what is the pythagorean theoream?
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They're both correct.
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This is a right isosceles triangle.  It's right because the Pythagorean Theorem works and it's isosceles because 2 sides are equal.
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It was believed that he saw the shadows of the pillars on the tiled temple floor, that looked like a right angled triangle with squares attached. \nFrom there he developed his theory that the square drawn on the hypotenuse of a right angled triangle, is equal in area to the sum of the squares drawn on the other two sides.\nPythagoras put forward a proof, but said proof contained errors to do with transient numbers, however it seemed the theory was sound.\nSome two centuries later the Great Euclid himself proved it, with his brilliant proof using congruent triangles, true poetry of geometry.
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ABOUT THE BINOMIAL THEOREM I AM TEEMING WITH A LOT OF NEWS\n\nWITH MANY CHEERFUL FACTS ABOUT THE SQUARE ON THE HYPOTENEUSE\n\n(W. S. Gilbert, I am The Very Model of a Modern Major General)\n\nI am assuming we are dealing with what we know is a PRIMITIVE Pythagorean triple ie one like (3,4,5) or (5,12,13) or (8,15,17) or (20,21,29) in which there is no common factor between the three sides, NOT one like (6,8,10) which is divisible by 2.\n\nThere are tables published of the smallest triples, in ascending order of hypooteneuse, so you may be able to look it up and not need to calculate it. \n\nhttp://www.math.rutgers.edu/~erowland/tripleslist-long.html\n\nHowever, two points.  \n\nNot every integer can be a hypoteneuse. There are gaps in the table eg after (3,4,5) the next smallesr hypoteneuse is 13 (in (5, 12, 13). There are no primitive triples with a hypoteneuse of 6, 7, 8, 9, 10, 11 or 12. And therefore if you were given a hypoteneuse of 12, there are no integer solutions to it.\n\nSecondly, there are cases where a number is hypoteneuse to more than one triple. 65 is the lowest such number and 85 is the next.\n\n(16, 63, 65) and (33, 56, 65)\n(13, 84, 85) and (36, 77, 85)\n\nThese are known as siblings. Both give correct solutions and you can't say one is right and the other is wrong. Here are a couple of special cases, with more than two solutions:\n\n(47, 1104, 1105)\n(744, 817, 1105)\n(576, 943, 1105) \n(264, 1073, 1105)\n\nThere are EIGHT siblings with a hypoteneuse of 32045, SIXTEEN siblings with a hypoteneuse of 1185665, THIRTY-TWO siblings with a hypoteneuse of 48612265, AND SIXTY FOUR siblings with a hypoteneuse of 2576450045.\n\nHere are rthe eight siblings with a hypoteneuse of 32045:\n\n(2277, 31964, 32045) \n(8283, 30956, 32045)\n(17253, 27004, 32045) \n(21093, 24124, 32045)\n(22244, 23067, 32045)\n(15916, 27813, 32045)\n(6764, 31323 , 32045)\n(716, 32037, 32045)\n\nLacking access to tables, for small numbers, you may be able to work it out, by trial and error, using the following properties\n\n(a) the hypoteneuse must be the biggest side and must be an odd number (to be a primitive triple)\n\n(b) one of the legs will be an odd number and the other an even number\n\n(c) As a result, the difference between the two small sides will always be an odd number. In fact it is either a prime number (eg 1, 7 and 17 feature, but 11, 13 and 19 do not) or a power of a prime number (49 is the lowest such).\n\n(d) the sum of the two legs must be bigger than the hypoteneuse\n\n(e) one and only one of the legs must be divisible by 3\n\n(f)  one and only one of the legs must be divisible by 4\n\n(g)  one and only one side (which might be the hypoteneuse) must be divisible by 5\n\n(h) follows from (e) (f) and (g)\n\n(h) the product of the three sides is always divisible by 60\n\nThus if you knew the hypoteneuse and could see it was not divisible by 5, then it follows the legs must, between them, be divisible BY 3 4 AND 5\n\nSo if you knew the hypoteneuse is 61, it seems obvious to try one of the legs as 60 and see if that works (it does, (11, 60, 61) is a triple. \n\nIn fact you can work out the third member is 11 by adding 60 and 61 to see if it is a perfect square (it is 121) and taking the square root of it, if so. (another property of triples whose two largest sides differ by 1).\n\nGiven that 17 is a hypoteneuse and knowing one of the other sides must be 3 x 4 or 3 x 5 (4 x 5 would be larger than 17) you have a choice of (12, x, 17) where x = 5 or 15 (has to be odd as 12 is even) or (y, 15, 17) where y is 4, 8, 12 or 16)\n\nIt turns out to be (8, 15, 17)\n\nAgain, knowing a property of all triples where the hypoteneuse and the larger leg differ by 2, there is a shortcut, that leads to 8 ... take the average of the two longer sides, multiply by 4 and take the square root. (thus 16 => 64 => 8)\n\nThis holds true for \n(12, 35, 37)\n(16, 63, 65)\n(20, 99, 101) etc\n\nSo it is possible to make intelligent guesses at the
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Is this the Pythagorean theorem? There are many proper ways of proving it. Check wikipaidia below.
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A pythagorean primitive is a triple in which their only common divisor is 1.\n\nFor example: 3,4,5 is a Pythagorean primitive and 6,8,10 is not.
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1. pythagoros theorem applies to any right angled triangle.\n\n2. The geometric mean of a set of positive data is defined as the nth root of the product of all the members of the set, where n is the number of members.\n\n3. in an isosceles triangle 2 sides and 2 angles are equal\n\n4.hypotenuse is opp. to the right angle.it is the largest side.\n\n5.true
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Pythagoras wasn't just a mathematician -- he was a cult leader, as well.  For more information, you could try a site like Wikipedia:
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I always found this one interesting:\n\nYou may have heard of the Ancient Greek society called the Pythagoreans (they were followers of Pythagorus - hence the Pythagorean Theorem).  Anyway, rumor says that they used the pentagon as their symbol whenever they wanted to correspond with each other via a letter.  Why did they do this?  Because it is VERY hard to draw a regular pentagon freehand.  Go ahead, try it, I would bet that it looks pretty funny when you're done.\n\nAnyway, they had a secret to drawing it that they only shared with others in the group:  They would take 5 triangles (probably equilateral - but isosceles would work also) and assemble them into the top of a pyramid-like figure.  When you look at the bottom of this figure, it makes a pentagon - and a regular one.  YOu can then trace around the outside edge of this pentagon, take apart your pyramid-like thingie, and you have your seal which others can't re-create.\n\nAnyway, that's the rumor about it... I think it's pretty cool...
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The Pythagorean Theorem definitely: a^2+b^2=c^2 \n\nPoint slope form: y - y1 = m(x-x1) \nStandard y equation: y=mx+b\n...Sorry, cant think of any more. \nSee, the rest of the equations that i could give you are not the ones EVERYONE should know :)
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Oh c'mon! Smiddy hasn't had geometry for 20 years! but this is basic. Pythagorean theorem states that on a triangle A-squared plus B-squared = C-squared where C is the hypotenuse. The hypotenuse of a triangle is the diagonal bit if it's two halves of a square and we know by definition that on a square A and B are equal. Therefore 2X squared = 8-squared or 2x^2 = 64\n\nx^2 = 32\nx = sq.rt 32\n\nthe area of the original square is x times x so you're back at 32 square centimeters
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You **did** say that the Pythagorean Theorem was a given, right? I think that your INTENT is to prove the converse of the Pythagorean Theorem, right?\n\nLet x, y, and z be positive integers such that\nx^2 + y^2 = z^2.\nClearly z > x and z > y (otherwise either x^2 or y^2 must be less than zero; prove this for yourself if you must).\n\nLet x and y be the sides of a right triangle that form the right angle.\nThen, by the Pythagorean Theorem the length of the hypotenuse h is\nh = √(x^2 + y^2)\nh = √(z^2) . . . . . . . . (given equation above)\nh = z . . . . . . . . . . . (given z is positive) \n        \nTherefore, by side-side-side congruence, any triangle whose sides are of lengths, x, y, and z, where x^2 + y^2 = z^2, must be a right triangle with hypotenuse z.\n\n\nASIDE TO HOOD...: You are using the Pythagorean Theorem to prove the Pythagorean Theorem. Although the questioner's intent is not perfectly clear, I don't think the intent was to prove "if x, then x". If so there are easier ways to do so.
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Think what you are looking for is Euclidian Space:\n\nAround 300 BC, the Greek mathematician Euclid laid down the rules of what has now come to be called "Euclidean geometry", which is the study of the relationships between angles and distances in space. Euclid first developed "plane geometry" which dealt with the geometry of two-dimensional objects on a flat surface. He then went on to develop "solid geometry" which analyzed the geometry of three-dimensional objects. All of the axioms of Euclid have been encoded into an abstract mathematical space known as a two- or three-dimensional Euclidean Space. These mathematical spaces may be extended to apply to any dimensionality, and such a space is called an n-dimensional Euclidean space or an n-space. This article is concerned with such mathematical spaces.\n\nIn order to develop these higher dimensional Euclidean spaces, the properties of the familiar Euclidean spaces must be very carefully expressed and then extended to an arbitrary dimension. Although the resulting mathematics is rather abstract, it nevertheless captures the essential nature of the Euclidean spaces we are all familiar with.\n\nAn essential property of a Euclidean space is its flatness. Other spaces exist that are not Euclidean. For example, the surface of a sphere is not a Euclidean space, nor is the four-dimensional spacetime described by the theory of relativity when gravity is present.
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It was Pythagoras who was the first to question the flat earth theory, although probably just the most famous....incidently it was considered heracy at the time.
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The Pythagorean Theorem deals with right triangles.  The two sides of the right triangle that are connected to the right angle are its legs, and the third (longest) side is called the hypotenuse.  The theorem says that\n\nleg squared plus other leg squared equals hypotenuse squared\n\nSo if one leg is 3 and the other is 4, the hypotenuse is 5 because 3 squared (9) plus 4 squared (16) = 25, which is 5 squared.
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This theorem has an amazing number of proofs but this theorem(like many other mathematical concepts) as you said was known way before pythogorean times but only to INDIANS who didnt share it on the global scale!!
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The formula is credited to Heron of Alexandria in the 1st century, and a proof can be found in his book Metrica. It is now believed that Archimedes already knew the formula, and it is of course possible that it has been known long before.\n\nLet the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that\n S2 = p(p - a)(p - b)(p - c)\n\nW.Dunham analyzes the original Heron's proof in his Journey through Genius.\n\nFor the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that\n p - a = (- a + b + c)/2, p - b = (a - b + c)/2, p - c = (a + b - c)/2\n\nIt takes a little algebra to show that\n \n(a + b + c)(- a + b + c)(a - b + c)(a + b - c)\n= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)\n\nFor the right triangle, this expression is equal to 16S2 = 4a2b2. So we have\n 4a2b2= 2a2b2 + 2a2c2 + 2b2c2 - (a4 + b4 + c4)\n\nTaking all terms to the left side and grouping them yields\n (a4 + 2a2b2 + b4) - 2a2c2 - 2b2c2 + c4 = 0\n\nWith a little more effort\n (a2 + b2)2 - 2c2(a2 + b2) + c4 = 0\n\nAnd finally\n [(a2 + b2) - c2]2 = 0\nRemark\n\nFor a quadrilateral with sides a, b, c and d inscribed in a circle there exists a generalization of Heron's formula discovered by Brahmagupta. In this case, the semiperimeter is defined as p = (a + b + c + d)/2. Then the following formula holds\n S2 = (p - a)(p - b)(p - c)(p - d)\n\nSince any triangle is inscribable in a circle, we may let one side, say d, shrink to 0. This leads to Heron's formula.\n\nFor more information you can go to www.wikipedia.com  \nhttp://mathworld.wolfram.com/HeronsFormula.html\nhttp://www.cut-the-knot.org/pythagoras/herons.shtml\n\nI hope this helps! =-)