Solutions to this question by accurate drawing will not be accepted. In the trapezium $OABC$, the point $A$ has coordinates $(5, 0)$ and the point $C$ has coordinates $(-2, 6)$. The sides $OC$ and $AB$ are parallel, and $BC$ is perpendicular to $OC$.

1. Find the gradient of $OC$.

Gradient of $OC = \frac{6 - 0}{-2 - 0} = -3$

2. Find the equation of line $AB$.

Since $AB$ is parallel to $OC$, gradient of $AB = -3$.
Equation of $AB$ passing through $A(5, 0)$:
$y - 0 = -3(x - 5)$
$y = -3x + 15$

3. Find the gradient of $BC$.

Since $BC$ is perpendicular to $OC$, gradient of $BC = -\frac{1}{-3} = \frac{1}{3}$.

4. Find the equation of line $BC$.

Equation of $BC$ passing through $C(-2, 6)$:
$y - 6 = \frac{1}{3}(x - (-2))$
$y = \frac{1}{3}x + \frac{2}{3} + 6$
$y = \frac{1}{3}x + \frac{20}{3}$

5. Solve the equations of $AB$ and $BC$ simultaneously to find the coordinates of $B$.

Substitute $y = -3x + 15$ into the equation of $BC$:
$-3x + 15 = \frac{1}{3}x + \frac{20}{3}$
Multiply by 3:
$-9x + 45 = x + 20$
$10x = 25$
$x = 2.5 = 2\frac{1}{2}$
Substitute $x = 2.5$ into $y = -3x + 15$:
$y = -3(2.5) + 15 = -7.5 + 15 = 7.5 = 7\frac{1}{2}$
Therefore, the coordinates of $B$ are $\left(2\frac{1}{2}, 7\frac{1}{2}\right)$.

1. Find the midpoint of $OC$

Midpoint of $OC = \left(\frac{0 + (-2)}{2}, \frac{0 + 6}{2}\right) = (-1, 3)$

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

Gradient of $OC = \frac{6 - 0}{-2 - 0} = -3$
Gradient of perpendicular bisector $= -\frac{1}{-3} = \frac{1}{3}$

3. Find the equation of the perpendicular bisector

Using $y - y_1 = m(x - x_1)$ with point $(-1, 3)$ and gradient $\frac{1}{3}$:
$y - 3 = \frac{1}{3}(x - (-1))$
$y - 3 = \frac{1}{3}x + \frac{1}{3}$
$y = \frac{1}{3}x + \frac{10}{3}$

1. Find the equation of the perpendicular bisector of $OC$ (from part iii).

Midpoint of $OC = \left(\frac{0-2}{2}, \frac{0+6}{2}\right) = (-1, 3)$
Gradient of $OC = \frac{6-0}{-2-0} = -3$
Gradient of perpendicular bisector = $\frac{-1}{-3} = \frac{1}{3}$
Equation of perpendicular bisector: $y - 3 = \frac{1}{3}(x - (-1)) \implies y = \frac{1}{3}x + \frac{10}{3}$

2. Express the coordinates of $E$ in terms of a single variable.

Since $E$ lies on the perpendicular bisector, let $E$ be $\left(x, \frac{1}{3}x + \frac{10}{3}\right)$.

3. Set up the area equation for quadrilateral $OAEC$ using the shoelace formula.

Vertices in order: $O(0,0)$, $A(5,0)$, $E\left(x, \frac{1}{3}x + \frac{10}{3}\right)$, $C(-2,6)$.
Area = $\frac{1}{2} \left| \begin{matrix} 0 & 5 & x & -2 & 0 \\ 0 & 0 & \frac{1}{3}x + \frac{10}{3} & 6 & 0 \end{matrix} \right|$
$15 = \frac{1}{2} \left| (0 + 5(\frac{1}{3}x + \frac{10}{3}) + 6x + 0) - (0 + 0 - 2(\frac{1}{3}x + \frac{10}{3}) + 0) \right|$
$30 = \left| \frac{5}{3}x + \frac{50}{3} + 6x - \left(-\frac{2}{3}x - \frac{20}{3}\right) \right|$
$30 = \left| \frac{25}{3}x + \frac{70}{3} \right|$

4. Solve for the positive $x$-coordinate of $E$.

$\frac{25}{3}x + \frac{70}{3} = 30 \quad \text{or} \quad \frac{25}{3}x + \frac{70}{3} = -30$
$25x + 70 = 90 \quad \text{or} \quad 25x + 70 = -90$
$25x = 20 \implies x = 0.8$
$25x = -160 \implies x = -6.4$
Since the $x$-coordinate of $E$ is positive, $x = 0.8$.

5. Find the corresponding $y$-coordinate of $E$.

$y = \frac{1}{3}(0.8) + \frac{10}{3} = \frac{0.8 + 10}{3} = \frac{10.8}{3} = 3.6$
Coordinates of $E$ are $(0.8, 3.6)$.