In the diagram below, A(−3; 0), B, C and D(−8; 5) are the vertices of a quadrilateral with BĈD = 105°. E(−11; 0) and F are points on the x-axis. ED is a straight line. FG || AD with G on ED.

The coordinates are $A(-3; 0)$ and $E(-11; 0)$. Since both points lie on the x-axis, the distance is the absolute difference of their x-coordinates.

The coordinates are $A(-3; 0)$ and $D(-8; 5)$.
Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:

The coordinates are $D(-8; 5)$ and $E(-11; 0)$.

Perimeter $= AE + AD + DE$
Perimeter $= 8 + 5\sqrt{2} + \sqrt{34}$
Perimeter $\approx 8 + 7.0710678 + 5.8309519$
Perimeter $\approx 20.9020197$
Rounded to two decimal places, the perimeter is $\boxed{20.90 \text{ units}}$.

3.2 It is given that F is the midpoint of AE.

3.2.1 Give a reason why G is the midpoint of DE.
In $\triangle ADE$, F is the midpoint of AE (given). We are also given that $FG \parallel AD$. According to the Midpoint Theorem (or Converse of Midpoint Theorem), if a line passes through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side. Therefore, G is the midpoint of DE.

3.2.2 Hence, determine the coordinates of G.

The coordinates are $D(-8; 5)$ and $E(-11; 0)$.
The midpoint formula is $M = \left(\frac{x_1 + x_2}{2}; \frac{y_1 + y_2}{2}\right)$.

The coordinates of G are $\boxed{\left(-\frac{19}{2}; \frac{5}{2}\right) \text{ or } (-9.5; 2.5)}$.

3.2.3 Write down the length of FG.

According to the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is half the length of the third side.
$FG = \frac{1}{2} AD$.
From 3.1, $AD = 5\sqrt{2}$ units.

3.2.4 Determine the equation of FG.

F is the midpoint of AE. The coordinates are $A(-3; 0)$ and $E(-11; 0)$.

Since $FG \parallel AD$, their gradients are equal.
The coordinates are $A(-3; 0)$ and $D(-8; 5)$.

So, $m_{FG} = -1$.

Using point $F(-7; 0)$ and $m_{FG} = -1$:

3.3 Calculate the size of $D\hat{A}O$.

The coordinates are $A(-3; 0)$ and $D(-8; 5)$.

Let $\alpha_{AD}$ be the inclination of AD.

Since the gradient is negative, $\alpha_{AD}$ is in the second quadrant.

A is $(-3; 0)$ and O is the origin $(0; 0)$. The line AO lies on the x-axis. Its inclination is $0^\circ$.

The angle $D\hat{A}O$ is the angle formed by the line segment AD with the negative x-axis (line segment AO).

3.4 It is given that ABCD is a cyclic quadrilateral and that the equation of CB is $y = \frac{-12+5\sqrt{3}}{3}x + \frac{24+5\sqrt{3}}{3}$. Determine the coordinates of B.

In a cyclic quadrilateral, opposite angles are supplementary.
Given $B\hat{C}D = 105^\circ$.
Therefore, $B\hat{A}D = 180^\circ - B\hat{C}D = 180^\circ - 105^\circ = 75^\circ$.

Let $m_{AB}$ be the gradient of AB.
From 3.3, the inclination of AD is $\alpha_{AD} = 135^\circ$.
The angle $B\hat{A}D = 75^\circ$. From the diagram, the line AB has an inclination less than that of AD.
So, $\alpha_{AB} = \alpha_{AD} - B\hat{A}D = 135^\circ - 75^\circ = 60^\circ$.
The gradient of AB is $m_{AB} = \tan(60^\circ) = \sqrt{3}$.

Line AB passes through $A(-3; 0)$ and has a gradient $m_{AB} = \sqrt{3}$.
Using the point-slope form $y - y_A = m_{AB}(x - x_A)$:

Equation of AB: $y = \sqrt{3}x + 3\sqrt{3}$
Equation of CB: $y = \frac{-12+5\sqrt{3}}{3}x + \frac{24+5\sqrt{3}}{3}$
Set the y-values equal:

Multiply the entire equation by 3 to clear the denominators:

Rearrange to solve for x:

Factor out common terms:

Divide both sides by $2(6 - \sqrt{3})$:

The coordinates of B are $\boxed{(2; 5\sqrt{3})}$.