The vertices of a rectangle?
336458
the sides are 2 units (3-1=2) \nand the ends are 10 units. (6-(-4)=10)\n\narea = side times end = 2*10 = 20\nperimeter = 2 * side + 2 * end = 2*2 + 2*10=4+20=24
846631
Construct meaning with a compass and straightedge? And by "adjacent" they're already at right angles? Just measure each side with the compass, move it so one leg is at the other corner, and draw a small arc approximately where the other leg extends. Do this for both sides, then the point of intersection is your missing point, and you can draw out the rectangle with the straightedge.\n\nIf the sides are not at 90 degrees, just turn one of them (measure one with the compass, draw a circle based on it, then take the two points of intersection and construct the perpendicular bisector, then take the point where that intersects....)
614253
In a rectangle, the opposite sides are equal.  Therefore, two of the sides = 2x-5 and the other two = x.\n\nPerimeter is the total (add) of all the sides.  So, your equation is\n2x-5 + 2x-5 + x + x =50  (Then you combine like terms and solve)\n\n6x -10 = 50\n6x = 60\nx = 10\n\nTo find the length and width, plug in the number you got.\n\n2x-5 = 2 (10) -5 = 20 -5 = 15\nx = 10\n\nTo check:  Add the perimeter with your numbers.\n15+15+10+10 = 50
143210
fancy description of a rectangle.\n\nIf it had four congruent sides, it would be a square.
566727
I'm afraid that with the given information, this would be impossible to determine. The only thing one could conclude from this information is that this polygon has an even number of sides. \n\nIt could be a square with sides of 3cm each\nIt could be a hexagon with sides of 2cm each\nIt could be an octagon with sides of 1.5cm each\nIt could be a decagon with sides of 1.2cm each\nIt could be a 12-gon with sides of 1cm each\n\nThe list is infinite unless you are given more specific information, such as the length of each side or the angles within the shape.
507349
In geometry, a rectangle is defined as a quadrilateral polygon in which all four angles are right angles.\n\nFrom this definition, it follows that a rectangle has two pairs of opposite sides of equal length; that is, a rectangle is a parallelogram. A square is a special kind of rectangle where all four sides have equal length; that is, a square is both a rectangle and a rhombus. A rectangle that is not a square is colloquially known as an oblong.\n\nOf the two opposite pairs of sides in a rectangle, the length of the longer side is called the length of the rectangle, and the length of the shorter side is called the width. The area of a rectangle is the product of its length and its width; in symbols, A = lw. For example, the area of a rectangle with a length of 5 and a width of 4 would be 20, because 5 × 4 = 20. See the picture above right.
774030
There are in infinate number of 2d shapes, but here are diagrams of some of them --> http://www.mathleague.com/help/geometry/polygons.htm
736325
Area of a rectangle is length XWidth. So if you have the area of a rectangular field or courtyard and either length or breadth/width,just divide the area by length to get the width and similarly just divide the area by width/breadth to get the length it is very easy\nWe know that Length XBreadth=area of the rectangular yard.\nTherefore,length=360.15/15\n=24.01m
1144014
draw a picture of the graph over the specified interval.  divide the graph into vertical rectangles.  the area under the curve will be the sum of the areas of the rectangles.  saying that youve made "n" rectangles, the only way to get the exact area is to find the sum for the limit as n approaches infinity.  work from that concept.  if you already knew that, or need more help or something, email me and i can give you more steps and a much more detailed description.  that would take a really long time though so im gonna see if this helps first.
282503
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3
441947
pentagon DUH! lol sry (Hexagon is SIX!)im in geometry rite now so yea
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
459931
a nine-sided polygon???  can't wait to see other answers...
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.
400659
Well... a slightly advanced method is that you should fist realize that a triangle drawn through these three points will have an area = 0\n\nformula for area of a triangle is:\n        \n| a1 b1 1 |\n| a2 b2 1 |  x 1/2 = A\n| a3 b3 1 |\n\nWhere (a1,b1) ; (a2,b2) and (a3,b3) are the 3 vertices of the triangle.\n{Sorry --- no better way to represent a matrix here...}\n\nor; A = 1/2 [ a1(b2-b3) - b1(a2-a3) + (a2.b3 - a3.b2) ]\n\nHere; A = 0 ; a1=-5; b1=1 : a2=1; b2=p and a3=4; b3=-2\n\ntherefore; 0 = 1/2 [ -5(p+2) - (1-4) + (-2 - 4p)]\nor;        0 = [-5p -10 + 3 -2 -4p]\nor;        0 = -9p -9\nor;       9p = -9\n\n=> p = -1\n\nAnother method would be to find the line through two given points and pass the line through the third point... but that is slightly lengthy.... \n\nIf you have any doubt; feel free to ask me.
328495
now, if you add that it has 4 sides and all right angles, you probably can pass that off as the definition of a rectangle.\n\notherwise you just have a parallelogram.
24869
List is a bit long, so heres the link:\nhttp://en.wikipedia.org/wiki/List_of_geometric_shapes