Answer: perimeter \(\approx143\text{ cm}\); area \(\approx494\text{ cm}^2\).

The angle at the centre is

\(\frac{7\pi}{9}.\)

The line \(AC\) bisects the angle by symmetry, so in right triangle \(ABC\),

\(\angle BAC=\frac{7\pi}{18}.\)

The tangent length is

\(BC=18\tan\frac{7\pi}{18}.\)

(a) The perimeter of the shaded region is the arc \(BD\) plus the two tangent lengths:

So

\(\text{perimeter}=14\pi+36\tan\frac{7\pi}{18}.\)

Numerically, this is approximately

\(143\text{ cm}.\)

(b) The shaded area is the area of kite \(ABCD\) minus the area of sector \(DAB\).

The kite is made from two congruent right triangles, so its area is

\(18\left(18\tan\frac{7\pi}{18}\right).\)

The sector area is

\(\frac12(18)^2\cdot\frac{7\pi}{9}.\)

Therefore

Numerically, this is approximately

\(494\text{ cm}^2.\)