how do you solve a regular polygon?
149999
A regular n-gon has the centers of its inscribed and circumscribed circles the same point. If you draw segments from this center to each of the vertex, you will get n number of radii and n number of triangles, which are all congruent. Using derivation you will arrive at the conclusion that\nA = .5aP or A = .5ans where\nA is the area, a is the apothem, s is the side and P is the perimeter\na =2A/ns\nThe radius is acquired by\nsin [180degrees(n - 2)/n] = a/r\nr = a/sin[180degrees(n - 2)/n]
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 ;)
1092713
triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
957591
line = 1 side\nparallel lines = 2 sides\ntriangle = 3 sides\nsquare/ rectangle = 4 sides\npentagon = 5 sides\nhexagon = 6 sides\nheptagon = 7 sides\noctagon = 8 sides\nnonagon = 9 sides\ndecagon = 10 sides\n\ni hope i have helped you.... good luck
797386
Consider a regular n-gon in a circle of radius r with side length l\nDraw in all radii from vertices to centre.\nThis forms n isosceles triangles with the sides of the n-gon as their bases\n\nThen angle at centre = 360/n° and each angle at base = (90 - 180/n)°\n\nBy sine rule: sin(90 - 180/n)° /r = sin(360/n)° / l \nWhence l = r * sin(360/n)°/ sin(90 - 180/n)°\n= r * sin(2 * 180/n)°/ cos(180/n)°\n= r * 2 sin(180/n)°cos(180/n)°/cos(180/n)°\n=2rsin(180/n)°\n\nSo the perimeter of the polygon is 2nrsin(180/n)°\n\nNote: As n →∞ Perimeter  → circumference of circle\nie lim  n →∞ 2nrsin(180/n)° = 2πr\nSo lim  n →∞ nsin(180/n)° = π\nOr for large n, sin(180/n)° ≈ π/n; which makes sense given that for small x (in radians) sinx ≈ x\n\nGiven area of a triangle = ½absinC\nthen area of 1 isosceles triangle = ½r²sin(360/n)°\nSo the area of the polygon is ½nr²sin(360/n)°\n\nThus for n = 9 and r = 10;\nl = 2rsin(180/n)° \n= 20 sin 20°\n≈ 6.84 in
16578
6\nSee http://en.wikipedia.org/wiki/Tilings_of_regular_polygons
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
744397
Hi Karen,\n\nIt's good that you tried to do your problems. Some ppl just post stuff on yahoo! answers, without trying, to get their answer.\n\nAnyway, This is how you would do you problem. Belive me it's that tough.\n\nDefinition of Perimeter=sum of all sides.\n \nX=one side of pentagon\nY= one side of octagon\n\nGiven-\n5X=8Y\nX=3Y-2\n\nNow substitute X=3Y-2 into 5X=8Y to find Y.\n\nSo...5(3Y-2)=8Y\n15Y-10=8Y\n7Y=10\nY=10/7\n\nSubstitute Y=10/7 into X=3Y-2 to find what X is. \nso...X=3(10/7)-2\nX = (30/7)-2\nX=16/7\n\nEnjoy!
47718
So you've got an eight-sided figure and each side is 72-feet.  If you draw horizontal and vertical lines from each vertex (the angles where sides meet) to the other side of the octagon, you'll see that the octagon can be split up into one large square (in the center), four rectangles (up, down, left and right), and four triangles (in the corners).  You following me?  Get out a pencil and get to it!  Okay.  So what are lengths of your square's sides?  You'll note that the square has 72-foot sides, just like the original octagon.  Since you know how to find the area of the square, you're on your way.\n\nNext.  you have four rectangles, each with a length of 72-feet, but the short end - the width, how do we figure out the width of the rectangle?  Okay.  Look at your triangles.  Each triangle has one side of 72 feet - the side which is the outside of the octagon - and then two equal sides - it's an isosceles triangle with a 90-degree angle.\n\nOkay, time to call on pythagoras.  Pythagorean theorem states that a^2 + b^2 = c^2.  What that means is this: If you take the length of sides a and b (the ones we don't know) and square them, and add them together, the answer will be 72-squared (side C).  Too bad I'm not there with you to make this part easier to understand.  We know that a and b are equal in this case because our triangle is isosceles.  Therefore, 2x^2 is equal to 72^2.  This gives us a messy equation wherein our answer is, x is ROUGHLY equal to 51 feet.\n\nSo.  You have a square with sides 72 feet long.  You have 4 rectangles each with length 72 and width 51.  You have 4 triangles each with base 51 feet and height 72 feet.  You know that squares and rectangles have an area equal to length times width, and triangles have an area equal to one-half base times height.\n\nCalculate the area for your 9 figures and add them together.  THAT will be the square footage of your octagon.\n\nWhat... did you think I was going to do it ALL for you?
642032
The only regular polygons that can tessellate are triangles, squares, and hexagons. There are other polygons that can tesselate, and I believe there are 12 or so patterns that can repeatedly and uniformly tessellate using regular polygons.
820962
I think you can do this.  Did you know that octahedron means 8-sided figure?  and the fact that it is regular means that all faces are the same size (area).  So the answer would be 32 / 8 = ?
185652
since your ddoing coordinate geometry you should at least tell what shape you're trying to get.\nI'm assuming it's a parallelogram so u should make it (1,8)
239138
well you could split up the thing into triangles. you know what the base is (1m) and the height is unknown, so lets call it h for now. so for each side, the area of the triangle is h/2 (bh/2 and b=1 so h/2). so for every 2 sides, the area is h. so the total area is 18h. \n\nso to find h, you have to use some trig. the equation to find the internal angles is (n-2)*180/n. cuz the sum of the internal angles is (n-2)*180. and then to find each angle, you have to divide by number of angles (which is same as number of sides). so in this case 34*180/36 = 170 degrees each angle. so now for each side, if you make a triangle with the center of the polygon, u should notice that the triangle is isosceles and the base angle is half of the side angle. \nas in the bottom 2 angles are equal and they are equal to 170/2. which is um 85. you can try this with any polygon if you dont believe it. try a square, (n-2)*180 = 2*180 = 360 is the sum of the internal angles. so each angle is 360/4 = 90 degrees. so if u connect each vertex to the center of the circle, you get 4 isosceles triangles and the base angles are 45 degrees each. \nso you have 85 degrees as the angle. so now if you split the isosceles triangle into 2 right angle triangles, you have 3 sides in each right angle triangle: the h, 0.5m and the line from the vertex to center. so you have the angle 85 degrees and you need h and you have the adjacent side = 0.5m. so you can use tan 85 = h/0.5 and you get h. \nso the total area is 18h. \n\ni really hope you understand that. its kinda hard to explain without drawing it out and showing you. if you need more help, then message me or somethin. later
302936
Without knowing the measurements of the sides, you can only figure this out if all sides of the polygon are equal in length. And this is how...\n\nSee, first you draw any polygon, except a triangle. So before doing it, you should know that all the angles of a triangle must add up to 180 degrees. (You can do it with a triangle too, but you would be using what you’re hoping to discover.) From the center of the polygon, draw a line to each vertex. This will create a number of triangles, inside the polygon, equal to its number of sides. Knowing that the angles inside each triangle must add up to 180 degrees, you simply multiply 180 by the number of triangles, which is also the number of sides. Then, realizing that the angles in the center of the polygon, which do not contribute to the angles of that polygon, form a complete circle, and therefore must add up to 360 degrees, you subtract 360 from the result. So the angles in any polygon must add up to 180x-360, or 180(x-2), where x is the number of sides. To find the measure of each angle, in a polygon where all sides and angles are equal, you divide that by the number of sides. The equation then works out to (180(x-2))/x.
366751
make x the width, and solve the two equations (the one given, and the one assumed about the rectangle where l*w=A)
24746
I think the sum of all internal angels of a polygon is given by (n-2)*180. So, for a 40 sided polygon, it should be (40-2)*180=38*180=6840 I guess.
133602
Naming a polygon by its number of sides requires that we know the appropriate Greek prefixes along with some rules for their use.\n       For polygons with 3 through 20 sides, simply add "gon" to the prefixes at the left (although a trigon is more commonly called a triangle and a tetragon, a quadrilateral!). \n       For more than 20 sides, we "construct" the name by using so-called combining prefixes:\n              3 = tri \n4 = tetra \n5 = penta \n6 = hexa or sexa \n7 = hepta \n8 = octa \n9 = nona or ennea \n10 = deca \n11 = undeca or hendeca \n12 = dodeca \n13 = triskaideca \n14 = tetradeca \n15 = pentadeca \n16 = hexadeca \n17 = heptadeca \n18 = octadeca \n19 = enneadeca \n20 = icosa \n3 - 99 SIDES\nTo "construct" a name, we start with the prefix for the tens digit, follow it by "kai" (the Greek word for "and"), then follow it with the prefix for the units digit, and finally add "gon."\nExample: A 35-sided polygon is called a "triacontakaipentagon."\n\n30           and        5              gon \ntriaconta    kai       penta           gon\n \n100 - 999 SIDES\nFor numbers from 100 to 999, we need one more combining prefix and another rule.\n\nTo "construct" the name, we start with the prefix for the hundreds digit taken from the "Ones Digit" table above, follow it by "hecta," then proceed as before.\n\nExample: A 672-sided polygon is called a "hexahectaheptacontakaidigon\n100 = hecta \n600            70           and    2    gon\n hexa + hecta  heptaconta   kai    di   gon\n\nI think its enough. Satisfied!