The diagram shows a trapezium $OABC$ inside a circle with centre $O$ and radius $r$ cm. Given that $OC$ is parallel to $AB$ and angle $OAB = \theta$ radians.

1. Find the angles in triangle OAB.

Since $OA$ and $OB$ are radii of the circle, $OA = OB = r$.
Therefore, $\triangle OAB$ is an isosceles triangle, which means $\angle OBA = \angle OAB = \theta$.
The angle at the centre is $\angle AOB = \pi - 2\theta$.

2. Find the angle COB using parallel lines.

Given that $OC$ is parallel to $AB$, $OB$ acts as a transversal.
Using alternate interior angles, $\angle COB = \angle OBA = \theta$.

3. Calculate the area of the trapezium by summing the areas of triangle OAB and triangle OCB.

The area of the trapezium $S$ is the sum of the areas of $\triangle OAB$ and $\triangle OCB$.
Area of $\triangle OAB = \frac{1}{2}(OA)(OB)\sin(\angle AOB) = \frac{1}{2}r^2\sin(\pi - 2\theta) = \frac{1}{2}r^2\sin 2\theta$
Area of $\triangle OCB = \frac{1}{2}(OC)(OB)\sin(\angle COB) = \frac{1}{2}r^2\sin \theta$

$S = \frac{1}{2}r^2\sin 2\theta + \frac{1}{2}r^2\sin \theta$
$S = \frac{r^2}{2}(\sin \theta + \sin 2\theta)$ (shown)