Geometry questions..... need help?
923619
What is the locus of points in space that are equidistant from two parallel planes?\nIt is a plane in the middle of the two parallel planes.\n\nWhat is the locus of points in a plane that are equidistant from points A and B in the plane ?\n\nIt is a line in the middle the points\n\nWhat is the locus points in the plane of an angle that are equidistant from the sides of the angle\n\nIt is an angle bisector of the angle.
489663
Reviewing the facts:\n\n1.  You're constructing a pentagon and two trapezoids by using the center of the square.  Since the sides are 1, the short parallel side of the trapezoid is 1/2.\n\n2.  Since the two trapezoids are congruent, the short parallel side is midway in the square.  So the height of each trapezoid is 1/2.\n\n3.  Since there are three areas now, and each area is equal, the area of the trapezoid is 1/3.\n\nThe area of a trapezoid is A = (1/2)h(b1 + b2).  We have everything we need now to solve for b2, the longer parallel side:\n\nA = (1/2)h(b1 + b2)\n1/3 = (1/2)(1/2)(1/2 + b2)\n1/3 = (1/4)(1/2 + b2)\n4/3 = (1/2 + b2)\nb2 = 5/6
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Do a Yahoo search.  Enter "Homework help, math."  There are bunches of web sites.  Good luck.
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go to a math website off yahoo. it should help.
541010
1.  Find the line that goes through d that is perpendicular to line l.  The slope of the perpendicular line is -1/m, where m is the slope from line l.  Since you have the slope for that line, then just plug in d to get the intercept, which I will call c.\n\n2.  Find the point (x,y) where the two lines meet by solving the system of equations.\n\ny = mx + b\ny = (-1/m)x + c\n\nI will call this point e.\n\n3.  Find the distance between d and e.  If d = (x1,y1) and e = (x2,y2) then the distance between d and e is\n\nsqrt((x1-x2)^2 + (y1-y2)^2)
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r=radius\nd=diameter\n\n1 C=2r(pi)\n2 C=d(pi)\n\n(Same thing because 2r=d)
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Why do some questions not have answers? I guess that is what you mean. I went to answer one just a minute ago and it said I need help with my geometry but that was all it said. Now how can you answer that. If the question is really vulgar I don't answer them I figure someone will.
579191
Area of a triangle:\n\nA = 1/2 (B)(A)\nwhere B = base which could be any of the three sides\nand A = altitude (or height), which is the perpendicular distance to the base B from the opposite vertex.
100993
how do you din the height of a trapezoid if one of the non parallel sides is straight, and the other is slanted. Which makes the two parallel sides different lengths? Also, how would you find the width of a trapezoid in this situation?\n\nIf the length of the shorter parallel side is A and the length of the longer parallel side is B and the length of the straight side is C and the length of the slanted side is D then:\n\nthe height is:\nh = C (the length of the straight side.\n\nThe width depends on where you measure the width.\n\nI am imagining that your trapezoid looks like this:\n.__\n/__|
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1.rectangle\n\n2.trapezoid
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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
780311
You are on the right track.\nThis a geometrical problem that requires drawings.\nCheck this link http://departments.kings.edu/chemlab/chemlab_v2/packgeo.html\n\nI'll get back to you later \n\nso later it is\n\n3--------R=r(1+2/3^.5)\n7--------R=3r\n19------R=5r\n37------R=7r\n61------R=9r
120928
ok, let the triangle be ABC with sides AB, BC, AC.\ndraw a typical line, get too point on in with the distance of BC, name them B and C. then draw an arc with the center of B and the length of AC and another one eith the center of C and the length of AB. if they had a common point, it would be A and your triangle is there, but if not, there is no triangle with those three sides.
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try sosmath.com  they helped me before. if they don't have the answer you are looking for they have a list of links to tons of other math sites.  \n\nand there probably is someone on your campus that can help as well.  One on one help is always best...\n\nGood luck!
91628
Also, when working out the surface area of the triangle sides, ensure that you are NOT usng the perpendicular height to calculate this. In some instances, you may need to use pythagoras' theorem to get the height measurement of the triangle sections
182640
I'm not sure if this is what you're asking, but if...\nd = length\nd-3 = width\n\nthen the area is d(d-3), a.k.a. d^2 - 3d .\n\nHope this helps!
427811
You can try these websites:\nwww.math.com/homeworkhelp/Geometry.html \nwww.mathleague.com/help/geometry/geometry.htm