WAEC 2014 · Paper 2 · Q8

(a)
Using a ruler and a pair of compasses only, construct: (i) a parallelogram $PQRS$ with $RS$ as the base such that $|PQ| = 7.8\text{ cm}$ , $|QR| = 5.6\text{ cm}$ and $\angle QRS = 120^\circ$ ; (ii) a rectangle $ABRS$ equal in area to the parallelogram $PQRS$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw the base $RS = 7.8$ cm. Construct $120^\circ$ at $R$ and mark $Q$ with $RQ = 5.6$ cm. Through $Q$ draw the line parallel to $RS$ and mark $P$ with $QP = 7.8$ cm. For the rectangle, draw perpendiculars to $RS$ at $R$ and $S$ to meet the line $QP$ (produced) at $B$ and $A$ . It has the same base and lies between the same parallels, so it has the same area. Measured: $|AP| = 2.8$ cm and $|AS| \approx 4.85$ cm.
(b)
Measure: (i) $|AP|$ ; (ii) $|AS|$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw the base $RS = 7.8$ cm.
Construct $120^\circ$ at $R$ and mark $RQ = 5.6$ cm on the arm.
With centre $Q$ and radius 7.8 cm, and centre $S$ and radius 5.6 cm, draw arcs meeting at $P$ .
Join $PQ$ and $PS$ : $PQRS$ is the parallelogram.

A rectangle on the same base between the same parallels has the same area.
Produce $QP$ and drop perpendiculars to it from $R$ and $S$ , meeting it at $B$ and $A$ .
$ABRS$ is the rectangle.

$|AS| \approx 4.9$ cm.
Check: $|AS|$ is the height, $5.6\sin 60^\circ \approx 4.85$ cm, and $|AP| = 5.6\cos 60^\circ = 2.8$ cm.