WAEC 2018 · Paper 2 · Q8

(a)
Using a ruler and a pair of compasses only, construct: (i) a trapezium $PQRS$ such that $|PQ| = 6.8\text{ cm}$ , $\angle PQR = 120^\circ$ , $QR \parallel PS$ , $|PS| = 10.6\text{ cm}$ and $|PR| = 9.3\text{ cm}$ ; (ii) the locus $l_1$ of points equidistant from $P$ and $R$ ; (iii) the locus $l_2$ of points equidistant from $Q$ and $R$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $PS = 10.6$ cm. $QR \parallel PS$ and $\angle PQR = 120^\circ$ , so $\angle QPS = 60^\circ$ : construct $60^\circ$ at $P$ and mark $Q$ with $PQ = 6.8$ cm. Draw the line through $Q$ parallel to $PS$ , and cut it with an arc of radius 9.3 cm from $P$ to get $R$ . Join $RS$ . Then bisect $PR$ ( $l_1$ ) and $QR$ ( $l_2$ ) perpendicularly; they meet at $Y$ , the centre of the circle through $P$ , $Q$ and $R$ . Measured: $|QR| \approx 3.8$ cm, $\angle PSR \approx 60^\circ$ , and $|QY| \approx$ 5.4 cm.
(b)
Measure: (i) $|QR|$ ; (ii) $\angle PSR$ ; (iii) $|QY|$ , where $Y$ is the point of intersection of $l_1$ and $l_2$ .
Show the answer
$|QR| \approx 3.8\text{ cm}$ , $\angle PSR \approx 60^\circ$ , $|QY| \approx 5.4\text{ cm}$

The accurate construction: P(0, 0), Q(6.8, 0), R(8.70, 3.29), S(5.3, 9.18); Y is the circumcentre of PQR.

Worked solution (try it first)

Draw $PQ = 6.8$ cm and construct $120^\circ$ at $Q$ .
With centre $P$ and radius 9.3 cm, cut the arm at $R$ .
$PS \parallel QR$ : through $P$ construct a line parallel to $QR$ , and mark $PS = 10.6$ cm on it.
Join $RS$ .

$|QY| \approx 5.4$ cm.
Check: the cosine rule gives $9.3^2 = 6.8^2 + |QR|^2 + 6.8|QR|$ , so $|QR| \approx 3.80$ cm.
$|QY|$ is the radius of the circle through $P$ , $Q$ , $R$ : $\frac{9.3}{2\sin 120^\circ} \approx 5.37$ cm.