Solutions to this question by accurate drawing will not be accepted. The diagram shows a trapezium $ABCD$ in which $AB$ is parallel to $DC$. The coordinates of the points $A$, $B$ and $D$ are $(2, 0)$, $(3a, 2a+4)$ and $(-2, 2)$ respectively, where $a$ is a positive integer. The length of $AB$ is $4\sqrt{5}$ units.

1. Set up an equation using the distance formula for the length of $AB$.

Length of $AB = \sqrt{(3a - 2)^2 + (2a + 4 - 0)^2}$
Given $AB = 4\sqrt{5}$, we have:
$(3a - 2)^2 + (2a + 4)^2 = (4\sqrt{5})^2$

2. Expand and simplify the equation to form a quadratic equation in $a$.

$(9a^2 - 12a + 4) + (4a^2 + 16a + 16) = 16 \times 5 13a^2 + 4a + 20 = 80 13a^2 + 4a - 60 = 0$

3. Solve the quadratic equation and reject the invalid root to show $a = 2$.

$(13a + 30)(a - 2) = 0$
$a = -\frac{30}{13}$ or $a = 2$
Since $a$ is given as a positive integer, we reject $a = -\frac{30}{13}$.
Therefore, $a = 2$ (shown).

1. Find the coordinates of $B$ and the gradient of $AB$.

From part (i), $a = 2$.
Coordinates of $B = (3(2), 2(2)+4) = (6, 8)$.
Gradient of $AB = \frac{8 - 0}{6 - 2} = \frac{8}{4} = 2$.

2. Find the equation of the line $DC$.

Since $AB$ is parallel to $DC$, the gradient of $DC$ is also $2$.
The line $DC$ passes through $D(-2, 2)$.
Equation of $DC$:
$y - 2 = 2(x - (-2))$
$y = 2x + 4 + 2$
$y = 2x + 6$

3. Determine the coordinates of $C$.

From the diagram, point $C$ lies on the $y$-axis, so its $x$-coordinate is $0$.
Substituting $x = 0$ into the equation of $DC$:
$y = 2(0) + 6 = 6$
Therefore, the coordinates of $C$ are $(0, 6)$.

1. Find the coordinates of $B$ and the midpoint of $AB$

Since $a = 2$, the coordinates of $B$ are $(3(2), 2(2)+4) = (6, 8)$.
Midpoint of $AB = \left(\frac{2+6}{2}, \frac{0+8}{2}\right) = (4, 4)$.

2. Find the gradient of $AB$ and the gradient of the perpendicular bisector

Gradient of $AB = \frac{8-0}{6-2} = \frac{8}{4} = 2$.
Let the gradient of the perpendicular bisector be $m_{\perp}$.
$m_{\perp} \times 2 = -1 \implies m_{\perp} = -\frac{1}{2}$.

3. Find the equation of the perpendicular bisector

Using the point $(4, 4)$ and gradient $-\frac{1}{2}$:
$y - 4 = -\frac{1}{2}(x - 4)$
$y = -\frac{1}{2}x + 2 + 4$
$y = -\frac{1}{2}x + 6$

1. Substitute the $x$-coordinate of $C$ into the equation of the perpendicular bisector of $AB$ to check if it yields the $y$-coordinate of $C$.

From (ii), $C = (0, 6)$.
From (iii), the equation of the perpendicular bisector of $AB$ is $y = -\frac{1}{2}x + 6$.
Substitute $x = 0$ into the equation:
$y = -\frac{1}{2}(0) + 6 = 6$
Since the $y$-value matches the $y$-coordinate of $C$, the point $C$ lies on the line.

1. State the coordinates of the vertices of the trapezium in order

From earlier parts, the coordinates are $A(2, 0)$, $B(6, 8)$, $C(0, 6)$ and $D(-2, 2)$.

2. Apply the shoelace formula to find the area of the trapezium

$\begin{aligned} \text{Area} &= \frac{1}{2} \left| \begin{matrix} 2 & 6 & 0 & -2 & 2 \\ 0 & 8 & 6 & 2 & 0 \end{matrix} \right| \\ &= \frac{1}{2} |(2(8) + 6(6) + 0(2) + (-2)(0)) - (0(6) + 8(0) + 6(-2) + 2(2))| \\ &= \frac{1}{2} |(16 + 36 + 0 + 0) - (0 + 0 - 12 + 4)| \\ &= \frac{1}{2} |52 - (-8)| \\ &= \frac{1}{2} (60) \\ &= 30 \end{aligned}$