Use the arc length formula:

\(\text{arc length}=r\theta\)

Here \(r=8\) and \(\theta=0.65\).

So

\(BC=8\times0.65=5.2\)

Therefore, the length of arc \(BC\) is \(5.2\text{ m}\).

Since \(D\) is the midpoint of \(AC\),

\(AD=\dfrac{8}{2}=4\text{ m}\).

The shaded area \(BCD\) is:

\(\text{area of sector }ABC-\text{area of triangle }ABD\).

Area of sector \(ABC\):

\(\dfrac{1}{2}r^2\theta=\dfrac{1}{2}\times8^2\times0.65=20.8\text{ m}^2\).

Area of triangle \(ABD\):

\(\dfrac{1}{2}ab\sin C=\dfrac{1}{2}\times8\times4\times\sin(0.65)\)

\(=16\sin(0.65)\approx9.683\text{ m}^2\).

So the shaded area is

\(20.8-9.683\approx11.12\text{ m}^2\).

Therefore, the area of region \(BCD\) is approximately \(11.12\text{ m}^2\).