WAEC 2020 · Paper 2 · Q13

(a)
Using a ruler and a pair of compasses only: (i) construct (α) $\triangle ABC$ with $|AB| = 7\text{ cm}$ , $|AC| = 13.5\text{ cm}$ and $\angle ABC = 120^\circ$ ; (β) the locus $l_1$ of points equidistant from $A$ and $B$ ; (γ) the locus $l_2$ of points equidistant from $B$ and $C$ . (ii) Using $N$ , the point of intersection of $l_1$ and $l_2$ , as centre, draw a circle to pass through points $A$ , $B$ and $C$ .
Model answer
Draw $AB = 7$ cm, construct $120^\circ$ at $B$ , and with centre $A$ and radius $13.5$ cm cut the arm at $C$ ( $|BC| \approx 8.6$ cm). The perpendicular bisectors $l_1$ of $AB$ and $l_2$ of $BC$ meet at $N$ , outside the triangle beyond $AC$ , because the angle at $B$ is obtuse. The circle with centre $N$ and radius $|NA| \approx 7.8$ cm passes through $A$ , $B$ and $C$ .
(b)
Using the method of completing the square, solve $4x^2 - 4\sqrt3x + 3 = 0$ , leaving the answer in surd form.

Worked solution (try it first)

(α)** Draw $AB = 7$ cm and construct $120^\circ$ at $B$ (two $60^\circ$ angles).
With centre $A$ and radius 13.5 cm, cut the arm at $C$ .
Join $AC$ .

$N$ , where $l_1$ and $l_2$ cross, is the same distance from $A$ , $B$ and $C$ .
With centre $N$ and radius $NA$ , draw the circle: it passes through $A$ , $B$ and $C$ .
(Its radius is $\frac{13.5}{2\sin 120^\circ} \approx 7.8$ cm.)

Divide by 4: $x^2 - \sqrt3x + \frac34 = 0$ , so $x^2 - \sqrt3x = -\frac34$ .
Add $\left(\frac{\sqrt3}{2}\right)^2 = \frac34$ to both sides: $\left(x - \frac{\sqrt3}{2}\right)^2 = 0$ .
So $x - \frac{\sqrt3}{2} = 0$ and $x = \frac{\sqrt3}{2}$ (a repeated root).