how do i find te perimeter around an object?
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add up the lengths of all the sides.....\n\nemail me if you need anything else\n\nnghttymguy112@yahoo.com\n\n~Brendon
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if you mean an ellipse....\n\narea is equal to pi * minor radius * major radius
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The area and perimeter would both be found using the same formula as finding it for a rectangle. Because the angles and lengths must be equal to be a parallegram if you cut off the angled part (trianle) from one end and put it on the other you would have a regular rectangle.\n\n2l+2h = perimeter\nl*w = area\n\nDo it the easy way!!!
826433
perimeter=4 times the side\n=4*64=256 foot
50370
Draw a circle.  From the centre point to the edge draw a straight line.  This is the radius.  The area of a circle is pi times (radius squared.)  The Perimetre of a circle is 2 times pi times radius.
44499
Try www.objectdesktop.com it's great !!!
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There is no simple formula for the perimeter of an ellipse. The exact value is given by an ELLIPTIC INTEGRAL OF THE SECOND TYPE -- in the past people used extensive tables to find approximate answers, but today one gets greater accuracy using a calculator to approximate the integral. When the ellipse has semi-axes of length a and b (so that its equation is x^2/a^2 + y^2/b^2 = 1\nA simple approximate formula for perimetre is\n2pi*sqrt[(a^2+b^2)/2]
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Looks to me like you are trying to figure out a "PERIMETER" rather than an "AREA".  If so, you need to ADD the lengths of all the sides together.  If you ARE looking for an area, it is simply Length x Width (for a rectangle or square), 1/2 the base x height (for a triangle), or Pi x radius squared (Pi is approximately =3.1415926) for a circle.  If the "shape" of the object you are trying to find the area of is irregular, then you must divide it up into these familiar shapes first, find EACH individual area, then add those areas together for the total area.
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Find the dementions and multiply.
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Area & Perimeter of Parallelogram\n\nA Parallelogram is a four-sided quandrilateral with both pairs of opposite sides parrallel.\nA area of parallelogram is calculated using the following formula:\n\nA=(base)*(height)=bh\n\nThe perimeter of a parallelogram is calculated using the following formula:\n\nP=2a+2b               (3-4)\n\nThe area of a parallelogram is always expressed in square units, and the perimeter of a parallelogram is always expressed in the original units.\n\nExample:\n\nCalculate the area and perimeter of a parallelogram with base (b) = 4', height (h) = 3', a = 5' and b = 4'. Be sure to include units in your answer.\n\nSolution:\nA= bh \nA=(4)(3)\nA=12 square feet \n\nP= 2a+2b\nP=2(5)+2(4)\nP=10+8\nP=18 feet
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Half of the diameter.\n\nor\n\nMeasure around it.  Divide that number by 3.14.  Then divide by 2.
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I am not sure if I understand your question correctly, or else your question is not complete.\nFor example, a usual question of this sort is to find the maximum area of a rectangle for a given perimeter, the classical fence problem, as follows:\n\nYou are given a specific length of fence L to make en enclosure of maximum area.  What would be the dimensions of the rectangle.\n\nHere we take the length of one side as x, then the other side will be (L-2x)/2.  The area is thus A(x)=x(L-2x)/2.\n\nTo obtain the value of x for a maximum area, we differentiate A with respect to x, \ndA(x)/dx = d(Lx/2-x^2)=L/2-2x\nfor maximum area, dA(x)/dx=0, so we equate\nL/2-2x=0\nand solving for x gives x=L/4.\n\nThat means that the maximum area rectangle for a given perimeter is a square!\n\nHope that helps with your project.\nPost if you need other information, but please post details, especially how far you are with your project, so we can help you better.
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The perimeter is twice the sum of the length and the width. Given the ratio of length to width, the perimeter is necessarily eight times the width. Therefore the width is eight and the length is twenty four.
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Let x be the radius of the semicircle, then the rectangle has base 2r and height h.  The total perimeter of the window is\n\npi*r + 2r + 2h = 20. \n\nSo h = (20 - pi*r - 2r) / 2\n\nThe area is (pi*r^2)/2 + 2rh.  Replace h with the above so that the area is a function of r.  Differentiate to find the maximum.
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Assuming you have a rectangular object, then L x H x W will work. For other shapes you need something else. See sources.
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Let r be the radius of the circle and s the length of the square's sides.\ncircle perimeter=2*pi*r, square perimeter=4*s\n2*pi*r=4*s => s=pi*r/2\ncircle area=pi*r^2, square area=s^2\ns^2=(pi*r/2)^2=pi^2*r^2/4\nBut pi<4 => pi^2*r^2/4<pi*r^2 => square area < circle area
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So am I answering a homework question for you? lol\n\nCircles have the minimal perimeter for the area contained.\nArea of a circle is pi x Radius x Radius. (pi R squared)\n\nNext best would be something close to a circle, so consider a hexagon or octagon, with the corners of the square intersecting the vertices of the outer figure.\n\nLet's try a circle first...\n\nDiameter of the circle would be the diagonal of the square, or five times the square root of two.  5 x 1.414 = 7.07.  So the Radius is 3.535 meters.\n\nPi x 3.535 x 3.535 = 3.14 x 12.5 = 39.34\n\nSo the circle solution is too big.\n\nNow work out on paper what an octagon or hexagon surrounding the square would look like.  One of those should be about 34 m square.  Remember the area of the square is already 25 m2, so you are looking for only 9 more m2.\n\nI played around for minute or so in paint, and came up with a parallelogram that worked too.  Finally, a smooth curving perimeter shape get you benefits similar to the "circle" solution, so an ellipse would be a good comparison solution.  Try that, and compare the perimeter result against the perimeter of the parallelogram, which has a perimeter of 24.23 meters.\n\nHave fun!\nOC