if every vertex of a regular pentagon is connected to every other vertex, how many triangles are formed?
956378
There are 35 triangles.  Count them this way:  All triangles whose vertices are vertices of the pentagon.  There are exactly 10 of hese, one for every different set of three vertices.  Next count the triangles using exactly two adjacent vertices of the pentagon, plus one vertex inside the figure.  Each pair of adjacent vertices has three such triangles, so we get 15 more.  Now count the triangles which use two NON-adjacent vertices of the pentagon plus an inside vertex.  There is just one triangle for each such pair, and there are five pairs, so 5 more triangles.  Now count the triangles using one pentagon vertex and two inside vertices.  There is only one for each vertex, so 5 more.  There are NO triangles using only inside vertices, so we are done, and it adds up to 35.\n\nI copied and pasted this directly from http://mathforum.org/library/drmath/view/54888.html
953741
Just break the pentagon up into 5 triangles.  You can then add together the area of the 5 triangles to get the total area.
476849
a pentagon, or a regular pentagon, depending on if all the sides & angles are equal or not. hope that helped.
776798
Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
147899
Here is a graph and some explanation:\nhttp://mathworld.wolfram.com/Tetrahedron6-Compound.html
441947
pentagon DUH! lol sry (Hexagon is SIX!)im in geometry rite now so yea
747730
use Eulers formula\n\nF+ V = E+2\n\nhere F = faces = 8\nV = vertices =12\nE = edges = 18\n\ntherefore it is possible to construct the solid.\n\nit is a hexagonal cylinder
280903
Generally you create a regular hexagon by drawing a circle and reporting on the circumference six consecutive segments of the same lenght of the circle radius. \nThe side of a regular hexagon is in fact equal to the radius of the external circle. You can use compassus to do a sharp work.\n\nAnyway, by this property, it comes out that every triangle from the centre of the hexagon to one of its sides is an equilateral triangle; so all its angles are 60°.\nEach internal angle of the hexagon is made by the sum of two of these 60° angles, so each internal angle of a regular hexagon is 120°.\n\nThe general formula for the angles of a regular n-agon is:\n\n(n-2)*180°/n
542325
Triangle: 180\nSquare: 360\nPentagon: 540\nHexagon: 720\nHeptagon: 900\n\nIn general:\nn-agon: (n-2) * 180\n\nWith seven angles in a heptagon, each angle is 128.581427... degrees.  The total is 900 degrees.\n\nDespite what others have posted here, the answer is *not* 360 degrees in every polygon.  Even if you don't know the right answer for a heptagon, intuitively you know that a triangle is 180, a square is 360.  If every polygon was 360, then the angles in a pentagon would be less than 90 degrees, and the angles would get smaller and smaller... but actually, the individual angles get closer and closer to 180.  And the heptagon doesn't need to be a regular heptagon for the angles to still add up to 900, any heptagon will do.
583148
Well, lets see if we can generate a pattern.\n\nA 3-sided polygon (aka a triangle) has no diagonals\nA 4-sided has 2\nA 5-sided has 5 (1-3, 2-4, 3-5, 4-1, 5-2)\nA 6-sided has 9 (1-3, 1-4, 1-5, 2-4, 2-5, 2-6, 3-5, 3-6, 4-6)...\nand so on.\n\nA polygon with n sides has n(n-3)/2 diagonals. The answer for 42 is left as an exercise for the reader ;)
922060
My suspicion upon seeing this question is that it can be solved using mathematical induction.  After all, couldn't we simply 'build' such a (maximally connected) graph starting with n=2 vertices and 'grow' it from there?\n\nRather than take on the proof myself -- mostly because I am unsure of the definition of 'components' -- I leave it as an exercise for you.  However, it bears noting that the formula shown is the commonly accepted "sum of the first n integers" formula used in almost every text on induction.\n\nGood luck!
786624
Pentagon It has a bottom and then two lines that go up and a point at the top. Kinda like a triangle without a bottom.
598966
1- line segment\n2- angle\n3 -triangle\n4 - square, rectangle, rhombus, diamond, parallelogram, or quadrilateral\n5-pentagon\n6-hexagon\n7-heptagon
910330
This will happen for only two irregular pentagons in general:\n\nAll triangles as the angle sum of the three angles is a straight angle so the the angles so placed will always make straight lines ... and thus tessellate\n\nAll quadrilaterals as the angle sum of the four angles is a straight angle so the the angles so placed will make a complete revolution ... and thus tessellate\n\nOther particular polygons MAY tessellate BUT no more general cases will occur
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."
419835
Euler's formula for any polyedra claims that\n\nEuler characteristic χ = V − E + F = 0 for anchor ring.\nHere V is a number of vertices,\nE - number of edges,\nF - number of faces.\n\nIn this task F=12, V=15.\nE = V + F = 27.\n\nSo, the answer is 27.
316050
square pyramid...if square as base\n\ntriangular pyramid...if triangle as base\n\nBasically just look at the base...and add pyramid to it.\n\nEx. 5 sided regular polygon base...pentagonal pyramid.\n\nIf it isn't a regular polygon...look it up...or just leave it as pyramid I guess. LOL!\n\nTetrahedron...sound molecular...like the shape of methane molecule. I'm thinkin' this is geometry...so stick with the simplistic "pyramid".\n\nOH...BTW...if it only has one base...and is composed of multiple triangular sides...then it's a pyramid. If it has 2 bases...and any other figure for it's sides...then it's a prism.