how to calculate the radii of two parallel curved lines?
117809
Now you're getting into Calc III with binormal planes, curls, divergents, and such.  There is a Hell-ish formula somewhere.
310080
Area = pi times radius squared, so divide the diameter by 2 and go from there
956315
The simplest way is to cut it into five slices, each with an arc of 72 degrees (or 2*pi/5 radians). You could probably also figure out a way to cut it into strips so that each strip has equal area, but that would be significantly more complicated.
579127
what is the diameter....? Just multiply the diameter times pi. (3.14)
949810
Use the distance formula:\n\nD = √ [ (y2-y1) ² + (x2-x1) ² ]\n\n1)  D = √ [ (-2+4) ² + (-2+4) ² ] = √(4+4) = √8 = 2√2\n\n2)  D = √ [ (3+1) ² + (-2+7) ² ] = √(16+25) = √41\n\n3)  D = √ [ (4-5) ² + (6+3) ² ] = √(1+81) = √82\n\nNRT
861159
You need to know the base and the height of the shapes. \n\ntrapeziod: average of two bases x height\nparallelogram: base x height\ntriangle: 1/2 x base x height\n\nOr if you know how to use the sin, cos, tan, then you only need to know a few sides to calculate the base and the heights.
950608
get some graph paper. plug in numbers for x and calculate its equivalent y value. do that 5 times, and you should have your curved line. If not, use higher or lower numbers than you used the first time ( I like starting with 0) Then  draw a line perpendicular to the most bent part (aka axis of symmetry, aka cut the line in two so both sides of the new line look like mirror images), find 2 points on that cutting line, and there you go.
529200
Good that you care to ask. The answer to this should be evident to you by integration (hope you have been introduced to calculus? If not, you will be this year.). \n\nI assume that you don't have much of a background.\n \nOK, let the cone be such that its height is H, base radius R. \n\nWhat you need to do is consider an infinitesimally thin section of the cone (a diagram should be so helpful here, unfortunately not possible) at height h from the bottom and of height (along the axis) 'dh', such that the radius of the section is a uniform 'r'. \n\n(Note that you have a uniform radius r to the section because you have chosen a very thin section of the cone.)\n\nNow, the surface area of the section considered is \n2*pi*r*dh*cosec x. (x is the cone half-angle)\n\nThe cosec term comes from basic trigonommetry, when you realise that the section is tilted and not exactly parallel to the axis. \n\nThe surface area  of the entire cone, indeed the curved surface area, is a summation of all such infinitesimal sections, each of surface area 2*pi*r*dh* cosec x \n(where r is the radius of each section). \n\nThis above sum can be found by \nintegrating 2*pi*r*dh*cosec x (that is a word in the language of calculus for finding out the sum of many small parts) over the entire height of the cone. \n\nalso realise that r = h tan x. \n\nso an integration of 2*pi*htan x*cosecx*dh gives us,\n\nCSA = 2*pi*sec x* (H^2/2)  \n    = pi*R* H (sec x/tan x)\n    = pi*R* H cosec x\n    = pi*R*L\n\nHope this satisfies.
303304
Use the well known formulas that relate the radius and diameter of a circle to its circumfrence.\n\nC = pi * D\nC = 2 * pi * R\n\nFor pi, you can use 3.14 or 22/7, or whatever value your teacher wants you to use.
385689
You want arc length or straight line?\n\n934.59 nm\n\naccording to the reference below.
970313
r=radius\nd=diameter\n\n1 C=2r(pi)\n2 C=d(pi)\n\n(Same thing because 2r=d)
493520
The length of the curve between the two points according to my calculation was:\n\nd = (3/2a) + (a*ln(2)/4)\n\nThe length is minimum under the condition that\n\na = √(6/ln(2))\n\n\n\nI don't have the time to show the full calculations, but I'll give you some key points:\n\n1. Find the line element 'ds' using the formula:\n   ds = √(dx^2 + dy^2)\n   -or-\n   ds = dx * √(1 + (dy/dx)^2)\n\n2. Evaluate (dy/dx) and substitute in above equation. \n\n3. Integrate the line element 'ds' determined above from x=1 to x=2 - this should give you the length of the curve in terms of 'a'\n\n4. To find the minimum length, differentiate the equation for length with respect to 'a', make the equation equal zero & solve for 'a'. This should give you the values for 'a' that is either for minimum or maximum. Now, differentiate the new equation once more & substitute the different values of 'a'. For each case, check if the equation yields a positive value - this means that it is a minimum case.
470650
diameter = 9 inches\n\nRadius = 4.5 inches\n\npi * r^2 = 3.14159 * 4.5^2 = 63.61725...
149565
cool! I'll think about it\n\n---\n\nhere I am.\nUse the two perpenducular segments as a coordinate system with O on their intersection.\n\nSay: the 4-3 segment defines the x axis, and the 2-6 segment defines the y axis.\nYou know four points belonging to the circle: (-4;0),(0;-2),(3;0) and (0,6)\nThe center of the circle has the mean value between -4 and 3 as its x and the mean value between -2 and 6 as its y (property of chords: the center is on the chord's axis)\n\nso the center's coordinates are C(-1/2;2)\n\nNow you can easily calculate the distance between C and anyone of the four points you know to get the radius.\n\nThen finish finding the diameter\n\n----
563113
radius is half of the diameter.\n\nyou can use the circumference equation; c = pi x d\n\nd = 2r so then c = pi x 2r\nc / pi = 2r\nc / pi / 2 = r
504335
Draw two lines from any tow angles to two opposite base
135768
Area = Pi times the radius squared (A=pr2)\nIf you need to know how to find the radius, go here:\n\nhttp://math.about.com/library/blcircle.htm