Solutions to this question by accurate drawing will not be accepted. The diagram (not drawn to scale) shows a trapezium OPQR in which $PQ$ is parallel to $OR$ and $igangle ORQ = 90^\circ$. The coordinates of $P$ and $R$ are $(-4,\,3)$ and $(4,\,2)$ respectively and $O$ is the origin.

1. Find the gradient of the line OR

Gradient of $OR = \frac{2 - 0}{4 - 0} = \frac{1}{2}$

2. Find the equation of the line PQ

Since $PQ \parallel OR$, gradient of $PQ = \frac{1}{2}$.
Using point $P(-4, 3)$:
$y - 3 = \frac{1}{2}(x - (-4))$
$y = \frac{1}{2}x + 2 + 3$
$y = \frac{1}{2}x + 5$

3. Find the gradient of the line RQ

Since $\angle ORQ = 90^\circ$, $RQ \perp OR$.
Gradient of $RQ = -\frac{1}{\frac{1}{2}} = -2$

4. Find the equation of the line RQ

Using point $R(4, 2)$:
$y - 2 = -2(x - 4)$
$y = -2x + 8 + 2$
$y = -2x + 10$

5. Solve the equations simultaneously to find the coordinates of Q

Equating the two expressions for $y$:
$\frac{1}{2}x + 5 = -2x + 10$
$\frac{5}{2}x = 5$
$x = 2$

Substitute $x = 2$ into $y = -2x + 10$:
$y = -2(2) + 10 = 6$

The coordinates of $Q$ are $(2, 6)$.

1. Find the gradient of OR and the equation of line PQ.

Gradient of $OR = \frac{2 - 0}{4 - 0} = \frac{1}{2}$.
Since $PQ \parallel OR$, gradient of $PQ = \frac{1}{2}$.
Equation of $PQ$ passing through $P(-4, 3)$:
$y - 3 = \frac{1}{2}(x - (-4))$
$y = \frac{1}{2}x + 2 + 3$
$y = \frac{1}{2}x + 5$

2. Find the coordinates of T.

Since $T$ is the y-intercept of $PQ$, let $x = 0$.
$y = \frac{1}{2}(0) + 5 = 5$.
Coordinates of $T$ are $(0, 5)$.

3. Calculate the lengths of OT, OR, and RT to show two sides are equal.

Length of $OT = 5$ units.
Length of $OR = \sqrt{(4-0)^2 + (2-0)^2} = \sqrt{16 + 4} = \sqrt{20}$.
Length of $RT = \sqrt{(4-0)^2 + (2-5)^2} = \sqrt{16 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Since $OT = RT = 5$, triangle $ORT$ is isosceles.

1. Equate the midpoints of the diagonals of the parallelogram

In parallelogram $ORPS$, the diagonals $OP$ and $RS$ bisect each other. Therefore, their midpoints are the same.
Let the coordinates of $S$ be $(x, y)$.
Midpoint of $OP$ = Midpoint of $RS$
$\left( \frac{0 + (-4)}{2}, \frac{0 + 3}{2} \right) = \left( \frac{4 + x}{2}, \frac{2 + y}{2} \right)$

2. Solve for the coordinates of S

Equating the $x$-coordinates:
$\frac{-4}{2} = \frac{4 + x}{2}$
$-4 = 4 + x \implies x = -8$

Equating the $y$-coordinates:
$\frac{3}{2} = \frac{2 + y}{2}$
$3 = 2 + y \implies y = 1$

The coordinates of $S$ are $(-8, 1)$.

1. Apply the shoelace formula using the coordinates of the vertices in order

Using the coordinates $O(0,0)$, $R(4,2)$, $Q(2,6)$ (from part (i)), and $P(-4,3)$ in counter-clockwise order:
$\text{Area} = \frac{1}{2} \left| \begin{matrix} 0 & 4 & 2 & -4 & 0 \\ 0 & 2 & 6 & 3 & 0 \end{matrix} \right|$
$\text{Area} = \frac{1}{2} |(0 \times 2 + 4 \times 6 + 2 \times 3 + (-4) \times 0) - (0 \times 4 + 2 \times 2 + 6 \times (-4) + 3 \times 0)|$

2. Evaluate the expression to find the area

$\text{Area} = \frac{1}{2} |(0 + 24 + 6 + 0) - (0 + 4 - 24 + 0)|$
$\text{Area} = \frac{1}{2} |30 - (-20)|$
$\text{Area} = \frac{1}{2} (50) = 25$