$ABCD$ is a parallelogram and triangle $CDE$ is isosceles.
$E$ and $F$ are points on $AD$ and $BC$ respectively.
$AB$, $EF$ and $CD$ are parallel lines.
Angle $BAE = 3x^\circ$ and angle $CEF = 1.5x^\circ$. Find the value of $x$. State your reasons clearly. The ratio of an interior angle of a regular $n$-sided polygon to an interior angle of a regular $(n + 2)$-sided polygon is 15 : 16. Find the value of $n$. Continuation of working space for question 9(b) Answer $n =$ ... $ABCD$ is a plot of a garden. A playground, $P$, was planned to be built in this garden, equidistant from side $AB$ and $BC$, and equidistant between point $A$ and point $D$. By construction, mark the position of $P$. Given that all profit earned from this charity event will be donated to the Voluntary Welfare Organisation, is it true that her donation is more than 900% of her cost price?
Justify your answer and show all your workings clearly.

1. Find $\angle CDE$ using interior angles between parallel lines.

Since $ABCD$ is a parallelogram, $AB \parallel CD$.
$\angle CDE = 180^\circ - \angle BAE$ (interior angles, $AB \parallel CD$)
$\angle CDE = 180^\circ - 3x^\circ$

2. Express $\angle CED$ using the isosceles triangle property.

Triangle $CDE$ is isosceles with $CD = CE$.
$\angle CED = \angle CDE = 180^\circ - 3x^\circ$ (base angles of isosceles triangle)

3. Use parallel lines $EF$ and $CD$ to find $\angle FED$ and form an equation to solve for $x$.

Since $EF \parallel CD$,
$\angle FED = 180^\circ - \angle CDE$ (interior angles, $EF \parallel CD$)
$\angle FED = 180^\circ - (180^\circ - 3x^\circ) = 3x^\circ$

From the diagram, $\angle FED = \angle CEF + \angle CED$.
$3x^\circ = 1.5x^\circ + (180^\circ - 3x^\circ)$
$3x = 180 - 1.5x$
$4.5x = 180$
$x = 40$

1. Express the interior angles of both polygons in terms of $n$.

Interior angle of a regular $n$-sided polygon = $\frac{(n-2) \times 180^\circ}{n}$
Interior angle of a regular $(n+2)$-sided polygon = $\frac{(n+2-2) \times 180^\circ}{n+2} = \frac{n \times 180^\circ}{n+2}$

2. Set up the equation based on the given ratio.

$\frac{\frac{(n-2) \times 180}{n}}{\frac{n \times 180}{n+2}} = \frac{15}{16}$
$\frac{n-2}{n} \times \frac{n+2}{n} = \frac{15}{16}$
$\frac{n^2 - 4}{n^2} = \frac{15}{16}$

3. Solve the equation for $n$.

$16(n^2 - 4) = 15n^2$
$16n^2 - 64 = 15n^2$
$n^2 = 64$
Since $n > 2$, $n = 8$.

1. Construct the locus of points equidistant from sides $AB$ and $BC$.

Using a pair of compasses and a straightedge, construct the angle bisector of $\angle ABC$. Draw a line from $B$ through the intersection of the arcs.

2. Construct the locus of points equidistant from points $A$ and $D$.

Using a pair of compasses and a straightedge, construct the perpendicular bisector of the line segment $AD$. Draw a line through the intersection of the arcs above and below $AD$.

3. Locate and mark point $P$.

Mark the point of intersection of the angle bisector of $\angle ABC$ and the perpendicular bisector of $AD$ as $P$.