Answer: (i) \(1.48\), (ii) \(0.436\), (iii) \(54.7\text{ cm}\), (iv) \(178\text{ cm}^2\).

For part (i), use arc length \(s=r\theta\).

So

\(14.8=10\theta.\)

Hence

\(\angle AOB=1.48\text{ radians}.\)

For part (ii), use the sector area formula

\(\text{area}=\dfrac12r^2\theta.\)

So

\(21.8=\dfrac12(10)^2\theta.\)

Therefore

\(\theta=0.436.\)

Thus

\(\angle DOC=0.436\text{ radians}.\)

Since \(AB\parallel DC\), the remaining angles at the centre on the two sides are equal. Therefore

\(\angle BOC=\dfrac{2\pi-1.48-0.436}{2}=2.18\text{ radians, approximately}.\)

Now find the chord \(BC\):

\(BC=2(10)\sin\left(\dfrac{2.184}{2}\right).\)

So

\(BC\approx17.75\text{ cm}.\)

The perimeter of the shaded region is made from arc \(AB\), arc \(DC\), and the two equal sides \(AD\) and \(BC\).

The arc \(DC\) has length

\(10(0.436)=4.36\text{ cm}.\)

Hence

\(\text{perimeter}=14.8+4.36+2(17.75)=54.66\text{ cm},\)

which rounds to

\(\boxed{54.7\text{ cm}}.\)

For the area, add the sector \(AOB\), the sector \(DOC\), and the two triangles \(AOD\) and \(BOC\).

The sector \(AOB\) has area

\(\dfrac12(10)^2(1.48)=74.\)

The sector \(DOC\) has area \(21.8\).

Each side triangle has area

\(\dfrac12(10)^2\sin(2.184).\)

So the shaded area is

\(74+21.8+2\left(\dfrac12(10)^2\sin(2.184)\right).\)

This gives

\(\boxed{178\text{ cm}^2}\) to 3 significant figures.