Can someone please help me?!\n\nI need the properties of:\n\nheptagon\ntriangular pyramid\nscalene triangle\nisosles triangle\nequilateral triangle\nright triangle\nsphere\ncone\ndecagon\ndodecagon\n\n\n\nH-E-L-P please even if you can answer all of them, anything would help me
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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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A equilateral triangle has rotational symmetry because there is a center point that you can rotate it about such that it will have the same positional appearance three times.  You need to rotate it one third of a circle to get each identical view.\n\nThere should be infinitesimal degrees of rotation for a circle such that it will still appear to be a circle.  You could move a point on its circumference to any other point on the circle and it will still be a circle.  I guess you knew that.
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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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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.
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yes can you also, email me the answer. It would be helpful in my polysci. class.
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a) "Suppose Triangle ABC and Triangle DEF are congruent."\nb) "Suppose there exists a line segment with at least two right bisectors."\nc) "Suppose there is an isosceles triangle for which the angles opposite two of its congruent sides are not congruent."\nd) "Suppose there is a line segment with at least two midpoints."\ne) "Suppose there are two intersecting lines such that opposite angles are not congruent."\nf) "Suppose AB is not equal to CD."
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Nor sure if I understand your question.  Are you looking for 3 dimensional and 2 dimensional shapes?\n\n2 dimensional\n-square\n-circle\n\n3 dimensional\n-cube\n-sphere
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I did some research... here are some websites that might help...\n\nThe 2nd one is a map quiz that will help you learn more about the physical features.
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That’s because the slope of the triangle with legs 2 and 5 is different from the slope with legs 7 and 3.  Therefore, the “triangle” isn’t a real triangle.  It just looks like one.\nThe triangles are supposed to look like a 5 by 12 triangle.  But the actual area sum of the parts is only 29.5 compared to 30.
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the smaller 1 being right u get the hypotenuse as 5..\n\nthe bigger triangle n the smaller 1 are similar...so sides are proportional...(the sides r acually duble of tht of the smaller 1) so the perpendicluar is 6... n the hypotenuse of the biggger triangle is 10..(duble of the smaller 1)...\n\nso perimeter is 8+6+10=24\n\nperimeter of the shaded region is indeed 19..4+6+5=19..got it?
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http://math.arizona.edu/~lega/UG/Colonies/Colonies.htm\n\nhttp://www.magma.ca/~scimat/B_subtil.htm\n\nhttp://www.pnas.org/cgi/content/full/0600829103/DC1
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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.
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Bob, I used to enjoy your show before you started doing all those really expensive houses.  Then, of course, you left.\n\nOh, well.\n\nDon't you know, Bob, that isosceles triangles don't have "right sides?"  An isosceles triangle has two "legs" of equal length.  If it is a right isosceles triangle, the side opposite the right angle is called the hypotenuse.\n\nSay hi to Norm for me!
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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!
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Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
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ok: i hope you sketched this out: forget for a minute the solid's diagonal is 26cm. let's use the ratios as actual units. so the one side with a 3*4 rectangle has a 5unit diagonal (from Pythagoras).\nnow-picture the right triangle with the two adjacent sides to the right angle, one side is 5 and one is 12 (that's the solid's diagonal). using OUR units for now, THAT diagonal is 13units (Pythagorean again). luckily, 26 is 2 * 13, so hopefully the measurements are 6,8,24, would such a solid yield a diagonal of 26?
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Ive noticed you were asking quite alot of these geometry questions. I think you should pay more attention in class and you wouldnt need to ask home work questions, and plus, just knowing the answer wont help, youll have to know how to do it, especially if this is for a final exam...