Area between Two Non-Intersecting Chords

Theorem
Let $AB$ and $CD$ be two chords of a circle whose center is at $O$ and whose radius is $r$.


 * Circle with chords and area.png
 * Circle with chords and area 3.1.png

Let $\alpha$ and $\theta$ be respectively the measures in radians of the angles $\angle COD$ and $\angle AOB$.

Then the area $\mathcal A$ between the two chords is given by:
 * $\mathcal A = \dfrac {r^2} 2 \left({\theta - \sin \theta - \alpha + \sin \alpha}\right)$

if $O$ is not included in the area, and:


 * $\mathcal A = r^2 \left({\pi - \dfrac 1 2 \left({\theta - \sin \theta + \alpha - \sin \alpha}\right)}\right)$

if $O$ is included in the area.

Proof

 * First case: the center $O$ is included in the area.

Then to obtain the area between the two chords we can subtract from the area of the whole circle the areas of the segments $\mathcal S_α$ (whose base subtends the angle $α$) and $\mathcal S_θ$ (whose base subtends the angle $θ$). In formula:


 * $\mathcal A = r^2 π - \mathcal S_α - \mathcal S_θ$

where $r^2 π$ is the Area of Circle.

By Area of Segment of Circle we have that $\mathcal S_θ = \dfrac 1 2 r^2 \left({\theta - \sin \theta}\right)$ and $\mathcal S_α = \dfrac 1 2 r^2 \left({\alpha - \sin \alpha}\right)$.

Then


 * $\mathcal A = r^2 π - \dfrac 1 2 r^2 \left({\theta - \sin \theta}\right) - \dfrac 1 2 r^2 \left({\alpha - \sin \alpha}\right)$
 * $= r^2 \left({\pi - \dfrac 1 2 \left({\theta - \sin \theta + \alpha - \sin \alpha}\right)}\right)$.

Hence the result.


 * Second case: the center $O$ is not included in the area.

Let $\theta \ge \alpha$.

We can see that the area between the two chords is the difference between the area of the segment in $θ$ and the area of the segment in $α$. In formula:


 * $\mathcal A = \mathcal S_θ - \mathcal S_α$.

By Area of Segment of Circle we have that $\mathcal S_θ = \dfrac 1 2 r^2 \left({\theta - \sin \theta}\right)$ and $\mathcal S_α = \dfrac 1 2 r^2 \left({\alpha - \sin \alpha}\right)$.

Then


 * $\mathcal A = \dfrac 1 2 r^2 \left({\theta - \sin \theta}\right) - \dfrac 1 2 r^2 \left({\alpha - \sin \alpha}\right)$
 * =$ \dfrac {r^2} 2 \left({\theta - \sin \theta - \alpha + \sin \alpha}\right)$.

Hence the result.