The plan of a triangular-shaped garden, PQR, is such that $QR = 9.5\text{ cm}$ and $\angle PQR = 30^\circ$. PQ has been drawn for you.

1. Construct the direction of $QR$ using the given angle at $Q$.

At $Q$, use a protractor to draw a ray making $30^\circ$ with $QP$ on the required side of the given line $PQ$. This ray is the possible direction of $QR$.

2. Locate $R$ and complete the triangle.

On the $30^\circ$ ray, measure $9.5\text{ cm}$ from $Q$ and mark this point as $R$, so that $QR=9.5\text{ cm}$. Join $R$ to $P$ using a ruler.

1. Construct equal arcs from $R$ and $P$ to locate two points equidistant from $R$ and $P$.

Set the compass radius to more than half of $RP$. With centre $R$, draw arcs above and below $RP$. Using the same radius and centre $P$, draw arcs to intersect the first pair of arcs.

2. Draw the perpendicular bisector through the arc intersections.

Join the two arc intersection points with a straight line. This line is the perpendicular bisector of $RP$.

1. Mark two equal-distance points on the arms of $\angle QPR$.

With centre $P$, draw an arc cutting $PQ$ and $PR$ at two points.

2. Locate a point equidistant from the two arms and draw the angle bisector.

Using the same compass radius, draw arcs from the two points marked on $PQ$ and $PR$ so that the arcs intersect inside $\angle QPR$. Draw a straight line from $P$ through this intersection point.

1. Identify the half of the triangle where points are nearer to $PR$ than to $PQ$.

The bisector of $\angle QPR$ was constructed in part (b)(ii). Points on this bisector are equidistant from $PR$ and $PQ$. Therefore, points nearer to $PR$ than to $PQ$ are in the half of the triangle on the $PR$ side of this angle bisector.

2. Apply the distance condition $SR \ge 3\text{ cm}$ and shade the required intersection region.

Draw a circle or arc with centre $R$ and radius $3\text{ cm}$. Since $SR \ge 3\text{ cm}$, the possible positions of $S$ must be outside or on this circle. Shade only the region that is:

• inside triangle $PQR$,
• on the $PR$ side of the angle bisector of $\angle QPR$,
• outside or on the circle centred at $R$ with radius $3\text{ cm}$.