A rectangle is defined by the vertices J(10,0) K(-8,6) L(-12,6) M(6,-12). Without solving it, show what steps you would need to follow to show that the diagonals bisect each other. Be specific, but DO NOT SOLVE!
130295
PROCEDURE:\n\n1) Plot the so-called vertices.\n2) Connect any two adjacent points with a segment.\n3) Determine if the resulting plane figure is a rectangle.\n(It seems that the vertices that you have mentioned do not form a rectangle.)\nNote: A rectangle is a quadrilateral with opposite sides being parallel and interior angles all being right angles.\n4) Connect opposite vertices with a line segment. You will form two diagonals.\n5) determine the equation of each line (segment) that constitute the diagonal. You will have two equations.\n6) Find the point of intersection between the two diagonals by solving for the unknowns (x & y) using the two equations.\n7) Find the length of each diagonal (the two diagonals should have the same length).\n8) Find the length from a vertex to the point of intersection between the two diagonals.\n9) If the length from any vertex to the point of intersection between the diagonals is half that of the length of any diagonal, then the diagonals are said to bisect each other.
538729
The question apparently is "How many diagonals does a polygon with n sides have?"  \n\nThe first formula is n(n-3)/2.  \nOne way to see this is to notice that you can draw (n-3) diagonals from every vertex of the polygon. This is because there are (n-1) other vertices, but two of them are adjacent vertices and so don't count towards making diagonals. This seems to give n(n-2) diagonals, but this way of counting counts every diagonal twice since each diagonal connects to two vertices, so you have to divide by 2.\n\nBy a recursive formula, we mean a way of expressing the answer for n vertices in terms of the answer for n-1 vertices. Suppose, for example, that you already know the answer for a polygon with n-1 vertices. Now if you add another vertex between two of the vertices of the original polygon, then all the diagonals of the original polygon will still be diagonals of the new polygon, and so will the side joining the two vertices that you added a new vertex between, and so will the line segments joining the new vertex to all the other vertices of the original polygon. (Got all that?) So if we let diag(n) be the number of diagonals for a polygon with n sides, we get the formula:\n\n        diag(n) = diag(n-1) + n - 3 + 1 or\n\n                  diag(n-1) + n - 2\n\nThe first formula is better, since it actually gives you the answer.  \nBut sometimes it's easier to get a recursive formula first and use \nthat to get an explicit formula (your first formula is an explicit one \nsince you only need the number of vertices in the polygon to get the number of diagonals in that polygon).  This is called "solving the recursion."  Sometimes a recursive formula is the best you can do because there simply is no explicit formula.\n\nI hope this helps.
614924
Its been a long time since I took Geometry but maybe I can give you a small pointer.\n\nTry dividing the rectangle into 3 segments lengthwise and 3 segments widthwise\n\nA--B--C--D\nE--F--G--H\nJ---K--L--M\nN--O--P--Q\n\nDraw the diagonal through NKGD\n\nProve that the segments NK, KG, and DG are equal (prove that the triangles are the same)\n\nNow draw the segments from the vertex Q to the mid point of AD and AN.  Prove that the mid point of AD is the same as the mid point of BC and the same for AN and EJ.\n\nNow we have to prove that the line segment we drew in the last step instersect points K and G. You can probably do this with similar triangles (the triangle NQ and mid point of EJ is similar to JK and mid point of EJ therefore the point K is in the triangle)\n\nSo if you prove that the lines drawn from Q to the midpoints intersect K and G and we proved that NK KG and GD are congruent we have proved the original hypothesis.\n\nYou can turn this gabably gook into a real mathematical proof.\n\nGood Luck
426466
I drew the picture as you described it and it looks like a parallelogram with its diagonals drawn in.  MX is half of one diagonal and MR is one side of the parallelogram.  If this is indeed the case, there is no reason why MX is congruent to MR.  So, either you need a better description of what the figure looks like, or you left out a given, such as DN is congruent to XN or something like that.
23724
A rectangle...equillateral means all equal sides, equiangular means all equal angles\n\nthe rectangle has all equal angles, 90 degress but teh sides are different, if the sides were the same it woudl be a square\n\na rhombus has equal sides but not necessarily equal angles\n\na parallelogram does not have to have equal angles or equal sides, it needs to have to sets of equal opposite angles and equal opposite sides\n\nhope this helps!!
526041
Yes, graph this. (0,0),(1,0),(1,1),(0,2)\nThere are two right angles and two sides of length 1, but the figure is not a parallelogram. It is only a trapezoid.
184301
Not all Parallelograms are rectangles, But all rectangles are parallelograms.
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.
352289
You know that the diagonal forms a right triangle with the length and the width. So you can use the Pythagorean Theorem.\nx=width=a\nx+1=length=b\n4=diagonal=c\n(a^2)+(b^2)=(c^2)\n(x^2)+(x+1)^2=(4^2)\n(x^2)+(x^2)+2x+1=16\n2(x^2)+2x+1=16\n2(x^2)+2x-16=0\nIf you use the quadratic formula thing (-b plus or minus, etc.)\nthen you get:\nx=2.2838\nor\nx=-7.5\nBut you know that a dimension cannot be negative, so it must be the first one.\nSo if \nx=2.2828\nthat means that that's the width. If you add one to that, you'll get the length.\nSo, the dimensions are:\n2.2838 x 3.2838
48857
Let:\n\nL = length of the rectangle\nW = width of the rectangle\nP = perimeter of the rectangle\n\nGiven:\n\nW = x\nL = 2x - 4\nP = 2L + 2W from geometry\n\nSolution:\n\nP = 2L + 2W\n64 = 2[2x - 4] + 2x\n\n2[2x - 4] + 2x = 64\n\n[4x - 8] + 2x = 64\n\n4x - 8 + 2x = 64\n\n4x + 2x = 64 + 8\n\n6x = 72\n\nx  =  72/6\n\nx  =  12 cm.\n\nTherefore:\n\nW = width = x = 12 cm. ANS.\n\nL = length = 2x - 4 = 2(12) - 4 = 24 - 4 = 20 cm. ANS.
99914
im a geo student, but i havent done proofs for a long time sorry. ok\n2. false, because squares are parallelograms with 4 congruent sides and 4 right angles, while kites have 2 pairs of congruent sides and only 1 pair of congruent angles that are NOT right angles\n3. false, because a rhombus is a parallelogram with 4 congruent sides, while a kite is any thing BUT a parallelogram. enuff said\n4. false, because kites are quadrilaterals with only two pairs of congruent sides, while the quadrilateral mentioned has 3 congruent sides\n5. false, for example- do trapezoids have 4 congruent sides, and 4 right angles, and is a parallelogram (characteristic of squares)? i think NOT\n6. read above- trapezoids are not parallelograms.
114451
If an object is a parallelogram, we define its area to be its length multiplied by its height. Rectangles and squares are common parallelograms.\nIf you cut a parallelogram in half along its diagonal, you get a triangle with the same width and height, but half the area. Thus the formula for the area of a triangle is (1/2)*base*height where the name "base" has replaced the now meaningless label of "length".\nUsing these two formulas, you should be able to find the area of any polygon (a closed shape with straight sides) by cutting it up into these shapes of known area. For example, a trapezoid can be cut up into a rectangle and two triangles.\nThe perimeter of a polygon is the easiest measure. Just add up the lengths of all the sides.
79099
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.
33525
Main article: Mathematics of Sudoku\n\nThe general problem of solving Sudoku puzzles on n2 x n2 boards of n x n blocks is known to be NP-complete [8]. This gives some indication of why Sudoku is difficult to solve, although on boards of finite size the problem is finite and can be solved by a deterministic finite automaton that knows the entire game tree.\n\nSolving Sudoku puzzles (as well as any other NP-hard problem) can be expressed as a graph colouring problem. The aim of the puzzle in its standard form is to construct a proper 9-colouring of a particular graph, given a partial 9-colouring. The graph in question has 81 vertices, one vertex for each cell of the grid. The vertices can be labelled with the ordered pairs (x,&#92;, y), where x and y are integers between 1 and 9. In this case, two distinct vertices labelled by (x,&#92;, y) and (x',&#92;, y') are joined by an edge if and only if:\n\n    * x = x'&#92;, or,\n    * y = y'&#92;, or,\n    * &#92;lceil x/3 &#92;rceil = &#92;lceil x'/3 &#92;rceil and &#92;lceil y/3 &#92;rceil = &#92;lceil y'/3 &#92;rceil\n\nThe puzzle is then completed by assigning an integer between 1 and 9 to each vertex, in such a way that vertices that are joined by an edge do not have the same integer assigned to them.\n\nA valid Sudoku solution grid is also a Latin square. There are significantly fewer valid Sudoku solution grids than Latin squares because Sudoku imposes the additional regional constraint. Nonetheless, the number of valid Sudoku solution grids for the standard 9×9 grid was calculated by Bertram Felgenhauer in 2005 to be 6,670,903,752,021,072,936,960 [9] (sequence A107739 in OEIS). This number is equal to 9! × 722 × 27 × 27,704,267,971, the last factor of which is prime. The result was derived through logic and brute force computation. The derivation of this result was considerably simplified by analysis provided by Frazer Jarvis and the figure has been confirmed independently by Ed Russell. Russell and Jarvis also showed that when symmetries were taken into account, there were 5,472,730,538 solutions [10] (sequence A109741 in OEIS). The number of valid Sudoku solution grids for the 16×16 derivation is not known.\n\nThe maximum number of givens that can be provided while still not rendering the solution unique is four short of a full grid; if two instances of two numbers each are missing and the cells they are to occupy form the corners of an orthogonal rectangle, and exactly two of these cells are within one region, there are two ways the numbers can be assigned. Since this applies to Latin squares in general, most variants of Sudoku have the same maximum. The inverse problem—the fewest givens that render a solution unique—is unsolved, although the lowest number yet found for the standard variation without a symmetry constraint is 17, a number of which have been found by Japanese puzzle enthusiasts [11] [12], and 18 with the givens in rotationally symmetric cells.
335700
the answer is:\n1\n\nThis is Triangle in which j-th entry in i-th row is (i+1-j)^j.\n\nYour particular row is the 7th row of the triangle.\nVisually it looks it looks like this:\n 1\n 1, 1,\n 1, 2, 1,\n 1, 3, 4, 1,\n 1, 4, 9, 8, 1,\n 1, 5, 16, 27, 16, 1\n 1, 6, 25, 64, 81, 32, 1,\n 1, 7, 36, 125, 256, 243, 64, 1,                       \netc
433022
Let's look at the rectangle problem:\n\nThe original is 12 x 8, for an area of 96.\n\nThe new rectanlge's sides are 3/4 the sides of the original, or 9 x 6 (since 9 is 3/4 of 12, and 6 is 3/4 of 8).  So the area of the new rectangle is 9*6 = 54.\n\nSo the ratio of the areas is 54/96 = 9/16 = (3/4)^2\n\nNotice that the ratio of the sides (new to old) = 3/4, and the ratio of the area (new to old) = (3/4)^2\n\n~ ~ ~ ~\n\nLet's look at the circles now.\n\nCircle 1 = radius 5\nArea = pi r² = 25pi\n\nCircle 2 = radius 15\nArea = pi r² = 225pi\n\nSo, when we multiply the radius by x, we are multiplying the area by x²\n\n~ ~ ~ ~\n\nIf every dimension of a 2-D figure is multiplied by k, then the areas can be expressed like this:\n\nFigure 1 Area = l * w = lw\n\nFigure 2 Area = lk * wk = lwkk = lwk²\n\nSo multiplying each side by k in a 2-D figure increases the area by k²  (that's k^2 in case you can't read the little 2)\n\n~ ~ ~ ~ \n\nIf every dimension of a 3-D figure is multiplied by k, then the volumes can be expressed like this:\n\nFigure 1 Volume = l * w * h = lwh\n\nFigure 2 Volume = lk * wk * hk = lwhkkk = lwhk³\n\nSo multiplying each side by k in a 3-D figure increases the volume by k³ (that's k^3 in case you can't read the little 3)\n\n~ ~ ~ ~ \n\nBeach ball volume = 100 in³\n\nLet r = radius of beach ball.  If we multiply the radius by 3, new radius is 3r, and the new volume is 4/3 * pi * (3r)³ = \n4/3 pi*27r³ = \n27 * 4/3 pi r³\n\nBut 4/3 pi r³ = 100 (our original beach ball), so 27 * 4/3 pi r³ = 2700 in^³.  (Notice this is 27 times our original ball, or 3³.)\n\n~ ~ ~ ~\n\nHope that helped you some!\n\n~ ♥ ~
473447
The triangle is a right triangle of sides 3,4,5.\nThe maximum rectangle occurs when a perpendicular is raised from the midpoint of the short side to the hypotenuse (side that equals 5) and then a perpendicular from point  to the side   that is of length 4. The dimensions of this rectangle are 2* 1.5 = 3 square units.