WAEC 2025 · Paper 2 · Q9✱✱

(a)
Using a ruler and a pair of compasses only, construct $\triangle ABC$ with $|AB| = 7.5\text{ cm}$ , $|BC| = 8.1\text{ cm}$ and $\angle ABC = 105^\circ$ .
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. Draw $BC = 8.1$ cm. $105^\circ = 90^\circ + 15^\circ$ : construct $90^\circ$ at $B$ , then bisect the $30^\circ$ between the $90^\circ$ and $120^\circ$ lines to add $15^\circ$ . Mark $A$ with $BA = 7.5$ cm and join $AC$ .
(b)
Locate $D$ on $BC$ such that $|BD| : |DC| = 3 : 2$ , and through $D$ construct the line perpendicular to $BC$ .
Model answer
To divide $BC$ in the ratio $3 : 2$ , draw a line from $B$ at any angle and step off 5 equal lengths with the compasses. Join the fifth mark to $C$ , and draw a parallel through the third mark to cut $BC$ at $D$ . Then $BD = \frac35 \times 8.1 = 4.86$ cm. At $D$ , construct the perpendicular to $BC$ . It meets $AC$ at $P$ , and $|BP|$ measures about 5.4 cm.
(c)
If the perpendicular meets $AC$ at $P$ , measure $|BP|$ (cm).

Worked solution (try it first)

Draw $BC = 8.1$ cm.
At $B$ construct $105^\circ$ : construct $90^\circ$ and $120^\circ$ , then bisect the angle between them.
Mark $BA = 7.5$ cm on the arm and join $AC$ .

$|BD| : |DC| = 3 : 2$ is 5 equal parts.
Draw a line from $B$ and step off 5 equal lengths.
Join the 5th mark to $C$ , and through the 3rd mark draw a line parallel to it, cutting $BC$ at $D$ ( $|BD| = \frac35 \times 8.1 = 4.86$ cm).
At $D$ construct the perpendicular to $BC$ : mark equal distances either side of $D$ on $BC$ and bisect the line between them.

The perpendicular meets $AC$ at $P$ .
Measure $|BP| \approx 5.4$ cm.
(Check: with $B$ at the origin, $A \approx (-1.94, 7.24)$ and $C = (8.1, 0)$ .
The perpendicular $x = 4.86$ meets $AC$ at height about 2.34, so $|BP| = \sqrt{4.86^2 + 2.34^2} \approx 5.4$ cm.)