Answer: perimeter \(94.1\text{ cm}\); area \(350\text{ cm}^2\).

(a) The diagonals of the rectangle bisect each other, so from \(O\) to the sides of the rectangle the half-lengths are \(3\text{ cm}\) and \(2\text{ cm}\).

In triangle \(AOD\), the angle at \(O\) is split symmetrically by the line through the centre of the rectangle. Therefore

\(\angle AOD=2\tan^{-1}\left(\frac23\right).\)

Hence

\(\angle AOD=1.1760\ldots\)

so \(\angle AOD=1.176\) radians correct to 3 decimal places.

(b) The major arc \(MN\) has angle

\(2\pi-1.176\)

at the centre, so its length is

\(12(2\pi-1.176).\)

Also,

\(OA=\sqrt{3^2+2^2}=\sqrt{13}.\)

Since \(OM=ON=12\),

\(MA=ND=12-\sqrt{13}.\)

The straight parts \(AB+BC+CD\) have total length

\(6+4+6=16.\)

Therefore the perimeter is

\(12(2\pi-1.176)+2(12-\sqrt{13})+16.\)

This gives

\(94.1\text{ cm}\)

to 3 significant figures.

(c) The shaded region is the major sector \(MON\), with the part of the rectangle inside that major sector removed.

The major sector area is

\(\frac12(2\pi-1.176)(12)^2.\)

The rectangle has area \(6\times4=24\). Triangle \(AOD\) has area

\(\frac12(4)(3)=6.\)

So the part of the rectangle inside the major sector has area \(24-6=18\). Therefore the shaded area is

\(\frac12(2\pi-1.176)(12)^2-18.\)

This gives

\(350\text{ cm}^2\)

to 3 significant figures.