phythagoras theoram question?
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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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You need to use Pythagorean theorem.\nMeasure the lenght of the lotus plant which is above the water.(X)\nBend the plant to any one side until it is just submerged under the surface of water.\nMeasure the Dist. between the initial and final positions of the plant(Y)(i.e. the initial position and final position where it is submerged)\n\nLet the actual lenght be L.\nThen (L x L) = (Y x Y) + (L - X)(L - X)
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What the Pythagoras theorem states is that the square of the hypotenuse is equal to the sum of the squares of the two other sides (lets call them "a" and "b").\nSo you have to satisfy this condition:\n \nhypotenuse^2 = a^2 + b^2\n \nIn the case hypotenuse = sqrt(3),\n \n3 = a^2 + b^2\n\nSince there are an infinite amount of pairs (a, b) that satisfy this equation, there are infinite triangles.\n \nIf you´re thinking of the special cases (the ones you memorize for geometry and trig) I can only remember the case of sides 1, sqrt(2), sqrt(3). But I´m pretty sure there were 2 other cases (with different hypotenuses).
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sorry i made a mistake....this is correct...\ndivide the base by 2.  Now you have a right triangle.  the height is the same, but the base is now 3.  pygathora (spelling, whatever) theorm is A squared + B squared = C squared.  c is the side opposite the right angle, a and b are the other sides.  so 3 squared + 4 squared is 9 + 16 = 25. The square root of 25 is 5, so that is the length of the side.  the volume = area of base times height.  Area of a triangle is 1/2 base times height.  so 3 (one half of 6) times 4 = 12.  times this by height of 8 = 96 cubic units. largest rectangular side is the 6 (the longest side) times 8 = 48 units squared. hope this helped.  the lateral area is the area of the sides added together. so C times 8 = 40 (area of one side)  add the other side of equal area (the C from the second triangle we got when we split the first one) 40 + 40 = 80.  Add the third side of length 6 times 8 = 48.  80 + 48 is 128 (in units squared).  or all sides added up 5 + 5 + 6 = 16 times 8 = 128.
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Integer solutions to the Pythagorean equation are called Pythagorean Triples. These are more interesting if they are primitive Triples, ie ones in which the 3 sides have no common factor. eg (3, 4, 5) is primitive but (6, 8, 10) is non-primitive.\n\nHere are three sequences of triples\n\n(3, 4, 5), (5, 12, 13), (7, 24, 25), (9, 40, 41) in which the two largest sides differ by 1. See if you can work out the next ones in the series.\n\n(8, 15, 17), (12, 35, 37), (16, 63, 65), (20, 99, 101), which the two largest sides differ by 2. See if you can work out the next ones in the series.\n\n(3, 4, 5), (20, 21, 29), (119, 120, 169). (696, 697, 985)  in which the two smaller sides differ by 1. This is how you can work out the next ones in the series:\n\ngiven (x, y, z) is a triple with this property (y-x = 1)\ncreate (X, Y, Z) as follows\n\nX = 2x + y + 2z \nY = x + 2y + 2z\nZ = 2x + 2y + 3z \n\nworking it through for (3, 4, 5) that leads to\nX = 6 + 4 + 10 = 20\nY = 3 + 8 + 10 = 21\nZ = 6 + 8 + 15 = 29\n\nworking it through for (119, 120, 169) that leads to\nX = 238 + 120 + 338 = 696\nY = 119 + 240 + 338 = 697\nZ = 238 + 240 + 507 = 985\n\nhave fun with numbers, good luck!
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Is this the Pythagorean theorem? There are many proper ways of proving it. Check wikipaidia below.
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since this starts as a square.  when we place the diagonal line from the bottom right corner to the upper left corner we end up with two right triangles.  This is the perfect set up to use the Pythagorean theorem, which states "the square of the hypotenuese is qual to the sum of the squares of the base and height. you may have seen it as \n       h^2 = x^2 + y^2\nwe started with a square so we know the base (x) and height (y) are each 7 sq rt 2.\nsquare the values then add them together.  this will give you that diagonal's square. ****remember to take the sqrt of the answer to find the length of the diagonal.
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hope this helps\n\nhttp://www.thea.nesinc.com/ptmath.htm
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Actually I had something to help you but x=0 which makes it pretty difficult.\n\nI guess you could try to talk about tomatoes or something.  Say Tom (X) has 1/2 the amount of tomatoes that Jill has, plus four.  Or in other words, if you take Jill's amount of tomatoes, multiply it by two, and take Tom's amount and multiply it by three, there total amount of tomatoes will equal 12.  How many tomatoes does Jill have?  How many does Tom have?
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1.   13\nused pythagorean theorem.\nleg^2 + leg^2 = hypotenuse^2\n\n2.    30 \npythgorean theroem\n\n3. 82.7\nsame thing\n\n\n4. just plug in the numbers into the equaions\nRange = {-20, -5, 1}\n\nhttp://en.wikipedia.org/wiki/Pythagorean_theorem\nu shuld practice ur pythagorean theorem
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Pythagoreanism is a term used for the esoteric and metaphysical beliefs held by Pythagoras and his followers, the Pythagoreans, who were much influenced by mathematics and probably a main inspirational source for Plato and platonism.\n\nLater resurgence of ideas similiar to those held by the early Pythagoreans are collected under the term neopythagoreanism.\n\nPythagorean thought was dominated by mathematics, but it was also profoundly mystical. In the area of cosmology there is less agreement about what Pythagoras himself actually taught, but most scholars believe that the Pythagorean idea of the transmigration of the soul is too central to have been added by a later follower of Pythagoras. On the other hand it is impossible to determine the origin of the Pythagorean account of substance. It seems that the Pythagorean account begins with Anaximander's account of the ultimate substance of things as "the boundless," by Anaximander called the "apeiron." On the other hand, the Pythagorean account says that it is through the notion of the "limit" that the "boundless" takes form.\n\nPythagoras wrote nothing down, and relying on the writings of Parmenides, Empedocles, Philolaus and Plato (people either considered Pythagoreans, or whose works are thought deeply indebted to Pythagoreanism) results in a very diverse picture in which it is difficult to ascertain what the common unifying Pythagorean themes were. Relying on Philolaus, whom all scholars agree should be highly representative of the Pythagorean school, one has a very intricate picture. Aristotle explains how the Pythagoreans (by which he meant the circle around Philolaus) developed Anaximander's ideas about the apeiron and the peiron, the unlimited and limited, by writing that:\n\n"… for they [the Pythagoreans] plainly say that when the one had been constructed, whether out of planes or of surface or of seed or of elements which they cannot express, immediately the nearest part of the unlimited began to be drawn in and limited by the limit."\n\nContinuing with:\n\n"The Pythagoreans, too, held that void exists, and that it enters the heaven from the unlimited breath – it, so to speak, breathes in void. The void distinguishes the natures of things, since it is the thing that separates and distinguishes the successive terms in a series. This happens in the first case of numbers; for the void distinguishes their nature."\n\nWhen the apeiron is inhaled by the perion it causes separation, which also apparently means that it "separates and distinguishes the successive terms in a series." Instead of an undiffrentiated whole we have a living whole of inter-connected parts separated by "void" between them. This inhalation of the apeiron is also what makes the world mathematical, not just possible to describe using math, but truly mathematical since it shows numbers and reality to be upheld by the same principle: both the continuum of numbers (that is yet a series of successive terms, separated by void) and the field of reality, the cosmos - both are a play of emptiness and form, apeiron and peiron. What really sets this apart from Anaximander's original ideas is that this play of apeiron and peiron must take place according to harmonia (harmony), about which Stobaeus commentated:\n\n"About nature and harmony this is the position. The being of the objects, being eternal, and nature itself admit of divine, not human, knowledge – except that it was not possible for any of the things that exist and are known by us to have come into being, without there existing the being of those things from which the universe was composed, the limited and the unlimited. And since these principles existed being neither alike nor of the same kind, it would have been impossible for them to be ordered into a universe if harmony had not supervened – in whatever manner this came into being. Things that were
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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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That's the trouble with answers that give no explanation. Who can you believe?\n\nSince we go in each direction once, the ORDER of the moves doesn't matter. \n\nDraw a grid. Start at 0,0\n10 North means 0,10 You are due north 10 miles. Then we come back south 6 miles, so we're 4 miles north of where we started. \n4 miles east and then 7 west means we're 3 miles west of the starting point. \n\nDraw a right triangle 4 miles up and then 3 miles to the left of the starting point. We want the hypotenuse of the triangle whose sides are 3 and 4. \nTime for Pythagoras: a^2 + b^2 = c^2 \n3^2 + 4^2 = c^2 \n9 + 16 = 25 Take the square root of that for your final distance. Direction: NW
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First distance: Pythagorean Theorem is a squared + b squared = c squared.  So (2*2) + (3*3) = ???  Then find the square root of ??? \n\nThen the bearing: We first need to find the measure of angle A.  Since we know the length of all 3 sides, any of the SohCahToa formulas work.  Tangent is easiest here.  tan(A) = opposite/adjacent (length of opposite leg divided by length of adjacent leg).  tan(A) = 3/2.  Use the tan-1 key (probably will require shift key).  \n\nWe know the first leg of the trip is south which represents 180 degrees.  So add the found angle to 180 degrees.  (Draw it out, it should make sense.)
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64 proofs\nexplanation\nProof #1\nThis is probably the most famous of all proofs of the Pythagorean proposition. It's the first of Euclid's two proofs (I.47). The underlying configuration became known under a variety of names, the Bride's Chair likely being the most popular.\n\n    \n\nThe proof has been illustrated by an award winning Java applet written by Jim Morey. I include it on a separate page with Jim's kind permission.\n\nFirst of all, ABF = AEC by SAS. This is because, AE = AB, AF = AC, and\n\n  BAF = BAC + CAF = CAB + BAE = CAE.\n \n\nABF has base AF and the altitude from B equal to AC. Its area therefore equals half that of square on the side AC. On the other hand, AEC has AE and the altitude from C equal to AM, where M is the point of intersection of AB with the line CL parallel to AE. Thus the area of AEC equals half that of the rectangle AELM. Which says that the area AC2 of the square on side AC equals the area of the rectangle AELM.\n\nSimilarly, the are BC2 of the square on side BC equals that of rectangle BMLD. Finally, the two rectangles AELM and BMLD make up the square on the hypotenuse AB.\n\nThe configuration at hand admits numerous variations. B. F. Yanney and J. A. Calderhead (Am Math Monthly, v.4, n 6/7, (1987), 168-170 published several proofs based on the following diagrams\n\n   \n\n\n\nProof #2\n    \n\nWe start with two squares with sides a and b, respectively, placed side by side. The total area of the two squares is a2+b2.\n\n    \n\nThe construction did not start with a triangle but now we draw two of them, both with sides a and b and hypotenuse c. Note that the segment common to the two squares has been removed. At this point we therefore have two triangles and a strange looking shape.\n\n    \n\nAs a last step, we rotate the triangles 90o, each around its top vertex. The right one is rotated clockwise whereas the left triangle is rotated counterclockwise. Obviously the resulting shape is a square with the side c and area c2.\n\n(A variant of this proof is found in an extant manuscript by Thâbit ibn Qurra located in the library of Aya Sofya Musium in Turkey, registered under the number 4832. [R. Shloming, Thâbit ibn Qurra and the Pythagorean Theorem, Mathematics Teacher 63 (Oct., 1970), 519-528]. ibn Qurra's diagram is similar to that in proof #27. The proof itself starts with noting the presence of four equal right triangles surrounding a strangenly looking shape as in the current proof #2. These four triangles correspond in pairs to the starting and ending positions of the rotated triangles in the current proof. This same configuration could be observed in a proof by tesselation.)\n\n\n\nProof #3\n    \n\nNow we start with four copies of the same triangle. Three of these have been rotated 90o, 180o, and 270o, respectively. Each has area ab/2. Let's put them together without additional rotations so that they form a square with side c.\n\n    \n\nThe square has a square hole with the side (a-b). Summing up its area (a-b)2 and 2ab, the area of the four triangles (4·ab/2), we get\n\n  c2 = (a-b)2+2ab \n  = a2-2ab+b2+2ab \n  = a2+b2 \n \n\n\n\nProof #4\n    \n\nThe fourth approach starts with the same four triangles, except that, this time, they combine to form a square with the side (a+b) and a hole with the side c. We can compute the area of the big square in two ways. Thus\n\n  (a + b)2 = 4·ab/2 + c2  \n\nsimplifying which we get the needed identity.\n\n\n\nProof #5\n\nThis proof, discovered by President J.A. Garfield in 1876 [Pappas], is a variation on the previous one. But this time we draw no squares at all. The key now is the formula for the area of a trapezoid - half sum of the bases times the altitude - (a+b)/2·(a+b). Looking at the picture another way, this also can be computed as the sum of areas of the three triangles - ab/2 + ab/2 + c·c/2. As before, simplifications yield a2+b2=c2.\n\nTwo copies of the same trapezoid can be combined in two ways by attac
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Listen to the first guy.\nThe distance on the y axis is 1 ((-5)-(-4)) and on the x axis it is 3 ((-3)-0).\nThis gives you a right angles triangle with sides of 3 & 1. \nUnder Pythagoras theoem the sum of the squares is equal to the square of the hypoteneuse so\n1^2 + 3^2 = 10.\nTherefore the distance is the square root of 10.
91628
Also, when working out the surface area of the triangle sides, ensure that you are NOT usng the perpendicular height to calculate this. In some instances, you may need to use pythagoras' theorem to get the height measurement of the triangle sections