# Given the area and height of a rectangle, what is the width of the base of a circular segment with the same height and area?

Given a rectangle of height $h$ and area $A$, what is the width $c$ of the chord at the base of a circular segment with the same height and area?

I've made a diagram of the problem:

My progress so far has come from manipulating equations from here. The best equation I have is $(\frac{1}{2}) (\frac{c^2}{8 h}+\frac{h}{2})^2 \left(2 \arccos\left[\frac{\frac{c^2}{8 h}-\frac{h}{2}}{\frac{c^2}{8 h}+\frac{h}{2}}\right]-\sin\left[2 \arccos\left[\frac{\frac{c^2}{8 h}-\frac{h}{2}}{\frac{c^2}{8 h}+\frac{h}{2}}\right]\right]\right)=h w$, which according to Mathematica is not solvable for $c$.

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Have you made any progress yourself? Please show it. – Henning Makholm Sep 6 '11 at 17:16
Done. It's not much to go on though. – Alexander Gruber Sep 6 '11 at 17:27
Yep, you have a transcendental equation over there. Barring special cases, you need Newton-Raphson or something. – J. M. Sep 6 '11 at 17:38
Well, I'm also not sure if that equation is right. (Or if there is a better method through calculus or by cases or something.) The arccos within a sin in particular is yucky, seems like something that could be simplified. – Alexander Gruber Sep 6 '11 at 17:47
Actually, yes. Since $\sin\,2x=2\sin\,x\cos\,x$ and $\sin\arccos\,x=\sqrt{1-x^2}$... what's making your problem not amenable to a symbolic solution is that you have a $c$ inside and outside the arccosine. – J. M. Sep 6 '11 at 17:50

Area of a circle segment is $r^{2}/2(\theta-\sin(theta))$, where $\theta$ is the angle in radians between the corners of the segment and the centre of the circle. The height=$r-\sqrt{ r^{2}-c^{2}/4}$. c is the chord length, $r\sqrt{2-2\cos(\theta)}$. Therefore the height is $r\sqrt{(1/2)-\cos(\theta)/2}$. Given an area a and a height h, $r=h/\sqrt{(1/2)-\cos(\theta)/2}$ and substituting into the other equation gives $(h^{2}/(1-\cos(\theta))^{2}(\theta-\sin(\theta))-a=0$. $1-h^{2}/2=\cos(\theta)-\sin(\theta)+\theta$ hence $2\sin(\theta)=\theta+h^{2}/a$. $\sin(\theta)$ is transcedential when $\theta$ is algebraic therefore all solutions have transcedential $\theta$ and are not easy to find.