How many diagonals are there in an 11-sided figure?
922356
There are 88 or 44 depending on how you define what a unique diagonal is.  ( Sorry, not trying to be "egghead-y" here, just trying to give you a complete explanation )\n\nANSWER OF 88:\n\nLet's look at an n-sided figure.  For each point (where two sides meet), you can't draw a diagonal from the point to the same point itself, nor from the point to the two points next to it on either side.  But, you can draw a diagonal to every other point, or to n-3 points.  You can do this for each of the n points.\n\nSo, in an n-sided figure, you can have n(n-3) diagonals.\n\nSo, in an 11-sided figure, you can have 11(11-3) diagonals = 11*8 diagonals = 88 diagonals.\n\nNotice this answer assumes that the diagonal from point A to point B is different than the diagonal from point B to point A!\n\nANSWER OF 44:\n\nTo eliminate such "duplicates", we need to look at it this way for an n-sided figure:\n\nPoint 1 has n-3 diagonals (same reasoning as above)\nPoint 2 has n-3 diagonals (same reasoning as above)\nPoint 3 has n-4 diagonals (since 1 of its diagonals was already counted under Point 1)\nPoint 4 has n-5 diagonals (since 2 of its diagonals have already been counted)\n... etc ... till we get to ....\nPoint n-2 has 1 diagonal\nPoint n-1 has 0 diagonals\nPoint n has 0 diagonals\n\nSo an 11-sided figure would have 44 diagonals.\n\nThe general formula being n(n-3)/2.\n\nIf you try this general formula for something that you can easily draw and count the diagonals of, such as a square or a pentagon, you will see it holds true.  (A square has 2 diagonals = 4(4-3)/2 = 4/2 = 2.  A pentagon has 5 diagonals = 5(5-3)/2 = 10/2 = 5.)\n\n~ ♥ ~
459931
a nine-sided polygon???  can't wait to see other answers...
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...
397805
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
889348
As the previous poster said, you don't have enough information.  If you have n-sided polyhedra, are the sides numbered 1 to n?  Is the expected value in reference to a throw of the dice?  If so, are we talking about two dice?\n\nFor two dice with the sides numbered 1 to n, the answer is n + 1.
602038
Short Answer:\nThe diagonal line or hypotenuse is not a straight line.  It bends either way thus both figures are NOT triangles.\n\nLong answer:\n\nThe two diagrams are not perfect triangles; they are quadrilaterals. Why is that significant? Look at the "diagonal line" on each of the diagrams. You may be able to see, if you look closely, that the supposed diagonal line is not really a straight line. Rather, it's two connected segments (two straight line segments) with a slight angle to it where the two segments meet.\n\nThe top diagram bends in a convex manner, so that there is slightly more area contained below the diagonal. In the bottom diagram, the bend is in the opposite direction.\n\nTo help you see this better, look at the green traingle in the top diagram. It's dimensions are 2 units high, and 5 units long. The Red triangle is 3 units high. In order for the two line segments that make up the combined diagonal to be a straight line, the Red triangle would therefore need to be 7.5 units long (2/5 = 3/x; therefore x=7.5). Instead, it's 8 units long. If that Red Triangle were only a half a unit shorter along the base, then the diagram would all line up, and we wouldn't be trying to solve this problem!\n\nThe four pieces that make up the upper diagram "appear to fit exactly" on a triangle that is half of a rectangle 13 units long and 5 units high. (Count the squares: 13 across, 5 down.) If "appearances are correct", then that would imply that the total area of the four pieces is half of 13x5, or half of 65. So you would think that the total area of these four pieces is 32.5 square units. Let's total them up and see:\n\nIn the top diagram, you can see that the two irregular pieces combine to make a 3x5 rectangle. So these two pieces are 15 square units (you can count the squares, as well! One is 7 squares, the other is 8.) That's part of our total.\n\nThe Red Triangle is half of a 3x8 rectangle. So it's 12 square units. So add 12 to our total.\n\nThe Green Triangle is half of a 2x5 rectangle. So it's 5 square units. So add 5 to our total.\n\nTotal Area of the four pieces is 15+12+5 square units, or 32 square units. So the top quadrilateral is covering only 32 squares out of the 32.5 square units that a triangle would cover. It LOOKS LIKE it covers 32.5 squares, but really it only covers 32!\n\nLet's check the bottom diagram. Obviously the four pieces only cover 32 square units, using the same math as above. There's a 33rd square which is empty.\n\nIf you take the top diagram, and rotate it 180 degrees, so that it fits perfectly on the bottom diagram, you'll see that the red triangles are butted against each other, as well as the green traingles. Therefore, if the top diagram has a slight angle caused by the line segments, the bottom diagram has a slight angle exactly complimentary to the top one. Therefore, if the top diagram undershoots the 32.5 square units by just a half of a square, then the bottom one OVERshoots it by half a square.\n\nAdd up the two half squares, and you have the whole hole.\n\nFour pieces that cover 32 squares, plus four pieces that cover 32 squares, plus one empty square equals 65 squares, which is the area of rectangle 13 units long and 5 units high.\n\n(7+8+12+5) + (7+8+12+5) + 1 = 13 x 5\n\nClear as mud, eh?
1092713
triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
189564
A right rectangular prism has six faces.
599733
I believe it would be one-half times the length of each diagonal.\n\nIn other words, if the diagonals are 4 & 6 the area would be:\n\nA = 1/2 * 4 * 6 = 12\n\nIf you don't know the diagonals, but you know the length of the sides, you could construct the diagonals and calculate their dimensions using the Pythagorean theorem:  c^2 = a^2 + b^2, and then solve from there.  Keep in mind that you'll have to do some extra geometric analysis to find the length of each leg of the right triangles that are created by the intersection of the diagonals before you can use the theorem.
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 ;)
917109
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
366847
An eleven sided polygon is called a hendecagon or also an undecagon.\n\nHendecagon is derived from the Greek word hendeca meaning eleven.
530421
decagon: S x 10\nrhombus: S x 4\nparallelogram: (L x2)+(W x2)\n\nragular fugures are figures that have the same lengths in all the sides ( which is the case with the decagon) so you just add up all the sides or multiply the side by the number of sides the figure has.\n\na rhmbus has all the sides of equal length so again you just add all the sides or multiply the side by 4.\n\nthe opposite sides of the parallelograms are of the same length so you multiply the lenght and width of the parallelogram by 2 and add the results and thats it.\n\ngood luck.
96075
you are talking about "Pascalles (not sure about spelling) Triangle." It goes 1,11,121,1331 and so on.
630757
A pentagon has 5 sides of equal length if that is what you are asking.
730795
Here are "drawings" or descriptions of all 11 trees:\n\n_ _ _ _ _ _\n\n_ _ _ _ L\n\n_ _ _ L _\n\n_ L _ L\n\n_ _ L L\n\nX _ _\n\n_ X _\n\nY  (where each of the three spokes consists of 2 edges)\n\n X.L\n\nA star with 5 spokes: one spoke consists of 2 edges:\n\n1-2 1-3 1-4 1-5 1-6 6-7\n\nA start with 6 spokes:\n\n1-2 1-3 1-4 1-5 1-6 1-7\n\nStart with the one tree with 0 edges. After drawing all the trees with n edges, add an extra edge to get all the trees with n+1 edges, removing any duplicates ("isomorphic" trees).
73554
ZERO (O)  NO RECTANGLES..\n\n\nCHESS BOARD HAVE ONLY SQUARES..