Using ruler and a pair of compasses only, (a) construct (i) a quadrilateral PQRS with \(PS = 6\text{ cm}\), \(\angle RSP = 9\text{ cm}\), \(QR = 8.4\text{ c...

Given quadrilateral \(PQRS\): \(|PS| = 6\text{ cm}\), \(\angle RSP = 90^\circ\), \(|SR| = 9\text{ cm}\), \(|QR| = 8.4\text{ cm}\) and \(|PQ| = 5.4\text{ cm}\).

(a)(i) Constructing quadrilateral \(PQRS\)

1. Draw the base line \(PS = 6\text{ cm}\).
2. At \(S\), erect a perpendicular to \(SP\) (construct \(\angle RSP = 90^\circ\) with ruler and compasses). Along this perpendicular mark off \(|SR| = 9\text{ cm}\) to fix \(R\).
3. With centre \(P\) and radius \(5.4\text{ cm}\) draw an arc; with centre \(R\) and radius \(8.4\text{ cm}\) draw a second arc. The two arcs cut at \(Q\).
4. Join \(PQ\), \(QR\) and \(RS\) to complete quadrilateral \(PQRS\).

(a)(ii) Bisectors of \(\angle RSP\) and \(\angle SPQ\) meeting at \(X\)

1. Bisect \(\angle RSP\): with centre \(S\) draw an arc cutting \(SR\) and \(SP\); from those two intersections draw equal arcs that cross, and join \(S\) to the crossing point. This bisector runs at \(45^\circ\) to \(PS\) (since \(\angle RSP = 90^\circ\)).
2. Bisect \(\angle SPQ\) in the same way, with centre \(P\).
3. Produce both bisectors until they intersect; label this point \(X\).

(a)(iii) The perpendicular \(XT\)

From \(X\), drop a perpendicular to \(PS\) (with centre \(X\) mark two points on \(PS\), then bisect the distance between them). The foot of this perpendicular on \(PS\) is \(T\).

(b) Measurement

Measuring the perpendicular from the accurate construction:

Check by calculation

Take \(S\) as origin with \(SP\) along the horizontal and \(SR\) vertical (\(\angle RSP = 90^\circ\)):

Point \(Q\) lies \(5.4\text{ cm}\) from \(P\) and \(8.4\text{ cm}\) from \(R\), giving \(Q=(7.49,\,5.19)\). \(X\) is equidistant from lines \(PS\), \(SR\) and \(PQ\) (it lies on both bisectors), and this common distance is the required perpendicular:

This agrees with the measured value, so \(|XT| = 3.4\text{ cm}\).