The shape

I know the length of the arc A, the sides L, the bottom W and the maximum height H. I would like to calculate the maximum width. What would I derive a formula for it?


The whole shape is symmetrical on both sides of H.

There is no guarantee that L are radii of the circle that would be formed from A.

Johan Biemans, who posted below, first posed the question to me.

  • 1
    $\begingroup$ Are $W$, $H$ and $A$ in the center? $\endgroup$
    – LL 3.14
    Feb 6 at 0:22
  • $\begingroup$ Yes they are. Thanks for checking. $\endgroup$ Feb 6 at 0:28
  • $\begingroup$ If you can get the radius $r$, then use the formular to find chord length: $2 × \sqrt(r^2− d^2)$ where $d$ is perpendicular distance from the centre $\endgroup$
    – Abel
    Feb 6 at 0:31
  • $\begingroup$ Is $A$ a circular arc? $\endgroup$
    – AKemats
    Feb 7 at 16:39
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    $\begingroup$ There seems to be a lot of confusion about your diagram. So first of all, is the figure symmetric about H? And second, are the lines L normal to the arc A, that is, are the lines L along radii of A? $\endgroup$ Feb 8 at 15:58

6 Answers 6


I will begin this by saying that the following is only true if $L\parallel L'$ for each pair of legs, $L+L'=R$ and if $A$ is a circular arc.

I've added a bit to your diagram, as you can see below.

enter image description here

For the sake of clarity, I have extended the sides to meet at a point. Additionally, I have notated the distance from the point to the middle of the arc as $R$ and the distance from the top of the arc to the midpoint of $D$ as $P$.

From here, we have a few basic geometric theorems we can use. Namely, if we allow $\theta$ to be the angle between the two $L'$ legs of our newly formed traingle,

  • $A=R\theta$
  • $D=2R\sin{(\frac{A}{2R})}$
  • $P=R-\sqrt{R^2-(\frac{D}{2})^2}=R(1-\sqrt{1-(\frac{D}{2R})^2})$

To start, we must make our first, most useful substitution: $R=\frac{A}{\theta}$. Swapping this out in our definition of $D$ and our formula for $P$, we get

  • $D=\frac{2A}{\theta}\sin{(\frac{\theta}{2})}$
  • $P=\frac{A}{\theta}(1-\sqrt{1-(\frac{D\theta}{2A})^2})$

Plugging our new formula for $D$ into our one for $P$, we get

  • $P=\frac{A}{\theta}(1-\sqrt{1-\sin^2{(\frac{\theta}{2})}})$

Here's where everything comes into play. If, as we said, $L\parallel L'$, then the triangle with sides $L', L',$ and $W$ will be similar to the triangle with sides $L+L', L+L',$ and $D$. If these two triangles are similar, then the following relation will also be true:

  • $\frac{R-H}{R-P} = \frac{W}{D}$

Expanding this out, we get

  • $\frac{\frac{A}{\theta}-H}{\frac{A}{\theta}-\frac{A}{\theta}(1-\sqrt{1-\sin^2{(\frac{\theta}{2})}})} = \frac{W}{\frac{2A}{\theta}\sin{(\frac{\theta}{2})}}$

This can be massively simplified if we multiply both sides by $\frac{\theta}{\theta}$. Doing so yields us

  • $\frac{A-H\theta}{A\sqrt{1-\sin^2{(\frac{\theta}{2})}}} = \frac{W\theta}{2A\sin{(\frac{\theta}{2})}}$

We can rearrange this to give us

  • $\frac{A-H\theta}{W\theta}=\frac{\sqrt{1-\sin^2 (\frac{\theta}{2})}}{2\sin{(\frac{\theta}{2})}}$

And now, squaring each side, we see

  • $(\frac{A-H\theta}{W\theta})^2 = \frac{1-\sin^2 (\frac{\theta}{2})}{4\sin^2{(\frac{\theta}{2})}}$

It is at this point I must concede that I don't know how you would solve for $\theta$ here, or if, in fact, you can solve for $\theta$. However, if you can, then the solution follows very naturally. If $\theta = f(A,H,W)$, then very simply, the width of the shape will be $D=\frac{2A}{f(A,H,W)}\sin(\frac{f(A,H,W)}{2})$.

I understand that this does not solve the problem in general and only solves it if a certain requirement is met, but at the very least, I hope that this can help you get on the right path.

  • $\begingroup$ Thank you! I was hoping for a general solution but I'll study yours and learn some things for sure. $\endgroup$ Feb 8 at 6:42
  • $\begingroup$ An update after thinking on it some more - that last equality is actually equal to $(\frac{A-H\theta}{W\theta})^2 = \frac{1}{4}\cot^2{(\frac{\theta}{2})}$, which further simplifies down to $\frac{A-H\theta}{W\theta} = \frac{1}{2}\cot{(\frac{\theta}{2})}$. This looks easier to solve, but I'm still not sure how to approach it. $\endgroup$
    – AKemats
    Feb 8 at 13:23

I am making some assumptions about your diagram with this solution. First of all, that it is symmetric about $H$ and second that the lines $L$ are normal to the arc $A$, that is, they are radial lines.

Here are the basic relations you need:

The arc $s=A=R\theta$, where $R$ is the radius of the arc and $\theta$ is the angle of the sector (both as yet unknown).

The unknown chord $c=2R\sin(\theta/2)$.

Extending the lines $L$ so that they intersect at the origin for the arc, we'll have two similar triangle for which we can say

$$ \frac cW=\frac{R}{R-L} $$

Substituting the above relations we can find the following equation for $\theta$ in terms of known quantities, to wit

$$ 2R\sin\frac{\theta}{2}=\frac{RW}{R-L}\\ 2\sin\frac{\theta}{2}\big(\frac{s}{\theta}-L\big)=W $$

You will have to solve for $\theta$ by iteration, then get $R=A/\theta$ and $c=2R\sin(\theta/2)=2A\sin(\theta/2)/\theta$.


I can't post replies to comments because I am a new user, but in the diagram in the OP:

  • Line H is perpendicular to W
  • Line H bisects W, Unknown and A
  • Line W and Unknown are parallel
  • Arc A is a circular arc
  • The L lines are not along the Radii of A, but they are symmetrical. (The shape is mirrored along H)


Here is a working version of the calculator if anyone is interested:


And the source code:


My solution was a little less orthadox but I thought I would share to see if it could spark someone else. I couldn't find a way to make a forumla that worked, so instead, I used a trial and error approach.

Unfortunately, because L is not on a radius line of A, I can't use the solutions above.

I created a triangle on the corner and made a guess at the angle (starting at 45). I then used the information I had to calculate the unknown and the arc length. I compared my arc length to the A value I was looking for, and then kept adjusting the angle until I got a matching arc length. Once I did, I found the unknown. It usually took the program between 10-20 attempts before it could narrow it down to the correct answer. I can share the code if anyone wants. The basic math though was:

//Use the triangle to find the height of the arc and length of the unknown

P (Height Of Arc) = H - (L * (Sin(90-aA) / Sin(90) ))

U (Cord Length - Unknown) = W + (L * (Sin(aA) / Sin(90) )) * 2

//Use the Cord length and height of arc to find the radius and the arclength

RA = (((U * U) / (P * 4)) + P) / 2

A = ((Asin(U / (RA * 2))) * 2) * RA

( I tried to simplify it out to make it clean. The calculation part of the code is:)

float topAngle = angleToTry; float bottomAngle = 90-topAngle;

float topAngleRadians = ConvertDegreeToRadians(topAngle);
float bottomAngleRadians = ConvertDegreeToRadians(bottomAngle);
float rightAngleRadians = ConvertDegreeToRadians(90);

float hypotenuseC = lengthOfSides;
float heightOfTriangle = (hypotenuseC * (Mathf.Sin(bottomAngleRadians) / Mathf.Sin(rightAngleRadians) ));
float baseOfTriangle = (hypotenuseC * (Mathf.Sin(topAngleRadians) / Mathf.Sin(rightAngleRadians) ));

float heightOfArc = heightOfShape-heightOfTriangle;
float cordLength = lengthOfBottom + baseOfTriangle + baseOfTriangle;
finalWidthOfPiece = cordLength;

float radiusOfArc = (((cordLength*cordLength) / (heightOfArc*4)) + heightOfArc) / 2;

float arcLength = ((Mathf.Asin(cordLength / (radiusOfArc * 2))) * 2) * radiusOfArc;

Now that I am advised that the $L$'s are not radial to the arc, I'll take a different view. This is not a solution, per se, but is too long for a comment. So, the unknown chord, $c$, can be expressed in terms of both the arc and the trapezoid. Let's define $\alpha$ as the obtuse angle ($LW$) of the trapezoid, $R$ as the radius of the arc, $h$ as the height of circular segment, and $\theta$ as the angle of the circular sector.

Now, for the circular arc,

$$ c=2R\sin\frac{\theta}{2}=2h\tan\frac{\theta}{2} $$

and for the trapezoid,

$$ c=W+2L\cos(\pi-\alpha)=W-2L\cos\alpha,\quad \alpha\in (\pi/2,\pi) $$

We can express the segment height in terms of $H,L,\alpha$,

$$ h=H-L\sin(\pi-\alpha)=H-L\sin\alpha $$

Combining the above, we find

$$ \tan\frac{\theta}{2}=\frac{W-2L\cos\alpha}{2(H-L\sin\alpha)} $$

This is as far as I can go without specific values for $L,W,H$, but it suggests that there are multiple possible solutions.


Edit: Without the assumption that the $L$ segments are along the radius of the arc, the given information is not enough to identify a unique solution to the problem. See an example in the figure below. In this example, for ease of verification, the angles of the arcs in the large and small green circles are $\frac{\pi}{4}$ and $\frac{\pi}{2}$, respectively, and the proportion of their radii is $\frac21$. Therefore it is easy to see that the two green arcs are of equal length. The radii of the dashed blue circles are $L$. It can be seen that there are an infinite number of solutions to this problem.

enter image description here

On the other hand, with the assumption that the $L$ segments are along the radius of the arc (thank you @DavidK and OP) the problem can be solved as described in my pre-edit answer:

Extend the sides to meet at point $O$. This is the center of the circle that includes your arc. Let the radius of this circle be $R$ and the angle between the two radii be $\alpha$. Then $$\sin \frac{\alpha}{2} = \frac{\frac W2}{R-L} $$ $$\cos \frac{\alpha}{2} = \frac{R-H}{R-L} $$ (Why?)

From here solve the equation $\sin^2\frac{\alpha}{2} + \cos^2\frac{\alpha}{2} = 1$ to find $R$. Can you take it from here?

  • 1
    $\begingroup$ You assume that the segments labeled $L$ lie along radii of the arc. I see nothing in the problem statement to guarantee that this is true. $\endgroup$
    – David K
    Feb 7 at 4:22
  • $\begingroup$ @DavidK Good observation, thanks. $\endgroup$
    – Saeed
    Feb 7 at 19:32
  • $\begingroup$ There is a comment under the problem about $W$, $H$ and $A$ being in the center, which was confirmed by the OP. Reading that, I jumped to the assumption that the orange segment is the perpendicular bissector of segment $W$. $\endgroup$
    – Saeed
    Feb 7 at 19:41
  • $\begingroup$ Actually I'm fairly sure that the orange segment is the perpendicular bisector of the bottom edge. But the question is the angle between the side edges and the arc. Suppose you have $A,L,W,H$ so that the $L$ segments are radii of the arc. Reduce $A.$ The figure still exists, but the edges $L$ now end inside the old figure and the arc has a smaller radius. The center of the arc has moved upward but the intersection of the $L$ lines (when extended) has moved downward, so those segments can no longer lie on radii. $\endgroup$
    – David K
    Feb 8 at 2:46
  • $\begingroup$ It's true that I was hoping for a solution that does not constrain the angle where the arc meets the sides at all. But I'm learning from these solutions! $\endgroup$ Feb 8 at 6:44

Cye Waldman got me thinking about this in a slightly different way. I was always focused on solving one shape before the other, but I should have been thinking of both shapes at the same time.

The following mess works to solve the arc length given $W$, $L$, $H$ and the top angle of the triangle (I labeled it $a_B$ on my previous response.)

However, in this case, we don't have $a$ (the angle), but we do have the other side $A$ (the arc length).

Is there any way to isolate a in the equation below to solve for it given $L$, $W$, $H$ and $A$? Or is it way too much of a mess to do so?

$$A = \sin^{-1}{\!\left(\frac{W + L\sin{\!(a)} + L\sin{\!(a)}}{\left(\frac{(W + L\sin{\!(a)} + L\sin{\!(a)})(W + L\sin{\!(a)} + L\sin{\!(a)})}{4(H - L\sin{\!(90^{\circ}-a)})} + H - L\sin{\!(90^{\circ}-a)}\right) / 2 \cdot 2}\right)} \cdot 2\left(\frac{(W + L\sin{\!(a)} + L\sin{\!(a)})(W + L\sin{\!(a)} + L\sin{\!(a)})}{4(H - L\sin{\!(90^{\circ}-a)})} + (H - L\sin{\!(90^{\circ}-a)})\right) / 2$$

I was able to clean it up a bit with the help of a math site:

$$A = \sin^{-1}{\!\left(\frac{W + 2L\sin{\!(a_B)}}{H + \frac{(W + 2L\sin{\!(a_B)})^2}{4H + 4L\sin{\!(a_B - 90^{\circ})}} + L\sin{\!(a_B - 90^{\circ})}}\cdot\left(H + \frac{(W + 2L\sin{\!(a_B)})^2}{4H + 4L\sin{\!(a_B - 90^{\circ})}} + L\sin{\!(a_B - 90^{\circ})}\right)\right)}$$

  • $\begingroup$ Please review the way I've used MathJax to format your question and use similar techniques in the future. $\endgroup$ Feb 15 at 1:36
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    $\begingroup$ Thank you, I couldn't figure out how to display it like that. I will try to learn more before future posts. $\endgroup$ Feb 15 at 13:15
  • $\begingroup$ @JacobManaker, do you have access to the initial equation I posted still ? I think there might be a formatting error in the MathJax version. Thank you. $\endgroup$ Feb 25 at 14:47
  • $\begingroup$ If you click on "edited Feb 15 at 1:35" it will show your original version and the changes I made. (That's what I meant by "review"; my comment may have seemed a little critical of you if you didn't know that, for which I apologize.) $\endgroup$ Feb 25 at 20:44
  • $\begingroup$ Thank you, I didn't know how to find that information. It turned out that my "cleaned up" version, had some errors in it, and when you fixed the formatting it made that error much more apparent. I just trusted the software that I ran it through without first double checking that it simplified it correctly. Also don't worry about the comment, I am glad that it was critical because I do need to learn. It was constructive criticism and it was also done with the corrections (not just telling me to fix it), so as far as I am concerned it was well done. $\endgroup$ Feb 28 at 15:34

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