Are there two Pythagorean triangles such that the ratio of their hypotenuses is the square root of three?
119688
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).
197687
Try this:\nAssuming the rectangle is centered within the half circle, draw a line from the center of the flat edge of the half circle (should also be the center of one edge of the rectangle) to one of the corners of the rectangle that is touching the edge of the circle.  That line has a length equal to the radius of 10.  By drawing this line you have also formed a 30-60-90 triangle inside the rectangle.  So, the triangle you have just created has a hypotenuse equal to 10 but the sides are unknown.  However, knowing that a 30-60-90 triangle has sides with a relationship of 1, 2, square root of 3, we use the method of similar triangles to  solve for the sides.  The shortest side of the 30-60-90 triangle is thus equal to 5 (this is also equal to the width of the rectangle), the other side is equal to 5*square root of 3 (the length of the rectangle will be two times this value), and the hypotenuse is 10. So the width of the rectangle is 5 and the length is 10*square root 3, so the area of the rectangle is 50*square root 3=86.6
415065
If it's a right triangle, use the Pythagorean Theorem.\n\na^2 + b^2 = c^2\n\n\nA and B are the legs (the ones that touch the right angle), C is the hypotenuse which is opposite the right angle.\n\n10^2 + x^2 = 26\n\n100 + x^2 = 676\n\nx^2 = 576\n\nx = 24
349402
Draw axis of symmetry to the base.  Look at the rightangled triangle on the left side.  It has hypotenuse s, short side s/2.  Use Pythagoras' to show the height is (sqrt(3)/2)*s, then use\narea = (1/2)*base*height
186666
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.
177203
Draw a picture, it's easier. \n\nSo the picture shows that from the triangle with point A, tan40° = x÷100\nso:\nx = 100tan40° = 83.91\n\nand in the triangle with point B, \ntan20° = 83.91÷y\nso:\ny = 83.91÷(tan20°) = 230.54\n\nthe triangle on the ground, containing  point A, point B, and the base of the tower, is a right triangle because one side points south and the other one points east, so just use pythagorean theorem to get the length of the hypotenuse, which is the length from A to B.
400981
you can figure out the hypotenuse using the pythagorean theorem (a^2 + b^2 = c^2), so the hypotenuse is 10.\n\nAfter that, you just add up the areas of all the surfaces - For the triangles, its 6x8 divided by 2 = 24\n\nFor the rectangular faces, you do the following\n4x8 = 32\n4x6 = 24\n10x4 = 40\n\nAdd all these up and you get as surface area of 120.
282115
Correct! It's a right triangle! Use the pythagorean theorem!\nc^2 = a^2 + b^2\nWhere c is the length of the hypotenuse (opposite the right angle), a and b are the lengths of the legs of the triangle.
301217
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
474700
It helps if you know something about Pythagorean triples (integer solutions to the Pythagorean equation for right-angled triangles).\n\nIf you do, you will know that there are no primitive triples in which the difference between the legs is 3, The smallest such differences are 1, 7 and 17 (always prime, but not all primes will serve).\n\nTherefore this must be a non-primitive triple which is a multiple by three of a primitive triple with the legs differing by 1. There is a series of such triples, starting with (3, 4, 5) which can be generated from one another.\n\nSo we divide 87 by 3 and get 29 and recognise that this must be the triple (20, 21, 29), the second in the series.\n\nSo plainly the answer is length = 63 cms and width = 60 cms (i.e. 3 x 21 and 3 x 20)\n\nTHE ALGEBRA \n\nL = B + 3\nL^2 + B^2 = 87^2\n2 B^2 + 6B + 9 = 7569\nB^2 + 3B - 3780 = 0\n(B - 60) (B + 63) = 0\n\nAs rectangles in the real world have sides of positive value, ignore B = -63 and take B = 60, yielding L = 63,\n\nBut none of that is necessary if you know your triples!
603561
The Pythagorean Theorem says that if you have a right triangle, then a squared + b squared = c squared.  The converse says that if you have a triangle in which a squared + b squared = c squared, then your triangle must be a right triangle.
295962
Just draw a right triangle with short sides of 10 and 24.\n\nThe hypotenuse can be calculated by the Pythagorean theorem as the square root of 24^2 + 10^2.  That works out to 576 + 100 or 676.  The square root of 676 is 26.\n\nThe direction is roughly ENE (East North-East).
581497
D\n\nSeriously, what class are you in? Algebra? Geometry? You chould know this!! It's the pythagorean theorm- [a2 +b2 =c2] (2's = squared).  You have a and b, the two adjacent sides of the triangle.  Therefore, it is the square root of 50, or approx. 7.07.  Thats SUPER important, make sure you remember it!
24009
There is actually a little controversy surrounding whether or not Pythagoras was actually the one who discovered the theorem, or if it was another of his associates, the Pythagoreans.  This website: http://www.geom.uiuc.edu/~demo5337/Group3/hist.html has some interesting history of other cultures illustrating an understanding of this, without expressing it in literal mathematical language.\n\nAs far as your second question, a great illustration can be found at http://mathforum.org/library/drmath/view/53937.html    The pythagorean theorem is actually a specific instance of the law of cosines, which is represented as a^2 = b^2 + c^2 - 2ab*cos(a).  As you can see, the equation only becomes the pythagorean theorem if 2ab*cos(a) = 0, which only happens if the angle is 90 degrees.
322539
http://en.wikipedia.org/wiki/Pythagorean_theorem#Proofs\n\nCheck below for the "proof of converse".  It's the most straightforward there is.\n\nBasically, you try to construct a triangle with x,y, and z, only to discover that it is congruent to a right triangle!
477852
let \nc=hypotenuse\na,b = legs\n\ntherefore:\nc=25\na=2b-10\n\nform pythagorean theorem;\n\na^2+b^2 = c^2\n(2b - 10)^2 + b^2 = 25^2\n4b^2 - 40b + 100 + b^2 = 625\n5b^2 - 40b - 525 = 0\n\ndividing both sides by 5\nb^2 - 8b - 105 = 0\n\nfactoring the left side\n\n(b - 15)(b + 7) = 0\n\nb + 7 = 0\nb = -7 absurd (no side can be negative)\n\nb - 15 = 0\nb = 15\n\nsolve for a:\n\na = 2b - 10\na = 2*15 - 10\na = 20\n\ntherefor the two legs of the right triangle are\n20 cm and 15 cm\n\nchecking:\nc^2 = 20^2 + 15^2\nc^2 = 400 + 225\nc^2 = 625\nc = 25 cm.\n\n\nthat's it
73984
The Pythagorean Theorem can only be applied to Right Triangles only.\n\nIn a right triangle, the side opposite the 90 degree angle is called the HYPOTENUSE. The other two sides are called LEGS.\n\nTHEOREM:\nThe square of the hypotenuse is equal to the sum of the squares of the legs. \n\nEQUATION: Supposing,\nc = hypotenuse\na = first leg\nb = second leg\n\nThen:\n\nc^2 = a^2 + b^2\n\nThis is the conventional way of assigning the letters to the sides of the right triangle. However, if other letters will be assigned, then determine which letter will represent the hypotenuse so that its square can be placed on the left side of the equation. The others will be placed on the right side.