why do all quadrilaterals tesselate?
451315
The correct answer was given above; the sum of the interior angles is 360, which is a necessary property for a shape to be able to tessellate. That this property is sufficient for quadrilaterals is shown in the source.
150758
to indicate a solid shape wich has the same rectangular section on each planar parallele to its rectangular base.\nIt has two congruent rectangular faces and four lateral sides shaped as parallelograms (eventually rectangles)\n\nIf all the faces are rectangles, then it's called a parallelepiped
949904
I suppose that it means all the lines that you can draw from every vertex to every other vertex.  For a rectangle it would be 2.  For a 5 sided figure.( I got up and sketched it) it would b 5.  You only count a line once, not both directions.\n\nAnother sketch suggests 9 for the hexagon.  Why don't you sketch a hexagon, draw and count the lines between vertices.
979529
OK, I think what you're saying is that all the corners have to be colored and that if you can rotate a square to superimpose it on another square, or if they're mirror images, then they're the same.  You didn't say that all colors have to be used, so I'll assume they don't.  That said...\n\nStart with squares in which all four corners are the same color.  There are obviously 4 of these.\n\nNow take squares in which three corners are one color and the fourth is different.  There are 12 of these (three squares can be any of four colors, and the odd one can be any of the other three.)\n\nNext, take squares in which two corners are one color and two are another.  There are six color combinations (red-white, red-blue, red-green, white-blue, white-green, blue-green) and each one can have two patterns:  either the same-colored corners are next to each other, or they're diagonally opposite.  So there are 12 of these.\n\nNext, consider square in which two corners are the same and the others are two different colors.  For each combination of a major (2-corner) color and two minor colors, there are only two patterns:  same colors adjacent and same colors opposite.  Whichever way you arrange the odd colors, it will be the mirror image of the other way and therefore identical.  So for each major color, you can pick any two of three minor colors.  This is the same as leaving out any one, so there are three options for minor colors.  4*3*2 patterns is 24 different squares.\n\nFinally, consider squares in which each corner is a different color.  Now, pick a color and a corner.  Two of the other colors will be adjacent to that corner, and one will be diagonally opposite.  For any choice of opposite colors, either arrangement of adjacent colors will just make another mirror image.  So the only choice here is which color will be opposite.  That means there are three different 4-color squares.\n\nNow, add it all up: 4 one-color squares, (12 + 12 =) 24 two-color squares, 24 three-color squares, and 3 four-color squares, equals 55 unique squares.
662883
Geometrically, each letter has at least one axis of symmetry.
196942
wow, that's pretty amazing.  i was convinced that was wrong until i tried drawing the "counterexample" i had in my head.  other than reading through that proof (props to the guy who posted that link, btw) i guess the best way to realize this is true is to show that for any five closed shapes in two dimensions, there will always be at least two shapes that do not touch one another.  because of this, you can color those two shapes the same color, so you will have a maximum of four colors.\n\nthis problem is reducible to a graph with five nodes, where each node represents a shape and each edge represents a shared boundary.  (showing that this is reducible to this is probably the difficult part of the proof, which is why i'm glossing over it).  you cannot connect each node to every other node without there being at least one set of intersecting edges.  if edges intersect, it indicates that the shapes (or countries) represented by the two nodes connected by one of the intersecting edges do not share a border.\n\nhere's my suggested proof (assuming we accept the reduction to this graph problem):\nstart with a graph G with three nodes, {A, B, C}.  There are also edges {AB, BC, AC}.  So, G is a closed graph with three nodes and three edges where each node is connected to each other node in the graph (i.e. G is a triangle).  If our reduction of the coloring problem is correct, then there exists a map reducible to this graph that requires three colors to be colored in the described way.\n\nNext, construct a graph H by adding a node D to graph G.  There are two possible ways to add D: 1) D can be on the interior of closed graph G; or 2) D can be outside the interior of G.\n\nIn case 1, let the new graph with nodes A,B,C,D be called H.  We need to show that we can make H such that D is connected to A, B, and C without any intersecting edges.  This is simple enough; straight edges to A, B, and C from D do not intersect each other or the existing edges of the graph G.  So, H demonstrates that a map can require four colors.  H now has four separate enclosed interior regions: triangles ABD, ACD, BCD, and ABC.\n\nFrom H, construct a graph I by adding another node E.  Repeating the above procedure, E can be placed in four places: a) in the interior of ABD; b) in the interior of ACD; c) in the interior of BCD; or d) on the exterior of ABC.\n\nCASE a: As with constructing H, edges can be drawn from E to A, B, and D.  However, C is on the exterior of ABD, so any edge from E to C must cross either AB, BD, or AD.  So, in case a, there must be an intersection, so four colors is still an option.\nCASES b and c follow similarly.\nCASE d: E is on the exterior of ABC, but D is on the interior of ABC.  Any edge between E and D must intersect AB, BC, or AC.  So, coloring with four colors is possible in this case.  Since it is possible in all cases, it is possible for any map with five regions with a sub-map reducible to a graph like H.\n\nCase 2) Instead of placing D in the interior of G, make a graph H' by placing D on the exterior of triangle ABC.  Any quick drawing will show that D is easily connected to A, B, and C.  So there is a map with four regions that requires four colors.  To make the rest of this easy, note that H' also produces four enclosed regions: ABD, ABC, BCD, and ADC.  I won't go through the trouble to prove this, but clearly H' is isomorphic to H.  That being said, whatever is true of the general graph H is also true of the general graph H'.  Therefore, a graph I' made by adding a node to H' cannot be complete without there being at least one set of intersecting edges.\n\nSince this is true for all cases, any map reducible to a graph with five nodes is colorable with just four colors.  QED\n\nSorry this is so long.  I hope it's helpful.
67852
this is difficult to do without drawing.  but here's the gist of it.\n\n1. <a + <b + <c + <d = 360  (definition of quadrilateral)\n2. <a + <b + <a + <b = 360  (since a = c and b = d)\n3. 2(a + b) = 360\n4. a + b = 180\n5. <a and <b are supplementary angles\n\nnow extend the line AB and include a new point X, so you have a picture that looks like:\n\nX\n\nA   D\n\n\nB   C\n\n<XAD and <BAD are supplementary since XAB is a line\n\nabove we proved that <ABC and <BAD are also supplementary\n\nso <XAD + <BAD = 180\nand <ABC + <BAD = 180\nwhich means <XAD = <ABC\n\nso now we can conclude that AD || BC\nbecause the corresponding angles <XAD and <ABC are congruent.\n\nyou can use this same logic to show that AB || DC\n\nhope this makes sense..
774873
False for all quadrilaterals.\nBut all squares and isosceles trapezoids can be inscribed in a circle.
493884
In any quadrilateral the sum of the angles is 360 degrees.  Its an axiom that has been proven.  It's similar to the axiom that all angles in a triangle add up to 180.\n\nNote: the pattern is:\n3 angles = 180\n4 angles = 360\n5 angles = 540\n6 angles = 720\n7 angles = 900\netc.\n\nAnother fact about parallelograms is that the opposite sides have the same lengths and the opposite angles have the same measure.  This also means that two consecutive angles will add up to 180 (since the other pair must be equal and the total must be 360, as noted before).
642032
The only regular polygons that can tessellate are triangles, squares, and hexagons. There are other polygons that can tesselate, and I believe there are 12 or so patterns that can repeatedly and uniformly tessellate using regular polygons.
240497
Here is some information about tiling regular polygons (tiling is also called tessellation):\nhttp://mathforum.org/sum95/suzanne/whattess.html
633798
Mosaic: a picture made of tile.\nTesselation: shapes that fit together with no gaps between (eg. squares tesselate, but circles don't)
838083
I like tessellations.org.. they have a "Do It Yourself" section that explains it pretty well.
414775
Not all Parallelograms are rectangles, But all rectangles are parallelograms.
1173874
Technically, a square is only a part of a plane. Planes go to infinity in all directions (in 2 dimensions), so a square is therefore only part. Similarly, a line continues on to infinty in both directions. What is commonly called a line (e.g. ___) is actually a 'line segment'
43907
Because geometry is part of life. Just look around you and see all the geometric shapes that you just take for granted. Not only man-made objects, but in nature also.
492854
ok, lets make this simple.\n\nIf a shape can be folded over itself and cover itself exactly, the article is said to be symmetrical and the "fold line" is said to be an axis of symmetry.\n\nA square is a symmetrical shape with 4 axis of symmetry, the two diagnals and one line drawn across it half way up, one line drawn down through it, half way across it.\n\nA circle has an infinate nubber of axis of symmetry, each one being a diameter.\n\nOne way to look at this axis is to think of it as a mirror placed on a plain geometrical shape, the reflection is the other half of the shape.