NECO 2024 · Paper 2 · Q11

(a)
Construct a quadrilateral $ABCD$ such that $|AB| = 6.0\text{ cm}$ , $|DB| = 7.5\text{ cm}$ , $\angle DAB = 60^\circ$ , $|DC| = 7.0\text{ cm}$ and $AD \parallel BC$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $AB = 6$ cm and construct $60^\circ$ at $A$ . With centre $B$ and radius $7.5$ cm, cut the arm at $D$ (so $AD \approx 8.4$ cm). Through $B$ draw a line parallel to $AD$ (copy the $60^\circ$ angle at $B$ ). With centre $D$ and radius 7 cm, cut that line at $C$ , taking the crossing beyond $B$ , then join $DC$ .
(b)
Construct the circumcircle through $A$ , $B$ and $C$ .
Model answer
Bisect $AB$ and $BC$ perpendicularly. The bisectors meet at the circumcentre $O$ . Draw the circle through $A$ , $B$ and $C$ with centre $O$ : its radius is about 8.1 cm, so the circumference is about $2\pi \times 8.14 \approx 51$ cm.
(c)
Calculate the circumference of the circumcircle, correct to the nearest whole number.
Show the answer
51 cm

Worked solution (try it first)

Draw $AB = 6.0$ cm and construct $60^\circ$ at $A$ .
With centre $B$ and radius 7.5 cm, cut the arm at $D$ (so $|AD| \approx 8.4$ cm).
Through $B$ construct a line parallel to $AD$ .
With centre $D$ and radius 7.0 cm, cut it at $C$ , on the far side from $A$ so that $ABCD$ is a trapezium.
Join $DC$ .

Measure the radius: about 8.1 cm.
(Check by calculation: with $A$ at the origin and $B$ at $(6, 0)$ , $C \approx (11.05, 8.75)$ , and the circumradius is about 8.14 cm.)
Circumference $= 2\pi r$
$\approx 2 \times \frac{22}{7} \times 8.14$
$\approx 51$ cm.