WAEC 2012 · Paper 2 · Q8

(a)
Using a ruler and a pair of compasses only: (i) construct a parallelogram $PQRS$ with diagonals $|PR| = 8\text{ cm}$ and $|QS| = 10\text{ cm}$ , which intersect at $O$ with $\angle POS = 75^\circ$ ; (ii) construct the perpendicular from $R$ to $\overline{PQ}$ produced, meeting it at $C$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the marks are for the construction. The diagonals of a parallelogram bisect each other: draw $PR = 8$ cm and mark its midpoint $O$ . Construct $75^\circ$ at $O$ ( $60^\circ + 15^\circ$ , bisecting a $30^\circ$ angle) and mark $OS = OQ = 5$ cm on the line through $O$ . Join $PQRS$ . Then drop the perpendicular from $R$ to $PQ$ (produced if necessary): arcs from $R$ cut the line twice, and bisecting between those points gives $C$ . With these measurements the angle at $Q$ is about $77^\circ$ (acute), so $C$ actually lands on $PQ$ itself, about 1.2 cm from $Q$ . Measured: $|PQ| \approx 7.2$ cm, $|PS| \approx 5.5$ cm, $\angle QRS \approx 103^\circ$ .
(b)
Measure: (i) $|PQ|$ ; (ii) $|PS|$ ; (iii) $\angle QRS$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

The diagonals of a parallelogram bisect each other, so $|OP| = |OR| = 4$ cm and $|OQ| = |OS| = 5$ cm.
Draw $PR = 8$ cm and mark its midpoint $O$ .
At $O$ construct $75^\circ$ ( $60^\circ$ plus half of the $30^\circ$ between $60^\circ$ and $90^\circ$ ), and draw the line $SOQ$ through $O$ at that angle.
Mark $|OS| = 5$ cm on one side and $|OQ| = 5$ cm on the other, with $\angle POS = 75^\circ$ .
Join $PQRS$ .

$\angle QRS \approx 103^\circ$ .
Check with the cosine rule in triangles $POS$ and $POQ$ : $|PS|^2 = 4^2 + 5^2 - 40\cos 75^\circ$ , so $|PS| \approx 5.5$ cm.
$|PQ|^2 = 4^2 + 5^2 - 40\cos 105^\circ$ , so $|PQ| \approx 7.2$ cm.