WAEC 2008 · Paper 2 · Q10

(i)
Using a ruler and a pair of compasses only, construct a quadrilateral $PQRS$ such that $|PQ| = 7\text{ cm}$ , $\angle QPS = 60^\circ$ , $|PS| = 6.5\text{ cm}$ , $\angle PQR = 135^\circ$ and $|QS| = |QR|$ .
Model answer
Draw $PQ = 7\text{ cm}$ . Construct $60^\circ$ at $P$ and mark $S$ with $PS = 6.5\text{ cm}$ . Construct $135^\circ$ at $Q$ ( $90^\circ$ plus half of $90^\circ$ ). Open the compasses to $QS$ (about $6.8\text{ cm}$ ) and, centre $Q$ , cut the $135^\circ$ arm at $R$ . Join $RS$ .
(ii)
Construct the locus $l_1$ of points equidistant from $P$ and $Q$ .
Model answer
$l_1$ is the perpendicular bisector of $PQ$ : with centres $P$ and $Q$ and the same radius (more than $3.5\text{ cm}$ ), draw arcs on both sides of $PQ$ and join the two crossing points. It passes through the midpoint of $PQ$ , $3.5\text{ cm}$ from $P$ .
(iii)
Construct the locus $l_2$ of points equidistant from $P$ and $S$ .
Model answer
$l_2$ is the perpendicular bisector of $PS$ , constructed the same way with centres $P$ and $S$ . It crosses $PS$ at right angles at its midpoint, $3.25\text{ cm}$ from $P$ .

Worked solution (try it first)

Draw $PQ = 7\text{ cm}$ with a ruler.
At $P$ , construct $60^\circ$ : an arc centred at $P$ , then the same radius from where it meets $PQ$ .
Mark $S$ on this arm with $PS = 6.5\text{ cm}$ .
At $Q$ , construct $135^\circ$ with $QP$ : construct $90^\circ$ , then bisect the $90^\circ$ between the perpendicular and the extension of $PQ$ .
Join $QS$ .
With the compasses set to $|QS|$ (about $6.8\text{ cm}$ when measured), centre $Q$ , cut the $135^\circ$ arm at $R$ .
Join $RS$ to complete $PQRS$ .

The points equidistant from $P$ and $Q$ lie on the perpendicular bisector of $PQ$ .
Construct it with equal arcs from $P$ and $Q$ and label it $l_1$ .