How do you prove that the midpoint quadrilateral of a trapezoid is a parallelogram?
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First we note that the two bases of the trapezoid are parallel and the segment joining the midpoints of the altitudes is parallel and equidistant from the bases. Therefore the vertical displacement of the midpoints of the bases from the midpoints of the altitudes is equal.\n\nIf we also contruct the segment joining the midpoints of the bases we see this intersects our other segment at its midpoint. We have now cut our quadrilateral into four triangles and it is an easy task to show that opposite triangles are congruent using SAS.
797921
"A trapezoid is a 4-sided figure with one pair of parallel sides. To find the area of a trapezoid, take the sum of its bases, multiply the sum by the height of the trapezoid, and then divide the result by 2." The height is the perpendicular distance between the two parallel bases.\n\nArea (trapezoid) = 1/2 h (b1 + b2) , where b1 and b2 are the lengths of the two parallel sides, and h is the perpendicular distance between them.
764103
No. The diagonals of a parallelogram bisect each other, but they don't bisect the angles.
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The only formula necessary is the definition of the area of a parallelogram, which is width*height (Choose one side as the width. The height is the length of a line segment perpendicular to this side that connects this side to its parallel).\nNote that every triangle forms a parallelogram with a copy of itself. Thus the area of a triangle is half the area of that parallelogram. Note that two sides of the triangle coincide with the two unique sides of the parallelogram, so follow the same instructions as above, choosing one side of the triangle as the width (in this case called a "base") and measuring a line segment drawn perpendicular to this base to the opposite vertex of the triangle as the height of the triangle (really the height of the parallelogram). Take the product and divide by 2.\nA trapezoid has only two parallel sides. Choose one of these sides as the "base", and it is easy to see that you can divide the trapezoid into two (possibly different) triangles and a rectangle (a parallelogram), all of the same height. Simply add these areas together keeping in mind what they have in common to get the formula for the area of a trapezoid.
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I'm going to give you 2 methods. \n\nMethod 1: Pure geometry\n\n* E is the midpoint of [OA]. That means that, vectorially, OE=1/2OA. \n\n* D is the midpoint of [BC]. Therefore, from what you know from vector operations you will have OB+OC=2OD, so OD=1/2(OB+OC) ("parallelogram rule")\n\n* Now G is the midpoint of [DE]. So, again, just as above, you should have the relation: OD+OE=2OG, so OG=1/2(OD+OE). \n\nBut you know how to expres OD and OE in terms of OA, OB and OC: OG=1/2(1/2OA+1/2(OB+OC))\n\nThence, OG=1/4(OA+OB+OC). From there on, it should be straitforward to show that G is indeed the midpoint of the other 2 bimedians. Define midpoints for AB, OC and AC, and simply repeat what we did above.\n\nMethod 2: Using a coordinate system\n\nDefine a coordinate system where O is the origin, OA is the x axis, OB is the y axis, OC is the z axis. Then, algebraically, we have:\n\nOA=<a, 0, 0>\nOB=<0, b, 0>\nOC=<0, 0, c>\n\nE is the midpoint of OA, so OE=<a/2, 0, 0>\n\nD is the midpoint of BC. So OD=1/2(OC+OC)=<0, b/2, c/2>\n\nNow G is the midpoint of DE, so again you have OG=1/2(OD+OE)\n\nSo OG=<a/4, b/4, c/4>\n\nSo just use whichever method you prefer to finish the exercise.
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i don;t know all of them, but here are a few:\n-rectangle\n-square\n-rhombus\n-parallelogram\n-kite\n-trapezoid\n-and closed abject with four sides!\n-cyclic quaDrilateral     \n-cyclic-inscriptable \n\ncheck out the site dr. math for more info on quadrilaterals!
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False for all quadrilaterals.\nBut all squares and isosceles trapezoids can be inscribed in a circle.
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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.
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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!!
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Draw a horizontal staight line to be the base of the trapezoid.\nMark a point to be a corner of the trapezoid. Label it A.\nSet your compass to the length of the base.\nMeasure that distance from the first mark and make a second mark on the line. Lable it B\n\nWithout changing the width of the compass, put the point at A and swing an arc above the line. Move the point of the compass to B and swing and arc. Lable the point where the arcs cross C.\n\nDraw the lines AC and BC. You now have an equilateral triangle. The angles are all 60 degress.\n\nSet your compass to the distance of the leg length.\nPut the point at A and swing an arc across AC. Label the intersection D.\nMove the point to B and swing an arc across BC. Label the intersection E.\nDraw the line DE.\nDone.
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er that formula is false and you cant prove it. i can disprove it for ya.\nyou have a trapezoid with legs a and b, and c and d being the bases. you know that a formula for a trapezoid is the average of the bases times the height: (h)(b1+b2)/2. (c+d)/2 is already the average of the bases, since c = b1 and d = b2. now you have to multiply it by the height. if the formula you're supposed to prove is true, then (a+b)/2 has to be equal to the height. how can that be possible? that will never happen unless you have BOTH of the of the legs are perpendicular to the bases. think about it. if they arent perpendicular to the bases, no matter how you draw the trapezoid, the average of the non parallel legs (a+b)/2 will always be greater than the height.
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decagon: S x 10\nrhombus: S x 4\nparallelogram: (L x2)+(W x2)\n\nragular fugures are figures that have the same lengths in all the sides ( which is the case with the decagon) so you just add up all the sides or multiply the side by the number of sides the figure has.\n\na rhmbus has all the sides of equal length so again you just add all the sides or multiply the side by 4.\n\nthe opposite sides of the parallelograms are of the same length so you multiply the lenght and width of the parallelogram by 2 and add the results and thats it.\n\ngood luck.
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One proof often seen uses two squares, one inside the other, with the points of the smaller square on the sides of the larger square.  This leaves you with four congruent right triangles that are inside the larger square but outside the smaller square.  Label the short sides of the right triangles a and b and call the hypotenuse c.\n\nNow note that the sides of the large square have length (a + b).  Therefore, one way to calculate the area of the large square is (a + b)^2 = a^2 + 2ab + b^2.        \n\nYou'll see that the area of the large square is equal to the area of the small square plus the four triangles.  Also note the sides of the small square have length c.  So, another way to calculate the area of the large square is to add up the area of the small square plus the four trianges, i.e. c^2 + 4 * ab/2 = c^2 + 2ab.  \n\nCombining the two ways to calculate of the area of the large square, we get:\n\na^2 + 2ab + b^2 = c^2 + 2ab\n\nwhich simplifies to:\n\na^2 + b^2 = c^2.\n\nand there is your proof!
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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
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Not all Parallelograms are rectangles, But all rectangles are parallelograms.
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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.
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I would use the trapezoid rule myself. For seven vertices:\nV0 = (x(0),y(0))\nV1 = (x(1),y(1))\nV2 = (x(2),y(2))\n...\nV6 = (x(6),y(6))\n\nFor notational convenience, define V7 = V0.\nThat is, x(7) = x(0) and y(7) = y(0).\n\nThen the area is:\n\n 6\nSUM ( x(i+1)*y(1) - x(i)*(y(i+1) ) / 2\ni=0\n\n= ( x(1)*y(0) - x(0)*y(1)\n + x(2)*y(1) - x(1)*y(2)\n + x(3)*y(2) - x(2)*y(3)\n + x(4)*y(3) - x(3)*y(4)\n + x(5)*y(4) - x(4)*y(5)\n + x(6)*y(5) - x(5)*y(6)\n + x(0)*y(6) - x(6)*y(0) ) / 2\n\n\nIf the heptagon is regular, you don't need to know the coordinates of the vertices, just the length of the regular side(s):\n\nA = (7/4) * s^2 * cot(π/7)