how many sides does an octagon,parallelogram,rhombus.trapezium have?
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click here:http://en.wikipedia.org/wiki/Polygon
590231
Doesn't a rhombus have equal diag.?\n\nIf you mean a parallelogram, the formula is:\n\n(diag. 1 x diag. 2) / 2.  \n\nSo:\n\n(8x10)/2\n\n80/2\n\n40 u^2
139968
false. a quadrilateral is a four sided polygon. so a square and rectangle are also quadrilaterals. a trapezoid is a four sided polygon with 2 opposite sides that are parallel to each other.
651972
A nonagon has 9 sides and 9 vertices.
622265
a trapezoid ia a quadrilateral with only two parallell sides.\n\nas to figuring out what CD is, i'd need to know what MA and HT are (are they lengths of sides? measurements of angles?)  so, if you post those, then i'll see what i can do
949599
Solution #1.\nAs its a parallelogram , the opposite sides are equal hence \nJK = OM;\nOJ = KM\nAnd as the diagnols bisect the angles it is a rhombus and has equal sides.\n\nx +5 = y - 3         (opposite sides are equal)\n2x - 4 = x + 5                            (all sides are equal)\nx = 9\ny = 17\nthe perimeter is 4(OJ)  or any other side as they are all equal.\nthus \nPerimeter = 4(x+5) = 4(9+5) = 56\n\nSolution #2.\nAs the sum of all the angle is 360º so ,\nlet one angle be mº and the other be nº so.\nm = 40 + 3n ...........eq.1\nand 2m + 2n = 360º\n\nm + n = 180º\nm = 180º - n ...........eq.2\ntherefore equating equations 1 & 2.\n40 +3n = 180 - n\n4n = 140\nn = 35º\nso,  m = 180º - 35º \n m = 145º\n\nso your angles are 145º,35º,145º,35º.
514188
(1) Straight-sided shapes are called polygons \n\n(2) Others are classified according to the number of sides they have.\na) triangles ( three sides) \nb) quadrilaterals ( four sides) \nc) pentagons (five sides)\nd) hexagons (six sides),\ne) octagons (eight sides)
332611
No - if a trapezoid has 3 congruent sides, the fourth side will also have to be congruent in order for you to close the shape.  In this case, your trapezoid will be a square.\n\nBy definition, a trapezoid has to have one pair of opposite sides that are parallel.  If you try to construct a trapezoid with three congruent sides, the fourth side will have to be the same length - try it with pencils or popsicle sticks to see.  If you do, remember, one pair of sides HAS to be parallel.\n\nA trapezoid can have 2 congruent sides - as long as they are adjacent (i.e. a side and a top).  If the opposite sides are congruent, then you'll end up with a parallelogram.
570600
Ok, first just so you get the figure right, draw dots n name them like i have n at the places like m showing them below\n      A                B\n      \n    E                   F     \n\n   D                      C             G\n\nNow, join AB, D-C-G, EF , A-E-D and B-F-C  also.. join\n A-F-G .. GET IT?? and you'll have AE equal to ED as well as BF equal to CF as the median of the trapezium joins the mid=points of the non-parallel sides...\n\nNOW IN triangle BFA and triangle CFG ,you have\nBF congruent to CF ----   Proved tht F midpoint of BC\nangle BFA congruent to angle CFG--vertically opposite angles\n and as AB is parallel to CD (opposite sides of a trapezoid) it is also parallel to CG ( construction was to join D,C and G , remember??\n\ntherefore, abgle ABF congruent to angle FCG --- alternate angles for parallel lines AB and CG ..\n\nSO YOU NOW HAVE triangle ABF congruent to triangle CFG --- Angle-Side-Angle Test of congruency...\n\ntherefore, AF congruent to FG --- corresponding sides for congruent triangles\n\nNOW, IN THE BIG TRIANGLE ADG,\n  E is the midpoint of side AD - by construction\n  F is the midpoint of AG  --- Proved AF congruent to FG\n \ntherefore... EF parallel to DG --- By the Midpoint Theorem which states that-- A line passing through the midpoints of any two sides of the triangle is parallel to the third side (and has half the length of the third side )\n\nSO WE'VE NOW PROVED EF // DG\n .\n. . , EF IS ALSO // TO DC --- D,C AND F LIE ON THE SAME LINE.. I.E. THEY ARE COLLINEAR\n\n SO, EF//DC\nBUT DC IS PARALLEL TO AB ---- OPPOSITE SIDES OF THE TRAPEZIUM ABCD\n \n .\n. . , EF IS ALSO PARALLEL TO AB\n\n .\n. ., WE CAN PROVE THAT ----> AB // EF // DC  (REASON- both AB and DC are parallel to the same linE EF)  \n                     \n\n .\n. . WE HAVE PROVED THAT THE MEDIAN OF THE TRAPEZIUM (in this case GE)  IS PARALLEL TO BOTH THE BASES (i.e AB and CD )
729594
Rhombus and square (c)\n\n\nThis can be proved using geometry, analytic geometry, etc.  But it is the case.  Surprisingly, a general reactangle and parallelograms don't have perpendicular diagonals.\n\nGeneral kites do, and a rhombus is a kite with all equal sides.  A square can be thought of as a type and rhombus and rectangle, or just a rhombus when each angle is 90 degrees.  Then it makes sense that it would have perpendicular diagonals.
819602
If the bases are in the ratio of k:1, then we can say they have lengths of kx and x (kx and x have ratio k:1). Let's say the trapezoid has height h.\n\nThe mid-segment of the trapezoid cuts the trapezoid into two similar trapezoids, and has length (1/2)(kx + x)\n\nThe area of a trapezoid is (1/2)(height)(b1 + b2).\n\nThe top trapezoid has height h/2, and bases kx and (1/2)(kx + x).  It has an area \nA1 = (1/2)(h/2)(kx + (1/2)(kx + x))\n     = (h/4)(kx + kx/2 + x/2)\n     = (h/4)(3kx/2 + x/2)\n     = (hx/8)(3k + 1)\n\nThe bottom trapezoid has height h/2 and bases (1/2)(kx + x) and x.  It has an area\nA2 = (1/2)(h/2)((1/2)(kx + x) + x)\n     = (h/4)(kx/2 + x/2 + x)\n     = (h/4)(kx/2 + 3x/2)\n     = (hx/8)(k + 3)\n\nIf A1:A2 = 5:2, then \n[(hx/8)(3k + 1)]/[(hx/8)(k + 3)] = 5/2\n(3k+1)/(k+3) = 5/2\n2(3k+1) = 5(k+3)\n6k + 2 = 5k + 15\nk = 13\n(C)
424627
11 - undeca\n\n12 - dodeca\n\nEX:\n\nundecagon - polygon with 11 sides.\n\ndodecagon - polygon with 12 sides.
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How to find the area of an octagon:\n\nMeaning of words\n\nAn octagon is a polygon having eight sides.\nA polygon is a closed plain figure bounded by three or more straight sides that meet in pairs in the same number of vertices and do not intersect other than at these vertices.\nVertices is the plural of vertex.\nVertex means the point opposite the base of a figure or the point of intersection of two sides of a plain figure or angle.\n\nDifferent methods or formulas are used for calculating the area of polygons. The following method is used for calculating the area of an octagon.\n\nFormula: 2 × r2× 1.4142135 (the last number is the square root of 2)\n\nr = radius\n\nThe radius is a straight line from the centre to any point on the outside of the shape.\n\nFind the radius of the octagon, then perform the formula.\ne.g. if the radius of an octagon was 4 cm, then 2 × 42× 1.4142135 = 45.25 cm2, which is the area.
583176
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.
740926
Use graph paper and graph out this parallelogram..\nA(-6,0) B(0,10) C(6,0) D(0,-10)\nAngle A = ang C (definition of parallelogram)\nAngle B = ang D (def. of para)\nThe diagonals bisect the angles(given)\nThe diagonals bisect each other(def. of Para)\nUsing side-angle-side you can prove all the triangles formed by the diagonals are congruent. So that means all sides of parallelogram are equal and you have a rhombus.
809433
1.rectangle\n\n2.trapezoid
530421
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.