(a) To find the perimeter of the shaded region, we first find the angle \(\angle APQ\) using the cosine rule:

\(\cos^{-1} \left( \frac{5}{6} \right) = 0.5857 \text{ radians}\)

The perimeter of the shaded region is given by:

\(4 \times r \times 0.5857 = 2.34r\)

Alternatively, it can be expressed as \(0.745\pi r\) or \(\frac{293}{125}r\).

(b) To find the area of the shaded region, we use the sector formula for \(\text{Sector APB}\):

\(\frac{1}{2}r^2 \times (2 \times 0.5857)\)

Using the appropriate formula for the area of a triangle and combining it with the sector, we find:

\(\text{Shaded area} = 0.250r^2\)

Alternatively, it can be expressed as \(0.0796\pi r^2\).

(i) The area of parallelogram ABCD can be found using the formula for the area of a parallelogram:

Area = AB × BD × sin(ABD)

Given AB = BD = 10 cm and ABD = 0.8 radians, the area is:

Area = 10 × 10 × sin(0.8) = 100 × sin(0.8) = 71.7

(ii) The area of the complete figure ABQCDP includes the area of parallelogram ABCD and the areas of sectors APD and BQC. Each sector has a radius of 10 cm and angle 0.8 radians.

Area of one sector = \(\frac{1}{2} \times 10^2 \times 0.8 = 40\)

Total area of sectors = 2 × 40 = 80

Total area of figure = Area of parallelogram + Total area of sectors = 71.7 + 8.3 = 80

(iii) The perimeter of the complete figure ABQCDP includes the sides AB, BC, CD, DA, and the arcs APD and BQC.

Length of one arc = 10 × 0.8 = 8

Total length of arcs = 2 × 8 = 16

Perimeter = AB + BC + CD + DA + Total length of arcs = 10 + 10 + 10 + 10 + 16 = 36

(i) The perpendicular distance from \(D\) to \(AX\) is \(6 \sin \frac{\pi}{3} = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\).

The perpendicular distance from \(E\) to \(AX\) is \(10 \sin \theta\).

Equating these distances gives \(10 \sin \theta = 3\sqrt{3}\), so \(\sin \theta = \frac{3\sqrt{3}}{10}\).

Thus, \(\theta = \sin^{-1}\left(\frac{3\sqrt{3}}{10}\right)\).

(ii) The arc \(DX\) is \(6 \times \frac{1}{3}\pi = 2\pi\).

The arc \(EX\) is \(10 \times 0.5464 = 5.464\).

The horizontal steps are \(6 \cos \frac{\pi}{3}\) and \(10 \cos \theta\).

\(DE = 10 + 6 - 6 \cos \frac{\pi}{3} - 10 \cos \theta\).

The perimeter is \(\text{arc } DX + \text{arc } BX + DE = 16.20\).

(i) The perimeter of the plate consists of the arc AB and the sides OC, CB, and OA. The length of the arc AB is given by \(r\theta\). The side OC is \(r \sin \theta\) and CB is \(r \cos \theta\). The side OA is r. Therefore, the perimeter is:

\(r\theta + r \sin \theta + r \cos \theta + r = r(1 + \theta + \cos \theta + \sin \theta)\)

(ii) The area of the plate consists of the area of the sector OAB and the area of the triangle OCB. The area of the sector OAB is \(\frac{1}{2} r^2 \theta\). For r = 10 and \(\theta = \frac{1}{5}\pi\), this becomes:

\(\frac{1}{2} \times 10^2 \times \frac{\pi}{5} = 31.42\)

The area of the triangle OCB is \(\frac{1}{2} \times OC \times CB\). With OC = \(10 \sin \frac{\pi}{5}\) and CB = \(10 \cos \frac{\pi}{5}\), the area is:

\(\frac{1}{2} \times 10 \cos \frac{\pi}{5} \times 10 \sin \frac{\pi}{5} = 23.78\)

Thus, the total area of the plate is:

\(31.42 + 23.78 = 55.2\)

(i) Since AX is tangent to the circle at A, \(\angle OAX = \frac{\pi}{2}\). Using the tangent formula, \(AX = 6 \tan \frac{\pi}{3} = 6 \sqrt{3}\).

(ii) The area of triangle OAX is \(\frac{1}{2} \times 6 \times 6 \sqrt{3} = 18 \sqrt{3}\).

The area of sector OAB is \(\frac{1}{2} \times 6^2 \times \frac{\pi}{3} = 6\pi\).

Thus, the area of the shaded region is \(18 \sqrt{3} - 6\pi\).

(iii) The arc length AB is \(6 \times \frac{\pi}{3} = 2\pi\).

Using trigonometry, \(OX = \frac{6}{\cos \frac{\pi}{3}} = 12\), so \(BX = 6\).

The perimeter of the shaded region is \(6 \sqrt{3} + 2\pi + 6\).