WAEC 2017 · Paper 2 · Q10

(a)
Three dormitories $D_1$ , $D_2$ and $D_3$ are such that $D_1$ and $D_3$ are $60\text{ m}$ and $80\text{ m}$ from $D_2$ respectively. The bearings of $D_1$ and $D_3$ from $D_2$ are $315^\circ$ and $060^\circ$ respectively. A dining hall is located at a point $M$ such that students of the three dormitories walk equal distances to the hall. Using a ruler and a pair of compasses only, and a scale of 1 cm to 10 metres, illustrate by construction the given information.
Model answer
Use only a ruler and a pair of compasses: leave every construction arc visible, because the examiner looks for them. At 1 cm to 10 m, draw a north line at $D_2$ . Measure $060^\circ$ clockwise from north and mark $D_3$ 8 cm away. $315^\circ$ is $45^\circ$ west of north: mark $D_1$ 6 cm along it. The angle $D_1D_2D_3 = 105^\circ$ . Equal distances from all three dormitories means $M$ is on the perpendicular bisector of each pair: bisect $D_1D_2$ and $D_2D_3$ , and they meet at $M$ . Measured: $|D_1D_3| \approx 11.2$ cm (about 112 m), and $|MD_2| \approx$ 5.8 cm (about 58 m).
(b)
(i) Measure $|D_1D_3|$ . (ii) Find the distance from $M$ to $D_2$ .
Separate values with commas, e.g. 3, −2

Scale 1 unit = 10 m: D₂ at the origin, D₁(−4.24, 4.24), D₃(6.93, 4), M the circumcentre.

Worked solution (try it first)

Mark $D_2$ and draw a north line there.
Draw $D_2D_1$ , 6 cm (60 m) on $315^\circ$ , and $D_2D_3$ , 8 cm (80 m) on $060^\circ$ .
The angle between them is $360^\circ - 315^\circ + 60^\circ = 105^\circ$ .
$M$ is the same distance from all three dormitories, so it lies on the perpendicular bisector of $D_1D_2$ and on the perpendicular bisector of $D_2D_3$ .
Construct both bisectors with compasses.
$M$ is where they cross.

Measure $|D_1D_3| \approx 11.2$ cm, which is about 112 m.
(By the cosine rule: $|D_1D_3|^2 = 60^2 + 80^2 - 2(60)(80)\cos 105^\circ$
$\approx 12\,485$ , so $|D_1D_3| \approx 111.7$ m.) (ii) Measure $|MD_2| \approx 5.8$ cm, which is about 58 m.