WAEC 2012 · Paper 2 · Q9

(a)
Three towns $X$ , $Y$ and $Z$ are such that $Y$ is 20 km from $X$ and 22 km from $Z$ . Town $X$ is 18 km from $Z$ . A Health Centre is to be built to serve the three towns, located such that patients from $X$ and $Y$ always travel equal distances to it, while patients from $Z$ travel exactly 10 km. Using a scale of 1 cm to 2 km, find by construction, using a pair of compasses and ruler only, the possible positions of the Health Centre.
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 2 km the sides are $XY = 10$ cm, $YZ = 11$ cm and $XZ = 9$ cm: draw $XY$ , then arcs of 9 cm from $X$ and 11 cm from $Y$ to fix $Z$ . "Equal distances from $X$ and $Y$ " is the perpendicular bisector of $XY$ ; "exactly 10 km from $Z$ " is the circle centre $Z$ , radius 5 cm. They meet at two points, $H_1$ and $H_2$ : about 28 km and 12.7 km from $X$ . The nearer one, $H_2$ , is more convenient for all three towns.
(b)
(i) In how many possible locations can the Health Centre be built? (ii) Measure and record the distances of the locations from town $X$ . (iii) Which of these locations would be convenient for all the three towns?
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

With 1 cm to 2 km: $XY = 10$ cm, $YZ = 11$ cm, $XZ = 9$ cm, and 10 km is 5 cm.
Draw $XY = 10$ cm.
With centre $X$ and radius 9 cm, and centre $Y$ and radius 11 cm, draw arcs meeting at $Z$ .
Join $XZ$ and $YZ$ .
Equal distances from $X$ and $Y$ : construct the perpendicular bisector of $XY$ .
Exactly 10 km from $Z$ : draw the circle with centre $Z$ and radius 5 cm.
The Health Centre is where they cross.

Measure each from $X$ and change back to km: about 14.0 cm $= 28$ km and 6.3 cm $= 12.7$ km.

The location 12.7 km from $X$ , inside the triangle, is convenient for all three towns.