Geomerty help please!?
469289
3x^2 = 12\n\nFirst, divide by three.\n\nx^2 = 4\n\nTake the square root.\n\nx = +2, -2
503725
Been a while, so I'm not sure what you're looking for exactly in "Geometric Shapes".. but these might help somewhat.\n\nSun/Plannets - Sphere\nTree Trunk - Cylindar\nPine Tree - Cone\n\n\n...and my brain stops working.  but just for humor's sake...\nGood Looking Woman - hourgalss\n\n=D good luck with the homework
151606
if all the lines are equal in every way (duplicate) then you can put a measurment (dimension) from the existing line to a datum or border whichever you choose and then put a callout like 3@12.00' (3 lines at 12.00 feet)in the dimensionwith a arrow pointing at the line.\n\njeff
430201
Its tricky work, i need digrams to explain to you so your better just following this link, I think it will help you out.
247013
this is not so easy to answer without doing some work: usually, when you speak of tessellation, it's more likely to deal with symmetries.\n\nHere is a good site which has a couple of interesting articles and softwares on tessellations:\n\nhttp://www.scienceu.com \n(click on the "Geometry center" icon)\n\nAnyway, I'll think better about it: it's a good question.
716347
Of three points, A,B,C\n Check the slope of A-B,  B-C, A-C.  If any of the slopes are the same, you have parallel lines and thus no triangle.  So, to do (2), do (4) first.\n  If you know the distance between A-B, B-C, A-C, you can calculate the area with the formula, so (5) also depends on (4).\n\n You write your own code.  yA isn't a site to do your homework, since if you don't do it yourself, you don't pass the class!
700886
sorry i am no good at math, but this may help:\nhttp://www.fourmilab.ch/earthview/pacalc.html
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
191332
Find the dementions and multiply.
224079
try this site... it shows step by step how to go about sloving these equations, good luck:
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
409326
I haven't used this, but it looks fairly simple (as opposed to the professional mapping products):\n\nhttp://www.smartdraw.com/specials/geography.asp?type=12195&id=12195\n\nGood luck
780884
As with many problems, it helps to draw a diagram.  I made a triangle, and then put in each of the altitudes, since they are my only known values.  With these in place, we now have 6 principal right triangles, of which we know the value one leg (the altitude).  In each of these triangles, the hypoteneuse is a complete side of the original triangle, and the angle opposite the known leg is also of the original triangle.  Personally, I chose to work with cosines and tnagents, but you could just as easily have used sines (though I think it might add a step or two).  Remember that the cosine of an angle in a right triangle is merely the ratio of the side opposite the angle to the hypoteneuse.  Don't worry that we don't know the value of any of these angles; I went through the entire problem without ever needing it.\n\nSo, since each angle, in its entirety, is in two of our right triangles, they give us two ratios to compare.  That is:\n\nIf the sides of the original triangle are x,y,z\ncos a1 = 20/x = 15/y\ncos a2 = 20/z = 12/y\ncos a3 = 12/x = 15/z\n\nWith a little cross multiplication and rearrangement of terms, we find that:\n12z = 15x = 20y.  \n\n(Your x, y, z may be in a different order depending on how you labeled your triangle, but if you manipulate your triangle a bit, you should be able to get it to match.  This will keep your coefficients identical to mine.  But even if you don't match it up, in the end you will get the same answer.  That's the beauty of math.)\n\nThe tangents are a little trickier, because they involve a fraction of the original side.  This involves the addition of a new variable on each, such that one portion of the line is "a", and the remainder is x - a; and then b and y - b; c and c - z.  The tangents then follow:\n\ntan a1 = 20/(y - b) = 15/a\ntan a2 = 12/(z - c) = 20/b\ntan a3 = 15/(x - a) = 12/c\n\nRearranging each gives us the three equations:\n20a = 15(y - b)\n12b = 20(z - c)\n15c = 12(x - a)\n\nFrom there, you can solve for either a, b, or c.  It's your choice.  I opted to solve for a.  Substituting the third equation into the second, and then that new equation into the first, we get:\n\n20a = 15(y - (20/12)(z - (12/15)(x - a)))\nwhich reduces to\n8a = 4x + 3y - 5z\n\nNow, side a is in a right triangle with hypoteneuse y and second leg 15, meaning that:\na^2 + 15^2 = y^2   or\na = sqrt(y^2 - 225)\n\nUsing this equation, plus the ratios we received from the cosine functions, we solve for y:\n\n8 * sqrt(y^2 - 225) = (16/3)y + 3y - (25/3)\n\nConveniently, the right half is equal to zero, so dividing by 8 and squaring:\n\ny^2 - 225 = 0\ny^2 = 225\ny = 15\n\nEnter this into our ratio:\n\n15x = 20y\n15x = 20*15\nx = 20\n\nand\n\n12z = 20y\n12z = 20*15\nz = 25\n\nSo finally we have the lengths of all three sides, and adding them together, the perimeter is\n15 + 20 + 25 = 60\n\nThis turns out to be a right triangle, but that is merely happenstance.  This process can be used to solve for any triangle with three known altitudes.\n\nHope this helps.
297357
http://www.webmath.com/geo_cylinder.html\nif u want examples go to this link or find another link on yahoo
1008897
try using a jpeg or png.  and reimport it inot your library.  Also when you convert it to a symbol make sure not to move the center achor point.
971893
write the equation of the distance from a point to the center,\nthat way it should work\nalso writing in english might get you more answers\ns
681882
I can't see your image, but if it is like any of these other pictures (see link), the problem is with the slopes of the triangles.  They may appear the same, but they aren't and that is where you lose a fraction of a square.  These small differences eventually add up to a full square if you rearrange things differently.\n\nThe key is in the slopes.\n\nI wish I could see the picture because I like area paradox puzzles like these.\n\n-----\n\nI'm back and can see your picture now... it is just like the first paradox in my link.\n\nThe answer, as I said is the slopes of the triangles.  One triangle has a rise of 2 for a run of 5 (slope = 0.40) and the other has a rise of 3 for a run of 8 (slope = 0.375).  So despite how it looks in the picture, that is not a straight line along the top.\n\nIn the top case, the slope causes a slight bend inward on that line.  In the bottom case, the slope causes a slight bend outward on that line.  If you take the difference between these two you get 1 sq. unit which accounts for the "missing" square.  The second link shows this visually.