The diagram shows part of the trapezium after the semi-circle was cut. AB = 26.8cm, BC = 14.4cm and diameter DC = 16cm. (i) Show that AD = 18cm. (ii) Calculate the perimeter of the shaded region.

Let $E$ be the point on $AB$ such that $DE$ is perpendicular to $AB$. This forms a rectangle $EBCD$ and a right-angled triangle $ADE$.

Since $EBCD$ is a rectangle, $DE = BC = 14.4$ cm and $EB = DC = 16$ cm.
Then, $AE = AB - EB = 26.8 - 16 = 10.8$ cm.

In the right-angled triangle $ADE$:

This shows that $AD = 18$ cm.

The perimeter of the shaded region consists of the lengths $AB$, $BC$, $AD$, and the arc length of the semi-circle.

$AB = 26.8$ cm
$BC = 14.4$ cm
$AD = 18$ cm (from part i)

The diameter of the semi-circle is $DC = 16$ cm.
The radius $r = \frac{16}{2} = 8$ cm.
The arc length of a semi-circle is given by the formula $\pi r$.

Add all the lengths together:

Using the value of $\pi \approx 3.14159$:

Rounding to one decimal place:

The perimeter of the shaded region is $\boxed{\text{84.3 cm}}$.