WAEC 2011 · Paper 2 · Q9

(a)
Using ruler and a pair of compasses only, construct a rhombus $PQRS$ of side $7\text{ cm}$ and $\angle PQR = 60^\circ$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $QR = 7$ cm and construct $60^\circ$ at $Q$ . Mark $P$ with $QP = 7$ cm. With centres $P$ and $R$ and radius 7 cm, draw arcs that meet at $S$ , then join $PS$ and $RS$ .
(b)
Locate point $X$ such that $X$ lies on the locus of points equidistant from $PQ$ and $QR$ and also equidistant from $Q$ and $R$ .
Model answer
Points equidistant from $PQ$ and $QR$ lie on the bisector of angle $PQR$ , which in a rhombus is the diagonal $QS$ . Points equidistant from $Q$ and $R$ lie on the perpendicular bisector of $QR$ . $X$ is where they cross, and $|XR| = \frac{3.5}{\cos 30^\circ} \approx 4.0$ cm.
(c)
Measure $|XR|$ .

Worked solution (try it first)

Points equidistant from the lines $PQ$ and $QR$ lie on the bisector of $\angle PQR$ : construct it (in a rhombus it is the diagonal $QS$ ).
Points equidistant from the points $Q$ and $R$ lie on the perpendicular bisector of $QR$ : construct it.
$X$ is where the two lines cross.

Measure $|XR| \approx 4.0$ cm.
(Check: $X$ is above the midpoint of $QR$ and $\angle XQR = 30^\circ$ , so $|XR| = |XQ|$
$= \frac{3.5}{\cos 30^\circ}$
$\approx 4.04$ cm.)