WAEC 2010 · Paper 2 · Q8

Using ruler and a pair of compasses only:

(a)
Construct a triangle $PQR$ with $|PQ| = 10\text{ cm}$ , $\angle QPR = 90^\circ$ and $\angle PQR = 30^\circ$ .
Model answer
Leave all the construction arcs showing. Draw $PQ = 10$ cm, construct $90^\circ$ at $P$ , and at $Q$ construct $60^\circ$ and bisect it to get $30^\circ$ . The two arms meet at $R$ , with $|PR| \approx 5.8$ cm and $\angle PRQ = 60^\circ$ .
(b)
(i) Construct $l$ , the locus of all points equidistant from $PR$ and $QR$ ; (ii) locate $M$ , the point where $l$ intersects $PQ$ .
Model answer
Points equidistant from the lines $PR$ and $QR$ lie on the bisector of the angle between them, so $l$ is the bisector of $\angle PRQ$ ( $60^\circ$ ), constructed with arcs from $R$ . It meets $PQ$ at $M$ , with $|PM| \approx 3.3$ cm (exactly $\frac{10}{3}$ cm).
(c)
(i) With $M$ as centre and radius $MP$ , draw a circle; (ii) calculate the area of the circle, correct to one decimal place. $\left[\text{Take }\pi = \frac{22}{7}\right]$

Worked solution (try it first)

Draw $PQ = 10\text{ cm}$ .
At $P$ construct $90^\circ$ .
At $Q$ construct $60^\circ$ (an equilateral-triangle arc) and bisect it to get $30^\circ$ .
The two arms meet at $R$ .

Points equidistant from the lines $PR$ and $QR$ lie on the bisector of the angle between them, so bisect $\angle PRQ$ (which is $60^\circ$ ).
This line is $l$ .

$l$ cuts $PQ$ at $M$ .
Measure: $|PM| \approx 3.3\text{ cm}$ .
By calculation: $|PR| = 10 \tan 30^\circ = 5.77\text{ cm}$ , and in triangle $PRM$ , $|PM| = 5.77 \tan 30^\circ$
$= \frac{10}{3}$
$= 3.33\text{ cm}$ .

With centre $M$ and radius $MP$ , draw the circle.
It touches $PR$ at $P$ and also touches $QR$ , because $M$ is the same distance from both lines.

Area $= \pi r^2$
$= \frac{22}{7} \times \left(\frac{10}{3}\right)^2$
$= \frac{22}{7} \times \frac{100}{9}$
$= 34.9\text{ cm}^2$ .