PYTHAGORAS THEOREM In mathematics, the Pythagorean theorem—or Pythagorass theorem—is a relation in Euclidean geometry among the three sides of a right triangle. It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. 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^2 + b^2 = c^2\!\, where c represents the length of the hypotenuse, and a and b represent the lengths of the other two sides. The Pythagorean theorem is named after the Greek mathematician Pythagoras (ca. 570 BC—ca. 495 BC), who by tradition is credited with its proof,[2][3] although it is often argued that knowledge of the theorem predates him. There is evidence that Babylonian mathematicians understood the formula, although there is little surviving evidence that they used it in a mathematical framework.[4][5] Mesopotamian, Indian and Chinese mathematicians have all been known for independently discovering the result, some even providing proofs of special cases. The theorem has numerous proofs, possibly the most of any mathematical theorem. These are very diverse, including both geometric proofs and algebraic proofs, with some dating back thousands of years. The theorem can be generalized in various ways, including higher-dimensional spaces, to spaces that are not Euclidean, to objects that are not right triangles, and indeed, to objects that are not triangles at all, but n-dimensional solids. The Pythagorean theorem has attracted interest outside mathematics as a symbol of mathematical abstruseness, mystique, or intellectual power; popular references in literature, plays, musicals, songs, stamps and cartoons abound. The Pythagorean Theorem was known long before Pythagoras, but he may well have been the first to prove it.[6] In any event, the proof attributed to him is very simple, and is called a proof by rearrangement. The two large squares shown in the figure each contain four identical triangles, and the only difference between the two large squares is that the triangles are arranged differently. Therefore, the white space within each of the two large squares must have equal area. Equating the area of the white space yields the Pythagorean Theorem, Q.E.D.[7] That Pythagoras originated this very simple proof is sometimes inferred from the writings of the later Greek philosopher and mathematician Proclus.[8] Several other proofs of this theorem are described below, but this is known as the Pythagorean one. As pointed out in the introduction, if c denotes the length of the hypotenuse and a and b denote the lengths of the other two sides, the Pythagorean theorem can be expressed as the Pythagorean equation: a^2 + b^2 = c^2\, If the length of both a and b are known, then c can be calculated as follows: c = \sqrt{a^2 + b^2}. \, If the length of hypotenuse c and one side (a or b) are known, then the length of the other side can be calculated with the following equations: a = \sqrt{c^2 - b^2}. \, or b = \sqrt{c^2 - a^2}. \, The Pythagorean equation relates the sides of a right triangle in a simple way, so that if the lengths of any two sides are known the length of the third side can be found. Another corollary of the theorem is that in any right triangle, the hypotenuse is greater than any one of the other sides, but less than their sum. A generalization of this theorem is the law of cosines, which allows the computation of the length of any side of any triangle, given the lengths of the other two sides and the angle between them. If the angle between the other sides is a right angle, the law of cosines reduces to the Pythagorean equation.This theorem may have more known proofs than any other (the law of quadratic reciprocity being another contender for that distinction); the book The Pythagorean Proposition contains 370 proofs.[9] Proof using similar triangles This proof is based on the proportionality of the sides of two similar triangles, that is, upon the fact that the ratio of any two corresponding sides of similar triangles is the same regardless of the size of the triangles. Let ABC represent a right triangle, with the right angle located at C, as shown on the figure. Draw the altitude from point C, and call H its intersection with the side AB. Point H divides the length of the hypotenuse c into parts d and e. The new triangle ACH is similar to 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, marked as θ in the figure. By a similar reasoning, the triangle CBH is also similar to ABC. The proof of similarity of the triangles requires the Triangle postulate: the sum of the angles in a triangle is two right angles, and is equivalent to the parallel postulate. Similarity of the triangles leads to the equality of ratios of corresponding sides: \frac{BC}{AB}=\frac{BH}{BC} \text{ and } \frac{AC}{AB}=\frac{AH}{AC}.\, The first result equates the cosine of each angle θ and the second result equates the sines. These ratios can be written as: {BC}^{2}={AB}\times {BH} \text{ and }{AC}^{2}={AB}\times {AH}. \, Summing these two equalities, we obtain {BC}^{2}+{AC}^{2}={AB}\times {BH}+{AB}\times {AH}={AB}\times({AH}+{BH})={AB}^{2} ,\,\! which, tidying up, is the Pythagorean theorem: {BC}^{2}+{AC}^{2}={AB}^{2} \ .\,\! The role of this proof in history is the subject of much speculation. The underlying question is why Euclid did not use this proof, but invented another. One conjecture is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the Elements, and that the theory of proportions needed further development at that time.[10][11] Euclids proof In outline, here is how the proof in Euclids Elements proceeds. The large square is divided into a left and right rectangle. A triangle is constructed that has half the area of the left rectangle. Then another triangle is constructed that has half the area of the square on the left-most side. These two triangles are shown to be congruent, proving this square has the same area as the left rectangle. This argument is followed by a similar version for the right rectangle and the remaining square. Putting the two rectangles together to reform the square on the hypotenuse, its area is the same as the sum of the area of the other two squares. The details follow. 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). The area of a triangle is half the area of any parallelogram on the same base and having the same altitude. The area of a rectangle is equal to the product of two adjacent sides. The area of a square is equal to the product of two of its sides (follows from 3). Next, each top square is related to a triangle congruent with another triangle related in turn to one of two rectangles making up the lower square.[12] 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. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.[13] 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. 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 is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC. Since A-K-L is a straight line, parallel to BD, then rectangle BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. (lemma 2) 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, which appears in Euclids Elements as that of Proposition 47 in Book 1,[14] demonstrates that the area of the square on the hypotenuse is the sum of the areas of the other two squares.[15] This is quite distinct from the proof by similarity of triangles, which is conjectured to be the proof that Pythagoras used.[11][16] Proof by rearrangement We have already discussed the Pythagorean proof, which was a proof by rearrangement. The same idea is conveyed by the leftmost animation below, which consists of a large square, side a + b, containing four identical right triangles. The triangles are shown in two arrangements, the first of which leaves two squares a2 and b2 uncovered, the second of which leaves square c2 uncovered. The area encompassed by the outer square never changes, and the area of the four triangles is the same at the beginning and the end, so the black square areas must be equal, therefore a2 + b2 = c2. A second proof by rearrangement is given by the middle animation. A large square is formed with area c2, from four identical right triangles with sides a, b and c, fitted around a small central square. Then two rectangles are formed with sides a and b by moving the triangles. Combining the smaller square with these rectangles produces two squares of areas a2 and b2, which must have the same area as the initial large square.[17] The third, rightmost image also gives a proof. The upper two squares are divided as shown by the blue and green shading, into pieces that when rearranged can be made to fit in the lower square on the hypotenuse – or conversely the large square can be divided as shown into pieces that fill the other two. This shows the area of the large square equals that of the two smaller ones.[18]
Posted on: Sat, 26 Jul 2014 10:57:44 +0000
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