THE PYTHAGOREAN THEOREM
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You are attempting to find the diagonal of a square with 5 on a side. The answer is obviously 5*√(2). C, the hypotenuse of a right triangle, is the longest side of the said triangle. Therefore, C must be larger than 5 in your particular problem. Hence, the first two answers are incorrect.\n\nExample:\nIn general, a square with sides of length x has a diagonal of length x*√(2).\n\na^2 + b^2 = c^2\nx^2 + x^2 = c^2\n2*x^2 = c^2\n√(2*x^2) = √(c^2)\n√(x^2)*√(2) = c\nx*√(2) = c\n\nIf x = 5, then c = 5*√(2).
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Fermat's Last Theorem states that \nx^n + y^n = z^n \n\nhas no non-zero integer solutions for x, y and z when n > 2. \n\nHis proposition was about an equation which is closely related to Pythagoras' equation. Pythagoras' equation gives you:\n\nx^2 + y^2 = z^2\n\nYou can ask, what are the whole number solutions to this equation, and you can see that:\n3^2 + 4^2 = 5^2\nand\n5^2 + 12^2 = 13^2\n\nAnd if you go on looking then you find more and more such solutions. Fermat then considered the cubed version of this equation:\nx^3 + y^3 = z^3\n\nHe raised the question: can you find solutions to the cubed equation? He claimed that there were none. In fact, he claimed that for the general family of equations:\nx^n + y^n = z^n where n is bigger than 2\n\nit is impossible to find a solution. That's Fermat's Last Theorem.\n\n\nProof : \n\nn = 3\n\nx^3 + y^3 = z^3 has integer solutions -> xyz = 0\n\n(1) Let's assume that we have solutions x,y,z to the above equation.\n\n(2) We can assume that x,y,z are coprime. \n\n(3) First, we observe that there must exist p,q such that \n\n(a) gcd(p,q)=1\n(b) p,q have opposite parities (one is odd; one is even)\n(c) p,q are positive.\n(d) 2p*(p2 + 3q2) is a cube.\n\n(4) Second, we know that gcd(2p,p2+3q2) is either 1 or 3. \n\n(5) If gcd(2p,p2+3q2)=1, then there must be a smaller solution to Fermat's Last Theorem n=3. \n\n(6) Likewise, if gcd(2p,p2+3q2)=3, then there must be a smaller solution to Fermat's Last Theorem n=3.\n\n(7) But then there is necessarily a smaller solution and we could use the same argument on this smaller solution to show the existence of an even smaller solution. We have thus shown a condition of infinite descent.\n\n\nI can not have much explaination about in general n. Anyways, that was a real nice question.\n\nI read this somewhere, its not my creation. \n\nAll the best and have nice time
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the answer is pi 3.14. use the pythagoran therom with x= pi(r2) then compensate for the remainder and it'll work itself out. If that doesn't work, try FOIL. first, outside, inside, last.
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First of all, you don't have to use the Pythagorean theorem for this. Find x, so 6x=54. x=9. Then, divide 45 by 9, because the lines are congruent, and you get 5 ft.
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Wow, what an interesting question. I'm surprised no one has solved it yet, but that may be a consequence of the fact that no one has understood it yet. Clearly, schools need to focus more on reading comprehension ^_^.\n\nTo answer your question - not only can we prove that there exist integers a and b satisfying your assumptions, but we can even show a method of finding them. The proof is long, so sit back, relax, grab your favorite drink, and prepare to read.\n\nFirst, consider that any solutions of these equations must obey the quadratic formula: x=(-b±√(b²-4ac))/2a. Substituting the coefficients of the  actual equations in, we find that the solutions of the first equation are (-p±√(p²-4q))/2 and the solutions of the second equation are (-p±√(p²+4q))/2. Let us consider only the greater of the solutions in each pair - that is, replace each ± with a +. The difference between these solutions is then (-p+√(p²-4q))/2 - (-p+√(p²+4q))/2. Call this number K. We can then perform some simple algebra as follows:\n\n(-p+√(p²-4q))/2 - (-p+√(p²+4q))/2=K\nmultiply both sides by 2 and simplify\n√(p²-4q)-√(p²+4q)=2K\nsquare both sides\n(p²-4q) - 2√((p²-4q)(p²+4q)) + (p²+4q)=4K²\nsimplify\n2p²-2√(p^4-16q²)=4K²\ndivide both sides by 2\np²-√(p^4-16q²)=2K²\nsubtract 2K² and add √(p^4-16q²) to both sides\np²-2K²=√(p^4-16q²)\nsquare both sides\np^4-4K²p²+4K^4=p^4-16q²\nmultiply both sides by -1 and simplify\n16q²=4K²p²-4K^4\ndivide both sides by 4\n4q²=K²p²-K^4\nadd K^4 to both sides\nK^4+4q²=K²p²\ndivide both sides by K² and rearrange\np²=K²+4q²/K²\n\nThus, if we let a=K and b=2q/K, then p²=a²+b² as required by the problem. All that remains is to show that a and b are nonzero integers.\n\nLemma 1: K is an integer\nProof: All solutions of both quadratic equations are integers. K is the difference of two of these solutions - that is, K is the difference of two integers. Therefore, K is an integer.\n\nLemma 2: K is nonzero\nProof: K is the difference between a solution of the first equation and a solution of the second equation. Suppose K=0. Then these equations have a common solution - that is, for some x, x²+px+q=0=x²+px-q. This would imply that q=-q. This can only be true if q=0. But q≠0. Therefore, the equations do not have a common solution and K≠0.\n\nLemma 3: 2q/K is an integer\nProof: Earlier we showed that √(p²-4q)-√(p²+4q)=2K. We can perform some simple algebra as follows:\n\n√(p²-4q)-√(p²+4q)=2K\nmultiply both sides by (√(p²-4q)+√(p²+4q)) and simplify\n2K(√(p²-4q)+√(p²+4q)) = (p²-4q)-(p²+4q)\nsimplify further and rearrange:\n-8q=2K(√(p²-4q)+√(p²+4q))\ndivide both sides by -4:\n2q=-K(√(p²-4q)+√(p²+4q))/2\ndivide both sides by K:\n2q/K=-(√(p²-4q)+√(p²+4q))/2\n\nNote that 2K=√(p²-4q)-√(p²+4q) Adding 2√(p²+4q) to both sides of that identity gives 2K+2√(p²+4q) = √(p²-4q)+√(p²+4q). We can then substitute to get:\n\n2q/K=-(2K+2√(p²+4q))/2\nSimplifying at last yields:\n2q/K=-K-√(p²+4q)\n\nWe already know K is an integer, so if √(p²+4q) is also an integer, then 2q/K is the difference of two integers and therefore an integer. Now we know that the solutions of the second equation are (-p+√(p²+4q))/2 and (-p-√(p²+4q))/2. The difference between them is √(p²+4q). However, since both of the solutions to the second equation are integers, their difference must be an integer. Therefore √(p²+4q) is an integer, and by extension, 2q/K is an integer. Q.E.D\n\nLemma 4: 2q/K is nonzero\nWe know 2q/k=-K-√(p²+4q). Suppose 2q/K=0. Then -K-√(p²+4q)=0 From this we would get:\n\n-K-√(p²+4q)=0\nadd K to both sides and rearrange:\nK=-√(p²+4q)\nmultiply both sides by 2\n2K=-2√(p²+4q)\nremembering 2K=√(p²-4q)-√(p²+4q), we substitute:\n√(p²-4q)-√(p²+4q)=-2√(p²+4q)\nadd √(p²+4q) to both sides and rearrange:\n-√(p²+4q)=√(p²-4q)\n\nHowever we specified in the beginning t
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1.  (0,y)\n2.  (x,x) or (y,,y)\n3.  they are three whole numbers that will satisfy the pythagorean equation, a^2+b^2=c^2. ("^2" means raising the number to 2).  an example would be 3,4,5.\n4. 12\n\nFOrmulas\n\n1.volume:sphere= (4/3)*pi*r*r*r\n\ncone = (1/3)*pi*r*r*h\n\n2.circumference is 2*pi*r
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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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it has already been proved for n > 2. \n\n2^n = t^n - w^n => t^n = 2^n + w^n\n\nFamous Format's Last theorem\n\nIt need to be proved for 2\n\n4= t^2-w^2\nor (t-w)*(t+w) = 4\n\nt-w and t+w have to be both odd or both even\n\nonly solution = t-w = t+w = 2 or w = 0\nso no solution
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Let x be the length of the shorter leg.\nLet 5 be the length of the longer leg.\nLet 3x be the length of the hypotenuse.\n\nUsing the Pythagorean theorem:\n5² + x² = (3x)²\n\nMultiply out:\n25 + x² = 9x²\n\nSubtract x² from both sides:\n25 = 8x²\n\nDivide both sides by 8:\nx² = 25/8\n\nTake the sqrt of both sides:\nx² = 50/16\nx = 5 sqrt(2) / 4\n\nx ≈ 1.76776695 ft.
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1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
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Yes!!!!!!!!! Email me, and I'll try to shed more light on this. Parabolas are where my genius comes out :)
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this is probably not a perfect answer but this site may give you some clues.\n\nhttp://www.gomath.com/Questions/question.php?question=4751\n\naltrnatively type 'prove v*v = 2gr*r/s+v0*v0-2gr' into a search engine and see what comes up.
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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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the pythagorean theorem is "a squared + b squared = c squared"\n\nyou have a right triangle. the two short sides are "a" and "b" and the long side is "c." if you have two of those lengths, you can find the third.\n\nIn your example:\nCall the 34 meters "a" and 41 meters "b." You're trying to find "c."\n\nc squared = 34 squared plus 41 squared = 1156 + 1681 = 2837\n\nc squared = 2837\nc = approximately 53.26\n\nto find the amount saved, you add the two short sides together and subtract the long side: 34+41-53.26
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hint: mean value theorem
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You might try asking this question under "Mathematics."
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Yes. Many. Here's a way to get more:\n\nPick and two numbers A and B with A more than B.\n\nLet x=A^2 -B^2\nLet y=2*A*B\nLet z=A^2 +B^2.\n\nThen it will turn out that x^2+y^2=z^2.\n\nAll primitive Pythagorean triples are of this form, by the way. Primitive means that there are no common divisors between the three numbers.\n\nFor example, if A=5 and B=2, then you get x=21, y=20 and z=29.