The resulting shape would be comprised of a total of 4 sides (including the base), and all 4 sides are triangular.  Think of it as a 3-sided pyramid.
133189
square pyramid...if square as base\n\ntriangular pyramid...if triangle as base\n\nBasically just look at the base...and add pyramid to it.\n\nEx. 5 sided regular polygon base...pentagonal pyramid.\n\nIf it isn't a regular polygon...look it up...or just leave it as pyramid I guess. LOL!\n\nTetrahedron...sound molecular...like the shape of methane molecule. I'm thinkin' this is geometry...so stick with the simplistic "pyramid".\n\nOH...BTW...if it only has one base...and is composed of multiple triangular sides...then it's a pyramid. If it has 2 bases...and any other figure for it's sides...then it's a prism.
179770
The oblique answer is correct. As for the slanted 3-D rectangle, I believe you are referring to a parallelepiped: http://mathworld.wolfram.com/Parallelepiped.html .
555946
Actually the only way it would be possible is if you use Elliptic or Riemannian geometry, and would require a manifold.  A representation of such a construct would be similar to drawing a 4 dimensional shape in 3 or even 2 dimensions.\n\nEuclidian geometry doesn't allow triangles with a sum of all angles to be greater than 180 degrees.
400396
.............6\n........______\n....../.|...........|&#92;\n...../..|...........|..&#92; 2\n..../_|_____|__&#92;\n\n\nThe triangle on the right is a 30-60-90, so the shorter side must be 2/1 = 1.  So the length of the bottom part is 6+1+1 or 8.  The height must be 1√3, so the area is 1/2(6+8)√3 = 7√3.\n\nGopal screwed up when he did 2*2cos60.  That's 2*2*1/2 = 2, not 4.
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!
509593
The figure we are talking about is an isosceles trapezoid. The lower base is equal to the two diagonals and the upper base and the two sides are all equal in length.\n\nCall the trapezoid ABCD with AB being the lower base and cd bbeing the upper base. The two sides which are both equal to the upper base are BC and AD. Also notte that AB parallel to CD.\n\nThe diagonals are AC and BD. \n\nIt turns out that the two base angles are bisected by each of the diagonals.Since the base angles are equal then they are each divided into two equal angles. Call each of them angle 1. also By the parallel postulate, angle CDB and angle must also be equal to angle 1.\n\nNow for the base to be the same length as the diagonals, triangle\nABC and Triangle must both be isosceles. This will be true only if angle ADB=.twice angle 1 and angle ACB = twice angle 1.\n\nTherefore the two base angles add up to 4 angle 1s and the agles C and D are both = 3 angle ones. Thus the four vertex angles add up to 10 angle 1s. Since the total of all angles in a quadrilateral is 360 degrees, the angle 1 must 360/10 =36 degrees.\n\nThus each base angle = 2 angle 1s = 2*36 =72 degrees and each of the upper angles must be equal to 3 angle 1s =3*36=108 degrees.\n\nIn summary angle A and angle B = 72 degrees and\nangles C and D both = 108 degrees.\n\nHope this was not too confusing. It will help if you draw a picture.
607144
Because the tangram is a set of physical shapes, it's difficult to demonstrate how to construct shapes with it in the limited capacity of a Yahoo! answer. See the wikipedia web page I list in Source(s) below as a start. This page does show a rectangle, comprising two squares, composed out of the tangram pieces. Tangrams are a great way to put your knowledge of geometry to use. If you understand the properties of isosceles trianges, squares and parallelograms, it's easier to put the pieces together in the shapes you want. But trial and error works, too.\n\nGood luck and have fun!
214490
When you put your parts together, make them in the shape of triangles. Triangles are the strongest shape for building a simple construction like the one you're making. Make your structure with many triangles. For example, if you can, make your entire structure in the shape of a triangle, and also connect the sides of the big triangle with lots of smaller ones.
91120
It depends on the object.  You find the area of all the sides and then add them together.
289067
Yes form a hexagon with 6 of the boards and then place the other six boards so they meet at the center. Each pig will then have an equilateral triangle as a pen.
519962
I am only using visual techniques and not math theorems here:\n\nThe first plane will divide space into 2 parts.\n\nIf totally random the second plane will probably not be parallel with the first.  This will divide the 2 parts into 4.  If it did happen to be parallel with the first, then the space is divided into 3 parts.\n\nThe third plane will most likely not be parallel with either of the other two.  It would divide the space into 8ths.  If it was parallel to either of the other two then the minimum parts is 4 and the maximum is 6 parts.\n\nThe fourth plane will most likely not be parallel with any of the first 3, and can divide the space into maximum of 15 parts.  The fifth will make that 26.  Parallel planes will reduce this number.\n\nThe space will be divided into a minimum of 6 and a maximum of 26 parts with 5 planes.\n\nI just found a great description at the site below:
223476
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
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
338138
It's a face that is not flat.  A cone has two faces: the base which is flat and the rest of the cone which is curved.
11194
You can't have a polygon of that size.  A polygon is an object with three or more straight sides.  One property of a polygon is that no one side can be longer than all the other sides added together.  You listed 6 dimensions, implying a six sided polygon.  But since one side is much longer than the other five together, this is not a real object.\n\nTry taking a yardstick and five toothpicks and put them together to make a polygon shape.  It won't work.\n\nIncidentally, even if you gave realistic dimensions, that's not enough information.  Consider this 6-sided object:\n._\n/ .&#92;\n&#92;_/\nand this one:\n||\n||\n||\n\nEven if the sides were all the same length, they have different areas because the angles are different.
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.
551725
area of the base times the height.