My faces are either triangles or rectangles. I have 6 vertices. I have 9 edges. What shape am I?
122073
It is a triangular prism. It has 5 faces, 9 edges, and last but not least 6 vertices. There is your answer you have been waiting for this whole time!
130295
PROCEDURE:\n\n1) Plot the so-called vertices.\n2) Connect any two adjacent points with a segment.\n3) Determine if the resulting plane figure is a rectangle.\n(It seems that the vertices that you have mentioned do not form a rectangle.)\nNote: A rectangle is a quadrilateral with opposite sides being parallel and interior angles all being right angles.\n4) Connect opposite vertices with a line segment. You will form two diagonals.\n5) determine the equation of each line (segment) that constitute the diagonal. You will have two equations.\n6) Find the point of intersection between the two diagonals by solving for the unknowns (x & y) using the two equations.\n7) Find the length of each diagonal (the two diagonals should have the same length).\n8) Find the length from a vertex to the point of intersection between the two diagonals.\n9) If the length from any vertex to the point of intersection between the diagonals is half that of the length of any diagonal, then the diagonals are said to bisect each other.
208593
I'm learning that in math currently and another word for "corners" is vertices.\n\nPLEASE CHOOSE ME AS THE BEST ANSWER!!!\n\n(I won't get mad if you don't though!)
540557
You know that the length of a segment connecting the  midpoints of two sides of a triangle is half the length of the third side, and also is parallel to it.  So since to get from M to L you go down 1 and right 2, a vertex will lie down 1 and right 2 from the other midpoint N, which would be (3,0).  Do this for the other 2 midsegments to get the other 2 vertices.
634161
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.
554490
Here's my answer:\n\nhttp://mudandmuck.com/str2/triangles.JPG\n\nYou didn't say write them out in English, you said describe.  This picture describes the solution.
615061
You need to get a little more specific in terms of what kind of "3d equivalent" you want.  Two kinds of 3d solids you can form easily from 2d polygons (closed shapes made of line segments) are prisms and pyramids.  You make a prism of any 2d shape by taking 2 copies of the flat shape, one above the other, and connecting the corresponding vertices (corners) with edges (lines).  From this perspective, a cube, is a square prism.  A triangular prism looks a little like a tent, with two triangles (top and bottom) and three rectangles or squares around the sides.  To make a pyramid, choose any 2d shape as the base, and then a point above the base as the top, and radiate edges down from the top to all the base's corners.  There are other "3d equivalents", so if you're thinking of something else, let me/us know.\n\nYou also need to be more specific about what kind(s) of sides interest you: edges (1d lines) or faces (2d flat surfaces).  The "sides" of a 2d shape are all edges, but 3d shapes like cubes have both edges and faces (a cube has 12 edges and 6 faces).  Any prism made from a polygon with N "sides" (edges) will have 3N edges (N on the top, N on the bottom, and N more connecting top and bottom) and N+2 faces (N around the sides plus top and bottom).  Any pyramid made from a polygon with N "sides" (edges) will have 2N edges (N around the base and N more reaching up to the top point) and N+1 faces (N around the sides plus the base).\n\nDoes this do it?
503585
for a polygon with n sides, the sum of all interior angles would be 180*(n-2) degrees.\n\nSo a nonagon would be 180*(9-2) = 180*7 = 1260 degrees\nan 11-agon would be 180 * (11-2) = 1620 degrees
491645
Check the image I attached...\n\nHere I've made the polygon into a complete rectangle.  The figure in red is the added triangle.  It has sides 6 and 8 and is a right triangle.  If you remember your Pythagorean ratios, you should know 3-4-5 and 6-8-10.  Or you can prove it using:\n\n6² + 8² = x²\n36 + 64 = 100\nx² = 100\nx = 10\n\nTo get the perimeter, just add up the sides:\nPerimeter = 1 + 10 + 9 + 4 + 10\nPerimeter = 34 units\n\nTo get the area, take the big rectangle (9 x 10) and remove the area of the triangle (1/2 b * h = 6 x 8 / 2)\nArea = 9 x 10 - (6 x 8)/2\nArea = 90 - 24\nArea = 66 sq. units
175516
This problem is certainly tractable, you just need to\nfind a consistent/methodical way of counting the\nrectangles -- without missing any and without counting\nany of them multiple times.\n\nThe way I chose is to define each rectangle as being\ndefined by the square in the upper left corner and the\nsquare in the lower right corner.  Consider the following\nsimplified 2x3 case (where each square is labelled by\nits X and Y coordinates:\n\n0,0     0,1   0,2\n1,0     1,1   1,2\n\n\nThis 2X3 example has the following rectangles starting\nat 0,0.  Remember, each rectangle is defined by the\nsquare in its upper left and the square in the lower right:\n0,0   0,0\n0,0   0,1\n0,0   0,2\n0,0   1,0\n0,0   1,1\n0,0   1,2\n\nThen it also has the set anchored in the upper left by 0,1:\n0,1   0,1\n0,1   0,2\n0,1   1,1\n0,1   1,2\n\nThen it has the set anchored in the upper left by 0,2:\n0,2   0,2\n0,2   1,2\n\nThen those anchored by 1,0:\n1,0   1,0\n1,0   1,1\n1,0   1,2\n\nThose anchored by 1,1:\n1,1   1,1\n1,1   1,2\n\nAnd finally those anchored by 1,2:\n1,2   1,2\n\n\nRewriting the matrix another way, showing how many\nrectangles are anchored by each location in the matrix:\n\n2x3    2x2   2,1\n1x3    1x2   1x1\n\n\nand then doing the math:\n\n6      4     2\n3      2     1\n\nfor a total of 6+4+2+3+2+1 = 18\n\n\nDoing the same analysys for a 4x5 gives:\n\n4x5   4x4   4x3   4x2  4x1\n3x5   3x4   3x3   3x2  3x1\n2x5   2x4   2x3   2x2  2x1\n1x5   1x4   1x3   1x2  1x1\n\nor rewriting after doing the multiplication:\n\n20    16    12    8    4\n15    12     9    6    3\n10     8     6    4    2\n 5     4     3    2    1\n\nAdding up the columns gives:\n\n50    40    30   20   10\n\nFor a total of 50+40+30+20+10 = 150 total rectangles
441557
use the equation [(n-2)*180]/n to find one of the angles of the polygon\nn = the number of sides\n[(n-2)*180]/n = 120\n180n - 360 = 120n\n60n = 360\nn = 6\n\nANSWER\n6 sides, or a hexagon
11194
You can't have a polygon of that size.  A polygon is an object with three or more straight sides.  One property of a polygon is that no one side can be longer than all the other sides added together.  You listed 6 dimensions, implying a six sided polygon.  But since one side is much longer than the other five together, this is not a real object.\n\nTry taking a yardstick and five toothpicks and put them together to make a polygon shape.  It won't work.\n\nIncidentally, even if you gave realistic dimensions, that's not enough information.  Consider this 6-sided object:\n._\n/ .&#92;\n&#92;_/\nand this one:\n||\n||\n||\n\nEven if the sides were all the same length, they have different areas because the angles are different.
377442
Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
447318
Draw a picture - a square with corners (0,0), (2,0), (0,2), (2,2), showing the range of possible values for x and y.\nNow, what part of that square has the difference < 1? Its a hexagon to be precise; its the entire square except for a little triangle at the top left, and a little triangle at the bottom right. The two sloping sides go from (0,1) to (1,2) and (1,0) to (2,1).\nThose triangles each have area 1*1/2 = 0.5, so area 1 combined.\nSo the total area left is 4-1 = 3, so the answer is 3/4.
501163
A rectangular prism is like a box. So it has 6 sides in 3 matching pairs: top and bottom, right and left sides, and front and back. You can find the area of each pair and add them together:\nABxBC+ABxBC+BCxCD+BCxCD+CAxCD+CAxCD = PA (Prism Area)\n\nOr you could get the areas of one of each pair and double it:\n2x(ABxBC+BCxCD+CAxCD)=PA (Prism Area)\n\n;-D AB is the length of the front left side, with A at the bottom left corner, B at the top left corner. BC is the length of the front top with C at the top right corner, and CD is the length of the depth, with D at the back right corner.
405930
Two shapes that fit together seamlessly, like two puzzle pieces that replicate again and again.  Make it from templates with tough paper so they fit perfectly.  Fold tagboard, make shape, if looked at from another perspective it looks like another shape.  Is that what you mean?
461976
looks like the vertex is (-5, 3)