help find facts abot the pythagorean theorem?
97476
More than you would ever want to know can be found at:
142078
Don't you mean the Pythagorean Theorem?  That theorem is a basic tenent of Geometry that states: In any right triangle, the square of the length of the hypotenuse (side opposite the right angle) is equal to the sum of the squares of the other two sides.   A^2 + B^2 = C^2.
1063957
you can't go wrong with :\n\nThe quadratic formula\nThe Pythagorean theorem\nlaw of sin/ law of cos (for trig)\narea and circumference of a circle\n\nThese are some general formulas that appear on exams\nI could be more precise if i knew the level you are aiming for.
290528
The statement of the Theorem was discovered on a Babylonian tablet circa 1900-1600 B.C. Whether Pythagoras (c.560-c.480 B.C.) or someone else from his School was the first to discover its proof can't be claimed with any degree of credibility. Euclid's (c 300 B.C.) Elements furnish the first and, later, the standard reference in Geometry. In fact Euclid supplied two very different proofs: the Proposition I.47 (First Book, Proposition 47) and VI.31. The Theorem is reversible which means that a triangle whose sides satisfy a2 + b2 = c2 is necessarily right angled. Euclid was the first (I.48) to mention and prove this fact.
641500
may be this site can help you\n\nhttp://www.perseus.tufts.edu/GreekScience/Students/Jordana/MATH.html
646347
This is a *VERY* famous math problem called "Fermats Last Theorem".  It took over 250 years to prove, but it has been proven to be true.\n\nHere is a snippet from the link below:\n"Using sophisticated tools from algebraic geometry (in particular elliptic curves and group schemes), and number theory (modular forms, Galois representations, Hecke algebras, and Iwasawa theory), the English-born mathematician Andrew Wiles, from Princeton University, with help from his former student Richard Taylor, devised a proof of Fermat's last theorem that was published in 1995 in the journal Annals of Mathematics."
1130099
(√3)² + 1 ² = 2²\nsin α = √3/2 α = 60º;  ß = 90 - α;  ß = 30º\n\nAnd your second question is\naⁿ = c² - b².  I'm going to assume that you wanted to put in a, b and c are all integers and a > 1.  Because if not, your theorem is not true and cannot be proven.  With that stipulation, this theorem passes the "intuitive" test.  It makes sense.\nWe know for certain that, for all even n, it must be true by the Pythagorean theorem.  So only odd n are of consequence to this theorem.\n\nI'm going to take a moment here and say that if you're trying to use this theorem as a FLT lemma, you must understand that it won't fly.  The lemma must use the carry the same stipulations as the successor theorem.  Said the other way, the corollary is bound by the restrictions of the theorem.\n\na=2, n=3 gives us c=3 and b=1.  8=9-1.  I took some time trying to find a counter-example and could not.  I did find some neat patterns emerging.\n\nThe proof to your theorem is simple: aⁿ = a²[(½a+½)² - (½a-½)²] because the expression in brackets will always be equal to a (a is odd prime; composite a can be simplified to prime a).\n\nI also found that 2ⁿ= [2^(n-2)+1]² - [2^(n-2)-1]².  So, in addition to all the solutions from the above expression, there will be another set of solutions for 2ⁿ; n is odd integer > 5.
1084262
use pythagoras theorm,if left hand side is equal to right hand side then it is a right triangle.\n\nsince 8 is the biggest no here that means the length of hypotenues is 8. \n\nlet side 1 be a=4\nlet side 2 be b=4 sqrt(3)\nlet hypotenuese be c=8\n\naccording to pythagoras theorm,\n\n(a)*(a)+(b)*(b)=(c)*(c)\n\nL.H.S\n\n=4*4+4sqrt(3)*4sqrt(3)\n=16+48\n=64\n\nR.H.S\n=8*8\n=64\n\nsince L.H.S=R.H.S\n\ntherefore, it is a right triangle\n\nhope u understand it\nif u hv any prob u can contact me
2767
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 :)
76447
p1v1t1=p2v2t2\n\nSee Bernoulli's theorem and Pascals Law
785502
This is a right isosceles triangle.  It's right because the Pythagorean Theorem works and it's isosceles because 2 sides are equal.
126450
President Garfield came up with a proof for the Pythagorean Theorem.
893767
Pythagoras Theorem. a^2=b^2+c^2\n\nSo b^2 = 58.37\nc^2 = 88.55\n\nb^2 + C^2 = 146.92\nSq Rt of 146.92 = 12.12.\nSo A(your diagonal line) is 12.12mtrs
1099275
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
190806
You would think Pythagoras, but actually not originally.\nHe studied many years at the library of Alexandria (Ancient Egypt), whereupon he came upon this theorem (now named after him) first recorded by an ancient Egyptian mathematician known as (I think) Tet.\nSince it was not permitted to take things from the library, or even copy them he memorized it, then later taught it to his disciples as his own.
730817
Define the lengths of the smallest triangle:\nAP = ½ AC\nAQ = ½ AB\n\nSquare the equations:\nAP^2 = ¼ AC^2\nAQ^2 = ¼ AB^2\n\nPythagorean theorem for smallest triangle:\nAP^2 + AQ^2 = PQ^2\n\nSubstitute:\n ¼ AC^2 + ¼ AB^2= PQ^2\n\nMultiply by 4:\nAB^2 + AC^2 = 4PQ^2\n\nPythagorean theorem for triangles ABP and ACQ:\nAP^2 + AB^2 = BP^2\nAQ^2 + AC^2 = CQ^2\n\nAdd the two equations:\nBP^2 + CQ^2 = AP^2 + AQ^2 + AC^2 + AB^2 \n\nSubstitute:\nBP^2 + CQ^2 = PQ^2 + AC^2 + AB^2\n\nSubstitute:\nBP^2 + CQ^2 = PQ^2 + 4PQ^2\n\nTherefore:\nBP^2 + CQ^2 = 5PQ^2\n\n\nIf you get stumped with this type problem, sometimes it helps to start at the end and work backwards...
560099
Is this the Pythagorean theorem? There are many proper ways of proving it. Check wikipaidia below.