Answer: (a) \(18\sqrt7+\sqrt{22}-14\); (b) \(9\sqrt7+\dfrac{87}{2}\); (c) \(\cot DBC=\dfrac{\sqrt7+3}{18}\).

We start with the main method. Use the measurements and relationships shown in the diagram, then translate them into algebraic or trigonometric equations. Use radian formulae for arc length and sector area: \(s=r\theta\) and \(A=\frac12r^2\theta\).

Since \(AB\) and \(CD\) are parallel and \(BC\) is perpendicular to both of them, the vertical height is

\(BC=\sqrt7+2.\)

The horizontal difference between the parallel sides is

\((9\sqrt7-9)-(8\sqrt7-7)=\sqrt7-2.\)

Therefore

\(AD^2=(\sqrt7-2)^2+(\sqrt7+2)^2.\)

Now

\((\sqrt7-2)^2=11-4\sqrt7\)

and

\((\sqrt7+2)^2=11+4\sqrt7.\)

So

\(AD^2=22\), giving \(AD=\sqrt{22}\).

The perimeter is

\((8\sqrt7-7)+(\sqrt7+2)+(9\sqrt7-9)+\sqrt{22}.\)

Hence

\(\text{perimeter}=18\sqrt7+\sqrt{22}-14.\)

For the area, use the trapezium formula:

\(\text{area}=\frac12(AB+CD)\times BC.\)

Thus

\(\text{area}=\frac12\{(8\sqrt7-7)+(9\sqrt7-9)\}(\sqrt7+2).\)

So

\(\text{area}=\frac12(17\sqrt7-16)(\sqrt7+2).\)

Expanding,

\((17\sqrt7-16)(\sqrt7+2)=119+34\sqrt7-16\sqrt7-32=87+18\sqrt7.\)

Therefore

\(\text{area}=9\sqrt7+\frac{87}{2}.\)

For \(\cot DBC\), use right triangle \(BCD\).

At angle \(DBC\), the adjacent side is \(BC=\sqrt7+2\), and the opposite side is \(CD=9\sqrt7-9\).

So

\(\cot DBC=\frac{BC}{CD}=\frac{\sqrt7+2}{9\sqrt7-9}.\)

Rationalise the denominator:

\(\frac{\sqrt7+2}{9(\sqrt7-1)}\cdot\frac{\sqrt7+1}{\sqrt7+1} =\frac{(\sqrt7+2)(\sqrt7+1)}{9(7-1)}.\)

The numerator is

\((\sqrt7+2)(\sqrt7+1)=9+3\sqrt7.\)

Hence

\(\cot DBC=\frac{9+3\sqrt7}{54}=\frac{\sqrt7+3}{18}.\)

This completes the solution and gives the required result.