WAEC 2016 · Paper 2 · Q11

(a)
Using a ruler and a pair of compasses only, construct: (i) a parallelogram $ABCD$ with diagonals $|AC| = 10\text{ cm}$ and $|BD| = 7\text{ cm}$ intersecting at $K$ , with $\angle BKC = 60^\circ$ ; (ii) the locus $l_1$ of points equidistant from $B$ and $C$ ; (iii) the locus $l_2$ of points $5\text{ cm}$ from $B$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. The diagonals bisect each other: draw $AC = 10$ cm and mark its midpoint $K$ . Construct $60^\circ$ at $K$ and mark $KB = KD = 3.5$ cm on the line through $K$ . Join $ABCD$ . Then (ii) bisect $BC$ perpendicularly for $l_1$ , and (iii) draw the circle centre $B$ , radius 5 cm, for $l_2$ .
(b)
Locate the points of intersection $M$ and $N$ of loci $l_1$ and $l_2$ . Measure: (i) $|MN|$ ; (ii) $|BC|$ .
Separate values with commas, e.g. 3, −2
Model answer
$M$ and $N$ are where the perpendicular bisector cuts the circle. Measured: $|MN| \approx$ 9.0 cm and $|BC| \approx$ 4.4 cm.

Worked solution (try it first)

The diagonals bisect each other: $|AK| = |KC| = 5$ cm and $|BK| = |KD| = 3.5$ cm.
Draw $AC = 10$ cm and mark its midpoint $K$ .
At $K$ construct $60^\circ$ and draw the line $BKD$ at that angle, with $\angle BKC = 60^\circ$ .
Mark 3.5 cm each side of $K$ for $B$ and $D$ .
Join $ABCD$ .

$l_1$ , equidistant from $B$ and $C$ : construct the perpendicular bisector of $BC$ .

$|BC| \approx 4.4$ cm.
Check: by the cosine rule, $|BC|^2 = 5^2 + 3.5^2 - 2(5)(3.5)\cos 60^\circ$
$= 19.75$ , so $|BC| \approx 4.44$ cm.
$M$ and $N$ are $\sqrt{5^2 - 2.22^2} \approx 4.48$ cm either side of the midpoint of $BC$ , so $|MN| \approx 9.0$ cm.