How can I find the height of the arced portion of a circular segment ('h' in this link) when I only know the radius of the circle and the area of the segment.

The link above gives me an equation for calculating the area of the segment. So I can set this to fraction of the total circle, 'f'.

$$f\pi r^2 = r^2 cos^{-1} \left(\frac{r-h}{r}\right)-(r-h)\sqrt{2rh-h^{2}} $$

I would then need to solve this for 'h', which I'm struggling to do.

I have also tried using the angle instead, which gives a simpler equation. I set the area to the proportion of the total area again.

$$f\pi r^2 = \left(\frac{\theta\pi}{360} - \frac{\sin(\theta)}{2}\right)r^{2}$$

But I have not managed to solve this for $\theta$ either. Ultimately, I want to be able to mark out segments on a circle that correspond with specific proportions of the circle, and need to know where to place the markers. In this particular case the circle has a radius of 23.5mm and the segment's area is $\frac{3}{8}$ of the total circle area.

  • $\begingroup$ This post may help math.stackexchange.com/questions/2619158/…. I don't think it will be possible to get an exact answer for h. $\endgroup$
    – S34NM68
    Commented Mar 30, 2022 at 16:33
  • $\begingroup$ For simplicity put $r=1$ and resort to numerical solution by Runge-Kutta. $\endgroup$
    – Narasimham
    Commented Apr 2, 2022 at 20:01

2 Answers 2


I think you're overcomplicating it. Since the ratio $f$ of sector area to circle area always equals the ratio of $θ$ to 2π radians, we have:

$θ = 2\pi f$ in radians or $360^\circ\cdot f$ in degrees.

If we bisect $θ$, we have two right triangles, the hypotenuse being $R$, and the height, $r$. We want to find $r$ since $h + r = R$. We know that $\cos(θ/2) = r/R$ by definition, so:

$r = R\cos(θ/2)$.


$h = R - r$,

$= R - R\cos(θ/2)$,

$= R\,( 1- \cos(\pi f) )$ in radians,

or $R\,(1-\cos(180^\circ\cdot f)$ in degrees.

For $f = \frac{3}{8}$ and $R = 23.5\,\text{mm}$, $h = 14.5\,\text{mm}$.

Please do follow up in the comments if needing further clarification.


$\begin{array}\\ f\pi r^2 &= r^2 \cos^{-1} \left(\frac{r-h}{r}\right)-(r-h)\sqrt{2rh-h^{2}}\\ &= r^2 \cos^{-1} \left(1-h/r\right)-r^2(1-h/r)\sqrt{2h/r-(h/r)^{2}}\\ &= r^2\left( \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{2h/r-(h/r)^{2}}\right)\\ f\pi &= \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{2h/r-(h/r)^{2}}\\ &= \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{1-1+2h/r-(h/r)^{2}}\\ &= \cos^{-1} \left(1-h/r\right)-(1-h/r)\sqrt{1-(1-h/r)^2}\\ &= \cos^{-1}(u)-u\sqrt{1-u^2} \qquad u=1-h/r\\ &= \cos^{-1}(\cos(v))-\cos(v)\sqrt{1-\cos^2(v)} \qquad u=\cos(v)\\ &= v-\cos(v)\sin(v)\\ &= v-\sin(2v)/2\\ &= (2v-\sin(2v))/2\\ 2f\pi &= 2v-\sin(2v)\\ \end{array} $

Solve this for $v$ in terms of $f$ (has to be done numerically), then get $u=\cos(v), h=r(1-u) $.

For small $v$, $2v-\sin(2v) \approx 2v-(2v-4v^3/3) =4v^3/3 $ so, for small $f$,

$\begin{array}\\ v &\approx (3f\pi/2)^{1/3}\\ u &=\cos(v)\\ &\approx 1-v^2/2\\ &\approx 1-(3f\pi/2)^{2/3}/2\\ h &\approx r(3f\pi/2)^{2/3}/2\\ \end{array} $


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