true of false. a quadrilateral is always a trapeziod. explain your answer.?
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.
70580
Well, I don't really understand you. This is the best I could.\n\nI think it is true. For example, the cube root of -8, which is x, is -2, which is y. y^3 is -8, which is x. So, yes, I think it's true.
342120
equilangular in a trianle. if a triange is equilangular, it is equilateral, meaning all sides are the same length
243764
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.
796936
ALWAYS ... \n\nTo qualify as an isosceles, you must have two sides equal.\nTo qualify as an equilateral, you must have three sides equal.\n\nJust because you have more than what you need, doesn't mean you aren't that too.
818692
They are both base times height; however, in a rectangle the base and height are always sides of the rectangle and in non-rectangular parallelograms, one of them will not be the same as a side.
549388
Yes.\nAny square is a rhombus.\nSome rhombuses are squares (the ones that have all angles of 90 degrees).
442031
If a rhombus is inscribed in a circle, then the rhombus is a square.  (Remember that a square is a rectangle with congruent sides, or a rhombus with congruent angles.)\n\nIf a trapezoid is inscribed in a circle, then the trapezoid is an isosceles trapezoid.
523203
a trapazoid has 2 parallel sides and two non parallel sides.\n\nArea=( 1/2 sum of the parallel sides) * height\n\nremember the height must be the perpendicular height.\ni.e the height perpendicular to the parallel sides
572376
You **did** say that the Pythagorean Theorem was a given, right? I think that your INTENT is to prove the converse of the Pythagorean Theorem, right?\n\nLet x, y, and z be positive integers such that\nx^2 + y^2 = z^2.\nClearly z > x and z > y (otherwise either x^2 or y^2 must be less than zero; prove this for yourself if you must).\n\nLet x and y be the sides of a right triangle that form the right angle.\nThen, by the Pythagorean Theorem the length of the hypotenuse h is\nh = √(x^2 + y^2)\nh = √(z^2) . . . . . . . . (given equation above)\nh = z . . . . . . . . . . . (given z is positive) \n        \nTherefore, by side-side-side congruence, any triangle whose sides are of lengths, x, y, and z, where x^2 + y^2 = z^2, must be a right triangle with hypotenuse z.\n\n\nASIDE TO HOOD...: You are using the Pythagorean Theorem to prove the Pythagorean Theorem. Although the questioner's intent is not perfectly clear, I don't think the intent was to prove "if x, then x". If so there are easier ways to do so.
829392
just a piont of view, but normally to divide something symmetrically you have to do it at the widest piont along the centre line. where as assymetricaly is off the centre line at a narrower piont and therefore not a symmetry,this only applies to poloygons and symmetric shapes.  hope this helps
689719
The mode of presentation of data in the question is not sufficient to draw a single quadrilateral, there are two such quadrilaterals and hence two different answers.\n\nFIRST ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are adjacent.\nLet the quadrilateral be ABCD. Then AB = BC = 5*sqrt(2), CD = 6, <DAB and <ABC are right angles. Drop a perpendicular to side AD from C intersecting AD at P.\n\nTotal area = area of rectangle ABCP + area of triangle CPD\n= [5*sqrt(2) x 5*sqrt(2)] + [0.5*{5*sqrt(2)-6}*5*sqrt(2)]\n= 50 + 3.7868 sq units\n= 53.7868 sq units\n\nSECOND ANSWER: \nEqual sides i.e. sides of length 5*sqrt(2) are opposite to each other.\nLet the quadrilateral be ABCD. Then AB = CD = 5*sqrt(2), BC = 6, <BCD and <ABC are right angles. \nClearly, quadrilateral ABCD is a rectangle. \nSo area = product of lengths of adjacent sides = AB x BC = 5*sqrt(2) x 6\n= 42.4264 sq units
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.
494278
a trapezoid.\n   __\n  /__&#92;
387942
All squares are parallelograms, but not vice versa.  All squares are rhombii (rhombusses?), but not vice versa.  But I am not sure what you mean by the congruent question.  Congruent to what? All squares are similar to each other, but not necessarily congruent.
13252
Generally, non-euclidean geometries are geometries that use a different fourth postulate, that, in Euclidean geometry, says that corresponding angles of a transversed parrallel lines are equal.  This basically means that parrallel lines never meet.  There are two other possible postulates.  One is that parrallel lines do meet, i.e. that space is like a globe where longitudinal lines meet at the poles.  The other is that other lines, besides parallel lines don't meet. In this situation space is shaped like a saddle, and lines get farther and farther apart.  Those are the two I know.  There are three other Euclidean postulates (e.g. you can draw exactly one circle through a point in the center and one on the rim) that someone might be able to change, but I am not aware that anyone has done that.
1080546
a rectangle, by definition is a quadrilateral where all four angles are right-angles, and an oblong is that which is right angled but not equilateral. I know its a bit obscure, but really, an oblong is an old fashioned term for a rectangle, without equilateral sides, if that's possible?!\n\nhope it helps