i.20
.5 : The sides AB and AC of a triangle are 6 cm and 8 cm. Let the angle bisector AD meet BC in D.18
Repeating the same activity with other triangles. Solution : According to the above result. Let us solve some examples to illustrate this. 36 cm and 48 cm.19
BC = BD + DC = 28 cm ∴ DC + 3 DC = 28 4
Fig. according to the above result. A opposite to it are 36 cm and 48 cm. Find the length of segment CD. We will find that BD AB = DC AC
Fig. if AD is the internal bisector of ∠A of ∆ABC..
Draw a ∆ABC. ∴ ∆ADB ~ ∆CDA ~ ∆CAB
Also.24
∆ADB and ∆CDA are similar. As ∠ADB = 90°. ∠BAD = 90° – α ∠BAC = 90° and ∠BAD = 90° – α . meeting it in D.e. 90° and 90° – α Another important result is about relation between sides and areas of similar triangles. the triangles on each side of the perpendicular are similar to each other and to the triangle. the angles of ∆BAC are α .106
Mathematics
17. as it has all the corresponding angles equal. Draw two triangles ABC and PQR which are similar i. ∠DAC = α ∠DCA = 90° – α
Fig. Let ∴ As Therefore Similarly.
Fig. Let us try of verify this by an activity.. 17. 17. their sides are proportional. right angled at A. We state the result below and try to verify the same. ∴ ∠DBA = α . It states that The ratio of the areas of similar triangles is equal to the ratio of the squares on their corresponding sides Let us verify this result by the following activity.8 SOME MORE IMPORTANT RESULTS Let us study another important result on similarity in connection with a right triangle and the perpendicular from the vertex of the right angle to the opposite side. Draw AD ⊥ to the hypotenuse BC. If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse.25
.
If AB = 6 cm and AD = 2 cm. if the corresponding altitudes AD and PS are in the ratio of 4 : 9. Q and R are the mid-points of the sides AB. 17. ABC is a triangle in which DE || BC.27.
. 17. P. If the ratio of the areas of two similar triangles is 16 : 25. find the ratio of the areas of ∆ABC and ∆PQR.
Fig. find the ratio of the area of ∆ADE and trapezium DBCE. In Fig. 17. ABC is a right triangle with ∠A = 90° and ∠C = 30°.27
2. find the ratio of their corresponding sides. Find the ratio of the areas of two similar triangles if the corresponding sides are of lengths 3 cm and 5 cm.28.28
4. 3. BC and CA of the ∆ABC respectively. In Fig.4 1. 25
CHECK YOUR PROGRESS 17.108
Mathematics
⇒
AB 5 = AP 2
⇒
AP 2 = AB 5
2 2
∴
Area ∆APQ AP 2 = = Area ∆ABC AB2
b b
g g
FG AP IJ = FG 2 IJ H ABK H 5 K
=
4 .
Fig.
LMHint : Use AB = AD = BC = CA OP PQ PS QR PR Q N
6. Show that the area of ∆PQR is one-fourth the area of ∆ABC. 5. Show that ∆DAB ~ ∆DCA ~ ∆ACB. 17. In two similar triangles ABC and PQR.
Draw a triangle ABC with side 3 cm.e. called Baudhayan/Phythagorus Theorem using the concept of similarity.1 Converse of Pythagoras Theorem The conserve of the above theorem states : In a triangle.9 BAUDHAYAN/PYTHAGORAS THEOREM We know prove an important theorem.29
..DC AB2 + BC2 = AC (AD + DC) = AC.. 4 cm and 5 cm.. we get
.(ii)
..Similarity of Triangle
109
17. BC = 4 cm and AC = 5 cm. if the square on one side is equal to sum of the squares on the other two sides. Theorem : In a right triangle. 17. From B. Baudhayan about 200 years before Pythagoras.. 17. To Prove : AC2 = AB2 + BC2 Construction. This result can be verified by the following activity. i. You can see that AB2 + BC2 = (3)2 + (4)2 = 9 + 16 = 25
Fig. we get ⇒ From (ii).. we get ⇒ ∆ADB ~ ∆ABC ∆BDC ~ ∆ABC AB AD = AC AB AB2 = AC..29) Proof : BD ⊥ AC ∴ and From (i).(B)
Adding (A) and (B). draw BD ⊥ AC (See Fig. AC = AC2 The theorem is known after the name of famous Greek Mathematician Pythagoras.. AB = 3 cm.(A)
Fig. 17. 17. in which ∠B = 90°. the angle opposite to first side is a right angle.AD BC DC = AC BC BC2 = AC.30
. Given.(i) . This was originally stated by the Indian Mathematician.9.. A right triangle ABC. the square on the hypotenuse is equal to the sum of the squares on the other two sides.
find the length of the ladder.34
5. 8. DC
Fig. Show that AB2 = AC2 + BC2 – 2BC.
. with corresponding angles equal and corresponding sides proportional. 4. Find the length of the diagonal of a square of side 10 cm. Two triangles are said to be similar. LET US SUM UP
z z
Objects which have the same shape but different sizes are called similar objects. Show that 4LC2 = AB2 + 4BC2 6. Prove that PR2 = 2PQ2.34. right angled at C. P and Q are points on the sides CA and CB respectively of ∆ABC. 17. L and M are the mid-points of the sides AB and AC of ∆ABC. A ladder is placed against a wall such that its top reaches upto a height of 4 m of the wall. Any two polygons. right angled at B.112
Mathematics
3. PQR is an isosceles right triangle with ∠Q = 90°. ∠C is acute and AD ⊥ BC. If the foot of the ladder is 3 m away from the wall. if (a) their corresponding angles are equal and (b) their corresponding sides are proportional
z
z
Criteria of similarity – AAA criterion – SSS criterion – SAS criterion
z
If a line is drawn parallel to one-side of a triangle. In Fig. Prove that AQ2 + BP2 = AB2 + PQ2 7. 17. are similar. it divides the other two sides in the same ratio and its converse.
then the angle opposite to the first side is a right angle – converse of (Baudhayan) Pythagoras Theorem. From the dimension given in the figure. CA = 3. RP = 7. In Fig. In Fig. ∠C = 60°.5 cm. QR = 7. The ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. DB = 4. 17. ∆'s ABC and PQR are similar (i) ∠A = 40°. 2.Similarity of Triangle
113
z
The internal bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle. ∠Q = 60° and ∠R = 70° (iii) AB = 2. BC = 4.
Fig. ∠B = 70°. In a right triangle. 15. ∠C = 80°.0 cm (iv) AB = 3 cm.35
Fig. Enumerate different criteria for the similarity of the two triangles.
. RP = 6. ∠P = 40°.36. QR = 9. 4. the square on the hypotenuse is equal to sum of the squares on the remaining two sides – (Baudhyan) Pythagoras Theorem In a triangle. 17. given that DE || BC.0 cm. AD = 3 cm. DE || AC.35.0 cm.0 cm.5 cm. 3. AE = 4. CA = 5. ∠Q = 60° and ∠R = 80° (ii) ∠A = 50°.5 cm. the triangles so formed are similar to each other and to the given triangle. ∠P = 50°. ∠B = 60°.5 cm. 17. find CE. TERMINAL EXERCISE
z
z
z
z
1. if the square on one side is equal to the sum of the squares on the remaining two sides.5 cm.0 cm. In which of the following cases. find the value of x. BC = 4 cm.36
5. If a perpendicular is drawn from the vertex of the right angle of a right angled triangle to the hypotenuse.0 cm PQ = 4. Write the criteria for the similarity of two polygons.5 cm PQ = 5.
17. BC = 2ab.50 cm and EC = 1. BC = 6 cm. BC = 12 cm. 8. Find the ratio of area of ∆ABC to that of ∆PQR.39. 15. show that AB2 = AC2 + BC2 + 2BC. CA = 6 cm (iv) AB = 25 cm. A ladder is placed against a wall and its top reaches a point at a height of 8 m from the ground. 17. Find the area of an equilateral triangle of side 2a. Find the distance between their tops.5 cm and 3.114
Mathematics
6. DB = 3 cm. CA = 13 cm (ii) AB = 8 cm. Is DE || BC ? Give reasons for your answer.39
.
Fig. CA = 7 cm (v) AB = a2 + b2. In Fig. The altitudes AD and PS of two similar ∆'s ABC and PQR are of length 2. 12. 13. stand on a plane ground and the distance between their feet is 12 m. find x. find the length of the ladder. In an equilateral triangle.38
7.37 is shown a ∆ABC in which AD = 5 cm. 17.38. BC = 24 cm. In Fig.5 cm. BC = 5 cm. 17. From the given dimension. Two poles of height 12 m and 17 m. AE = 2. CA = 10 cm (iii) AB = 10 cm.CD 14. CA = a2 – b2 11. Find the ratio of the area of ∆ABC to that of ∆DEF. 17.37
Fig. Which of the following are right triangles ? (i) AB = 5 cm. 17. The perimeter of two similar ∆'s ABC and DEF are 12 cm and 18 cm. 9. show that three times the square of a side equals four times the square on medians. If the distance between the wall and foot of the ladder is 6 m. 10.
Fig. In Fig. AD is the internal bisector of ∠A of ∆ABC.5 cm.
Perimeter of a closed figure Region bounded by a closed figure Congruence of closed figures
18. alphabet O.2 OBJECTIVES After studying this lesson.1 INTRODUCTION You are already familiar with geometrical figures such as a line segment. the learner will be able to :
z z z z z z
define a circle give examples of various terms related to a circle illustrate congruent circles and concentric circles identify and illustrate terms connected with circles like chord. etc. an angle. In this lesson we shall study in some details about the circle and related concepts. a bangle. 18.1 Circle : A circle is a collection of all points in a plane which are at a constant distance from a fixed point in the same plane. Examples. segment. quadrilaterals polygons. etc verify experimentally results based on arcs and chords of a circle use the results in solving problems
18. of a circle are a wheel.116
Mathematics
18
Circles
18.
.4 CIRCLE AND RELATED TERMS 18. a quadrilateral and a circle. etc.4. a triangle. sector.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z z z z
Line segment and its length Angle and its measure Parallel and perpendicular lines Closed figures such as triangles. arc.
Such a chord is called a diameter of the circle. In Figure 18.3 Fig. Diameter is usually denoted by 'd'. the diameter of a circle = twice the radius of the circle. 18. Activity for you (a) Take a point Q in the inner part of the circle (See Figure 18. 18. It is customary to write radius instead of the length of the radius. Activity for you : Measure the length d of PQ.2 Chord A line segment joining any two points of a circle is called a chord. the inner part of the figure. Thus All radii (plural of radius) of a circle are equal The length of the radius of a circle is generally denoted by the letter 'r'. In Figure 18.3).Circles
117
Radius : A line segment joining the centre of the circle to a point on the circle is called its radius. the radius r and find that d is the same as 2r. In Fig.4 Fig.1
. 18. there is a circle with centre O and one of its radius is OA.2 the shaded portion is the inner part of the circle. The inner part of the circle is called the interior of the circle. the boundary is the circle and the unshaded portion is the outer part of the circle. Activity for you : Measure the length OA and OB and observe that they are equal.4 AB and PQ and CD are three chords of a circle with centre O and radius r. (b) Now take a point P in the outer part of the circle (Figure 18.2
Fig. 18. The outer part of the circle is called the exterior of the circle. A chord passing though the centre of circle is called its diameter.4. 18.
Fig.1.3).e. 18. Measure OQ and find that OQ < r. Measure OP and find that OP > r. OB is another radius of the same circle. A closed geometric figure in the plane divides the plane into three parts namely. Thus we have d = 2r i. the figure and the outer part. The chord PQ passes through the centre O of the circle.
118
Mathematics
Measure the length PQ. PAQP is called a minor segment and PBQP is called a major segment. 18. If this point moves along the circle once and comes back to its original position then the distance covered by P is called the circumference of the circle
. In Figure 18. the shaded portion is a sector formed by the arc PRQ and the unshaded portion is a sector formed by the arc PTQ. PQ is a diameter and 18.
Fig.4. the shaded region PAQP and the unshaded region PBQP are both segments of the circle. each known as a semicircle. In Figure 18.5
(b)
18. or
(a) Fig.4.4. 18. 18. 18.6 Segment A chord divides the interior of a circle into two parts.4. we may conclude Diameter is the longest chord of a circle. In Figure 18.5(a) ABC is an arc and is denoted by arc ABC .6
Choose a point P on a circle. 18.7.5 Sector The region bounded by an arc of a circle and two radii at its end points is called a sector. In Figure 18. each called a segment.4 Semicircle A diameter of a circle divides a circle into two equal arcs.5(b).7 Circumference
Fig. 18.7
is a semicircle and so is
.6. AB and CD and find that PQ > AB and PQ > CD.3 Arc A part of a circle is called an arc.4.
The ratio of the circumference of a circle to its diameter is always a constant. To measure its length we put a thread along PAQ and then measure the length of the thread with the help of a scale. MEASUREMENT OF AN ARC OF A CIRCLE Consider an arc PAQ of a circle (Fig 18. Measure the distance between the Ist and last position of P along the line. Rotate the wheel along a line till the point P comes back on the ground. 18. In fact this number π is an irrational number. Therefore. 18.Circles
119
Fig. The length of the boundary of a circle is the circumference of the circle. where c is the circumference of the circle.8
Activity for you : Take a wheel and mark a point P on the wheel where it touches the ground. Similarly.9). This constant is universally denoted by Greek letter π . Observe that in each case the ratio of the circumference to diameter turns out to be the same.9
. This distance is equal to the circumference of the circle.5.5. Aryabhata–I (476 AD).1416. 18. you may measure the length of the arc PBQ. Activity for you Consider different circles and measure their circumference(s) and diameters. 18.1 Minor arc An arc of a circle whose length is less than that or a semicircle of the same circle is called a minor arc. a famous Indian Mathematician 7 gave a more accurate value of π which is 3. radius. 18. 22 An approximate value of π is . PAQ a minor arc (See Fig. Thus.9) c = c =π . d its diameter and r is its d 2r
Fig.
18. arc PBQ is a major arc. 18.12 if arc PAQ = arc RBS then ∠POQ = ∠ROS and conversely if ∠POQ = ∠ROS then arc PAQ = arc RBS. In Figure 18.9.13.8 SOME IMPORTANT RULES Activity for you : (i) Draw two circles with centre O1 and O2 and radius r and s respectively (See Fig.10
Two circles (or arcs) are said to be congruent if we can superpose (place) one over the other such that they cover each other completely.10). 18.120
Mathematics
18.2 Major arc An arc of a circle whose length is greater than that of a semicircle of the same circle is called a major arc. 18. In Figure 18.6 CONCENTRIC CIRCLES Circles having the same centre but different radii are called concentric circles (See Fig. 18. Two arcs of a circle are congurent if and only if the angles subtended by them at the centre are equal. 18.5.7 CONGRUENT CIRCLES OR ARCS
Fig.11)
(i) Fig. 18. In Figure 18.11
(ii)
(ii) Supeimpose the circle (i) on the circle (ii) so that O1 coincides with O2 (iii) We observe that circle (i) will cover circle (ii) if and only if r = s Two circles are congurent if and only if they have equal radii. if arc PAQ = arc RBS then PQ = RS and conversely if PQ = RS then
Fig. 18.12
.
Activity for you : (i) Draw a circle with centre O (See Fig. 18.17). (ii) Draw a chord PQ. (iii) Draw ⊥ ON from O on the chord PQ. (iv) Measure PN and NQ You will observe that . PN = NQ. The perpendicular drawn from the centre of a circle to a chord bisects the chord. Activity for you : (i) Draw a circle with centre O (See Fig. 18.18). (ii) Draw a chord PQ. (iii) Find the mid point M of PQ. (iv) Join O and M. (v) Measure ∠OMP or ∠ OMQ with set square or protractor We observe that ∠OMP = ∠ OMQ = 90°.
Fig. 18.18 Fig. 18.17
The line joining the centre of a circle to the mid point of a chord is perpendicular to the chord. Activity for you : Take three non collinear points A, B and C. Join AB and BC. Draw perpendicular bisectors MN and RS of AB and BC respectively. Since A, B, C are not collinear, MN is not parallel to RS. They will intersect only at one point O. Join OA, OB and OC and measure them. We observe that OA = OB = OC
Fig. 18.19
Now taking O as the centre and OA as radius draw a circle which passes through A, B and C. Repeat the above procedure with another three non-collinear points and observe that there is only one circle passing through there given non-collinear points. This gives us a method to draw a circle passing through three non-collinear points. There is one and only one circle passing through three non-collinear points. Note. It is important to note that a circle can not be drawn to pass through three collinear points.
Circles
123
Activity for you : (i) Draw a circle with centre O [Fig. 18.20(a)]. (ii) Draw two equal chords AB and PQ of the circle. (iii) Draw OM ⊥ AB and ON ⊥ PQ. (iv) Measure OM and ON and observe that they are equal. Equal chords of a circle are equidistant from the centre. In Fig. 18.20(b) OM = ON. Measure and observe that AB = PQ. Thus, Chords that are equidistant from the centre of a circle are equal. The above results hold good in case of congurent circles also. We now take a few examples using these properties of circles. Examples 18.3 : In Figure 18.21, O is the centre of the circle and ON ⊥ PQ. If PQ = 8 cm and ON = 3 cm, find OP. Solution : ON ⊥ PQ (given) and since perpendicular drawn from the centre of a circle to a chord bisects the chord. ∴ PN = NQ = 4 cm In a right triangle OPN, ∴ OP2 = PN2 + ON2 OP2 = 42 + 32 = 25 ∴ OP = 5 cm. Examples 18.4 : In Figure 18.22, OD is perpendicular to the chord AB of a circle whose centre is O and BC is a diameter. Prove that CA = 2OD. Solution : Since OD ⊥ AB (Given) ∴ D is the mid point of AB
Fig. 18.22 Fig. 18.20b Fig. 18.20a
Fig. 18.21
(Perpendicular through the centre bisects the chord) Also O is the mid point of CB (Since CB is a diameter) Now in ∆ABC, O and D are mid points of the two sides BC and BA of the triangle ABC. Since the line segment joining the mid points of any two sides of a triangle is parallel and half of the third side. ∴ i.e. OD = 1 CA 2 CA = 2OD.
124
Mathematics
Example.18.5 : A regular hexagon is inscribed in a circle. What angle does each side of the hexagon subtend at the centre ? Solution : A regular hexagon has six sides which are equal. Therefore each side subtends the same angle at the centre. Let us suppose that a side of the hexagon subtends an angle x° at the centre. Then, we have 6x° = 360° ⇒ x = 60
Fig. 18.23
Hence, each side of the hexagon subtends an angle of 60° at the centre. Example 18.6 : In Fig. 18.24, two parallel chords PQ and AB of a circle are of lengths 7 cm and 13 cm respectively. If the distance between PQ and AB is 3 cm, find the radius of the circle. Solution : Let O be the centre of the circle. Draw perpendicular bisector OL of PQ which also bisects AB at M. Join OQ and OB (Fig. 18.24). Let OM = x cm and radius of the circle be r cm Then ∴ OB2 = OM2 + MB2 and OQ2 = OL2 + LQ2 13 r2 = x 2 + 2 r2
Fig. 18.24
FH IK
2
...(i)
2
and
2 7 = x+3 + 2
b g FH IK
2
...(ii)
Therefore from (i) and (ii) , x 2 + 13 2 ∴ ∴ ∴ ∴
FH IK = bx + 3g + FH 7 I 2K
2
2
6x = x= r2 = r=
169 − 9 − 49 4 4 7 2
FH 7 I + FH 13 I 2K 2K
2
2
=
49 + 169 = 218 4 4 4
218 2 218 cm. 2
Hence the radius of the circle is
Circles
125
CHECK YOUR PROGRESS 18.1 In questions 1 to 5, fill in the blanks to make each of the statements true. 1. In Figure 18.25, (i) AB is a ... of the circle. (ii) Minor arc corresponding to AB is ... . 2. A ... is the longest chord of a circle.
Fig. 18.25
3. The ratio of the circumference to the diameter of a circle is always ... . 4. The value of π as 3.1416 was given by great Indian Mathematician ... . 5. Circles having the same centre are called ... circles. 6. Diameter of a circle is 30 cm. If the length of a chord is 20 cm, find the distance of the chord from the centre. 7. Find the circumference of a circle whose radius is (i) 7 cm (ii) 11 cm.
FH Take π = 22 I 7K
8. In the Figure 18.26, RS is a diameter which bisects the chords PQ and AB at the points M and N respectively. Is PQ || AB ? Give reasons.
Fig. 18.26
Fig. 18.27
9. In the Figures 18.27, a line l intersects the two concentric circles with centre O at points A, B, C and D. Is AB = CD ? Give reasons. LET US SUM UP
z z
The circumference of a circle of radius r is equal to 2 πr . Two arcs of a circle are congurent if and only if either the angles subtended by them at the centre are equal or their corresponding chords are equal. Equal chords of a circle subtend equal angles at the centre and vice versa.
z
126
Mathematics
z z
Perpendicular drawn from the centre of a circle to a chord bisects the chord. The line joining the centre of a circle to the mid point of a chord is perpendicular to the chord. There is one and only one circle passing through three non-collinear points. Equal chords of a circle are equidistant from the centre and the converse. TERMINAL EXERCISE
z z
1. If the length of a chord of a circle is 16 cm and the distance of the chord from the centre is 6 cm, find the radius of the circle. 2. Two circles with centre O and O' (See Fig. 18.28) are congurent. Find the length of the arc CD.
Fig. 18.28
3. A regular pentagon is inscribed in a circle. Find the angle which each side of the pentagon subtend at the centre. 4. In Figure 18.29, AB = 8 cm and CD = 6 cm are two parallel chords of a circle with centre O. Find the distance between the chords.
19
Angles in a Circle and Cyclic Quadrilateral
19.1 INTRODUCTION You must have measured the angles between two straight lines, let us now study the angles made by arcs and chords in a circle and a cyclic quadrilateral. 19.2 OBJECTIVES After studying this lesson, the learner will be able to :
z z z z z z
prove that angles in the same segment of a circle are equal cite examples of concyclic points define cyclic quadrilaterals prove that sum of the opposite angles of a cyclic quadrilateral is 180° use properties of a cyclic quadrilateral solve problems based on Theorems (proved) and solve other numerical problems based on verified properties.
19.3 EXPECTED BACKGROUND KNOWLEDGE
z z z
Angles of a triangle Arc, chord and circumference of a circle Quadrilateral and its types
19.4 ANGLES IN A CIRCLE Central Angle. The angle made at the centre of a circle by the radii at the end points of an arc (or a chord) is called the central angle or angle subtended by an arc (or chord) at the centre. In Figure 19.1, ∠POQ is the central angle made by arc PRQ.
Fig.19.1
The length of an arc is closely associated with the central angle subtended by the arc. Let us define the "degree measure" of an arc in terms of the central angle.
Angles in a Circle and Cyclic Quadrilateral
131
The degree measure of a minor arc of a circle is the measure of its corresponding central angle. In Figure 19.2, Degree measure of PQR = x° The degree measure of a semicircle in 180° and that of a major arc is 360° minus the degree measure of the corresponding minor arc. Relationship between length of an arc and its degree measure. Length of an arc = circumference × If the degree measure of an arc is 40° 40° = 2 πr then length of the arc PQR = 2 πr. 360° 9 Inscribed angle : The angle subtended by an arc (or chord) on any point on the remaining part of the circle is called an inscribed angle. In Figure 19.3, ∠PAQ is the angle inscribed by arc PRQ at point A of the remaining part of the circle or by the chord PQ at the point A. 19.5. SOME IMPORTANT PROPERTIES ACTIVITY FOR YOU : Draw a circle with centre O. Let PAQ be an arc and B any point on the circle. Measure the central angle POQ and an inscribed angle PBQ by the arc at remaining part of the circle. We observe that ∠POQ = 2∠PBQ Repeat this activity taking different circles and different arcs. We observe that The angle subtended at the centre of a circle by an arc is double the angle subtended by it on any point on the remaining part of the circle. Let O be the centre of a circle. Consider a semicircle PAQ and its inscribed angle PBQ ∴ 2 ∠PBQ = ∠POQ
Fig.19.5 Fig.19.4 Fig.19.2
degree measure of the arc 360°
Fig.19.3
132
Mathematics
(Since the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle) But ∴ ∠POQ = 180° 2∠PBQ = 180° ∠PBQ = 90° Thus, we conclude the following : Angle in a semicircle is a right angle. Theorem : Angles in the same segment of a circle are equal. Given : A circle with centre O and the angles ∠PRQ and ∠PSQ in the same segment formed by the chord PQ (or arc PAQ) To prove : ∠PRQ = ∠PSQ Construction : Join OP and OQ. Proof : As the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle, therefore we have ∠POQ = 2 ∠PRQ and ∠POQ = 2∠PSQ 2∠PRQ = 2∠PSQ ∴ ∠PRQ = ∠PSQ We take some examples using the above results Example 19.1 : In Figure 19.7, O is the centre of the circle and ∠AOC = 120°. Find ∠ABC. Solution : It is obvious that ∠x is the central angle subtended by the arc APC and ∠ABC is the inscribed angle. ∴ But ∴ ∴ Solution : ∠x = 2∠ABC ∠x = 360° – 120° 2∠ABC = 240° ∠ABC = 120° ∠POQ = 2∠PAQ = 70° ...(i)
Fig.19.7
(Since PQ is a diameter of the circle)
...(i) ...(ii)
Fig.19.6
From (i) and (ii), we get
Example 19.2 : In Figure 19.8 O is the centre of the circle and ∠PAQ = 35°. Find ∠OPQ. (Angle at the centre is double the angle on the remaining part of the circle)
(Angles in the same segment of a circle are equal) Example 19.4 : In Figure 19.10, O is the centre of the circle, ∠POQ = 70° and PS⊥OQ . Find ∠MQS. Solution : 2∠PSQ = ∠POQ = 70° (Angle subtended at the centre of a circle is twice the angle subtended by it on the remaining part of the circle) ∴ ∠PSQ = 35° Fig.19.10 Since ∠MSQ + ∠SMQ + ∠MQS = 180° (Sum of the angles of a triangle) ∴ 35° + 90° + ∠MQS = 180° ∴ ∠MQS = 180° – 125° = 55°. CHECK YOUR PROGRESS 19.1 1. In Figure 19.11, ADB is an arc of a circle with centre O, if ∠ACB = 35°, find ∠AOB.
Fig. 19.11
19. AOB is a diameter of a circle with centre O. passing through the points.134
Mathematics
2.14.6 CONCYCLIC POINTS Definition : Points which lie on a circle are called concyclic points. If ∠PTR = 35°. you can draw not only one but many circles passing through it as in Fig. Is ∠APB = ∠AQB = 90°? Give reasons. Find ∠ADB. In Figure 19. In Figure 19. 19.
Fig.12
3. Let us now find certain conditions under which points are concyclic.15. 19. In Figure 19. 19. If you take a point P. 19. find ∠PSR.13
4.12. (Fig.15
Now take two points P and Q on a sheet of a paper.
. O is the centre of a circle and ∠AOB = 60°.19.
Fig.16).
Fig.13. PQR is an arc of a circle with centre O.
Fig. You can draw as many circles as you wish.14
19.
You will see that it is not always possible to draw a circle passing through four non-collinear points.18
(b)
Note.16
Let us now take three points P. 4.
. 19.17). Q and R which do not lie on the same straight line. In Fig 19.Angles in a Circle and Cyclic Quadrilateral
135
Fig.18 (a) and (b) points are noncyclic but concyclic in Fig 19.18(c). Given one or two points there are infinitely many circles passing through them. If the points P. Q and R are collinear then it is not possible to draw a circle passing through them. Three collinear points are not concyclic (or noncyclic).
(a)
(b) Fig.17
Further let us now take four points P. Three non-collinear points are always concyclic and there is only one circle passing through all of them. and S which do not lie on the same line. Thus we conclude 1. 2. Four non-collinear points may or may not be concyclic. 3. R.
Fig. 19. Q. In this case you can draw only one circle passing through these three non-colinear points (Figure 19. 19.
If a pair of opposite angles of a quadrilateral is supplementary.19 shows a cyclic quadrilateral PQRS. We solve some examples using the above results.6. Show that it is a rectangle.1 CYCLIC QUADRILATERAL A quadrilateral is said to be a cyclic quadrilateral if there is a circle passing through all its four vertices.19
Adding ∠ABC on both the sides. Given : A cyclic quadrilateral ABCD To prove : ∠BAD + ∠BCD = ∠ABC + ∠ADC = 180° Construction : Draw AC and DB Proof : and ∴ ∠ACB = ∠ADB ∠BAC = ∠BDC [Angles in the same segment] ∠ACB + ∠BAC = ∠ADB + ∠BDC = ∠ADC
Fig. So we conclude that quadrilateral PQRS is a cyclic quadrilateral. then the quadrilateral is cyclic. Verification : Draw a quadrilateral PQRS Since in quadrilateral PQRS. Theorem.22
. 19.19. Solution : ∠A + ∠C = 180° (ABCD is a cyclic quadrilateral) Since ∠A = ∠C [Opposite angles of a parallelogram]
Fig. Fig. Converse of this theorem is also true.19.5 : ABCD is a cyclic parallelogram. we get ∠ACB + ∠BAC + ∠ABC = ∠ADC + ∠ABC But ∠ACB + ∠BAC + ∠ABC = 180° ∴ ∴ ∠ADC + ∠ABC = 180° ∠BAD + ∠BCD = 360° – (∠ADC + ∠ABC) = 180°.20 Fig. ∠P + ∠R = 180° and ∠S + ∠Q = 180°
Fig. Q and R and observe that it also passes through the point S.19. For example.
[Sum of the angles of a triangle]
Hence proved.19.21
Therefore draw a circle passing through the point P. Sum of the opposite angles of a cyclic quadrilateral is 180°. Example 19.136
Mathematics
19.
If a pair of opposite angles of a quadrilateral is supplementary. What angle does each side subtend at the centre O ? 2. Angles in the same segment of a circle are equal. A square PQRS is inscribed in a circle with centre O. If O1O2 intersect AB at M then show that (i) ∆O1AO2 ≅ ∆O1BO2 (ii) M is the mid point of AB (iii) AB⊥O1O 2
Fig. Sum of the opposite angles of a cyclic quadrilateral is 180°. Angle in a semicircle is a right angle. The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle. 19. Points lying on the same circle are called concyclic points. From (i) conclude that ∠1 = ∠2 and then prove that ∆AO1M ≅ ∆BO1M (by SAS rule)).
.30
(Hint.Angles in a Circle and Cyclic Quadrilateral
139
LET US SUM UP
z
The angle subtended by an arc (or chord) at the centre of a circle is called central angle and an angle subtended by it at any point on the remaining part of the circle is called inscribed angle. TERMINAL EXERCISES
z z
z z z z
1. C1 and C2 are two circles with centre O1 and O2 and intersect each other at points A and B. In Figure 19. then the quadrilateral is cyclic.30.
34
. In Figure 19. In Figure 19.33.34. ∠AOB = 80° and ∠PQB = 70°.32. Two circle intersect in A and B.
Fig.31
[Hint. B and D are collinear. 19. AC and AD are the diameters of the circles. Join CB. AB is a chord of a circle with centre O. BD and AB. If ∠ACB = 40°. Find ∠PBQ. find ∠OAB.
Fig. Find ∠POR. 19. In Figure 19. Since ∠ABC = 90° and ∠ABD = 90°] 4. Prove that C.
Fig. O is the centre of a circle and ∠PQR = 115°.
Fig.140
Mathematics
3. 19. O is the centre of a circle.32
5. 19.33
6.
Secants. the wheels of the moving cycle touch the road at a very limited area. What do you observe from the above situations ?
Fig. 20.1
If you consider a wheel or a coin as a circle and the touching surface (road or table) as a line. the above illustrations show that a line touches a circle. If you roll a coin on a smooth surface. only one point of the coin comes in contact with the surface it is rolled upon. you will find that at any instant of time. You will observe that at any instant of time. say a table or floor. more correctly a point. Tangents and Properties
8.1 INTRODUCTION Look at a moving cycle.
. Tangents and Properties
143
20
Secants. In this lesson. we shall study about the possible contacts that a line and a circle have and try to study their properties.
In other words. There can be three distinct possibilities as shown in Fig. whereas a line segment is a portion of a line bounded by two points. the line XY does not intersect the circle. 20.2.2 (ii).
Fig. 20. 20. the line XY intersects the circle in only one point and is said to touch the circle at the point P. Recall that a circle is the locus of a point in a plane which moves in such a way that its distance from a fixed point in the plane always remains constant.2 (iii).
20.144
Mathematics
20. the line XY intersects the circle in two distinct points A and B.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z z
Measurement of angles and line segments Drawing circles of given radii Drawing lines perpendicular and parallel to given lines Knowledge of previous results about lines and angles. You can see that in Fig. You also know that a line is a collection of points. The fixed point is called the centre of the circle and the constant distance is called the radius of the circle.4 SECANTS AND TANGENTS—AN INTRODUCTION You have read about lines and circles in your earlier lessons. with centre O. the learner will be able to :
z z z
define a secant and a tangent to the circle differentiate between a secant and a tangent verify and use important results (given in the curriculum) related to tangents and secants to circles.
20. extending indefinitely to both sides. 20. we say that the line XY and the circle have no common point.2 OBJECTIVES After studying this lesson.2(i).2
Now consider the case when a line and a circle co-exist in the same plane. congruence and circles. 20.
. and in Fig. Knowledge of Pythagoras Theorem. In Fig.
Secants. 20. We therefore define the following : Tangent A line which touches a circle at exactly one point is called a tangent line and the point where it touches the circle is called the point of contact. we can say that in case of intersection of a line and a circle.2 (ii). B'''. in Fig. R.2 (iii). the two points of intersection lie on the circle and the remaining portion lies in the exterior of the circle. S and T are points in the exterior of the circle and P is on the circle. R. we say that A tangent is the limiting position of a secant when the two points of intersection coincide. Thus. (iii) The line touches the circle in exactly one point. intersecting the circle in the points A and B. In that case. B''. Thus. i. when it becomes tangent to the circle at A. Join OP. Imagine that one point A.6 TANGENT AND RADIUS THROUGH THE POINT OF CONTACT Let XY be a tangent to the circle.e. (ii) The line intersects the circle at two distinct points.3 and ultimately attains the position of the line XAY. As Q. which lies on the circle.
. 20. Take points Q. B'''' as shown in Fig. XY is a tangent to the circle at P. Secant A line which intersects the circle in two distinct points is called a secant line (usually referred to as a secant). a part of the line lies in the interior of the circle. S and T on the tangent XY and join OQ. the following three possibilities are there : (i) The line does not intersect the circle at all. of the secant XY is fixed and the secant rotates about A. Tangents and Properties
145
Thus. OS and OT. 20.3
20. In Fig. 20.. 20.
Fig. the line lies in the exterior of the circle.5 TANGENT AS A LIMITING CASE OF A SECANT Consider the secant XY of the circle with centre O. which is called the point of contact. XY is a secant line to the circle and A and B are called the points of intersection of the line XY and the circle with centre O. intersecting the circle at B'. OR. with centre O. at the point P.
OR.5
If the point P lies on the circle. Consider two ∆s OPT and OPT′ ∠OTP = ∠OT'P OT = OT′ OP = OP (Each being a right angle) (Radii of the same circle) (Common) . 20. The above result can also be verified by measuring angles OPX and OPY and finding each of them equal to 90°.4
OP ⊥ XY
Thus. can there still be two tangents to the circle from that point ? You can see that only one tangent can be drawn to the circle in that case. we can say that From an external point. Some of these are shown as PT. 20.. (A) Now. You will again find the same result. PA. Thus.. measure the lengths of PT and PT′. What about the case when P lies in the interior of the circle ? Note that any line through P in that case will intersect the circle in two points and hence no tangents can be drawn from an interior point to the circle. we know that of all the segments that can be drawn from a point (not on the line) to the line.7 TANGENTS FROM A POINT OUTSIDE THE CIRCLE Take any point P in the exterior of the circle with centre O. "previous study of Geometry. Draw lines through P.
From our. PB.(i)
. 20. OS and OT. How many of these touch the circle ? Only two. is perpendicular to the tangent at that point.
Fig. two tangents can be drawn to a circle.5. PD and PT' in Fig. PC. the perpendicular segment is the shortest" : As OP is the shortest distance from O to the line XY
∴
Fig. we can state that A radius. You will find that PT = PT′ (B) Let us see this logically also.146
Mathematics
∴ OP is less than each of OQ.20. Repeat the activity with another point and a circle. through the point of contact of tangent to a circle.
Let us now take some examples to illustrate..8.6
Thus.8
. 20. we can say that The tangents drawn from an external point to a circle are equally inclined to the line joining the point to the centre of the circle. Example 20.7.Secants. 20.(ii)
From (A) and (B).. Find the length of the tangent PT from P to the circle. in Fig.6 as ∆OPT ≅ ∆OPT′ ∴ ∠OPT = ∠OPT′
Fig. Solution : ∠OTP = 90°. OP = 5 cm and radius of the circle is 3 cm. we have OP2 = OT2 + PT2 or or ∴ 52 = 32 + x2 x2 = 25 – 9 = 16 x=4
Fig. 20.e.7
i. Find the lengths of PT and PT′. Let PT = x
∴ In right triangle OTP. 20. tangents PT and PT′ are drawn from a point P at a distance of 25 cm from the centre of the circle whose radius is 7 cm. Solution : Here OP = 25 cm and OT = 7 cm We also know that ∠OTP = 90° ∴ ∴ PT2 = OP2 – OT2 = 625 – 49 = 576 = (24)2 PT = 24 cm PT = PT′ ∴ PT′ = 24 cm We also know that
Fig. 20. the length of tangent PT = 4 cm Example 20. Tangents and Properties
147
∴ ∴
∆OPT ≅ ∆OPT′ PT = PT′ . Also. we can say that The lengths of two tangents from an external point are equal.2 : In Fig.1 : In Fig. with centre O. 20.
You will again find that PA × PB = PC × PD Let us now consider the case of chords intersecting outside the circle.Secants. If PX = 2. Draw two chords BA and DC intersecting each other outside the circle at P. Find the products PA × PB and PC × PD. Tangents and Properties
149
(iv) From an external point
tangents can be drawn to a circle. 20. 20.11. We will now verify some results regarding chords intersecting inside a circle or outside a circle. Repeat the above activity with another two circles after drawing chords intersecting inside.12. the incircle of ∆PQR is drawn. PC and PD. Measure the lengths of line segments PA. RZ = 3.8 INTERSECTING CHORDS INSIDE AND OUTSIDE A CIRCLE. Let us perform the following activity : Draw a circle with centre O and any radius.
2.13
. Find the ∠OYP and ∠OYT. PB. 2012
4. 20.
Fig.5 cm and perimeter of ∆PQR = 18 cm . In Fig. PC. circle. You have read various results about chords in the previous lesson.5 cm. Find the products PA × PB and PC × PD. 20.
Fig. find the length of QY. 20. Draw two chords AB and CD intersecting at P inside the circle. 3. Measure the lengths of the line-segments PD. You will find that they are equal. In Fig. tangent(s) can be drawn to the
(v) From a point in the interior of the circle. Write an experiment to show that the lengths of tangents from an external point to a circle are equal. PA and PB. Let us perform the following activity : Draw a circle of any radius and centre O. ∠POY = 40°. when produced.11
Fig.
20. From an external point P. Thus. What do you find ? You will find that PA × PB = PT2
Fig. 20. AB and CD are two chords of a circle intersecting at a point P inside the circle. You will again find the same result. we can say If PAB is a secant to a circle intersecting the circle at A and B.e. 20. 20.14
Repeat the above activity with two other circles. You will again find that PA × PB = PC × PD. 20.16
. Solution : It is given that PA = 3 cm. PB = 2 cm and PC = 1. If PA = 3 cm.16. Find the products PA × PB and PT × PT or PT2..5 cm. PB = 2 cm.8 INTERSECTING SECANTS AND TANGENTS OF A CIRCLE To see if there is some relation between the intersecting secant and tangent outside a circle.150
Mathematics
You will see that the product PA × PB is equal to the product PC × PD. PC = 1. find the length of PD. Thus. then PA × PB = PT2 Let us illustrate these with the help of examples : Examples 20. Measure the length of the line-segments PA. i. we conduct the following activity: Draw a circle of any radius with centre O. PB and PT.5) × x
Fig.5 : In Fig.15 Fig. then PA × PB = PC × PD.5 cm Let we know that ⇒ PD = x PA × PB = PC × PD 3 × 2 = (1. and PT is a tangent to the circle at T. draw a secant PAB and a tangent PT to the circle. we can say that If two chords AB and CD of a circle intersect at a point P (inside or outside the circle). PA × PB = PC × PD Repeat this activity with two other circles with chords intersecting outside the circle.
and draw a chord PQ and let it form ∠PRQ in alternate segment as shown in Fig.9 ANGLES MADE BY A TANGENT AND A CHORD. if PA = 4 cm. 20.20.
Fig. Now measure ∠QPX and ∠QSP. with centre O. 20. You will again find that ∠QPY = ∠PRQ. OP = 8 cm. PA = 5 cm and PB = (x + 2) cm. Thus. Mark a point R on the major arc and let S be a point on the minor arc . Draw a circle.21. we can state that The angles formed in the alternate segments by a chord through the point of contact of a tangent to a circle is equal to the angle between the chord and the tangent. 20. Draw a chord PQ of the circle through the point P as shown in the Fig. 20. 20. draw ∠QPY = ∠QRP.22) What do you find ? You will see that ∠PRQ = ∠QPY Repeat this activity with another circle and same or different radius. if O is the centre of the circle. Let there be a circle with centre O and let XY be a tangent to the circle at point P.
. The segment formed by the major arc and chord PQ is said to be the alternate segment of ∠QPY and the segment and chord PQ is said to be the formed by the minor alternate segment to ∠QPX. This result is more commonly called as "Angle in the Alternate Segment".23
What do you observe ? You will find that ∠OPY = 90° showing thereby that XY is a tangent to the circle.152
Mathematics
3. In Fig. find the radius of the circle. PT = 2 7 cm. 20. Join QR and PR. Join OP and measure ∠OPY. PB = 10 cm. Let us now check the converse of the above result. 20. You will again find that these angles are equal. 5. find PD 4. Extend the line segment PY to both sides to form line XY. PC = 5 cm.22. At P. Measure ∠PRQ and ∠QPY (See Fig. PD = (x + 5) cm. Let us see if there is some relationship between angles in the alternate segments and the angle between tangents and chord.20. find x. In Fig. 20. 20. In Fig.23. if PC = 4 cm.22
Fig.
Solution : By the Alternate-segment theorem. 2. Let us now take some examples to illustrate. Example 20. If AOB is a diameter and ∠PAB = 40°. ∴ ∠BPY = 40° ∠APB = 90°
Fig.9 : In Fig. we can state that If a line makes with a chord angles which are equal respectively to the angles formed by the chord in alternate segments. 20. XY is tangent to a circle with centre O. TA and TB are joined and TM is the angle bisector of ∠ATB. If ∠OQP = 40°. In Fig.26. PT is a tangent to the circle from an external point P. 3. Thus. find ∠APX and ∠BPY. Example 20.24.25
(Angle in the Alternate segment)
CHECK YOUR PROGRESS 20. 20. Chord AB of the circle. Show that XY is parallel to base BC.8 : In Fig. ∠BPY + ∠APB + ∠APX = 180° ∠APX = 180° – (∠BPY + ∠APB) = 180° – (40° + 90°) = 50°. 20. AB = AC ∴ ∠1 = ∠2 Again XY is tangent to the circle at A. ∴ ∠3 = ∠2 ∴ ∠1 = ∠3 But these are alternate angles ∴ XY || BC
Fig. Explain with the help of a diagram. ABC is an isosceles triangle with AB = AC and XY is a tangent to the circumcircle of ∆ABC. when produced meets TP in P. we know that ∠BPY = ∠BAP ∴ Again. Tangents and Properties
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Repeat this activity by taking different circles and you find the same result.
. 20. XY is a tangent to the circle with centre O at a point P. In Fig. then the line is a tangent to the circle.24
[Q ∠BAP = 40° (Given)] [Angle in a semi-circle] (Angles on a line)
And.25. 20.3 1.Secants. find the value of a and b. 20. the angle formed by a chord in the alternate segment of a circle.27. Solution : In ∆ABC.
26 If ∠PAB = 30° and ∠ATB = 60°. 20. 2.27
A line which intersects the circle in two points is called a secant of the circle. A tangent to a circle is perpendicular to the radius through the point of contact. then PA × PB = PT2
z
The angles formed in the alternate segments by a chord through the point of contact of a tangent to a circle are equal to the angles between the chord and the tangent. 20. A tangent is the limiting position of a secant when the two points of intersection coincide. Show that a tangent is a line perpendicular to the radius through the point of contact.154
Mathematics
Fig. A line which touches the circle at a point is called a tangent to the circle. and PT is a tangent to the circle at T. TERMINAL EXERCISE
z
1. LET US SUM UP
z z z z z z
Fig.
. If a line makes with a chord angles which are respectively equal to the angles formed by the chord in alternate segments. with the help of an activity. show that PM = PT. Differentiate between a secant and a tangent to a circle with the help of a figure. then the line is a tangent to the circle. then PA × PB = PC × PD
z
If PAB is a secant to a circle intersecting the circle at A and B. which are of equal length. From an external point. If two chords AB and CD of a circle intersect at a point P (inside or outside the circle). two tangents can be drawn to a circle.
z
Construct rectilinear figures such as parallelograms.1 INTRODUCTION One of the aims of the studying Geometry is to acquire the skill of drawing figures accurately. Construct a quadrilateral from the given data (i) four sides and a diagonal (ii) three sides and both diagonals (iii) two adjacent sides and three angles
z
. the learner will be able to :
z z
divide a given line segment internally in a given ratio.Constructions
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21
Constructions
21. You have constructed angles of 30°. squares. 120° and 45°. You have also drawn perpendicular bisector of a line segment and bisector of an angle. rhombuses and trapeziums. Construct a triangle from the given data (i) SSS (ii) SAS (iii) ASA (iv) RHS (v) perimeter and base angles (vi) base. 21. 60°. In this lesson we will extend our learning to construct some other important geometrical figures. sum/difference of the other two sides and one base angle. 90°. You have learnt how to construct geometrical figures namely triangles. rectangles.2 OBJECTIVES After studying this lesson. squares and circles with the help of ruler and compasses.
(vii) two sides and a median corresponding to one of these sides.
C4 and C5 at equal distances from the point A.3 EXPECTED BACKGROUND KNOWLEDGE We assume that the learner already knows how to use a pair of compasses and ruler to construct
z z z z z
angles of 30°.
z z
Construct a triangle equal in area to a given quadrilateral. the right bisector of a line segment bisector of a given angle. mark off 5 points C1. 90°. We go through the following steps. Construct tangents to a circle from a point (i) outside it (ii) on it using the centre of the circle Construct circumcircle of a triangle Construct incircle of a triangle. parallelograms. C3. C2. Step 2 : Starting with A. 21. and squares a circle
21. rectangles.e. Step 3 : Join C5 and B. rhombuses.4 DIVISION OF A LINE SEGMENT IN THE GIVEN RATIO INTERNALLY Construction 1 : To divide a line segment internally in a given ratio.
Fig. the second point).
z z
21. 21. You are required to divide it internally in the ratio 2 : 3. 45°. 105°. Step 4 : Through C2 (i. 120°. Step 1 : Draw a ray AC making an acute angle with AB.1
. 60°.1
Then D is the required point which divides AB internally in the ratio 2 : 3 as shown in Fig. draw C2D parallel to C5B meeting AB in D. Given a line segment AB.158
Mathematics
(iv) three sides and two included angles (v) four sides and an angle.
Then ∆ABC is the required triangle. Draw a line segment PQ = 8 cm. To construct a triangle when two angles and the included side are given (ASA). Step 3 : With B as centre and radius 5 cm draw another arc intersecting the arc of Step 2 at C.6 cm Step 2 : At Q.5 cm as the base instead of PQ] Construction 4. 21. ∠C = 45° and BC = 4. Step 4 : Join AC and BC. AC = 4.5 cm and ∠PQR = 60° For constructing the triangle.3 Fig. 21. 4
. draw an arc.2
3 PQ .5 CONSTRUCTION OF TRIANGLES Construction 2 : To construct a triangle when three sides are given (SSS) Suppose you are required to construct ∆ABC in which AB = 6 cm.6 cm. Measure each part.Constructions
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CHECK YOUR PROGRESS 21. Let us construct a ∆ABC in which ∠B = 60°. we go through the following steps. Step 1 : Draw PQ = 5.
Fig. Divide it internally in the ratio 3 : 4.5 cm draw an arc cutting QX at R.8 cm. QR = 4. 2. Step 4 : Join PR Then ∆PQR is the required triangle. [Note : You may take QR = 4.8 cm and BC = 5 cm. [Note : You may take BC or AC as the base] Construction 3 : To construct a triangle.1 1. 21. construct an angle ∠PQX = 60° Step 3 : With Q as centre and radius 4. Also write the steps of construction.7 cm. Draw a line segment 7 cm long. We go through the following steps : Step 1 : Draw AB = 6 cm. Find the point R on it such that PR = [Hint : Divide the line segment PQ internally in the ratio 3 : 1]. when two sides and the included angle is given (SAS) Suppose you are required to construct a triangle PQR in which PQ = 5. Step 2 : With A as centre and radius 4.
4
Construction 5 : To construct a right triangle. we go through the following steps. 21.5
Fig. Step 2 : At B. 21. Note : To construct a triangle when two angles and any side (other than the included side) are given. Step 1 : Draw XY= 9. construct ∠BCR = 45° meeting BQ at A. To construct the triangle.7 cm. Construction 6 : To construct a triangle when its perimeter and two base angles are given.5 cm
Fig. side BC = 3 cm and hypotenuse AC = 5 cm. when its hypotenuse and a side are given. construct ∠CBP = 90°. Step 1 : Draw BC = 3 cm Step 2 : At B.
Fig. Step 3 : With C as centre and radius 5 cm draw an arc cutting BP in A. Then ∆ABC is the required triangle. 21. Step 4 : Join AC ∆ABC is the required triangle. construct ∠CBQ = 60° Step 3 : At C.6
. right angled at B.160
Mathematics
To construct the triangle we go through the following steps : Step 1 : Draw BC = 4. we find the third angle (using angle sum property of the triangle) and then use above method for constructing the triangle. Suppose we have to construct a triangle whose perimeter is 9.5 cm and base angle are 60° and 45°. we go through the following steps. Let us construct a right triangle ABC. To construct triangle.
9
. Step 5 : Join AC and BC. Step 1 : Draw AB= 6 cm. Note : You are also required to write the steps of construction in each of the remaining problems.
Fig. Write the steps of construction as well. Step 6 : Join AC ∆ABC is the required triangle. Suppose you have to construct a ∆ABC in which AB = 6 cm.2 cm. 21.5 cm draw an arc.2 1. Step 2 : Draw right bisector of AB meeting AB in D.
Fig. the third side and one of the angles on the third side are given. We go through the following steps.8
Construction 9 : To construct a triangle when its two sides and a median corresponding to one of these sides.162
Mathematics
Construction 8 : To construct a triangle when difference of two sides. Step 4 : With B as centre and radius 4 cm draw another arc intersecting the arc of Step 3 in C. Step 4 : Join CK Step 5 : Draw right bisector of CK meeting BP produced at A. Construct a ∆PQR. ∠P = 120° and PQ = 5. Step 3 : With D as centre and radius 3.1 cm. 21. Construct a ∆DEF. Step 3 : From BP cut off BK = 1. EF = 4 cm and DF = 5.5 cm. Then ∆ABC is the required triangle. CHECK YOUR PROGRESS 21. in which BC = 4 cm ∠B = 60°. given that PR = 6.2 cm. Step 2 : Construct ∠CBP = 60°. AB – AC = 1. given that DE = 5. Suppose we have to construct a ∆ABC. 2. To construct the triangle we go through the following steps : Step 1 : Draw BC = 4 cm. BC = 4 cm and median CD = 3.2 cm.6 cm. are given.5 cm.
8 cm.8 cm. QR = 4. Construct a ∆ABC given that AB + BC + AC = 10 cm. Construct a ∆ABC given that BC = 5. 6. Step 5. 8. Construct a ∆LMN. ∠B = 60°. 5.0 cm. ∠B = 75° and ∠C = 45°. Construct a triangle PQR in which PQ = 5 cm. Step 3 : From AK cut off AD = 3 cm. BC + AC = 9. Construct a ∆ABC in which AB = 5 cm. 4. Join CD and BC. To construct the required parallelogram we go through the following steps : Step 1 : Draw AB = 4 cm Step 2 : At A.
Fig.2 cm and median RS = 3. when ∠M = 30°. To construct the rectangle we go through the following steps. Recall that in a rectangle. 7.6 CONSTRUCTION OF RECTILINEAR FIGURES You are advised to draw rough sketch for the given data in each of the following constructions. construct ∠BAK = 60°. each angle is 90° and opposite sides are equal. Suppose that you have to construct a rectangle ABCD in which AB = 4 cm and AC = 5.Constructions
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3.5 cm. 9. that it helps you to visualise/understand the steps of construction. Construction 10 : To construct a parallelogram when two adjacent sides and the included angle are given. 21. Construction 11 : To construct a rectangle when one of its diagonal and a side are given.5 cm. ∠A = 60°. Step 4 : With B and D as centres and radii equal to 3 cm and 4 cm respectively draw two arcs cutting each other at C. MN = 5 cm and LM – LN = 1. Then ABCD is the required parallelogram. Construct a right angled isosceles triangle in which one of equal side is 4. You will observe. Suppose that you have to construct a parallelogram in which the adjacent sides are 4 cm and 3 cm and included angle is 60°. ∠C = 30°. Construct a right triangle in which one side is 3 cm and hypotenuse is 7. 21.5 cm. Step 1 : Draw AB = 4 cm.10
.8 cm.
Step 3 : Construct ∠COP = 60° and produce PO to Q. Step 5 : Join AB. Step 5 : Join PS and RS. We have to follow the following steps to construct the square : Step 1 : Draw PQ = 4. draw two arcs of radii 4.13
. Step 3 : With A as centre and radius 5 cm. draw ∠ABK = 90°. Recall that diagonals of a parallelogram bisect each other. we go through the following steps : Step 1 : Draw AC = 8 cm. draw an arc cutting the arc drawn in Step 4 at D. Step 3 : From QT cut off QR = 4. Step 2 : Draw right bisector of AC meeting it at O. draw an arc cutting BK at C. draw an arc.4 cm each to cut each other at S. Step 6 : Join DC and AD ABCD is the required rectangle. 21.
Fig. To construct the parallelogram.
Fig. AD and CD. Step 4 : Cut off OB = OD = 3 cm (1/2 × 6. Step 2 : Construct ∠PQT = 90° at Q.164
Mathematics
Step 2 : At B. PQRS is the required square. Suppose you have to construct a square PQRS in which PQ = 4. BC.4 cm.11
Construction 13 : To construct a parallelogram when two diagonals and the angle between them is given. 21. Suppose that the lengths of two diagonals are 8 cm and 6 cm and the angle between them is 60°. Step 4 : With C as centre and radius 4 cm. Step 5 : With A as centre and radius = BC. length of second diagonal) from OP and OQ.4 cm. 21. ABCD is the required parallelogram.4 cm.12 Fig. Step 4 : From P and R. Construction 12 : To construct a square when its side is given.
when one of its diagonal is 5. Step 3 : From AP cut off AK = 3 cm. draw AP ⊥ AB. 21. BC. Step 5 : With A and B as centres and radii 4 cm and 5 cm respectively draw two arcs cutting KL at D and C respectively.Constructions
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Construction 14 : To construct a rhombus when one diagonal and side are given. Note : You are also required to write the steps of construction in each of the following problems. Step 2 : At A.5 cm and 4 cm and the included angle is 75°. To construct the rhombus.3 cm draw two arcs one above AC and the other below AC intersecting the arcs of Step 2 in B and D respectively. Suppose you have to draw a trapezium in which one of parallel sides is 6 cm. To construct the trapezium we go through the following steps : Step 1 : Draw AB = 6 cm. Step 3 : With C as centre and radius 3.13
Construction 15 : To construct a trapezium in which one of parallel sides. Construct a parallelogram if the lengths of its adjacent sides are 5. ABCD is the required rhombus.15
. Step 6 : Join AD and BC. Then ABCD is the required trapezium.3 1.
Fig.5 cm and the side is 3. Step 4 : Join AB. you have to construct a rhombus. CHECK YOUR PROGRESS 21. two non-parallel sides and the distance between parallel sides are given.5 cm. CD and AD. Write steps of construction as well.3 cm. Step 2 : With A as centre and radius 3. we go through the following steps : Step 1 : Draw AC = 5. draw KL ⊥ AK. 21. two non-parallel sides are of length 4 cm and 5 cm and distance between parallel sides is 3 cm. Suppose. Step 4 : At K. draw two arcs one above AC and the other below AC.
Fig.3 cm.
AD = 3cm. 5. Construct a trapezium one of whose parallel side is 7 cm. AD = 3. Construct a parallelogram if its sides are 4 cm and 6 cm and one of its diagonal is 7 cm. Step 3 : With B as centre and radius 5 cm draw another arc intersecting the arc of Step 2 in D. we go through the following steps. Step 4 : Join AD and BD. 3.1 cm.16
Step 5 : With B and D as centres and radii 3. We go through the following steps : Step 1 : Draw AB = 3 cm.5 cm. 21.8 cm draw an arc. non-parallel sides are 4. Construction 17 : To construct a quadrilateral when three sides and both diagonals are given. To construct the required quadrilateral. Construct a rectangle whose sides are 4.5 cm.2 cm and 3.3 cm and the distance between the parallel sides is 3.7 cm. Construction 16 : To construct a quadrilateral when four sides and one diagonal are given.1 cm. Step 1 : Draw AB = 3. CD = 4. Construct a square whose side measures 5. 6.7 CONSTRUCTION OF QUADRILATERALS You are advised to draw rough sketch from the given data in each of the following constructions. AC = 4.6 cm. ABCD is the required quadrilateral.6 cm.4 cm. Construct a parallelogram if its diagonals are 8 cm and 5 cm and the angle between them is 45°.8 cm and BD = 5 cm.2 cm and 5.166
Mathematics
2. BC = 3. 21. 8. draw two arcs intersecting each other at C. Suppose you are required to construct a quadrilateral ABCD when AB = 3. Step 2 : With A as centre and radius 3.5 cm and 4. 7. Step 6 : Join BC and DC. 4.1 cm respectively. You will observe that it helps you to visualize/understand the steps of construction.
Fig.8 cm and diagonal BD = 5 cm. Construct a rhombus whose side is 5 cm and one diagonal is 8 cm.
. Suppose you have to construct a quadrilateral ABCD in which AB = 3 cm. BC = 2. Construct a rhombus whose diagonals measure 5 cm and 4 cm.
21
Step 3 : Join AE.5 cm respectively draw two arcs intersecting each other at D. DA = 4. in which AB = 3 cm. Then ∆ABE is the required triangle.6 cm. 21. Step 2 : Construct ∠ABK = 135° and cut off BC = 4. Step 6 : Through C draw CL || DB meeting AB produced in E.2 cm from BK. We go through the following steps : Step 1 : Draw AB = 3 cm. ABCD is the quadrilateral.5 cm and ∠B = 135° and then to construct a triangle equal in area to this quadrilateral.22
. Step 3 : With C and A as centres and radii 3. Step 7 : Join DE Then ∆DAE is the required triangle. BC = 4. 21.2 cm. For that we go through the following steps : Step 1 : Join AC. Step 4 : Join AD and CD.
Fig.Constructions
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We have to construct a triangle equal in area to the quadrilateral ABCD. Step 5 : Join DB. CD = 3. Construction 22 : To construct a quadrilateral and to construct a triangle equal in area to this quadrilateral. Suppose you have to construct a quadrilateral ABCD.6 cm and 4. Step 2 : Through D. draw a line segment DE || AC intersecting BC produced at E.
Fig.
Take a point A on the circle. Also write steps of construction. We go through the following steps. From a point P outside the circle. You have to draw a tangent to the circle. Let R be mid point of OA. DA = 3 cm and AC = 6 cm. Draw a circle of 3 cm radius. 21. draw two tangents PQ and PR to the circle. For that. CHECK YOUR PROGRESS 21.5 cm. Step 3 : With R as centre and radius equal to RO. draw PT ⊥ OP. draw a circle intersecting the given circle at P and Q. 3. Then AP and AQ are the required tangents. draw a tangent to the circle by using the centre of the circle. Also write the steps of construction.6 1. Draw a quadrilateral PQRS and construct a triangle equal in area to this quadrilateral. Construct a triangle equal in area to this quadrilateral. Step 2 : At P. AB = 4 cm. 21. 2. Suppose C be the given circle with centre O and a point P on it.9 CONSTRUCTION OF TANGENTS TO A CIRCLE Construction 23 : To draw a tangent to a given circle at a given point on it using the centre of the circle. Draw a circle of radius 2. 21.24
.2 cm. Step 3 : Produce TP to Q. Also write steps of construction. equal in area to the rectangle on AB as base.170
Mathematics
CHECK YOUR PROGRESS 21. Then TPQ is the required tangent.5 cm. Suppose C be the given circle and a point A outside it. Construct a rectangle ABCD in which AB = 5 cm and BC = 3.5 cm. Construct a quadrilateral ABCD in which. Note : You are required to write the steps of construction in each of the following problems. 2. Construct a triangle. we go through the following steps : Step 1 : Join OA. At A. CD = 4.
Fig. BC = 3. You have to draw tangents to the circle from the point A. Step 1 : Join OP. Verify that lengths of PQ and PR are equal. Step 2 : Draw the right bisector of OA.5 1. Step 4 : Join AP and AQ.
Fig.23
Construction 24 : To draw tangents to a circle from a given point outside it.
Construction 26 : Construct a triangle with sides 4 cm.10. Construction 27 : To construct incircle of a triangle. You have to draw a circumcircle of this triangle. Step 5 : Join OB. We go through the following steps Step 1 : Draw BC = 3. 21. Suppose a ∆ABC is given. Step 3 : Join O with any one vertex say B. BC = 3. We go through the following steps. 21.26
.Constructions
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21. 5 cm and 6 cm. Step 4 : With O as centre and radius OB. ABC is the required triangle. Step 6 : With O as centre and radius OB. Step 3 : Join AB and AC. Step 4 : Draw right bisectors of AB and BC which meet each other at O. draw a circle. This is the required circumcircle.25
Fig. CONSTRUCTION OF CIRCUMCIRCLE AND INCIRCLE OF A TRIANGLE. 21.5 cm and ∠B = 60° and draw its incircle. Construction 25 : To construct circumcircle of a triangle. we go through the following steps. This is the required circumcircle. Step 2 : With B and C as centres and radii 4 and 6 cm respectively draw two arcs intersecting each other at A. Step 2 : Draw right bisectors of any two sides ray BC and AC which meet each other at O. cut off BA = 4 cm
Fig. Step 1 : BC = 5 cm. Step 1 : Draw the given ∆ABC. draw a circle.27 Fig. Suppose you have to construct a triangle ABC with AB = 4 cm. Draw a circumcircle of this triangle.5 cm Step 2 : Draw ∠CBM = 60° Step 3 : From BM. To construct it.
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Mathematics
Step 4 : Join AC ABC is the required triangle. draw a circle. Construct a ∆ABC in which BC = 3. Construct a square with one side 4. 5. Construct a ∆ABC in which AB = 4 cm. Construct a rectangle with sides 8 cm and 6 cm.7 1. Step 5 : Draw bisectors of ∠B and ∠C meeting each other at I. Measure the length of its two diagonals. Construct a parallelogram having its diagonals as 5 cm and 6 cm and angle between them is 45°. Construct a right angled triangle whose hypotenuse is 8 cm and one if its other two sides is 5. TERMINAL EXERCISE 1. 2. Draw a line segment AB = 6 cm. 4. 9. Construct a triangle with base 4 cm and base angles 60° and 75° and draw its incircle. Measure AC and CB. Also write the steps of construction. and radius = ID. Step 6 : From I. 7. ∠A = 45° and AC – BC = 1 cm.
. 6. 2.5 cm. draw IK perpendicular to BC meeting BC in D. 3. Step 7 : With I as centre. Construct a ∆ABC with AB = 5 cm. Measure its diagonals.5 cm. 4. AB + AC = 8 cm and ∠B = 60°. 8. Construct a triangle with perimeter 14 cm and base angle 60° and 90°. Also write steps of construction. Note : You are required to write the steps of construction in each of the following problems. This is the required incircle. Divide it internally in the ratio 3 : 5. ∠B = 75° and draw its circumcircle. Construct an equilateral triangle of side 5 cm and draw its incircle. Draw a line segment PQ = 8 cm long. 3. CHECK YOUR PROGRESS 21. Find a point C on AB such that AC : CB = 3 : 2. Construct an isosceles ∆ABC with base BC = 4 cm and one of the equal sides AB = 3 cm and draw its circumcircle. Note : You are also required to write the steps of construction in each of the following problems.2 cm. BC = 4.5 cm.
or of two numbers. 22. one of which refers to a vertical division of the map into columns.
4 3 2
1
A
B
C (i)
D
E
F
3
2
1
1
2
3 (ii) Fig. and the other to a horizontal division into rows.1
4
5
6
. But the task can be made easier by dividing it into squares of managable size.1 INTRODUCTION The problem of locating a village or a road on a large map can involve a good deal of searching. Each square is identified by a combination of a letter and a number.174
Mathematics
22
Co-ordinate Geometry
22.
distance between two points in a plane.3 EXPECTED BACKGROUND KNOWLEDGE
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Idea of number line Fundamental operations on numbers. 22. 22. The point O.r.1(ii))]. section formula and co-ordinates of the centroid of a triangle.2)or (4.e.e. called the Cartesian co-ordinates of a point. 22. In this lesson. roughly we can indicate its location inside the shaded square on the map.2 OBJECTIVES After studying this lesson. Similarly. Properties of a right triangle..1(i). we can identify the shaded square on the map by the coding.. we will learn more about cartesian co-ordinates of a point. at right angles to each other at O (See Fig. The method of finding the position of a point in a plane very precisely was introduced by the French Mathematician and Philosopher. The horizontal number line XOX′ is called x-axis and the vertical number line YOY′ is called y-axis.4 CO-ORDINATE SYSTEM Recall that you have learnt to draw the graph of a linear equation in two variable in Lesson 6. solve problems based on the above concepts. the portion of y-axis above the origin O. 22. In this. Rene Descates (1596–1650).Co-ordinate Geometry
175
In the above Fig.
22. find the co-ordinates of the centroid of a triangle with given vertices. a point in the plane is represented by an ordered pair of numbers. the learner will be able to :
z z z
fix the position of different points in a plane. find the distance between two different points whose co-ordinates are given. It may be noted that. find the co-ordinates of a point.t to its distances from two axes of a reference.
. the positive direction of x-axis is taken to the right of the origin O. If we know the coding of a particular city. where both axes intersect each other called the origin. The two axes together are called rectangular coordinate system. 2) [See Fig. in a given ratio internally.e. OX and the negative direction is taken to the left of the origin O. the side OY′ is taken as negative. which divides the line segment joining two points. i. (B.. i. the side OY is taken as positive and the portion below the origin O. The position of a point in a plane is fixed w. find the co-ordinates of the mid-point of the join of two points. the side OX′.2). The pair of numbers used for coding is called ordered pair. which are usually drawn by the two graduated number lines XOX′ and YOY′.
z z z
22. i. But still we do not know its precise location.
2
22. 2) and (–2.
Fig.e. (x co-ordinate. the co-ordinates of the points A and B are (3. the y-coordinate (or ordinate) indicates the distance from the x-axis.3. –4) respectively.176
Mathematics
Fig. called co-ordinates which refer to the distances of the point from these two axes. 2) from the y-axis is 3 units and from the x-axis is 2 two units. the x-co-ordinate (or abscissa). always indicates the distance from the y-axis and the second number. It is customary to write the co-ordinates of a point as an ordered pair i. By convention the first number. You can say that the distance of the point A(3. y co-ordinate). 22.
. 22.3
In the above Fig 22.5 CO-ORDINATES OF A POINT The position of a point is given by two numbers..
You may also note that the position of points (x. b) and (0. (d) (2.1 : Write down x and y co-ordinate for each of the following points : (a) (1. 0). In general. Similarly. y) imply that distance of P from the y-axis is x units and its distance from the x-axis is y units. 4) and (4. (d) x co-ordinate is 2 y co-ordinate is –6. 2) (c) (–7. You may note that the co-ordinates of the origin O are (0.4
Example 22. For example position of points (3. 0). The y co-ordinate of every point on the x-axis is 0 and the x co-ordinate of every point on the y-axis is 0. –6)
Solution : (a) x co-ordinate is 1 y co-ordinate is 1 (c) x co-ordinate is –7 y co-ordinate is –5. –b) respectively where 'b' is a non-zero positive number. co-ordinates of a point P(x. y) and (y. co-ordinates of any point on the x-axis to the right of the origin is (a. where 'a' is a non-zero positive number. 3) are shown in Fig 22. x) in the rectangular co-ordinate system is not the same. 0) and that to left of the origin is (–a. 2) that its x co-ordinate is 3 and the y-co-ordinate is 2.4.
Fig. –5) (b) x co-ordinate is –3 y co-ordinate is 2.Co-ordinate Geometry
177
It is clear from the point A(3. 22.
. Similarly x co-ordinate and y co-ordinate of the point B(–2. y co-ordinates of any point on the y-axis above and below the x-axis would be (0. In general. 1) (b) (–3. –4) are –2 and –4 respectively.
Solution : (a) The distance of the point A from the y-axis is 3 units and from the x-axis is 4 units. (b) The distance of the point B from the y-axis is 5 units to the left of the origin and from the x-axis is 1 unit. (c) The distance of the point C from the y-axis is 3 units to the left of the origin and from the x-axis is also 3 units below the origin. (d) The distance of the point D from the y-axis is 8 units to the right of the origin and from the x-axis is 9 units below the origin. 22.6 QUADRANTS The two axes XOX′ and YOY′ divide the plane into four parts called quadrants.
We have discussed in Section 22.4 that (i) along x-axis, the positive direction is taken to the right of the origin and negative direction to its left. (ii) along y-axis, portion above the x-axis is taken as positive and portion below the x-axis is taken as negative (See Fig. 22.6).
Co-ordinate Geometry
179
Fig. 22.6
Fig. 22.7
Therefore, co-ordinates of all points in the first quadrant are of the type (+, +) (See Fig. 22.7) Any point in the II quadrant has x co-ordinate negative and y co-ordinate positive [(–, +)]. Similarly, in III quadrant, a point has both x and y co-ordinates negative [(–, –)] and in IV quadrant, a point has x co-ordinate positive and y co-ordinate negative [(+, –)]. For example : (a) P(5, 6) lies in the first quadrant as both x and y co-ordinates are positive. (b) Q(–3, 4) lies in the second quadrant as its x co-ordiante is negative and y co-ordinate is positive. (c) R (–2, –3) lies in the third quadrant as its both x and y co-ordinates are negative. (d) S(4, –1) lies in the fourth quadrant as its x co-ordinate is positive and y coordinate is negative.
22.7 PLOTTING OF A POINT WHOSE CO-ORDINATES ARE GIVEN The point can be plotted by measuring its distances from the axes. Thus, any point (h, k) can be plotted as follows : (i) Measure OM equal to h along the x-axis (See Fig. 22.8). (ii) Measure MP perpendicular to OM and equal to k. Follow the rule of sign in both cases. For example points (–3, 5) and (4, –6) would be plotted as shown in Fig 22.9.
Fig. 22.8
Fig. 22.9
22.8 DISTANCE BETWEEN TWO POINTS The distance between any two points P(x1, y1) and Q(x2, y2) in the plane is the length of the line segment PQ. From P, Q draw PL and QM perpendiculars on the x-axis and PR perpendicular on QM. Then,
∴
Corollary : The distance of the point (x1, y1) from the origin (0, 0) is
bx − 0g + by − 0g
1 2 1
2
=
x12 + y12
Let us consider some examples to illustrate. Example 22.3 : Find the distance between each of the following points : (a) P(6, 8) and Q(–9, –12) (b) A(–6, –1) and B(–6, 11) Solution : (a) Here the points are P(6, 8) and Q(–9, –12) By using distance formula, we have PQ = = Hence,
b−9 − 6g + mb−12 − 8gr
2
2
152 + 202 =
225 + 400 =
625 = 25
PQ = 25 units.
(b) Here the points are A(–6, –1) and B(–6, 11) By using distance formula, we have AB = = Hence,
m−6 − b−6gr + m11 − b−1gr
2
2
02 + 12 2 = 12
AB = 12 units.
Example 22.4 : The distance between two points (0, 0) and (x, 3) is 5. Find x. Solution : By using distance formula, we have the distance between (0, 0) and (x, 3) is
Example 22.6 : Find the radius of the circle whose centre is at (0, 0) and which passes through the point (–6, 8) Solution : Let A(0, 0) and B(–6, 8) be the given points. Now, radius of the cirlce is same as the distance of the line segment AB.
∴
2. Find the radius of the cirlce whose centre is at (2, 0) and which passes through the point (7,–12) 3. Show that the points (–5, 6), (–1, 2) and (2, –1) are collinear. 22.9 SECTION FORMULA To find the co-ordinates of a point, which divides the line segment joining two points, in a given ratio internally. Let A(x1, y1) and B(x2, y2) be the two given points and P(x, y) be a point on AB which divides it in the given ratio m : n. We have to find the co-ordinates of P. Draw the perpendiculars AL, PM, BN on OX, and, AK, PT on PM and BN respectively. Then, from similar triangles APK and PBT, we have
Fig. 22.12
∴ The co-ordinates of the mid-point joining two points (x1, y1) and (x2, y2) are :
FG x H
2
+ x1 , 2
y 2 + y1 2
IJ K
Let us take some examples to illustrate : Example 22.7 : Find the co-ordinates of a point which divides the line segment joining each of the following points in the given ratio : (a) (2, 3) and (7, 8) in the ratio 2 : 3 internally (b) (–1, 4) and (0, –3) in the ratio 1 : 4 internally. Solution : (a) Let A(2, 3) and B(7, 8) be the given points. Let P(x, y) divide AB in the ratio 2 : 3 internally. Using section formula, we have x=
2 × 7 + 3 × 2 20 = =4 5 2+3
∴ C(4, 8) are the co-ordinates of the mid-point of the line segment joining two points (3, 4) and (5, 12).
Example 22.9 : The co-ordinates of the mid-point of a line segment are (2, 3). If co-ordinates of one of the end points of the line segment are (6, 5), find the co-ordiants of the other end point. Solution : Let other the end point be A(x, y) It is given that C(2, 3) is the mid point
∴ We can write,
Fig. 22.13
2= or or
∴
x+6 2
and or or
3=
y+5 2
4=x + 6 x = –2
6=y+5 y=1
A(–2, 1) be the co-ordinates of the other end point.
186
Mathematics
22.10 CENTROID OF A TRIANGLE To find the co-ordinates of the centroid of a triangle whose vertices are given. Definition : The centroid of a triangle is the point of concurrency of its medians and divides each median in the ratio of 2 : 1. Let A(x1, y1), B(x2, y2) and C(x3, y3) be the vertices of the triangle ABC. Let AD be the median bisecting its base. Then, using mid-point formula, we have D=
Example 22.10 : The co-ordinates of the vertices of a triangle are (3, –1), (10, 7) and (5, 3). Find the co-ordinates of its centroid. Solution : Let A(3, –1), B(10, 7) and C(5, 3) be the vertices of a triangle. Let G(x, y) be its centroid then, x=
3 + 10 + 5 =6 3
Co-ordinate Geometry
187
and
y=
−1 + 7 + 3 =3 3
Hence G(6, 3) are the co-ordinates of the centroid. CHECK YOUR PROGRESS 22.3 1. Find the co-ordinates of the point which divides internally the line segment joining the points : (a) (1, –2) and (4, 7) in the ratio 1 : 2 (b) (3, –2) and (–5, 4) in the ratio 1 : 1 2. Find the mid-point of the line joining : (a) (0, 0) and (8, –5) (b) (–7, 0) and (0, 10) 3. Find the centroid of the triangle whose vertices are (5, –1), (–3, –2) and (–1, 8). LET US SUM UP
z
If (2, 3) are the co-ordinates of a point, then x co-ordinate (or abscissa) is 2 and the y co-ordinate (or ordinate) is 3. In any co-ordinates (x, y), 'x' indicates the distance from the y-axis and 'y' indicates the distance from the x-axis. The co-ordinates of the origin are (0, 0) The y co-ordinate of every point on the x-axis is 0 and the x co-ordinate of every point on the y-axis is 0. The two axes XOX′ and YOY′ divides the plane into four parts called quadrants. The distance of the line segment joining two points (x1, y1) and (x2, y2) is given by :
z
z z
z z
bx
z z
2
− x1
g + by
2
2
− y1
g
2
.
The distance of the point (x1, y1) from the origin (0, 0) is
x12 + y12
The co-ordinates of a point, which divides the line segment joining two points (x1,y1) and (x2, y2) in a ratio m : n internally are given by :
FG mx + nx H m+ n
2
1
,
my 2 + ny1 m+ n
IJ K
188
Mathematics
z
The co-ordinates of the mid-point of the line segment joining two points (x1, y1) and (x2, y2) are given by :
FG x H
z
2
+ x1 y 2 + y1 , 2 2
IJ K
IJ K
The co-ordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) are given by :
FG x + x + x H 3
1 2
3
,
y1 + y 2 + y 3 3
TERMINAL EXERCISE 1. In Fig 22.15, AB = AC. Find x.
Fig. 22.15
2. The length of the line segment joining two points (2, 3) and (4, x) is x.
13 units. Find
3. Find the lengths of the sides of the triangle whose vertices are A(3, 4), B(2, –1) and C(4, –6). 4. Prove that the points (2, –2), (–2, 1) and (5, 2) are the vertices of a right angled triangle. 5. Find the co-ordinates of a point which divides the join of (2, –1) and (–3, 4) in the ratio of 2 : 3 internally. 6. Find the centre of a circle, if the end points of a diameter are P(–5, 7) and Q(3, –11). 7. Find the centroid of the triangle whose vertices are P(–2, 4), Q(7, –3) and R (4, 5).
In this text. are famous among all these sutras. the human beings were faced with the following types of things in their immediate environment : part of the earth on which they live (plane). Brahmgupta (598 AD).C. The desire to know more about them. Egyptians. and circles made of cuboidal. we shall study about the perimeter and area of plane figures and surface area. With this came into being the science of measurement of land and knowledge about rectilinear and solid figures. to have a measure of sizes and capacity and to acquire them was a natural phenomenon. stone crests and other solids or plane objects.
. In the old Indian scriptures. there are rules laid down to perform special types of yajnas to fulfil specific desires in specially built altars (vedis) These vedis were in the shape of squares. After that even the Indian mathematician like Aryabhatt (476 AD). One of the oldest text "Surya Sidhant" deals with the shapes (mostly spherical) and orbits (mainly elliptical) of heavenly bodies. total surface area and volumes of solids. Old civilization like Greek. birds and animals. cubical and spherical bricks. which was given the name of mensuration. Arabian and especially Indian contributed a lot to the development of this science. "Baudhayan Sulba" Apastamba Sulba and Katyayan Sulba which date back to 8th century B. rectangles. Bhaskaracharya (1114 AD) and Varahmihir (1120 AD) devoted much of their works in providing rules for finding perimeters and areas of plane figures and volumes of solids. fruit and vegetables trees.Area of Plane Figures
193
Module 4
Mensuration
With the advancement of civilization on this planet earth. The whole lot of "Sulba Sutras" and Srauta Sutras contain the knowledge of methods of construction of such vedis and the properties of the used shapes.
parks.2 OBJECTIVES After studying this lesson the learner will be able to :
z z z
identify rectilinear and non-rectilinear plane figures. rectangle. find the perimeter and area of rectilinear figures. find the area of rectilinear paths of different types in a rectangle (or a rectangular enclosure). we shall study the methods of finding perimeters and areas of plane figures. In this lesson. find the area of a triangle using Hero's formula. a rectangle.3 EXPECTED BACKGROUND KNOWLEDGE
z z
Measurement of line-segments Knowledge and conversion of units of measurements
. a quadrilateral. triangle. especially square. a parallelogram and a rhombus. Solve problems from day-to-day life situations based on the above concepts
z z
z z z z
23. a trapezium. like a square. a triangle. special quadrilaterals and circle.1 INTRODUCTION In our day-to-day life. explain the meaning of perimeter and area of closed figures. We will also study methods to find areas of rectangular or circular paths. we have to estimate the amount of wood required for a rectangular table top or the area of cloth required for table cover. plots and floor etc. 23.194
Mathematics
23
Area of Plane Figures
13. find the circumference and area of a circle. find area and perimeter of a sector. or to find the area of paths. find the area of circular paths inside or outside a circle (or a circular enclosure).
The perimeter is measured in terms of linear units.1
Similarly a square of side 1 m is called a unit meter square and its area is taken to be 1 m2 and its perimeter as 4 m.
Fig.1 A Unit Square A square of side 1 cm is called a unit cm square and its area is taken to be 1 cm2 and its perimeter is taken to be 4 cm. In connection with the circle.Area of Plane Figures
195
z z z
Four fundamental operations on numbers Drawing plane geometric figures Properties of special quadrilaterals
23.1 : Find the area of a square whose perimeter is 80 meters. 23. unit and its perimeter is 4a unit.5.4 PERIMETER OF PLANE FIGURES The distance covered to walk along the boundary of a plane figure is called its perimeter. 23. this distance covered is called circumference of the circle. m
= 20 × 20 sq. 23.5 AREA OF PLANE FIGURES The measure of the planar region enclosed by a plane figure is called its area. This is measured in terms of square units. m or 400 sq m. Example 23. Solution : Perimeter of square = 4a = 80 meters
∴
a=
80 or 20 metres 4
∴ Area of square = a2 sq.
. area of a square of side 'a' unit is a2 sq. In general.
Solution.6. Diagonal of the square ABCD = AC = =
∴
AB2 + BC 2
a 2 + a 2 or
2a m
2 a = 15 or a =
15 m 2
225 or 112. 23.3(ii). 2
Area of the square ABCD = a2 sq m = 23. Area of square park (a2) = 625 sq. 23. Example 23. Solution. Add another row of four centimetre square over the and above the first row to form the rectangle ABCD as shown in Fig 23. triangle. trapezium. quadrilateral. Find the length of the wire required to fence around the park. parallelogram and rhombus are rectilinear figures while a circle and a sector are non-rectilinear figures.3
(ii)
. The area of a square park is 625 sq m.7 PERIMETER AND AREA OF A RECTANGLE Take 4 centimetre squares and join them in a row as shown in Fig.5 sq m.
length of wire = Perimeter of square = 4a = 4 × 25 m = 100 m
i.3. 23.2. length of the wire required to fence the park is 100 m. RECTILINEAR FIGURES
Fig.3(i). rectangle. For example square.196
Mathematics
Example 23.
(i) Fig. 23.2
The figures whose boundaries are formed by line-segments are called rectilinear figures. m
∴
Side of square (a) =
625 m =
25 × 25 m or 25 m
Now. Find the area of a square whose diagonal is 15 m.e.
4 is (5 + 4 + 5 + 4) cm or 18 cm. 23. we see that it contains 20 or (5 × 4) unit cm squares.3(ii). Solution : (i) Area of the field = Length × breadth
.4 : The length and breadth of a rectangular field are 23. the distance covered to go around the rectangle ABCD is (4 + 2 + 4 + 2) cm or 12 cm. the perimeter of a square = 4 × side linear units rectangle = 2 (length + breadth) linear units Example 23. 23. Similarly. Do you observe that in each case the distance covered is obtained by adding 2 × length and 2 × breadth? i.e.Area of Plane Figures
197
Rectangle ABCD is formed by 8 small unit squares. Find : (i) the area of the field (ii) the length of wire required to put a fence around the boundary of the field. or
From here we can generalise that the area of a rectangle of length l cm and breadth b cm can be written as (l × b) cm2 or lb cm2. Thus.7 m and 14. 23. the distance covered to go around the rectangle PQRS in Fig. Perimeter of a Rectangle = 2(l + b) A square is a special rectangle in which length and breadth are equal. Therefore the planar region enclosed by it is 8 cm2. In case of Fig 23. If we take another rectangle PQRS with base formed by 5 unit squares and has four such rows (See Fig.4
i.5 m respecively.e.4).
∴
Its area is 20 cm2 (5 × 4) cm2 20 cm2
Fig. the area of a square = (side)2 square units rectangle = (length × breadth) square units And. You can see that 4 × 2 = 8.
∴ Area of trapezium ABCD
= Area of ∆ ABD + Area ∆ BCD = = 1 1 .h + .8 : Find the area of a trapezium lengths of whose parallel sides are 20 cm and 12 cm and the distance between them is 5 cm.7. Solution : Area of the field = 20250 sq m 15
= 1350 sq m Let the altitude of the triangular field be x m
∴ ∴
Base = 3x m Area of the field = 1 3 . the base of the field is 90 m and its altitude is 30 m.x = x 2 sq m 2 2
As given in the question.b. 23. 23. find the base and the altitude of the field. If the cost of cultivating the field at the rate of Rs 15 per square metre is Rs 20250. 23. the area of a trapezium = 1 (sum of the parallel sides) × Distance between them square units 2
Example 23.3x.10 AREA OF A TRAPEZIUM In Fig.7
Thus.7 : The base of a triangular field is three times its altitude. Solution : We know that Area of Trapezium = 1 (Sum of the Parallel sides) × Distance between them 2
. ABCD is a trapezium in which AB || DC.h 2 2 1 (a + b). 3 x2 = 1350 2 or or x2 = 1350 × 2 = 900 3
x = 30
Thus.Area of Plane Figures
199
Example 23.a.h 2
Fig.
Find the area of the rhombus. 23. Example 23.23.12 : The sides of a triangular field are 165 m. The formula was also obtained by the Indian Mathematicians Brahmagupta and Aryabhata. let ABCD be the given rhombus with AB = BC = CD = AD = 10 cm and the diagonal BD = 12 cm.11. Find the area of the field. The diagonals of a rhombus are of length 12 cm and 8 cm. b and c is given by
∆ =
s s− a s− b s− c
b gb gb g
where s =
a+b+c 2
This formula is known as Hero's formula after the name of Greek Mathematician Heron of Alexendria. 23.13 AREA OF A TRIANGLE USING HERO'S FORMULA The area of triangle ABC whose sides are given by a. Solution. AC and BD bisect (at right angles) at O (See Fig. 143 m and 154 m. Find the area of a rhombus.Area of Plane Figures
201
Example 23. Solution : In Fig. one of whose diagonals is 12 cm and the side is 10 cm.11
Area of rhombus = =
= 96 cm2 23.11)
∴ ∴ ∴
OB = OD =6 cm OA =
102 − 62 cm = 8 cm
Diagonal AC = 16 cm 1 (product of diagonals) 2 1 × 12 × 16 cm2 2
Fig. Solution : Here we use the Hero's formula for finding the area of the field
. We know that Area of a Rhombus = = 1 × Product of Diagonals 2 1 (12 × 8) cm2 2
= 48 cm2 Example 23.10.11.
From the centre of each side a path 5 m wide runs across up to the opposite side.5 m.1 1. Find the area of the paths. 9. Find the area of a quadrilateral one of whose diagonals is 30 metres long and the lengths of perpendiculars from the other two vertices to this diagonal are 10 m and 14 m respectively. 2. AD = 85 m and the diagonal AC = 154 m. 23. The sides of a rectangular field of area 726 sq metres are in the ratio 3 : 2. Find the area of a triangular field whose sides are 50 m. 6. 7. 10. Find its area. Find the area of a parallelogram with base and altitude of length 20 cm and 12 cm respectively.
. Also. Find (i) the area of the field. Solution : Area of paths = Area of rectangle ABCD + Area of rectangle PQRS – Area of square QLMN
Fig. find the length of the perpendicular from the opposite vertex to the side measuring 112 m.16
= [(100 × 5) +(80 × 5) – (5 × 5)] sq m = [500 + 400 – 25] sq m = 875 sq m CHECK YOUR PROGRESS 23. Find the length of its diagonal correct up to one decimal place. The area of a square field is 225 square metres. CD = 85 m. (ii) the length of the wire required to put a fence along the boundary of the field. Find its perimeter. 8.204
Mathematics
Example 23. The perimeter of a rhombus is 146 cm and one of its diagonals is 48 cm.5 m and 12. 4.15 : A rectangular piece of land measures 100 m by 80 m. 5. Find the area of a field in the form of a quadrilateral ABCD in which AB = 165 m. What will be the diagonal of a square whose perimeter is 60 m? 3. The length and breadth of a rectangular field are 22. The parallel sides of a trapezium are 20 metres and 16 metres long respectively and the distance between them is 12 m. find the other diagonal and the area of the rhombus. 78 m and 112 m. BC = 143 m.
A C = 2 = constant.16 : Find the circumference and area of a circle of radius 3.
Thus. A rectangular courtyard 120 m long and 90 m broad has a path of uniform width 5 m on the inside running round it. We state without any logical proof that the ratio of the circumference of a circle to its diameter is always constant which is also equal to the ratio of the area enclosed by a circle to the square of its radius (r).15 CIRCUMFERENCE AND AREA OF A CIRCLE Recall that a circle is the locus of a point which moves in a plane in such a way that its distance from a fixed point (in the same plane) always remains constant. Any straight line drawn through the centre of circle.5 cm2
=
. one parallel to the length and the other parallel to the breadth. 2r r 1 The value of this constant is nearly 3 and is usually 7 denoted by the Greek letter π (pie). each 2 metres wide running in the middle of it. the area of the circle = πr2 square units.5 cm [Use π = 22/7] Solution : Circumference of the circle = 2 πr 22 × 35 .1416. 12. correct to four places of decimals is 3. 23.e. Example 23.Area of Plane Figures
205
11. The fixed point (O) is called the centre of the circle and the constant distance is called the radius of the circle. whose end point lie on the circle is called a diameter. circumference of a circle = 2πr linear units and. which is an irrational number but its value.5 × 3. A rectangular lawn 80 metres by 60 metres has two roads. whose length is equal to twice the radius of the circle. 7 C = π or C = 2 πr ∴ 2r
i.17
or
A = π r2 .5 cm2 7 = 38. cm = 2× 7 = 22 cm Also. Find the area of the path. Find the area of the roads. and
A = π r2
Fig. 23. to avoid lengthy calculations the value of π 22 is often taken to be . area of the circle = π r 2
22 × 3.
Find the area of the road. 4. Find the areas of sectors of a circle with radius 3.5 m having central angle (i) 60° (ii) 90°. b anc c are sides of the triangle and s=
z
a+b+c units 2
Area of a parallelogram = Base × Altitude sq units Area of a trapezium = Area of a rhombus = 1 (sum of parallel sides) × Distance between them sq units 2 1 (Product of the diagonals) sq units 2
z
z
z
Area of rectangular path = Area of outer rectangular enclosure – Area of inner rectangular enclosure Area of perpendicular paths in the middle of a rectangular field = Area of path parallel to length + Area of path parallel to breadth – Area of common square.Area of Plane Figures
209
3. Use π = 7 . A circular park of radius 21 m has a road 7 m wide on its outside all around. LET US SUM UP
z z z
LM N
OP Q
Area of a square of side a = a2 sq units and perimeter = 4a units Perimeter of rectangle = 2(length + breadth) units Area of rectangle = (length × breadth) sq units Area of a triangle = 1 × base × height sq units 2 =
s s − a s − b s − c sq units
z
b gb gb g
where a. 6. The length of the minute hand of a clock is 7 cm. The difference between the area of a circle and the square of its radius is 105 sq m. 1. Find the speed of the train in km/hr.
7. The driving wheel of a locomotive engine. Find the area covered by the minute 22 hand in 6 minutes.4 m in radius. makes 70 revolutions in one minute. 5.
z
. Find the circumference of circle.
5 m. Find the area of each of the following parallelograms : (i) side 20 m and the corresponding altitude 12 m. Find the cost of wire required to go four times round the field at the rate of Rs 7 per 10 metres of length of wire. If its breadth is 4.6 m 12 m
9. The area of a rectangular field is 27000 sq meters and the ratio between its length and breadth is 6 : 5.5 m and 10. second side is 14 m and the diagonal is 15 m.5 m 14.210
Mathematics
z z z
Circumference of a circle = 2 π r units Area of a circle = π r2 sq units Area of circular path = π (R2 – r2). Find its area.5 m (iii) 17 m and 40 m (iv) 40 m and 22 m Perpendicular distance between them 15 m 7. 360° θ Perimeter of a sector = 2 r + 2 πr 360°
z
z
LM N
OP units Q
TERMINAL EXERCISE 1. 8.5 m.
. Find its area. What is its area ? 3. 6. Find the area of the following trapezia : Lengths of parallel sides (i) 30 m and 20 m (ii) 15. How long will a man take to walk round the boundary of a square field of area 40000 sq meters at the rate of 4 km an hour ? 4. The perimeter of a rectangle is 980 meters. The length of a room is three times its breadth. where R > r θ sq units Area of a sector of a circle = π r 2 . (ii) one side is 13m. and the length is to breadth is 5 : 2. 7. 5. The side of a square park is 37. The perimeter of a square is 480 m. Find the area of a piece of land in the shape of a quadrilateral one of whose diagonals is 20 m and lengths of perpendiculars from the other two vertices on the diagonal are of length 12 m and 18 m respectively. 2. find the area of its floor.
153 m 13. A circular plot with a radius of 15 m has a road 2 m wide running all around inside it. DA = 85 metres and DB = 154 metres. 52 m. Find the areas of the following triangles whose sides are : (i) 25 m. the perpendicular between them is 24 metres and the area of the trapezium is 312 sq metres. 111 m. The sides of a triangle are 51 m. BC = 143 metres. one of whose diagonals measures 8 m and the side is 5 m. 60 m. Find the perpendicular from the opposite side on the side of length 52 m. CD = 165 metres. Find the lengths of the two parallel sides. From the circular piece of cardboard with radius 1. A rectangular plot of grass measures 65 m by 40 m. 22. 17. a sector with central angle 60° has been removed. A path 3 metres wide runs around a circular park whose radius is 9 meters. 20. 21. Find the total area of the two paths. From the centre of each side a path of 10 m wide runs across up to the opposite side. Find the area of paths.
. A rectangular plot of land measuring 30 m by 20 m has two paths each 2 m wide on both the sides (inside and outside) of the boundary. Find the area of the portion removed. Find the area of the path. 19. Find the area of a field in the form of a trapezium whose parallel sides measure 48 m and 160 m and the non-parallel sides measure 50 m and 78 m respectively. Also find the areas of the two triangles into which the original triangle is divided. Find the area of a quadrilateral ABCD in which AB = 85 meters. The difference between the circumference and diameter of a circle is 30 m. 65 m (ii) 60 m. It has a path of uniform width 8 m all around inside it. Find the area of a rhombus.47 m.Area of Plane Figures
211
10. A rectangular piece of land measures 200 m by 150 m. 18. and 53 m. 16. Find the area of the road. 12. The difference between the two parallel sides of a trapezium is 8 metres. 14. Find the cost of spreading red sand stone on the path at the rate of Rs 5. 11. 15.25 per sq m. Find the radius of the circle.
we have a paper cut in the form as shown. are not plane figures. a football. find the area of four walls of a room explain the meaning of volume of a solid find the volume of a cube..1 INTRODUCTION In the previous lesson you have studied about plane figures i. etc. rectangles. figures which completely lie in a plane like squares.
z z z
z
24. It is a plane figure. an ice cream cone.2 OBJECTIVES After studying this lesson.4. circles etc. cuboid. cylinder.3 EXPECTED BACKGROUND KNOWLEDGE
z z z
Area of plane and rectilinear figures Circumference and area of a circle Four fundamental operations on numbers
24. cuboid. the learner will be able to :
z z z
identify different solids explain the meaning of surface area of a solid find the surface area of a cube. In this lesson we will study about these types of solids.
. a glass tumbler. cylinder. SOLID FIGURES In Fig 24(i). cone. cone. sphere and a hemisphere using the respective formulae. a box. triangles. sphere and a hemisphere using respective formulae solve problems from day to day life situations based on the above concepts. 24. They are called Solids. But the objects like a brick.e.Surface Area and Volume of Solids
213
24
Surface Area and Volume of Solids
24.
It can be easily seen from the figure that a cuboid has six rectangular plane surfaces called faces. C. The sum of the areas of the plane figures making up the boundary of a solid figure is called its surface area. 24. A cuboid has 12 edges in all.
(i) Fig. match box and a book are all examples of cuboids. For example. The line-segment joining the vertex A to the vertex H is called a main diagonal of the cuboid.2
It can be seen that at every vertex. A.1(i) is the surface area of box] and the measure of part of space occupied by a solid is called its Volume.5 CUBOID A brick. a cone and a sphere and learn to find their surface areas and volumes.
. A cuboid has 8 corners called the vertices. ABEF. [For example. E. faces ABCD and BCHF meet along the edge BC. One of these edges is taken as length. it has more than two dimensions. Fig 24.1(ii). BCHF. ADGE and DCHG]. the other the breadth and the third as height and are denoted by 'l'. a cylinder. they have three dimensions) are called solids. [ABCD. there are three edges meeting [called coterminous edges].
Fig. 'b' and 'h' respectively. geometrical box. B.2 represents a cuboid. we shall take some solids like a cuboid. 24. Two adjacent faces meet along a line segment called an edge. G and H are the vertices of the cuboid represented by Fig 24. and the opposite faces [like ABCD and EFHG] are congurent.2. F.1
(ii)
As the box occupies some part of the space. Such objects which occupy space (i. Now.e.. chalk box. 24. D.214
Mathematics
But when we fold the paper along the dotted lines. the area of paper in Fig 24. we can make a box (like a chalk box) as shown in Fig 24. EFHG.
Let us have a cuboid with sides 5 cm. the volume is 1 cu m. 4 cm and 3 cm. = 5 cm × 4 cm × 3 cm = length × breadth × height
.e. you can easily find that the number of unit cubes in the cuboid are 5 × 4 × 3 = 60. cm) and if the side is 1 m. A unit cube is the volume of a cube of side 1 unit. ADGE.Surface Area and Volume of Solids
215
Let AB= l. So. which is unit cube.3. its volume is 1 cubic centimeter (or 1 cu. if the side of a cube is 1 cm. cm. 24.3
You can see that Volume = 60 cu. l = b = h = a (say)
∴ The surface area of a cube = 6a2
Main Diagonal of a cube =
3a
To find the volume of a cuboid. BCHF. So. we first define the unit of measurement of volume. In Fig. cm. To find the volume of a cuboid. then the total surface area of cuboid = sum of the areas of six faces ABCD. 24.
Fig. EFGH. ABFE and DCHG = (lh + lh + bh + bh + lb + lb) = 2(lb + bh + hl)
∴ Total surface area of a cuboid = 2(lb + bh + hl)
Main diagonal of a cuboid =
l 2 + b2 + h2
Cube : A cube is a special case of a cuboid when all the edges are equal i. the volume of the cuboid = 60 cu. we are to find the number of unit cubes contained in it. AE = b and AD = h.
24. Three cubes each of side 8 cm are joined end to end.6. 24.7
. 24. 24. It its volume is V. it describes the cylinder shown in Fig 24. Find the surface area of the resulting cuboid. we come across many solids like water pipes. 9. 27 dm. 24.6.6
Fig. thin cans.75 cubic meters. r is the radius of the base and h is the height of the cylinder. powder box.1 Volume of a Cylinder = (area of the base) × height = π r2h cubic units where.
Fig.75 m and of volume 33. We note that the ends (or bases) of a right circular cylinder are congurent circles and the line joining the centres A and B of these circles is perpendicular to the two ends. 12 dm is made of wood 1 dm thick. In our daily life. beakers in the laboratory. The areas of three adjacent faces of a cuboid are a. What is the capacity of the box and what is the volume of wood used in it ? 8. A cuboidal box whose external dimensions including the lid are 32 dm. Thus if the rectangle ABCD revolves about the side AB.6.6 RIGHT CIRCULAR CYLINDER A right circular cylinder is a solid generated by the revolution of a rectangle about one of its sides which remains fixed. Find the total surface area of a wooden plank of width 3 m. 10. which are right circular cylinders.2 Curved Surface of a Cylinder Area of the rectangle obtained by cutting a hollow cylinder along any line on its surface parallel to the axis is called its curved surface area. b and c. prove that V2 = abc.220
Mathematics
7. thickness 0.
open at both ends. A rectangular piece of paper 33 cm long and 16 cm wide is rolled along its breadth to get a cylinder of height 16 cm. If the external diameter be 50 cm and the length of tube be 140 cm. and spread it into a sector ABC as shown in Fig 24. In Fig 24. find its volume. let us cut it along its slant height. Find the curved surface area of the cylinder. The volume of right circular cylinder is 3080 cu cm and the radius of base is 7 cm. 9. Find the height of the embankment. Calculate the ratios of their volumes and of the cured surface areas. Find the whole surface area of a hollow cylinder open at the ends.7.1 Surface Area of a Right circular cone To find the total surface of a cone... The diameter of a garden roller is 2... 7. Earth taken out of it has been spread all-around it to a width of 5 meters to form an embankment. 24..5 m long. A hollow cylindrical tube. find volume of iron used in making the tube. A well.. 24..9.8
.. Find the volume of the cylinder.. 4. How much area will it cover in 100 revolutions ? 10. 6..8 m and it is 1. 8. A water storage tank has a cylindrical shape.....5 m 1.... the external diameter is 10 cm and the thickness is 2 cm ( π = 3...8 right triangle AOB revolves along AO to generate the cone of height 'h' and radius 'r' and slant height l = AC.4 m
3.. The radii of two cylinders are in the ratio 3 : 2 and their heights are in the ratio 7 : 4.. (iv) Surface area of a cylinder open at both ends = ... is made of iron 2 cm thick. Find the volumes and total surface area of the following cylinders : Radius (i) (ii) (iii) 7 cm 10 cm 5m Height 12 cm 3. 24.. 5..7 SURFACE AREA AND VOLUME OF RIGHT CIRCULAR CONE A right circular cone is a solid generated by the revolution of a right angled triangle about one of the its sides containing the right angle as axis.224
Mathematics
(iii) Surface area of a cylinder open at the top = . is dug 14 meters deep.. 2. If it is 2. if its length is 8 cm. with 10 meters inside diameter.
Fig.1 m high and has a diameter 1 m...1416).
2πr = π rl sq. units 2
Since area of the base is πr2.Surface Area and Volume of Solids
225
Fig.8 SURFACE AREA AND VOLUME OF A SPHERE A sphere is a solid generated by the revolution of a semicircle about its diameter. 24. 24. so total surface area = πrl + πr2 = πr(l + r) 24. l. It can also be defined as under : A sphere is the locus of a point which moves in space such that its distance from a fixed point in space remains constant. arc BC. radius AC . O is the centre and OP = r is the radius. we use the following formulae.7. Surface area of sphere = 4 π r2 Volume of sphere = 4 πr 3 3
Fig 24. To find the surface area and the volume of a sphere. 3
24. In Fig.9
Thus curved surface area = =
1 .10
.2 Volume of a Right circular cone Volume of a cone = =
1 × area of base × height 3 1 2 πr h cu units. The fixed point is called the centre of sphere and the constant distance is called the radius of sphere.10. 2 1 .
. 10. Find the diameter of a sphere of volume 4851 cu cm....5 m Height (h) 28 cm 12 cm 12 m
3.14] 7......... 12.... A conical tent is 6 m high with radius of base as 8 m (a) Find the cost of cloth required to make the tent.. The water is poured into a cylinder of base radius 7 cm. the radius of whose base is 5 m and whose height is 12 m ? 4... 8.... (v) Volume of a sphere = . Find the slant height and curved surface area of cone whose volume is 12936 cu cm and the diameter of whose base is 42 cm....Surface Area and Volume of Solids
229
(iv) Surface area of a sphere = . It is melted to form smaller spheres of diameter 1. Find the volume. Show that their volumes are inthe ratio 1 : 2 : 3.... How many meters of cloth 3 m wide will be required to make a conical tent. 5.. [Use π = 3. 9.. Find the volume and the surface area of a sphere of radius 2.1 cm. if one square meter of cloth costs Rs 30.....75 cm.. 11. (b) How many persons can sit in the tent if each person requires 4 sq m of space on the ground and 20 cu meter of air to breath in.. a hemisphere and a cylinder stand on equal bases and have same height. The radius of an iron sphere is 3..... 6... A cone.. The cup is full of water..
. A conical cup of height 15 cm has a base radius of 12. 2..5 cm. Find the height to which water will rise in the cylinder. Find the number of smaller spheres formed. Find the radius of the base of a right circular cone of height 21 cm and volume 550 cu cm....... Find its volume. The area of the base of a right circular cone is 616 sq cm and its height is 9 cm. (vii) Total surface area of a solid hemisphere = ..6 cm... the curved surface area and the total surface area of right circular cone with the following dimensions : Radius (r) (i) (ii) (iii) 21 cm 14 cm 3. (vi) Volume of a hemisphere = ....
Find the cost of papering the remaining portion of the walls at the rate of Rs 1. b = 12 m. Volume of a cuboid = l × b × h. where a is its side. The doors and windows in the room occupy a space of 5 square meters.4 1. Find the cost of painting the walls and the ceiling of a room measuring 10 m × 6m ×3 m at the rate of Rs 1. Find the area of four walls of the room in each of the following cases : (i) l = 8 m.25 × 1 m. A right circular cylinder is a solid generated by revolution of a rectangle about one of its sides. each having an area of 3 square meters and 4 windows each measuring 1. 4. and the area of the floor is 4875 sq dm. LET US SUM UP
z
The figures. Total surface area of a cuboid = 2(lb + bh + hl) Total surface area of a cube = 6a2. which remains fixed. It has three doors. Find the cost of papering the remaining portion of the walls with paper 75 cm wide.50 per square meter. b = 6 m. (ii) l = 20 m. are called Solid figures. A room is 6 m long. h = 3m. 5. which occupy space and have more than two dimensions. h = 8 m 2. 3. at the rate of Rs 1. A room measures 9 m × 7m × 3m.232
Mathematics
b2 = or
∴
77 × 7 11
b=7 m l= 11 × 7 44 = 11 m and h = =4m 7 11
∴ The length of the room is 11m.
z
z z z z z z
.50 per square meter.20 per meter. The amount of space occupied by the solid object is called its Volume. The area of two side walls of a room is 5250 sq dm and the area of the two end walls 4550 sq dm. Volume of a cube = a3. 5 m wide and 4 m high.
CHECK YOUR PROGRESS 24. The sum of the areas of the plane figures making up the boundary of a solid object is called its surface area. Find the dimensions of the room. its breadth is 7 m and its height is 4 m.
z
Surface area of a sphere = 4 π r2 Volume of a sphere =
4 3 πr 3
z
z z
Curved surface area of a hemisphere = 2 π r2 Total surface area of a solid hemisphere = 3 π r2 Volume of a hemi-sphere =
2 3 πr 3
z
z
Area of four walls of a room = 2(l + b) × h. How many bricks 20 cm × 10 cm × 7. If the room is to be 14 m long. so as to allow 2. 4. where l2 = r2 + h2 Total surface area of a solid cone = π rl + π r2 Volume of a cone =
1 2 πr h 3
z z
z
A sphere is a solid generated by the revolution of a semicircle about its diameter. If there is no waste in the process. Find the number of 5 cm cubes that can be cut out of a 15 cm cube. [1 metric ton = 1000 kg]
. 2.2 sq m of floor area and 11 cu m of space for each child. TERMINAL EXERCISE
1. Find the edge of a cube of volume equal to the volume of a cuboid of dimensions 63 cm × 56 cm × 21 cm. find the edge of the new cube so formed. 4 and 5 cm respectively are melted and formed into a single cube. Three cubes of metals whose edges are 3.5 cm be carried by a truck whose capacity to carry load is 6 metric tons ? One cubic meter of bricks weighs 2000 kg. Total surface area of an cylinder open at one end = 2 π rh + π r2 Total surface area of a cylinder closed at both ends = 2 π rh + 2 π r2 = 2 π r (h + r) A right circular cone is a solid generated by the revolution of a right triangle about one of its sides containing the right angles as axis. 3.Surface Area and Volume of Solids
233
z z z z
Volume of a cylinder = π r2h. Curved surface area of a cone = π rl. A school room is to be built to accommodate 70 children. what must be its breadth and height ? 5.
A room 12 meters long. If the internal length. If this water is poured into a cylinder with internal radius 21 cm. How much is the level of field raised ? 7. 10 cm ad 8 cm. 4 meters broad and 3 meters high has two windows 2 m × 1 m and a door 2. 16. A conical vessel of internal radius 14 cm and height 36 cm is full of water.5 cm thick and bottom is 1 cm thick. 8. Find the dimensions of the room. A well with 8.5 times its breadth. A field is 200 m long and 75 m broad. Earth taken out of it has been spread all around it to a width of 4 meters to form an embankment. Four cubes each of sides 5 cm are joined end to end. and the earth taken out of it is spread evenly over the field. Find the diameter of a sphere whose volume is 606. A cubic cm of gold is drawn into a wire 1/5 mm in diameter.14) 11. 17.4 meter inside diameter is dug 10 meter deep. find the height to which the water rises in the cylinder. The length of a room is 1. Find the whole surface area of a hollow cylinder open at the ends. Find the area of the floor. find the quantity of material used in the construction of the box. 15. 9. Find the slant height of a cone whose volume is equal to 12936 cubic meters and the diameter of whose base is 42 meters. 20 meter broad and 10 meter deep is dug in the field. The volume of a cone is 616 cubic meters. whose length is 15 m and breadth is twice its height. Find the surface area of the resulting cuboid. Find the cost of papering the walls with paper 50 cm wide at Rs 20 per meter. If the height of cone is 27 meters.234
Mathematics
6.375 cubic meter. takes 250 meters of paper 2 meters wide for its four walls. A hall. breadth and depth are respectively 14 cm. find the radius of its base.
. The cost of carpeting it at Rs 150 per square meter is Rs 14400 and the cost of white washing the four walls at Rs 5 per square meter is Rs 625. the internal diameter is 8 cm and the thickness is 1 cm [Use π = 3. 14. if its length is 10 cm. 18.5 m × 2 m. and a tank 40 meter long. ( π = 3. 12. The sides of an open box are 0.14] 10. find the length of wire. Find the height of the embankment. 13.
in terms of rotation of a ray from its initial position to its final position. But it goes to the credit of Neelkanth Somstuvan (1500 A.
. we shall define an angle-positive or negative. In that. trigonometric ratios of complementary angles and solve simple problems on height and distances. We know that problems of this and related problems can be solved only with the help of a science called trigonometry. The first introduction to this topic was done by Hipparcus in 140 B.. 45° and 60°.D.D. he used the words Jaya.D. In this chapter. Kotijya and "sparshjya" which are presently used for sine. he wants to have an approximation of the distance between him and the temple.) while writing his work on Goladhayay. depression and gave examples of some problems on heights and distance.C. in terms of its sides develop some trigonometric identities. But it was Aryabhatta (476 A. Before venturing to start climbing the hill. The subject was completed by Bhaskaracharya (1114 A. when he hinted at the possibility of finding distances and heights of inaccessible objects. using at the most two right triangles. using angles of 30°. Tolemy again raised the same possibility and suggested the use of a right triangle for the same.) who developed the science and used terms like elevation.240
Mathematics
Module 5
Trigonometry
Imagine a man standing near the base of a hill. cosine and tangent (of an angle). In 150 A.) whose introduction to the name "Jaya" lead to the name "sine" of an acute angle of a right triangle.D. looking at the temple on the top of the hill. define trigonometric ratios of an acute angle of a right triangle.
astronomy etc. trigonometry is that branch of Mathemetics which deals with the measurement of sides and angles of a triangle.
z z z
25. the learner will be able to
z z
write the trigonometric ratios of an acute angle of a right triangle. Thus. navigation. This branch of Mathematics has been instrumental in the development of architecture.Introduction to Trigonometry
241
25
Introduction to Trigonometry
25.2 OBJECTIVES After studying this lesson. The great astronomer Hipparchus is said to have developed this branch of Mathematics.
. The Indian traditional texts like Surya Sidhant have developed and used the knowledge of trigonometry extensively. 25. find the sides and angle of a right triangle when some of its sides and trigonometric ratios are known. surveying. equilateral. solve problems based on trigonometric ratios and identities. establish trigonometric identities. write the relationships among trigonometic ratios. obtuse and right Types of triangles–acute.3 EXPECTED BACKGROUND KNOWLEDGE
z z z z z z
Concept of an angle Construction of right triangles Drawing parallel and perpendicular lines Types of angles–acute.1 INTRODUCTION The word 'Trigonometry' is derived from the word "Trigon" meaning "a triangle" and "metron" meaning "measurement". obtuse and right Types of triangles–isosceles.
25. 25. (See Fig. Let ∠POB = θ. (θ is a Greek letter. an angle POB is formed with x-axis. Let PB = p. Draw PB ⊥ OX.1
Side opposite to angle θ PB p tangent of θ = Adjacent side to angle θ = OB = b Adjacent side to angle θ OB b cotangent θ = Side opposite to angle θ = PB = p
secant θ = cosecant θ =
Hypotenuse = OP = h Adjacent side to angle θ OB b Hypotenuse = OP = h Side opposite to angle θ PB p
The above trigonometric ratios are written below in an abbreviated form : sine of θ is abbreviated as sin θ cosine of θ is abbreviated as cos θ tangent of θ is abbreviated as tan θ cotangent of θ is abbreviated as cot θ secant of θ is abbreviated as sec θ and cosecant of θ is abbreviated as cosec θ Notes 1. Let A be a point on OX.4 TRIGONOMETRIC RATIOS OF AN ACUTE ANGLE Let XOX' and YOY' be rectangular axes of co-ordinates. Thus. Throughout the study of trigonometry we shall be using only abbreviated form of these trigonometric ratios. we define the following trigonometric ratios for angle θ. sine of θ = cosine of θ =
Side opposite to angle θ PB p = = Hypotenuse OP h Adjacent side to angle θ OB b = = Hypotenuse OP h
Fig.
. 2. sin θ is an abbreviation for "sine of angle θ"and not the product of sin and θ.1). Let the ray OA start rotating in the plane in an anti-clockwise direction from the initial position OA about the point O till it reaches its final position OP after some interval of time.242
Mathematics
25. OB = b and OP = h. Now clearly ∆PBO is right angled triangled. and we read it as "theta").
45° and 60°. a triangle. in a right-angled triangle ABC right-angled at B. find trigonometric ratios of complementary angles. using trigonometric ratios.1 INTRODUCTION In the previous lesson. a triangle with three sides equal. solve daily life problems of heights and distances.
26.2 OBJECTIVES After studying this lesson. the angle formed with x-axis is positive. We will also use the knowledge of trigonometry to solve simple problems based on heights and distances taken from day to day life.3 EXPECTED BACKGROUND KNOWLEDGE The student must know before starting the lesson that :
z
when a ray rotates in an anti-clockwise direction about the origin. 26. we have defined trigonometric ratios for acute angles and developed some relationship between them. with one angle of 90°.262
Mathematics
26
Trigonometric Ratios of Some Special Angles
26. AC2 = AB2 + BC2. if the sum of two angles is 90°. then the angles are said to be complementary. right-angled at B sin C =
z z z z z z
opposite side hypotenuse
hypotenuse cosec C = opposite side
. in a right triangle ABC. is said to be an equilateral triangle. the learner will be able to :
z z z
find geometrically the trigonometric ratios for the angles of 30°. is called a right-angled triangle. 45° and 60° by using our knowledge of geometry. In this lesson. we shall find the values of trigonometric ratios of angels of 30°. recalls the conditions of congruency of two triangles.
26. is looking at an object at a lesser height. we have learnt to determine the values of trigonometric ratios of the angles of 30°.1 Angle of Elevation Whenever an observer is looking at an object which is at a greater height than the observer.2 Angle of Depression On the contrary.Trigonometric Ratios of some Special Angles
275
3. we will learn how trigonometry can be used to determine the distance between the objects (particularly inaccessible ones) or the heights of the objects by taking some examples from day-to-day life. In this lesson. We shall first define some terms which will be needed in the study of heights and distances. and the horizontal line. 26. (i) cos 55° + sin 68° (ii) cot 75° + cosec 75° (iii) sec262° + sec269° 26. α is the angle of depression.5
Fig. The length of the ladder is 8 m. 45° and 60°.
. the angle formed between the line of sight and the line joining eye of the observer to the object is called an angle of depression. 26. ∠θ is the angle of elevation.6
Example 26. Also. Find the distance of the foot of the ladder from the well. 26. In Fig.10. he has to lift his eyes to see the object.10. Solution : Let AC be a ladder leaning against the wall AB making an angle of 60° with the level ground BC. 26. at a height.10 APPLICATIONS OF TRIGONOMETRY We have so far learnt to define trigonometric ratios of an angle. if the observer.5 . and an angle of elevation is formed between the line of sight joining the observers eye to the object.
Fig.6. Prove that (i) sin θ cos(90° – θ) + cos θ sin (90° – θ) = 1 (ii) cos θ cos (90° – θ) – sin θ sin (90° – θ) = 0 cos 90°−θ 1 + sin 90°−θ (iii) 1 + sin 90°−θ + cos 90°−θ = 2 cosec θ
b
b
g
g
b
b
g
g
tan 90°−θ (iv) sin (90° – θ) cos (90° – θ) = 1 + tan 2 90°−θ
b
b
g
g
4.15 : A ladder leaning against a window of a house makes an angle of 60° with the ground. 26. In Fig. Express each of the following in terms of angles between 0° and 45°.
Assume that there is no slack in the cable.
Fig. the foot of the ladder is 4 m away from the wall.16 : A balloon is connected to a meteorological ground station by a cable of length 100 m inclined at 60° to the horizontal.
Example 26. The distance between the base of a tree and the point where it touches the ground is 10 m. attached to a string AC of length 100 m which makes an angle of 60° with the level ground BC. Solution : Let A be the position of the balloon. 26. 26.276
Mathematics
Let BC = x m Now. Determine the height of the balloon from the ground. Find the height of the tree. in right-angled ∆ABC sin 60° = AB = x AC 100
3 = x 100 2 ⇒ x = 50 × 3 m = 86.6 m
Fig.7
Example 26. the balloon is at a height of 86. Solution : Let AB represent a tree.17 : The upper part of a tree is broken by the action of wind. The top of the tree makes an angle of 30° with the horizontal ground. in right-angled ∆CBD tan 30° = ⇒ BC = x BD 10
1 x = 10 3
Fig.8
Hence. Let AB = x m Now. Let C be a point from where the tree was broken by the action of the wind in two parts upper part makes an angle of 30° at D with level ground such that BD = 10 m Let BC = x m Now.6 m. 26. in right-angled triangle ABC cos 60° = ⇒ ⇒
BC x = AC 8
1 x = 2 8
x=4
Hence.9
.
. 12 4
Now. Swati observes two cars on the opposite side of the tower.(i)
Again.278
Mathematics
Example 26. and ∠RPA = ∠PAB = 60° ∠SPB = ∠PBQ = 45°
. We are given that tan θ = 5 3 and tan C = .. Let AB = h m and BC = x m. in right-angled ∆ABC tan C = h x
Fig. D be the two positions of the observer such that CD = 192 m.12 Now. Example 26. If their angles of depression are 45° and 60°. Let A and B be the position of the two cars. 26. 26. On walking 192 m towards the foot of the tower. Solution : Let PQ be a tower which is 100 m high. Find the height of the tower.20 : Standing on the top of a tower 100 high. the height of the tower is 180 m. Solution : Let AB be a tower and C.19 : At a point on level ground.11
3 h = 4 x ⇒ x= 4h 3 . find the distance between the two cars. Let the angle of depression of the car at A be 60° and for the car at B be 45° as shown in Fig. the angle θ made by the top of the tower with it is found to be such that tan θ = 5 12 . the tangent of the angle becomes 3/4. in right-angled ∆ABD tan θ = h 192 + x
5 h = 12 192 + 4 h 3 = ⇒ ⇒ ⇒
3h 576 + 4 h
36 h = 2880 + 20 h 16 h = 2880 h = 180
Hence.
26. If the angle of its vertex is 90°.732 = = = 57. Find the height of the tower. The foot of the ladder is at a distance of 3 m from the wall. After going 40 m towards the foot of the tower. an observer measures the angle of elevation of the top of the tower to be 60°.
. the distance between the two cars = AQ + QB = (100 + 57. One of the equal sides of an isosceles triangle is 18 2 m. find the length of the string. find the length of the base. Find the height of the tower. The angle of elevation of the top of the tower is 30° from a point 150 m away from its base. Two men are on either side of a cliff which is 80 m high.3 1. At a point 50 m away from the base of a tower. It makes an angle of 60° with the horizontal ground.Trigonometric Ratios of some Special Angles
279
In right-angled triangle PQB. 3. in right-angled ∆PQA PQ 100 tan 60° = QA = QA
Fig. Find the height of the kite.74 3 3 3
Hence. 4. assuming that there is no slack in the string.74 m CHECK YOUR PROGRESS 26. The angle of elevation of tower at a point is 45°. 7. Find the height of the tower. 6. If the string of kite makes an angle of 60° with a point on the ground. Find the distance between the two men.12
100 3 = QA
⇒ QA = 100 100 3 100 × 1. A ladder leaning against a vertical wall makes an angle of 60° with the ground. The string of a kite is 100 m long. They observe the angles of elevation of the top of the cliff to be 30° and 60° respectively. Find the length of the ladder. 2. PQ 100 tan 45° = QB = QB ⇒
100 1 = QB
⇒ QB = 100 m
Also. 8. A kite is flying at a height of 100 m from the level ground. assuming that there is no slack in the string.74) m = 157. 5. the angle of elevation becomes 60°.
The foot of the ladder is kept fixed on the same point of the level ground. LET US SUM UP 1. Ratio sin θ cos θ tan θ cot θ cosec θ sec θ θ 0° 30° 45° 60° 90°
0 1 0 not defined not defined 1
1 2 3 2 1 3
1 2 1 2 1 1
3 2 1 2
1 0 not defined 0 1 not defined
3
1 3 2 3 2
3
2 2 3
2 2
. The following are the relation between trigonometrical ratios for the complementary angles : (i) sin (90° – θ) = cos θ (ii) cos (90° – θ) = sin θ (iii) tan (90° – θ) = cot θ (iv) cosec (90° – θ) = sec θ (v) sec (90° – θ) = cosec θ (vi) cot (90° – θ) = tan θ 2. Find the height of the tower and its distance from the building.280
Mathematics
9. the angles of depression of the top and bottom of a tower are observed to be 45° and 60° respectively. From the top of a building 60 m high. Find the distance between these two rooms. 10. making an angle of 60° with the level ground. A ladder of length 4 m makes an angle of 30° with the level ground while leaning against a window of a room. The following table illustrates the values of trigonometric ratios for the angle θ such that 0° ≤ θ ≤ 90° : Table of Values of Trigonometric Ratios Trig. It is made to lean against a window of another room on its opposite side.
8. Two pillars of equal height stand on either side of a roadway which is 150 m wide. An observer standing 40 m from a building notices that the angles of elevation of the top and bottom of a flagstaff. 10. the elevation of the top of the pillars are 60° and 30°. between a kite and a point on the ground is 150 m. the angle of depression of two consecutive kilometer stones due east are found to be 30° and 60°. Find the height of the hill.282
Mathematics
6. determine the height of the kite. 9.
. If the string makes an angle θ with the horizontal plane such at sin θ = 4 15 . At a point on the roadway between the pillars. From the top of a hill. Determine the length of the ladder. 7. which is surmounted on the building. Find the height of the pillars and the position of the point. A ladder leaning against a vertical wall makes an angle θ with the ground such that tan θ = 4 3 . The length of a string without slack. The foot of the ladder is 3 m away from the wall. Find the height of the tower and the flag staff. are 60° and 45° respectively.
mean height of a group. In order to read and interpret these correctly. median score of a class or modal shoe size of a group.. To make readers acquainted with the methods of recording. etc. etc. we come across data in the form of tables. this record keeping was started by ancient kings to keep account of their warriors. They would also learn about the characteristics and limitations of these measures. etc. Sometimes. armoury and other fighting materials. throwing a die. the learner will get acquainted to the concept of probability as measure of uncertainty. on different aspects of a number of variables. Everyday through communication media like radio. expenditure and other resources. charts. the learners would be introduced to lesson on "Graphical Representation of Data". graphs. we are required to describe the data arithmetically. etc. etc.Data and Their Representation
287
Module 6
Statistics
Since ancient times. television. In the last lesson of this module. magazines. like describing mean age of a class. through games of chance like tossing a coin. The information presented is eye-catching and important. newspapers. to keep records of their income. In fact.
. they would be exposed to the lesson on "Data and their Representation". it has been the practice by the house holders. individuals. periodicals. shopkeepers. The learner will be introduced to these in the lesson on "Measures of Central Tendency". condensing and taking out relevant information from given records (or data).
3 EXPECTED BACKGROUND KNOWLEDGE
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knowledge of writing numbers in increasing/decreasing order finding averages of given numbers classification of numbers into different classes making/using tools to collect data.
27.1 INTRODUCTION In this lesson. recording and condensing data to take out relevant information from them. classwise literacy rate in different
. we shall first learn about statistics as science dealing with collection. class interveral. frequency.
27. frequency table. We shall also learn about some important concepts like classes.288
Mathematics
27
Data and Their Representation
27.4 STATISTICS AND STATISTICAL DATA Statistics is the science which started with collection of data about different aspects.2 OBJECTIVES After studying this lesson. etc. the learner will be able to :
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define statistics – both in singular and plural form differentiate between primary and secondary data differentiate between raw/ungrouped and grouped data and cite examples define frequency and cumulative frequency of a class condense raw data into frequency table form cumulative frequency tables define class-mark. cumulative frequency table. 27. class-interval. raw/ungrouped and grouped. We shall learn about different types of data-primary and secondary. class mark. class limits and true class limits in case of grouped data. It may be data regarding number of students presented in a school.
5 PRIMARY AND SECONDARY DATA Data are of two types – primary and secondary. the investigator uses the data collected by other investigators in some other context or published data from government records. In that case.1 1. income-tax rates etc. Let us try to arrange the data in ascending order.. 14. collected for some definite purpose. 12. 17. 10.. CHECK YOUR PROGRESS 27. 20. while in plural sense it means the numerical facts or data/observations collected with some specific purpose in view. 19. The data given in (i). The word statistics is used both in plural and singular sense.. 25. 19. 20. 12. 14. Since the data collected from other sources. (d) The data taken from other sources and not collected by the investigator himself are called ... 9. With the passage of time besides collection of data. data. 18. 25.. they are called secondary data. 25 . may have been collected with an objective different from what it is being used. Fill in blanks with suitable word(s) so that the following sentences give the proper meaning : (a) Statistics in plural sense refers to . (b) Statistics in singular sense refers to the . 10. 17. (c) The data which are collected by the investigator himself are called . do not give much information about the standard achievement of students..6 RAW/UNGROUPED AND GROUPED DATA Consider the marks obtained by 20 students of a class in a class test (out of 25): 15. 27. 25.. finances and other reasons. 15. books etc. 9.Data and Their Representation
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states of the country.. they are called primary data. In any study. 16. data. it is not possible for the investigator to collect data himself due to paucity of time. unemployment data. 25. 22. 8. 14. 27. Sometimes.(ii)
. interpretation and drawing of inferences came under the purview of statistics. journals. 12. it refers to the science of statistics. 25 . 22. 20. 6. 18. We will have 6. . 20... 25. 12. 14. their tabulations. 16. 8.. if the investigator himself is responsible for the collection of data according to the desired plan and objective. 25. In its singular form.(i)
The data in this form are called raw/ungrouped data.
eleven students have got 16 or more marks. To further simplify and overcome this difficulty. and so on. and so on. The above data is called a frequency distribution table for ungrouped data. The difference between the maximum and minium observations of the data is called the range. The arrayed data gives somewhat better perspective of the situation.
. we can arrange the data in tabular form as follows : Table–1 Marks 6 8 9 10 12 14 15 16 17 18 19 20 22 25 Total Number of Students 1 1 1 1 2 2 1 1 1 1 1 2 1 4 20
Table–1 shows the number of students getting a particular number of marks. We can also say that seven students have got 20 or more marks. A point may be made about arrayed data. two have secured 20 marks each. We can say that four students have secured 25 marks each. it is very difficult and time consuming to arrange data in arrayed form. When the number of observations is very large. They present the minimum and maximum of the data at a glance. Similarly. The quantity about which the observations (or data) are collected is called variable (or variate) and the number of times an observation is repeated is called the frequency of that observation.290
Mathematics
Data arranged in the above form (ii) are called arrayed data.
which admit fractional measurements like 2.5 and upper limit increased by 0. 13.5–16.5-10. height.5.5 is the True Lower class limit and 10. to have same class-interval (i. 20 is called the Lower Class Limit and 22 is called the Upper Class Limit.5. Thus for the class 7. 7. From this type of presentation. This can be done as follows for data in Table 2.65 m. the lower limit of each class is decreased by 0.5–19.5 16. etc. Some of these conclusions are : (i) The number of students getting marks between 23 and 25 is 4.5–10. This may happen with marks. 1. we see that the classes are non-overlapping because there are no fractional marks here. we can draw better conclusions about the data than before. 10.5 19.Data and Their Representation
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To bring out some more salient features of data.5 7.5 13.5–22.5 10. The changed limits are called True class Limits. length of the class)..5.5–13. Let us form the groups as follows : Table – 2 Classes Frequencies 5–7 1 8–10 3 11–13 2 14–16 4 17–19 3 20–22 3 23–25 4
Thus. Table 2 is called a frequency table for Grouped Data. The data presented in classes (or groups) is called Grouped Data. we need to have continuous classes so that all observations can be entered. 22.5..5 kg.e.e.5 – 13.5 – 16.5 – 19.7.5 is the True upper class limit.5–25. we further simplify the presentation of data in condensed form into classes (or groups). 19. etc. (ii) No student has got marks less than 5 (iii) Four students have got marks between 14 and 16 and so on.5 TOTAL Frequency 1 3 2 4 3 3 4 20
.5 – 10. but there may be variables like weight.5 – 7.5.5 22. Classes : 4. 27.7 NEED FOR CONTINUOUS CLASSES From Table 2.5
i. This changes the frequency distribution table as follows : Table – 3 Classes 4.5 – 22.5 – 25.5.5.5. for the class 20–22. 16. In such cases.5–7.
5–45. 12.. 15.5 is not included and in the class 22.5 are taken in respective next higher classes. . Write the lower and upper class limits for each of the following classes : (a) 11–15.. called class-mark of that class
∴ Class mark of a class =
Upper Class Limit + Lower Class Limit 2
or
True Upper Class Limit + True Lower Class Limit 2
∴ The class marks of distribution in Table 3 are
6. 27 3.5 7. 24 (ii) In each class.5 is not included. Explain one of the reasons for condensing raw data to grouped data. 3. 12. Thus.5–20. Differentiate between raw data and arrayed data 2.5.5–25. 25. 41–45 (b) 10.292
Mathematics
For making Table 3.. 16–20.5–10.5. the following two assumptions have been made: (i) The frequency in a class is centred at its mid-point. 27. 10. 9. 22.5 and 25. Give one example each for frequency distribution table for ungrouped data and grouped data.8 CONDENSING RAW DATA INTO FREQUENCY TABLE For condensing raw data into grouped data we follow the following steps : Step 1 : Arrange the data in ascending order and find the range of raw data. 4. Find the range of the following data : 6. Define : (a) class limits (b) true class limits (c) class interval (d) class-mark 5.5–15. 25. 40. in the class 7. CHECK YOUR PROGRESS 27. The observations 10.5. the upper class boundary is not included.
. 8.5.2 1. 18. What assumptions do you make while making a frequency table for grouped data ? 6. 14. 9.
Frequency Table for Marks obtained by Students
Total
. 15. we get the number classes as 7. 8–12. 20–24. till all observations are over. 19. 29. 7. the fifth one crossing the other four diagonally. 14. 12. 29. Table 4 Classes 4– 8– 12 – 16 – 20 – 24 – 28 – 8 12 16 20 24 28 32 Tally Marks ||| ||| |||| | |||| — | ||| Frequency 3 3 6 4 — 1 3 20 24–28. 14. 18. For example. Step 3 : Take each observation from raw data. 14. 12. 28–32
Let the classes be 4–8. 6 is denoted by tally marks as |||| |) Step 4 : Count the tally marks in each class to get the frequency of that class. 16. 30. 10. 4. 9. 30 Here the range of data is 26. 13. 19. Let us take an example to illustrate.1 : The marks obtained by 20 students in a class test are given below: 10. 15. 14. one at a time and put a tally mark (|) against the class to which the observation belongs.Data and Their Representation
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Step 2 : We decide upon number of classes to be formed. we took 7 classes. Solution : The data in ascending order is : 4. For that we remember the following : (a) There should be classes to accommodate the minimum and maximum of data (b) The classes should not be open ended (c) There is no definite rule for number of classes. the class interval being 3. 16. Taking the class size as 4. 16–20. The golden rule mostly used is "Not fewer than 5 and not more than 10 classes" For our example. 16. Step 5 : The resulting table is called Frequency Table. 9. 8. (Next integral value of 26 ) 4 12–16. 15. 25. This is obtained by dividing the range by class-interval and increase that to next integral value. 29. 29. (The tally marks are taken in bunches of five. 18. 25. 7 Construct a frequency distribution tables with a class size of 4. Example 27. 7. 16. 13. 8. 7. 15.
is called raw data. 52. classification and interpretation of data.
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. The data collected from the respondents "as it is". 5. 2.3 1.296
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CHECK YOUR PROGRESS 27. The number of observations lying in a particular class is called its frequency and the table showing classes with frequencies is called a frequency table. 25. In that case. The class marks and their corresponding frequencies are given below : Class mark : Frequency : 23 1 28 2 33 5 38 8 43 14 48 6 53 3 58 1
Form a cumulative frequency table from the above data. Form a frequency table and cumulative frequency table for the data on runs scored by a batsman in 20 innings. 72. one class being 0–20 12. 71. Enumerate the steps needed to condense raw data to grouped data. 9 3. 27. 45. Sometimes the classes have to be changed to make them continuous.11 LET US SUM UP
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Statistics is that branch of Mathematics which deals with collection. The data arranged in ascending/descending order is called "arrayed data" When raw data are arranged with frequencies. 70 102. In a class 10–15. The mid-points of classes are called class-marks. they are called grouped data. When the data are divided into groups/ classes. they are said to form a frequency table for ungrouped data. 15. 35. 69. 15. 15. The data collected by experimenter himself through his own designed tools is called primary data. Statistics has meaning both in singular and plural sense. the class limits are called true class limits. The data taken from other sources and not collected by the experimenter is called secondary data. 28. The classes have to be decided according to the range of data and size of class. 10 is called the lower limit and 15 is called the upper limit of that class. 2. 75. 79. 16. The total of frequencies of a particular class and of all classes prior to that is called the cumulative frequency of that class and the table showing cumulative frequencies is called a cumulative frequency table.
. data. Enumerate different steps needed to convert a given raw data to grouped data and to form a frequency table. 13. 21.. . 4 (i) Form a frequency table. (ix) The sum of frequencies of a class and all classes prior to that class is called ... 3... they are called . 4.. (v) The mid-points of a class are called . 27.. 18. (viii) The difference between the upper limit and lower limit is called the . 5. 9. the class limits are renamed as .. Find the range of the following raw data and put it as arrayed data : 7. 34. 13. data. 4. (vi) When the class limits are adjusted to make them continuous. . frequency of that class.. 48. is called . data and the table is called .. 18. .. (iii) When the raw data are arranged in ascending/descending order. etc.. 47. Define statistics as a term used in singular and a term used in plural sense. 8. 4. 23. magazines. 3. (x) Class size = Difference between successive . 31. The marks obtained by 30 students in a class test are given below : 10.. 4. 28. with one of the classes being 14–21 (21 not included) (ii) Form a cummulative frequency table for table formed in (i) (iii) Can you interpret the data and give some salient observations on it ?
.. 8.. 12. 24 Also change the above data into a frequency distribution of class size 3. 2.... 19.. which are not collected by investigatory himself should be used . 19. 21. 4. Fill in the blanks by appropriate words/phrases to make each of the following statements true : (i) The data which are taken from government records. 19. . . (iv) When the data are condensed in classes of equal size with frequencies... table. 13.. 46. 13.. 17.. 18. 16. 19 15. . 4. (ii) The data. Give examples to clarify the difference... 18.. they are called . (vii) The number of observations falling in a particular class is called its . 43. 42.. 19..Data and Their Representation
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TERMINAL EXERCISES 1.... What are primary data ? Why are they more reliable than secondary data ? 3. 25. 12... 12. one of the classes being 6–9 (9 not included) 6. 17 25. 5. 17...
velocity-time graph. it makes easier for the learner to understand the problem and attempt its solution. like (i) temperature-time graph
. etc. it makes the presentation eye-catching and more intelligible. They are (i) Bar graphs (ii) Histograms (iii) Frequency polygons (iv) Ogive (v) Pictographs (vi) Pie charts In this lesson. Histograms and Frequency polygons. we shall learn to read and draw Bar graphs. 28.Graphical Representation of Data
301
28
Graphical Representation of Data
28. Other form of graphs are beyond the scope of the present lesson.2 OBJECTIVES After studying this lesson. The learners can easily see the salient features of the data and interpret them. pressure-volume graph. when the data are presented pictorially (or graphically) before the learners. There are many forms of representing data graphically. and graphs related to day-to-day use. Similarly.1 INTRODUCTION Whenever verbal problems involving a certain situation is presented visually before the learners. the learner will be able to
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draw bar charts for given data draw a histogram and frequency polygon for given data read and interpret given bar charts and histograms read the relevant information from graphs relating to day-to-day activities. like temperature-time graph.
Step 6 : Mark the axes with proper labelling Let us take some examples to illustrate : Example 28. like the ones above. we choose the uniform (equal) width of bars and the uniform gap between the bars. 28.
28. Step 3 : Along the horizontal axis.
28.1 Construction of Bar Graphs For the construction of bar graphs.3 EXPECTED BACKGROUND KNOWLEDGE
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Knowledge of drawing and marking axes Knowledge of drawing rectangles and plotting points Practice of reading graphs. Step 5 : Calculate the heights of the bars.1 :
. Step 4 : Choose a suitable scale to determine the heights of the bars.1 : The number of trees planted by an agency in different years is given below : Years Number of trees planted 1997 400 1998 450 1999 700 2000 750 2001 900 2002 1500 Total 4700
Solution : The bar graph is given below in Fig. The scale is chosen according to the space available.302
Mathematics
(ii) velocity-time graph (iii) pressure volume graph.
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draw graph relating to day-to-day activities.4. we take the frequencies. It is a pictorial representation of the numerical data by a number of bars (rectangles) of uniform width erected vertically (or horizontally) with equal spacing between them. we take the values of the variables and along the vertical axis.4 BAR GRAPHS A bar graph is a graphical representation of frequency distributions of ungrouped data. we go through the following steps : Step 1 : We take a graph paper and draw two lines perpendicular to each other and call them horizontal and vertical axes. according to the space available. Step 2 : Along the horizontal axis. according to the scale chosen and draw the bars. 28. etc.
we represent years.
. from 1997–2002 and on OY we represent the number of trees planted. Step 4 : The height of the bars are calculated according to the number of trees. Examples 28. 28. 28. we start with 400 and marks points at equal intervals of 200. Step 3 : On OY.1
Step 1 : We draw two perpendicular lines OX and OY.2 : The data below shows the number of students present in different classes on a particular day : Classes Number of students present VI 35 VII 40 VIII 30 IX 40 X 50
Represent the above data by a bar graph. A kink (~) has been shown on the vertical axis showing that the marking on the vertical axis starts from zero but has been shown to start from 400 as the data needs. Step 2 : On OX.Graphical Representation of Data
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Fig.2. Solution : The bar graph for the above data is shown in Fig.
Graphical Representation of Data
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28. as in that year the height of the bar is maximum.5 : Read the bar graph given in Fig.3 and answer the following questions : (i) Which cause is responsible for maximum accidents in factories ? Which is for minimum ? (ii) Can you think of one of the "other" causes ? (iii) How many percent of accidents could have been avoided by timely action? Solution : (i) Delay in repairs is responsible for maximum (35%) of accidents.1. 2. 28. Let us take some examples and do the same. and answer the following questions : (i) In which year the maximum number of trees were planted ? (ii) What trend the number of trees planted show ? (iii) In which years the number of trees planted differ by 50 only ? Solution : After reading the bar graph.1 1. which is called interpreting bar graphs. (ii) The number of trees planted kept on increasing year after year (iii) (a) The years 1997 and 1998 (b) The years 1999 and 2000 Examples 28. the answers to the above questions are as follows : (i) The maximum number (1500) of trees were planted in the year 2002. Example 28. What are the steps needed to represent a data by a Bar Graph ?
. "Other causes" are responsible for minimum number of accidents (ii) Carelessness of workers (iii) (35 + 20)% or 55% accidents could have been avoided by taking steps for timely repairs and provision of equipment which can control electrical disturbances.4 : Read the bar graphs given in Fig.2 Interpretation of Bar graphs After drawing a bar graph. Enlist the possible forms of representing a data graphically. we can draw some conclusions. 28. CHECK YOUR PROGRESS 28.4.
28.4
(i) The names of two steel plants which produced maximum steel in the country during the time period. answer the following questions : Percentage 45 20 20 10 5
Fig. Given below are data on causes of strikes in mills : Causes (i) Non fulfillment of economic demands (ii) Overwork (iii) Rivalry in unions (iv) Non-congenial working conditions (v) Others Draw a bar graph depicting the above data.306
Mathematics
3. 5. draw a bar graph : Head Salary of Employees Travelling Allowance Rent of Premises Machinery and materials Other expenditure Percentage of Expenditure 45% 15% 20% 10% 10%
4. For the data on expenditure of a company over different heads. (ii) What percentage of steel was produced in "other" plants ?
. From the bar graph given below.
As the axis starts from 45–50. 6.. The mid-points of the first and the last classes are joined to the mid-points of the classes preceding and succeding respectively at zero frequency to complete the polygon. grouped frequency distributions.. we take one interval 40–45 before it and put a kink on axis before that Step 3 : Choose a suitable scale on the vertical axis to represent the frequency. 14 Step 4 : Draw the rectangles as shown in Fig. Step 2 : Along the horizontal axis. 50–55... Classes : 45–50 3 50–55 7 55–60 12 60–65 5 65–70 3 Total 30
Frequency :
Solution : For drawing a histogram we go through the steps similar to those of a bar graph. The area of the rectangles must be proportional to the frequencies of the respective classes. 2. For each class. we draw two perpendicular lines and call them horizontal and vertical axes. 4. we take classes of equal width : 45–50.e. a rectangle is drawn with base as width of the class and height as the class frequency.5...5 HISTOGRAMS AND FREQUENCY POLYGONS A histogram is a graphical representation of a continuous frequency distribution i.. 12. Let us illustrate these with the help of examples. . A frequency polygon is the join of the mid-points of the tops of the adjoining rectangles.6 : The following is the frequency distribution of weights of 30 students of class IX of a school. The class-intervals are taken along the horizontal axis and the respective class frequencies on the vertical axis using suitable scales on each axis. . i. 0. 28. Draw a histogram to represent the data. including vertical rectangles.. It is a graph. Durgapur and Bokaro steel plants ? 28... with a step of 2. with no space between the rectangles.
. They are given below : Step 1 : On a paper. It can start from 0 to 12.Graphical Representation of Data
307
(iii) The steel plant at Durgapur produced how much less steel than at Bokaro? (iv) What percentage of total steel under discussion was produced at Bhalai. Examples 28.e.
15) and F(275. 28. 16). 125.1 Reading a Histogram Let us explain it with the help of an example Example 28. 225. 175. E and F and complete the polygon as explained before The frequency polygon is given below :
Fig. D (175. 75. B. The following histogram shows the monthly wages (in rupees) of workers in a factory
. C(125. E(225. They are 25. 275 Step 3 : Plot the ordered pairs A (25. 5) Step 4 : Join the points A. Step 2 : Find the class-marks of different classes.5.Graphical Representation of Data
309
Example 28. 13).8 : Draw a frequency polygon for the following data : Pocket allowance (in rupees) Number of students 0–50 50–100 100–150 150–200 200–250 250–300
16
25
13
26
15
5
Solution : To draw a frequency polygon without-drawing a histogram we go through the following steps : Step 1 : Draw two lines perpendicular to each other. 26). B (75. 25).7
28.9. D. C.
Write various steps in the construction of a Histogram.310
Mathematics
Fig. (ii) The least wage is between Rs (4000 – 5000) and 4 workers are getting that. of workers earning them (iii) How many workers get a monthly wage of Rs. The corresponding figures for highest wage are Rs (9000 – 10000) and four workers get that (iii) 50 workers get a wage of Rs 8000 or less as Rs (4000 – 5000) – 4 workers Rs (5000 – 6000) – 10 workers Rs (6000 – 7000) – 12 workers Rs (7000 – 8000) – 24 workers Total – 50 CHECK YOUR PROGRESS 28. 28.2 1.8
(i) Find the maximum number of workers getting a wage. 8000 or less ? Solution : (i) The maximum number of workers is 25 getting wages between Rs (7000 – 8000). (ii) Find the least wage and highest wage with no. What is the difference between a bar graph and a histogram ? 2.
.
Interpret the data represented by the following histogram by answering the following questions :
Fig. 5.9
Shirt sale in a week in a shop.Graphical Representation of Data
311
3. of families in a locality 4-6 25 6-8 20 8-10 10-12 12-14 14-16 16-18 18-20 15 15 13 7 3 2
6. 28. Draw a histogram and a frequency polygon for the following grouped data: Annual income (in ten thousand rupees) No. Draw a frequency polygon for the data in Question 3 on a separate paper. of students 3 5 12 7 5 3
Also draw a frequency polygon for the above data on the same sheet 4. Draw a histogram for the following frequency distribution : Height of students 135–140 140–145 145–150 150–155 155–160 160–165 (in cm) No. (i) The least number of shirts were sold in which class ? (ii) The maximum number of shirts were sold in which class ? (iii) How many shirts were sold upto the 42 shirt size ? (iv) How many shirts of size 44–66 were sold ?
.
which can be referred to any time for reference. 28. We shall learn to draw these graphs and interpret them in the sections below : 28. 103). 99) in the rectangular system of coordinates. we are sometimes faced with graphs of other types... by line-segments. (23.. The graph has been obtained by joining the points corresponding to pairs.. When a patient is admitted in a hospital with fever the doctor/nurses prepare a temperature-time graph.
.10 : The body temperature of a patient admitted in a hospital with typhoid fever at different times of a day are given below : Time of the day Temperature (in °F) 7 hrs 102 9 hrs 103 11 hrs 104 13 hrs 103 15 hrs 101 17 hrs 100 19 hrs 99 21 hrs 100 23 hrs 99
Draw a graph to represent the above data. . 102). Similarly. the same trend was present.. 28.10
Note : While drawing the graph it has been assumed that during the time interval in between times. Solution : The graph of the above data is given in Fig.10..
Fig.6.1 Temperature-Time Graph-Reading and Construction Example 28. like (7.312
Mathematics
28.. the velocity time graph and pressure-volume graph are of day-to-day use. (9.6 GRAPHS RELATED TO DAY-TO-DAY ACTIVITIES In addition to histograms and frequency polygons...
(iii) The administered medicine suited the patient as the temperature constantly fell after that.. (7. During a journey from city A to city B by car the following data regarding the time and velocity of the car was recorded :
Time of the day (in hours) Velocity (in km/hour)
6
7
8
9
10
11
12
13
14
15
16
17
60
60
45
50
60
50
45
60
50
65
40
50
Represent the above data by a velocity time graph.
. the action of medicine had started ? (iii) What trend do you observe from the above graph ? Solution : (i) The temperature of the patient was highest at 11 hours and lowest at 19 hrs and 23 hrs. 60). Solution : As before the graph can be obtained by plotting the ordered pairs (6. whose temperature-time graph is shown in Fig. .12.Graphical Representation of Data
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Example 28. 28.2 Velocity Time Graph During a journey from one place to other. with the exception of period between 19 hrs and 21 hrs when it became slightly higher at 100°F but again fell after that 28.10.11 : If the medicine was given to the patient at 9 hours. (17.6. 50) in the rectangular system of coordinates and then by joining them by line-segments. answer the following questions : (i) At what time of the day was the temperature highest ? At what time lowest? (ii) After how much time. This can be very well shown by a velocity-time graph.. 65).. the speeds of vehicles keep on changing according to traffic congestions. (ii) The action of the medicine started 2 hours after the medicine was given as the temperature started falling after that. Let us illustrate it with the help of example : Example 28... 60). . (15.
At 15 hours (ii) The velocity was constant at 60 km/hour between 6 hours and 7 hours (iii) Between 8 hours to 10 hours (iv) Between 10 hours to 12 hours 2.92 km/hour. 28. The average speed of the car was
FG 60 + 60 + 45 + 50 + 60 + 50 + 45 + 60 + 50 + 65 + 40 + 50IJ H K 12
=
635 or 52. At what time duration of the day. What was the average speed of the car in the journey ? Solution : 1 (i) At 16 hours.11
Example 28. 28.13 : Read the velocity-time graph given in Fig. the velocity of the car (i) was lowest ? was highest ? (ii) constant (iii) went on increasing (iv) went on decreasing 2. 12
km/hour
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Fig.11 and reply the following questions : 1.
the corresponding pressure . (75. the volume . is there any relation between pressure and volume of the gas ? Let us see that from the following example : Example 28.6..3 Pressure-Volume Graph For a fixed quantity of a gas at a constant temperature. Solution :
Fig. Full in the blank : (i) As volume increases. 28.14 : The following data pertains to pressure and volume of a fixed quantity of gas : Pressure (p) (in Newton) Volume (v) (in cm3) 60 90 90 60 45 120 30 180 75 72
Draw a graph to represent the above data... 60). 90).12
The graph is obtained by joining the plot of the ordered pairs (60. . and answer the following questions : 1.. (90.15 : Read the above graph...12. (iii) Pressure × Volume = ... 28.Graphical Representation of Data
315
28. Example 28... 72) by free hand curve. given in Fig. (ii) As pressure decreases.
.
p = 54 Newtons
as can be seen from the graph at point A 3. The body temperatures of a patient admitted in a hospital are given below: Time of the day (in hours) Temperature (in °F) 8 9 10 11 12 13 14 99 15 99 16 100 17 98
103 104 105 102 102 100
3. when the pressure is 100 Newton ? Solution : 1 (i) Decreases (ii) Increases (iii) Constant = 5400 2. When p = 100 Newtons.3 Represent the data given in each of the questions below graphically : 1. What will be the volume. What will be the pressure when volume is 100 cm3 ? 3. The speeds of a car going from station A to station B at different times of the day are given below : Time of the day (in hours) Speed (in km/hour) 7 45 8 45 9 50 10 60 11 60 12 75 13 60 14 60 15 50
4. CHECK YOUR PROGRESS 28. The data on pressure and volume of a gas are given below : Pressure (in Newtons) Volume (in cm3) 60 40 80 30 50 48 30 80 40 60 20 120
. v = 54 cm3 as can be seen from the graph at the point B. For a town. We know that pv = 5400
∴ When volume = 100.316
Mathematics
2. the maximum temperature for the following months are given below : Months Maximum temperature (in °C) March April 35 38 May 38 June July August September October 42 45 40 38 35
2.
The graphical representation of data from day-to-day life is the join of points corresponding to ordered pairs represented by the data. For question No. 2 and answer the following question : (i) At what time of the day was the temperature of the patient maximum? (ii) If the medicine takes at least two hours to show the effect.00 40 10 8 60 12 11 5 40 8 12 4 50 7 15 3
4.Graphical Representation of Data
317
5. Interpret the data given in Question 1 and 2. Number of parcels received in a post office Weight of parcels (in kg) 1 2 3 4 5 6 15 7 4 120 0.0 25 9 7 80 1. TERMINAL EXERCISE
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Draw the bar graph for the following data in each case : 1. Read the graph of Question No.25 1. 1. Weight (in kg) No. Height of samplings (in m) 0. read the graph and reply the following questions : (i) Find the range of temperature for given months (ii) Which month had the least temperature ? (iii) Which month had the highest temperature ? (iv) In which month was the temperature less than 40°F ? (v) Can you predict the temperature for the next two months ? 6.75 2.50 1. of samplings 2. The graphical representations show the trends readily and at a glance only. of baskets of apples 3. at what time of the day was the medicine given ? LET US SUM UP
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Bar graphs are the graphical representation of ungrouped frequency data.75 18 8 5 150 1.
. Histograms and frequency polygons are the graphical representation of continuous grouped frequency data.5 No.
318
Mathematics
Draw a histogram and frequency polygon for the data in each case below: 5. Daily earning (in rupees) No. Read the graphs for Question Nos. The speed of the car at different times of the day is given below : Time of the day (in hour) Velocity (in km/hour) 7 45 9 50 11 60 13 65 15 70 17 60 19 55 21 40
Represent the above data by a velocity time graph and answer the following questions At what time was the velocity of the car (i) Maximum (ii) Minimum (iii) Between (50–60) km/hour (iv) Can you give hypothesis regarding the places where speed is extreme ? 10. The minimum temperatures of a town for a year are given below : Month Min. A man left New Delhi for Lucknow by car at 7 AM. Weight (in kg) 40–45 No. 8. of students in the class 6. 5 and 6 and interpret them. of workers 4 45–50 7 50–55 8 55–60 9 60–65 6 65–70 3
100-120 5
120-140 7
140-160 8
160-180 180-200 3 2
7. (in °C) Jan Feb March April 12 14 16 20 May June July 20 24 25 Aug Sep Oct Nov Dec 24 22 18 16 12
Draw a graph to represent the above data and interpret it. temp. 9. The following data pertains to a gas in a container : Pressure (in Newton) Volume (in cm3) 100 40 80 50 50 80 40 100 125 32 200 20
Represent the above data by a pressure volume graph. What relation do you find between pressure and volume from the data ?
.
(i)
. The name itself suggests that these measures describe the data arithmetically— for example.2 OBJECTIVES After studying this lesson. We will learn to calculate them by the formulae and will also learn properties of some of them. we will study about different measures of central tendency for ungrouped data and some of them for grouped data also.. xn are n observations. Represented graphically.. x2.. the graph of the observations of the group will be around and close to the measure of central tendency. ungrouped and grouped data calculate mean of raw. The measures with which we do that are called Arithmetical Descriptors of Data. . the learner will be able to
z z z z z
define mean of raw.3 MEAN OF RAW DATA If x1. In this lesson. x3. + x n n
. then their mean x is defined as Sum of all observations x = Number of observations or x =
x1 + x 2 + x 3 + ... 29.. they are called Measures of Central Tendency –central because these are the measures around which the measures of all the members of the group gather around. average height of a class.. median score of the group or modal collar size of a team. Because of the fact that these measures are representative of the group they represent. ungrouped and grouped data by ordinary and short-cut-methods define mode and median of raw data calculate mode and median of raw data cite properties of mean and median
29. average age of a group.320
Mathematics
29
Measures of Central Tendency
29.1 INTRODUCTION Sometimes we are required to describe the data arithmetically.
equal number of observations on its both sides.e. 5 2
th
and 6th observations
16 + 17 = 16. 9 . 6. 17. x. Example 29. 17. 9. 16. 6 . 17.6. 22. we have 4.. 21 Solution : As the median is 20. 12.10 : If the median of the following data is 20. 4. 25. 20.5 2 Example 29.
MEDIAN OF RAW DATA
Median is that value of the variate which divides the raw data into two equal halves. 22. 23. 16. 18. 15. 18. 23.e. find x : Median = 19. 24
. 20. 5. x. we have 16.Measures of Central Tendency
327
29. 18 The median is
∴ Median = 9
+ 1I FH 7 2 K th or 4th observation
(b) Here n is even (10) Arranging the data is ascending order.9 : Find the median of the following data : (a) 5. 7. the median is observations Let us illustrate it with examples. we first arrange the observations in ascending (or descending) order of magnitude and then follow the following : + 1I FH n 2 K th observation F nI Fn I (ii) If the number of observations n is even. 16. 3. we have 3. Arranging the data in ascending order. 21. 19. 19. 24. 18. 18. 16. when arranged in order (ascending or descending) For finding the median of raw data. i. the median is the average of H 2 K th and H 2 + 1K th (i) If the number of observations n is odd. and there are four observations more than 20 and four observations less than 20. 18. 10 (b) 16. 19. 15. 7 Solution : (a) Here the number of observations is odd (7 here). 12. 25 The median is average of 10 th and 2
∴
FH IK
FH 10 + 1I K th observations i. x must be 20 Arranging the data in ascending order. 17. 10.
. 13. that is more than 3.2 1.. 46 (c) 15.. 10 (b) 18. 16. 2 25. 11.
. 18. 3 12.. 20. 25. If the median of the following data arranged in ascending order is 17..328
Mathematics
CHECK YOUR PROGRESS 29. 25... 28.. 16. 18.. 12 Solution : Here the distinct observation are 8 1 11 3 12 2 13 1 15 4 16 2 18 1 19 1 with their respectively frequencies as Thus. (ii) When the number of observations n is odd. 25..7 MODE OF RAW DATA Mode is that value of the variable which occurs most frequently in the data. observation... x... 15. 19. 11. 13. find x 12.. 10. 13. 19.. 17.. 19.. 8... the observation 15 has maximum frequency (4 here)
∴ Mode of the data is 15
Example 29. 16... x = 11. 12. 13.. Fill in the blanks suitably : (i) Median is that value of the variable . 12. The method involved is to find the distinct values of observations in the data and find the frequency of occurrence of each. 20. 11.. The observation with the maximum frequency is called mode. 15. 28. 18. 16. Example 29.11 : Find the mode of the following data : 15. 11. 16. 13. 18. 1 20. 11. Therefore. 20 (d) 6. 25. 18. median is given by . 12....12. observation (iii) When the number of observations n is even... 15.. 52... 13. 20.. 19.. 12. 15. 20 Solution : The distinct values of observations are 11. 72.. 15.. 42. 11... 9. 1 3 18. 2. 80. Let us take some examples to illustrate. median is given by .. 18. 7. x.. 22 29. 15. 2 19. 69.. then 11 has to have maximum frequency. 17 3. Find the value of x so that the mode of the data is 11 12.. Find the median of the following : (a) 5. 1 x with frequencies The mode of data is 11.
it divides the data exactly in two halves. LET US SUM UP
z z
The three measures of central tendency are mean.e. where k is the number of classes and xi′s are class marks
i =1 (ii) x = A + k
∑ fi d i
i =1
×c
∑ fi
xi − A c
where A is the assumed mean di = and c is the length of interval.
z
Median of raw data is given by nI FH 2 K th observation if the number of observation (n) is odd.. Median is the middlemost observation of the data i. 29. unlike mean. Median may not lie in the data itself. Median. 2. is not affected by extreme values. 4. median and mode. F nI Fn I (ii) average of H 2 K th and H 2 + 1K th observations of n is even. If each observation in the data is divided by some non-zero constant. Median can be determined graphically also while mean can not. If each observation in the data is multiplied by some constant. Mean is that value in the group of observations which describes it arithmetically.
z z
Mean x of raw data is given by x = Mean x of grouped data is given by
i =1 k
i =1
∑ xi
n
n
(i) x =
∑ fi x i
i =1
k
∑ fi
k
.9 PROPERTIES OF MEDIAN 1. 3. then the mean gets multiplied by the same constant. 6. (i)
z
Mode is that value of variate which occurs most frequently.330
Mathematics
5.
. then the mean gets divided by the same non-zero constant. when arranged in order (ascending or descending).
is not certain to occur. the origin of the theory of probability has been the study of games of chance and is connected with events like tossing a coin. probability measures the degree of uncertainty involved and hence measures the degree of certainty of occurrence of events. If we go by the dictionary meaning of the word "probable" it says "likely though not certain". "likely" or "probably" convey the meaning that the event. Thus. selecting a card at random from a well shuffled pack of cards. throwing a die. Under the head of probability.2 OBJECTIVES After studying this lesson. throwing a die. he will come today.Introduction to Probability
333
30
Introduction to Probability
30. Interestingly. Ram is likely to get good marks in his examination. we sometimes make the following statements : It may be foggy today. we are talking about. we will study the measure of this uncertainty under given conditions. The train may reach in time. Probably.
. an outcome and an event define different types of events define probability of occurrence of an event solve problems based on tossing a coin.1 INTRODUCTION In our day-to-day life. The usage of terms of the type "may". the learner will be able to :
z z z z
define an experiment. The word probability has been derived from the same and is intended to be used to have a measure of uncertainty in making statements or taking decisions in such situations. and drawing a card from a well shuffled pack of cards. 30.
5 TYPES OF EVENTS Let us now study about some types of events and sample space : 30.4 Event All the possible outcomes of a trial are called events. 4. • •• ••• four fundamental operations on numbers terminology connected with "playing cards"
30. • . when a coin is tossed.4.4.1 Sample space The collection of all possible outcomes of an experiment constitute its Sample Space. If both the faces of a coin have head. We will take them one by one below 30.• • . 30. When a die is thrown.2 Trial and Outcome Performing an experiment once is called a trial and the result of the trial is called as outcome. For example. ••.1 Experiment An activity which ends in some well defined result is called an experiment. tossing a fair coin or throwing a fair die Note : By the word fair.• • . 30. the result will always be head. • • • .5. 3. (sometimes •••
. ••.4. it will be desirable to have knowledge of basic terms being used in the text. 2. •••.3 Random Experiment An experiment in which all possible outcomes are known before but will be the result of a trial can not be surely predicted.4 SOME DEFINITIONS Before saying something about probability.3 EXPECTED BACKGROUND KNOWLEDGE We assume that the learner is already familiar with :
z z z z
the terms connected with coin-head and tail •• •• •• ) the terms connected with a dice the dots on six faces as (•. Drawing a card from a pack of cards. Any specific of these is called an event 30. 5 and 6) . the sample space consists of Head (H) and Tail (T). •••. otherwise we can always predict the result.334
Mathematics
30. we mean that no trick has been played with the object being tossed. the sample space consists of •.4. For example. 30. throwing a die and tossing a coin are examples of experiments. is called a random experiment. ••• •• •• called 1.
For example. 30.5. when P(X) = 1.
.4 Equally Likely Events Two events are said to be equally likely events if they have equal chance of occurrence For example. the favourable cases to the event "occurrence of an odd number. called probability of non-occurrence of an event. the event is said to be a Sure event.. getting a number less than 7. we can say that
When do we say that the probability of occurrence of event x is zero ? When there is no outcome favourable to the event. in a fair toss of a coin. For example. In that case.. are called favourable to that event. 3. 5.3 Mutually Exclusive Events If two events cannot occur together.(ii)
m≤n 0 ≤ P(X) ≤ 1
From (i) and (ii). For example getting a number greater than 6 when a die is thrown once is an impossible event. For example. In a throw of a coin. all the possible out comes are favourable to the event. Let us now introduce another concept.Introduction to Probability
335
30. head and tail are mutually exclusive events.(i) . In the case.6 PROBABILITY OF OCCURRENCE OF AN EVENT
The probability P(E) of occurrence of an event E is defined as P(E) = Number of outcomes favourable to the event Total number of possible outcomes 1 . they are said to be mutually exclusive events. when a die is thrown once is a sure event. 30...5. 2
Thus. the probability of occurrence of a head (H) when a coin is tossed is
Suppose there are m cases (outcomes) which are favourable to an event capital X and n is the total number of outcomes. occurrence of head and tail are equally likely events. Such an event is called an Impossible event.5. then the probability of occurrence of the event X is given by P(X) = We know that
m n
. when a die is thrown are 1. 30.2 Favourable Event The cases which ensure the occurrence of an event.
Example 30. F2. Example 30. F2. 3. Then the probability of non-occurrence of the event X. M3.4 : A die is tossed once. F3.1 : Write the sample space if a die is thrown once. the probability of non-occurrence of an event equals [1 – Probability of occurrence of the event] Note : X is called an event complementary to event X.2 : 3 males and 4 females appear for an interview.. The sample space consists of M1. Write the sample space. •••. M2. ••. ••• • • • • ••
This is the required sample space.
∴ The probability of occurrence of a tail = 1 2
Example 30. 5.336
Mathematics
m If P(X) = n .
Solution : We know that when a die is thrown once. Examples 30. the sample space is H. Solution : Let the three male candidates be denoted by M1. of which one candidate is to be selected. F4.e. the likely outcomes are •• •• or 1.3 : Find the probability of occurrence of a tail (T) when an unbiased coin is tossed once. 2. •• . 4. . . called P( X) is given by P( X) = n−m = n − m n n n
m = 1− = 1 – P(X) n i. F1. 6
. then there are (n – m) outcomes which are not favourable to the occurrence of X. M3 and four female candidates be denoted by F1. the sample space is 1. 4. find the probability of occurrence of (i) 2 (ii) an even number Solution : When a die is tossed. 6 •. 2. M2. 5. Let us now take some examples to illustrate. T There is only one outcome favourable to a tail (T). 3. F3 and F4. Solution : When a coin is tossed once.
A group of people have following distribution of members according to age Number 5 10 15 Age 30 years and below (31 – 40) years (41 – 60) years
A person is selected at random from the group. Find the probability of selection of (a) a boy (b) a girl for Question 2(iii) 5. An urn contains 6 black and 4 red balls. Find the probability of selecting a (i) black ball (ii) red ball 6.Introduction to Probability
339
(iii) Event and outcome (iv) Types of events (v) Probability fo occurrence of an event 2. Write the sample space for the following : (i) when a fair coin is tossed once (ii) when a fair die is tossed once (iii) when 6 boys and 5 girls appear for a selection. 3.
. Find the probability of occurrence of (a) a multiple of 2 when a die is tossed (b) a multiple of 3 when a die is tossed (c) a number greater than 4 when a die is tossed (d) a number greater than 6 when a die is tossed (e) a multiple of 4 which is less than 3 4. One ball is selected from the urn at random. Find the probability of getting a person with age (i) less than or equal to 40 yrs (ii) beyond 40 years (iii) less than or equal to 30 years.
Find the probability of the following events when a die is tossed once: (i) occurrence of 4 (ii) occurrence of a number greater than 2 (iii) occurrence of a number less than 3 (iv) occurrence of multiple of 4 (v) occurrence of a number greater than 6 (vi) occurrence of a number greater than or equal to 1 (vii) occurrence of a number which is prime. An experiment in which the sample space of outcomes is known but which one will occur cannot be predicted is called a random experiment.340
Mathematics
LET US SUM UP
z z
Probability is a measure of index of uncertainty of occurrence of an event. Outcomes which ensure the occurrence of an event are called favourable cases to the occurrence of that event The events which cannot occur together are called mutually exclusive events Events which have equal chances of occurrence are called equally likely events.
. Write the sample space for (i) selection of a candidate having 8 male and 6 female candidates (ii) when a die is tossed (iii) An urn with 4 red and 3 black balls 3. Probability of occurrence of an event. =
Number of outcomes favourable to the event Total Number of outcomes
z z z
z z z
TERMINAL EXERCISE 1. Performing an experiment once is called a trial and its result is called an outcome. 2. Define probability of occurrence of an event. The collection of all possible outcomes is called sample space. An activity that ends in some well defined result is called an experiment. Define sample space.
Find the probability of selection of a (i) red ball (ii) black ball (iii) green ball (iv) non-red ball (v) non black ball 5.Introduction to Probability
341
4. 6 boys and 2 girls have been called. Find the probability of selection of a (i) boy (ii) girl (using the concept of complementary events)
. A ball is selected at random. Find the probability of a non-leap year (365 days) containing 53 Sundays. An urn contains 6 red. In an interview for the selection of a candidate. 6. 4 black and 5 green balls.
. probability (Jack of spades) = 1 52 1 52
♠ ♥ ♦ ♣
— Black colour — Red colour — Red colour — Black colour
. we roll a die and ask for a number. if that number comes on the top. 2. The plural of the word "die" is "dice". ∴ Probability (of selection of a card) = Particularly.
Fig. Jack.1). each of the 52 cards have equal chance of being drawn. each number 1 to 6 is equally likely to come on the top (occur) A PACK OF CARDS A pack of card consists of 52 cards divided into 4 suits (colours) called (i) Spades (ii) Hearts (iii) Diamonds (iv) Clubs
Each suit consists of 13 cards Ace. If you ) has carefully see the figure on the side. Queen and King are called face cards If we draw a card from a well shuffled pack of cards. Whenever.1
Remmeber. the attempt is described as "win". 3. he would have won in that trial. 30. . for an unbiassed die. Jack. (See Fig. the number 6 ( come on the top and if the player had asked for that number.. 30. Queen and King.. 10.Introduction to Probability
343
APPENDIX A DIE A die is a stable cube with its six faces marked with dots from 1 to 6.
1. Give the following informations on your answer sheet.Practice Work
205
Secondary Course Mathematics
Practice Work–Algebra
Maximum Marks : 25 Instructions : 1. Which of the following is not an integer ? (A) – 4 (B) – 25 (C) 36 (D)
4 5
1
1
. Answer all the questions on a separate sheet of paper. Do not send your practice work to National Institute of Open Schooling. 2 Fractions are called : (A) positive integers (B) positive rationals (C) negative rationals (D) negative integers 2.
The sum of the first four and the first five terms of an A. find the common difference.P. Simplify :
3. nth and rth term of an A. 2 x −1 − 2 x− 2
2
10. are 26 and 40 respectively. prove that x(n – r) + y(r – m) + z(m – n) = 0. If mth. are x.2 x − 2 x −1 .P. If the length is increased by 2 m and breadth decreased by 2 m.Practice Work
207
9. 4
12. Find the dimensions of the rectangle. If the first term is 2. y and z respectively. the area is decreased by 8 m2. 2 11. The area of a rectangle gets increased by 4 m2 if its length is increased by 4 m and breadth decreased by 2 m. 6
.
1. Get your practice work checked by the subject teacher at your study centre so that you get positive feedback about your performance. The marked price of the item is : 1 (A) Rs 420 (B) Rs 400 (C) Rs 380 (D) Rs 385
.Practice Work
289
Secondary Course Mathematics
Practice Work–Commercial Mathematics
Maximum Marks : 25 Instructions : 1. A customer has to pay Rs 20 more. Answer all the questions on a separate sheet of paper. Give the following informations on your answer sheet. Rate of discount given by the shop is : 1 (A) 15% (B) 20% (C) 25% (D) 30% 2. if the discount is reduced from 20% to 15%.
z z z z z
Time : 45 Minutes
Name Enrolment number Subject Topic of practice work Address
3. 2. A saree is available for Rs 450 whose marked price is Rs 600.
Do not send your practice work to National Institute of Open Schooling.
5 : 2.50 (C) Rs 150. (C) 3% per quarter interest compounded quarterly. Which one is a better investment ? (A) 12% per annum interest compounded yearly. A shopkeeper allows 25% discount on his articles in a sale. A T. 6 1
. However he has to charge 10% sales tax on goods sold. A watch is sold for Rs 405 at a loss of 10%. The population of a village in the year 2000 was 8000. Find the total population at the end of the year 2002. if the interest is compounded yearly. The difference between the compound interest and simple interest as a certain sum of money at 10% per annum for 2 years is Rs 155.5 cm. Its population increased by 5% during the year 2001 and 4% during the year 2002. If the perimeter is 13. Find the marked price of an article. set is available for Rs 7500 cash or Rs 2000 as cash down payment followed by 6 monthly instalments of Rs 1000 each. if a customer has to pay Rs 330 inclusive of sales tax. (D) 1% per month interest compounded monthly. Find the sum. Find the profit percent. the cost price of the article was : 1 (A) Rs 175. 2 8.50 on transportation and had a gain of 25%. find the length of each side. 6. The cost price of 15 pens is equal to selling price of 12 pens. (B) 6% per half year interest compounded half yearly.00 4. 4 12. A man sold an article for Rs 187. If he had spent Rs 12. A sum of money becomes Rs 2000 in 2 years and Rs 2250 in 4 years at the same rate of simple interest.290
Mathematics
3. The sum is : 1 (A) Rs 2000 (B) Rs 1850 (C) Rs 1750 (D) Rs 1600 5.50. Another watch is sold for Rs 540 at a gain of 8%. Find the total gain or loss in both transactions.V. The sides of a triangle are in the ratio 1 : 1. Find the rate of interest charged under instalment plan.00 (D) Rs 140. 2 9. 2 11. 2 10. 2 7.00 (B) Rs 137. to buy it.
B have co-ordinates (2. PT is a tangent to the circle at T. x) respectively. then prove that 4AB2 = AC2 + BD2. Find AD. 3) and (4. The quadrilateral formed by joining the mid points of the pair of adjacent sides of a rectangle is a : 1 (A) rectangle (B) square (C) rhombus (D) trapezium 4. In ∆ABC. Then ∠ABT is : 1 (A) 110° (B) 70° (C) 45° (D) 25° 5. the possible value of x is : 1 (A) – 6 (B) 0 (C) 9 (D) 12 6. AB = 10 cm and DE is parallel to BC such that AE =
1 AC.Practice Work
191
3. If ABCD is a rhombus. If ∠BTA = 45° and ∠PTB = 70°. In the adjacent figure. Two points A. 2 4
7.
2
. If AB2 = 13.
2 9. (10. Find the co-ordinates of the point on x-axis which is equidistant from the points whose co-ordinates are (3. Find the co-ordinates of its centroid. 3). 8) and (9. If co-ordinates of one of the end points of the segment are (6. The co-ordinates of the vertices of a triangle are (3. 5). Prove that parallelograms on equal (or same) bases and between the same parallels are
. –1). 7) and (5. AD⊥BC. 2 10. In an acute angled triangle ABC. 4 6
12. BD equal in area. then find the co-ordinates of the other end point. Prove that AC2 = AB2 + BC2 – 2BC. 2 11.192
Mathematics
8. The co-ordinates of the mid-point of a line segment are (2. 3). 5).
2.Secondary Course Mathematics
Practice Work–Mensuration
Maximum Marks : 25 Instructions : 1. Do not send your practice work to National Institute of Open Schooling. Answer all the questions on a separate sheet of paper. then its altitude is : (A)
3 2 a 2
1
(B)
3 2a 2
3 a 2
(C)
(D)
3 2a
. If 'a' is the side of an equilateral triangle. 1 Give the following informations on your answer sheet.
The sides of a triangle are 30 cm. whose volume equals that of a cuboid of dimensions 63 cm × 56 m × 21 cm is : 1 (A) 21 cm (B) 28 cm (C) 36 cm (D) 42 cm 6. Find the area of the rhombus. m at the rate of 6 km an hour ? 2 7. The radii of two right circular cylinder are in the ratio 3 : 4 and their heights are in the ratio 5 : 2. The height of the triangle is : 1 (A) 5 cm (B) 4 cm (C) 3 cm (D) 2 cm 5. One of the diagonals and a side of a rhombus are 8 cm and 5 cm respectively. 2 8. Find the ratio of their curved surface areas. The base of an isosceles triangle is 8 cm and one of the equal sides is 5 cm. How long will a man take to walk round the boundary of a square field of area 90000 sq. Three equal cubes are placed end-to-end in a row. The edge of a cube. 2
. Find the ratio of the total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. Its area is : (A) 120 cm2 (B) 600 cm2 (C) 750 cm2 (D) 1200 cm2 3. 2 9.238
Mathematics
2. The perimeter of a square of side 'l' is given by : (A) l2 (B) 4l (C) l 2 (D) 2l
1
1
4. 40 cm and 50 cm.
Find the height of the room.5 mm Find the length of the wire. 4 12. Find the number of marbles that should be dropped.4 cm diameter are dropped into a cylindrical vessel of diameter 7 cm containing water. A room is 7 m long and 4 cm wide. if the water level in the cylinder rises by 5. A cubic meter of iron is melted to form a wire of diameter 3.Practice Work
239
10.
FG Use π = 22 IJ H 7K
6
. It has two doors of 2 m × 1 m each and two windows 1 m × 1m each. The cost of painting the walls at the rate of Rs 5 per square meter is Rs 300.6 cm. Spherical marbles of 1. 2 11. | 677.169 | 1 |
An Overview of our Calculus Material
We divide our calculus material into two categories: The first is the everyday classroom activities. This includes teaching and testing. The second is preparing for the AP exam. While teachers are encouraged to incorporate AP exam prep in their everyday teaching, sometimes you have to have students do many mundane problems to nail down differentiation and integration techniques. These problems are covereded in the everyday materials.
Everyday classroom materials:
AB student manual and BC student manual. These manuals, described below, are great for an everyday workbook for students. You teach directly from it. It effectively will replace the textbook.
AB and BC exams and quizzes. These are exams and quizzes that you give that will cover and entire topic. While some AP type problems are included, most of the exams are what you traditionally give to determine whether students have mastered a topic.
On typical AP questions, many concepts can be addressed. A problem giving a function might a) ask for the equation of a tangent line to the function at a point as well as b) ask for the area under the curve. While students might be able to part a) in October, then cannot do part b) until they have studied integration which might be in February. Still, these are realistic AP type problems, so we offer a good number of problems and examples with this approach.
Demystifying the AP Exam - Revision coming soon
You can DOWNLOAD any version of Demystifying for FREE
Demystifying the AB Exam - free response
Demystifying the AB Exam - multiple choice
Demystifying the BC Exam - free response
Demystifying the BC Exam - multiple choice
Ripped From the Headlines - 38 free response type questions that use real-life type problems inspired from news articles from newspaper and the Internet.
Review Topics for the exam - a series of phrases that students see on the AP exam. These focus on what to do rather than how to do them. For instance, if a student is asked to find a critical point, he must know what that means. These are in both written form and classroom slides.
These problems use the approach that nothing is tested before it is taught. For example, problems involving differentiation techniques will not involve integrals or applications of differentiation. So it is a continually building processs. While students will be asked to problems in the global approach on the AP exam, this approach allows teachers to focus in on a specific calculus topic.
Have some fun while reviewing for the AP exam with our very popular Clue Game and Jeopardy.
Clue. Based on the old Parkers Brothers game that most kids played when they were kids, solve a mystery of a buried treasure (who buried it, where it was buried, and what the treasure is) by a process of elimination. Students have to correctly solve 87 calculus problems to solve the mystery. The game can be played individually or in teams.
Jeopardy. Play the popular game show with topics like Limits, Differentiation, What's my Area, and Taylor's World to name a few. There is a Jeopardy, Double Jeopardy, Triple Jeopardy, and BC Jeopardy and the answers and questions are also available in written form. This game uses a browser like Firefox or Chrome, and many others, but does not require an internet connection.
Evolution of the A.P. Calculus Manuals
I started teaching A.P. calculus AB in 1993 after years of teaching a college preparatory calculus course. The textbook was Larson and Hostetler, 3rd edition. It has been my experience that we buy expensive and heavy textbooks for the students but find that students simply have trouble learning from them. Students see the textbook as a source of problems, but not a teaching tool.
As is my practice, I developed worksheets that would lead students through certain types of problems. If a student was absent and wanted the work that we missed, I found it easier just to hand the student the worksheet rather than refer him to the text.
Since our periods were short, I also found that rather than assign many problems from the textbook, I would assign what I felt to be essential problems I needed to get through and simply added them to the worksheet. Over a period of six years, I developed worksheets for just about every topic in A.P. calculus.
However, I found myself spending most of free periods in front of the copy machine. One year, I got the brainstorm to put all of these worksheets together in one comprehensive manual so the reproducing could be done all at once and the students would receive the manual at the start of school. I did so, adding a table of contents, and explaining where the concepts taught in the manual were to be found in the textbook we were using.
Over a summer, I copied these 225 page manuals and put them together. I used them first in the 1999-2000 school year and had unbelievable testimonials from students. I have been using them ever since. A textbook is distributed at the beginning of the school year and students are encouraged to take it home and use it as a reference if they need more insight to a proof of a theorem or simply want more practice problems.
Taking over the BC calculus class in 2000, I spent the summer writing a BC manual and my students have used it for seven years as well. The reactions were even stronger.
My school moved to copying service. At the start of the summer, I requested that these manuals be published. They came back to me in August, double sided, and with holes punched in them. They were given to the students on day one and we complete the entire manual. In using a text, we rarely complete half of it. The students are allowed to keep the manual at the end of the course. Many have told me that they have taken it to college with them and it remains a valuable source of review for them. If students are diligent and do all the classwork and homework in the appropriate places in the manual, they have a complete record of the entire AP calculus course.
Several years later, I decided to write an accompanying solution manual. I do not like answer keys as there is too much fumbling trying to find the correct page. My thought was to have a solution manual that was exactly like the manual I give the students, except that the solutions to the problems are written in the spaces. So the teacher can teach directly out of the solution manual and not have to handle two separate documents.
The manuals changed little over the years with the exception of correcting some errors. But I felt that the look of the manual, using much older technology, had become dated. Graphics were blocky and hard to understand. Because the manual used a much older version of Microsoft Word that was no longer supported, I ended up typing the manual again and it took more than 6 months to re-create it. But it is a huge improvement in terms of looks and content. Both manuals now cover all topics in the revised AB and BC curriculum for 2016-2017. Each manual has an Essentials section and a Non-Essential section that covers topics if you have more time to spend. The AB manual has a brief review of important precalculus topics and the BC manual has a review of essential AB topics. There is a Combination manual available as well for schools that teach AB/BC as one course.
Use of the A.P. Calculus Manuals
The manual is not a textbook. It is not intended to be one. It simply presents topics and walks students through a series of classwork problems. The teacher does the teaching of the problem. The students take notes in the space provided. There is little or no attempt to prove theorems. You can still do that on your own. The manual provides you and your students sample problems, which will touch upon every concept that is covered in the A.P. curriculum. (and some that has been eliminated from the curriculum as well). To reiterate, this manual still needs a good teacher - you. You cannot just hand the student the manual and expect him or her to read it and understand it.
There are also homework problems for every concept. Again, space is provided for students to complete them. There are enough challenging problems but the idea is to give students enough problems to test whether they understand a concept without overburdening them. I am not smart enough to write a math book. In coming up with sample problems and homework problems, I used a number of calculus textbooks for ideas. While one cannot copyright "take the derivative of 2x," there are word problems that I adapted from these sources. In most cases, I attempted to change the problem slightly by altering a given value or the context of the problem.
By downloading the free manuals (or purchasing paper copies), you have the right to make as many copies as you wish as long as you use them for face-to-face classroom use.
18. Geometric and p-series
19. The Integral Test
20. Comparison of series
21. Alternating series
22. Ratio and root tests
23. Series Convergence Review
24. Power Series
25. Taylor & Maclaurin Series
26. Manipulation of Series
II. Non-Essentials
0a. Marginal Analysis/Demand Elasticity
1a. Epsilon Delta Definition of Limits
2a. Power of Sines & Cosines
3a. Partial Fractions Advances
5a. Simpson's Method
7a. Work
7b. Fluid Force
7c. Center of Mass
27. First Order Differential Equations
III. AB Review
R1. Basic Differentiation
R2. Linear Approximation
R3. Limits/Continuity/Differentiation
R4. Related Rates
R5. Function Analysis
R6. Integration Techniques
R7. Definite Integral/Accumulation
R8. Riemann Sums
R9. Fundamental Theorem of Calculus
R10. Straight-Line Motion
R11. Area and Volume
R12. Differential Equations/Growth & Decay
Pricing
We are happy to offer these manuals for no charge. Go to the Calc AB Manual or Calc BC Manual page to download any or all sections of either the AB or BC manual. Each section has approximately 5 topics. They are in PDF format.
The manual is also available in paper format as well as the solution manual for both AB and BC calculus. Go to the page on paper manual and solution manuals. These do have a cost associated with them. You can order them from the same pages. Go to Purchase Options to order them in combination for less money.
Every teacher needs an answer key. You can either solve the problems in your own copy of the student manual, or purchase the answer key in paper format. The answer key has the same page numbering as the student manual to make it easy to keep your students 'on the same page'. | 677.169 | 1 |
Scoring Guide and Sample Student Work Select a score point in the table below to view the sample student response.
The student response demonstrates an exemplary understanding of the Patterns, Relations, and Algebra concepts involved in representing real situations and mathematical relationships with tables and with rules using words and symbols. The student determines if a scenario is possible based on a model of a real situation.
The student response demonstrates a good understanding of the Patterns, Relations, and Algebra concepts involved in representing real situations and mathematical relationships with tables and with rules using words and symbols. Although there is significant evidence that the student was able to recognize and apply the concepts involved, some aspect of the response is flawed. As a result the response merits 3 points.
The student response demonstrates a fair understanding of the Patterns, Relations, and Algebra concepts involved in representing real situations and mathematical relationships with tables and with rules using words and symbols. While some aspects of the task are completed correctly, others are not. The mixed evidence provided by the student merits 2 points.
The student response demonstrates a minimal understanding of the Patterns, Relations, and Algebra concepts involved in representing real situations and mathematical relationships with tables and with rules using words and symbols.
The student response contains insufficient evidence of an understanding of the Patterns, Relations, and Algebra concepts involved in representing real situations and mathematical relationships with tables and with rules using words and symbols to merit any points. | 677.169 | 1 |
This text uses the concepts usually taught in the first semester of a modern abstract algebra course to illuminate classical number theory: theorems on primitive roots, quadratic Diophantine equations, and the Fermat conjecture for exponents three and four. The text contains abundant numerical examples and a particularly helpful collection of exercises,... more...
An excellent introduction to the basics of algebraic number theory, this concise, well-written volume examines Gaussian primes; polynomials over a field; algebraic number fields; and algebraic integers and integral bases. After establishing a firm introductory foundation, the text explores the uses of arithmetic in algebraic number fields; the fundamental... more...
An imaginative introduction to number theory, this unique approach employs a pair of fictional characters, Ant and Gnam. Ant leads Gnam through a variety of theories, and together, they put the theories into action?applying linear diophantine equations to football scoring, using a black-magic device to simplify problems in modular structures, and... more...
Classic two-part work now available in a single volume assumes no prior theoretical knowledge on reader's part and develops the subject fully. Volume I is a suitable first course text for advanced undergraduate and beginning graduate students. Volume II requires a much higher level of mathematical maturity, including a working knowledge of the theory... more...
Unusually clear and interesting classic covers real numbers and sequences, foundations of the theory of infinite series and development of the theory (series of valuable terms, Euler's summation formula, asymptotic expansions, other topics). Includes exercises. more...
This text covers the basics of algebraic number theory, including divisibility theory in principal ideal domains, the unit theorem, finiteness of the class number, and Hilbert ramification theory. 1970 edition. more...
An engaging treatment of an 800-year-old problem explores the occurrence of Fibonacci numbers in number theory, continued fractions, and geometry. Its entertaining style will appeal to recreational readers and students alike. more...
About This Book This book will help high school math students at all learning levels understand basic mathematics. Students will develop the skills, confidence, and knowledge they need to succeed on high school math exams with emphasis on passing high school graduation exams. More than 20 easy-to-follow lessons... more... | 677.169 | 1 |
Synopsis
Galois theory is a fascinating mixture of classical and modern mathematics, and in fact provided much of the seed from which abstract algebra has grown. It is a showpiece of mathematical unification and of "technology transfer" to a range of modern applications.
Galois Theory, Second Edition is a revision of a well-established and popular text. The author's treatment is rigorous, but motivated by discussion and examples. He further lightens the study with entertaining historical notes - including a detailed description of Évariste Galois' turbulent life. The application of the Galois group to the quintic equation stands as a central theme of the book. Other topics include the problems of trisecting the angle, duplicating the cube, squaring the circle, solving cubic and quartic equations, and the construction of regular polygons
For this edition, the author added an introductory overview, a chapter on the calculation of Galois groups, further clarification of proofs, extra motivating examples, and modified exercises. Photographs from Galois' manuscripts and other illustrations enhance the engaging historical context offered in the first edition.
Written in a lively, highly readable style while sacrificing nothing to mathematical rigor, Galois Theory remains accessible to intermediate undergraduate students and an outstanding introduction to some of the intriguing concepts of abstract algebra.
eBook Details
Chapman and Hall/CRC, May 1990
ISBN:
9780203489307
Language:
English
Download options:
PDF (Adobe DRM)
You can read this item using any of the following Kobo apps and devices: | 677.169 | 1 |
Schaum's Outline of Advanced Mathematics for Engineers and Scientists
Murray Spiegel
This book is available for download with iBooks on your Mac or iOS device, and with iTunes on your computer. Books can be read with iBooks on your Mac or iOS device.
Description
Tough Test Questions? Missed Lectures? Not Enough Time?
Fortunately for you, there's Schaum's.
More than 40 million students have trusted also get hundreds of examples, solved problems, and practice exercises to test your skills.
This Schaum's Outline gives you: Practice problems with full explanations that reinforce knowledgeCoverage of the most up-to-date developments in your course fieldIn-depth review of practices and applications Fully compatible with your classroom text, Schaum's highlights all the important facts you need to know. Use Schaum's to shorten your study time-and get your best test scores! | 677.169 | 1 |
Keep Learning
The books that Diophantus authored contain the earliest known use of syncopated notation. An entire area of study is named after the Greek mathematician, called Diophantine analysis and Diophantine equations. The analysis portion is where whole number solutions are looked for as the answer to equations. Diophantine equations are polynomial equations with whole number coefficients and only solutions with whole number answers are sought to solve them. It has been debated if Diophantus is truly the father of algebra or if the honor should go to one of the older mathematicians in history.
Related Questions
Introductory algebra is a fundamental mathematics course. It is essential to master this course before moving on to more advanced material. Key concepts in introductory algebra involve the study of variables, expressions and equations.
An intermediate algebra rational expression is a math problem for the intermediate level that is expressed as a ratio of two polynomials p (x) and q (x). A student at this level may be required to multiply, divide, add and subtract rational expressions.
Algebra has been developed over thousands of years in several different countries. The earliest methods for solving mathematical problems with one or more unknown quantities come from ancient Egypt. The word "algebra" itself is derived from the title of Baghdad mathematician Al-Kwarizmi's 9th century book, "Hidab al-jabr wal-muqubala."
In addition to finding it useful in a wide variety of everyday applications, Americans need algebra, because those who learn it are better able to compete in the international marketplace. Students from countries in Europe, Asia and elsewhere are generally well-versed in the mathematical form. | 677.169 | 1 |
Featured Articles
To successfully master basic math, you need to practice addition, subtraction, multiplication, and division problems. You also need to understand order of operations, fractions, decimals, percents, ratios, weights and measures, and even a little geometry.[more…]
What follows are over three dozen of the most important geometry formulas, theorems, properties, and so on that you use for calculations. If you get stumped while working on a geometry problem and can't[more…]
In addition to finding the volume of unusual shapes, integration can help you to derive volume formulas. For example, you can use the disk/washer method of integration to derive the formula for the volume[more…] | 677.169 | 1 |
Grading
(details TBA)
Course description
Many problems we have to solve in day-to-day business, engineering, and science practice require the simultaneous study of several different but interrelated factors. Although problems of this form have been studied throughout the long history of mathematics, only in the early 20th century did the systematic approach we now refer to as linear algebra based on matrices emerge. Matrices and linear algebra are now recognized as the fundamental tool for foundational methods in statistics, optimization, quantum mechanics, and many other fields, and are an essential component of most subfields of mathematics. Linear algebra provides students their first introduction to the concept of dimension in an abstract setting where things with 4, 5, or even more dimensions are often encountered. MATH 220 is a 2 credit course that teaches the core concepts of matrix arithmetic and linear algebra. It is a required course for many students majoring in engineering, science, or secondary education. In past coursework, students should have gained practice solving pairs of equations like 3 x + 4 y = 10, x - y = 1. This is a system of two linear equations with two unknowns and as a unique solution students can find by isolating and substituting. In linear algebra, this system is represented as A x = b, where x is a vector of unknowns, A is a matrix, and b is a vector of constants. Linear algebra is the field of mathematics that grew out of a need to solve systems like these and related problems with many unknown variables. Topics covered in MATH 220 include matrix algebra, vectors, linear transformations, solution to systems of linear equations, determinants, matrix inverses, concepts of rank and dimension, eigenvalues, eigenvectors, and others as time permits. | 677.169 | 1 |
Details about Trigonometry for College Students:
Known for its student-friendly approach, this book starts with a unit-circle approach. With a focus on applications, it helps students investigate and understand the trigonometric functions, their graphs, their relationships to one another, and ways in which they can be used in a variety of applications.
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Rent Trigonometry for College Students 7th edition today, or search our site for other textbooks by Ali Smith. Every textbook comes with a 21-day "Any Reason" guarantee. Published by Brooks Cole.
Need help ASAP? We have you covered with 24/7 instant online tutoring. Connect with one of our Trigonometry tutors now. | 677.169 | 1 |
Prealgebra, 4th Edition
Offering a uniquely modern, balanced approach, Tussy/Gustafson/Koenig's PREALGEBRA, Fourth Edition, integrates the best of traditional drill and practice with the best elements of the reform movement. To many developmental math students, algebra is like a foreign language. They have difficulty translating the words, their meanings, and how they apply to problem solving. Emphasizing the "language of algebra," the text's fully integrated learning process is designed to expand students' reasoning abilities and teach them how to read, write, and think mathematically. It blends instructional approaches that include vocabulary, practice, and well-defined pedagogy with an emphasis on reasoning, modeling, communication, and technology skills. | 677.169 | 1 |
Catalog Number:
10804107
Credits:
3.00
Description:
This course is designed to review and develop fundamental concepts of mathematics pertinent to the areas of: 1) arithmetic and algebra; 2) geometry and trigonometry; and 3) probability and statistics. Special emphasis is placed on problem solving, critical thinking and logical reasoning, making connections, and using calculators. Topics include performing arithmetic operations and simplifying algebraic expressions, solving linear equations and inequalities in one variable, solving proportions and incorporating percent applications, manipulating formulas, solving and graphing systems of linear equations and inequalities in two variables, finding areas and volumes of geometric figures, applying similar and congruent triangles, converting measurements within and between U.S. and metric systems, applying Pythagorean Theorem, solving right and oblique triangles, calculating probabilities, organizing data and interpreting charts, calculating central and spread measures, and summarizing and analyzing data. Prerequisite: Accuplacer Math score of 65 and Accuplacer Algebra score of 30 or higher or Pre-Algebra 10834109 with a grade of "C" or better
Requirements:
Prerequisites: Accuplacer Math score of 65 and Accuplacer Algebra score of 30 or higher or Pre-Algebra 10834109 with a grade of "C" or better | 677.169 | 1 |
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