Famous Theorems of Mathematics/Pythagoras theorem - Wikibooks, open books for an open world Famous Theorems of Mathematics/Pythagoras theorem From Wikibooks, open books for an open world Jump to: navigation , search The Pythagoras Theorem or the Pythagorean theorem, named after the Greek mathematician Pythagoras states that: In any right triangle, the area of the square whose side is the hypotenuse (the side opposite to the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle). This is usually summarized as follows: The square of the hypotenuse of a right triangle is equal to the sum of the squares on the other two sides. If we let c be the length of the hypotenuse and a and b be the lengths of the other two sides, the theorem can be expressed as the equation: a {\displaystyle a^{2}+b^{2}=c^{2}\,} or, solved for c: .

{\displaystyle c={\sqrt {a^{2}+b^{2}}}.\,} If c is already given, and the length of one of the legs must be found, the following equations can be used (The following equations are simply the converse of the original equation): c {\displaystyle c^{2}-a^{2}=b^{2}\,} or . {\displaystyle c^{2}-b^{2}=a^{2}.\,} This equation provides a simple relation among the three sides of a right triangle so that if the lengths of any two sides are known, the length of the third side can be found. A generalization of this theorem is the law of cosines, which allows the computation of the length of the third side of any triangle, given the lengths of two sides and the size of the angle between them. If the angle between the sides is a right angle it reduces to the Pythagorean theorem. Contents History[ edit ] The history of the theorem can be divided into four parts: knowledge of Pythagorean triples, knowledge of the relationship between the sides of a right triangle, knowledge of the relationship between adjacent angles, and proofs of the theorem.

Megalithic monuments from circa 2500 BC in Egypt, and in Northern Europe, incorporate right triangles with integer sides. Bartel Leendert van der Waerden conjectures that these Pythagorean triples were discovered algebraically. Written between 2000 and 1786 BC, the Middle Kingdom Egyptian papyrus Berlin 6619 includes a problem whose solution is a Pythagorean triple. During the reign of Hammurabi the Great, the Mesopotamian tablet Plimpton 322, written between 1790 and 1750 BC, contains many entries closely related to Pythagorean triples. The Baudhayana Sulba Sutra, the dates of which are given variously as between the 8th century BC and the 2nd century BC, in India, contains a list of Pythagorean triples discovered algebraically, a statement of the Pythagorean theorem, and a geometrical proof of the Pythagorean theorem for an isosceles right triangle. The Apastamba Sulba Sutra (circa 600 BC) contains a numerical proof of the general Pythagorean theorem, using an area computation.

Van der Waerden believes that "it was certainly based on earlier traditions". According to Albert Bŭrk, this is the original proof of the theorem; he further theorizes that Pythagoras visited Arakonam, India, and copied it. Pythagoras, whose dates are commonly given as 569–475 BC, used algebraic methods to construct Pythagorean triples, according to Proklos's commentary on Euclid. Proklos, however, wrote between 410 and 485 AD. According to Sir Thomas L. Heath, there is no attribution of the theorem to Pythagoras for five centuries after Pythagoras lived. However, when authors such as Plutarch and Cicero attributed the theorem to Pythagoras, they did so in a way which suggests that the attribution was widely known and undoubted. Around 400 BC, according to Proklos, Plato gave a method for finding Pythagorean triples that combined algebra and geometry. Circa 300 BC, in Euclid's Elements, the oldest extant axiomatic proof of the theorem is presented.

Written sometime between 500 BC and 100 AD, the Chinese text Chou Pei Suan Ching (周髀算经), (The Arithmetical Classic of the Gnomon and the Circular Paths of Heaven) gives a statement of the Pythagorean theorem — in China it is called the "Gougu Theorem" (勾股定理) — for the (3, 4, 5) triangle. A visual proof is recorded in a Ming dynasty text though it is unclear as to when it was originally provided. During the Han Dynasty, from 202 BC to 220 AD, Pythagorean triples appear in The Nine Chapters on the Mathematical Art, together with a mention of right triangles. The first recorded use is in China, known as the "Gougu theorem" (勾股定理) and in India known as the Bhaskara Theorem. There is much debate on whether the Pythagorean theorem was discovered once or many times. Boyer (1991) thinks the elements found in the Shulba Sutras may be of Mesopotamian derivation.

Proofs[ edit ] This is a theorem that may have more known proofs than any other; the book Pythagorean Proposition, by Elisha Scott Loomis, contains 367 proofs. Proof using similar triangles[ edit ] Proof using similar triangles. Like most of the proofs of the Pythagorean theorem, this one is based on the proportionality of the sides of two similar triangles. Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. We draw the altitude from point C, and call H its intersection with the side AB. The new triangle ACH is similar to our triangle ABC, because they both have a right angle (by definition of the altitude), and they share the angle at A, meaning that the third angle will be the same in both triangles as well. By a similar reasoning, the triangle CBH is also similar to ABC. The similarities lead to the two ratios..: As B Euclid's proof[ edit ] Proof in Euclid's Elements In Euclid's Elements, Proposition 47 of Book 1, the Pythagorean theorem is proved by an argument along the following lines.

Let A, B, C be the vertices of a right triangle, with a right angle at A. Drop a perpendicular from A to the side opposite the hypotenuse in the square on the hypotenuse. That line divides the square on the hypotenuse into two rectangles, each having the same area as one of the two squares on the legs. For the formal proof, we require four elementary lemmata: If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles included by those sides equal, then the triangles are congruent. (Side - Angle - Side Theorem) The area of a triangle is half the area of any parallelogram on the same base and having the same altitude. The area of any square is equal to the product of two of its sides. The area of any rectangle is equal to the product of two adjacent sides (follows from Lemma 3). The intuitive idea behind this proof, which can make it easier to follow, is that the top squares are morphed into parallelograms with the same size, then turned and morphed into the left and right rectangles in the lower square, again at constant area.

The proof is as follows: Let ACB be a right-angled triangle with right angle CAB. On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order. From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively. Join CF and AD, to form the triangles BCF and BDA. Illustration including the new lines Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H. Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC. Since AB and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC. Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD. Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC. Therefore rectangle BDLK must have the same area as square BAGF = AB2.

Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2. Adding these two results, AB2 + AC2 = BD × BK + KL × KC Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC Therefore AB2 + AC2 = BC2, since CBDE is a square. This proof appears in Euclid's Elements as that of Proposition 1.47. Garfield's proof[ edit ] James A. Garfield (later President of the United States) is credited with a novel algebraic proof [1] using a trapezoid containing two examples of the triangle, the figure comprising one half of the figure using four triangles enclosing a square shown below. Proof using area subtraction. Similarity proof[ edit ] From the same diagram as that in Euclid's proof above, we can see three similar figures, each being "a square with a triangle on top". Since the large triangle is made of the two smaller triangles, its area is the sum of areas of the two smaller ones.

By similarity, the three squares are in the same proportions relative to each other as the three triangles, and so likewise the area of the large square is the sum of the areas of the two smaller squares. Proof by rearrangement[ edit ] A proof by rearrangement is given by the illustration and the animation. In the illustration, the area of each large square is (a + b)2. In both, the area of four identical triangles is removed. The remaining areas, a2 + b2 and c2, are equal. Q.E.D. Animation showing another proof by rearrangement. Proof using rearrangement. A square created by aligning four right angle triangles and a large square. This proof is indeed very simple, but it is not elementary, in the sense that it does not depend solely upon the most basic axioms and theorems of Euclidean geometry. In particular, while it is quite easy to give a formula for area of triangles and squares, it is not as easy to prove that the area of a square is the sum of areas of its pieces. In fact, proving the necessary properties is harder than proving the Pythagorean theorem itself and Banach-Tarski paradox.

Actually, this difficulty affects all simple Euclidean proofs involving area; for instance, deriving the area of a right triangle involves the assumption that it is half the area of a rectangle with the same height and base. For this reason, axiomatic introductions to geometry usually employ another proof based on the similarity of triangles (see above). A third graphic illustration of the Pythagorean theorem (in yellow and blue to the right) fits parts of the sides' squares into the hypotenuse's square. A related proof would show that the repositioned parts are identical with the originals and, since the sum of equals are equal, that the corresponding areas are equal. To show that a square is the result one must show that the length of the new sides equals c. Note that for this proof to work, one must provide a way to handle cutting the small square in more and more slices as the corresponding side gets smaller and smaller. [1] Algebraic proof[ edit ] An algebraic variant of this proof is provided by the following reasoning. Looking at the illustration which is a large square with identical right triangles in its corners, the area of each of these four triangles is given by an angle corresponding with the side of length C. 1 .

{\displaystyle c^{2}=a^{2}+b^{2}.\,} As can be seen, the squares are due to the particular proportion between the changes and the sides while the sum is a result of the independent contributions of the changes in the sides which is not evident from the geometric proofs. From the proportion given it can be shown that the changes in the sides are inversely proportional to the sides. The differential equation suggests that the theorem is due to relative changes and its derivation is nearly equivalent to computing a line integral. These quantities da and dc are respectively infinitely small changes in a and c. But we use instead real numbers Δa and Δc, then the limit of their ratio as their sizes approach zero is da/dc, the derivative, and also approaches c/a, the ratio of lengths of sides of triangles, and the differential equation results. Converse[ edit ] The converse of the theorem is also true: For any three positive numbers a, b, and c such that a2 + b2 = c2, there exists a triangle with sides a, b and c, and every such triangle has a right angle between the sides of lengths a and b. This converse also appears in Euclid's Elements.

It can be proven using the law of cosines, or by the following proof: Let ABC be a triangle with side lengths a, b, and c, with a2 + b2 = c2. We need to prove that the angle between the a and b sides is a right angle. We construct another triangle with a right angle between sides of lengths a and b. By the Pythagorean theorem, it follows that the hypotenuse of this triangle also has length c. Since both triangles have the same side lengths a, b and c, they are congruent, and so they must have the same angles. Therefore, the angle between the side of lengths a and b in our original triangle is a right angle. A corollary of the Pythagorean theorem's converse is a simple means of determining whether a triangle is right, obtuse, or acute, as follows. Where c is chosen to be the longest of the three sides: If a2 + b2 = c2, then the triangle is right. If a2 + b2 > c2, then the triangle is acute. If a2 + b2 < c2, then the triangle is obtuse.

Consequences and uses of the theorem[ edit ] A Pythagorean triple has 3 positive numbers a, b, and c, such that a 2 {\displaystyle a^{2}+b^{2}=c^{2}} . In other words, a Pythagorean triple represents the lengths of the sides of a right triangle where all three sides have integer lengths. Evidence from megalithic monuments on the Northern Europe shows that such triples were known before the discovery of writing. Such a triple is commonly written (a, b, c). Some well-known examples are (3, 4, 5) and (5, 12, 13).

List of primitive Pythagorean triples up to 100[ edit ] (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (12, 35, 37), (13, 84, 85), (16, 63, 65), (20, 21, 29), (28, 45, 53), (33, 56, 65), (36, 77, 85), (39, 80, 89), (48, 55, 73), (65, 72, 97) The existence of irrational numbers[ edit ] One of the consequences of the Pythagorean theorem is that irrational numbers, such as the square root of 2, can be constructed.

A right triangle with legs both equal to one unit has hypotenuse length square root of 2. The Pythagoreans proved that the square root of 2 is irrational, and this proof has come down to us even though it flew in the face of their cherished belief that everything was rational. According to the legend, Hippasus, who first proved the irrationality of the square root of two, was drowned at sea as a consequence. Distance in Cartesian coordinates[ edit ] The distance formula in Cartesian coordinates is derived from the Pythagorean theorem. If (x0, y0) and (x1, y1) are points in the plane, then the distance between them, also called the Euclidean distance, is given by ( The Pythagorean Theorem The Pythagorean Theorem by Stephanie J. Morris The Pythagorean Theorem was one of the earliest theorems known to ancient civilizations. This famous theorem is named for the Greek mathematician and philosopher, Pythagoras. Pythagoras founded the Pythagorean School of Mathematics in Cortona, a Greek seaport in Southern Italy. He is credited with many contributions to mathematics although some of them may have actually been the work of his students.

The Pythagorean Theorem is Pythagoras' most famous mathematical contribution. According to legend, Pythagoras was so happy when he discovered the theorem that he offered a sacrifice of oxen. The later discovery that the square root of 2 is irrational and therefore, cannot be expressed as a ratio of two integers, greatly troubled Pythagoras and his followers. They were devout in their belief that any two lengths were integral multiples of some unit length. Many attempts were made to suppress the knowledge that the square root of 2 is irrational. It is even said that the man who divulged the secret was drowned at sea. The Pythagorean Theorem is a statement about triangles containing a right angle. The Pythagorean Theorem states that: "The area of the square built upon the hypotenuse of a right triangle is equal to the sum of the areas of the squares upon the remaining sides." Figure 1 According to the Pythagorean Theorem, the sum of the areas of the two red squares, squares A and B, is equal to the area of the blue square, square C.

Thus, the Pythagorean Theorem stated algebraically is: for a right triangle with sides of lengths a, b, and c, where c is the length of the hypotenuse. Although Pythagoras is credited with the famous theorem, it is likely that the Babylonians knew the result for certain specific triangles at least a millennium earlier than Pythagoras. It is not known how the Greeks originally demonstrated the proof of the Pythagorean Theorem. If the methods of Book II of Euclid's Elements were used, it is likely that it was a dissection type of proof similar to the following: "A large square of side a+b is divided into two smaller squares of sides a and b respectively, and two equal rectangles with sides a and b; each of these two rectangles can be split into two equal right triangles by drawing the diagonal c. The four triangles can be arranged within another square of side a+b as shown in the figures. The area of the square can be shown in two different ways: 1. As the sum of the area of the two rectangles and the squares: 2.

As the sum of the areas of a square and the four triangles: Now, setting the two right hand side expressions in these equations equal, gives Therefore, the square on c is equal to the sum of the squares on a and b. (Burton 1991) There are many other proofs of the Pythagorean Theorem. One came from the contemporary Chinese civilization found in the oldest extant Chinese text containing formal mathematical theories, the Arithmetic Classic of the Gnoman and the Circular Paths of Heaven. The proof of the Pythagorean Theorem that was inspired by a figure in this book was included in the book Vijaganita, (Root Calculations), by the Hindu mathematician Bhaskara. Bhaskara's only explanation of his proof was, simply, "Behold". These proofs and the geometrical discovery surrounding the Pythagorean Theorem led to one of the earliest problems in the theory of numbers known as the Pythgorean problem. The Pythagorean Problem: Find all right triangles whose sides are of integral length, thus finding all solutions in the positive integers of the Pythagorean equation: The three integers (x, y, z) that satisfy this equation is called a Pythagorean triple.

Some Pythagorean Triples: 9 40 41 11 60 61 The formula that will generate all Pythagorean triples first appeared in Book X of Euclid's Elements: where n and m are positive integers of opposite parity and m>n. In his book Arithmetica, Diophantus confirmed that he could get right triangles using this formula although he arrived at it under a different line of reasoning. The Pythagorean Theorem can be introduced to students during the middle school years. This theorem becomes increasingly important during the high school years. It is not enough to merely state the algebraic formula for the Pythagorean Theorem. Students need to see the geometric connections as well. The teaching and learning of the Pythagorean Theorem can be enriched and enhanced through the use of dot paper, geoboards, paper folding, and computer technology, as well as many other instructional materials. Through the use of manipulatives and other educational resources, the Pythagorean Theorem can mean much more to students than just and plugging numbers into the formula. The following is a variety of proofs of the Pythagorean Theorem including one by Euclid.

These proofs, along with manipulatives and technology, can greatly improve students' understanding of the Pythagorean Theorem. The following is a summation of the proof by Euclid, one of the most famous mathematicians. This proof can be found in Book I of Euclid's Elements. Proposition: In right-angled triangles the square on the hypotenuse is equal to the sum of the squares on the legs. Figure 2 Euclid began with the Pythagorean configuration shown above in Figure 2. Then, he constructed a perpendicular line from C to the segment DJ on the square on the hypotenuse. The points H and G are the intersections of this perpendicular with the sides of the square on the hypotenuse. It lies along the altitude to the right triangle ABC. See Figure 3. Figure 3 Next, Euclid showed that the area of rectangle HBDG is equal to the area of square on BC and that the are of the rectangle HAJG is equal to the area of the square on AC. He proved these equalities using the concept of similarity. Triangles ABC, AHC, and CHB are similar.

The area of rectangle HAJG is (HA)(AJ) and since AJ = AB, the area is also (HA)(AB). The similarity of triangles ABC and AHC means and therefore or, as to be proved, the area of the rectangle HAJG is the same as the areaof the square on side AC. In the same way, triangles ABC and CHG are similar. So and Since the sum of the areas of the two rectangles is the area of the square on the hypotenuse, this completes the proof. Euclid was anxious to place this result in his work as soon as possible. However, since his work on similarity was not to be until Books V and VI, it was necessary for him to come up with another way to prove the Pythagorean Theorem. Thus, he used the result that parallelograms are double the triangles with the same base and between the same parallels. Draw CJ and BE. The area of the rectangle AHGJ is double the area of triangle JAC, and the area of square ACLE is double triangle BAE. The two triangles are congruent by SAS. The same result follows in a similar manner for the other rectangle and square.

(Katz, 1993) Click here for a GSP animation to illustrate this proof. The next three proofs are more easily seen proofs of the Pythagorean Theorem and would be ideal for high school mathematics students. In fact, these are proofs that students could be able to construct themselves at some point. The first proof begins with a rectangle divided up into three triangles, each of which contains a right angle. This proof can be seen through the use of computer technology, or with something as simple as a 3x5 index card cut up into right triangles. Figure 4 Figure 5 It can be seen that triangles 2 (in green) and 1 (in red), will completely overlap triangle 3 (in blue). Now, we can give a proof of the Pythagorean Theorem using these same triangles. Proof: Figure 8 We have proved the Pythagorean Theorem. The next proof is another proof of the Pythagorean Theorem that begins with a rectangle. It begins by constructing rectangle CADE with BA = DA. Next, we construct the angle bisector of <BAD and let it intersect ED at point F.

Thus, <BAF is congruent to <DAF, AF = AF, and BA = DA. So, by SAS, triangle BAF = triangle DAF. Since <ADF is a right angle, <ABF is also a right angle. Figure 9 Next, since m<EBF + m<ABC + m<ABF = 180 degrees and m<ABF = 90 degrees, <EBF and <ABC are complementary. Thus, m<EBF + m<ABC = 90 degrees. We also know that m<BAC + m<ABC + m<ACB = 180 degrees. Since m<ACB = 90 degrees, m<BAC + m<ABC = 90 degrees. Therefore, m<EBF + m<ABC = m<BAC + m<ABC and m<BAC = m<EBF. By the AA similarity theorem, triangle EBF is similar to triangle CAB. Now, let k be the similarity ratio between triangles EBF and CAB. . Figure 10 Thus, triangle EBF has sides with lengths ka, kb, and kc. Since FB = FD, FD = kc.

Also, since the opposite sides of a rectangle are congruent, b = ka + kc and c = a + kb. By solving for k, we have and By comparing these two triangles, we see that By adding equations 3 and 4 we get: From Figures 11 and 12, with CD, we have that (p + q) = c. By substitution, we get The next proof of the Pythagorean Theorem that will be presented is one in which a trapezoid will be used. Figure 15 By the construction that was used to form this trapezoid, all 6 of the triangles contained in this trapezoid are right triangles. Thus, Area of Trapezoid = The Sum of the areas of the 6 Triangles And by using the respective formulas for area, we get: We have completed the proof of the Pythagorean Theorem using the trapezoid. The next proof of the Pythagorean Theorem that I will present is one that can be taught and proved using puzzles. These puzzles can be constructed using the Pythagorean configuration and then, dissecting it into different shapes.

Before the proof is presented, it is important that the next figure is explored since it directly relates to the proof. Figure 16 In this Pythagorean configuration, the square on the hypotenuse has been divided into 4 right triangles and 1 square, MNPQ, in the center. Since MN = AN - AM = a - b. Each side of square MNPQ has length of a - b. This gives the following: Area of Square on the hypotenuse = Sum of the Areas of the 4 triangles and the Area of Square MNPQ As mentioned above, this proof of the Pythagorean Theorem can be further explored and proved using puzzles that are made from the Pythagorean configuration. Students can make these puzzles and then use the pieces from squares on the legs of the right triangle to cover the square on the hypotenuse. This can be a great connection because it is a "hands-on" activity. Students can then use the puzzle to prove the Pythagorean Theorem on their own. Figure 17 To create this puzzle, copy the square on BC twice, once placed below the square on AC and once to the right of the square on AC as shown in Figure 17.

Proof Using Figure 17: Area 6 (and Area 5): Area 7 (and Area 4): By adding all of these areas together we get the following result: and Thus, we have proved the Pythagorean Theorem for the puzzle. The proofs presented here are just a few of the many proofs of the Pythagorean Theorem. The Pythagorean Theorem is a very important concept for students to learn and to understand. It cannot be stressed enough that students need to understand the geometric concepts behind the theorem as well as its algebraic representation. This can be accomplished through the use of technology, manipulatives, and proofs. Students who are taught the Pythagorean Theorem using these methods will see the connections, and thus, benefit greatly. Bibliography Beamer, James E. . Using Puzzles to Teach the Pythagorean Theorem: Mathematics Teacher. May 1989; NCTM : Reston,VA. Burton, David M. . Burton's History of Mathematics: An Introduction (Third edition). 1991; William C. Brown Publishers: Dubuque, IA. deLemos, Jamie. The Pythagorean Theorem: Mathematics Teacher.

January 1995; NCTM : Reston,VA. Hoehn, Larry. A New Proof of the Pythagorean Theorem: Mathematics Teacher. February 1995; NCTM: Reston, VA. Katz, Victor J. . A History of Mathematics. 1993; Harper Collins: New York, New York. Lufkin, Dan. . The Incredible Three-by-Five Card! : Mathematics Teacher. February 1996: Reston, VA. Pythagorean Theorem Essay - 1149 Words Essay about math pythagorean theorem ... Pythagorean Theorem Pythagoras was born in Samos, Greece around 570 BCE. From there he emigrated to Croton, Italy where most of his most important ideas and theories would develop. Early on, Pythagoras founded a society of disciples where he introduced the idea eternal recurrence into Greek thought, and it was Pythagoras’ ambition to reveal the basis of divine order. This is how Pythagoras came to mathematics, and he saw math as a purifier of the soul, and...

1017 Words | 3 Pages Triangles and Pythagorean Theorem Essay ...triangles have special properties which make it easier to conceptualize and calculate their parameters in many cases. The side opposite of the right angle is called the hypotenuse. The sides adjacent to the right angle are the legs. When using the Pythagorean Theorem, the hypotenuse or its length is often labeled with a lower case c. The legs (or their lengths) are often labeled a and b. Either of the legs can be considered a base and the other leg would be... 2381 Words | 10 Pages Pythagoras Theorem Essay ...Pythagorean Theorem: Proof and Applications Kamel Al-Khaled & Ameen Alawneh Department of Mathematics and Statistics, Jordan University of Science and Technology IRBID 22110, JORDAN E-mail: kamel@just.edu.jo, ———————————————————————————————— Idea Investigate the history of Pythagoras and the Pythagorean Theorem. Also, have the opportunity to practice applying the Pythagorean Theorem to several...

1518 Words | 9 Pages Pythagorean Theorem Essay ...ancient Greek mathematician and philosopher. Pythagoras was responsible for important developments in the history of mathematics, astronomy, and the theory of music. The thing that Pythagoras is probably the most famous for is the Pythagorean Theorem. The Pythagorean Theorem is used in the field of mathematics and it states the following: the square of the hypotenuse of a right triangle is equal to the sum of the squares of the two... 735 Words | 2 Pages Scarecrow s Pythagorean Theorem Essay ...THE WIZARD OF OZ 2 The Wizard of Oz Scarecrow’s Speech on Pythagorean Theorem The Pythagorean theorem is one of the earliest theorems known to ancient civilization. The well-known theorem is named after the Greek mathematician and philosopher, Pythagoras. In the Wizard of Oz, after the Scarecrow gets a brain, he states the Pythagorean theorem. However, he... 392 Words | 2 Pages The History of the Pythagorean Theorem Essay ...A Brief History of the Pythagorean Theorem Just Who Was This Pythagoras, Anyway?

Pythagoras (569-500 B.C.E.) was born on the island of Samos in Greece, and did much traveling through Egypt, learning, among other things, mathematics. Not much more is known of his early years. Pythagoras gained his famous status by founding a group, the Brotherhood of Pythagoreans, which was devoted to the study of mathematics. The group was almost cult-like in that... 577 Words | 2 Pages Pythagoras Theorem Essay ...------------------------------------------------- Pythagorean Theorem In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle (right-angled triangle). In terms of areas, it states: In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares... 1825 Words | 6 Pages Pythagorean Theorem Essay ...PYTHAGOREAN THEOREM More than 4000 years ago, the Babyloneans and the Chinese already knew that a triangle with the sides of 3, 4 and 5 must be a right triangle.

They used this knowledge to construct right angles. By dividing a string into twelve equal pieces and then laying it into a triangle so that one side is three, the second side four and the last side five sections long, they could easily construct a right angle. A Greek scholar named Pythagoras, who... 2616 Words | 9 Pages Pythagoras Theorem Pythagoras' Theorem Over 2000 years ago there was an amazing discovery about triangles: When a triangle has a right angle (90°) ... ... and squares are made on each of the three sides, ... ... then the biggest square has the exact same area as the other two squares put together! It is called "Pythagoras' Theorem" and can be written in one short equation: a2 + b2 = c2 c is the longest side of the triangle a and b are the other two sides Definition The longest side of the triangle is called the "hypotenuse", so the formal definition is: In a right angled triangle: the square of the hypotenuse is equal to the sum of the squares of the other two sides. Sure ... ? Let's see if it really works using an example.

Example: A "3,4,5" triangle has a right angle in it. Let's check if the areas are the same: 32 + 42 = 52 It works ... like Magic! Why Is This Useful? If we know the lengths of two sides of a right angled triangle, we can find the length of the third side. (But remember it only works on right angled triangles!) How Do I Use it? Write it down as an equation: Now you can use algebra to find any missing value, as in the following examples: Example: Solve this triangle. You can also read about Squares and Square Roots to find out why √169 = 13 Example: Solve this triangle. Take 81 from both sides: b2 = 144 Example: What is the diagonal distance across a square of size 1? c2 = 2 c = √2 = 1.4142... It works the other way around, too: when the three sides of a triangle make a2 + b2 = c2, then the triangle is right angled. Example: Does this triangle have a Right Angle?

a2 + b2 = 102 + 242 = 100 + 576 = 676 c2 = 262 = 676 Yes, it does have a Right Angle! Example: Does an 8, 15, 16 triangle have a Right Angle? Does 82 + 152 = 162 ? 82 + 152 = 64 + 225 = 289, but 162 = 256 So, NO, it does not have a Right Angle Example: Does this triangle have a Right Angle? So this is a right-angled triangle And You Can Prove The Theorem Yourself ! Get paper pen and scissors, then using the following animation as a guide: Draw a right angled triangle on the paper, leaving plenty of space.

Draw a square along the hypotenuse (the longest side) Draw the same sized square on the other side of the hypotenuse Draw lines as shown on the animation, like this: Cut out the shapes Arrange them so that you can prove that the big square has the same area as the two squares on the other sides Another, Amazingly Simple, Proof Here is one of the oldest proofs that the square on the long side has the same area as the other squares. Watch the animation, and pay attention when the triangles start sliding around. You may want to watch the animation a few times to understand what is happening. The purple triangle is the important one. becomes Requirements The Pythagorean Theorem The Pythagorean Theorem or the Pythagoras’ theorem as referred to by some mathematicians is arguably one of the most useful formulae not only in math but science at large. Attributed to the mathematician, Pythagoras, who found it; this theorem shows the relationship that exists between different sides of a right-angled triangle.

Though the theorem is interpreted as a relation that shows the relationship between the different sides of a right-angled triangle, the initial finding showed that the sum area of two small square blocks placed to form a right angle is equal to the area of a square block placed across the two small blocks as shown: In the simplest terms, the area of the two small squares (in blue) is equal to the area of larger square in maroon colour. i.e. (a × a) + (b × b) = (c × c) a2 + b2 = c2 The Pythagorean Theorem states that in any right-angled triangle, the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares that form the other two sides (legs) of the triangle. Although Pythagorean Theorem has been understood to work only for right triangles, this theorem actually applies to any type of a triangle as long as you can identify two shorter sides (legs) and one longer side which represent the hypotenuse. The best way however, would be to use the law of cosines for non right-angle triangles if you are looking to find the most accurate results.

If the lengths of the two shorter sides of a triangle are known, the length of the hypotenuse can be found by using the formula shown: On the other hand, if the size of the hypotenuse is given with one of the shorter sides, then the formulae used to find one of the sides are as shown: Let us now look at how we can put these concepts into practice. We are going to consider a right triangle whose sides are presented as a pair at any given time and we are supposed to find the third side. Concepts of area will also come in handy here. Example 1 Find the hypotenuse of a right triangle whose sides measure 5 cm and 12 cm respectively. Solution The hypotenuse of the triangle is 13 cm Example 2 The area of a right triangle is 64 cm2. If the height of the triangle measures 10 cm, find the base and the hypotenuse of the triangle. Pythagoras Theorem Geometry Pythagoras' Theorem Pythagoras of Samos (c570-495BC) was a philosopher and mathematician and formed a cult relating mathematics to everything. Actually, very little is known about him.

We cannot be sure that he actually discovered the proof of the relationship between the squares of the sides of a right angled triangle, or whether it was one of his successors. Like many labels in mathematics, calling the theorem (or identity) Pythagoras' Theorem is a convenient way of referring to it. Pythagoras Theorem states: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. A useful right angled triangle is that with sides 3, 4, 5. Of course most right angled triangles have sides that are not integers, but irrational numbers. For instance: 1, 1, √2; and √3, 2 1. Proof 1 and Proof 2 below are basically the same, with Proof 2 using more modern labelling. Proof 1 The sketch on the right can be used as a guide to proving the theorem geometrically. Another proof is here . As with many geometric proofs, we need to make the appropriate construction lines. Some assumptions First, ACDE is the square on the hypotenuse, AC. The square BCFG is the square on the side BC.

And the square ABJH is the square of the side AB. BXY is a line from B at right angles to AC and DE. BXY is therefore parallel to CD and AE. (Because the sum of the internal angles of the diagonal DE, and those of the diagonal AC are 180°) Prove the area of the square on BC is equal to the area of the rectangle CDYX We note that the area of ΔACF is equal to half the area of the square BCFG, because the area of the square BCFG is BC·CF. And the area of ΔACF, which has a height, BC, is BC·CF/2. Similarly, we note that the area of the triangle BCD is equal to half the area of the rectangle CDYX. This is because the area of ΔBCD, height CX, is CD·CX/2 and the area of the square CDYX is CD·CX Now if ΔACF is equal in area to ΔBCD, then the area of square CB is equal to the area of rectangle CDYX.

This is indeed the case because ΔACF and ΔBCD each have sides equal to the length of AC, the hypotenuse and sides equal to CB. The angle between these is 90° + angle ACB, so the triangles are congruent (side angle side, SAS). So the area of the rectangle CDYX is equal to the area of the square on CB. Prove the area of the square on AB (the other side) is equal to the area of the rectangle AXYE We draw the lines JC and BE. As before we note the area of ΔACJ is half the area of the square on AB, because (as before) the height of the triangle is equal to the side of the square and the base is the same as the side of the square. Similarly, ΔABE is equal to half the area of AXYE, because the area of ΔABE is AX·AE/2, which is half the area of the rectangle AXYE. As before, if ΔACJ is equal to ΔABE, then the area of AXYE is equal to the area of the square on AB.

Also, as our previous argument, the triangles have equal sides and equal enclosed angles (both are 90° + angle BAX). Conclusion Because the area CDYX is equal to the square on BC and the area AXYE is equal to the square on AB, the square on AC is equal to the sum of the squares on AB and BC ■ Proof 2 This is really a repetition of the previous proof, using different labelling. The sketch is made exactly as before. So, we represent the sides of the triangle ABC using the small letters, as shown. So the side AC is b. We wish to show that: a2+c2=b2 [2.01] Drop a perpendicular from B, through X and Y, cutting AC at X, and CX=b-x and AX=x. (First, considering the blue lines) The area of ΔACF=a2/2 (Half the base, a, times the height). [2.01] The area of ΔBCD=b(b-x)/2 (Half the base, b, times the height, b−x).

[2.02] ΔACF and ΔBCD are congruent because, CB=CF (=a), and CD=AC (=b), and angle ACD=angle ACF (=ACB+90°) So Algorithm Implementation/Mathematics/Pythagorean theorem - Wikibooks, open books for an open world Algorithm Implementation/Mathematics/Pythagorean theorem Jump to: navigation , search The Pythagorean theorem: The sum of the areas of the two squares on the legs (a and b) equals the area of the square on the hypotenuse (c). In mathematics, the Pythagorean theorem or Pythagoras' theorem is a relation in Euclidean geometry among the three sides of a right triangle (right-angled triangle). In terms of areas, it states: In any right triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle). The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation: [1] a

Question: In mathematics, whose theorem states that the square on the hypotenuse of a right-angled triangle is equal in area to the sum of the squares on the other two sides?
Answer: Pythagoras