WAEC 2018 · Paper 2 · Q10

(a)
Using a ruler and a pair of compasses only, construct: (i) a quadrilateral $PQRS$ with $|PQ| = 6\text{ cm}$ , $|PS| = 8\text{ cm}$ , $\angle PSR = 90^\circ$ , $|SR| = 12\text{ cm}$ and $|QR| = 11\text{ cm}$ ; (ii) a perpendicular from $Q$ to cut $\overline{SR}$ at $K$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $SR = 12$ cm and construct $90^\circ$ at $S$ ; mark $P$ with $SP = 8$ cm. With centre $P$ and radius 6 cm, and centre $R$ and radius 11 cm, draw arcs that meet at $Q$ (on the side away from $S$ ). Join $PQ$ and $QR$ . Then drop the perpendicular from $Q$ to $SR$ to meet it at $K$ . Measured: $|QK| \approx$ 9.1 cm and $\angle QRS \approx 56^\circ$ .
(b)
Measure: (i) $|QK|$ ; (ii) $\angle QRS$ .
Show the answer
$|QK| \approx 9.1\text{ cm}$ , $\angle QRS \approx 56^\circ$

The accurate construction: S(0, 0), R(12, 0), P(0, 8), Q(5.89, 9.15), K(5.89, 0).

Worked solution (try it first)

Draw $SR = 12$ cm.
Construct $90^\circ$ at $S$ and mark $SP = 8$ cm on the arm.
With centre $P$ and radius 6 cm, and centre $R$ and radius 11 cm, draw arcs meeting at $Q$ , on the side of $PR$ away from $S$ .
Join $PQ$ and $QR$ .

$\angle QRS \approx 56^\circ$ .
Check: with $S$ at the origin, $R$ at $(12, 0)$ and $P$ at $(0, 8)$ , the two arcs meet at $Q \approx (5.89, 9.15)$ , so $|QK| \approx 9.15$ cm and $\angle QRS = \tan^{-1}\frac{9.15}{12 - 5.89}$
$\approx 56^\circ$ .