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 $r$.

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

Then the area $\AA$ between the two chords is given by:

$\AA = \dfrac {r^2} 2 \paren {\theta - \sin \theta - \alpha + \sin \alpha}$

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

$\AA = r^2 \paren {\pi - \dfrac 1 2 \paren {\theta - \sin \theta + \alpha - \sin \alpha} }$

if $O$ is included in the area.

Proof

Let $\SS_\alpha$ be the area of the segment whose base subtends $\alpha$.

Let $\SS_\theta$ be the area of the segment whose base subtends $\theta$.

Case $(1)$: Center included in Area

Let the center $O$ be included in the area.

The area between the two chords is given by:

the area of the whole circle

minus:

the areas of the segments $\SS_\alpha$ and $\SS_\theta$ .

Thus:

\(\ds \AA\)

\(=\)

\(\ds \pi r^2 - \SS_\alpha - \SS_\theta\)

Area of Circle: $\pi r^2$

\(\ds \)

\(=\)

\(\ds \pi r^2 - \frac 1 2 r^2 \paren {\theta - \sin \theta} - \frac 1 2 r^2 \paren {\alpha - \sin \alpha}\)

Area of Segment of Circle

\(\ds \)

\(=\)

\(\ds r^2 \paren {\pi - \frac 1 2 \paren {\theta - \sin \theta + \alpha - \sin \alpha} }\)

rearranging

$\Box$

Case $(2)$: Center not included in Area

Let $\theta \ge \alpha$.

The area between the two chords is given by:

the area of the segment $\SS_\theta$

minus:

the area of the segment $\SS_\alpha$.

Thus:

\(\ds \AA\)

\(=\)

\(\ds \SS_\theta - \SS_\alpha\)

\(\ds \)

\(=\)

\(\ds \frac 1 2 r^2 \paren {\theta - \sin \theta} - \frac 1 2 r^2 \paren {\alpha - \sin \alpha}\)

Area of Segment of Circle

\(\ds \)

\(=\)

\(\ds \frac {r^2} 2 \paren {\theta - \sin \theta - \alpha + \sin \alpha}\)

rearranging

$\blacksquare$