How do you find the area of a 9 sided object with all sides of unequal length?
156289
Yes, follow the suggestion of x60157. In addition, if you can determine the:\n\n1. height ,h, of each triangle you can use the formula for area: A = 0.5bh where b = side prependicular to the height\n\n2. sides of the triangle, then this formula can be used:\nA = Sq. root of (s(s-a)(s-b)(s-c)) where a,b,c are the sides of each triangle; s = semi-perimeter = 0.5(a+b+c)
520472
Let s = length of the cube\nsurface area of a cube = 6s^2\nThe diagonal is found by using the pythagorean theorem.\nLet c = length of the diagonal of a side, then\ns^2 + s^2 = c^2\nLet d = length of the diagonal, then\nd^2 = s^2 + c^2\nReplace c^2 with s^2 + s^2, then\nd^2 = s^2 +(s^2+s^2) \nd^2 = 3s^2\nSurface area(A) is twice the square of the diagonal means\nA = 2d^2\nNow A = 6s^2 and d^2 = 3s^2\nThus,\n6s^2 = 2(3s^2)\n6s^2 = 6s^2\n\n2) V = s^3\nWe want to show that\nV = (1/9)sqrt(3)*d^3\nd^2 = 3s^2, then d = sqrt(3s^2) = s*sqrt(3)\nd^3 = d^2*d = 3s^2(s*sqrt(3)\nd^3 = 3s^3*sqrt(3)\n\ns^3 = (1/9)sqrt(3)*3s^3*sqrt(3)\ns^3 = (1/9)(3)(3)*s^3\ns^3 = s^3
87753
Just add up the lengths of the three sides. Works for any triangle - wether it's equilateral, isosceles or obtuse.
991928
Assuming you mean a regular hexagon, then with the 30-60-90 relationship, you should get that half the length of a side is 2.\n\nSo, the area of that yellow triangle in the link below is\n.5 * 2 * 2 * sqrt(3) = 2 * sqrt(3)\n\nThere are 12 triangles, so the area is 24 * sqrt (3).
955861
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.
522898
Imagine you unraveled the cylinder, the middle would turn into a rectangle, right? so to find the surface area, take the height of the cylinder and multiply it by the circumference of the circles at the bases, then add in the area of the circles. Circumference is 2(pi)r and area is (pi)r^2.
182640
I'm not sure if this is what you're asking, but if...\nd = length\nd-3 = width\n\nthen the area is d(d-3), a.k.a. d^2 - 3d .\n\nHope this helps!
285497
the hieght times length so (5x+2)*say y for the height so area =5x+2 * y and there are too many unknowns y would equal area/(5x+2) or so say x is 2 and y is 3 the area would be12*3 or 36 in this case x*18
398507
since the aspect ratio is 16:9, then the length and width are in proportion. Let width=9*x and length=16*x. Then area=length*width= (9*x)*(16*x)= 144*x^2=190000\n\nx^2=190000/144=1319.44444\nx=sqrt(1319.4444)=36.324\n\nwidth=9*x=9*36.324=326.917\nlength=16*x=16*36.324=581.187\n\nCHECK: 326.324*581.187 =190000
23529
that is one of the answers, if the length of one side is 5. Use the Phytagorous theorem.
378913
Do you mean like an object that has a semi-circle on one end?\n\nIf yes, you can find the area of the semi-circle by figuring the area of the entire circle and dividing by half. Then you just find the area of the rectangular portion and add the two answers.
282503
Let's break the area into many vertical rectangular strips between x=1 and x=2, each strip so narrow that we can assume that the top of the strip is horizontal. Thus each strip will be a very narrow and tall rectangle.\n\nLet's say the number of vertical strips we use is n. The first strip will have its left-hand side at x = 1, the next will have its left-hand side at x = 1 + 1/n, the third will have its left-hand side at x = 1 + 2/n, and in general the kth strip will have its left-hand side at x = 1 + (k-1)/n.\n\nNow, what is the y value at each of these x points? Using f(x)=x² we get, for the kth point,\n\ny = (1 + (k-1)/n)² = 1 + 2(k-1)/n + ((k-1)/n)² \n\nNow the area of each strip is equal to its base times its height. The base of each strip is 1/n. So the area of the kth strip is\n\nA = 1/n * (1 + 2(k-1)/n + ((k-1)/n)²)\n\nNow we have to add the areas of all these strips from k = 1 to k = n and rearrange the terms.\n\n.....k=n\nA = Σ (1/n + 2(k-1)/n² + (k-1)²/n^3)\n.....k=1   \n\n.....k=n\nA = Σ (1/n + (2k - 2)/n² + (k² - 2k + 1)/n^3)\n.....k=1\n\nGiving all fractions the common denominator of n^3,\n\n.....k=n\nA = Σ ((n² + 2kn - 2n + k² - 2k + 1)/n^3)\n.....k=1\n\n.....k=n\nA = Σ ((n² - 2n + 1)/n^3  + (2kn + k² - 2k)/n^3)\n.....k=1\n\nThe first term in the summation sign appears n times in the sum, so we can take it out by multiplying it by n. \n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ (2kn + k² - 2k)/n^3\n\nNow let's clear the parentheses in the summation:\n\nSum going from k=1 to k=n:\nA =  (n² - 2n + 1)/n² + Σ 2k/n² + Σ k²/n^3 - Σ 2k/n^3)\n\nThe first summation contains k=1, 2, 3, ... n, so we use the formula for the sum of the arithmetic series:\n\nS = n(n+1)/2\n\nso the first summation becomes\n\nn(n+1)/n² = (n² + n)/n² = 1 + 1/n\n\nThe last summation also contains k=1, 2, 3, ... n, so we use the same formula for the sum of a series, and we get\n\nn(n+1)/n^3 = (n² + n)/n^3 = 1/n + 1/n²\n\nFor the middle summation, we have to compute the sum of k² for k=1, 2, 3, ... n, or that is the sum of 1, 4, 9, 16, ... n².\n\nThe site\n\nhttp://mathworld.wolfram.com/PowerSum.html\n\ngives the formula for the sum of the squares of a series as\n\nS = (2n^3 + 3n² + n)/6\n\nso the middle summation becomes\n\n1/3 + 1/2n + 1/6n²\n\nNow, we are almost done.\n\nLet's let the number of strips get very large, go to infinity as a limit. Let's see what each term becomes:\n\n(n² - 2n + 1)/n² becomes 1 as n gets very large,\n1 + 1/n becomes 1 as n gets very large,\n1/n + 1/n² becomes 0 as n gets very large, and\n1/3 + 1/2n + 1/6n² becomes 1/3 as n gets very large.\n\nThe area under the curve is 1 + 1 + 1/3, or 7/3.\n\nYou can see why integral calculus was invented. Using integral calculus, this problem can be solved in two steps. The integral of f(x)=x² is x^3/3. Inserting the limits of x=1 and x=2 gives\n\nA = 2^3/3 - 1^3/3 = 7/3
306739
4 times the length of one side
147295
add up the lengths of all the sides.....\n\nemail me if you need anything else\n\nnghttymguy112@yahoo.com\n\n~Brendon
224714
Perimeter for a square is 4*side\nSo side of square = 36/4=9 inches\n\nAnd area of square is side*side\nthus area is 9*9=81 sq inches !!!
277948
as the first answer said divide it into triangles then find the area of each triangle using K(area)=(1/2)ab sin C, that is two sides of the triangle and the angle in between.  Then you simply add together the different triangles.  This is assuming the polygon is regular and every side is of equal length otherwise alot of other trig must be done to find the angle and distances.
327427
Each side would have length sqrt(3), so the area would be 3 m^2.