a figure having a polygon base and triangular sides which meet in a point  at the top.?
409945
The figure you are asking about is a pyramid.  If you think of the "traditional pyramid" it has a square base (which is a polygon), with 4 triangular sides that meet in a point at the top.  Yet in fact, a pyramid can have any polygon for a base.
990247
A sphere.  It has an inside and an outside.
189752
sphere, cube, pyramid, cilinder
847030
Create a triangle whose sides are x,y and z and whose angles opposite those sides are X, Y and Z respectively.  Draw a perpendicular line from one angle to the opposite side.  This segment is the height(h) of the of the triangle and the opposite side is the base.  If you picked /_ Z to draw the height then side z is your base and the area of triangle XYZ is 1/2 the base times the height or 1/2(z)(h).  There are 2 right triangles formed in triangle XYZ by the height segment to side z.  The point on side z I labeled W.  Triangle XZW and YZW have shared side ZW(h).  W divides side z into two segments XW and WY.  I let XW = d and WY = (z-d).  Using the Pythagorean Theorem we get two equations:\nh^2 + d^2 = y^2  and  h^2 + (z-d)^2 = x^2  I expaneded the second equation to h^2 + z^2 -2dz + d^2 = x^2.  Next I solved the equation for h^2 + d^2 = x^2 + 2dz - z^2 .  Using the two equations with \nh^2 + d^2 you can set  y^2 = x^2 + 2dz - z^2.  You can now substitute x = (a^2 + c^2)^.5, y=(a^2+b^2)^.5 and z=(b^2+c^2)^.5 which were given as the sides fo the original triangle into the last equation solved for d as (y2+z^2-x^2)/2z.  This will give you a value for d in terms of a,b an c.  I got d = b^2/(b^2 + c^2)^.5 \n\nAfter getting this value you can substitute for d and y in the equation h^2+d^2=y^2 to solve for h.  \n\nUsing the value of h and z both in terms of a,b and c you can calculate the area of triangle XYZ in terms of a, b and c using \n\narea of a triangle = 1/2(height)(base) \n\nwhere the height = h and the base = z.  \n\nWhen I solved for h, I got \nh = [(a^2b^2 + a^2c^2 + b^2c^2)/(b^2 + c^2)]^.5\nz = (b^2 + c^2)^.5 was given\n\nArea Triangle XYZ = (1/2)(h)(z) = (1/2) times \n[(a^2b^2 + a^2c^2 + b^2c^2)/(b^2 + c^2)]^.5 times\n(b^2 + c^2)^.5)\n\nThe numerator and denominator have a common factor\n(b^2 + c^2)}^.5.  when the ratio is reduced you get\n\nArea = [(a^2b^2 + a^2c^2 + b^2c^2)^.5]/2
759341
If the figure is a square or rhombus, p = 4L\n\nIf it is an equilateral polygon with n sides, it is p = nL
459931
a nine-sided polygon???  can't wait to see other answers...
685476
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.
659736
Not a site, but I suggest you invest in something called Mathcad.\nTeach, please note, it is a site not a sight! Unless you were 'looking' for the answer... sorry!
128133
A hexakaidecagon (hexadecagon).\n\nFor the rest of the polygon questions you'll have, try the link below:
809433
1.rectangle\n\n2.trapezoid
668644
"triangular numbers" they're called
204082
i think it might be called a triangular prism\ni am srry that was tough i was trying to get it rite so i could get the ten points for best answer. I AM SORRY!
1085259
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?
142784
an isoceles triangle pointing up\nand another pointing down right on top of it
100955
Cube, pyramid, sphere, cone, cylinder.  There are others.
743787
infinitly as many... there are millions of figures... triangle (3 sides) square (4 sides) pentagon (5 sides)... I wonder what you call a figure with 100,000 sides... hehe
113217
There are two names for a seven sided shape, heptagon and septagon.  Both answers are equally correct