Is there a mathematical formula for figuring area based on perimeter (one number, not two.)?
548292
Assuming that an area may be dissected into triangles, then the area of each triangle is given by \n\nArea = (s(s-a)(s-b)(s-c))^(1/2)\n\nwhere s = (1/2)(a + b + c)\n\nand a, b, and c are the lengths of sides of each triangle -\n\nThe sought for area is (finally) the sum of all the areas of the triangles.\n\nGiven sufficient patience, any arbitrary area will be found.\n\n(Drawing a diagram while slogging through the exercise and looping around the separate triangles in the same direction can yield a shortcut whereby all of the internal perimeters eliminate each other, leaving only the perimeter of the original figure.)
164837
Perimeter = (2*L) + (2*W)\nArea = L*W\n\nFor example, if P=30 and A=50, L could be 5 and W could be 10.\n\nBut this might also work...\n\n((2*L) + (2*W))/Perimeter = 1 = (L*W)/Area\n\nFrom there, you can multiply both sides by the area (and/)or the perimeter to eliminate one of the fractions.  Just massage the numbers around a bit until the answer jumps out at you.
353664
find the area of the cross section.... then times the area by the length... that is of course if it's a prism
551690
T= L + 2B\n\nT is total surface area\nL is Perimeter x Height\nB is base area
554315
First pick what you want your area to be.\nLet A = 100.\nThe first one is easy, just let it be a square with all side lengths of 10\nNow pick two other numbers that when you multiply them you get 100.\n4 times 25 = 100\nA rectangle with width 4 and length 25.\nKeep picking more\n2*50 = 100\nwidth = 2 and length = 50\nYou can find more.\nAlso, you don't have to use 100, just pick any number you want for the area.
310080
Area = pi times radius squared, so divide the diameter by 2 and go from there
241144
to calculate the area of a 2-dimensional space you always multiply length x width, or in this case height\n\nC
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.
264303
For a trapezoid? its a 2D figure so there are only area and perimeter.  Area is A= 1/2(base1 + base2)height or A= ((base1 + base2)height)/2
818428
perimeter = 2w + 2l = 440\nw + l = 220\n\nA = lw = w(220-w) = 220w - w^2\ndA/dw= 220-2w\nmaximum point occurs at dA/dw=0\n2w=220\nw=110\n\nMaximum area= 110(220-110) = 12100sq. ft\n\nI'm not quite sure whts the string of numbers about
921006
Area = πab where a and b are the semi-major and semi-minor axis length
863802
area=(1/2)d1*d2\n=(1/2)*28*13\n=14*13\n=182 sft
426495
It depends on what shape you want to know the area.\n\nSimple shapes have formulas. (such as a circle's area is Pi*r*r)\n\nComplex shapes which have pattern.(you know the equation of lines that form the shape), calculus is used to find the area.\n\nIf a shape is not ordinary. You should draw lines to grid it and count the area square by square.\n\n-Hope this would help-
768235
surface area = (2 * pi * r^2) + (2 * pi * r * h)\nh = 5, surface area = 100 (put in units later)\n\n100 = (2 * pi * r^2) + (2 * pi * r * 5)\n2pi*r^2 + 10pi*r - 100 = 0\n\nNow bring in the best equation in the world. (Quadratic Formula)\nx = (-b +- sqrt(b^2 - 4ac)) / 2a\n\nr = (-10pi +- sqrt(100pi^2 + 800pi) / 4pi\n\nThrow the negative out, because length cannot be negative.\n\nr ~ 2.21cm
609947
Perimeter is the sum of the sides.  For example, if you have a 4 cm x 6 cm rectangle, you would add the 2 sides of 4 cm with the 2 sides of 6 cm to get 20 cm.  So the perimeter is 20 cm.\n\nFor area, it depends on the object you have.  If you have a kind of rectangle, you'd multiply the length times the width to get the area.  For a triangle, you'd multiply the base and the  height, then divide it by 2.\n\nTo see a drawing of a pentagon, go here:\n\nhttp://en.wikipedia.org/wiki/Pentagon
405573
you need to use some calculus.\nget an expression for the area \ndifferentiate it\nthen find when it is a maximum.
850259
here:http://www.cut-the-knot.org/Curriculum/Geometry/AreaOfParallelogram.shtml