I have a question about the Pythagorean Theorem...?
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House builders (for angles of the roof)\nMountain climbers (to see how far they need to go up a mountain)\nteachers (to teach thier students)
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1.  Area of rt triangle whose leg is 6ft and whose\n    hypotenuse is 10ft?\n\n    Area = 24 sq. ft.  \n\n  Explanation:\n\n    First we need to find the other leg's length...\n    Using Pythagorean Theorem,\n    \n    c**2=a**2+b**2 \n       \n    where c is the hypotenuse, a is one leg, and \n    b is the other leg.\n\n    So, 100 = 36 + b**2;  b**2 = 100-36 = 64; therefore b=8.\n    Now, Area = 1/2(base*height) since the base is one leg\n    (we'll say it is a) and the height is the other leg \n    (therefore we'll name it b) \n    the Area = 1/2(8*6) = 1/2(48)= 24 sq. ft.\n\n2.   The hypotenuse is approx. 4.9 meters.\n    \n  Explanation:\n     \n     Using Pythagorean Theorem: c**2=a**2+b**2 we find that\n     c**2 = (3.5)(3.5) + (3.5)(3.5)\n     c**2 = 12.25 + 12.25\n     c**2 = 24.25\n     c = sq. rt. (24.25)\n     c = 4.9244 meters\n     hypotenuse is approx. 4.9 meters.\n\n3. hypotenuse is approx. 20 inches long.\n\n  Explanation: Area = 1/2(x)**2 \n               where x is length of one leg.\n\n     x**2 = 2*Area;\n     x**2 = 2*98\n     x**2 = 196\n     x = sq. rt. (196)\n     x = 14 inches\n\n     Now, to find length of hypotenuse, we use the\n     Pythagorean Theorem:\n\n     c**2 = a**2 + b**2; since a=b this can be rewritten so\n     c**2 = 2(a)**2\n     c**2 = 2(14)**2 -- Remember 14 is the length of a leg.\n     c**2 = 2(196)\n     c**2 = 392\n     c    = sq. rt. (392)\n     c = 19.7989\n     c is approx. 20 inches long.\n\n\n4.   The researcher will climb approx. 10 meters \n     to reach the crack.\n\n   Explanation:\n      \n      Using the Pythagorean Theorem: c**2 = a**2 + b**2\n      If we set a = 24, then c**2 = (24)**2 + b**2,\n      but we also know that c + b = 36 \n      which means c = 36 - b.\n\n      So to solve for this we need to set\n      (36-b)**2 = (24)**2 + b**2\n      1296-72b + b**2 = (24)**2 + b**2\n      1296 - 72b = (24)**2\n      1296 - 72b = 576\n      -72b = 576 - 1296\n      -72b = -720\n         b = -720/-72\n         b = 10 \n\n      And now we have our 10 meters the researcher \n      must climb to reach the crack in the tree.\n\n\n6.    Approximately a 42 inch stick is the longest stick\n      that will fit in a box that is 24 in X 30 in X 18 in.\n\n  Explanation:\n\n      First use Pythagorean Theorem to find the diagonal\n      inside the rectangle of 24 in X 30 in.\n\n      c**2 = (24)**2 + (30)**2\n      c**2 = 576 + 900\n      c**2 = 1476\n      c = approx. 38.418 in. -- we will call it 38 in.\n\n      Now, use Pythagorean Theorem to find the diagonal\n      inside the "new and invisible" rectangle that is \n      38 in. X 18 in.\n\n      c**2 = (38)**2 + (18)**2\n      c**2 = 1476 + 324\n      c**2 = 1800\n      c = sq. rt. (1800)\n      c is approx. 42.43 in.  \n      \n      We will call "c" 42 inches long.
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My response is to use the pythagoras theory to answer the question. After all, the question was set for you, so it wouldn't be right to answer it for you, now would it?
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Fermat's Last Theorem states that no integer solution exists for the equation\n\na^n = (b^n) + (c^n)  \n\nwhere n is greater than 2.\n\nYour equation has n = 2, so it does not contradict (or prove) Fermat's last theorem - it is one example of it.
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Pythagoras:\nAC squared = AB squared + BC squared\nTherefore\nAC = square root of quantity (AB^2 + BC^2)\nYou can do the math.
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The Pythagorean Theorem? It's a^2 + b^2 = c^2.\n\nIf there's a right triangle, and a and b are the two "legs" (the shorter sides), and c is the hypotenuse (the longest side).
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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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A Pythagorean triple consists of three positive integers a, b, and c, such that a^2 + b^2 = c^2. Such a triple is commonly written (a, b, c), and a well-known example is (3, 4, 5). If (a, b, c) is a Pythagorean triple, then so is (ka, kb, kc) for any positive integer k. A primitive Pythagorean triple is one in which a, b and c are coprime.\n\nThe name is derived from the Pythagorean theorem, of which every Pythagorean triple is a solution. The converse is not true. For instance, the triangle with sides a = b = 1 and c = √2 is right, but (1, 1, √2) is not a Pythagorean triple because √2 is not an integer. Moreover, 1 and √2 don't have an integer common multiple because √2 is irrational.\n\nThere are 16 primitive Pythagorean triples with c ≤ 100:\n(3, 4, 5) (20, 21, 29) (11, 60, 61) (13, 84, 85)\n(5, 12, 13) (12, 35, 37) (16, 63, 65) (36, 77, 85)\n(8, 15, 17) (9, 40, 41) (33, 56, 65) (39, 80, 89)\n(7, 24, 25) (28, 45, 53) (48, 55, 73) (65, 72, 97)
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There is a theorem (I think you remember it better). I just put it with my own words. Let points A, B, C & D belong to the same circle. Then the angle ABC=(Arc ADC)/2\nThus the Sum of 7 Arcs under 7 Angles will be = to 2pi radians, that means that the Sum of those Angles = pi radians, the main demand being they belong to the circle.
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Without the knowledge of where point e is, this problem is insolvable.\n\nI suspect, however, from a brief diagramming session, that you will need to solve for the length of ab using the Pythagorean Theorem, and set up similar triangles.
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Pythagoras? As in the Pythagorean theorem. (In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.)
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it's the Pythagorean Theorem\n\nIn any right triangle, the area of the square whose side is the hypotenuse (the side of a right triangle opposite the right angle) is equal to the sum of the areas of the squares on the other two sides.
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Jan 17 Note: If this is hard to understand, after sleeping on it, I may have an easier construction.\nIf you need further explanation, add another note to your question.\n\n-----------\n\nI think I have a solution.\nI'm making the assumptions that the Babylonians didn't have an acurate value for pi and used the value three. If I do that, every thing falls out.\n\nDraw a cirlce. Mark off 2 arcs. Make one the top portion of a circle say from 10 o'clock to 2 o'clock, the other the matching arc from 4 o'clock to 8 o'clock. Note: These arcs are just to get a representation of the problem and are not an acurate length.\n\nNext draw the two chords for the arcs, that is, the straight line from 10 to 2 and the other from 8 to 4.\nNow draw in the diameter of the circle from 12 o'clock to 6 o'clock.\nThe two chords cut off pieces on the ends of the diameter. Label these as length 2 (2 is the perpendicular) "Double 2 and get 4" That's the length of the two end pieces of the diameter.\n"Take 4 from 20 and get 16" This is where using 3 as a value for pi comes in. If pi 3, then a circle of circumference 60 will have a diameter of 20, so let the length of the diameter be 20. Taking 4 from 20 gives you the length of the diameter that is between the 2 chords, 16.\n\n"Square 20 get 400, square 16 get 256, take 256 from 400 get 144. Find the square root of 144. 12, the square root, is the chord."\n\nWe're obviously doing some Pythagorean theorem here so where is the triangle? \nDraw the diameter of the circle from 8 o'clock to 2 o'clock, that's the hypoteneuse as has length 20.\nThe distance between the two cords is 16. That's also the distance from 8 o'clock to 10 o'clock. Draw that line and label it 16. (Remember the drawing isn't accurate just enough to show us how things are layed out).\nThat leave the 3rd side of the triangle as the chord between 10 o'clock and 2 o'clock. \n\n20^2 - 16^2 = 12^2  so 12 is the length of the chord.\n\nHow does that sound?\n\nIn modern English, the statement of the problem would be something like this:\n\nLet AB be an arc of a circle of circumference 60. Let X be the chord connecting A and B.\nLet Y be the perpendicular bisector of the chord. If the distance from X to the circle along Y is 2, find the length of X. (You may assume that pi = 3 for the purposes of solving this problem.)
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theorm of Pythagorus states that if one of the angle of triangle is right angle{90degrees} than the hypotenuese square{the side opposite to right angle} is egual to the sum of squares of the sides forming the right angle \ni.e. (Hyp)square =one side square + second side square \n a2=b2+c2
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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
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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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One. \n\nThis relation is really just Pythagoras' theorem, hiding in a new form. Recall:\n\nH^2 = A^2 + B^2 \n\nThat is, the square of the hypotenuse H is the sum of the squares of the other two sides A and B, for a right triangle with sides A, B and hypotenuse H.  \n\nOf course, A=H*cosx and B=H*sinx for some angle x, so that \n\nH^2= (H^2)*(sinx*sinx + cosx*cosx)\n\nleading to the result:\n\n1=sinx*sinx + cosx*cosx\n\nThis result also is nicely connected to the simple fact that \n\nx - x = 0 \n\nbecause \n\nexp(i*x)=cosx + i*sinx and\nexp(-i*x)=cosx -i*sinx\n\nwhere i=sqrt(-1). \n\nThen multiplying these two, \n\nsinx*sinx + cosx*cosx=exp(i*x)*exp(-i*x)=\nexp(i*[x-x])=exp(0)=1