is it possible to draw a geometric figure with four dimensions
203662
Sure, why not.  I have a map that tells elevation and is color-coded for the geologic age of the exposed rocks.  That's four dimensions.
138216
U can see the whole world as Geo-metric figures. The Design of a house by civil Engineer is full of Geometry. Followed by lands Planned to use for building purpose and for Ariel view. And with the help of ice in winters if u go to the north and south pole u can find natural and Man made designs which r Geometric figures which u have mentioned above.\n\n  Geometry is not the different types of angles and triangles it includes figures like cone, circle, cylinders , Pentagon and u can go on  Infinity
468169
well you can use 3d max or maya ( its betta to use 3d max if its not very high weighted model ) in case of 3d max you'll have to get 3 drawings of the model. 1 from front/back, 1 from side & 1 from top or bottom. Scan them & then edit it using 3d max as you'll have to play with volume & vertex in it.
646355
I think so, actually...\n\n1. Draw 2 perp. lines.\n2. Draw 2 hypotenuses to each base such that the triangles made are 45-45-90 triangles (extend the bases as well, if you need to).\n3.  You have just made an isoceles triangle since the base angles are congruent.\n4.  The line that is perp. to the base is now the perpendicular bisector because the triangle is isoceles.\n\nViola!
445041
It would really help to draw a diagram here, but...\n\nThere are 3 sets of 5 guy wires, so we really only have to deal with one set and multiply by 3 at the end.\n\nThe base of each triangle will be 1/4 x 100m = 25m.\n\nThe heights of the triangles will be 20, 40, 60, 80, and 100.\n\nCalculate the hypotenuse in each case, add on 2 meters to each hypotenuse, add all the answers together, and multiply by 3.\n\nNeed me to go further?
615061
You need to get a little more specific in terms of what kind of "3d equivalent" you want.  Two kinds of 3d solids you can form easily from 2d polygons (closed shapes made of line segments) are prisms and pyramids.  You make a prism of any 2d shape by taking 2 copies of the flat shape, one above the other, and connecting the corresponding vertices (corners) with edges (lines).  From this perspective, a cube, is a square prism.  A triangular prism looks a little like a tent, with two triangles (top and bottom) and three rectangles or squares around the sides.  To make a pyramid, choose any 2d shape as the base, and then a point above the base as the top, and radiate edges down from the top to all the base's corners.  There are other "3d equivalents", so if you're thinking of something else, let me/us know.\n\nYou also need to be more specific about what kind(s) of sides interest you: edges (1d lines) or faces (2d flat surfaces).  The "sides" of a 2d shape are all edges, but 3d shapes like cubes have both edges and faces (a cube has 12 edges and 6 faces).  Any prism made from a polygon with N "sides" (edges) will have 3N edges (N on the top, N on the bottom, and N more connecting top and bottom) and N+2 faces (N around the sides plus top and bottom).  Any pyramid made from a polygon with N "sides" (edges) will have 2N edges (N around the base and N more reaching up to the top point) and N+1 faces (N around the sides plus the base).\n\nDoes this do it?
540583
I'm not sure, but I have an idea.  The surface area of a sphere is 4 pi r^2.  You have 40 points spread out over that area, so each point is "entitled" to 1/40th of the total area, or (pi r^2)/10 square units.\n\nNow think about how these points, or actually, small areas around points, are going to be packed.  They won't be circles ... if you put several pennies together on a table, there's be small gaps wherever three pennies join together.\n\nHexagons are more efficient.  Draw a regular hexagon, and you'll see that you can surround it with six more hexagons, completely filling the space without any gaps.\n\nNow consider just one hexagon -- the space surrounding just one point, which is located in the center.  This hexagon, which has an area of (pi r^2)/10, is composed of six equilateral triangles.\n\nSuppose the side of the hexagon is "a" (which is also the side of one of the six small equilateral triangles).  Then the height of the small triangle is (a/2) sqrt(3), and the area of the small triangle is 1/2 bh = (a/2)^2 sqrt(3).  Since there are six small triangles in the hexagon, the area of the hexagon is 6(a/2)^2 sqrt(3) = 3/2 a^2 sqrt(3).\n\nBut that must equal (pi r^2)/10, so\n\n3/2 a^2 sqrt(3) = (pi r^2)/10\n\nSolving for a, we get a = sqrt[pi sqrt(3)/45] r = 0.3477 r\n\n[One cautionary note here:  I'm doing plane geometry, and all these hexagons are plane figures.  This assumes the sphere is circumscribed around all the vertices of the 40 hexagons ... something like a soccer ball.\n\n[Total area, though, will be equal to the sphere because we divided the sphere's total surface area by 40.  Although I don't want to do the work, I think it's possible to include the curvature into this analysis.]\n\nNow we want to know how far apart two adjacent points are located.  Each point is at the center of a hexagon, and the distance between two hexagon centers is twice the height of any one of the small triangles.\n\nThis height is (a/2) sqrt(3), so the distance between points is a sqrt 3 = sqrt[pi sqrt(3)/15] r = 0.6023 r (and all the angles are 60 degrees).  (And that 0.6023 r is probably the answer to your question.)\n\nNow let's get the curvature in there and figure out the central angles.  Assume the hexagons are all stretched so that they form part of the sphere.  Draw a great circle passing through two adjacent points.  We want the central angle of the sector where the arc length is 0.6023 r.  But because arc length is just r theta (in radians), the central angle is 0.6023 radians, or 34.5 degrees apart.
47757
It is not as easy as the last person suggested.\nFirst it depends on the versions of both ACAD and Corel Draw.\nSecond it depends on whether or not you want to be able to manipulate the ACAD Drawings in Corel Draw.\n\nIf all you want to do is place the image into Corel than yes you can export them from ACAD as JPG's.\n\nIf you want to be able to manipulate them that is a whole new ball game.\n\nI believe from as far back as Corel Draw 5 there is the option to import ACAD files. Unfortunately it doesn't work that well in any of the versions. yes it can be done but the more complicated the drawing the worse it will translate. The biggest problem bring that Corel Draw will bread ever line segment into individual pieces. That means that all of your curves and and each dash of a dashed line etc... will all come in as individual pieces. That makes for a huge file and lots of crashes because of memory issues.\n\nI have done this a LOT... In my experience it is best to Open your ACAD file. Select only as much as you absolutely need and then cut it out of ACAD.\nIn Corel Paste the object in using "Paste Special" and as a "Picture (Metafile)".\nThen you have to ungroup the ACAD file.\nDelete the Black background and change the line weights and line properties to meet your needs.\n\nMind you depending no how close your are zoomed in, in ACAD will dictate the size your drawings is pasted into Corel Draw. So a drawing scale can change by simply zooming in and zooming out.\n\nHope that helps.\n\nfeel free to let me know if I have confused you.\n\nTodd
352379
No, if u draw two right triangles of a common( extended base ) snd then try drawing another triangle ( any type ) by extending the perpendiculars, its impossible as the perpendicualrs are parallel to each other.
605893
hi \nquestions are easy but need to draw it. it seems not posible here. \nphonecardgalery@yahoo.com
55522
Something that has height (length) and width . . . and no depth.\n\nDraw a square on the piece of paper. That square is two dimensional . . . if you draw a cube . . . it shows a third dimension (depth).
522278
This is freeware. Enjoy.\n\nhttp://www.sketchup.com/
521943
Geometry Session 2, Part B: Linkage-Strip Constructions  \n... (or the hands-on version) to try to build quadrilaterals with the given lengths. ... Can a set of four lengths make two different quadrilaterals? ...learner.org/channel/courses/learningmath/.../part_b/constructing.html \nEasy Construction of Bicentric Quadrilateral II  \nEasy Construction of Bicentric Quadrilateral ... By construction, four quadrilaterals MPND, MPLC, LPKB, and KPNA are also cyclic. ...www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri2.shtml \nBrahmagupta Quadrilaterals  \nIn this paper we give a construction of Brahmagupta. quadrilaterals from rational Heron triangles. ... Construction of Brahmagupta quadrilaterals ...forumgeom.fau.edu/FG2002volume2/FG200221.pdf \nEasy Construction of Bicentric Quadrilateral  \nEasy Construction of Bicentric Quadrilateral ... appears as a consequence of another construction of bicentric quadrilaterals. ...www.cut-the-knot.org/Curriculum/Geometry/BicentricQuadri.shtml \nOnline Lesson On Geometrical Construction Using iMovies  \n... On Geometrical Construction Using iMovies. LESSON ... Module 3 - Construction of Quadrilaterals. After this module, the students should be able to. 1. ...ali.apple.com/ali_sites/sgli/exhibits/1001179/Lesson_Objectives.html
349402
Draw axis of symmetry to the base.  Look at the rightangled triangle on the left side.  It has hypotenuse s, short side s/2.  Use Pythagoras' to show the height is (sqrt(3)/2)*s, then use\narea = (1/2)*base*height
333728
As I understand from your question, a point fill set will be one that contains ONLY regular (or suicide) polygons. I am assuming that each polygon will have a vertex affixed to the so-said point. And that a complete set will be one of having the vertices (i.e. interior angles at each vertex-on-point) add to a circle (360 degrees).\n\nIf such are the conditions, then the triangle would have a "vertex-on-point" angle of 60 degrees; while the pentagon will possess 108 degrees. This brings to a total of 168 degrees and a remaining of 192 degrees to fill up.\n\nConsidering this, it is impossible to obtain a single regular polygon with such an interior angle.\n\nNow to consider multiple polygons to fill up this space..\n\nThe regular polygon with the least number of sides is an equilateral triangle (already in question). This has 60 degrees for its interior angle, also its angle of vertex-on-point.\n\nYou'll realise as you increase the number of sides, the interior angles will be as follows:\n n -->> angle (in degrees)\n==========================\n 3 -->> 60\n 4 -->> 90\n 5 -->> 108\n 6 -->> 120\n 7 -->> 128.6\n 8 -->> 135\n 9 -->> 140\n 10 -->> 144\nand so on..\n\nAs you'll realise, the trend is that the angle sizes increase with the number of sides.\n\nNotice how from these figures that it is impossible to obtain a combination that totals to 192 degrees? (Note: I've not worked on this conjecture to prove that it holds true for all of n)\n\nSo since that there is no such single polygon with an angle-on-point value of 192 and that no multiple combinations of polygons can add up to that figure, does this hold true that a point fill set cannot have both a triangle and a pentagon? I think so.\n\nI sincerely hope that the above has helped in aiding your understanding of the question.
287759
It's called ASCI art. There's a software that can produce them using software called bg_asci
462476
Well it depends on the problem specifically, but usually you have to find the ratio of "two corresponding lengths in two similar geometric figures."(mathworlds)  Actually that is what the scale factor is.