what are the special lines and segments that are related to triangles?
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Theres quite a few, so check out the source below for a good summary.
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Not all triangles are congruent.
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For Sine 105- Perpendicular/Hypotenuse\n\nFor Cosine 105- Base/Hypotenuse\n\nFor Tangent 105- Perpendicular/Base\n\n\n* Perpindicular, base and hypotenuse are the three sides of the triangle.
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its the middle of a angle like if da two lines combine the pointwere they combine is da vertex..........and you cant graph vertexs
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circular segment\n\nhttp://en.wikipedia.org/wiki/Circular_segment
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Enjoy:\nhttp://unmuseum.mus.pa.us/triangle.htm
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HEPTAGON - a seven-sided polygon.\n\nTRIANGULAR PYRAMID - a pyramid with a triangular base and 3 triangular faces.\n\nSCALENE TRIANGLE - a triangle none of whose sides have equal length. The angles don't have the same measures, too.\n\nISOSCELES TRIANGLE - a triangle 2 of whose sides have equal measures. Therefore, the angles opposite the congruent sides have equal measures also.\n\nEQUILATERAL TRIANGLE - a triangle all of whose sides have equal measures. The angles have equal measures, too.\n\nRIGHT TRIANGLE - a triangle 1 of whose angles has a measure of 90 degrees. The other 2 angles have measures that are less than 90 degrees. The side opposite the 90-degree angle is the longest side and is called the hypotenuse. The other sides are called legs.\n\nSPHERE - is shaped like a ball (round).\n\nCONE - it is a pyramid with a circular base.\n\nDECAGON - a polygon with 10 sides.\n\nDODECAGON - a polygon with 12 sides.
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First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.
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A hexakaidecagon (hexadecagon).\n\nFor the rest of the polygon questions you'll have, try the link below:
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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?
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Look at this site and you can find out.  It has several different kinds of polygons listed.\n\nhttp://www.mathgoodies.com/lessons/toc_vol1.html\n\nEDIT:  Sorry.  It looks like you have to sign up for something on that site.  Try this one instead:\n\nhttp://www.math.com/school/subject3/lessons/S3U2L4GL.html#
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You might be thinking of a Reuleaux (or Rouleaux) triangle, but it has rounded sides, not rounded corners.
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This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
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Try these;\nhttp://www.mathleague.com/help/geometry/geometry.htm\nhttp://www.coolmath.com/home.htm\nhttp://www.ics.uci.edu/~eppstein/geom.html
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Hope this helps\n\n\nhttp://argyll.epsb.ca/jreed/math9/strand3/triangle_congruent.htm\n\nhttp://mathforum.org/geopow/fullsolutions/102596.solutions.html or www.mathforum.org
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The answer to the question depends on exactly what kind of geometry you are dealing with and what "points" and "lines" mean. \nIf you are talking about ordinary lines and ordinary geometry, then parallel lines do not meet. For example, the line x=1 and the line x=2 do not meet at any point, since the x coordinate of a point cannot be both 1 and 2 at the same time. \n\nIn this context, there is no such thing as "infinity" and parallel lines do not meet. \n\nHowever, you can construct other forms of geometry, so-called non-Euclidean geometries. For example, you can take the usual points of the plane and attach to them an additional point called "infinity" and consider all lines to also include this additional point. In this context, there is a single "infinity" location where all lines meet. In a geometry like this, all lines intersect at infinity, in addition to any finite point where they might happen to meet. \n\nOr, you could attach not just one additional point, but a whole collection of additional points, one for each direction. Then you can consider two parallel lines to meet at the extra point corresponding to their common direction, whereas two non-parellel lines do not intersect at infinity but intersect only at the usual finite intersection point. This is called projective geometry
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Yes. Right triangles have them all the time. it's called the orthocenter i think....it's been a while since Geometry.