how can pascal's triangle be used in real life situations?
481887
If you are an achitect.
726169
It is possible only when the given triangle is isosceles or equilateral. It's also possible for right-angled triangles but only when the given angle isn't 90 degrees.
330849
http://en.wikipedia.org/wiki/Pascal_programming_language#Syntax\nhttp://www.allegro.com/papers/htpp.html\n\nThey will help
440758
Pascal (Same as Newton per Meter sqaure)\nmm of Hg (Mercury)
849360
If you feel comfortable with geometry the golden ratio is good.  Fibonacci is good if you like sequences.  Pascal's triangle is good if you like factorials, binomials, and patterns.\n\nNote that the ration of consequtive fibonacci numbers converges to the golden ration as n-> infinity.\n\nHow about looking into how they are related?  I've shown you one.  I don't know how you'd relate Pascal's triangle to the other two, though.
255722
It is discovered by Yang Hui, and there is no mistake.  It is identical to Pascal's triangle, except discovered 400 years earlier.
725701
It's used in geometry to determine how many elements an object is composed of.\n\n"Pascal's triangle can be used as a lookup table for the number of arbitrarily dimensioned elements within a single arbitrarily dimensioned version of a triangle (known as a simplex). For example, consider the 3rd line of the triangle, with values 1, 3, 3, 1. A 2-dimensional triangle has one 2-dimensional element (itself), three 1-dimensional elements (lines, or edges), and three 0-dimensional elements (vertices, or corners). The meaning of the final number (1) is more difficult to explain (but see below). Continuing with our example, a tetrahedron has one 3-dimensional element (itself), four 2-dimensional elements (faces), six 1-dimensional elements (edges), and four 0-dimensional elements (vertices). Adding the final 1 again, these values correspond to the 4th row of the triangle (1, 4, 6, 4, 1). Line 1 corresponds to a point, and Line 2 corresponds to a line segment (dyad). This pattern continues to arbitrarily high-dimensioned hyper-tetrahedrons (simplices).\n\nTo understand why this pattern exists, one must first understand that the process of building an n-simplex from an (n − 1)-simplex consists of simply adding a new vertex to the latter, positioned such that this new vertex lies outside of the space of the original simplex, and connecting it to all original vertices. As an example, consider the case of building a tetrahedron from a triangle, the latter of whose elements are enumerated by row 3 of Pascal's triangle: 1 face, 3 edges, and 3 vertices (the meaning of the final 1 will be explained shortly). To build a tetrahedron from a triangle, we position a new vertex above the plane of the triangle and connect this vertex to all three vertices of the original triangle.\n\nThe number of a given dimensional element in the tetrahedron is now the sum of two numbers: first the number of that element found in the original triangle, plus the number of new elements, each of which is built upon elements of one fewer dimension from the original triangle. Thus, in the tetrahedron, the number of cells (polyhedral elements) is 0 (the original triangle possesses none) + 1 (built upon the single face of the original triangle) = 1; the number of faces is 1 (the original triangle itself) + 3 (the new faces, each built upon an edge of the original triangle) = 4; the number of edges is 3 (from the original triangle) + 3 (the new edges, each built upon a vertex of the original triangle) = 6; the number of new vertices is 3 (from the original triangle) + 1 (the new vertex that was added to create the tetrahedron from the triangle) = 4. This process of summing the number of elements of a given dimension to those of one fewer dimension to arrive at the number of the former found in the next higher simplex is equivalent to the process of summing two adjacent numbers in a row of Pascal's triangle to yield the number below. Thus, the meaning of the final number (1) in a row of Pascal's triangle becomes understood as representing the new vertex that is to be added to the simplex represented by that row to yield the next higher simplex represented by the next row. This new vertex is joined to every element in the original simplex to yield a new element of one higher dimension in the new simplex, and this is the origin of the pattern found to be identical to that seen in Pascal's triangle."
174124
This geometric trick is very involving:\nYou can use it as a starting point for geometric construction of square roots, equivalence of areas and isometric transforms\nhttp://answers.yahoo.com/question/?qid=1006030501625\n\nThis other about divisibility and prime factoring of integers has deep mathematical concepts beyond it: starting from it you can speak about prime factoring, bynary count, combinatory calculus and mathematical induction\nhttp://answers.yahoo.com/question/?qid=1006022300298&r=w\n\nI'm going to use them in my school classes
481919
It is impossible to trisect an arbitrary angle using only compass and straightedge (but bisection is possible). No there is no mathematical technique (i.e a theoretical construction technique to do so, even if impractical). It cannot be done exactly. The best that can be done is approximation. One can imagine, I suppose, a radial arm (our radius rotating about a fixed  point) whose rotation speed was controlled very precisely by a computer and the speed could be extremely slow and a laser could be used with a super accurate quartz timer to 'print' a dot at fixed intervals thus dividing an arbitrary angle as desired. Certainly, technologically feasible, but I haven't heard anything about it, and it would be a technological technique, not a mathematical "breakthrough".
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
1063
Typically the 3-4-5 triangle method is used during contruction of some sort where you need to ensure a right angle.  You often know the length of one side of something and want to know where to place the other side so that it is at right angles.  I used this to frame our basement walls so they would be "square" for drywalling and laying the floor.
454825
We can do this 2 different ways.\n\nFirst: Use the Pythagorean Theorem. Since all angles are right angles, the diagonal (measure from corner to corner) is the hypotenuse of the right triangle formed.\n\nBy the Pythagorean Theorem, 10^2 + 10^2 =x^2, where x=diagonal\n\nx^2=200\nx=sqrt(200)\nx=10 sqrt 2\n\n2nd way: use 45-45-90 Triangles\n\nWhen you draw any diagonal in a square, you get a 45-45-90 degree right triangle. This is because the 2 sides are the same, so the 2 base angles must be also. The top is 90, so the others must be 45 and 45 to equal each other.\n\nThe sides to a 45-45-90 right triangle are always in the ratio 1:1: sqrt(2) where sqrt(2) is the hypotenuse.\n\nYou know that the sides are 10 feet. Therefore the hypotenuse is 10 sqrt2 feet
403281
take a straw (cylinder) and ram it through a potato.  If you do it right, the straw will go right through the potato!  The cylinder is the strongest and most stable of the shapes and you can see why soups, veggies, fruits, etc. all come in cans!\n\nRight triangels, or right angles are all around you.  The floor and the wall of you room for example.  The bottom of a pool is a good example of an obtuse triangle, suspension bridges - trapezoid, if you look at the tip of a pencil - it's a cone but the shape of a cone when you look at it head on is a scalene triangle
1116606
The two smaller sides must always add up to more than the largest side (otherwise you can't make a triangle).  Since the sides must be *integer* sides, the only valid triangle is 2, 6 and 7.\n\nStated another way, here are the combinations:\n2-3-7, impossible triangle (try connecting the 2 and 3 sides together...)\n2-4-7, impossible triangle (try connecting the 2 and 4 sides together...)\n2-5-7, straight line, not a triangle\n2-6-7, valid triangle\n\nSo there is only one combination (2-6-7) that makes a triangle when x is an *integer* greater than 2 and less than 7.
375103
It's called Pascal's Wager.
972787
Heron's formula can only be used when the sides are given. The formula is:\nArea = square root of[(s)(s - a)(s - b)(s - c)]\n\ns = (a + b + c)/2\nwhere a,b and c are the sides of the triangle.\n\nSorry, but there really isn't enough information to work with.
902207
I'm assuming you are referring to the disguised triangles in the attached link... (link 1).  I'm sure there are some rigorous mathematical proofs you could supply, or automated ways to count the triangles, but I basically decided to just use the "brute force" method of counting.  The hard part is recognizing all the different triangles while not double counting...\n\nWhat I finally did was to create several copies of the diagram and tried to color the triangles.  You can color the individual triangles (24), right off the bat.  Now if you try to pair up triangles, you can pair them one way (12) or another using the flipped image (12).  Grouping them further in groups of three or four triangles, I counted another set of (7) and the flip for another (7).  Then I noticed four other ways to group four triangles (3), (3), (3), (3).  Lastly, there are two large triangles composed of the inner sixteen triangles (1) and (1).\n\nAltogether I got seventy (76) triangles.  You can double check my work using the attached image (link 2).  I think I got them all and didn't double count...