WAEC 2022 · Paper 2 · Q12

(a)
Using a ruler and a pair of compasses only: (i) construct $\triangle XYZ$ such that $|XY| = 7.2\text{ cm}$ , $|YZ| = 8.4\text{ cm}$ and $\angle XYZ = 60^\circ$ ; (ii) locate, by construction, a point $M$ on $\overline{XY}$ such that $|XM| = |MY|$ ; (iii) construct $\overline{MN} \parallel \overline{YZ}$ such that $MNZY$ is a parallelogram.
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $XY = 7.2$ cm, construct $60^\circ$ at $Y$ and mark $Z$ with $YZ = 8.4$ cm; join $XZ$ . Bisect $XY$ perpendicularly to find its midpoint $M$ . Through $M$ draw the line parallel to $YZ$ , and through $Z$ the line parallel to $XY$ . They meet at $N$ , completing parallelogram $MNZY$ . Measured: $|XZ| \approx 7.9$ cm and $\angle NZX \approx 68^\circ$ .
(b)
Measure: (i) $|XZ|$ ; (ii) $\angle NZX$ .
Show the answer
$|XZ| \approx 7.9\text{ cm}$ ; $\angle NZX \approx 68^\circ$

The accurate construction: X(0, 0), Y(7.2, 0), Z(3, 7.27), M(3.6, 0), N(−0.6, 7.27).

Worked solution (try it first)

Draw $YZ = 8.4$ cm, construct $60^\circ$ at $Y$ and mark $YX = 7.2$ cm on the arm.
Join $XZ$ .

With centre $M$ and radius $|YZ| = 8.4$ cm, and with centre $Z$ and radius $|YM| = 3.6$ cm, draw arcs meeting at $N$ .
Join $MN$ and $NZ$ : $MNZY$ is a parallelogram, with $MN \parallel YZ$ .

$\angle NZX \approx 68^\circ$ .
Check: by the cosine rule $|XZ|^2 = 7.2^2 + 8.4^2 - 2(7.2)(8.4)\cos 60^\circ$
$= 61.92$ , so $|XZ| \approx 7.87$ cm.
$ZN \parallel XY$ , so $\angle NZX = \angle ZXY$ , and the sine rule gives $\sin\angle ZXY = \frac{8.4\sin 60^\circ}{7.87}$ , about $68^\circ$ .