NECO 2023 · Paper 2 · Q11

Using a ruler and a pair of compasses only:

(a)
Construct a triangle $ABC$ such that $|AB| = 5\text{ cm}$ , $|AC| = 7\text{ cm}$ and $\angle BAC = 120^\circ$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $AC = 7$ cm. At $A$ , construct $120^\circ$ (two $60^\circ$ steps along the same arc from $AC$ ). Mark $B$ on that arm with $AB = 5$ cm and join $BC$ .
(b)
Construct (i) the locus $l_1$ of points equidistant from $A$ and $C$ ; (ii) the locus $l_2$ of points $4.5\text{ cm}$ from $C$ .
Model answer
(i) Points equidistant from $A$ and $C$ lie on the perpendicular bisector of $AC$ : with a radius more than half of $AC$ , draw arcs from $A$ and from $C$ that cross above and below the line, and join the crossings. (ii) Points $4.5$ cm from $C$ lie on the circle centre $C$ , radius $4.5$ cm.
(c)
Locate the points of intersection, $N_1$ and $N_2$ , of $l_1$ and $l_2$ .
Model answer
$N_1$ and $N_2$ are where the perpendicular bisector cuts the circle. By calculation they are $2\sqrt{4.5^2 - 3.5^2}$ apart, so $|N_1N_2| \approx$ 5.7 cm, and $|BC| = \sqrt{109} \approx 10.4$ cm.
(d)
Measure (i) $|N_1N_2|$ ; (ii) $|BC|$ (cm).
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Draw $AC = 7$ cm.
At $A$ construct $120^\circ$ (two $60^\circ$ angles side by side), and with the compasses set to 5 cm cut the arm at $B$ .
Join $BC$ .

$l_1$ , the points equidistant from $A$ and $C$ , is the perpendicular bisector of $AC$ : equal arcs from $A$ and $C$ , and the line through their crossings.

$l_2$ , the points 4.5 cm from $C$ , is the circle with centre $C$ and radius 4.5 cm.

$|BC| \approx 10.4$ cm.
Check by calculation: the bisector is 3.5 cm from $C$ , so $|N_1N_2| = 2\sqrt{4.5^2 - 3.5^2}$
$= 2\sqrt8$
$\approx 5.7$ cm.
And by the cosine rule $|BC|^2 = 5^2 + 7^2 - 2(5)(7)\cos 120^\circ$
$= 109$ , so $|BC| \approx 10.4$ cm.