THE AREA OF A PLANE SURFACE WITH FOUR STRAIGHT UNEQUAL SIDES.LENGTHS OF ALL FOUR SIDES ARE KNOWN.BUT HEIGHT IS NOT KNOWN.WHAT IS THIS SHAPE CALLED AND HOW TO FIND OUT THE AREA OF THIS SHAPE.IS THERE AN ONLINE CALCULATOR TO DO THIS.WHAT IS THE FORMULA.
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www.onlineconversion.com
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You can use Heron's formula if you know the length of all three sides. \nHeron's Formula:\nLet s be the semiperimeter (which is half the perimeter) and a,b,c be the length of the sides, then the area is\nA = sqrt[s*(s-a)*(s-b)*(s-c)]\nperimeter = 4+4+6 = 14\ns = 14/2 = 7 (because s is one-half the perimeter)\nA = sqrt[7(7-4)(7-4)(7-6)]\n= sqrt(7*3*3*1)\n= sqrt(63)\n= 3*sqrt(7) which is approximately 7.93725\n\nYou can also use the Pythagorean Theorem to find the height.
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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.
577106
None of those are "planes".\n\nA plane is drawn as a square simply because of the limitations of a sheet of paper.\n\nA plane is infinite in two directions.  It has no start or end point on the xy axis.  A sheet of paper of course does have start end end points.  Usually 8x11\n\nNone of rectangels, squares, or triangles are planes.  They are all subsets of them.  And darned small ones at that.\n\n-Dio
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C=2(Pi)r or diameter times Pi\nI don't know how to solve the problem cause not enough info is given. Does it say what C equals?\n\nv = lwh . H stands for the height. Does it say what v equals?\n\na= 1/2h(b1+b2) b1 is base 1. Does it say waht the area is?\n\nProvide more info please. :)
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Surface Area = 4*PI*r^2\n\nr = SQRT(SA/4/PI)
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let l be the length and w be the width\n lw=111\n\nbut it has a margin so Area= (l-11cm)(w-4)\ndifferntiate\nbut we have 3 unknowns so we use the first ea lw=111\nl=111/w\n\nA=(111/w-11)(w-4)\n\ndA/dw= (111/w-11)+(w-4)(-111/w^2)=0\n\n111/w-444/w^2= 111/w-11\ncancel 111/w\n-444/w^2=-11\ncrossmultiply\nw^2=-444/-11\nw=6.4 cm\nl=111/6.4=17.3
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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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To determine if it is a parallelogram:\nIf a quadrilateral has one pair of opposite sides which are parallel and equal in length, then the figure is a parallelogram.\n\nTo determine if it is a rectangle:\nIf a parallelogram has two pairs of opposite sides which are parallel and equal in length, then the figure is a rectangle.\n\nGraph the points and determine the distance between the points, or, in your case, two points have an X of 0 and two have X of 3\n\nBetween the (0, 2) and (0, 6), there is a distance of 4\nBetween the (3, -2) and (3,2) there is a distance of 4\n\nBetween the (0,6) and (3, 2) there is a distance of 3 horizontally.\nBetween the (0,2) and (3, -2) there is a distance of 3 horizontally.\n\nYour figure has two pairs of opposite sides that are equal in length and parallel.  You have a rectangle.
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area: l*w = 256\n-> 4w*w = 256\n->w(sqaure) = 64\n->w=8 (width)\n->l = 4w = 4*8 = 32(length). \n\nThere you go.
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the slant height is called the 'hypotenuse' only if the polygon whose height you are getting is a  triangle.   but parallelograms do not have hypotenuse, only height. \n\nyou might also mean there are ten 'letters' in the answer.
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Your question is too complicated to appreciate the math in it, which is supposed to be the real beauty in the first place. Anyway, I have tried figuring it out and come with equations. I do not even want to try and begin substitution and end up with more complications. I just thought once you saw the relation, you would throw those 10 points my way.\n\nAssuming a to be sides of the squares of the prism and b to be the length of the rectangles, we know that the sum of all dimensions is 14a+5b. Now I do not even want to get started why you brought it the story of the equilateral triangle. You can just say the sum  of the dimensions is 61. Anyway, since any side of the trianle, as it is equiangular hence equilateral, is twice 14a+5b and 122, the sum of the dimensions will be 61. So your first equation is 14a+5b=61\n\nYou know the volume of the prism is the length X width X heiht = a x a x b = (a^2)b This is equal to the surface area. The surface area is the sum of all the areas on the prism and that is a^2+4ab. (How about you calculate how that came up?)\n\nSo there is your second equation, (a^2)b = a^2+4ab So then from 14a+5b=61, express a in terms of b or the other way around and plug it into the second equation and experience brutal complications and figure out what the variables are. I think you know it is too easy to calculate each dimension and area after that.
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Every side of this triangle has a pair of ends and all the three pairs are cyclic. ie, if AB,BC & CA are the three sides of the given triangle , then the side AB has the ends (4,5),(1,1), BC has the ends  (1,1)and (-2,5) & CA has the pair of ends (4,5),(-2,5).\nBy simple inspection of these pairs we can find that the points C and A are at the same height from the abcissa hence parallel to it. By this we can conclude easily that the length of CA=6 units.\nFurther, to be an isocelous triangle in this case the point B should be on the mid line of CA, and it is normal to CA. The mid point of CA is three units away from its ends C&A. As CA is parallel to absissa we can clearly conclude that any point between (1,0)&(1,5) will be the vertex of an isosceles triangle. Here the point B is (1,1) on the said line. Thus proved that the given points form an isosceles triangle.\nWe alredy know that CA = 6\nAnd the other sides are equal to sqrt(4-1)^2 +(5-1)^2 = 5 units\nNow, we can find the area of this triangle by Heron's Formula \nie sqrt[8(8-6)(8-5)^2] =12 sq units
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I'm going to focus on the usual non-euclidean geometry due to Lobachevsky. In that geometry, there are infinitely many lines parallel to a given line and through a given point. Squares (i.e. figures with 4 equal sides and 4 equal angles) don't have right angles. There is a pentagon with all right angles. The area of a triangle can be determined by the sum of the angles in the triangle (it is proportional to the difference of 180 degrees and the sum of the angles). There is a 'natural' unit of distance obtained from angles between parallel lines and perpendicular lines.
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I would think that the area is proportional to the *square* of their corresponding sides.\n\nIf you had a rectangle that was 3 x 4 and the corresponding rectangle was 6 x 8, the ratio of their sides is 1 : 2.  But their areas are 1 : 4 (2²).
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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
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Area of a square = L * L\n\nArea of A = x * x ..........(1)\nArea of B = y *y  ..........(2)\n  \n y = x -1               ..........(3)\n\nA = 4 * B              ...........(4)\n\nSubstitute (1) & (2) into (4) gives:\n\nx^2 = 4 y^2           ...........(5)\n\nSub. (3) into (5)\n\nx^2 = 4 (x-1) ^2\n\nSolve:\n\nx^2 = 4( x^2 -2x +1)\nx^2 = 4x^2 -8x +4\nOr\n\n3x^2 - 8x +4 =0\n\nSecond order eq.\n\nx1,x2= [+8 +- Sqr(8^2 - 4*3*4)]/(2*3)\n\nx1,x2 = [8+-4]/6\n\nx1 = 2   &  x2 = 4/6\n\nThen\ny1 = 2-1 =1   & y2 = 4/6 -1 = -2/6 ( illogical )\n\nthen\n\nx= 2  and y = 1