WAEC 2010 · Paper 2 · Q8

Using ruler and a pair of compasses only,

(a)
Construct: (i) a quadrilateral $PQRS$ with $|PS| = 6\text{ cm}$ , $\angle RSP = 90^\circ$ , $|RS| = 9\text{ cm}$ , $|QR| = 8.4\text{ cm}$ and $|PQ| = 5.4\text{ cm}$ ; (ii) the bisectors of $\angle RSP$ and $\angle SPQ$ to meet at $X$ ; (iii) the perpendicular $XT$ to meet $PS$ at $T$ .
Model answer
Leave every construction arc showing. $PS = 6$ cm with a $90^\circ$ angle constructed at $S$ and $SR = 9$ cm; $Q$ is where the arc of radius $8.4$ cm from $R$ crosses the arc of radius $5.4$ cm from $P$ . The bisector of the right angle at $S$ and the bisector of $\angle SPQ$ (about $106^\circ$ ) meet at $X$ , and the perpendicular from $X$ meets $PS$ at $T$ , with $|XT| \approx 3.4$ cm.
(b)
Measure $|XT|$ .

Worked solution (try it first)

Draw $PS = 6\text{ cm}$ .
At $S$ construct $90^\circ$ (bisect a straight angle) and mark $R$ on that arm with $SR = 9\text{ cm}$ .
With centre $R$ , radius $8.4\text{ cm}$ , and centre $P$ , radius $5.4\text{ cm}$ , draw two arcs on the side away from $S$ .
They cross at $Q$ .
Join $RQ$ and $QP$ .

Bisect $\angle RSP$ (this gives a $45^\circ$ line from $S$ ) and bisect $\angle SPQ$ .
Label their meeting point $X$ .

From $X$ , construct the perpendicular to $PS$ : arcs from $X$ cut $PS$ twice, then bisect that chord.
The foot is $T$ .

Measure $XT$ with the ruler: $|XT| \approx 3.4\text{ cm}$ .
Check by calculation: $\angle SPQ \approx 106^\circ$ , so in triangle $SXP$ the angles at $S$ and $P$ are $45^\circ$ and $53^\circ$ , which gives $XT = \dfrac{6}{1 + \cot 53^\circ}$
$\approx 3.4\text{ cm}$ .