WAEC 2013 · Paper 2 · Q9

(a)
Using a ruler and a pair of compasses only, construct: (i) a trapezium $WXYZ$ such that $|WX| = 10.2\text{ cm}$ , $|XY| = 5.6\text{ cm}$ , $|YZ| = 5.8\text{ cm}$ , $\angle WXY = 60^\circ$ and $\overline{WX}$ is parallel to $\overline{YZ}$ ; (ii) a perpendicular from $Z$ to meet $\overline{WX}$ at $N$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $WX = 10.2$ cm and construct $60^\circ$ at $X$ ; mark $Y$ with $XY = 5.6$ cm. Through $Y$ draw a line parallel to $WX$ and mark $Z$ with $YZ = 5.8$ cm. Join $WZ$ . Then drop the perpendicular from $Z$ to $WX$ to meet it at $N$ . Measured: $|WZ| \approx 5.1$ cm and $|ZN| \approx 4.8$ cm.
(b)
Measure: (i) $|WZ|$ ; (ii) $|ZN|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw $WX = 10.2$ cm.
Construct $60^\circ$ at $X$ and mark $XY = 5.6$ cm on the arm.
$YZ \parallel WX$ , so the angle at $Y$ is $180^\circ - 60^\circ = 120^\circ$ (co-interior angles): construct $120^\circ$ at $Y$ , towards $W$ , and mark $YZ = 5.8$ cm.
Join $WZ$ .

$|ZN| \approx 4.8$ cm.
Check: $|ZN|$ is the height, $5.6\sin 60^\circ \approx 4.85$ cm.
$|WN| = 10.2 - 5.6\cos 60^\circ - 5.8$
$= 1.6$ cm, so $|WZ| = \sqrt{1.6^2 + 4.85^2} \approx 5.1$ cm.