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I'd like to calculate the height of a segment based on the area. I have the radius of the circle, the area of the segment and need to calculate the height of the segment.

I found the following formula at the following address but lack the mathematic ability to turn the equation around. Any help would be much appreciated.

$$A = r^2\arccos\left(\frac{1 - h}{r}\right) - (r-h)\sqrt{2rh - h^2}$$

http://www.had2know.com/academics/area-circular-segment-formula.html

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  • $\begingroup$ I take it that formula is supposed to look like $A=r^2\arccos(1-(h/r))-(r-h)\sqrt{2rh-h^2}$, and you want to solve it for $h$. But I think it's impossible to solve for $h$ in terms of $A$ and $r$ in closed form, using only the familiar functions like polynomials, square roots, exponentials, logarithms, trig functions; I think the only way to solve for $h$ is numerically. $\endgroup$ – Gerry Myerson Dec 11 '14 at 12:36
  • $\begingroup$ OK, looks like I may not have interpreted the formula correctly, but my point still stands; I think it's impossible to express $h$ in closed form in terms of elementary functions of $A$ and $r$. $\endgroup$ – Gerry Myerson Dec 11 '14 at 22:34
  • $\begingroup$ Thanks Gerry Myerson this gives me a better understand of why it's been so hard to find an answer elsewhere. So it is possible to solve it numerically - Does that mean that if we know the values of r and A we can get h? $\endgroup$ – Laurence Lord Dec 12 '14 at 15:15
  • $\begingroup$ It means you can get $h$ to however many decimal places you need. You know $h$ is between zero and $r$, so you can just try lots of values of $h$ in that range, and zero in on the value that works. There are some systematic ways to do this. There are also Calculus-based methods of getting these approximate solutions, such as Newton's Method. $\endgroup$ – Gerry Myerson Dec 12 '14 at 22:11
  • $\begingroup$ Great! Thanks again. I've managed to calculate some values for what I needed thanks to your help and this calculater (arndt-bruenner.de/mathe/scripts/kreissehnen) $\endgroup$ – Laurence Lord Dec 14 '14 at 19:20

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