Answer: Area \(=\frac{8r^2\theta}{9}-\frac{r^2\sin\theta}{2}\), and perimeter \(=r(\pi-\theta)+2r+2r\sin\frac{\theta}{2}\).

The shaded region \(BCDE\) is the larger sector \(OCD\) minus triangle \(OBE\).

The area of sector \(OCD\) is

\(\frac12\left(\frac{4r}{3}\right)^2\theta=\frac{8r^2\theta}{9}.\)

The area of triangle \(OBE\) is

\(\frac12 r^2\sin\theta.\)

Therefore

\(\text{area }BCDE=\frac{8r^2\theta}{9}-\frac{r^2\sin\theta}{2}.\)

For the second shaded region, the two equal arcs together have angle \(\pi-\theta\), so their total arc length is

\(r(\pi-\theta).\)

The straight lengths \(OA\) and \(OF\) contribute \(2r\).

Finally, \(BE\) is the chord in a circle of radius \(r\) subtending angle \(\theta\), so

\(BE=2r\sin\frac{\theta}{2}.\)

Hence the perimeter is

\(r(\pi-\theta)+2r+2r\sin\frac{\theta}{2}.\)