A study of Pythagoras Theorem

Similar documents
1 PYTHAGORAS THEOREM 1. Given a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

Part I: Study the theorem statement.

1 ELEMENTARY ALGEBRA and GEOMETRY READINESS DIAGNOSTIC TEST PRACTICE

Farey Fractions. Rickard Fernström. U.U.D.M. Project Report 2017:24. Department of Mathematics Uppsala University

set is not closed under matrix [ multiplication, ] and does not form a group.

Lecture 2 : Propositions DRAFT

Chapter 9 Definite Integrals

Things to Memorize: A Partial List. January 27, 2017

Triangles The following examples explore aspects of triangles:

Mathematics. Area under Curve.

7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus

Bases for Vector Spaces

1B40 Practical Skills

The practical version

Polynomials and Division Theory

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

Calculus Module C21. Areas by Integration. Copyright This publication The Northern Alberta Institute of Technology All Rights Reserved.

4 VECTORS. 4.0 Introduction. Objectives. Activity 1

Mathematics Number: Logarithms

2. VECTORS AND MATRICES IN 3 DIMENSIONS

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

Vectors , (0,0). 5. A vector is commonly denoted by putting an arrow above its symbol, as in the picture above. Here are some 3-dimensional vectors:

8Similarity UNCORRECTED PAGE PROOFS. 8.1 Kick off with CAS 8.2 Similar objects 8.3 Linear scale factors. 8.4 Area and volume scale factors 8.

p-adic Egyptian Fractions

2.4 Linear Inequalities and Interval Notation

20 MATHEMATICS POLYNOMIALS

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

I1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3

Chapters Five Notes SN AA U1C5

6.2 The Pythagorean Theorems

Bridging the gap: GCSE AS Level

Lecture 3: Equivalence Relations

378 Relations Solutions for Chapter 16. Section 16.1 Exercises. 3. Let A = {0,1,2,3,4,5}. Write out the relation R that expresses on A.

Quadratic Forms. Quadratic Forms

I do slope intercept form With my shades on Martin-Gay, Developmental Mathematics

8. Complex Numbers. We can combine the real numbers with this new imaginary number to form the complex numbers.

10. AREAS BETWEEN CURVES

AQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system

Shape and measurement

Beginning Darboux Integration, Math 317, Intro to Analysis II

3.1 Review of Sine, Cosine and Tangent for Right Angles

Trigonometry Revision Sheet Q5 of Paper 2

9.4. The Vector Product. Introduction. Prerequisites. Learning Outcomes

8Similarity ONLINE PAGE PROOFS. 8.1 Kick off with CAS 8.2 Similar objects 8.3 Linear scale factors. 8.4 Area and volume scale factors 8.

5: The Definite Integral

On the diagram below the displacement is represented by the directed line segment OA.

Torsion in Groups of Integral Triangles

PYTHAGORAS THEOREM WHAT S IN CHAPTER 1? IN THIS CHAPTER YOU WILL:

Intermediate Math Circles Wednesday, November 14, 2018 Finite Automata II. Nickolas Rollick a b b. a b 4

Section 6: Area, Volume, and Average Value

Section 4: Integration ECO4112F 2011

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

Improper Integrals, and Differential Equations

7.1 Integral as Net Change Calculus. What is the total distance traveled? What is the total displacement?

Chapter 8.2: The Integral

Review of Gaussian Quadrature method

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Number systems: the Real Number System

Project 6: Minigoals Towards Simplifying and Rewriting Expressions

An Overview of Integration

MA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.

Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1

Parse trees, ambiguity, and Chomsky normal form

1.2 What is a vector? (Section 2.2) Two properties (attributes) of a vector are and.

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

Chapter 0. What is the Lebesgue integral about?

Chapter 1: Logarithmic functions and indices

CM10196 Topic 4: Functions and Relations

Algebra II Notes Unit Ten: Conic Sections

The Regulated and Riemann Integrals

Linear Inequalities. Work Sheet 1

Infinite Geometric Series

Analytically, vectors will be represented by lowercase bold-face Latin letters, e.g. a, r, q.

Discrete Mathematics and Probability Theory Spring 2013 Anant Sahai Lecture 17

Exponentials - Grade 10 [CAPS] *

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Chapter 6 Techniques of Integration

9.1 Day 1 Warm Up. Solve the equation = x x 2 = March 1, 2017 Geometry 9.1 The Pythagorean Theorem 1

List all of the possible rational roots of each equation. Then find all solutions (both real and imaginary) of the equation. 1.

Homework 3 Solutions

MTH 505: Number Theory Spring 2017

= x x 2 = 25 2

Exercises with (Some) Solutions

SUMMER KNOWHOW STUDY AND LEARNING CENTRE

USA Mathematical Talent Search Round 1 Solutions Year 21 Academic Year

Math 8 Winter 2015 Applications of Integration

Lecture 1. Functional series. Pointwise and uniform convergence.

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

Geometry AP Book 8, Part 2: Unit 3

STRAND J: TRANSFORMATIONS, VECTORS and MATRICES

PHYS 1114, Lecture 1, January 18 Contents:

Coordinate geometry and vectors

APPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line

Lecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.

HW3, Math 307. CSUF. Spring 2007.

青藜苑教育 The digrm shows the position of ferry siling between Folkestone nd lis. The ferry is t X. X 4km The pos

MATH STUDENT BOOK. 10th Grade Unit 5

Matrix Algebra. Matrix Addition, Scalar Multiplication and Transposition. Linear Algebra I 24

Transcription:

CHAPTER 19 A study of Pythgors Theorem Reson is immortl, ll else mortl. Pythgors, Diogenes Lertius (Lives of Eminent Philosophers) Pythgors Theorem is proly the est-known mthemticl theorem. Even most nonmthemticins hve some vgue ide tht it involves tringles nd squring something known s the hypotenuse. Becuse the ides in the previous chpters on How to red theorem nd How to red proof re so importnt we will pply them to this fmous theorem to see them in ction. So in this chpter we will pull prt the theorem nd its proof, we ll see converse for it nd lso generliztion. Sttement of Pythgors Theorem As you re udding mthemticin, you proly hve etter ide thn nonmthemticin of wht the sttement is, ut here it is gin. Theorem 19.1 For right-ngled tringle the squre of the hypotenuse is equl to the sum of the squres of the other sides. Exercise 19.2 Use the ides from Chpter 16, How to red theorem, to nlyse the theorem. Compre your nlysis with the one given elow. Study of the theorem 126 We now nlyse the theorem s though we were meeting it for the first time. Oviously we would check wht ll the words men, for exmple, wht is hypotenuse? This is firly ovious, ut wht out the other techniques in Chpter 16? We shll pply them now.

Study of the theorem 127 Drw in this line Figure 19.1 Any tringle gives two right-ngled tringles Find the ssumptions nd conclusions The sttement given for Pythgors Theorem is good exmple of sttement not in the form A= B or if..., then.... We cn rewrite it in this form in numer of wys. For exmple, If T is right-ngled tringle with sides, nd hypotenuse c, then c 2 = 2 + 2. This mke it ovious tht the ssumptions concern ll right-ngled tringles: T is right-ngled tringle with sides, nd hypotenuse c nd tht the conclusion is n eqution relting the lengths: c 2 = 2 + 2. Rte the strength of the ssumptions nd conclusions Let s rte the ssumptions nd conclusions. The ssumption is out right-ngled tringles. Certinly, there re mny exmples of these, ut they re only smll suset of ll tringles. Thus we my e tempted to sy tht this is quite wek ut not too wek. But consider this. For ny tringle we cn produce two right-ngled tringles; see Figure 19.1. Thus, despite initil impressions, this theorem will tell us something out ll tringles. This mens it is very wek ssumption. Tht is good. The more exmples theorem pplies to the etter. The conclusion is very concrete nd definite sttement. It is n eqution. If we just hd n inequlity (sy c 2 2 + 2 ), then the conclusion would e less impressive. This is strong conclusion. Note lso tht the conclusion llows us to clculte: given the lengths of ny two sides of right-ngled tringle we cn clculte the third. Being le to clculte is lwys good for conclusion. From this we know we hve gret theorem which will e very useful s it llows us to clculte ttriutes of lrge numer of ojects. In fct, s you my know, it llows us to clculte distnces in Crtesin coordintes. Compre with previous theorems We don t hve mny theorems to compre with s this is very simple theorem from the foundtions of geometry. Another theorem from the foundtions is tht in tringle (not necessrily right-ngled) the ngles dd up to 180 degrees (or 2π rdins if you like).

128 CHAPTER 19 A study of Pythgors Theorem We cn see this is similr in the sense tht given two ngles we cn clculte the third. It is different in the sense tht it is out ngles rther thn out lengths. Thus they re quite complementry! Now tht we hve oserved theorem tht dels with lengths nd one tht dels with ngles we cn sk whether there exists one tht links oth. Indeed there is one. Exercise 19.3 Find theorem tht reltes ngles nd lengths in tringle. Oserve the detil There re not mny detils to oserve in Pythgors Theorem s it is rther simple. The detil to rememer perhps is tht the tringle is right-ngled not, for exmple, isosceles. (See the exercises t the end of the chpter for some infmous exmples of mistkes with the sttement.) Clssify wht the theorem does nd how it cn e used We hve lredy discussed ove wht the theorem does: it llows us to clculte the length of side of tringle given the lengths of the other two. Drw picture Since this is geometric theorem it is out tringles then we hve good reson to drw picture. In this cse drw lots of tringles nd mesure the lengths. Do the lengths stisfy the eqution? They should do ut rememer we cn only mesure lengths pproximtely. The clssic picture is the (3, 4, 5)-tringle given in Figure 19.2. This is often mistkenly given s proof of the theorem. However, it is only single exmple, not proof. 5 4 3 Figure 19.2 The clssic (3, 4, 5)-tringle

Study of the theorem 129 Apply the theorem to simple exmples In drwing the pictures we hve lredy pplied the theorem to simple exmples such s the (3, 4, 5)-tringle. Apply the theorem to trivil nd extreme exmples The ssumptions concern right-ngled tringles. Wht re trivil nd extreme exmples of such ojects? I would sy tht the trivil exmples re when = 1 nd = 1 so c = 2. This is tringle with two sides of rtionl lengths nd one side of irrtionl length. One could view this s counterexmple to the sttement If two sides of tringle re rtionl, then so is the third. Another useful exmple is = 1, = 2, nd so c = 5. This just shows us wht hppens when one side is twice the length of the other. Another interesting exmple is = 1, = 2, nd so c = 3. I like this, it involves the first three numers 1, 2 nd 3. The tringle = 1, = 3 nd c = 4 is good s it hs n ngle of π/6. Now for extreme exmples. Wht is n extreme right-ngled tringle? We cn view = s n extreme tringle ecuse it is very specil cse of right-ngled tringle. Here c 2 = 2 2, so c = 2. This mens tht if is rtionl, then c cnnot e rtionl (since rtionl numer times n irrtionl numer, in this cse 2, is irrtionl). Another extreme we cn go to is to mke one of the sides very ig or very smll compred with the other. If we let tend to zero, then c 2 must tend to 2 s 2 gets smll. In other words c tends to. We cn see this in digrm. Just drw tringle with very smll s compred with ; you cn see tht nd c re lmost equl. Thus we hve shown tht the lger nd the geometry re linked in the wy we would expect. The lger grees with the geometry. Note tht if equls zero, then we do not hve tringle! Apply the theorem to non-exmples Let us consider non-exmples of the theorem. Here non-exmples re tringles tht re not right-ngled. However, in this cse how do we even define hypotenuse? This is lwys defined s the length opposite the right-ngle. There is no wy of identifying specil length, so there is no wy to identify which should go on the right-hnd side of c 2 = 2 + 2. We need nother method of identifying c. Mye we hve to tke the longest side. This is resonle since, for c 2 = 2 + 2 to e true, we must hve c>nd c>. To see this, consider tht 2 > 0 is true, thus c 2 > 2, nd so c>. Similrly c>. If you try drwing few exmples, then you will proly see tht the eqution does not hold, even tking into ccount pproximtions in mesuring. This should led us to sk Are there ny non-right-ngled tringles for which c 2 = 2 + 2? We shll del with this in the next section when we sk Is the converse true? We shll lso look t non-exmples in Exercises 19.12.

130 CHAPTER 19 A study of Pythgors Theorem Rewrite in symols or words We hve lredy written the theorem in s mny symols s possile when we gve the if, then sttement. Note tht is not sufficient to sy only c 2 = 2 + 2. We hve to explin wht, nd c re, nd tht c is the hypotenuse in right-ngled tringle. Proof of Pythgors Theorem As mentioned erlier, it is not enough to show the theorem works pproximtely for some exmples or even perfectly in some specil cses such s the (3, 4, 5)-tringle. We need to prove tht it holds for ll tringles. Mye you were told tht the theorem ws true y some uthority figure in the pst nd tht is good enough for you. However, centrl im of this ook is to encourge you to think for yourself nd tht involves checking ny rgument given to you. We will now see proof of the theorem. There re literlly hundreds of proofs. 1 My fvourite proof is geometricl. Proof (of Pythgors Theorem). The proof cn e shown using the two squres in Figure 19.3. To drw the first squre egin y drwing generl tringle with sides nd nd then extend these edges y lengths nd respectively. Then we cn complete the drwing to get the squre on the left-hnd side of Figure 19.3. We cn drw nother squre like the one on the right-hnd side of the figure. From the figure we cn see tht oth squres hve equl re nd so we cn conclude tht Are of left squre = Are of right squre c 2 + (4 Are of (, )-tringle) = 2 + 2 + (4 Are of (, )-tringle) c 2 = 2 + 2. c c 2 2 c 2 Figure 19.3 Proof of Pythgors Theorem 1 There re over 360 in Elish Loomis, The Pythgoren Proposition, Ntionl Council of Techers of Mthemtics, 1968.

Exercise 19.4 Proof of Pythgors Theorem 131 Use the ides from Chpter 18, How to red proof, to nlyse the proof. Compre your nlysis with the one given elow. Not ll the suggestions in Chpter 18 re relevnt. Let us try some of them to help pull the proof prt. Find where the ssumptions re used Where hve we used the ssumption tht the tringle is right-ngled? It is used in constructing the first squre. If sides nd did not meet t right-ngle, then we could not construct squre. And hence could not conclude tht the re ws the sme s the squre on the right. Check the text If you check the text, then you my see tht one fct hs een used ut not explicitly stted. The picture is very convincing in tht the re lelled c 2 in the left-hnd picture certinly looks squre. But notice tht we hve not proved tht it is squre. We know tht ll the edges re the sme length ut this does not men tht the shpe is squre think of dimond or kite shpe. How do we know it is truly squre? Well, we need to show tht one of the internl ngles is right-ngled, once we get it for one, the sme proof will work for ll. (Cn you see why?) Cll the internl ngle γ. Suppose tht the tringle in Figure 19.4 hs smll ngle α nd lrger ngle β. Then we cn see tht α + β + γ = 180 since we hve stright line: g Figure 19.4 But we know tht α + β + 90 = 180 since the ngles in tringle dd up to 180 nd our tringle is right-ngled. Using these two equtions we deduce tht γ is 90 s required. If we dd this explntion, then the rgument will e even more convincing. True, the picture looks convincing lredy, ut we hve to e creful with pictures they cn misled. Exercise 19.5 How else could the picture fool us?

132 CHAPTER 19 A study of Pythgors Theorem Wht out the converse? Let s now look t n importnt technique for exploring theorems: is the converse true? To do this it helps to write the sttement in n If, then form: If T is right-ngled tringle with sides, nd hypotenuse c, then c 2 = 2 + 2. The converse will e If c 2 = 2 + 2, then T is right-ngled tringle with sides, nd hypotenuse c. Note tht this does not quite mke sense ecuse in the ssumptions we do not know wht, nd c re. Let s rewrite Pythgors Theorem s the following: Let T e tringle with sides of length, nd c with c the longest. If T is right-ngled tringle, then c 2 = 2 + 2. Now the converse ecomes: Theorem 19.6 (Converse of Pythgors Theorem) Let T e tringle with sides of length, nd c with c the longest. If c 2 = 2 + 2, then T is right-ngled tringle. Is this true or not? It is! But why? Proof. Let T e tringle with sides of length, nd c where C is the ngle opposite to the side of length c, then the Cosine Rule sttes tht c 2 = 2 + 2 + 2 cos C. Now suppose tht c 2 = 2 + 2, then we hve 2 cos C = 0. As nd cnnot e zero we must hve cos C = 0. This implies tht C = 90 + 180 n, where n is some integer. Since 0 <C<180 we must hve C = 90, i.e. T is right-ngled tringle. Exercise 19.7 Apply the methods of Chpter 18, How to red proof, to the ove proof. Remrk 19.8 The proof of the converse used the Cosine Rule, the proof of which uses Pythgors Theorem (see pge 22 nd following). Note tht the Cosine Rule is more generl thn Pythgors Theorem, since we cn deduce Pythgors from it: if we hve right-ngled tringle, then C = 90 nd so c 2 = 2 + 2 2 0 = 2 + 2. Thus, we cn see tht more generl sttement cn use the simpler theorem in its proof. Since Pythgors Theorem nd its converse re true we hve two equivlent sttements (i.e. If A = B nd B = A, then A B). This mens we cn stte the following theorem. Theorem 19.9 Let T e tringle with sides of length, nd c with c the longest. Then T is right-ngled tringle if nd only if c 2 = 2 + 2.

Exercises 133 Proof. We cn prove this y comining the proof of Pythgors we gve nd then using the converse rgument we just gve. Note tht we cn t just use the Cosine Rule in oth directions since Pythgors Theorem is used in the proof of the Cosine Rule! Did you spot tht? A it more on understnding converse Suppose tht we hve tringle with sides of length 2.0, 2.1 nd 2.9. We hve 2.0 2 + 2.1 2 = 2.9 2. The tringle is therefore right-ngled. Exercise 19.10 Did we use Pythgors Theorem or its converse to deduce this? Suppose tht we hve tringle of sides of length 3.6, 7.7 nd 8.4. In this cse, 3.6 2 +7.7 2 = 72.25 = 70.56 = 8.4 2. Thus the tringle is not right ngled. Exercise 19.11 Did we use Pythgors Theorem or its converse to deduce this? A common mistke in nswering this is to mke n rgument tht the sides do not stisfy the eqution c 2 = 2 + 2 so this cn t use Pythgors Theorem. Therefore, it must use the converse. This is wrong. The correct rgument is tht Pythgors Theorem sys tht if you hve right-ngled tringle, then it sides stisfy the eqution. Thus if the eqution is not stisfied, then there is no wy tht the tringle could e right-ngled. In effect, we re using the contrpositive sttement to Pythgors Theorem, i.e. n equivlent sttement: If c 2 = 2 + 2, then T is not right-ngled. Exercises Exercises 19.12 (i) (ii) Pythgors Theorem is often misquoted. In the film The Wizrd of Oz, the Screcrow is given diplom nd to show how clever he hs ecome he points his finger to his temple nd sys The squre root of the hypotenuse is equl to the sum of the squre roots of the other sides for n isosceles tringle. This is most definitely not Pythgors Theorem s it involves squre roots nd isosceles tringles. However, just ecuse it is not Pythgors Theorem does not men tht the sttement is flse! Apply our methods to nlyse this sttement. Is it true? If not, then give counterexmple. The theorem is lso misquoted in the long-running nimted comedy The Simpsons. In the episode $pringfield (or, how I lerned to stop worrying nd love leglized gmling) Homer finds some glsses nd puts them on nd à l Screcrow puts his finger to his temple nd sttes The sum of the squre roots of ny two sides

134 CHAPTER 19 A study of Pythgors Theorem of n isosceles tringle is equl to the squre root of the remining side. Another chrcter then shouts Tht s right-ngled tringle, you idiot! Why re Homer nd the other chrcter oth wrong? (iii) Mny theorems from lower-level mthemtics re given without proof. Find s mny exmples of this s you cn. Try to find proofs for them from other sources nd nlyse these using the methods of Chpter 18, How to red proof. Some exmples you my like to try: () The sum of ngles in tringle is 180 degrees. () Sutrcting negtive is equivlent to dding positive. (c) The vlue of π is 3.14159... (d) The definition of sine nd cosine don t depend on the tringle. (You will need to know out similr tringles.) (e) For ny ngle θ, sin 2 θ + cos 2 θ = 1. (f) The re of circle of rdius r is πr 2. (iv) Consider the tringles with sides of the following lengths. Decide which re rightngled nd stte whether you used Pythgors Theorem or its converse. () 7, 24, 25, () 28, 45, 52, (c) 36.9, 80.0, 88.0, (d) 0.8, 1, 4, 1.7. (v) Construct right-ngled tringle such tht the hypotenuse hs irrtionl length ut the other two sides hve rtionl length. (vi) Construct right-ngled tringle such tht the hypotenuse hs rtionl length ut the other two sides hve irrtionl length. (vii) Let X nd Y e distinct points in plne. Drw line etween them. At the midpoint drw line perpendiculr to the line. See Figure 19.5. This line is clled the perpendiculr isector of X nd Y. Show tht for every point p on the perpendiculr isector the distnce from X to p nd the distnce from Y to p re the sme. We sy p is equidistnt from X nd Y. (viii) Go ck to previous chpters nd exercises nd re-nlyse the theorems nd proofs. Did you oserve fcts tht you did not oserve efore? X Perpendiculr isector Y Figure 19.5 Perpendiculr isector

Summry 135 Summry Pythgors Theorem is: Let T e tringle with sides of length, nd c with c the longest. If T is right-ngled tringle, then c 2 = 2 + 2. The converse of Pythgors Theorem is: Let T e tringle with sides of length, nd c with c the longest. If c 2 = 2 + 2, then T is right-ngled tringle.